SOLUTIONS MANUAL for DSP First 2nd Edition by James McClellan, Ronald Schafer & Mark Yoder

SOLUTION MANUAL

SOLUTION MANUAL

SOLUTION MANUAL

SOLUTION MANUAL

Solutions Manual for DSP First Second Edition

J.H. McClellan, R.W. Schafer, M.A. Yoder

© 2022 Pearson Education.

Chapter

Sinusoids 2-1

Problems

1 © 2022 Pearson Education.

2

CHAPTER 2. SINUSOIDS

x.t /

P-2.1

9 6 3 0 -3 -6 -9 -20

DSP First 2e

-16

-12

-8

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-4

0 4 Time t (ms)

8

12

16

20

CHAPTER 2. SINUSOIDS P-2.2

DSP First 2e

In the plot the period can be measured, T = 12.5 ms ⇒ ω0 = 2π/(12.5 × 10−3 ) = 2π(80) rad.

Positive peak closest to t = 0 is at t 1 = 2.5 ms ⇒ ϕ = −2π(2.5 × 10−3 )/(12.5 × 10−3 ) = 2π/5 = −0.4π rad.

Amplitude is A = 8. x(t) = 8 cos(160πt − 0.4π)

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CHAPTER 2. SINUSOIDS P-2.3

DSP First 2e

(a) Plot of cos θ 1

cos

0.5 0 -0.5 -1 3

2

0 (rad)

2

3

(b) Plot of cos(20πt) cos.20 t /

1 0.5 0 -0.5 -1 -0.1

-0.08 -0.06 -0.04 -0.02

0 0.02 Time t (s)

0.04

0.06

0.08

0.1

3T0 =4

T0

(c) Plot of cos(2π/T0 + π/2) 1

x.t /

0.5 0 -0.5 -1 T0

3T0 =4

T0 =2

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T0 =4

0 T0 =4 Time t (s)

T0 =2

CHAPTER 2. SINUSOIDS P-2.4

DSP First 2e

( jθ) 2 ( jθ) 3 ( jθ) 4 ( jθ) 5 + + + +··· 2!

3!

4!

5! θ2 θ3 θ4 θ5 = 1 + jθ − −j + + j +··· 2!

3! 4!

5! ! !

2 4 θ θ θ3 θ5 = 1− + − · · · +j θ − + +··· 2! 4!

3! 5! | {z } | {z }

e jθ = 1 + jθ +

cos θ

Thus, e jθ = cos θ + j sin θ

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sin θ

CHAPTER 2. SINUSOIDS P-2.5

DSP First 2e

(a) Real part of complex exponential is cosine. ( ) ( ) cos(θ 1 + θ 2 ) = < e j (θ1 +θ2 ) = < e jθ1 e jθ2 = < {(cos θ 1 + j sin θ 1 )(cos θ 2 + j sin θ 2 )} = < {(cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) + j (sin θ 1 cos θ 2 + cos θ 1 sin θ 2 )} cos(θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 (b) Change the sign of θ 2 . ) ( ) ( cos(θ 1 − θ 2 ) = < e j (θ1 −θ2 ) = < e jθ1 e−jθ2 = < {(cos θ 1 + j sin θ 1 )(cos θ 2 − j sin θ 2 )} = < {(cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ) + j (sin θ 1 cos θ 2 − cos θ 1 sin θ 2 )} cos(θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2

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CHAPTER 2. SINUSOIDS P-2.6

DSP First 2e

n (cos θ + j sin θ) n = e jθ = e jnθ = cos(nθ) + j sin(nθ)

100 100

3 4 n = e j0.295167π = e j0.927 5 + j5 j29.5167π =e

1 *

e j28π = e j1.5167π

= cos(1.5167) + j sin(1.5167) = 0.0525 − j0.9986

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CHAPTER 2. SINUSOIDS P-2.7

DSP First 2e

(a) 3e jπ/3 + 4e−jπ/6 = 5e j0.12 = 4.9641 + j0.5981 10 √ √ (b) ( 3 − j3) 10 = 12e−jπ/3 = 248, 832 | e−j10π/3 {z } = −124, 416 + j215, 494.83 e+ j2π /3

−1 √ √ √ (c) ( 3 − j3) −1 = 12e−jπ/3 = (1/ 12)e+jπ/3 = 0.2887e+jπ/3 = 0.14434 + j0.25 1/3

√ √ (d) ( 3 − j3) 1/3 = 12e−jπ/3 e j2π` = (12) 1/6 e−jπ/9 e j2π`/3 for ` = 0, 1, 2.

There are 3 answers: 1.513e−jπ/9 = 1.422 − j0.5175 1.513e−j7π/9 = −1.159 − j0.9726 1.513e−j13π/9 = 1.513e+j5π/9 = −0.2627 + j1.49 ) ) ( ) ( ( √ (e) < je−jπ/3 = < e jπ/2 e−jπ/3 = < e jπ/6 = cos(π/6) = 3/2 = 0.866

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CHAPTER 2. SINUSOIDS P-2.8

DSP First 2e

The variable zz defines z(t), and xx defines x(t) = <{z(t)}. z(t) = 15e j (2π (7)(t+0.875)) ⇒ x(t) = 15 cos(2π(7)(t + 0.875)) The period of x(t) is 1/7 = 0.1429, so the time interval −0.15 ≤ t ≤ 0.15 is (0.3)(7) = 2.1 periods.

There will be positive peaks of the cosine wave at t = −0.1607 s and t = −0.0179 s.

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CHAPTER 2. SINUSOIDS P-2.9

DSP First 2e

A=9 T = 8 × 10−3 s ⇒ ω0 = 2000π/8 = 250π rad/s t 1 = −3 × 10−3 s ⇒ ϕ = −2π(−3/8) = 3π/4 rad z(t) = 9e j (250πt+0.75π) , X = 9e j0.75π , and x(t) = 9 cos(250πt + 0.75π)

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CHAPTER 2. SINUSOIDS P-2.10

DSP First 2e

(a) Add complex amps: 3e−j2π/3 + 1 = 2.646e−j1.761 (b) x(t) = <{z(t)} = <{2.646e−j1.761 e jω0 t }

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⇒ x(t) = 2.646 cos(ω0 t − 1.761)

CHAPTER 2. SINUSOIDS P-2.11

DSP First 2e

−jπ/3 −jπ/3 = e−jπ/3 Add complex amps: e−jπ + e jπ/3 + 2e−jπ/3 = | e−jπ + e jπ/3 {z + e } +e

⇒ x(t) = cos(ωt − π/3) Here is the Matlab plot of the vectors.

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=0

CHAPTER 2. SINUSOIDS P-2.12

DSP First 2e

Find angles satisfying −π < θ ≤ π; all others are obtained by adding integer multiples of 2π. <{(1 + j)e jθ } = 0 √ <{ 2e jπ/4 e jθ } = 0 √ <{ 2e j (θ+π/4) } = 0 √ 2 cos(θ + π/4) = 0   π/2 ⇒ θ + π/4 =   −π/2 

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  π/4 ⇒ θ=  −3π/4 

⇒ e

jθ

√   2 (1 + j)/ √ =  (−1 − j)/ 2 

CHAPTER 2. SINUSOIDS P-2.13

DSP First 2e

Three periods of the signal will be 3(1/250) = 12 ms. (a) Plot si (t) = <{ j s(t)} = <{0.8e jπ/2 e jπ/4 e j500πt } = 0.8 cos(2π(250)t + 3π/4).

0.8

si .t /

0.4 0 -0.4 -0.8 -3

-2

-1

0

1

2 3 Time t (ms)

4

-6

-5

-4

q.t /

d (b) Plot q(t) = <{ dt s(t)} = <{0.8e jπ/4 ( j500π)e j500πt } = <{400πe j3π/4 e j500πt } = 400π cos(500πt + 3π/4)

1200 800 400 0 -400 -800 -1200 -3

-2

-1

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0

1

2 3 Time t (ms)

4

-6

-5

-4

CHAPTER 2. SINUSOIDS P-2.14

DSP First 2e

√ (a) If z1 (t) = 5e−jπ/3 e j7t then x 1 (t) = <{z1 (t)}. √ (b) If z2 (t) = 5e jπ e j7t then x 2 (t) = <{z2 (t)}. √

√ √ (c) If z(t) = z1 (t) + z2 (t) = 5e j7t e−jπ/3 + e jπ = 5e−j2π/3 e j7t , then x(t) = <{z(t)} = 5 cos(7t − 2π/3).

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CHAPTER 2. SINUSOIDS P-2.15

DSP First 2e

Need to add complex amps: 2e j5 + 8e j9 + 4e j0 = 3.051e j2.673 Here is the plot of vectors representing the complex amplitudes:

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CHAPTER 2. SINUSOIDS P-2.16

DSP First 2e

(a) ϕ = −2π

(−2) 4π π t1 = −2π = = ⇒ True.

T 8 8 2

(b) ϕ = −2π

t1 3 3π = −2π = − ⇒ False.

T 8 4

(c) In this case a multiple of 2π must be added. t1 7 −7π −7π π ϕ = −2π = −2π = → + 2π = ⇒ True.

T 8 4 4 4

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CHAPTER 2. SINUSOIDS P-2.17

DSP First 2e

(a) Need to plot five vectors: {1, e j2π/5, e j4π/5, e j6π/5, e j8π/5 }.

Note: one is NOT missing; these are the five “5th roots of unity.” (b) The sum is zero: x(t) =

4 X

cos(ωt + 25 πk) = 0.

k=0

If the upper limit were 3 instead of 4, 3 4 X X cos(ωt + 25 πk) = x(t) = cos(ωt + 52 πk) − cos(ωt + 58 π) = − cos(ωt + 58 π) then x(t) = k=0

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k=0

CHAPTER 2. SINUSOIDS P-2.18

DSP First 2e

(a) Inverse Euler formula: ω = 8 rad/s,

A = 9/2,

ϕ = −2π/3

(b) 30-60-90 triangle: ω = 9 rad/s,

ϕ = 0,

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A = 8.66

CHAPTER 2. SINUSOIDS P-2.19

DSP First 2e

(a) 9e j0.5 = 3Ae j (−2+ϕ) + 4 (b) 9e j0.5 = 3 |{z} Ae jϕ e−j2 + 4 z

(c) z =

9e j0.5 − 4 j2 9e j0.5 − 4 = 3e j2.5 − (4/3)e j2 = 1.938e j2.836 = (1/3)e 3e−j2

(d) A = 1.938 and ϕ = 2.836

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CHAPTER 2. SINUSOIDS P-2.20

DSP First 2e

(a) Convert to complex amplitudes (phasors): 1 = A1 e jϕ1 + A2 e jϕ2 e−jπ/2 = 2A1 e jϕ1 + A2 e jϕ2 (b) Write complex amplitudes as z1 and z2 : 1 = z1 + z2 e

−jπ/2

= 2z1 + z2

√ (c) z1 = e−jπ/2 − 1 = 2e−j3π/4 and z2 = 2 − e−jπ/2 = 2.236e j0.464 √ √ (d) A1 = 2, ϕ1 = −0.75π rad, and A2 = 2.236 = 5, ϕ2 = 0.148π = 0.464 rad

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CHAPTER 2. SINUSOIDS P-2.21

DSP First 2e

(a) Convert to complex amplitudes (phasors): e−j1 = 4e−jπ/2 A1 e jϕ1 + A2 e jϕ2 e−jπ/2+j2 = 3e−jπ/2 A1 e jϕ1 + A2 e jϕ2 e−j1 = 4e−jπ/2 z1 + z2 e−jπ/2+j2 = 3e−jπ/2 z1 + z2 z1 = 1.2576 − j0.3690 = 1.311e−j0.285 z2 = 2.0163 + j4.1890 = 4.649e j1.122 A1 = 1.311, ϕ1 = −0.285 rad A2 = 4.649, ϕ2 = 1.122 rad (b) Should plot − j4z1 + z2 and − j3z1 + z2 . Here is the Matlab plot of the vectors.

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CHAPTER 2. SINUSOIDS P-2.22

DSP First 2e

Convert to phasors (complex amps): Me jπ/3 = 5e jψ − 4 The lefthand side is a ray from the origin at the angle of π/3 rad, or 60◦ when M > 0; and at −2π/3 when M < 0.

The righthand side is the set {z : z = 5e jψ − 4} which is a circle of radius 5 centered at z = −4 + j0. Since the origin is inside the circle, there must be two solutions.

For M > 0, ray at π/3: M = 5e j (ψ−π/3) − 4e−jπ/3

must be purely real

√ 0 = ={5e j (ψ−π/3) − 4e−jπ/3 } = 5 sin(ψ − π/3) − 4(− 3/2) √ ⇒ sin(ψ − π/3) = −2 3/5 ⇒ ψ − π/3 = −0.7654 ⇒ ψ = 0.2818 (or 16.1458◦ )

Then solve for M via : ={Me jπ/3 = 5e jψ − 4} √ √ ⇒ M ( 3/2) = 5 sin ψ ⇒ M = (10/ 3) sin ψ ⇒ M = 1.6056 For M < 0, ray at −2π/3: M = 5e j (ψ+2π/3) − 4e j2π/3

must be purely real

√ 0 = ={5e j (ψ+2π/3) − 4e j2π/3 } = 5 sin(ψ + 2π/3) − 4( 3/2) √ ⇒ sin(ψ + 2π/3) = 2 3/5 ⇒ ψ + 2π/3 = 0.7654 ⇒ ψ = −1.329 (or −76.146◦ )

Then solve for M via : ={Me−j2π/3 = 5e jψ − 4} √ √ ⇒ M (− 3/2) = 5 sin ψ ⇒ M = (−10/ 3) sin ψ ⇒ M = 5.6056 Another way to obtain M follows: Me jπ/3 = 5e jψ − 4 ⇒ Me jπ/3 + 4 = 5e jψ ⇒ |Me jπ/3 + 4| 2 = |5e jψ | 2 = 25 M 2 + 8M cos(π/3) + 16 = 25 M 2 + 4M − 9 = 0

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which has two roots: M = 5.6056 and M = 1.6056.

CHAPTER 2. SINUSOIDS P-2.23

DSP First 2e

−j2.1π j10πt −j0.1π j10πt (a) z(t − 0.24) = Ze j10π (t−0.24) = 7e j0.3π e j10πt e−j2.4π = 7e = 7e | {z } e | {z } e W

(b) z(t − t d ) = Ze j10π (t−td ) = 7e j0.3π e j10πt e−j10πtd must equal y(t) = Y e j10πt = 7e−j0.1π e j10πt ⇒ 7e j0.3π−j10πtd = 7e−j0.1π

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⇒ 0.3π − 10πt d = −0.1π

⇒ t d = (0.4/10) = 0.04 s

CHAPTER 2. SINUSOIDS P-2.24

DSP First 2e

(a) The frequency is the same for all terms, so ω̂0 = 0.22π rad in the expression for y[n]. (b) Perform phasor addition: y[n] = 7e j (0.22π (n+1)−0.25π) − 14e j (0.22πn−0.25π) + 7e j (0.22π (n−1)−0.25π) = 7e j (0.22πn−0.03π) − 14e j (0.22πn−0.25π) + 7e j (0.22πn−0.47π)

= 7e−j0.03π − 14e−j0.25π + 7e−j0.47π e j0.22πn | {z } = 3.213 e

Phasor Addition j0.75π j0.22πn

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e

⇒ A = 3.213, ϕ = 0.75π rad

CHAPTER 2. SINUSOIDS P-2.25

DSP First 2e

d d z(t) = dt Ze j2πt = ( j2π) Z e j2πt ⇒ Q = ( j2π)(e jπ/4 ) = 2πe j3π/4 (a) dt | {z } Q

(b) Need a plot. Angle of Q is greater by π/2 rad. (c) Compare the interchange of derivative and real part, which is always true. ( ) d < dt z(t) = <{2πe j3π/4 e j2πt } = 2π cos(2πt + 3π/4) ) d d ( d <{z(t)} = < e jπ/4 e j2πt = cos(2πt + π/4) = (2π)(− sin(2πt + π/4)) = 2π cos(2πt + 3π/4) dt dt dt (d) Integrating a complex exponential over one period should give zero.

Z0.5

0.5

e −0.5

jπ/4 j2πt

e

e jπ/4 e j2πt e jπ − e−jπ =0 dt = = e jπ/4 j2π j2π −0.5

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CHAPTER 2. SINUSOIDS P-2.26

DSP First 2e

Try x(t) = Ae jωt and solve for ω d x(t) = jω Ae jωt dt

and

d2 x(t) = ( jω) 2 Ae jωt |{z} dt 2 −ω 2

Plug x(t) into differential equation −ω2 Ae jωt = −289 Ae jωt ⇒ −ω2 = −289 Two solutions: x(t) = Ae

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⇒ ω = ±17 j17t

or x(t) = Ae−j17t

CHAPTER 2. SINUSOIDS P-2.27

DSP First 2e

! d 2 v(t) 1 d d dv d 2 v(t) ⇒ =− v(t) (a) v(t) = −L i(t) = −L C = −LC 2 2 dt dt dt LC dt dt (b) The frequency of oscillation will be ω0 = √

1 LC

(c) Starting with v(t) = A cos(ω0 t + ϕ), we obtain

1 d 2 v(t) = − ω02 A cos(ω0 t + ϕ) = − v(t) 2 | {z } |{z} LC dt 1/LC

v(t)

(d) v(t) = 5 cos(ω0 t + π/3) ⇒ i(t) = C dv dt = 5Cω0 sin(ω0 t + π/3) = 5Cω0 cos(ω0 t + π/3 − π/2) There is a 90◦ phase difference between the current and the voltage. (e) This is true in general: ! d 2i(t) 1 di d 2i(t) d d i(t) −L = −LC i(t) = C v(t) = C ⇒ =− 2 2 dt dt dt LC dt dt

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CHAPTER 2. SINUSOIDS P-2.28

DSP First 2e

In a mobile radio system a transmitting tower sends a sinusoidal signal, and a mobile user receives not one but two copies of the transmitted signal: a direct-path transmission and a reflected-path signal (e.g., from a large building) as depicted in the following figure. y

TRANSMITTER (0, dt )

REFLECTOR (dr , dt )

Velocity = c

0

VEHICLE

x

The received signal is the sum of the two copies, and since they travel different distances they have different time delays, i.e., r (t) = s(t − t 1 ) + s(t − t 2 ) The distance between the mobile user in the vehicle at x and the transmitting tower is always changing. Suppose that the direct-path distance is q d 1 = x 2 + d t2 (meters) where d t = 1000 meters, and where x is the position of the vehicle moving along the x-axis. Assume that the reflected-path distance is q d 2 = dr + (x − dr ) 2 + d t2 (meters) where dr = 55 meters. (a) The amount of the delay (in seconds) can be computed for both propagation paths, by converting distance into time delay by dividing by the speed of light (c = 3×108 m/s). q p x 2 + d t2 x 2 + 106 t 1 = d 1 /c = = secs. c 3×108 q q 2 2 dr + (x − dr ) + d t 55 + (x − 55) 2 + 106 t 2 = d 2 /c = = secs. c 3×108 (b) When the transmitted signal is s(t) = cos(300π×106 t), the general formula for the received signal is: r (t) = s(t − t 1 ) + s(t − t 2 ) = cos(300π×106 (t − t 1 )) + cos(300π×106 (t − t 2 ))

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CHAPTER 2. SINUSOIDS When x = 0 we can calculate t 1 and t 2 , and then perform a phasor addition to express r (t) as a sinusoid with a known amplitude, phase, and frequency. When x = 0, the time delays are p 02 + 106 = 3.3333×10−6 secs. t1 = 3×108 q 55 + (0 − 55) 2 + 106 t2 = = 3.5217×10−6 secs.

3×108 Thus we must perform the following addition: r (t) = cos(300π×106 (t − 3.3333×10−6 )) + cos(300π×106 (t − 3.5217×10−6 )) = cos(300π×106 t − 1000π)) + cos(300π×106 t − 1056.5113579π) As a phasor addition, we carry out the following steps (since 1000π and 1056π are integer multiples of 2π): R = 1e j0 + 1e j0.5113579π = 1 + j0 + (−0.035674 + j0.99936) = 0.9643 + j0.9994 = 1.389e j0.803 = 1.389e j0.256π = 1.389∠46.02◦ From the polar form of the phasor R, we can write r (t) as a sinusoid: r (t) = 1.389 cos(300π×106 t + 0.256π) (c) In order to find the locations where the signal strength is zero, we note that the phase angles of the two delayed sinusoids must differ by an odd multiple of π in order to get cancellation. Thus, (2` + 1)π = ∆ϕ = −ωt 1 − (−ωt 2 ) p = −300π×10 .

6* .

x 2 + 106 3×108

−

, = −π

p

x 2 + 106 − 55 −

q

55 +

q

(x − 55) 2 + 106 + // 3×108 !

(x − 55) 2 + 106

The general solution to this equation is difficult, involving a quartic. However, if we choose ` = 27 so that the left hand side becomes 55π, then the 55π term on the right hand side will cancel, and we obtain an equation in which squaring both sides will produce the answer. q p 2 6 π x + 10 = −π (x − 55) 2 + 106 =⇒ x 2 + 106 = (x − 55) 2 + 106 =⇒ x 2 = x 2 − 110x + 552 =⇒ 110x = 552 !

55 =⇒ x = 55 = 27.5 meters 110

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CHAPTER 2. SINUSOIDS The general solution would be done in the following manner: q p −(2` + 1) = x 2 + 106 − 55 − (x − 55) 2 + 106 q p ⇒ 55 − (2` + 1) = x 2 + 106 − (x − 55) 2 + 106 q p ⇒ 552 − 110(2` + 1) + (2` + 1) 2 = x 2 + 106 − 2 x 2 + 106 (x − 55) 2 + 106 + (x − 55) 2 + 106 q p 2 6 ⇒ 2 x + 10 (x − 55) 2 + 106 = −4` 2 + 216` + 109 − 552 + x 2 + 2×106 + (x − 55) 2 Squaring both sides would eliminate the square roots, but would produce a fourth-degree polynomial that would have to be solved for the vehicle position x. (d) Here is a Matlab script that will plot the signal strength versus vehicle position x, thus demonstrating that there are numerous locations where no signal is received (note the null at x = 27.5). xx = -100:0.05:100; d1 = sqrt(xx.*xx + 1e6); d2 = 55 + sqrt((xx-55).*(xx-55)+1e6); omeg = 300e6*pi; c = 3e8; phi1 = -omeg*d1/c; phi2 = -omeg*d2/c; RR = 1*exp(j*phi1) + 1*exp(j*phi2); subplot(’Position’,[0.1,0.1,0.6,0.3]); hp = plot(xx,abs(RR)); grid on, xlabel(’Vehicle Position (x)’); ylabel(’Signal Strength’); title(’Multipath Problem in SP-First’); set(hp,’LineWidth’,2); print -dpdf multipathResult.pdf

Multipath Problem in SP−First

Signal Strength

2 1.5 1 0.5 0 −100

−80

−60

−40

−20 0 20 Vehicle Position (x)

40

60

80

100

Over the range −100 ≤ x ≤ 100 the nulls appear to be equally spaced 36.4 meters apart, but they are not uniform. A plot over the range 0 ≤ x ≤ 1500 would demonstrate the non-uniformity.

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Chapter

Spectrum 3-1

Problem Solutions

32 © 2022 Pearson Education.

3

CHAPTER 3. SPECTRUM P-3.1

DSP First 2e

(a) x(t) = 11 + 14 cos(100πt − π/3) + 8 cos(350πt − π/2) (b) Since the gcd of 50 and 175 is 25, x(t) is periodic with period T0 = 1/25 = 0.04 s. (c) Negative frequencies are implicit in the cosine terms. They are needed to give a real signal when combined with their corresponding positive-frequency terms.

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CHAPTER 3. SPECTRUM P-3.2

DSP First 2e

(a) A single plot labeled with complex amplitudes is sufficient. The spectrum consists of the lines {(400, 5e jπ/4 ), (−400, 5e−jπ/4 ), (600, 3.5e−jπ/3 ), (−600, 3.5e jπ/3 ), (800, 1.5), (−800, 1.5)} where the frequencies are in Hz. 5e−jπ/4

5e jπ/4 3.5e−jπ/3

3.5e jπ/3 1.5 −800

1.5 −600

−400

0

400

600

800

-

f

(b) The signal x(t) is periodic with fundamental frequency 200 Hz or period 1/200 = 0.005 s since the gcd of {400, 600, 800} is 200. (c) The spectrum has the added components {(500, 2.5e jπ/2 ), (500, 2.5e−jπ/2 )}. Now we seek the gcd of {400, 500, 600, 800} so the fundamental frequency changes to 100 Hz and the period is 0.01 s.

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CHAPTER 3. SPECTRUM P-3.3

DSP First 2e

(a) x(t) = 2 cos(2t) + 4 cos(4t − π/3) + 2 cos(6t + π/4) (b) The spectrum is {(2, 1), (−2, 1), (4, 2e−jπ/3 ), (−4, 2e jπ/3 ), (6, e jπ/4 ), (−6, e−jπ/4 )} The frequencies are all in rad/s. e−jπ/4 −6

2e−jπ/3

2e jπ/3 1 −4

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−2

1 0

2

4

e jπ/4 6

ω

-

CHAPTER 3. SPECTRUM P-3.4

DSP First 2e

(a) Determine a formula for x(t) as the real part of a sum of complex exponentials.

Use Euler’s formula for the sine function obtaining !3 e j27πt − e−j27πt sin (27πt) = 2j

1 j27πt3 = e − 3e j27πt2 e−j27πt + 3e j27πt e−j27πt2 − e−j27πt3 −8 j 1 3 = sin(27πt) − sin(81πt) 4 4 3

(b) What is the fundamental period for x(t)?

The fundamental frequency is 27/2 so the fundamental period is 2/27. (c) Plot the spectrum for x(t).

The spectrum is {(27π, − j3/8), (−27π, j3/8), (81π, j/8), (−81π, − j/8)}, where the frequencies are in rad/s.

− j3/8

j3/8 − j/8 −81π

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j/8 −27π

0

27π

81π

ω

-

CHAPTER 3. SPECTRUM P-3.5

DSP First 2e

There are seven spectral components: {(−3, 1/(1 − j3)), (−2, 1/(1 − j2)), (−1, 1/(1 − j)), (0, 1), (1, 1/(1 + j)), (2, 1/(1 + j2)), (3, 1/(1 + j3))}, where the frequencies are all in rad/s.

Putting all the complex numbers in polar form gives the following plot: 1 0.707e−j0.25π

0.707e j0.25π

0.447e−j0.3524π 0.316e−j0.398π

0.447e j0.3524π 0.316e j0.398π −3

−2

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−1

0

1

2

3

ω

-

CHAPTER 3. SPECTRUM P-3.6

DSP First 2e

(a) In this case we need to find the gcd of 36 and 84, which is 12. Thus, the fundamental frequency is ω0 = 1.2π rad/s. (b) The fundamental period is T0 = 2π/ω0 = 1/0.6 = 5/3 s. (c) The DC value is −7. (d) The ak coefficients are nonzero for k = 0, ±3, ±7.

Here is the list of the nonzero Fourier series coefficients in a table. k ak

−7 3e−jπ/4

−3 4e jπ/3

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0 7e jπ

3

7

4e−jπ/3

3e jπ/4

CHAPTER 3. SPECTRUM P-3.7

DSP First 2e

(a) The phasor representation is z(t) = Ae j2π ( fc − f∆ )t + Be j2π ( fc + f∆ )t (b) z(t) = e j2π fc t ( Ae−j2π f∆ t + Be j2π f∆ t ) = e j2π fc t ( A cos(2π f ∆ t) − j A sin(2π f ∆ t) + B cos(2π f ∆ t) − j B sin(2π f ∆ t)) = e j2π fc t [( A + B) cos(2π f ∆ t) − j ( A − B) sin(2π f ∆ t)] Therefore, the real part is x(t) = <{z(t)} = ( A + B) cos(2π f ∆ t) cos(2π f c t) + ( A − B) sin(2π f ∆ t) sin(2π f c t) so C = A + B and D = A − B. If A = B = 1, C = 2 and D = 0, so using the trigonometric identity cos α cos β = 1 1 2 cos(α − β) + 2 cos(α + β), it follows that x(t) = 2 cos(2π f ∆ t) cos(2π f c t) = 2[ 12 cos(2π( f c − f ∆ )t) + 21 cos(2π( f c + f ∆ )t)] (c) The values are A = 1 and B = −1. In this case, ! ! e j2π f∆ t − e−j2π f∆ t e j2π fc t − e−j2π fc t x(t) = 2 2j 2j

j2π( fc + f∆ )t j2π ( fc − f∆ )t 1 −e − e−j2π ( fc − f∆ )t + e−j2π ( fc + f∆ )t = −2 e The spectrum is {(− f c − f ∆, −0.5), (− f c + f ∆, 0.5), ( f c − f ∆, 0.5), ( f c + f ∆, −0.5)}, and the plot is −0.5

−( f c + f ∆ )

0.5

− fc

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−( f c − f ∆ )

−0.5

0.5

0

( f c − f∆)

fc

( f c + f∆)

-

f

CHAPTER 3. SPECTRUM P-3.8

DSP First 2e

(a) Using Euler’s relation we get x(t) = 10 + 10e jπ/4 e j2π (100)t + 10e−jπ/4 e−j2π(100)t + 5e j2π (250)t + 5e−j2π (250)t The gcd of 100 and 250 is 50 so f 0 = 50 and therefore N = 5. The nonzero Fourier coefficients are, therefore, a−5 = 5, a−2 = 10e−jπ/4 , a0 = 10, a2 = 10e jπ/4 , and a5 = 5. (b) The signal is periodic because all the frequencies are multiples of 50 Hz. Therefore, the fundamental period is T0 = 1/50 = 0.02s. (c) Here is the spectrum plot of this signal versus f in Hz. 10e−jπ/4

10

10e jπ/4

5

−250

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5

−100

0

100

250

-

f

CHAPTER 3. SPECTRUM P-3.9

DSP First 2e

(a) Use phasors to show that x(t) can be expressed in the form x(t) = A1 cos(ω1 t + ϕ1 ) + A2 cos(ω2 t + ϕ2 ) + A3 cos(ω3 t + ϕ3 ) where ω1 < ω2 < ω3 ; i.e., find values of the parameters A1, A2, A3, ϕ1, ϕ2, ϕ3, ω1, ω2, ω3 .

Using Euler’s relation we get

g f x(t) = 12 − 3.5 je−jπ/3 e jπt + 3.5 je jπ/3 e−jπt 0.5e j13πt + .5e−j13πt = 12 cos(13πt) − 1.75 je−jπ/3 (e j14πt + e−j12πt ) + 1.75 je jπ/3 (e j12πt + e−j14πt ) = 12 cos(13πt) + 1.75e−jπ/2 e−jπ/3 (e j14πt + e−j12πt ) + 1.75e jπ/2 e jπ/3 (e j12πt + e−j14πt ) = 12 cos(13πt) + 1.75e−j5π/6 e j14πt + 1.75e−j5π/6 e−j12πt + 1.75e j5π/6 e j12πt + 1.75e j5π/6 e−j14πt = 12 cos(13πt) + 3.5 cos(14πt − 5π/6) + 3.5 cos(12πt + 5π/6) The requested parameters are easily picked off from this equation. (b) Sketch the two-sided spectrum of this signal on a frequency axis. Be sure to label important features of the plot. Label your plot in terms of the numerical values of Ai , ϕi , and ωi .

12

3.5e j5π/6

12

3.5e−j5π/6

−14π −13π −12π

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3.5e−j5π/6

3.5e j5π/6 0

12π

13π

14π

ω

-

CHAPTER 3. SPECTRUM P-3.10

DSP First 2e

(a) Assume without limitation that ω2 − ω1 > 0. For periodicity with period T0 we require that ω0 = 2π/T0 . This means that k 1 ω0 = ω2 − ω1 and k 2 ω0 = ω2 + ω1 , where k 1 and k 2 are integers and k 2 > k 1 . (b) Part (a) gives two equations for ω1 and ω2 . If we solve them in terms of ω0 we get ω1 = (k2 − k 1 )ω0 /2 and ω2 = (k 2 + k 1 )ω0 /2, so the main condition is that both ω1 and ω2 are integer multiples of ω0 /2. This is the most general condition.

Therefore, the relationship between ω2 and ω1 is ω2 =

k2 + k1 ω1 k2 − k1

if x(t + T0 ) = x(t). Thus, ω2 could be an integer multiple of ω1 if k 2 − k 1 divides into k 2 + k 1 with no remainder, but that is not necessary for periodicity of x(t).

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CHAPTER 3. SPECTRUM P-3.11

DSP First 2e

(a) The gcd of 40, 60, 120 is 20 so ω0 = 20π and the fundamental period is T0 = 2π/ω0 = 0.1s. The finite Fourier series has components indexed by 0, ±2, ±3, ±6 so N = 6. The coefficients are a0 = 2, a±2 = 2e∓jπ/5 , a±3 = 1.5e∓jπ/2 , a±6 = 2e∓π/3 (b) The spectrum is {(−120π, 2e jπ/3 ), (−60π, 1.5e jπ/2 ), (−40π, 2e jπ/5 ), . . .

(0, 2), (40π, 2e−jπ/5 ), (60π, 1.5e−jπ/2 ), (120π, 2e−jπ/3 )} 2e jπ/3

2e jπ/5

2

−40π

0

1.5e jπ/2

−120π

−60π

2e−jπ/5 1.5e−jπ/2

40π

60π

2e−jπ/3

120π

ω

-

(c) Now the fundamental frequency is 10π rad/s because the gcd of 20, 40, 50, and 120 is 10. Therefore, the period is T0 = 2π/10π = 1/5 = 0.2s. The spectrum is the same as in part (b) except there are two additional components at ±50π rad/s: (−50π, 5e jπ/6 ) and (50π, 5e−jπ/6 ).

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CHAPTER 3. SPECTRUM P-3.12

DSP First 2e

(a) Make a table of the frequencies of the tones of the octave beginning with middle C, assuming that the A above middle C is tuned to 440 Hz. Note name Note number Frequency Note name Note number Frequency

C 40 262 F# 46 370

C# 41 277 G 47 392

D 42 294 G# 48 415

E[ 43 311 A 49 440

E 44 330 B[ 50 466

F 45 349 B 51 494

F# 46 370 C 52 523

(b) The formula for the frequency f as a function of note number n is f = 440 · 2(n−49)/12 (c) The spectrum would have the form: {(−440, a3∗ ), (−370, a2∗ ), (−294, a1∗ ), (294, a1 ), (370, a2 ), (440, a3 )} To sound like a musical chord, the coefficients should have similar magnitudes, but the phases could be arbitrarily chosen. A chord from a real instrument would have overtones (higher harmonics) of each individual note.

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CHAPTER 3. SPECTRUM P-3.13

DSP First 2e

(a) The frequency of the DC component is by definition 0. The waveform is periodic with period 25 ms so the frequency is 1/0.025 = 40 Hz. (b) The DC level is (20 − 10)/2 = 5, the amplitude of the cosine is (20 + 10)/2 = 15, and the cosine is delayed by 0.005 s, so x(t) = 5 + 15 cos(2π(40)(t − .005)) = 5 + 15 cos(80πt − 0.4π) (c) x(t) = 5 + 7.5e j (80πt−0.4π) + 7.5e−j (80πt−0.4π) = 7.5e j0.4π e−j80πt + 5 + 7.5e−j0.4π e j80πt (d) Plot of the two-sided spectrum of the signal x(t).

7.5e−j0.4π

7.5e j0.4π 5

−80π

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0

80π

ω

-

CHAPTER 3. SPECTRUM P-3.14

DSP First 2e

(a) Using symmetry we obtain √ √ X−1 = 2 − j 2 = 2e−jπ/4

X2 = 8e jπ/3

ω1 = 70π

ω2 = −100π

(b) x(t) = 20 + 4 cos(70πt + π/4) + 16 cos(100πt + π/3) (c) The gcd of 70 and 100 is 10, so the fundamental frequency of the signal is f 0 = 5 Hz and the fundamental period is T0 = 1/5 = 0.2 s. (d) Note that the −20 ≤ 4 cos(70πt + π/4) +16 cos(100πt + π/3) ≤ 20 since the individual terms satisfy −4 ≤ 4 cos(70πt + π/4) ≤ 4 and −16 ≤ 16 cos(100πt + π/3) ≤ 16. The value ±20 would be attained only if the phases of the two cosines are such that 4 cos(70π(t − t 0 )) + 16 cos(100π(t − t 0 )).

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CHAPTER 3. SPECTRUM P-3.15

DSP First 2e

(a) x(t) = cos2 (7πt − 0.1π) We need to express x(t) in terms of complex exponentials 2

x(t) = 0.5e j (7πt−0.1π) + 0.5e−j (7πt−0.1π) = 0.25e j0.2π e−j14πt + 0.5 + 0.25e−j0.2π e j14πt 0.5 0.25e−j0.2π

0.25e j0.2π

−14π

0

ω

-

14π

(b) y(t) = cos2 (7πt − 0.1π) cos(77πt + 0.1π)

x(t) = 0.25e j0.2π e−j14πt + 0.5 + 0.25e−j0.2π e j14πt 0.5e j0.1π e j77πt + 0.5e−j0.1π e−j77πt = 0.125e j0.1π e−j91πt + 0.25e−j0.1π e−j77πt + 0.125e−j0.3π e−j63πt + 0.125e j0.3π e j63πt + 0.25e j0.1π e j77πt + 0.125e−j0.1π e j91πt

0.25e−j0.1π

0.25e j0.1π

0.125e j0.1π 0.125e−j0.3π

−91π −77π −63π

0.125e j0.3π

0

63π

0.25e−j0.1π

77π

91π

ω

-

Note: ω axis is not to scale.

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CHAPTER 3. SPECTRUM P-3.16

DSP First 2e

(a) Plotting spectrum of v(t) versus f in kHz, 0.5e−j0.3π

−3

0.5e j0.3π

0

-

f (kHz)

3

(b) The spectrum for x(t) versus f in kHz. 0.75

0.25e−j0.3π

0.75

0.25e j0.3pi

−683 −680 −677 Note: f axis is not to scale.

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0

0.25e−j0.3π

0.25e j0.3π

677

683

680

-

f (kHz)

CHAPTER 3. SPECTRUM P-3.17

DSP First 2e

(a) The gcd of 105 and 180 is 15, so the given frequencies are the 7th and 12th harmonics of f 0 = 15 Hz. (b) x(t) = 22 cos(2π(105)t − 0.4π) + 14 cos(2π(180)t − 0.6π) (c) Simplify the numerical values for the complex amplitudes, i.e., phases should be in [−π, π]. x 2 (t) = 22 cos(2π(105)(t − 0.05) − 0.4π) + 14 cos(2π(180)(t − 0.05) − 0.6π) = 22 cos(2π(105)t − 10.5π − 0.4π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 10π − 0.9π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 0.9π) + 14 cos(2π(180)t − 0.6π) Note that even multiples of 2π rad can be dropped from the equation. Thus, the spectrum is: {(−180, 7e j0.6π ), (−105, 11e j0.9π ), (105, 11e−j0.9π ), (180, 7e−j0.6π )} where the frequencies are in hertz. Therefore the plot of the spectrum looks just like Fig. ?? except the phase is different at frequencies ±105. (d) The effect of this operation is simply to increase all the frequencies by 105 Hz, or, in other words, to shift the spectrum of x(t) to the right by 105 Hz. Therefore the spectrum line at −105 Hz will move to f = 0. The DC component is therefore, the value of the spectrum at f = −105, or 11e j0.9π .

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CHAPTER 3. SPECTRUM P-3.18

DSP First 2e

(a) The spectrum of y(t) is the spectrum of x(t) with an added DC component of size 8.

11e−j0.4π

11e j0.4π 8

7e j0.6π

−180

−105

7e−j0.6π

0

180

105

-

f

(b) The spectrum of z(t) is the same as that of x(t) with the addition of components of size 9e±j0.8π at frequencies ±40 Hz. 11e j0.4π 7e j0.6π

−180

−105

9e−j0.8π

−40

9e j0.8π

0

40

11e−j0.4π 7e−j0.6π

180

105

-

f

(c) The fundamental frequency is the gcd of 40, 105, 180, which is 5 Hz. (d) The derivative operation multiplies each spectrum component by j2π f , where f is the frequency of the complex exponential component. So we get j2π(105)11e−j0.4π

− j2π(105)11e j0.4π

j2π(180)7e−j0.6π

− j2π(180)7e j0.6π

−180

−105

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0

105

180

-

f

CHAPTER 3. SPECTRUM P-3.19

DSP First 2e

(a) X1 = 8e−jπ/3

and ω1 = −10π

(b) Here is the plot of the spectrum of x(t).

8e−jπ/3

8e jπ/3

−10π

0

ω

-

10π

(c) The symmetry implies that ωb = 20π and B = +4 j. Furthermore, symmetry requires that ω a = 0. To find A, ωc , and ϕ we can write y(t) as y(t) = 0.5x(t)e jϕ e jωc t + 0.5x(t)e−jϕ e−jωc t , which shows that the spectrum of y(t) will consist of the sum of scaled copies of the spectrum of x(t) shifted right (up) by ωc and left (down) by ωc . In order to have only three components we must choose ωc = 10π so that two of the shifted spectrum lines over lap at ω = 0.

4e jπ/3 e jϕ + 4e−jπ/3 e−jϕ 4e jπ/3 e−jϕ

−20π

0

4e−j π /3 e jϕ

20π

ω

-

jϕ = 4e−jπ/2 so ϕ = −π/6. Finally, note that the DC value can be written as A = 8 cos(π/3 + ϕ) = Now, 4e−jπ/3 e √ √ 8 cos(π/6) = 8 3/2 = 4 3.

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CHAPTER 3. SPECTRUM P-3.20

DSP First 2e

(a) The gcd of 40 and 90 is 10, so f 0 = 10 Hz. (b) The fundamental period is T0 = 1/ f 0 = 1/10 = 0.1 s. (c) From the plot, the DC value is 0.5. (d) With f 0 = 10, the harmonics are k = 0, ±4, ±9.

© 2022 Pearson Education.

k

−9

−4

ak

0.4e−j2

0.6e j1.4

0

4

9

0.5

0.6e−j1.4

0.4e j2

CHAPTER 3. SPECTRUM P-3.21

DSP First 2e

(a) The instantaneous frequency is ωi (t) = dψ dt = 2αt + β, so ω1 = ωi (0) = β and ω2 = ωi (T2 ) = 2αT2 + β. (b) The instantaneous frequency versus time is ωi (t) = 80t + 27 (c) Here is the plot of the instantaneous frequency (in Hz) versus time over the range 0 ≤ t ≤ 1 sec. i (t) f i (t) = ω2π

6 107 2π

27 2π

-

0

© 2022 Pearson Education.

1

t

CHAPTER 3. SPECTRUM P-3.22

DSP First 2e

(a) The general form for the chirp signal is x(t) = cos(αt 2 + βt + ϕ). The instantaneous frequency of this signal is ωi (t) = 2αt + β. From this we observe that ω1 = 2π f 1 = 2π(4800) = ωi (0) = β. To obtain α, we note that ω2 = 2π(800) = ωi (2) = 2α(2) + β = 4α + 9600π so α = −2000π. Therefore, the signal is x(t) = cos(−2000πt 2 + 9600πt + ϕ) where ϕ is an arbitrary phase constant. (b) The instantaneous frequency is ωi = 800πt + 500π, so ω1 = ωi (0) = 500π and ω2 = ωi (3) = 800π(3) + 500π = 2900π rad/s.

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CHAPTER 3. SPECTRUM P-3.23

DSP First 2e

(a) The instantaneous frequency is ωi (t) = 2αt + β. Substituting the given parameters gives α = 4π and β = 2π, so the signal with the given parameters is x(t) = cos(4πt 2 + 2πt + ϕ).

frequency in Hz

(b–f) The solution to this problem is given in the following figure. Note that the times at which f i (t) is equal to 4 Hz and 8 Hz are indicated with dashed lines. Careful scrutiny of the plots confirms that the waveform of the chirp signal does match the waveforms of the 4 Hz and 8 Hz constant-frequency sinusoids at the corresponding two times.

Instantaneous Frequency of Chirp Signal

10 5 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

4 Hz Signal

1 0.5 0 -0.5 -1

0

0.2

0.4

0.6

0.8

1

Chirp Signal

1 0 -1

0

0.2

0.4

0.6

0.8

1

8 Hz Signal

1 0.5 0 -0.5 -1

0

0.2

0.4

0.6

0.8

1

time in seconds

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CHAPTER 3. SPECTRUM P-3.24

DSP First 2e

1 √

(a) f 1 (t) =

(c) f 3 (t) = e2t /π

frequency in Hz

0.8

frequency in Hz

2π (t) (b) f 2 (t) = t/π

1

(d) f 4 (t) = − sin(2πt) Solution to Problem 3.24

0.6 0.4 0.2 0

0

0.5

1

1.5

2

2.5

3

3.5

4

0.5 0 -0.5

frequency in Hz

150

frequency in Hz

-1 -3

1

-2

-1

0

1

2

3

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5

2

2.5

3

100 50 0

0.5 0 -0.5 -1

time in seconds

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CHAPTER 3. SPECTRUM P-3.25

DSP First 2e

(a) Let x(t) be given by the Fourier series x(t) =

∞ X

ak e j (2π/T0 )kt . Then it follows that

k=−∞

x(0) =

∞ X

∞ :1 X

(2π/T 0 )k (0)

e j = ak ak .

k=−∞

k=−∞

(b) Let f 3 = − f 2 = f 0 and f 4 = − f 1 = 3 f 0 so that from the spectrum we can write x(t) = 12 cos(2π f 0 t + π/4) + 4 cos(6π f 0 t + 3π/4) √ √ √ Therefore x(0) = 12 cos(π/4) + 4 cos(3π/4) = 6 2 − 2 2 = 4 2. Now if we add the coefficients of the Fourier series we get √ a1 + a2 + a3 + a4 = 2e−j3π/4 + 6e−jπ/4 + 6e jπ/4 + 2e j3π/4 = 12 cos(π/4) + 4 cos(3π/4) = 4 2

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CHAPTER 3. SPECTRUM P-3.26

DSP First 2e

The equations corresponding to the spectra are: The matches are

(a) (b) (c) (d) (e)

(3) (1) (2) (5) (4)

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x 1 (t) = 4 cos(4πt + π) + 4 cos(6πt + π/2) x 2 (t) = 2 cos(4πt + π/4) + 4 cos(6πt − 0.333π) x 3 (t) = −3 + 2 cos(4πt + π/4) x 4 (t) = −2 + 4 cos(4πt + π) x 5 (t) = 4 cos(2πt + π) + 4 cos(4πt + π)

CHAPTER 3. SPECTRUM P-3.27

DSP First 2e

(a) x(t) = cos(−250πt 2 ) Spectrogram (2)

(d) x(t) = cos(100πt) cos(400πt) Spectrogram (1)

(b) x(t) = cos(100πt − π/4) + cos(400πt) Spectrogram (5)

(e) x(t) = cos(200πt 2 ) Spectrogram (6)

(c) x(t) = cos(1000πt − 250πt 2 ) Spectrogram (4)

(f) x(t) = cos(30e2t ) Spectrogram (3)

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Chapter

Sampling and Aliasing 4-1

Problems

60 © 2022 Pearson Education.

4

CHAPTER 4. SAMPLING AND ALIASING P-4.1

DSP First 2e

(a) A = 11, ϕ = −π/3, and ω/ f s = 7π/9 = ω̂1 . (b) A = 11, ϕ = +π/3, and ω/ f s = 7π/6 → ω̂1 = 5π/6. (c) A = 11, ϕ = −π/3, and ω/ f s = 7π/3 → ω̂1 = π/3.

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CHAPTER 4. SAMPLING AND ALIASING P-4.2

DSP First 2e

x 1 [n] = x 3 [n] and x 2 [n] = x 4 [n] = x 5 [n]

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CHAPTER 4. SAMPLING AND ALIASING P-4.3

DSP First 2e

(a) y(t) = 8 cos(2160πt − π/3) (b) y(t) = 4 cos(2880πt + π/4) (c) y(t) = 2 cos(1440πt − π/5)

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CHAPTER 4. SAMPLING AND ALIASING P-4.4

DSP First 2e

(a) Sketch will have four spectral lines at ω̂ = ±4000π, ±12000π. The complex amplitude of each spectral line will be 0.25.

0.25

0.25

6

0.25

0.25

−12000π

−4000π

0

4000π

12000π

(b) The Nyquist rate for x(t) is 12,000 samples/s. (c) The Nyquist rate for r (t) is 15 × 106 samples/s. (d) The Nyquist rate for v(t) is 7 × 106 samples/s.

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ω

-

CHAPTER 4. SAMPLING AND ALIASING P-4.5

DSP First 2e

(a) The Nyquist rate for x(t) is 50 samples/s. (b) The Nyquist rate for r (t) is 72 samples/s. (c) The Nyquist rate for v(t) is 45 samples/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.6

DSP First 2e

(a) Sketch will have four spectral lines: at ω̂ = ±0.2π with complex amplitudes equal to 13e±jπ/4 , and at ±0.5π with complex amplitudes of 7e±j3π/4 .

There is no aliasing when f s = 2000 samples/s.

13e−jπ/4 6 13e jπ/4 7e−j3π/4

−π

−0.5π

7e j3π/4

−0.2π

0

0.2π

0.5π

π

-

ω̂

(b) The reconstructed output when the D-to-C rate is f s = 1600 samples/s will be y(t) = 26 cos(320πt+π/4)+14 cos(800πt+ 3π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.7

DSP First 2e

(a) No, because there is aliasing. (b) Sketch will have four spectral lines: at ω̂ = ±4π/9 with complex amplitudes equal to 13e±jπ/4 , and at ±2π/9 with complex amplitudes of 7e±j3π/4 .

When f s = 450 samples/s, there will be an alias for the lines at ±500 Hz, i.e., the spectral line first maps to ω̂ = ±20π/9 and then aliases to ±2π/9.

13e−jπ/4 7e−j3π/4

−π

13e jπ/4

6

−4π/9 −2π/9

7e j3π/4

0

2π/9

4π/9

π

-

ω̂

(c) The reconstructed output when f s = 450 samples/s is y(t) = 26 cos(400πt + π/4) + 14 cos(100πt + 3π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.8

DSP First 2e

(a) Yes. f Nyquist = 1000 Hz. (b) Sketch will have two spectral lines at ω̂ = ±4π/7 with complex amplitudes equal to 6e±jπ/4 .

When f s = 700 samples/s, there will be a folded alias for the lines at ±500 Hz, i.e., the spectral line first maps to ω̂ = ±10π/7 and then aliases (via folding) to ∓4π/7. The complex amplitudes of the coincident spectral lines must be summed.

−π

6e−jπ/4

6

6e jπ/4 = 13e jπ/4 + 7e−j3π/4

−4π/7

0

4π/7

π

-

ωˆ

(c) The reconstructed output when f s = 700 samples/s is y(t) = 12 cos(400πt + π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.9

DSP First 2e

(a) ω = 2π(15200) rad/s. (b) f s = 120 samples/s. (c) f s = 75 samples/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.10

DSP First 2e

(a) ω1 = 18π rad/s. (b) ω̂ = 0.6π rad. (c) ω̂1 = π/7 and ω̂2 = −π/7, but b[n] involves a folded alias so ω̂2 = π/7. (d) Unequal because the phases are different: a[n] = cos(πn/7 + 0.1π) and b[n] = cos(πn/7 − 0.1π).

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CHAPTER 4. SAMPLING AND ALIASING P-4.11

DSP First 2e

(a) Average is 9 samples per period. (b) ω = 8000π rad/s. (c) Average is 0.9 samples per period, or 9 samples taken every 10 periods.

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CHAPTER 4. SAMPLING AND ALIASING P-4.12

DSP First 2e

Frequencies will be: 1000 Hz, 4000 Hz (but x(t) will have a different phase), and 6000 Hz. x 1 (t) = 9.1 cos(2000πt + 0.4π) x 2 (t) = 9.1 cos(8000πt − 0.4π) x 3 (t) = 9.1 cos(6000πt + 0.4π)

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CHAPTER 4. SAMPLING AND ALIASING P-4.13

DSP First 2e

(a) f = 1600 Hz. Phase will be −2.03 rad. (b) fsamp = 9600 Hz. (c) f = 2200 Hz. Duration = 12.8 s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.14

DSP First 2e

(a) Sketch will have six spectral lines at ω = ±2000π, ±(2000π ± 6000) rad/s.

In hertz, these frequencies are f ≈ ±47.07, ±1000, ±1954.93 Hz. The complex amplitudes of the spectral lines are 14 e±jπ/2, 1.5e±j0, 41 e∓jπ/2 .

1.5

1.5 6

j 4

−1955

−1000

(b) Not periodic. (c) f s > 3909.859 Hz.

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− 4j

j 4

−47 0

47

− 4j 1000

1955

-

f (Hz)

CHAPTER 4. SAMPLING AND ALIASING P-4.15

DSP First 2e

(a) x 1 (t) = cos(2π(2800)t + 0.2π) and x 2 (t) = cos(2π(5200)t − 0.2π) (b) y(t) = cos(2π(1750)t + 0.2π)

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CHAPTER 4. SAMPLING AND ALIASING P-4.16

DSP First 2e

(a) x 1 (t) = cos(2π(8700)t + 0.2π) and x 2 (t) = cos(2π(7300)t − 0.2π) (b) x(t) = cos(2π(7300)t − 0.2π) (c) f = 8000 − f s

ω̂ Hz, and ϕ = −0.2π rad.

2π

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CHAPTER 4. SAMPLING AND ALIASING P-4.17

DSP First 2e

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CHAPTER 4. SAMPLING AND ALIASING P-4.18

DSP First 2e

(a) ω1 = 12π rad/s. (b) ω̂ = 0.8π rad. (c) ω1 = ω2 = 200π rad/s. ) ( (d) The three possible answers are 2π(14 27 ), 2π(28 74 ), 2π(33 13 ) rad/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.19

DSP First 2e

(a) Run the code to make the stem plot.

(b) Period of x[n] in the plot is 5 samples. (c) ω̂0 = 0.4π rad, and x[n] = 2 cos(0.4πn − 0.6π). (d) 2π(380/100) = 7.6π > 2π and 7.6π = 4(2π) − 0.4π.

Over the total duration of 0.3 s, a 380 Hz continuous-time signal would have 114 periods, but x[n] only has 6 complete periods.

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CHAPTER 4. SAMPLING AND ALIASING P-4.20

DSP First 2e

(a) Expect 34 zero crossings in 17 periods. Number of positive peaks = number of negative peaks = 17. (b) 11 zero crossings, 6 positive peaks and 6 negative peaks. (c) T = 0.2 s, so f = 5 Hz. Sampling a 15-Hz sinusoid at 20 Hz (Ts = 0.05) s. involves aliasing, and a 5-Hz sinusoid is a folded version of the 15-Hz sinusoid. (d) x[n] = 1.005 cos(0.4πn − 0.1). (e) To obey the Sampling Theorem, Ts < (1/30) s, but to make a smooth plot Ts < 0.05 s would be needed.

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CHAPTER 4. SAMPLING AND ALIASING P-4.21

DSP First 2e

x(t) = (3 − j0.5e jπt + j0.5e−jπt )(0.5e j13πt+jπ/2 + 0.5e−j13πt−jπ/2 ) = 1.5e j13πt+jπ/2 + 1.5e−j13πt−jπ/2 + 0.25e j14πt + 0.25e−j14πt + 0.25e j12πt+jπ + 0.25e−j12πt−jπ = 3 cos(13πt + π/2) + 0.5 cos(14πt) + 0.5 cos(12πt + π) (a) ω1 = 12π rad/s, A1 = 0.5, ϕ1 = π; ω2 = 13π rad/s, A2 = 3, ϕ2 = π/2; ω3 = 14π, A3 = 0.5, ϕ3 = 0; (b) Use the second line above to plot six spectrum lines. −3 j

3j

6

0.5

0.5

−14π −13π −12π (c) Nyquist rate is 28π rad/s, or 14 Hz.

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0.5 0

12π

0.5 13π

14π

-

ω (rad/s)

CHAPTER 4. SAMPLING AND ALIASING P-4.22

DSP First 2e

(a) Sampling rate is greater than 2(300π) rad/s, or 300 Hz. (b) x[n] = x(t) t=n/250 = 2 cos(2π(50)n/250 + π/2) + cos(2π(150)n/250) = 2 cos(2πn/5 + π/2) + cos(6πn/5) = 2 cos(0.4πn + 0.5π) + cos(0.8πn) (c) Plot four lines; two at ±0.4π with complex amps equal to 1e±j0.5π ; and two at ±0.8π with complex amps equal to 10.5.

Note: amplitudes are not to scale.

10.5

e−jπ/2 −π −0.8π

10.5

6

−0.4π

(d) f s = 300π rad/s, or 150 Hz.

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e jπ/2 0

0.4π

0.8π

π

-

ωˆ

CHAPTER 4. SAMPLING AND ALIASING P-4.23

DSP First 2e

(a) The discrete-time exponential when flashing at 15 flashes/s is p[n] = e+j2π (13)n/15+jπ/2 = e j2π (−2)n/15+jπ/2 . Thus, at each flash the spot moves −4π/15 rad, which is 48◦ clockwise. (b) p(t) = p[n]

n=15t

= e j2π (−2)(15t)/15+π/2

appears to rotate through an angle of −4π every second. This is 2 rev/s clockwise. (c) Thirteen flashes in one second; disk moving CCW which is negative. The disk moves 2π(−15/360) + 2π` rad per flash. Thus the disk moves 2π(13)(−1/24 + `) in one second. When ` = 0, the rotation speed is 13/24 = 0.54 rev/s CCW. When ` < 0, the rotation speed is |13` − 13/24| rev/s CCW. When ` > 0, the rotation speed is |13` − 13/24| rev/s CW. For example, when ` = 1, the speed is 12.46 rev/s CW. Complex Exponential Derivation of Rotation Speeds (r) in rev/s: The discrete-time exponential after flashing at f s flashes/s is p[n] = e+j2π (r )n/fs . Thus, to get the movement of 15◦ CCW, which is −2π/24 rad/s CCW, we must determine whether or not there is is an integer ` for which the following is true: * 1e j2π` e−j2π (15/360)

p[1] = e+j2π (r )(1)/fs = p[0]

e+j2π(r )/fs = e−jπ/12 e j2π`

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⇒ 2π(r)/ f s = 2π` − 2π/24

⇒ r = f s (` − (1/24))

CHAPTER 4. SAMPLING AND ALIASING P-4.24

DSP First 2e

(a) p(t) = e+j2π (12)t rotates through an angle of 24π every second. (b) The discrete-time exponential after flashing at n flashes/s is p[m] = e+j2π (12)m/n . Thus, given n, we must determine whether or not there is is an integer ` for which the following is true: p[1] = p[0] e

+j2π(12)(1)/n

= e j2π`

⇒ 24π/n = 2π`

⇒ 12/n = `

So, n must be a divisor of 12, which gives flashing rates of n = 1, 2, 3, 4, 6, 12 flashes/s. (c) When n = 13, p[1] = e+j2π (12)/13 = e−j2π (1)/13 , so the disk rotates clockwise by 2π/13 rad, or 360/13 = 27.69◦ , for every flash. The observed rotation is 1 rev/s clockwise.

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CHAPTER 4. SAMPLING AND ALIASING P-4.25

DSP First 2e

(a) Plot 8 unit-length vectors (with zvect). The angles will be {0.2π, 0.5π, 0.8π, 0, 0.3π, −0.5π, 0.5π, −0.5π}. (b) Period is 20. (c) Plot 5 unit-length vectors (with zvect). The angles will be {0, π/15, 4π/15, −14π/15, −14π/15}. (d) Shortest period is 30.

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CHAPTER 4. SAMPLING AND ALIASING P-4.26

DSP First 2e

(a) θ[n] = 0.07π, 1.75π, 7π are the angles of the vectors. When reduced to the −π to +π range, these angles are θ[n] = 0.07π, −0.25π, π. (b) The instantaneous frequency of the signal below is 2π(44.8 × 103 )t Hz. The plot runs from t = 0 to t = 200/8000 = 1/40 s, so it starts at f = 0 and ends at (1/40)44.8 × 103 = 1120 Hz. x(t) = x[n]

n=8000t

= cos(π(0.7 × 10−3 × 64 × 106 )t 2 ) = cos(π(44.8 × 103 )t 2 )

(c) 2800 Hz

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CHAPTER 4. SAMPLING AND ALIASING P-4.27

DSP First 2e

ALPHA = 437.5, BETA = 2000, fs = 8000

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Chapter

FIR Filters 5-1

Problems

88 © 2022 Pearson Education.

5

CHAPTER 5. FIR FILTERS P-5.1

DSP First 2e

y[n] = 7x[n] + x[n − 3] − 5x[n − 4]

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CHAPTER 5. FIR FILTERS P-5.2

DSP First 2e

G = 9e−j0.5π (3) − 9e−j0.5π (4)

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√ e−j0.5πn = 9 2 e+j3π/4

CHAPTER 5. FIR FILTERS P-5.3

DSP First 2e

(a) First of all, y[n] = 0 for n = −2, −1. The system is causal. Then y[n] = n + 1 for n = 0, 1, 2, 3; and y[n] = 4 for n ≥ 4. (b) Plot with Matlab.

(c)  0 n<0      y[n] =  n + 1 0 ≤ n < L   L n≥L 

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CHAPTER 5. FIR FILTERS P-5.4

DSP First 2e

(a) Use numerical convolution table: n n<0 0 1 2 3 4

5

6

7 n>7

1

1

1

1

1

2

2

2

2

2

x[n]

0

1

2

3

2

h[n]

0

2 −3

2

0

h[0]x[n]

0

2

6

4

h[1]x[n−1]

0

0 −3 −6 −9 −6 −3 −3 −3 −3

h[2]x[n−2]

0

0

0

2

4

6

4

2

2

2

y[n]

0

2

1

2 −1

2

3

1

1

1

4

(b) (c) h[n] = 2δ[n] − 3δ[n − 1] + 2δ[n − 2]

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CHAPTER 5. FIR FILTERS P-5.5

DSP First 2e

Here are the figures: 2 2

xŒn

xŒn

v2 Œn Unit Delay xŒn 1

Unit Delay

3 3

v1 Œn Unit Delay xŒn 2

Unit Delay

2

2

yŒn

yŒn

(a)

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(b)

CHAPTER 5. FIR FILTERS P-5.6

DSP First 2e

(a) P = N + M (b) Input signal is time-shifted by N1 . Thus, the output starts at N3 = N1 , and the last nonzero value in the output is at N4 = N2 + M. The length of y[n] is N4 − N3 + 1 = N2 − N1 + M + 1.

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CHAPTER 5. FIR FILTERS P-5.7

DSP First 2e

(a) Make a plot in Matlab.

(b) First of all, y[n] = 0 for n = −2, −1. The system is causal. Then y[n] = 1 for n = 0; and y[n] = (−0.5) n + (−0.5) n−1 for n = 1, 2, 3, 4, 5, 6, 7, 8. The nonzero values starting at n = 0 are 1.0, 0.5, −0.25, 0.125, −0.0625, 0.03125, −0.015625, 0.0078125, −0.00390625 (c) Causality implies that y[n] = 0 for n < 0.

Running onto the signal (partial overlap) region which is 0 ≤ n < L−1: sum (−0.5) k for k = 0 to k = n n X

(−0.5) k =

k=0

(−0.5) 0 − (−0.5) n+1 1 − (−0.5) n+1 = 1 − (−0.5) 1.5

Complete overlap region: sum (−0.5) k for k = n − L + 1 to k = n which is n ≥ L−1: n X

(−0.5) k =

k=n−L+1

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(−0.5) −L − 1 (−0.5) n−L+1 − (−0.5) n+1 = (−0.5) n+1 1 − (−0.5) 1.5

CHAPTER 5. FIR FILTERS P-5.8

DSP First 2e

Filter order is M = 4, and bk = {3, 7, 13, 9, 5}

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CHAPTER 5. FIR FILTERS P-5.9

DSP First 2e

Impulse response is h[n] = 13δ[n] − 13δ[n − 1] + 13δ[n − 2]. The output can be obtained via convolution: y[n] = h[n] ∗ x[n].

Assume that the input is nonzero for −∞ < n < ∞, which implies that x[n − 2] is equal to x[n]. Also, x[n − 1] will be equal to 0 for n odd, and equal to 1 for n even.   26 y[n] = 13x[n] − 13x[n − 1] + 13x[n − 2] = 26x[n] − 13x[n − 1] =   −13 

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n odd n even

CHAPTER 5. FIR FILTERS P-5.10

DSP First 2e

(a) y[n] = x[n] cos(0.2πn)

(1) linear YES, (2) time-invariant NO, and (3) causal YES.

(b) y[n] = x[n] − x[n − 1] (1) linear YES, (2) time-invariant YES, and (3) causal YES. (c) y[n] = |x[n]|

(1) linear NO, (2) time-invariant YES, and (3) causal YES.

(d) y[n] = Ax[n] + B, where A and B are nonzero constants.

(1) linear NO, (2) time-invariant YES, and (3) causal YES.

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CHAPTER 5. FIR FILTERS P-5.11

DSP First 2e

(a) Plot of y[n]

(b) Use LTI properties with x[n] = 7 (δ[n] − δ[n − 1]) +7 (δ[n − 1] − δ[n − 2]). | {z } first x[n] in table

Thus, y[n] = 7 (δ[n] − δ[n − 1] + 2δ[n − 3]) +7 (δ[n − 1] − δ[n − 2] + 2δ[n − 4]) | {z } first x[n] in table

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CHAPTER 5. FIR FILTERS P-5.12

DSP First 2e

Note that the Delay by 2 block in (b) is the cascade of two unit delays.

1 xŒn

3

v3 Œn

xŒn

Unit Delay xŒn 1

Delay by 2

4 2

v2 Œn Unit Delay xŒn 2

Unit Delay

2 4

v1 Œn Delay by 2 xŒn 4

Unit Delay

3

1

yŒn

yŒn

(a)

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(b)

CHAPTER 5. FIR FILTERS P-5.13

DSP First 2e

(a) Use the LTI properties. The output will be y2 [n] = y1 [n] − 2y1 [n − 2]. y2 [n] = δ[n] − 3δ[n − 1] − 3δ[n − 2] − 2δ[n − 2] + 6δ[n − 3] + 6δ[n − 4] = δ[n] − 3δ[n − 1] − 5δ[n − 2] + 6δ[n − 3] + 6δ[n − 4] (b) A = 1 and B = −1. (c) Using LTI properties, h[n] = ( A/3)y1 [n] + (B/3)y1 [n − 1] = (1/3) (δ[n] − 3δ[n − 1] − 3δ[n − 2]) − (1/3) (δ[n − 1] − 3δ[n − 2] − 3δ[n − 3]) = (1/3)δ[n] − (4/3)δ[n − 1] + δ[n − 3]

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CHAPTER 5. FIR FILTERS P-5.14

DSP First 2e

Do the convolutions with h[n] = δ[n] − aδ[n − 1] (a) x[n] ∗ h[n] = a n u[n] ∗ (δ[n] − aδ[n − 1]) = a n u[n] − aa n−1 u[n − 1] = a n (u[n] − u[n − 1]) = δ[n] | {z } δ[n]

(b) Use the result from the previous part. h[n] ∗ a n (u[n] − u[n − 8]) = h[n] ∗ a n u[n]) − h[n] ∗ a8 a n−8 u[n − 8] {z } | {z } | δ[n]

(c) Sketch the plot which has only two nonzero stems, 1 at n = 0, and a8 at n = 8.

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a8 δ[n−8]

CHAPTER 5. FIR FILTERS P-5.15

DSP First 2e

Use trial and error, and just do the convolutions (a) x[n] ∗ δ[n − 3] = x[n − 3], so we want x[n − 3] = 5u[n − 2] − u[n − 6]. Unshift to get x[n] = 5u[n + 1] − u[n − 3]. (b) y[n] = u[n − 2] − u[n − 3] which simplifies to y[n] = δ[n − 2] (c) x[n] − x[n − 1] = y[n] = δ[n − 2] − δ[n − 7].

Rewrite y[n] as y[n] = u[n − 2] − u[n − 3] − u[n − 7] − u[n − 8]. {z } | {z } | δ[n−2]

δ[n−7]

Rewrite again as y[n] = u[n − 2] − u[n − 7] − (u[n − 3] − u[n − 8]) . | {z } | {z } x[n]

x[n−1]

(d) x[n] + x[n − 1] = y[n] = −8δ[n] − 8δ[n − 3].

Mimic the previous part to obtain x[n] = −8u[n] + 8u[n − 3]

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CHAPTER 5. FIR FILTERS P-5.16

DSP First 2e

It is easy to verify these; just do the convolutions (a) h[n] = δ[n − 1] (b) h[n] = δ[n] − δ[n − 1] (c) h[n] = δ[n − 1] − 0.5δ[n − 2] because ( 21 ) n−1 u[n − 1] − 12 ( 12 ) n−2 u[n − 2] is zero for n ≥ 2.

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CHAPTER 5. FIR FILTERS P-5.17

DSP First 2e

If an FIR filter with impulse response h[n] is used to deconvolve, then y[n] = h[n] ∗ (δ[n] + δ[n − 1]) = h[n] + h[n − 1] *0

h[−1] ⇒ h[0] = 1 We want y[0] = δ[0] = 1 ⇒ 1 = y[0] = h[0] + 1 >

We want y[1] = δ[1] = 0 ⇒ 0 = y[1] = h[1] + h[0] ⇒ h[1] = 1 At this point we know that the order of the filter is at least M = 1. For an M th order FIR filter the coefficient of z −M must be nonzero, i.e., b M , 0. With that fact, the following shows that y[M + 1] will also be nonzero, so y[n] cannot be equal to δ[n]. h[n] =

M X

bk δ[n − k]

where b M , 0

k=0

*0

⇒ y[M + 1] = h[M + 1] + h[M] = h[M] = b M , 0

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CHAPTER 5. FIR FILTERS P-5.18

DSP First 2e

(a) Let x 1 [n] = δ[n], x 2 [n] = δ[n], and x 3 [n] = δ[n], S1 :

h1 [n] = δ[n] + δ[n − 1]

S2 :

h2 [n] = δ[n] − δ[n − 2]

S3 :

h3 [n] = δ[n − 1] − δ[n − 2]

(b) Carry out the convolutions: h[n] = h1 [n] ∗ h2 [n] ∗ h3 [n] h[n] = (δ[n] + δ[n − 1) ∗ (δ[n] − δ[n − 2]) ∗ (δ[n − 1] − δ[n − 2]) | {z } h[n] = (δ[n] + δ[n − 1) ∗ (δ[n − 1] − 2δ[n − 2] + δ[n − 3]) h[n] = δ[n − 1] − δ[n − 2] − δ[n − 3] + δ[n − 4] (c) Get the filter coefficients from h[n], and then y[n] = x[n − 1] − x[n − 2] − x[n − 3] + x[n − 4]

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CHAPTER 5. FIR FILTERS P-5.19

DSP First 2e

In this case, the three systems are as follows: S1 :

y1 [n] = x[n] + 2x[n − 1] + 3x[n − 2] + 4x[n − 3]

h1 [n] = δ[n] + 2δ[n − 1] + 3δ[n − 2] + 4δ[n − 3]

S2 :

y2 [n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3]

h2 [n] = δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3]

S3 :

y[n] = w[n] − w[n − 1]

h3 [n] = δ[n] − δ[n − 1]

where w[n] = y1 [n] + y2 [n]. The impulse responses are also given above (a) We need three blocks and one adder. Systems S 1 and S 2 are in parallel; their outputs are added and the sum is processed by S 3 . (b) We can get h[n] from the combination of the individual systems, and the write the difference equation from the filter coefficients. h[n] = (h1 [n] + h2 [n]) ∗ h3 [n] h[n] = (2δ[n] + 3δ[n − 1] + 4δ[n − 2] + 5δ[n − 3]) ∗ (δ[n] − δ[n − 1]) h[n] = 2δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3] − 5δ[n − 4] y[n] = 2x[n] + x[n − 1] + x[n − 2] + x[n − 3] − 5x[n − 4] If we had the z-transform at this point we would write

H (z) = 1 + 2z −1 + 3z −2 + 4z −3 + 1 + z −1 + z −2 + z −3 1 − z −1

H (z) = 2 + 3z −1 + 4z −2 + 5z −3 1 − z −1 = 2 + z −1 + z −2 + z −3 − 5z −4

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Chapter

Frequency Response 6-1

Problem Solutions

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6

CHAPTER 6. FREQUENCY RESPONSE P-6.1

DSP First 2e

y[n] = x[n] − x[n − 1] = e j (0.4πn−0.5π) − e j (0.4π(n−1)−0.5π) = (1 − e−j0.4π )e j (0.4πn−0.5π) = H (e j0.4π )e j (0.4πn−0.5π) H (e j0.4π ) = 1 − e−j0.4π = 1.1756e j0.3π Therefore, A = 1.1756, ϕ = 0.3π − 0.5π = −0.2π, and ω̂0 = 0.4π.

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CHAPTER 6. FREQUENCY RESPONSE P-6.2

DSP First 2e

(a) y[n] = x[n] + x[n − 1] = (e j0.1πn + e j0.4πn ) + (e j0.1π (n−1) + e j0.4π (n−1) ) = (1 + e−j0.1π )e j0.1πn + (1 + e−j0.4π )e j0.4πn = H (e j0.1π )e j0.1πn + H (e j0.4π )e j0.4πn where H (e j ω̂ ) = 1 + e−j ω̂ . (b) The system is LTI, so superposition applies. Each complex exponential is modified by the frequency response evaluated at its frequency. (c) H (e j0.1π ) = 1 + e−j0.1π = A1 e jϕ1 = 1.9754e−j0.05π H (e j0.4π ) = 1 + e−j0.4π = A2 e jϕ2 = 1.6180e−j0.2π

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CHAPTER 6. FREQUENCY RESPONSE P-6.3

DSP First 2e

2 (a) y[n] = e j0.1πn + e j0.4πn = e j0.2πn + 2e j0.5πn + e j0.8πn (b) Clearly, the squaring operation produces three complex exponential signals each with a different frequency that is not one of the original two frequencies. (c) The system is clearly not linear since the output cannot be written as the sum of two terms involving the two input frequencies. The principle of superposition is not satisfied, therefore the system is NOT linear.

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CHAPTER 6. FREQUENCY RESPONSE P-6.4

DSP First 2e

(a) H (e j ω̂ = 2 − 2e−j ω̂ + 2e−j ω̂2 = 2e−j ω̂ (e j ω̂ − 1 + e−j ω̂ ) = 2e−j ω̂ (2 cos(ω̂) − 1) Thus, R(e j ω̂ ) = 4 cos(ω̂) − 2 and n0 = 1. (b) The period = 2π

5 0 −π π

−π/2

0

π/2

−π/2

0

π/2

π

3π/2

2π

5π/2

3π

π

3π/2

2π

5π/2

3π

0

6

H(ej ω̂ )

|H(ej ω̂ )|

10

(c)

−π −π

ω̂

(d) The output is zero whenever the magnitude of the frequency response is zero; i.e., when |2 cos(ω̂) − 1| = 0 or when cos(ω̂) = 0.5. The solution is therefore ω̂ = π/3 rad. (e) We know that for a cosine input with frequency ω̂0 = 0.5, the output is y[n] = |H (e j0.5π )| cos(0.5πn + ∠H (e j0.5π )) where H (e j0.5π ) = 2e−j0.5π (2 cos(0.5π) − 1) = −2e−j0.5π = 2e j0.5π ⇒ y[n] = 2 cos(0.5πn + 0.5π) = 2 cos(0.5π(n + 1)).

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CHAPTER 6. FREQUENCY RESPONSE P-6.5

DSP First 2e

(a) H (e j ω̂ ) = 3 + 2e−j ω̂ + e−j ω̂2 (b) y[n] = H (e j0 ) + 2|H (e j0.5π )| cos(0.5πn + 0.1π + ∠H (e j0.5π )) √ where H (e j0 ) = 6 and H (e j0.5π ) = 3 − 2e−j0.5π + e−jπ = 2e−j0.25π . √ √ Thus, y[n] = 6 + 2 2 cos(0.5πn + 0.1π − 0.25π) = 6 + 2 2 cos(0.5πn − 0.15π) (c)  0      3 y[n] = 3u[n] + 2u[n − 1] + u[n − 2] =   5     6 

n<0 n=0 n=1 n≥2

(d) The unit step input is identical to a suddenly applied complex exponential of zero frequency. Thus, for an FIR filter like this, the steady state response is simply H (e j0 ) times 1. The steady state is reached at n = 2 because the filter order is 2.

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CHAPTER 6. FREQUENCY RESPONSE P-6.6

DSP First 2e

(a) H (e j ω̂ ) = 1 − e−j ω̂4 = e−j ω̂2 (e j ω̂2 − e−j ω̂2 ) = 2 je−j ω̂2 sin(2ω̂) (b) y[n] = 4H (e j0 ) + |H (e j0.25π )| cos(0.25πn − 0.25π + ∠H (e j0.25π )) where H (e j0 ) = 0 and H (e j0.25π ) = 2 j (− j) sin(π/2) = 2. Thus, y[n] = 2 cos(0.25πn − 0.25π) (c) Since the impulse response has length 5 samples, the output reaches the steady state at n = 4 so the output due to the suddenly applied signal is identical to the output in (b) for n ≥ 4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.7

DSP First 2e

(a) h[n] = 3δ[n] − δ[n − 2] (b) We need to expand H (e j ω̂ ) in terms of complex exponentials. ! j2ω̂ + e−j2ω̂ j ω̂ −j3ω̂ e H (e ) = 5e = 2.5e−j ω̂ + 2.5e−j5ω̂ 2 Therefore h[n] = 2.5δ[n − 1] + 2.5δ[n − 5]. (c) We should recognize this as the form of an L sample running sum whose frequency response is H (e

j ω̂

)=e

−j ω̂(L−1)/2 sin(ω̂L/2)

sin(ω̂/2)

=

L−1 X

e−j ω̂n

n=0

Thus, it follows that L = 10 and the impulse response is ( 1 0≤n≤9 h[n] = u[n] − u[n − 10] = 0 otherwise

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CHAPTER 6. FREQUENCY RESPONSE P-6.8

DSP First 2e

(a) To get the difference equation we need to expand the factors of H (e j ω̂ ) so that we get powers of e−j ω̂ . √ H (e j ω̂ ) = (1 − e−j ω̂ )(1 − 2 cos(3π/4)e−j ω̂ + e−j2ω̂ ) = (1 − e−j ω̂ )(1 − 2e−j ω̂ + e−j2ω̂ ) √ √ = 1 + ( 2 − 1)e−j ω̂ − ( 2 − 1)e−j2ω̂ − e−j3ω̂ = 1 + 0.4142e−j ω̂ − 0.4142e−j2ω̂ − e−j3ω̂ Therefore, the difference equation is y[n] = x[n] + 0.4142x[n − 1] − 0.4142x[n − 2] − x[n − 3] (b) 0        1  1.4142 y[n] = u[n] + 0.4142u[n − 1] − 0.4142u[n − 2] − u[n − 3] =      1     0

n<0 n=0 n=1 n=2 n≥3

(c) We need to find the values of ω̂ such that H (e j ω̂ ) = 0. These are easily found from the given factored form as ω̂ = 0 and ω̂ = ±3π/4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.9

DSP First 2e

(a) We need to multiply out the factors of H (e j ω̂ ) to get: H (e j ω̂ ) = 1 + 0.5e−j ω̂ − 0.25e−j2ω̂ + 0.25e−j3ω̂ Therefore, the difference equation is y[n] = x[n] + 0.5x[n − 1] − 0.25x[n − 2] + 0.25x[n − 3] (b) By definition, the impulse response is: h[n] = δ[n] + 0.5δ[n − 1] − 0.25δ[n − 2] + 0.25δ[n − 3] (c) We need to find the values of ω̂ such that H (e j ω̂ ) = 0. The factored form shows that this can be true for values of ω̂ such that 1 + e−j2ω̂ = 0 or ω̂ = ±π/2 rad.

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CHAPTER 6. FREQUENCY RESPONSE P-6.10

DSP First 2e

(a) Plot {3, 0, −2, 1} for 0 ≤ n ≤ 3. (b) The first entry in the table says that the impulse response is h[n] = δ[n] − δ[n − 3]. Therefore, y[n] = x[n] − x[n − 3].

For the given input the output is y[n] = 3δ[n] − 2δ[n − 2] + δ[n − 3] − 3δ[n − 3] + 2δ[n − 5] − δ[n − 6] = 3δ[n] − 2δ[n − 2] − 2δ[n − 3] + 2δ[n − 5] − δ[n − 6] Plot {3, 0, −2, −2, 0, 2, −1} for 0 ≤ n ≤ 6. (c) In this case we can use the third result in the table, which says that the frequency response at ω̂ = π/3 is H (e jπ/3 ) = 2; i.e., the phase shift is zero. Therefore, the output for this input is y[n] = 2 cos(π(n − 3)/3) (d) It is FALSE because the impulse response is δ[n] − δ[n − 3], so the frequency response is H (e j ω̂ ) = 1 − e−j3ω̂ which is nonzero at ω̂ = π/2.

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CHAPTER 6. FREQUENCY RESPONSE P-6.11

DSP First 2e

(a) The zeros marked with o are at multiples of 2π/8 with gaps at 0, ±2π, ±4π, . . ..

D8 (ω̂)

1 0

−1 −3π

−2π

−π

0 Figure P-6.11

π

2π

3π

(b) The period is 4π. (c) The maximum value occurs at ω̂ = 0. Using the small angle approximation for the sine functions we get D L (0) ≈

(L ω̂/2) =1 L(ω̂/2)

We get the same result using L’ Hospital’s Rule.

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CHAPTER 6. FREQUENCY RESPONSE P-6.12

DSP First 2e

(a) From the difference equation, we can write down H (e j ω̂ ) as H (e j ω̂ ) = 1 − 3e−j ω̂ + 3e−j2ω̂ − e−j3ω̂ = (1 − e−j ω̂ ) 3 = e−j ω̂3/2 (e j ω̂/2 − e−j ω̂/2 ) 3 = e−j ω̂3/2 (2 j) 3 (sin(ω̂/2)) 3 = 8e−j ω̂3/2 (− j)(sin(ω̂/2)) 3 = 8 sin3 (ω̂/2)e j (−π/2−3ω̂/2)

|H(ej ω̂ )|

10 5 0 −2π

−3π/2

−π

−π/2

0

π/2

π

3π/2

Figure P-6.12

(b) (c) We need to determine the frequency response at 0 and 0.5π. √ √ These are H (e j0 ) = 0 and H (e j0.5π ) = 8(1/ 2) 3 e j (−π/2−3π/4) = 2 2e−j7π/4 . Therefore, y[n] = 5H (e j0 ) + 6|H (e j0.5π )| cos(0.5π(n − 1) + ∠H (e j0.5π )) √ = 12 2 cos(0.5π(n − 1) − 7π/4) (d) h[n] = δ[n] − 3δ[n − 1] + 3δ[n − 2] − δ[n − 3] (e) y[n] = 5H (e j0 ) + 6|H (e j0.5π )| cos(0.5πn + ∠H (e j0.5π )) + 9h[n − 3] √ = 12 2 cos(0.5πn − 7π/4) + 9h[n − 3] where h[n] is as determined in part (d).

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2π

CHAPTER 6. FREQUENCY RESPONSE P-6.13

DSP First 2e

(a) S1 :

H1 (e j ω̂ ) = 1 − e−j ω̂

S2 :

H2 (e j ω̂ ) = 1 + e−j2ω̂

S3 :

H3 (e j ω̂ ) = e−j ω̂ + e−j2ω̂

(b) H (e j ω̂ ) = (1 − e−j ω̂ )(1 + e−j2ω̂ )(e−j ω̂ + e−j2ω̂ ) (c) Multiply out all the factors in H (e j ω̂ ) and then pick off the coefficients to get the difference equation.

H (e j ω̂ ) = (1 − e−j ω̂ + e−j2ω̂ − e−j3ω̂ )(e−j ω̂ + e−j2ω̂ ) = e−j ω̂ − e−j5ω̂ Therefore, y[n] = x[n − 1] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.14

DSP First 2e

(a) H (e j ω̂ ) =

4 X n=0

e−j ω̂n =

1 − e−j5ω̂ sin(ω̂5/2) = e−j ω̂2 sin(ω̂/2) 1 − e−j ω̂

(b) H (e j ω̂ ) = 0 for ω̂ = ±0.4π, ±0.8π (c) We need the frequency response at ω̂ = 0, 0.25π, and 0.4π.

H (e j0 ) = 5 H (e j0.25π ) = e−j0.5π

sin(5π/8) = 2.4142e−j0.5π sin(π/8)

H (e j0.4π ) = 0 Therefore, y[n] = 35 + 8(2.4142) cos(0.25πn − π/2) (d) Since the length of the impulse response is 5, the steady-state output is attained for n ≥ 4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.15

DSP First 2e

The sampled signal is x[n] = x(n/ f s ) = 7 + 8 cos(1000πn/4000) + 9 cos(1600πn/4000) = 7 + 8 cos(0.25πn) + 9 cos(0.4πn + 0.7π) This is the same input signal as in Problem ?? and the discrete-time system is the same as in that problem. Therefore, from the solution to Problem ??(c), the output sequence is y[n] = 35 + 8(2.4142) cos(0.25πn − π/2) The output of the D-to-C converter is therefore y(t) = 35 + 8(2.4142) cos(1000πn − π/2)

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CHAPTER 6. FREQUENCY RESPONSE P-6.16

DSP First 2e

(a) f s > 1000 Hz. Since h[n] = δ[n], y[n] = x[n]. In other words, the system is a C-to-D converter followed by a D-to-C converter, so the issue is sampling above the Nyquist rate. The highest frequency in the input signal is 500 Hz, so the Nyquist rate is 2(500) = 1000 Hz. (b) A delay of 10 samples must correspond to 0.001 s. After sampling x(t) at a rate of f s x[n] = 10 + 20 cos(ω0 (n/ f s ) + π/3) y[n] = x[n − 10] = 10 + 20 cos(ω0 (n − 10)/ f s + π/3) Reconstructing y(t) from y[n] with an ideal D-to-C converter gives y(t) = 10 + 20 cos(ω0 ( f s t − 10)/ f s + π/3) = 10 + 20 cos(ω0 t − 10ω0 / f s + π/3) We want y(t) = x(t − 0.001) = 10 + 20 cos(ω0 (t − 0.001) + π/3) which requires −10ω0 / f s = ω0 (−0.001)

⇒

f s = 10, 000 Hz

In addition, the output frequency must be the same as the input frequency, so there must be no aliasing, i.e., the relationship between ω0 and f s = 10, 000 Hz must obey the Nyquist condition.

2ω0 < 2π f s

⇒

ω0 < 2π(5000) rad/s

(c) To get y(t) = A, we need y[n] = constant. Since x[n] = 10 + 20 cos(ω0 (n/ f s ) + π/3), the filter must null out the cosine term; or the cosine term must alias to DC. sin(5ω̂/2) −j ω̂2 e are at ω̂ = 2π/5, −2π/5, 4π/5, −4π/5.

Nulling: The zeros of H (e j ω̂ ) = 5 sin(ω̂/2) For ω0 and f s = 2000 Hz, these frequencies are 2π(±400) and 2π(±800) rad/s.

The negative frequencies must be used when we include aliases. The complete list of input frequencies that alias to the zeros of H (e j ω̂ ) is 2π(400 + 2000`) = 2π(400), 2π(2400), 2π(4400), . . .

2π(800 + 2000`) = 2π(800), 2π(2800), 2π(4800), . . .

2π(−400 + 2000`) = 2π(−400), 2π(1600), 2π(3600), . . .

2π(−800 + 2000`) = 2π(−800), 2π(1200), 2π(3200), . . .

For these input frequencies, the cosine term will be filtered out, and only the DC remains. Since H (e j0 ) = 1 and the DC value of x(t) = 10, the output will be y(t) = A = 10.

Cosine aliases to DC: When ω0 is a frequency that aliases to DC, then ω̂0 = (ω0 + 2π` f s )/ f s = 0, and x[n] = 10 + 20 cos(0n + π/3) = 10 + 20 cos(π/3) = 20.

The frequencies for which this happens are 2π(0 + 2000`) = 2π(0), 2π(2000), 2π(4000), . . .

In this case, the DC value of x(t) = 20 and H (e j0 ) = 1, so the output will be y(t) = A = 20.

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CHAPTER 6. FREQUENCY RESPONSE P-6.17

DSP First 2e

(a) The constant component passes through the first system as a constant. Then the first difference system gives a zero output for a constant input. By superposition, the output is thus equal to y1 [n], the output due to the other input component x 1 [n]. (b) The overall frequency response function is the product of the frequency responses. Therefore, H (e

j ω̂

!

4 X 1 − e−j5ω̂ −j ω̂n + −j ω̂ * (1 − e−j ω̂ ) = 1 − e−j5ω̂ e )= (1 − e ) = −j ω̂ 1−e , n=0 -

(c) The frequency response is zero at the 5th roots of unity or ω̂ = 2πk/5 for k = 0, 1, 2, 3, 4. (d) From the frequency response, the overall difference equation is y[n] = x[n] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.18

DSP First 2e

(a) Need a plot of h[n] = −δ[n − 2] + δ[n − 12]

(b) H (e j ω̂ ) = −e−j2ω̂ +e−j12ω̂ = e−j7ω̂ (−e j5ω̂ +e−j5ω̂ ) = e−j7ω̂ (−2 j) −2 sin(5ω̂) and n0 = −7.

e j5ω̂ − e−j5ω̂ = j (−2 sin(ω̂5))e−j7ω̂ Therefore, R(e j ω̂ ) = 2j

2 1

0 −π 5

−π/2

0

π/2

π

0

π/2

π

0

6

H(ej ω̂ )

|H(ej ω̂ )|

(c,d) Plots of |H (e j ω̂ )| and the principal value of ∠H (e j ω̂ ) for −π < ω̂ ≤ π.

-5 −π

−π/2

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ω̂

CHAPTER 6. FREQUENCY RESPONSE P-6.19

DSP First 2e

(a) H (e j ω̂ ) = H1 (e j ω̂ )H2 (e j ω̂ ) = (1 + 2e−j ω̂ + e−j ω̂2 )(1 − e−j ω̂ + e−j ω̂2 − e−j ω̂3 ) = 1 + e−j ω̂ − e−j ω̂4 − e−j ω̂5 (b) h[n] = δ[n] + δ[n − 1] − δ[n − 4] − δ[n − 5]

(c) y[n] = x[n] + x[n − 1] − x[n − 4] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.20

DSP First 2e

(a) We need to have zeros of the frequency response at ω̂ = 0 and 0.2π. Thus, H (e j ω̂ ) = (1 − e−j ω̂ )(1 − e−j0.2π e−j ω̂ )(1 − e j0.2π e−j ω̂ ) = (1 − e−j ω̂ )(1 − 2 cos(0.2π)e−j ω̂ + e−j ω̂2 ) = 1 − 2.6180e−j ω̂ + 2.6180e−j ω̂2 − e−j ω̂3 Therefore, the difference equation is y[n] = x[n] − 2.6180x[n − 1 + 2.6180x[n − 2] − x[n − 3] (b) We need to null the frequencies ω̂ = 0, ±0.2π, ±0.4π, ±0.6π, ±0.8π, π. A system whose frequency response is zero at all these frequencies is H (e j ω̂ ) = 1 − e−j ω̂10 .

If we set H (e j ω̂ ) = 0 we get the equation 1 − e−j ω̂10 = 0 or e−j ω̂10 = 1 or, equivalently, e j ω̂10 = 1 = e j2πk where k = 0, 1, . . . , 9. If we then equate the exponents of the last equation we get ω̂10 = 2πk, or ω̂ = 2πk/10 = 0.2πk.

The difference equation for this system is y[n] = x[n] − x[n − 10].

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CHAPTER 6. FREQUENCY RESPONSE P-6.21

DSP First 2e

(a) The two normalized frequencies are 120π/ f s = 0.12π and 240π/ f s = 0.24π. (b) We need to place zeros at ω̂ = ±0.12π and ±0.24π. Therefore, choose H (e j ω̂ ) = (1 − e−j0.12π e−j ω̂ )(1 − e j0.12π e−j ω̂ )(1 − e−j0.24π e−j ω̂ )(1 − e j0.24π e−j ω̂ ) = 1 − 3.3175e−j ω̂ + 4.7111e−j ω̂2 − 3.3175e−j ω̂3 + e−j ω̂4 The corresponding difference equation is therefore: y[n] = x[n] − 3.3175x[n − 1] + 4.7111x[n − 2] − 3.3175x[n − 3] + x[n − 4]

|H(ej ω̂ )|

(c) Here is the Matlab plot of the frequency response (magnitude) versus ω̂, and also versus the analog frequency f in hertz. The frequencies 60 Hz and 120 Hz are marked and are snown to be nulled out.

Note: the plot of |H (e j ω̂ )| versus ω̂ is zoomed vertically so as to see the detail of the zeros.

1

|H(ej2πf /fs )|

0.5 0 −π

−π/2

0 normalized frequency ω̂

π/2

π

20 10 0 -500

-120-60 0 60 120 analog frequency f Figure P-6.21

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500

CHAPTER 6. FREQUENCY RESPONSE P-6.22

DSP First 2e

(a) To find y[n] we need the frequency response evaluated at frequencies 0 and 0.5π which are from the graph: H (e j0 ) = 0

H (e j0.5π ) = 1.4e−jπ/4

It is easy to get these analytically as well since H (e j ω̂ ) = 1 − e−j ω̂ = je−j ω̂/2 2 sin(ω̂/2) = 2 sin(ω̂/2)e j (π/2−ω̂/2) Thus, the output is y[n] = 10 √1 cos(0.5πn − π/4) = 7.0711 cos(0.5πn − π/4).

2

(b) The discontinuity of size π radians is due to the fact that H (e j ω̂ ) is zero at ω̂ = 0 so there is a sign change from one side of 0 to the other. This is evident in H (e j ω̂ ) equation above since the function sin(ω̂/2) is negative for −π < ω̂ < 0 and positive for 0 < ω̂ < π.

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CHAPTER 6. FREQUENCY RESPONSE P-6.23

DSP First 2e

(a) When the input is x[n] = 10 + 10 cos(0.2πn) + 10 cos(0.5πn), we need to determine the frequency response at frequencies 0, 0.2π and 0.5π. From the curves we get H (e j0 ) = 1

H (e j0.2π ) ≈ 0.9e−j0.4π

H (e j0.5π ) = 0

Therefore, y[n] = 10 + 9 cos(0.2πn − 0.4π). (b) There are two kinds of discontinuities in the phase. At ω̂ = 2π(0.17), the discontinuity of size 2π is due to taking the principal value. At ω̂ = 0.5π the frequency response is zero and the discontinuity is a phase jump of ±π due to the sign change as the frequency response crosses zero.

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Chapter

DTFT 7-1

Problem Solutions

132 © 2022 Pearson Education.

7

CHAPTER 7. DTFT P-7.1

DSP First 2e

This solution requires mostly table lookup and applying the delay property for the DTFT. DTFT

(a) x 1 [n] = 2δ[n − 3] ←→ X1 (e j ω̂ ) = 2e−j ω̂3 DTFT

(b) x 2 [n] = 3δ[n − 2] − δ[n − 3] + 3δ[n − 4] ←→ X2 (e j ω̂ ) = 3e−j ω̂2 − e−j ω̂3 + 3e−j ω̂4  7 1≤n≤8 (c) x 3 [n] = 7u[n − 1] − 7u[n − 9] =  0 otherwise  Note how the step sequence is used to create a finite-length sequence. Plug into the DTFT definition and set n = m + 1 in the sum. Then apply the formula for the sum of terms of a geometric series.

X3 (e j ω̂ ) =

8 X

7e−j ω̂n = 7e−j ω̂

n=1

sin(0.25πn) DTFT ←→ X4 (e j ω̂ ) = (d) x 4 [n] = 9πn

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7 X

e−j ω̂m = 7e−j ω̂

m=0

sin(4ω̂) −j ω̂9/2 1 − e−j ω̂8 =7 e −j ω̂ sin(ω̂/2) 1−e

|ω̂| < 0.25π 0 0.25π < |ω̂| ≤ π

( 1 9

CHAPTER 7. DTFT P-7.2

DSP First 2e

This solution requires table lookup and use of DTFT properties DTFT

(a) Y1 (e j ω̂ ) = 2π ←→ y1 [n] = 2πδ[n] DTFT

(b) Y2 (e j ω̂ ) = 5e−j3ω̂ ←→ y2 [n] = 5δ[n − 3] DTFT

(c) Y3 (e j ω̂ ) = 6 cos(3ω̂) = 3e j3ω̂ + 3e−j3ω̂ ←→ y3 [n] = 3δ[n + 3] + 3δ[n − 3] DTFT

(d) Y4 (e j ω̂ ) = j sin(7ω̂) = 21 e j7ω̂ − 12 e−j7ω̂ ←→ y4 [n] = 12 δ[n + 7] − 12 δ[n − 7]

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CHAPTER 7. DTFT P-7.3

DSP First 2e

This solution involves either evaluating the definitions of the direct and inverse DTFT or using table lookup and the properties of the DTFT. 1 (a) v1 [n] = 2π

0.3π Z

e j ω̂n d ω̂ =

e j0.3πn − e−j0.3πn sin(0.3πn) = j2πn πn

−0.3π

(b) V2 (e

j ω̂

)=

9 X

1 − e−j ω̂10 sin(ω̂5) −j ω̂4.5 e−j ω̂5 (e j ω̂5 − e−j ω̂5 ) e = = −j ω̂ −j ω̂/2 j ω̂/2 −j ω̂/2 1−e e (e −e ) sin(ω̂/2)

e−j ω̂n =

n=0

(c) Note that V3 (e j ω̂ ) = 1 − V1 (e j ω̂ ). Thus, v3 [n] = δ[n] − v1 [n] = δ[n] −

sin(0.3πn) πn

Alternatively, you can plug into the inverse DTFT integral as in part (a) obtaining 1 v3 [n] = 2π

−0.3π Z

e

j ω̂n

1 d ω̂ + 2π

−π

(d) V4 (e

j ω̂

)=

9 X

e

jπn −j ω̂n

e

n=0

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Zπ

e j ω̂n d ω̂ =

sin(0.3πn) sin(πn) sin(0.3πn) − = δ[n] − πn πn πn

0.3π

=

9 X n=0

e−j (ω̂−π)n = V2 (e j (ω̂−π) ) =

sin((ω̂ − π)5) −j (ω̂−π)4.5 e sin((ω̂ − π)/2)

CHAPTER 7. DTFT P-7.4

DSP First 2e

This solution requires using the linearity and delay properties of the DTFT and the following basic DTFT pair: ( sin(ω̂0 n) DTFT 1 |ω̂| < ω̂0 j ω̂ h[n] = ←→ H (e ) = 0 ωˆ 0 < |ω̂| ≤ π πn It might be helpful to sketch the different bandlimited DTFTs that are involved. ( ( sin(0.25πn) DTFT 1 |ω̂| < 0.25π 0 j ω̂ ←→ H1 (e ) = 5 − 5 = (a) h1 [n] = 5δ[n] − 0 0.25π < | ω̂| ≤ π 5 0.2πn (b) h2 [n] =

sin(0.4πn) sin(0.1πn) − 0.1πn 0.1πn

DTFT

←→ H2 (e

(c) h3 [n] =

|ω̂| < 0.25π 0.25π < |ω̂| ≤ π

j ω̂

 0 0 ≤ |ω̂| ≤ 0.1π         1 0 ≤ | ω̂| ≤ 0.4π 1 0 ≤ | ω̂| ≤ 0.1π   1 0.1π < |ω̂| ≤ 0.4π ) = 10   0 0.4π < |ω̂| ≤ π − 10  0 0.1π < |ω̂| ≤ π = 10     0 0.4π < |ω̂| ≤ π   

sin(0.4π(n − 8)) sin(0.1π(n − 8)) − = 0.1h2 [n − 8] π(n − 8) π(n − 8)

DTFT

←→ H3 (e

j ω̂

) = 0.1H2 (e

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j ω̂

)e

−j ω̂8

=e

    0 0 ≤ |ω̂| ≤ 0.1π 1 0.1π < |ω̂| ≤ 0.4π     0 0.4π < |ω̂| ≤ π 

−j ω̂8  

CHAPTER 7. DTFT P-7.5

DSP First 2e

Write down the system function directly from the difference equation obtaining H (e j ω̂ ) = 3

6 X

e−j ω̂k .

k=2

With a change of index in the sum we can apply the formula for the sum of terms in a geometric series.

H (e j ω̂ ) = 3

sin(ω̂5/2) −j ω̂4 e sin(ω̂/2)

Thus, the answers as requested are: L = 5, α = 3, and β = 4.

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CHAPTER 7. DTFT P-7.6

DSP First 2e

The exponential factors suggest a delay, but we cannot use the DTFT delay property directly because the delay would be a fraction of a sample. This means that we should use the inverse DTFT if we can evaluate it, and, of course, we can. (a) Inverse DTFT of G1 (e j ω̂ ) 0.4π Z

1 ←→ g1 [n] = 2π DTFT

5e

−j0.3ω̂ j ω̂n

e

−0.4π

0.4π Z

1 d ω̂ = 2π

5eω̂(n−0.3) d ω̂ = 5

e j0.4π (n−0.3) − e−j0.4π (n−0.3) 2π(n − 0.3)

−0.4π

sin 0.4π(n − 0.3) . π(n − 0.3) This looks like it was obtained using the delay property, but remember that the delay property only holds in general for integer shifts.

so g1 [n] = 5

(b) Using the approach of (a), the solution is g2 [n] = 4

sin π(n − 0.1) sin 0.7π(n − 0.1) −4 π(n − 0.1) π(n − 0.1)

If the delay had been an integer, the first function would reduce to an impulse, but in this case it does not. (c) Again using the direct evaluation of the inverse DTFT, the solution is g3 [n] = 9

sin 0.4π(n − 0.2) sin 0.7π(n − 0.2) −9 π(n − 0.2) π(n − 0.2)

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CHAPTER 7. DTFT P-7.7

(a)

DSP First 2e

sin(0.25πn) sin(0.44πn) sin(0.25πn) ∗ = 9πn 4πn 36πn The DTFT of this signal is the product of two ideal lowpass filter frequency responses. Therefore the result of the convolution is a sinc function corresponding to the lowest cutoff frequency and scaled by the product of the gains.

(b)

sin(0.25πn) sin(0.25πn) sin(0.25πn) sin(0.25πn) ∗ ∗ = 9πn 8πn 7πn (9 × 8 × 7)πn In this case we have the convolution of three sinc functions all with the same cutoff frequency so the result is the same signal but with the product of the gains.

(c) Considerable simplification is possible.

(0.3) n u[n] ∗ (10δ[n − 2] − 3δ[n − 3]) = 10(0.3) n−2 u[n − 2] − 3(0.3) n−3 u[n − 3] | {z } break into 2 terms

= 10δ[n − 2] + 10(0.3)(0.3) n−3 u[n − 3] − 3(0.3) n−3 u[n − 3] = 10δ[n − 2]

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CHAPTER 7. DTFT P-7.8

DSP First 2e

We will need the impulse response of the system to do parts (a) and (c). We can get it by expanding H (e j ω̂ ) using the inverse Euler relation; i.e., H (e j ω̂ ) = 21 e j ω̂3 + 12 e−j ω̂3 , so h[n] = 12 δ[n + 3] + 12 δ[n − 3]. (a) By definition, y1 [n] = h[n − 3] = 12 δ[n] + 21 δ[n − 6]. (b) Here we can use the frequency response directly if we express x 2 [n] as x 2 [n] = 1.5e j0.25πn + 1.5e−j0.25πn . Then y2 [n] = 1.5H (e j0.25π )e j0.25πn + 1.5H (e−j0.25π )e−j0.25πn

√ = 1.5 cos(0.75π)e j0.25πn + 1.5 cos(−0.75π)e−j0.25πn = −1.5 2 cos(0.25πn)

 1 0 ≤ n ≤ 8 (c) This signal is a length-9 pulse: x 3 [n] = u[n] − u[n − 9] =   0 otherwise  This one should be done by convolution because the impulse response is so simple and a frequency-domain solution would not be feasible.

y3 [n] = x 3 [n] ∗ h[n] = x 3 [n] ∗ 21 δ[n + 3] + 12 δ[n − 3] 1    2    1  = 12 (u[n + 3] − u[n − 6] + u[n − 3] − u[n − 12]) =  1    2    0 

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−3 ≤ n ≤ 2 3≤n≤5 6 ≤ n ≤ 11 otherwise

CHAPTER 7. DTFT P-7.9

DSP First 2e

(a) Expand cosine with the inverse Euler relation: x[n] = a n cos(ω̂0 n)u[n] = a n

! e j ω̂0 n + e−j ω̂0 n u[n] = 12 e j ω̂0 n a n u[n] + 12 e−j ω̂0 n a n u[n] 2

1 DTFT and the frequency-shift property of the DTFT we can write the DTFT of (b) Using the DTFT pair a n u[n] ←→ 1 − ae−j ω̂ x[n] as X (e j ω̂ ) =

1 1 2 2 + 1 − ae−j (ω̂−ω̂0 ) 1 − ae−j (ω̂+ω̂0 )

1 − 1 ae−j ω̂0 e−j ω̂ + 12 − 12 ae j ω̂0 e−j ω̂ = 2 2 j ω̂ −j ω̂ (1 − ae 0 e )(1 − ae−j ω̂0 e−j ω̂ )

= =

1 − 12 a(e j ω̂0 + e−j ω̂0 )e−j ω̂ 1 − a(e j ω̂0 + e−j ω̂0 )e−j ω̂ + a2 e−j2ω̂ 1 − a cos(ω̂0 )e−j ω̂ 1 − 2a cos(ω̂0 )e−j ω̂ + a2 e−j ω̂2

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CHAPTER 7. DTFT P-7.10

DSP First 2e

(a) The plot of |V (e j ω̂ )| 2 for −π < ω̂ ≤ π is below.

1 1 From the table, V (e j ω̂ ) = so |V (e j ω̂ )| 2 = . −j ω̂ 2 1 − ae 1 + a − 2a cos ω̂ 1 = 400 and it would fall off monotonically A sketch of this for a = 0.95 would show a maximum at ω̂ = 0 of size (1 − a) 2 1 with increasing ω̂ to a value of = 0.2630 at ω̂ = ±π.

(1 + a) 2 (b) If V (e j ω̂ ) were a frequency response, is the filter lowpass, highpass or bandpass? LOWPASS (c) Using the frequency shift property of the DTFT we obtain W (e j ω̂ ) = 21 V (e j (ω̂−0.3π) ) + 21 V (e j (ω̂+0.3π) ). Therefore, a plot of |W (e−j ω̂ )| 2 will look like two shifted versions of the plot in part (a) with everything scaled by 14 . This will place peaks at approximately ω̂ = ±0.3π with heights of approximately 100.

Note: the plots use a semilog scale to show details.

(d) If W (e j ω̂ ) were a frequency response, is the filter lowpass, highpass or bandpass? BANDPASS

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CHAPTER 7. DTFT P-7.11

DSP First 2e

(a) Using the definition of the ideal lowpass filter we can write     1 |ω̂| ≤ ω̂co    1 |ω̂| ≤ ω̂co  0 |ω̂| ≤ ω̂co Hhp (e j ω̂ ) =  − =  1 ω̂co < |ω̂| ≤ π  0 ω̂co < |ω̂| ≤ π  1 ω̂co < |ω̂| ≤ π    which is the frequency response of an ideal highpass filter.

Note: A plot of the function 1 − Hlp (e j ω̂ ) might make this easier to visualize. (b) The cutoff frequency of the ideal HPF is ω̂co . (c) From part (a) it follows that hhp [n] = δ[n] − hlp [n] so hhp [n] = δ[n] −

sin(ω̂co n) πn

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CHAPTER 7. DTFT P-7.12

DSP First 2e

(a) To get a bandpass filter, we need to subtract the low frequency band from a wider lowpass filter. Thus, we can write  0 |ω̂| < 0.3π         1 | ω̂| ≤ 0.6π 1 | ω̂| ≤ 0.3π j ω̂    Hbp (e ) =  − =  1 0.3π ≤ |ω̂| ≤ 0.6π   0 0.6π < |ω̂| ≤ π  0 0.6π < |ω̂| ≤ π  0 0.3π < |ω̂| ≤ π   Therefore, the cutoff frequencies of the two lowpass filters are ω̂lp1 = ω̂co2 = 0.6π, and ω̂lp2 = ω̂co1 = 0.3π. (b) hbp [n] =

sin(0.6πn) sin(0.3πn) − πn πn

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CHAPTER 7. DTFT P-7.13

DSP First 2e

(a) A sketch would show a negative exponential decaying to the left for n ≤ −1. (b) The DTFT exists for values of b such that |x[n]| is absolutely summable; i.e.,

−∞

This will be true for |b| > 1. (c) By definition, the DTFT of x[n] = −bn u[−n − 1] when |b| > 1 is X (e j ω̂ ) =

−1 X

−bn e−j ω̂n = −

n=−∞

=1− =

∞ X

b−n e j ω̂n = 1 −

n=1

−b−1 e j ω̂ 1 = 1 − b−1 e j ω̂ 1 − b−1 e j ω̂

−b−1 e j ω̂ 1 = −1 j ω̂ −j ω̂ −b e (1 − be ) 1 − be−j ω̂

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−1 X

∞ X n=0

b−n e j ω̂n

| − bn | < ∞.

CHAPTER 7. DTFT P-7.14

DSP First 2e

0.25π Z ∞ 2 X 1 1 sin(0.25πn) 1 (a) d ω̂ = = 3πn 2π 3 12 n=−∞ −0.25π

since Zπ (b) 4

 1/3 0 ≤ |ω̂| ≤ 0.25π sin(0.25πn) DTFT  ←→  0 3πn 0.25π < |ω̂| ≤ π  2

20 sin(ω̂5) −j ω̂9/2 e d ω̂= sin(ω̂/2) π

−π

since

9 X

DTFT

2δ[n − k] ←→ 2

k=0

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sin(ω̂5) −j ω̂9/2 e sin(ω̂/2)

CHAPTER 7. DTFT P-7.15

DSP First 2e

(a) This one is easy with the DTFT since X1 (e j ω̂ ) = e−j ω̂4 and Cx1 x1 (e j ω̂ ) = |X1 (e j ω̂ )| 2 = 1. Therefore, cx1 x1 [n] = δ[n].

In the index domain, it is also easy: cx1 x1 [n] = x 1 [n] ∗ x 1 [−n] = δ[n − 4] ∗ δ[−n − 4] = δ[n] (b) This one can be done easily by discrete convolution, but let’s use the DTFT. Since X2 (e j ω̂ ) = e−j ω̂4 − e−j ω̂6 , it follows that Cx2 x2 (e j ω̂ ) = |X2 (e j ω̂ )| 2 = (e−j ω̂4 − e−j ω̂6 )(e j ω̂4 − e j ω̂6 ) = −e j ω̂2 + 2 − e−j ω̂2 so cx2 x2 [n] = −δ[n + 2] + δ[n] − δ[n − 2]. (c) Let’s try this with the DTFT. First note that x 3 [n] is identical to the impulse response of a 10-point running sum system with additional delay of 6 samples. Therefore, !2 sin(ω̂5) sin(ω̂5) −j ω̂4.5 −j ω̂6 j ω̂ j ω̂ 2 j ω̂ e e and Cx3 x3 (e ) = |X3 (e )| = .

X3 (e ) = sin(ω̂/2) sin(ω̂/2) (Note that the exponential factors corresponding to delay disappear when we take the magnitude.)

To determine cx3 x3 [n] we have to determine the inverse DTFT. We cannot evaluate the integral with what we know and we have no entry in our table of transforms for this function. Therefore, we should try convolution, which is not too difficult. The result is   10 − |n| cx3 x3 [n] =  0 

−9 ≤ n ≤ 9 otherwise

This might be a good transform pair to know in a general case. In particular, you can show the following for any sequence x[n] (including an autocorrelation function):   L − |n| x[n] =  0 

|n| ≤ (L − 1) otherwise

DTFT

←→

X (e

j ω̂

sin(ω̂L/2) )= sin(ω̂/2)

!2

  1/9 |ω̂| < 0.25π sin(0.25πn) DTFT ←→ X4 (e j ω̂ ) =  0 9πn 0.25π < |ω̂| ≤ π  This one is a clear candidate for using the DTFT since it would be impossible to evaluate the discrete-time autocorrelation in the time domain.

(d) Recall the DTFT pair: x 4 [n] =

We should recognize that x 4 [n] is identical to the impulse response of an ideal lowpass filter with cutoff frequency 0.25π and gain 1/9. If we take the squared-magnitude of X4 (e j ω̂ ), we get the same cutoff but the gain is squared. sin(0.25πn) Therefore, it follows that cx4 x4 [n] = 81πn

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CHAPTER 7. DTFT P-7.16

DSP First 2e

(a) This is accomplished by evaluating the discrete-time convolution cxx [n] = x[n] ∗ x[−n] and the result is cxx [n] =    5 − |n| |n| ≤ 4 . 0 otherwise  (b) To do this, we don’t try to find the DTFT of cxx [n] in part (a) directly, but instead use the fact that Cxx (e j ω̂ ) = |X (e j ω̂ )| 2 .

We can look up the DTFT of x[n] in our table of DTFT pairs with L = 5. Thus Cxx (e

j ω̂

2

sin(ω̂5/2) sin(ω̂5/2) −j ω̂(L−1)/2 e )= = sin(ω̂/2) sin(ω̂/2)

!2

Cxx (e j ω̂ ) is clearly real and positive because it is a squared-magnitude of the complex DTFT X (e j ω̂ ). (c) Since Cyy (e j ω̂ ) = |Y (e j ω̂ )| 2 = |X (e j ω̂ )e−j ω̂2 | 2 = |X (e j ω̂ )| 2 , it follows that cyy [n] = cxx [n]. In fact the auto correlation function for a delayed signal is always the same as the autocorrelation function for the signal itself. Delay always cancels out when you evaluate a correlation.

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CHAPTER 7. DTFT P-7.17

DSP First 2e

(a) Plot shows details of the passband and stopband magnitudes.

Magnitude of Frequency Response of Lowpass Filter M = 32

1.5 1 0.5 0

6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0021663 ! s = 0.4932:

#10 -3

4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0029736 ! p = 0.7034:

1.004 1.002 1 0.998 0.996 0.994 0.6

0.65

0.7

0.75

0.8

!/:

0.85

0.9

0.95

1

Figure 7-1: Solution for Problem 7-17(a).

(b) From the plots the measured parameters are ω̂ p = 0.7034π, ω̂s = 0.4932π, δ p = 0.003 and δ s = 0.0022. (c) A good rule of thumb is that with a Hamming window, the transition region satisfies ∆ω̂ M = 8π/M, where M is the order of the filter. For M = 32, this formula gives ∆ω̂32 = 0.25π, whereas the measured value is ∆ω̂32 = (0.7034 − 0.4932)π = 0.2102π, so the formula over-estimates the transition width in this case. (d) However, the formula gives us a rationale for estimating the effect of going from M = 32 to M = 80. We should have ∆ω̂80 ≈ ∆ω̂32 32/80 = 0.2102π(32/80) = .0841π. The formula would give a value for ∆ω̂80 of 0.1π. The measured cutoff frequencies for M = 80 are ω̂ p = (.6 + .0841/2)π = 0.642π and ω̂s = (0.6 − .0841/2)π = 0.558π.

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CHAPTER 7. DTFT

Magnitude of Frequency Response of Lowpass Filter M = 80 1 0.8 0.6 0.4 0.2 0

5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0020404 ! s = 0.5584:

#10 -3

4 3 2 1 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0018343 ! p = 0.6418:

1.005 1 0.995 0.99 0.985 0.98 0.6

0.65

0.7

0.75

0.8

!/:

0.85

0.9

0.95

1

Figure 7-2: Solution for Problem 7-17d. The estimates of the cutoff frequencies from part (c) are seen to be quite accurate.

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CHAPTER 7. DTFT P-7.18

DSP First 2e

To solve this problem use Matlab to design the filter using filterdesign. Save the impulse response to the work space as h24 for use. The solution to each of the following parts requires that we evaluate the frequency response at a given frequency.

The frequencies are 0, 0.5π and 0.75π. This can be done in many ways, but the most straightforward way is to use Matlab’s freqz( ) function as follows: >> H24 = freqz(h24,1,[0,0.5*pi,0.75*pi]) H24 = Columns 1 through 2 0.000134504777373 + 0.000000000000000i Column 3 -0.996401749465568 - 0.000000000000001i

0.041265977941552 + 0.000000000000000i

We will need the magnitude and angle at these three frequencies so we do the following: >> zprint(H24) Z = X + jY 0.0001345 0 0.04127 7.164e-17 -0.9964 -1.142e-15

Magnitude 0.0001345 0.04127 0.9964

Phase 0.000 0.000 -3.142

Ph/pi 0.000 0.000 -1.000

Ph(deg) 0.00 0.00 -180.00

(a) If the input is x 1 [n] = 10 for all n, the output is y1 [n] = 10H24 (e j0 ), so we use Matlab as above to find H24 (e j0 ).H32=sum(h32)= so y1 [n] = 0.001345. This is a highpass filter so the DC is greatly attenuated. (b) When the input is a zero-phase cosine, the output is y2 [n] = 10|H24 (e j0.5π )| cos(0.5πn + ∠H24 (e j0.5π )), so from the Matlab analysis, y2 [n] = 0.4127 cos(0.5πn + 0). This frequency is in the transition zone near the stopband so its amplitude is much reduced. (c) As in part (b) the output in this case is y3 [n] = 10|H24 (e j0.75π )| cos(0.75πn + ∠H24 (e j0.75π )), so from the Matlab analysis, y3 [n] = 9.964 cos(0.75πn − π). This frequency is in the passband so the amplitude is barely reduced.

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CHAPTER 7. DTFT P-7.19

DSP First 2e

A

Matlab program to make the requested plots is as follows: % solution to problem 7-19 % assumes that h37 has been saved in workspace by filterdesign GUI % h37 = fir1(37,0.3,ones(1,38); n=0:99; x1=10*ones(1,100); x2=10*cos(0.5*pi*n); y1=filter(h37,1,x1); y2=filter(h37,1,x2); sn=x1+x2; snOutput=filter(h37,1,sn); % subplot(411) stem(n,y1,’filled’);hold on; plot([37,37],[-9,14.5],’r--’);hold off title(’Output y_1[n] due to x_1[n] = 10’) text(20,1,5,’Transient region’);hold off subplot(412) stem(n,y2,’filled’);hold on; plot([37,37],[-10,15],’r--’);hold off title(’Output y_2[n] due to x_2[n] = 10cos(0.5 \pi n)’) subplot(413) stem(n,y1+y2,’filled’);hold on; plot([37,37],[-10,15],’r--’);hold off title(’Sum of Outputs’) subplot(414) stem(n,snOutput,’filled’) title(’Output Due to Sum of Inputs’) xlabel(’time index n’) hold on han=stem(n,snOutput-y1-y2);hold on; plot([37,37],[-10,15],’r--’);hold off The resulting plot is shown in the figure above. (a) When the input x 1 [n] is DC with a DC level of 10, the output level y1 [n] converges to 10H (e j0 ) = 10(1) = 10 after a brief transient of less than 37 samples. (b) When the input is the sinusoid x 2 [n] = 10 cos(0.5πn) we must evaluate H (e j ω̂ ) at ω̂ = 0.5π rad. We obtain H (e j0.5π ) = 0.02814e j0.75π , so the output would be y2 [n] = 0.2814 cos(0.5πn + 0.75π). (c) Note that the bottom two figures are identical, thereby showing that superposition holds. This is also shown by the open circles, which show the difference between y1+y2 and snoutput.

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CHAPTER 7. DTFT

Output y [n] due to x [n] = 10 1

1

10 0 −10 0

Transient region is n=0 to 37 10

20

30 40 50 60 70 Output y [n] due to x [n] = 10cos(0.5 π n) 2

2 1 0 −1 0

10

20

30

80

90

100

70

80

90

100

70

80

90

100

70

80

90

100

2

40

50

60

Sum of Outputs 15 10 5 0 0

10

20

30

40

50

60

Output Due to Sum of Inputs 15 10 5 0 0

10

20

30

40 50 60 Time Index (n)

Figure 7-3: Solution for Problem 7-19.

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CHAPTER 7. DTFT P-7.20

DSP First 2e

(a) Use the GUI to find this frequency to be approximately 0.712π rad. The unwrapped phase at this frequency is ϕ ≈ −10.5 rad. The principal value of the phase is −10.5 + 2π ≈ 2.01 rad. (b) Here again, you can use the GUI to find the frequencies where |H (e j ω̂ )| = 1.

These are approximately {0.71π, 0.788π, 0.854π, 0.914π, 0.972π} (c) Use the GUI to find the frequencies where |H (e j ω̂ )| = 0.

These are approximately {0.03π, 0.091π, 0.151π, 0.212π, 0.272π, 0.334π, 0.398π, 0.486π}.

The following figure shows the places where the frequency response is one and zero more accurately than is possible with the GUI. This plot can be constructed by saving the impulse response to the work space and then using freqz to compute the frequency response.

Magnitude of Frequency Response of Lowpass Filter M = 32

1.5 1 0.5 0

6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0021663 ! s = 0.4932:

#10 -3

4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0029736 ! p = 0.7034: 1.002 1 0.998 0.996 0.994 0.6

0.65

0.7

0.75

0.8

!/:

0.85

Figure 7-4: Solution for Problem 7-20.

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0.9

0.95

1

Chapter

DFT 8-1

Problem Solutions

155 © 2022 Pearson Education.

8

CHAPTER 8. DFT P-8.1

DSP First 2e

(a) The DTFT of x 0 [n] is X0 (e j ω̂ ) = 1 for all ω̂. Therefore, X0 [k] = X0 (e j (2πk/10) ) = 1 for k = 0, 1, . . . , 9. In Matlab >> X0=fft([1,zeros(1,9)]) X0 = 1 1 1 1

(b) The DTFT of x 1 [n] is X1 (e j ω̂ ) =

9 X

1

1

e−j ω̂n =

1 − e−j ω̂10 . Thus, for k = 0, 1, . . . , 9 we get 1 − e−j ω̂

n=0

X1 [k] = X1 (e j (2π/10)k ) =

1

1

1

1

k=0 k = 1, 2, . . . , 9

  10 1 − e−j2πk = −j (2π/10)k 0 1−e 

1 − e−j2πk is indeterminant when we try to evaluate for k = 0. L’ Hôpital’s 1 − e−j (2π/10)k rule doesn’t work here because k is an integer variable. Therefore we need to go back to the sum form where we see 9 X that X1 [0] = X1 (e j (2π/10)0 ) = 1 = 10.

Note that the closed form X1 [k] =

n=0

Matlab verification: >> X1 = fft(ones(1,10)) X1 = 10 0 0

0

0

0

0

0

(c) The DTFT of x 2 [n] is X2 (e j ω̂ ) = e−j ω̂4 so X2 [k] = e−j (2π/10)k4 Matlab verification: >> X2 = fft([zeros(1,4),1,zeros(1,5)]) X2 = 1.0000 + 0.0000i -0.8090 - 0.5878i 0.3090 - 0.9511i -0.8090 + 0.5878i -0.8090 - 0.5878i 0.3090 + 0.9511i -0.8090 + 0.5878i (d) The DTFT of x 3 [n] is X3 (e j ω̂ ) =

9 X n=0

e j2πn/5 e−j ω̂n =

9 X n=0

0

0

k = 0, 1, . . . , 9.

0.3090 + 0.9511i 1.0000 + 0.0000i 0.3090 - 0.9511i

e−j (ω̂−2π/5)n =

1 − e−j (ω̂−2π/5)10 .

1 − e−j (ω̂−2π/5)

Therefore, X3 [k] = X3 (e

j (2π/10)k

 0     1 − e−j ((2π/10)k−2π/5)10 1 − e−j2π(k−2)  10 )= = =  1 − e−j ((2π/10)k−2π/5) 1 − e−j (2π/10)(k−2)   0 

As in part (b), we have to be careful in evaluating this expression for k = 2.

Matlab verification: >> X3 = fft(exp(j*2*pi*(0:9)/5)) X3 = -0.0000 + 0.0000i -0.0000 + 0.0000i 10.0000 - 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i

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k = 0, 1 k=2 k = 3, 4, . . . , 9

CHAPTER 8. DFT P-8.2

DSP First 2e

(a) Substituting into the inverse DFT definition we get, 9

x a [n] =

1 X δ[k]e j (2π/10)kn = 0.1 10 k=0

n = 0, 1, . . . , 9

Matlab verification: >> xa = ifft([1,zeros(1,9)]) xa = 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000

0.1000

0.1000

(b) This one is the first entry in Table 8-1, so we can write the answer down by inspection: x b [n] = δ[n]. Alternatively, you can plug into the inverse DFT definition and get 9 1 1 X j (2π/10)kn 1 − e j (2π/10)10n  = e = x b [n] = j (2π/10)n 0 10 k=0 1−e  Matlab verification: >> xb = ifft(ones(1,10)) xb = 1 0 0 0 0 0 0

n=0 = δ[n] n = 1, 2, . . . , 9

0

0

0

(c) Since there are only two nonzero terms, we can write down the answer directly and manipulate it if possible. x c [n] = 0.1(e j (2π/10)3n + e j (2π/10)7n ) = 0.1(e j (2π/10)3n + e j (2π/10)(10−3)n ) = 0.1(e j (2π/10)3n + e j (2π/10)10n e−j (2π/10)3n ) = 0.1(e j (2π/10)3n + e−j (2π/10)3n ) = 0.2 cos((2π/10)3n) Matlab verification: >> xc = ifft([0,0,0,1,0,0,0,1,0,0]) xc = 0.2000 -0.0618 -0.1618 0.1618 0.0618 0.1618 -0.1618 -0.0618

0.0618

-0.2000

(d) We can observe that X d [k] = cos(2πk/5)X b [k] = 21 e j (2π/10)2k X b [k] + 21 e−j (2π/10)2k X b [k] = 12 e j (2π/10)2k X b [k] + 21 e j (2π/10)(10−2)k X b [k] = 12 e j (2π/10)2k X b [k] + 21 e j (2π/10)8k X b [k] so by the time-shift property of Table 8-2, x d [n] = 12 x b [n − 2] + 21 x b [n − 8] = 12 δ[n − 2] + 12 δ[n − 8] Matlab verification: >> xd = ifft(cos(2*pi*(0:9)/5)) xd = -0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 - 0.0000i

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0.5000 + 0.0000i 0.0000 + 0.0000i 0.5000 - 0.0000i

CHAPTER 8. DFT P-8.3

DSP First 2e

(a) Express (−1) n as e jπn = e j (2π/12)6n and substitute into the DFT definition.

Y0 [k] = 3

11 X

e j (2π/12)6n e−j (2π/12)kn = 3

n=0

 0     = 36    0 

11 X

e−j (2π/12)(k−6)n = 3

n=0

1 − e−j2π (k−6) 1 − e−j (2π/12)(k−6)

k = 0, 1, . . . , 5 k=6 k = 7, 8, . . . , 11

Note that the numerator of the closed form is zero for all integer k, but when k = 6 we have an indeterminant form that evaluates to 3 × 12 = 36.

Matlab verification: >> Y0 = 3*fft((-1).^(0:11)) Y0 = 0 0 0 0

0

0

36

0

0

0

0

0

(b) This can be looked up in Table 8-1. It is the third entry with L = 4 and N = 12 so Y1 [k] =

sin(πk/3) −jπk/4 e sin(πk/12)

Matlab verification: >> Y1 = fft([ones(1,4),zeros(1,8)]) Y1 = 4.0000 + 0.0000i 2.3660 - 2.3660i 0.0000 + 0.0000i 1.0000 + 0.0000i 0.0000 + 0.0000i 0.6340 + 0.6340i 0.0000 + 0.0000i 0.0000 + 1.7321i

0.0000 - 1.7321i 0.6340 - 0.6340i 1.0000 + 0.0000i 2.3660 + 2.3660i

(c) In this problem, the even samples are 1 and the odd indexed samples are zero. This solution is facilitated if we write 11

X j (2π/12)6n 1 y2 [n] as y2 [n] = 2 1 + e δ[n−m]. Therefore, it follows from the linearity and frequency-shift properties m=0

in Table 8-2 and the third entry in Table 8-1 with L = N = 12 that Y2 [k] = 21

sin(πk) −jπk11/12 1 sin(π(k − 6)) −jπ (k−6)11/12 e +2 e sin(πk/12) sin(π(k − 6)/12)

The first term is zero except at k = 0 where its value is Y2 [0] = 6. The second term is zero except when k = 6 where its value is Y2 [6] = 6. Thus  6 Y2 =  0 

k = 0, 6 k = 1, 2, . . . , 5, 7, 8, . . . , 11

We can get another closed form expression by summing the DFT expression only over the even indices using a index m = 2n, n = 0, 1, . . . , 5. This gives us Y2 [k] =

5 X

e−j (2π/12)2km =

m=0

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1 − e−j (2π/12)6k 1 − e−jπk = 1 − e−j (2π/12)2k 1 − e−j (2π/6)k

CHAPTER 8. DFT

In this case, the expression is indeterminant for k = 0, 6. The value at these DFT indices is Y2 [0] =

11 X n=0

because the odd-indexed samples are zero, Y2 [6] =

11 X

y2 [n]e−j (2π/12)6n =

n=0

11 X

y2 [n]e−jπn = 6

n=0

Matlab verification: >> Y2 = fft([1,0,1,0,1,0,1,0,1,0,1,0]) Y2 = 6 0 0 0 0 0

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6

0

0

0

0

0

y2 [n] = 6, and

CHAPTER 8. DFT P-8.4

DSP First 2e

(a) The DFT of y[n] is Y [k] =

5 X

7e−j (2π/6)kn − 3W [k] = 7

n=0

1 − e−j2πk − 3W [k] = 7(6)δ[k] − 3W [k] 1 − e−j (2π/6)k

Therefore, Y [k] = {39, 0, −3 j, 0, +3 j, 0} (b) If we write v[n] = 4(−1) n w[n] as v[n] = 4e j (2π/6)3n w[n], we see that the new DFT is V [k] = 4W [k − 3] where the shift is periodic with period 6.

Therefore, V [k] = {0, −4 j, 0, 4, 0, 4 j}

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CHAPTER 8. DFT P-8.5

DSP First 2e

(a) X[0] =

255 X

x[n]e−j2π (0)n/256 =

n=0

X[128] =

255 X

x[n] is real since all x[n] are real-valued.

n=0

255 X

x[n]e−j2π (128)n/256 =

n=0

255 X

x[n](−1) n is real since all x[n] are real-valued.

n=0

(b) X[k] = 3 − e−j (2π/4)k + 3e−j (2π/4)2k − e−j (2π/4)3k = {4, 0, 8, 0} 3

1 X V [k]e j (2π/4)(2)k = 41 (3 + (5 j)(−1) + 0 + (−5 j)(−1)) = 43 (c) v[2] = N k=0 (d) Y [0] =

9 X

y[n] = 0

n=0

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Y [5] =

9 X n=0

7(−1) n e−j (2π/10)5n = 70

CHAPTER 8. DFT P-8.6

DSP First 2e

(a) Sketch should show x[n] = {0, 1, 2, 3, 4, 0, 0, 0} starting at n = 0. (b) Sketch should show y[n] = {2, 3, 4, 0, 0, 0, 0, 1} because y[n] = x[n − 6] evaluated with a period of 8.

Problem 8.6(a), x[n] 4 3 2 1 0 0

1

2

3

4

5

6

7

5

6

7

Problem 8.6(b), y[n] 4 3 2 1 0 0

1

2

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3 4 Time Index (n)

CHAPTER 8. DFT P-8.7

DSP First 2e

(a) Plot y[n] = {2, 1, 0, 0, −2, −1} starting at n = 0.

Problem 8.7, y[n] 2 1 0 −1 −2 0

1

2

3

4

5

Using the method of synthetic polynomial multiplication to evaluate the convolution, we get 1 2 2

1 -1 1

1 -1 0

1 so y[n] = 2δ[n] + δ[n − 1] − 2δ[n − 4] − δ[n − 5] 0

-2

-1

(b) Applying the definition of the DFT and writing out the individual terms gives X[k] = 1 + e−j (2π/N )k + e−j (2π/N )k (2) + e−j (2π/N )k3 + e−j (2π/N )k (4) H[k] = 2 − e−j (2π/N )k − e−j (2π/N )k (2) (c) Forming the product Y [k] = H[k]X[k] is the same as doing the polynomial multiplication as in part (a) so we get Y [k] = 2 + e−j (2π/N )k − 2e−j (2π/N )k (4) − e−j (2π/N )k (5) . (d) For N = 6, the DFT is Y [k] = 2 + e−j (2π/6)k − 2e−j (2π/6)k (4) − e−j (2π/6)k (5) . All we need to do is pick off the coeficients of Y [k] to obtain y[n] = {2, 1, 0, 0, −2, −1} (e) We need to adjust the exponents of W [k] so that any power of e−j (2π/4)k (n+4) is replaced by e−j (2π/4)kn .

Thus, Y [k] = 2 + e−j (2π/4)k − 2e−j (2π/4)k (4) − e−j (2π/4)k (5) is equivalent to Y [k] = 2 + e−j (2π/4)k − 2e−j (2π/4)k (0) − e−j (2π/4)k = 0. (f) Since the length of the convolution is L + M − 1, we need to be able to represent the sequence of this length by the DFT. Therefore, N ≥ L + M − 1 is required.

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CHAPTER 8. DFT P-8.8

DSP First 2e

In general, the 100-point DFT would be X[k] = X (e j (2π/100)k ) = 7e−j (2π/100)18k

sin(12.5(2π/100)k) sin(πk/100)

If this is evaluated for k = 13 we get X[13] = X (e j (2π/100)13 ) = 12.46 e j0.32π = 12.46 e j1.005

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CHAPTER 8. DFT P-8.9

DSP First 2e

(a) From Q(e j ω̂ ) it is clear that the peaks will occur around DFT indices corresponding to the frequencies ω̂0 and 2π − ω̂0 ; i.e. when (2π/32)k 1 = ω̂0 and (2π/32)(32 − k 1 ) = 2π − ω̂0 . Therefore, ω̂0 = (2π/32)10 = 0.625π. (b) The second peak occurs when (2π/32)(32 − k1 ) = (2π/32)k 2 or when k2 = 22. The third nonzero value occurs at k 3 = 0, corresponding to the DC component of 0.1 in s[n]. The DFT of that DC component is 0.1N δ[k], which is zero for all k except k = 0. The Dirichlet components are also zero except at k 1 = 10 for the first one and k 2 = 22 for the second one. (c) The peak magnitude at k = 10 is 12 ( A)(32) = 50 so A = 100/32 = 3.125.

Matlab verification: >> x=0.1+(100/32)*cos(2*pi*10*(0:31)/32); >> X=fft(x); >> abs(X) ans = 3.2000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

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0.0000 50.0000 0.0000 50.0000 0.0000

0.0000 0.0000 0.0000 0.0000 0.0000

CHAPTER 8. DFT P-8.10

DSP First 2e

(a) N = 32 + 48 = 80. (b) k peaks = 8, and −8 → (80 − 8 = 72) because (2π/80)k peaks = ±0.2π The peaks will occur around DFT indices corresponding to ω̂ = ±0.2π, or when (2π/80)k = 0.2π or k = 8 and when (2π/80)(80 − 8) = 2π − 0.2π or k = 72. (c) max{|R[k]|} = 118.4553. Evaluate in Matlab via k

r=1+7*cos(0.2*pi*(0:31)); R=fft(r,80); [mx,ix]=max(abs(R))

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CHAPTER 8. DFT P-8.11

DSP First 2e

The impulse response h[n] =

sin(0.15πn) corresponds to a frequency response 5πn

  1/5 |ω̂| ≤ 0.15π H (e j ω̂ ) =  0 0.15π < |ω̂| ≤ π  The input signal has fundamental frequency ω̂0 = 0.08π. Thus, only DC and the first harmonic are below the cutoff of the filter. Since the gain is 1/5, we get three terms in the output

y[n] = (1/5) 1 + 2e j0.08πn + 2e−j0.08πn = 0.2 + 0.4 cos(0.08πn)

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CHAPTER 8. DFT P-8.12

DSP First 2e

(a) In order to show the spectrum only for positive frequencies, we need to convert the frequencies by adding 2πn to the exponent of each of the components having negative frequency. Therefore, x[n] = 3 + 2e j0.2πn + 2e j (2π−0.2π)n − 7 je j0.7πn + 7 je j (2π−0.7π)n = 3 + 2e j0.2πn + 2e j1.8πn − 7 je j0.7πn + 7 je j1.3πn Therefore, the spectrum plot is as follows: −7 j

7j

7j

3

3

−π

−.7π

2

2

−.2π 0

.2π

2 .7π

π

1.3π

1.8π 2π

-

ω̂

The dashed lines show negative frequencies that differ by 2π from frequencies in the range 0 ≤ ω̂ < 2π. (b) If we carry out the multiplication and adjust any resulting negative frequencies as we did in part (a), we get x 1 [n] = e j0.4πn x[n] = 3e j0.4πn + 2e j0.6πn + 2e0.2πn − 7 je j1.1πn + 7 je j (−0.3π)n = 3e j0.4πn + 2e j0.6πn + 2e0.2πn − 7 je j1.1πn + 7 je j1.7πn This is an example of the frequency shift property since by adding 0.4π to each frequency we shift the spectrum to the right by that amount. The resulting spectrum is −7 j

7j

−7 j

7j

π 1.1π

1.7π

3 2 −.9π

−.3π

0

2

.2π .4π .6π

2π

-

ω̂

(c) Now we can write x 2 [n] = (−1) n x[n] = e jπn x[n] and we see that by the same process as in part (b), each of the original frequencies will be increased by π radians. If we choose the alias frequencies that are in the range 0 ≤ ω̂ < 2π, we can write x[n] = 3e jπn + 2e j1.2πn + 2e j0.8πn − 7 je j1.7πn + 7 je j0.3πn Therefore the spectrum plot is that of part (a) shifted to the right by π radians.

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CHAPTER 8. DFT −7 j

−7 j

7j

3

3 2 −π −.8π

−.3π

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0

.3π

2

2

.8π

π 1.2π

1.7π

2π

-

ωˆ ω̂

CHAPTER 8. DFT P-8.13

DSP First 2e

(a) The sampled signal is x[n] = x(n/60) = 6 cos(42πn/60) + 4 cos(18πn/60 − 0.5π) = 6 cos(2π(7/20)n) + 4 cos(2π(3/20)n − 0.5π) = 6 cos(0.7πn) + 4 cos(0.3πn − 0.5π) Since 7 and 3 have no common factors, the period is N = 20 and the fundamental frequency is ω̂0 = 2π/20. (b) x[n] = 2e−j0.5π e j (2π/20)3n + 2e j0.5π e−j (2π/20)3n + 3e j (2π/20)7n + 3e−j (2π/20)7n (c) From the answer to part (b), the DFS coefficients are c3 = 2e−j0.5π , c−3 = 2e j0.5π , c7 = 3, c−7 = 3. (d) Here is the spectrum plot of the DFS: 3

3 −2 j

2j

−π

−.7π

−.3π

3

0

.3π

3 −2 j

2j

π

1.3π

1.7π

2π

Dashed lines are aliases of the frequencies in the range −π < ω̂ ≤ π.

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2.3π

2.7π

3π

ωˆ

-

CHAPTER 8. DFT P-8.14

DSP First 2e

sin(4ω̂) −j ω̂7/2 e . The zeros of this function are at frequencies where sin(ω̂/2) 4ω̂ = πr where r is an integer; i.e., where ω̂ = 0.25π, 0.5π, 0.75π, π. The equivalent values of DFT index k are when (2π/N )k = 0.25πr or k = Nr/8. For N = 512, these values are multiples of 64.

(a) The frequency response of this filter is H (e j ω̂ ) =

Problem 8.14 Magnitude

8 6 4 2 0 0

32

64

32

64

96

128

160

192

224

256

96 128 160 192 DFT Frequency Index (k)

224

256

Phase (rad)

pi/2 0 −pi/2 −pi 0

(b) Note that the given frequencies are all multiples of 0.25π so all the frequencies except DC fall at the zeros of H (e j ω̂ ).

Therefore, these frequencies do not appear in the output. The gain of the filter at ω̂ = 0 is H (e j0 ) = 8, so the output is y[n] = 3(8) = 24 for all n. You can pick all these values off the plot, or evaluate H (e j ω̂ ) at individual frequencies with freqz.

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CHAPTER 8. DFT P-8.15

DSP First 2e

(a) We have two constraints. To avoid aliasing we need 2(1000) < f s . To meet the frequency spacing condition, we need f s /N ≤ 5, where N = 2ν .

The combined constraints are given by 2000 < f s ≤ 5 · 2ν , or equivalently 400 < f s /5 ≤ ·2ν .

Thus we require N = 512 = 29 . If we fix N at 512, the condition on f s is 2000 < f s ≤ 2560 Hz. (b) As discussed in Section 8-7.4.1, the width of the main lobe of the Hann and Hamming windows is approximately ∆ω̂ = 8π/L where L is the window length. Therefore, for L = N/2 = 256, we should expect a main lobe width of about 8π/256 = π/32 rad. This is equivalent to analog radian frequency of ∆ω = (π/32) f s or using analog frequency in hertz, ∆ f = f s /64.

Using the constraints on f s obtained in part (a), we get 2000/64 < ∆ f ≤ 2560/64, or 31.25 < ∆ f ≤ 40.

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CHAPTER 8. DFT P-8.16

DSP First 2e

(a) Here’s a Matlab plot of the ideal spectrogram. A sketch should look like this.

Problem 8-16a: Ideal Spectrogram

2000 1800 1600

frequency

1400 1200 1000 800 600 400 200 0

0

0.2

0.4

0.6

0.8

1

1.2

time axis

(b) The width of the main lobe of the DTFT of the Hann window is approximately ∆ω̂ = 8π/L, where L is the window length. Therefore the normalized main lobe width for the L = 200 window is ∆ω̂ = 0.04π. In terms of analog cyclic frequency, this would be equivalent to ∆ f = ∆ω̂ f s /(2π) = 0.04π8000/(2π) = 160 Hz. Thus, the spectrogram sketch should show bars about 160 Hz wide and there should be blurred regions of approximately 200 samples (or equivalently, 200/8000 = 0.025 s) duration at the beginning and end of each sinusoidal component. The following is a Matlab spectrogram of the signal.

Problem 8-16b: Spectrogram with 200-point Window

2000 1800 1600

frequency

1400 1200 1000 800 600 400 200 0

0

0.2

0.4

0.6

time axis

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0.8

1

1.2

CHAPTER 8. DFT P-8.17

DSP First 2e

Note that there are 8 equal duration notes in (approximately) 2 seconds of time. Therefore, each note has duration 2/8 s or 250 ms.

Spectrogram of C Major scale, L D 500; fs D 4 kHz

600 550

Frequency (Hz)

500 450 400 350 300 250 200 0

0.2

0.4

0.6

0.8

1 1.2 Time (s)

Figure P-8.17

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1.4

1.6

1.8

2

CHAPTER 8. DFT P-8.18

DSP First 2e

The ratio of the window lengths is 256/100 ≈ 2.5.

The sketch will look similar to the spectrogram in Fig. P-8.17, but it will have bars that are 2.5 times as wide in the frequency dimension. The frequency width of the horizontal bars in Fig. P-8.17 is approximately 16 Hz, so it will change to 40 Hz in the new figure.

In the time dimension, the fuzzy regions at the time of frequency change will be shorter—their duration will be 40% of the duration of those in Fig. P-8.17 which are approximately 0.03 s. Thus, in the new figure the duration of the fuzzy regions will be approximately 0.01 s.

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CHAPTER 8. DFT P-8.19

DSP First 2e

(a) The Hamming window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. For the given parameters, this would give a main lobe width equivalent to ∆ f = 4 f s /L = 4(8000)/256 = 125 Hz. Thus, two sinusoids whose frequencies are farther apart than 125 Hz will show up as two distinct peaks, but if the two frequencies are closer that 125 Hz they will tend to blend together in the spectrum. (b) The window length is equivalent to 256/8000 =32 ms so if there are abrupt changes in the waveform, the spectral characteristics will look blurred over approximately 32 ms around the abrupt change.

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CHAPTER 8. DFT P-8.20

DSP First 2e

(a) The Hamming window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. In terms of analog cyclic frequency this corresponds to ∆ f = 4 f s /L, where L is the window length. For the given parameters, we want 40000/L = 250 or L = 160. (b) Here is the plot of the DTFT magnitude:

DTFT of Windowed Sinusoids at Frequencies 3000 and 3250 Hz

45 40 35 30 25 20 15 10 5 0

0

500

1000

1500

2000

2500

3000

3500

analog frequency in Hz

Figure 8-1: Solution for Problem 8-20.

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4000

4500

5000

CHAPTER 8. DFT P-8.21

DSP First 2e

(a) The notes are in the octave above A-440, so the frequencies are given by the equation f = 440(2n/12 ) with n = 0, 1, . . . , 12 covering one octave from A-440 to A-880. Here are the notes and their theoretical frequencies. (Subscript denotes octave number.) note f

A4 440

B4 [ 466

B4 494

C5 523

C5 ] 554

D5 587

D5 ] 622

E5 659

F5 ] 740

F5 698

G5 784

G5 ] 831

A5 880

Now the first 9 notes of Für Elise are { E5 , D5 ], E5 , D5 ], E5 , B4 , D5 , C5 , A4 }. Therefore, the theoretical frequencies for the first nine notes are { 659, 622, 659, 622, 659, 494, 587, 523, 440 }.

Spectrogram of Fur Elise (9 notes), L D 1000; fs D 8 kHz

800 750 700

E

650

Frequency (Hz)

600

D

550 500

C B

450

A

400 350

E

300 250 200

6

6.25

6.5

6.75

7

7.25

7.5 7.75 Time (s)

8

8.25

8.5

8.75

9

(b) The spectrogram in the figure above is marked with the true frequencies. Observe that the passage of interest starts at t = 6.4 s (approximately). It is difficult to obtain very accurate estimates of the frequencies by measuring on the spectrogram because the spectral peaks have a width of about 10 Hz. However, careful examination of the figure above confirms that the theoretical note frequencies are close to the center of the spectral peaks. (c) Observe that the passage of interest starts at t = 6.4 s (approximately). From the score in Fig. 3-22, the first eight notes are 16th notes while note nine is an eighth note, which should have twice the duration.

The first five notes are easy to distinguish, but overlap in time due to the sliding window of the spectrogram. Their total duration is 7.7 − 6.4 = 1.3 s, so each 16th note has a duration of approximately 0.26 s. The sixth note (B4 ) appears shorter (≈ 0.15 s), but might be softer so its second half is hard to see; likewise, for note eight (C5 ). The seventh note (D5 ) is barely visible, but there is evidence of a spectral peak in the 0.2 s gap between B4 and C5 . The ninth note (A4 ) is easy to distinguish, and its duration seems a bit longer (≈ 0.3 s), but not twice as long.

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CHAPTER 8. DFT P-8.22

DSP First 2e

(a) From the spectrogram it appears that ω̂0 = 0.7π and ω̂1 = 0.75π. To estimate the chirp parameter α, recall that the instantaneous frequency is ω̂i [n] = 2αn. From the spectrogram we see that ω̂i [n] starts at 0 and increases linearly to ω̂i [2000] = 0.5π = 2α(2000). Therefore, α = π/8000. (b) The Hann window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. Thus, if we multiply the window length by 4, we divide the main lobe width by 4. Thus, the spectrogram with L = 400 will like Fig. ?? except that the bars will be much narrower—they will have 1/4 the width.

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Chapter

z -Transform 9-1

Problem Solutions

180 © 2022 Pearson Education.

9

CHAPTER 9. z-TRANSFORM P-9.1

DSP First 2e

X1 (z) = 7,

X2 (z) = z −1,

X3 (z) = 5z −4,

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X4 (z) = 7 − z −1 + 5z −4

CHAPTER 9. z-TRANSFORM P-9.2

DSP First 2e

Y (z) = X (z) − z −1 X (z) = (1 − z −1 )X (z) = H (z)X (z) where H (z) = 1 − z −1 is the z-transform of h[n] = δ[n] − δ[n − 1], the impulse response of the system.

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CHAPTER 9. z-TRANSFORM P-9.3

DSP First 2e

(a) y[n] = x[n] + 5x[n − 2] − 3x[n − 3] + 2x[n − 5] + 4x[n − 7], (b) h[n] = δ[n] + 5δ[n − 2] − 3δ[n − 3] + 2δ[n − 5] + 4δ[n − 7], Plot {1, 0, 5, −3, 0, 2, 0, 4} for 0 ≤ n ≤ 8.

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CHAPTER 9. z-TRANSFORM P-9.4

DSP First 2e

(a) H (z) = 0.1(1 − z −1 + z −2 ) (b) The roots of H (z) are e±π/3 so we can write H (z) = 0.1(1 − e jπ/3 z −1 )(1 − e−jπ/3 z −1 ) = 0.1

(z − e jπ/3 )(z − e−jπ/3 ) z2

Imaginary Part

1 0.5 2

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) H (e j ω̂ ) = 0.1(1 − e−j ω̂ + e−j ω̂2 ) 0.3 0.2π

0.1

H(ej ω̂ )

0 −π 5

−0.8π −0.6π −0.4π −0.2π

0

0.2π

0.4π

0.6π

0.8π

π

0

0.2π

0.4π

0.6π

0.8π

π

6

0

(d)

-5 −π

−0.8π −0.6π −0.4π −0.2π

ω̂

(e) For this we need the frequency response evaluated at ω̂ = 0, π/4, 2π/3.

H (e j0 = 1

H (e jπ/4 ) = 0.4142e−jπ/4

H (e j2π/3 ) = 2e jπ/3

Therefore, the output is y[n] = 9 − 8(0.4142) cos[0.25π(n − 1) − π/4] + 7(2) cos[(2π/3)n + π/3] = 9 − 3.3136 cos[0.25π(n − 2)] + 14 cos[(2π/3)n + π/3]

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CHAPTER 9. z-TRANSFORM P-9.5

DSP First 2e

(a) We have to multiply out the factors to get a polynomial in z −1 .

H (z) = 1 + 2z −1 + 2.64z −2 + 2.28z −3 + 1.28z −4 + 0.64z −5 Then the difference equation is: y[n] = x[n] + 2x[n − 1] + 2.64x[n − 2] + 2.28x[n − 3] + 1.28x[n − 4] + 0.64x[n − 5] (b) There are five poles at z = 0 and the five zeros are at z = −1, ± j0.9, and e±j2π/3 .

Imaginary Part

1 0.5 5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) We need to find the frequencies where H (e j ω̂ ) = 0.

These frequencies are the angles of the zeros that are on the unit circle; i.e., (1 + z −1 ) = 0, (1 − e j2π/3 z −1 ) = 0 and (1 − e−j2π/3 z −1 ) = 0, or ω̂ = π, ±2π/3.

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CHAPTER 9. z-TRANSFORM P-9.6

DSP First 2e

(a) H (z) = H2 (z)H1 (z) = 14 (1 + z −1 + z −2 + z −3 ) 13 (1 + z −1 + z −2 ) 1 (1 + 2z −1 + 3z −2 + 3z −3 + 2z −4 + z −5 ) = 12

(b) Y2 (z) = X (z)H2 (z) = (1 − z −1 ) 31 (1 + z −1 + z −2 ) = 13 (1 − z −3 ) W (z) = Y2 (z)H1 (z) = 13 (1 − z −3 ) 14 (1 + z −1 + z −2 + z −3 ) 1 = 12 (1 + z −1 + z −2 − z −4 − z −5 − 3z −3 )

1 (c) y[n] = 12 (x[n] + 2x[n − 1] + 3x[n − 2] + 3x[n − 3] + 2x[n − 4] + x[n − 5])

Clearly, the samples are not weighted equally although the sum of the weights (coefficients) is equal to one. (d) There are five poles at z = 0, and the five zeros are at z = −1, ± j, and e±j2π/3

Imaginary Part

1 0.5 5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

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1

CHAPTER 9. z-TRANSFORM P-9.7

DSP First 2e

>> z=roots([1,-.5,.5,-1]) z = 1.0000 -0.2500 + 0.9682i -0.2500 - 0.9682i >> magz=abs(z) magz = 1.0000 1.0000 1.0000 >> angz=angle(z)/pi angz = 0 0.5804 -0.5804 Thus, the zeros are at z = 1, e±j0.5804π and there is one pole at z = 0 for each of the zeros.

Imaginary Part

1 0.5 3

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

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1

CHAPTER 9. z-TRANSFORM P-9.8

DSP First 2e

(a) Impulse response: h[n] = δ[n] − δ[n − 1] + δ[n − 2] − δ[n − 3] + δ[n − 4] =

4 X

(−1) k δ[n − k]

k=0

(b) H (z) =

4 X

k −k

(−1) z

=

k=0

4 X

(−1z −1 ) k =

k=0

1 − (−z −1 ) 5 1 + z −5 z5 + 1 = = 1 − (−z −1 ) 1 + z −1 z 4 (z + 1)

(c) Plot the poles and zeros of H (z) in the complex z-plane.

Imaginary Part

1 0.5 4

0

-0.5 -1 -1

(d) H (e j ω̂ ) =

-0.5

0 0.5 Real Part

1

cos(ω̂2.5) 1 + e−j ω̂5 = e−j ω̂2 −j ω̂ cos(ω̂/2) 1+e

(e) Sketch of the frequency response (magnitude and phase) , or Matlab plot.

|H(ej ω̂ )|

6 4 2

H(ej ω̂ )

0 −π 5

0.2π

0.6π

π

0.2π 0 ω̂ Figure P-9.8

0.6π

π

−0.6π

−0.2π

−0.6π

−0.2π

0

6

0

-5 −π

(f) We need the frequency response evaluated at ω̂ = 0, 0.5π, 0.6π. These values are H (e jω̂ ) = 1, 1, 0.

Therefore, y[n] = 5 + 4 cos(0.5πn).

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CHAPTER 9. z-TRANSFORM P-9.9

DSP First 2e

(a) H (z) = (1 − z −1 ) 5 (b) Plot of the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 5

5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c)

5 H (e j ω̂ ) = (1 − e−j ω̂ ) 5 = e−j ω̂5/2 = 2e jπ/2 e−j ω̂/2 sin(ω̂/2) = 32e (π−ω̂)5/2 sin5 (ω̂/2) (d) H (z) = (1 − z −1 ) 5 = 1 − 5z −1 + 10z −2 − 10z −3 + 5z −4 − z −5 h[n] = δ[n] − 5δ[n − 1] + 10δ[n − 2] − 10δ[n − 3] + 5δ[n − 4] − δ[n − 5]

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CHAPTER 9. z-TRANSFORM P-9.10

DSP First 2e

(a) We can find N1 as the lowest power of z −1 of the product X (z)H (z). That will be N1 = 0 + 6 = 6. Similarly, N2 is the highest power of z −1 if the product X (z)H (z). That is N2 = 9 + 9 = 18. (b) Again, we can use the polynomials of X (z) and H (z) to get the answer. y[6] is the product of the coefficients of the two lowest order terms or y[6] = 2 × 1 = 2. Similarly, y[18] is the product of the coefficients of the highest order terms, or y[18] = −1 × 3 = −3. (c) y[n] = 2δ[n − 6] − 4δ[n − 7] + 8δ[n − 8] + δ[n − 9] − 6δ[n − 10] + 12δ[n − 11] − 11δ[n − 12] +16δ[n − 13] − 32δ[n − 14] + 11δ[n − 15] − 6δ[n − 16] + 12δ[n − 17] − 3δ[n − 18]

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CHAPTER 9. z-TRANSFORM P-9.11

DSP First 2e

(a) H (z) = 1 + z −1 + z −2 = (1 − e j2π/3 z −1 )(1 − e−j2π/3 z −1 ), so there are zeros at z = e±j2π/3 . (b) To solve this part we need the frequency response evaluated at ω̂ = π/6. That is, H (e j ω̂ ) = 1 + e−j ω̂ + e−j ω̂2 = e−j ω̂

sin(ω̂3/2) sin(ω̂/2)

so H (e jπ/6 ) = 2.7321e−jπ/6 and, therefore, the formula for the steady-state output is y[n] = 2.7321 cos((π/6)(n − 1) − π/4) (c) omega = 2*pi/3 gives zero for the steady-state output after n ≥ 2. (Until the last two samples.)

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CHAPTER 9. z-TRANSFORM P-9.12

DSP First 2e

(a) y[n] = x[n] − x[n − 4] (b) H (e j ω̂ ) = 1 − e−j ω̂4 = 2e j (π/2−ω̂2) sin(ω̂2) (c) |H (e j ω̂ )| = 2| sin(ω̂2)|

and

∠H (e j ω̂ ) =

(

π/2 − 2ω̂ π/2 − 2ω̂ + 2π

0 < ω̂ < π/2 π/2 < ω̂ < π

(d) The frequencies nulled are the angles of the zeros of H (z); ω̂0 = 0, ±π/2, π. (e) y[n] = |H (e jπ/3 )| cos(πn/3 + ∠H (e jπ/3 )) = 1.7321 cos(πn/3 − π/6)

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CHAPTER 9. z-TRANSFORM P-9.13

DSP First 2e

We should find the output as the sum of the response to DC plus the response to an impulse of size -70 plus the response to the sinusoid of frequency 0.5π. We need the impulse response and the frequency response function to do this.

From H (z) in expanded form, h[n] = δ[n] − 5δ[n − 2] + 4δ[n − 4] and H (e j ω̂ ) = (1 − e−j ω̂2 )(1 − 4e−j ω̂2 ) = 1 − 5e−j ω̂2 + 4e−j ω̂4 so H (e j0 ) = 0 and H (e j0.5π ) = 1 − 5e−j2π/2 + 4e−j4π/2 = 1 − 5e−π + 4e−j4π/2 = 10. Therefore, y[n] = −70δ[n] + 350δ[n − 2] − 280δ[n − 4] + 300 cos(0.5πn + π/4)

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CHAPTER 9. z-TRANSFORM P-9.14

DSP First 2e

After sampling, we get the sequence x[n] = 4+cos(1000πn/2000− π/4) −3 cos(500πn/2000). Thus, we need to evaluate the frequency response at ω̂ = 0, π/2, π/4 to find the outputs due to each component of the input. These values are H (e j0 ) = 0, H (e jπ/2 ) = 0, and H (e jπ/4 ) = 1.0824e jπ/8 . Therefore, y[n] = −3(1.0824 cos(πn/4 + π/8) = 3.2472 cos(πn/4 − π + π/8) = 3.2472 cos(πn/4 − 2.7489) and therefore the output of the D-to-C converter with f s = 2000 is y(t) = 3.2472 cos(π2000t/4 − 2.7489) = 3.2472 cos(500πt − 2.7489)

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CHAPTER 9. z-TRANSFORM P-9.15

DSP First 2e

(a) α = 4, β = 10 From H (z) we get H (e j ω̂ ) = 5(1 + .8e−j ω̂ + e−j ω̂2 . By factoring e−j ω̂ out and using the inverse Euler relation we get H (e j ω̂ ) = e−j ω̂ (4 + 10 cos(ω̂)) (b) In the plot the DC value is +14, H (e jπ ) = −6, and H (e j ω̂ ) = 0 at ω̂ = 0.631π = 1.9823.

|H(ej ω̂ )|

15 10 5

H(ej ω̂ )

0 −π 2

0.2π

0.4π

0.6π

0.8π

π

0.2π 0 ω̂ Figure P-9.15

0.4π

0.6π

0.8π

π

−0.8π −0.6π −0.4π −0.2π

0

6

0

-2 −π

−0.8π −0.6π −0.4π −0.2π

(c) We want the system function to have the form H (z) = (1−e j ω̂0 z −1 )(1−e−j ω̂0 z −1 ) = 1−2 cos ω̂0 z −1 +z −2 . Therefore, we must set b1 = −2 cos ω̂0 . The normalized frequency that we want to null is ω̂0 = 2π f 0 / f s = 2π(1600/8000) = 0.4π.

Thus, we need to set b1 to b1 = −0.618.

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CHAPTER 9. z-TRANSFORM P-9.16

DSP First 2e

(a) Factoring the two second-order factors gives H (z)(1 − z −1 )(1 + z −1 )(1 − 0.8e jπ/4 z −1 )(1 − 0.8e−jπ/4 z −1 )(1 + e jπ/2 z −1 )(1 + e−jπ/2 z −1 )

Imaginary Part

1 0.5 6

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-9.16

(b) To satisfy the condition, we must group the zeros at ±1 and e±jπ/2 together to form H1 (z) as in H1 (z) = (1 − z −2 )(1 + z −2 ) = (1 − z −4 ) . That leaves the remainder for H2 (z) as H2 (z) = (1 − 0.8e jπ/4 z −1 )(1 − 0.8e−jπ/4 z −1 ) = 1 − 1.6 cos(π/4)z −1 + 0.64z −2 . (c) y[n] is the output of H2 (z) when the input is y1 [n] so from H2 (z) we get y[n] = y1 [n] − 1.6 cos(π/4)y1 [n − 1] + 0.64y1 [n − 2]

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CHAPTER 9. z-TRANSFORM P-9.17

DSP First 2e

(a) h[n] = 7δ[n] − 14.35δ[n − 1] + 22.4175δ[n − 2] − 14.35δ[n − 3] + 7δ[n − 4] We need to determine the expanded version of H (z) with an additional gain factor.

H (z) = A(1 − 0.8e jπ/3 z −1 )(1 − 0.8e−jπ/3 z −1 )(1 − 1.25e jπ/3 z −1 )(1 − 1.25e−jπ/3 z −1 ) = (1 − 0.8z −1 + 0.64z −2 )(1 − 1.25z −1 + 1.5625z −2 ) Clearly, when expanded out, the H (z) polynomial will be of the form H (z) = A + . . . + A(0.8) 2 (1.25) 2 z −4 , so if h[0] = 7 then A = 7. (b) H (z) = 1.81406 − 3.71882z −1 + 5.80952z −2 − 3.71882z −3 + 1.81406z −4 If we expand H (z) into a polynomial with unknown gain we get H (z) = A(1 − 2.05z −1 + 3.2025z −2 − 2.05z −3 + z −4 ) The DC value of the frequency response is H (e j0 ) = A(1 − 2.05 + 3.2025 − 2.05 + 1) = 1.1025A = 2 so A = 1.8141 and H (z) = 1.8141(1 − 2.05z −1 + 3.2025z −2 − 2.05z −3 + z −4 )

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CHAPTER 9. z-TRANSFORM P-9.18

DSP First 2e

(a) H (1) = b0 + b1 − b1 − b0 = 0 (b) H (e j ω̂ ) = b0 + b1 e−j ω̂ − b1 e−j ω̂3 − b0 e−j ω̂4 = e−j ω̂2 (b0 e j ω̂2 + b1 e j ω̂ − b1 e−j ω̂ − b0 e−j ω̂2 ) = [2b0 sin(2ω̂) + 2b1 sin(ω̂)]e j (π/2−j ω̂

(from the inverse Euler formula)

(c) H (z −1 ) = b0 + b1 z − b1 z 3 − b0 z 4 = z 4 (b0 z −4 + b1 z −3 − b1 z −1 − b0 ) = −z 4 H (z) (d) In the general case with bk = −b M−k , the condition H (1/z) = −z M H (z) holds for M even or odd.

The frequency response has the form H (e j ω̂ ) = 2e j (π/2−ω̂ M/2)

M/2 X

b M/2−k sin(ω̂k)

M an even integer

k=1

and H (e

j ω̂

) = 2e

j (π/2−ω̂ M/2)

(M+1)/2 X k=1

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b(M+1)/2−k sin(ω̂(2k − 1)/2)

M an odd integer

CHAPTER 9. z-TRANSFORM P-9.19

DSP First 2e

(a) H1 (z) = H2 (z) = 1 + z −1 + z −2 + z −3 (b) H (z) = H1 (z)H2 (z) = (1 + z −1 + z −2 + z −3 ) 2 (c) H (z) = 1 + 2z −1 + 3z −2 + 4z −3 + 3z −4 + 2z −5 + z −6 so h[n] = δ[n] + 2δ[n − 1] + 3δ[n − 2] + 4δ[n − 3] + 3δ[n − 4] + 2δ[n − 5] + δ[n − 6] (d) y[n] = x[n] + 2x[n − 1] + 3x[n − 2] + 4x[n − 3] + 3x[n − 4] + 2x[n − 5] + x[n − 6] (e) We can use the formula for the sum of N terms of a geometric series to write H1 (z), or H2 (z) as H1 (z) = H2 (z) = 1 + z −1 + z −2 + z −3 =

1 − z −4 1 − z −1

(f) From part (e) we see that (1 + z −1 + z −2 + z −3 ) 2 so all the zeros at {−1, ± j} are double zeros, and there are six poles at z = 0.

2

Imaginary Part

1 0.5 0

2

6

-0.5 2

-1 -1

-0.5

0 0.5 Real Part

1

Figure P-9.19

(g) From the result in (e), we can write H1 (e j ω̂ ) = H2 (e j ω̂ ) =

sin(2ω̂) −j3ω̂/2 1 − e−j ω̂4 e−j ω̂2 (e j ω̂2 − e−j ω̂2 ) e = = −j ω̂ −j ω̂/2 j ω̂/2 −j ω̂/2 1−e e (e −e ) sin(ω̂/2)

(h) Note that H (e j ω̂ ) = H2 (e j ω̂ )H1 (e j ω̂ ) = [H1 (e j ω̂ )]2 = [H2 (e j ω̂ )]2 .

H (e

j ω̂

) = H1 (e

j ω̂

)H2 (e

j ω̂

sin(2ω̂) )= sin(ω̂/2)

!2

e−j3ω̂

Since the squared-term is nonnegative, it is the magnitude, and the system has linear phase with time delay equal to 3 samples. The sketch should approximate the Matlab plots of magnitude and phase shown below.

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CHAPTER 9. z-TRANSFORM

|H(ej ω̂ )|

20 10

H(ej ω̂ )

0 −π 5

0.2π

0.4π

0.6π

0.8π

π

0.2π 0 ω̂ Figure P-9.19

0.4π

0.6π

0.8π

π

−0.8π −0.6π −0.4π −0.2π

0

6

0

-5 −π

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−0.8π −0.6π −0.4π −0.2π

Chapter

IIR Filters 10-1

Problem Solutions

201 © 2022 Pearson Education.

10

CHAPTER 10. IIR FILTERS P-10.1

DSP First 2e

(a) The system function is H (z) =

1 1 − z −1 − z −2

=

1 (1 − p1 z −1 )(1 − p2 z −1 )

√ where, by the quadratic formula, p1,2 = (1 ± 5)/2 = 1.618, −0.618. The partial fraction expansion of H (z) is √ √ 5+1 5−1 p2 p1 √ √ p1 − p2 p2 − p2 2 5 2 5 + = + H (z) = 1 − p1 z −1 1 − p2 z −1 1 − p1 z −1 1 − p2 z −1 Therefore, the impulse response is √ n √ n √ √ 5 + 1+ *1 + 5+ 5 − 1+ *1 − 5+ * * h[n] = u[n] + u[n] √ √ , 2 5 -, 2 , 2 5 -, 2 Try a few terms of this closed form expression to see if it generates the Fibonacci sequence, 1, 1, 2, 3, 5, 8, 13, ... √ (b) The poles determined in part (a) are p1,2 = (1 ± 5)/2 = 1.618, −0.618. Since one of the poles is outside the unit circle, this is an unstable system.

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CHAPTER 10. IIR FILTERS P-10.2

DSP First 2e

First, find the system function by inspection as !

1 z −1 = z −1 = G(z)z −1 H (z) = 1 − 0.9z −1 + 0.9z −2 1 − 0.9z −1 + 0.9z −2 The z −1 factored out suggests that the impulse response will be zero prior to n = 1. Then we can find the partial fraction expansion of the rational function part and write G(z) as G(z) =

0.568e j0.4942 0.568e−j0.4942 + 1 − 0.9487e j1.0766 z −1 1 − 0.9487e−j1.0766 z −1

and n

n

g[n] = 0.568e−j0.4942 0.9487e j1.0766 u[n] + 0.568e j0.4942 0.9487e−j1.0766 u[n] = 1.136(0.9487) n cos(1.0766n − 0.4942)u[n] Therefore, the impulse response is h[n] = g[n − 1] = 1.136(0.9487) n−1 cos(1.0766(n − 1) − 0.4942)u[n − 1] which, by virtue of u[n − 1], is the desired formula showing h[n] being zero for n < 1 and being a decaying sinusoid for n ≥ 1.

It should be noted that we could have made a partial fraction expansion of H (z) directly. The resulting inverse transform would be a formula that is valid from n = 0 since the complex exponentials would be multiplied by u[n]. However, when the formula is evaluated at n = 0, the result would be zero.

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CHAPTER 10. IIR FILTERS P-10.3

DSP First 2e

The system function is H (z) =

5z −7 1 − 0.5z −1

Us the z-transform table and then the delay property, h[n] = 5(0.5) n−7 u[n − 7] Note that h[n] = 0 for n < 7. It decays exponentially as (0.5) n for n ≥ 7.

h[n]

6 4 2 0 0

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5

10

15 20 time index n Figure P-10.3

25

30

CHAPTER 10. IIR FILTERS P-10.4

DSP First 2e

The impulse response is h[n] = 5(0.5) n u[n], so the output for the given input is y[n] = 3h[n − 1] − 2h[n − 2] + 3h[n − 4] − h[n − 5] = 15(0.5) n−1 u[n − 1] − 10(0.5) n−2 u[n − 2] + 15(0.5) n−4 u[n − 4] − 5(0.5) n−5 u[n − 5] Evaluate with Matlab in order to make a plot.

y[n]

20 10 0 -10 0

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5

10

15 20 time index n Figure P-10.4

25

30

CHAPTER 10. IIR FILTERS P-10.5

DSP First 2e

The easiest way to work this is to write the output in terms of shifted impulse responses as y[n] = h[n] + βh[n − 2] + h[n − 4] √ and then examine y[n] for n ≥ 4. Define a = 1/ 2 so h[n] = 10a n u[n]. Then for n ≥ 4, the output is y[n] = 10a n + 10 βa n−2 + 10a n−4 = 10a n (1 + βa−2 + a−4 ) To make y[n] = 0 for n ≥ 4, we need 1 + βa−2 + a−4 = 0. Solving this equation for β gives β = −(1 + a4 )/a2 , or for the given value of a, β = −2.5. If you use this value of β, the output looks like the following

y[n]

5 0 -5 -10 0

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5

10

15 20 time index n Figure P-10.5

25

30

CHAPTER 10. IIR FILTERS P-10.6

DSP First 2e

(a) The system function is H (z) =

−1 + 3z −1 − 2z −2 1 − 13 z −1 − 12 z −2

The zeros are at z = 1, 2, and the poles at z = 0.8931, −0.5598

Imaginary Part

2 1 0 -1 -2 -2

-1

0 Real Part

1

(b) The system function is H (z) =

2

−1 + 3z −1 + 2z −2 1 − 13 z −1 + 12 z −2

The zeros are now at z = 3.5616, −0.5616, and the poles at z = 0.7071e±j1.3329 .

Imaginary Part

4

2

0

-2

-4 -4

-2

0 Real Part

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2

4

CHAPTER 10. IIR FILTERS P-10.7

DSP First 2e

(a) y[n] = 13 y[n − 1] − 21 y[n − 2] − x[n] H (z) =

−1 1 − 31 z −1 + 12 z −2

=

−z 2 z 2 − 13 z + 12

=⇒Poles: z = 0.7071e±j1.3329 Zeros: z = 0, 0 (b) y[n] = 13 y[n − 1] − 21 y[n − 2] − x[n − 2] H (z) =

−z −2 1 − 31 z −1 + 12 z −2

=

−1 z 2 − 13 z + 12

=⇒Poles: z = 0.7071e±j1.3329 Zeros: z = ∞, ∞ (c) y[n] = 13 y[n − 1] − 21 y[n − 2] − x[n − 4] H (z) =

−z −4 1 − 31 z −1 + 12 z −2

=

−1 z 2 (z 2 − 13 z + 21 )

=⇒Poles: z = 0.7071e±j1.3329, 0, 0 Zeros: z = ∞, ∞, ∞, ∞ In (c) there are four zeros at infinity balanced by two additional poles at the origin. (b) Two zeros at z = ∞

(a) Two zeros at origin

0.5 (2)

0 -0.5 -1

1 Imaginary Part

1 Imaginary Part

Imaginary Part

1

(c) Four zeros at z = ∞

0.5 0 -0.5 -1

-1

0 Real Part

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1

0.5 (2)

0 -0.5 -1

-1

0 Real Part

1

-1

0 Real Part

1

CHAPTER 10. IIR FILTERS P-10.8

DSP First 2e

(a) By inspection, H (z) =

1 1 + 21 z −1

=

z z − 21

H (z) has a pole at z = − 21 and a zero at z = 0. The corresponding impulse response is h[n] = (−0.5) n u[n]. (b) Write the input as x[n] = δ[n] + δ[n − 2] + δ[n − 5]. Therefore, it follows that for this input the output is y[n] = h[n] + h[n − 2] + h[n − 5]. If we substitute for h[n] we get y[n] = (−0.5) n u[n] + (−0.5) n−2 u[n − 2] + (−0.5) n−5 u[n − 5] Taking the step sequence into account we get  0      (−0.5) n y[n] =   5(−0.5) n     −27(−0.5) n 

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n<0 n = 0, 1 n = 2, 3, 4 n≥5

CHAPTER 10. IIR FILTERS P-10.9

DSP First 2e

(a) H (z) =

−15 + 20z −1 −15(z − 4/3) = z − 0.75 1 − 0.75z −1

(b) Need plot. Pole at z = 0.75, zero at z = 4/3. (c) H (e j ω̂ ) =

−15 + 20e−j ω̂ 1 − 0.75e−j ω̂

(d) |H (e j ω̂ )| 2 = 400 conjugate

}| {! !z j ω̂ −j ω̂ −15 + 20e −15 + 20e |H (e j ω̂ )| 2 = H (e j ω̂ )H ∗ (e j ω̂ ) = 1 − 0.75e−j ω̂ 1 − 0.75e j ω̂

© 2022 Pearson Education.

=

625 − 600 cos(ω̂) 1.5625 − 1.5 cos(ω̂)

=

400(1.5625 − 1.5 cos(ω̂)) = 400 1.5625 − 1.5 cos(ω̂)

CHAPTER 10. IIR FILTERS P-10.10

DSP First 2e

(a) By inspection, H (z) =

1 1 = √ √ −2 1 + 0.8z (1 − 0.8 e jπ/2 z −1 )(1 − 0.8 e−jπ/2 z −1 )

z2 √ √ (z − 0.8 e jπ/2 )(z − 0.8 e−jπ/2 ) √ Zeros at z = 0, 0 and poles at z = 0.8 e±jπ/2 . =

(b) To find the output for the given input, determine the impulse response.

H (z) =

0.5 0.5 1 = + √ √ 1 + 0.8z −2 1 − 0.8 e jπ/2 z −1 1 − 0.8 e−jπ/2 z −1

so h[n] = 12

√ √ n n √ n 0.8 e jπ/2 u[n] + 21 0.8 e−jπ/2 u[n] = 0.8 cos( π2 n)u[n]

Note: h[n] = 0 for n even, which could have been predicted from the fact that the initial H (z) has only z −2 in the denominator—an even power.

Therefore, y[n] = 5h[n] + 2.5h[n − 1] + 4h[n − 2] + 2h[n − 3] √ n−1 √ n = 5 0.8 cos( π2 n)u[n] + 2.5 0.8 cos( π2 (n − 1))u[n − 1] √ n−2 √ n−3 + 4 0.8 cos( π2 (n − 2))u[n − 2] + 2 0.8 cos( π2 (n − 3))u[n − 3]

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CHAPTER 10. IIR FILTERS P-10.11

DSP First 2e

(a) By inspection, H (z) =

z5 1 = 1 − 0.97z −5 z 5 − 0.97

(b) There are 5 zeros at z = 0, and the poles are at the 5th roots of 0.97, zk = (0.97) 0.2 e j2πk/5 = (0.9939)e j2πk/5 .

Imaginary Part

1 0.5 5

0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-10.11

(c) The z-transform of the output is Y (z) = H (z)X (z) =

1 (1 + βz −p ) 1 − 0.97z −5

If we choose β = −0.97 and p = 5, we get Y (z) = 1 or y[n] = δ[n], which satisfies the condition y[n] = 0 for n ≥ 5.

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CHAPTER 10. IIR FILTERS P-10.12

DSP First 2e

(a) By using the delay and linearity properties, h a [n] = (−0.77) n u[n] − (−0.77) n−1 u[n − 1] or, by making the partial fraction expansion, Ha (z) = −1/.77 +

1.77/0.77 ⇐⇒ x a [n] = −1/0.77 + (1.77/0.77) (−0.77) n u[n] | {z } | {z } 1 + 0.77z −1 −1.2987

2.2987

(b) By using the delay and linearity properties, hb [n] = (0.9) n u[n] + 0.8(0.9) n−1 u[n − 1] or, by making the partial fraction expansion, Ha (z) = −0.8/0.9 +

1.7/0.9 ⇐⇒ x a [n] = −0.8/0.9 + (1.7/0.9) (0.9) n u[n] | {z } | {z } 1 − 0.9z −1 −0.8889

(c) Using the delay property, hc [n] = (0.9) n−2 u[n − 2] (d) Using the transform of an impulse and the delay property, hd [n] = δ[n] − δ[n − 1] + 2δ[n − 3] − 3δ[n − 4]

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1.8889

CHAPTER 10. IIR FILTERS P-10.13

DSP First 2e

(a) Using residuez or hand calculation, determine the partial fraction expansion of H (z) as X a (z) =

1 − z −1 (1 − 12 z −1 )(1 + 13 z −1 )

=

−0.6 1 − 21 z −1

+

1.6 1 + 13 z −1

Therefore, x a [n] = −0.6( 12 ) n u[n] + 1.6(− 31 ) n u[n] (b) Determine the partial fraction expansion as 1 + z −2 (1 − 0.9e j2π/3 z −1 )(1 − 0.9e−j2π/3 z −1 ) 0.6556e j1.7506 0.6556e−j1.7506 = 1.2346 + + 1 − 0.9e j2π/3 z −1 1 − 0.9e−j2π/3 z −1

X b (z) =

Now the inverse z-transform is by table look up, x b [n] = 1.2346 δ[n] + 0.6556e j1.7506 (0.9e j2π/3 ) n u[n] + 0.6556e−j1.7506 (0.9e−j2π/3 ) n u[n] = 1.2346 δ[n] + 1.3112(0.9) n cos((2π/3)n + 1.7506)u[n] (c) The partial fraction expansion is X c (z) =

1 + z −1 −0.1176 1.1176 + = (1 − 0.9z −1 )(1 + 0.8z −1 ) 1 − 0.9z −1 1 + 0.8z −1

so the inverse transform is x c [n] = 1.1176(0.9) n u[n] − 0.1176(−0.8) n u[n]

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CHAPTER 10. IIR FILTERS P-10.14

DSP First 2e

S1 :

y[n] = 0.9y[n − 1] + 21 x[n] + 12 x[n − 1] #2

S2 :

y[n] = −0.9y[n − 1] + 9x[n] + 10x[n − 1] #3

S4 :

1 (1 − z −1 ) #6 H (z) = 2 1 + 0.9z −1 y[n] = 41 x[n] + x[n − 1] + 23 x[n − 2] + x[n − 3] + 41 x[n − 4]

S5 :

H (z) = 1 − z −1 + z −2 − z −3 + z −4

S6 :

y[n] =

S3 :

2 X

x[n − k]

not used

#5

#4

k=0

S7 :

y[n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4] + x[n − 5]

© 2022 Pearson Education.

#1

CHAPTER 10. IIR FILTERS P-10.15

DSP First 2e

(a) By inspection, H (z) =

1 1 − 12 z −1

. Therefore,

!

1 −1 1 2 * + Y (z) = H (z)X (z) = = + −1 1 −1 1 −1 1 − z −1 1 − 2z ,1 − 2 z - 1 − z Therefore, y[n] = 2u[n] − ( 12 ) n u[n] = (2 − ( 12 ) n )u[n]. (b) Use a partial fraction expansion:

!

1 1 1 −jπ/4 1 − 2e jπ/4 + = 1 − 2e 1 −1 1 − e jπ/4 z −1 1 − 2z

1 1 + Y (z) = H (z)X (z) = * jπ/4 1 −1 z −1 ,1 − 2 z - 1 − e !

1 1 + e j (π/4)n u[n] ( 21 ) n u[n] + * Therefore, y[n] = jπ/4 1 −jπ/4 1 − 2e e 1 − ,| 2{z }H (e j π /4 )

(c) Note that H (e j ω̂ ) =

1 1 − 12 e−j ω̂

so the second term in y[n] in part (b) is the steady-state component.

!

1 ( 1 ) n u[n] + H (e jπ/4 )e j (π/4)n u[n] y[n] = | {z } 1 − 2e jπ/4 2 steady state

© 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.16

DSP First 2e

(a) Multiply out the factors in numerator and denominator: H (z) =

1 − 0.1756z −1 − 0.1756z −2 + z −3 1 + 0.4944z −1 + 0.64z −2

Therefore, the difference equation is y[n] = −0.4944y[n − 1] − 0.64y[n − 2] + x[n] − 0.1756x[n − 1] − 0.1756x[n − 2] + x[n − 3] (b) Get the poles and zeros from the given factored form of H (z); then plot in the complex z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-10.16

(c) H (z) has zeros on the unit circle at ω̂ = 0 and ω̂ = ±0.3π. Therefore, any complex exponential with one of these frequencies will have zero output.

© 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.17

DSP First 2e

S1 :

y[n] = 0.9y[n − 1] + 21 x[n] + 12 x[n − 1] #1

S2 :

y[n] = −0.9y[n − 1] + 9x[n] + 10x[n − 1] #4

S4 :

1 (1 − z −1 ) #2 H (z) = 2 1 + 0.9z −1 y[n] = 41 x[n] + x[n − 1] + 23 x[n − 2] + x[n − 3] + 41 x[n − 4]

S5 :

H (z) = 1 − z −1 + z −2 − z −3 + z −4

S6 :

y[n] =

S3 :

2 X

x[n − k]

#6

#5

#3

k=0

S7 :

y[n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4] + x[n − 5]

© 2022 Pearson Education.

not used

CHAPTER 10. IIR FILTERS P-10.18

(A) #2 (B) #4 (C) #1 (D) #6 (E) #3

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DSP First 2e

CHAPTER 10. IIR FILTERS P-10.19

DSP First 2e

(a) Expand the numerator and denominator to get H (z) =

1 − z −1 + z −2 1 − 2 cos(π/3)z −1 + z −2 = 1 − 2r cos(π/3)z −1 + r 2 z −2 1 − r z −1 + r 2 z −2

Therefore, the difference equation is y[n] = r y[n − 1] − r 2 y[n − 2] + x[n] − x[n − 1] + x[n − 2]. (b) Plot the poles and zeros of H (z) for the case r = 0.9. Angles to poles and zeros are ±π/3.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) Magnitude of the frequency response plotted for the case r = 0.9. Notches are at ±π/3. |H(ej ω̂ )|

1.5 1

0.5 0 −π

−0.8π −0.6π −0.4π −0.2π

0

0.2π

0.4π

0.6π

0.8π

π

ω̂

(d) The notches get sharper and the magnitude at other frequencies approaches 1. (e) Frequency = Angle. Change the angle of the poles and zeros from π/3 to whatever frequency is desired.

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CHAPTER 10. IIR FILTERS P-10.20

(J) #6 (K) #5 (L) #4 (M) #2 (N) #3

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DSP First 2e

CHAPTER 10. IIR FILTERS P-10.21

(a) H (z) =

DSP First 2e

z − 1.25 −0.8 + z −1 = (−0.8) −1 z − 0.8 1 − 0.8z

(b) The pole is at 0.8 and the zero is at 1.25 = 1/0.8. (c) |H (e j ω̂ )| 2 = H (e j ω̂ )H ∗ (e j ω̂ ) =

(−0.8 + e−j ω̂ )(−0.8 + e j ω̂ ) 1.64 − 1.6 cos(ω̂) =1 = 1.64 − 1.6 cos(ω̂) (1 − 0.8e−j ω̂ )(1 − 0.8e j ω̂ )

(d) This results because the magnitude response is unity for all ω̂. Only the phases will change. (e) By evaluating H (e j ω̂ ) at the two frequencies (with Matlab), we get ϕ1 = −2.4826 and ϕ2 = −3.0285.

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CHAPTER 10. IIR FILTERS P-10.22

DSP First 2e

If we sample with f s = 4000, the input to the LTI system is x[n] = 9 + 8 cos(0.3πn) + 7 cos(0.15πn − 0.7π) so we need to evaluate the frequency response at ω̂ = 0, 0.3π, 0.15π. The frequency response is obtained by substituting z = e j ω̂ into the expression for H (z) and then evaluating at the required frequencies.

Since H (z) has zeros on the unit circle at ω̂ = ±0.3π, the component at that frequency is nulled out. Therefore, the output of the LTI system is y[n] = 9H (e j0 ) + 7|H (e j0.15π )| cos(0.15πn − 0.7π + ∠H (e j0.15π )) y[n] = 9(0.7725) + 7(0.6010) cos(0.15πn − 0.7π − 0.3190)) y[n] = 6.9525 + 4.2070 cos(0.15πn − 0.7π − 0.3190)) Now use the sampling frequency to write the output of the D-to-C converter as y(t) = 6.9525 + 4.2070 cos(600πt − 0.7π − 0.3190))

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CHAPTER 10. IIR FILTERS P-10.23

DSP First 2e

(a) Plot the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

(b) The frequency response is constant for all ω̂ at (1.25) 2 = 1.5625. The vector lengths should cancel out to a constant. |H(ej ω̂ )|

1.5625 1.5625 1.5625 −π

0

-5

5

π

ω̂

(c) The zeros of an all-pass filter are at reciprocals of the poles, so the angle stays the same but the radius of the zeros must be decreased to 1/0.9 =1.1111.

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CHAPTER 10. IIR FILTERS P-10.24

DSP First 2e

(a) Expand H (z) to H (z) =

(1 + z −2 ) (1 + z −2 ) = (1 − 0.9z −1 )(1 + 0.5z −1 ) 1 − 0.4z −1 − 0.45z −2

Therefore, the difference equation is y[n] = 0.4y[n − 1] + 0.45y[n − 2] + x[n] + x[n − 2]. (b) Plot the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) The pole at z = 0.9 dominates around ω̂ = 0 and the zeros pull down the frequency response around ω̂ = ±π/2. The pole at z = −0.5 is symmetrical with the other pole, but much farther away from the unit circle. Therefore the maximum value occurs at ω̂ = 0.

The estimate of the maximum value is obtained as the ratio of the product of the lengths of the vectors from the zeros √ 2 to z = 1 divided by the product of the lengths of the vectors from the poles to z = 1. The maximum value is ( 2) /(0.1 × 1.5) = 13.33. (d) At ω̂ = ±π/2, |H (e j ω̂ )| = 0.

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CHAPTER 10. IIR FILTERS P-10.25

DSP First 2e

(a) This system has a pole at z = −0.8 and a zero at z = 0.9. (b) |H (e j ω̂ )| 2 = H (e j ω̂ )H ∗ (e j ω̂ ) =

(1 − 0.9e−j ω̂ )(1 − 0.9e j ω̂ ) 1.81 − 1.8 cos(ω̂) = (1 + 0.8e−j ω̂ )(1 − 0.8e j ω̂ ) 1.64 + 1.6 cos(ω̂)

(c) This is a highpass filter since it has a zero near z = 1 or ω̂ = 0 and a pole near z = −1 or ω̂ = π.

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CHAPTER 10. IIR FILTERS P-10.26

DSP First 2e

! !

15z 15 10 + 5z −1 3 = = (a) H (z) = H1 (z)H2 (z) = −1 −1 −1 z − 0.7 2+z 1 − 0.7z 1 − 0.7z Therefore, H (z) has a zero at z = 0 and a single pole at z = 0.7. (b) Since H1 (z) =

H (z) =

1 + 21 z −1 1 − 12 z −1

, it follows that

3z −1 (1 + 12 z −1 ) (1 − 21 z −1 )(1 − 56 z −1 + 16 z −2 )

=

3z −1 + 32 z −2 7 −2 1 −3 1 − 34 z −1 + 12 z − 12 z

7 1 y[n − 2] + 12 y[n − 3] + 3x[n] + 23 x[n − 2 − 1] (c) y[n] = 43 y[n − 1] − 12

(d) For the systems in (b), plot the poles and zeros of H (z) in the complex z-plane.

Imaginary Part

1 0.5 2

0 -0.5 -1 -1

-0.5

0 0.5 Real Part

Figure P-10.26

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1

CHAPTER 10. IIR FILTERS P-10.27

DSP First 2e

(a) Since we require H (z) = H1 (z)H2 (z) = 1 for the output to equal the input, then H2 (z) = 1/H1 (z). (b) Since H1 (z) = 1 − 43 z −1 , H2 (z) =

1 1 = H1 (z) 1 − 34 z −1

(c) Since the poles of H2 (z) must be the zeros of H1 (z), both the poles and the zeros of H1 (z) must be inside the unit circle of the z-plane.

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CHAPTER 10. IIR FILTERS P-10.28

DSP First 2e

(a) Here is a stem plot of y[n] versus n, over the range 0 ≤ n ≤ 10.

y[n]

1 0 -1 0

5

10

15 20 time index n Figure P-10.28

25

30

(b) We need to find a system whose impulse response is h[n] = (0.95) n cos(0.2πn + 0.4π)u[n]. Do this by finding the corresponding system function, writing it as a ratio of polynomials in z −1 , and picking off the coefficients of the corresponding difference equation.

First write h[n] as h[n] = 12 e j0.4π (.95e j0.2π ) n u[n] + 21 e−j0.4π (.95e−j0.2π ) n u[n] The corresponding z-transform is 1 −j0.4π 1 j0.4π 2e 2e + H (z) = 1 − 0.95e j0.2π z −1 1 − 0.95e−j0.2π z −1

If we put this over a common denominator and collect terms using the inverse Euler formula, we get H (z) =

cos(0.4π) − 0.95 cos(0.2π)z −1 0.3090 − 0.7686z −1 = 1 − 1.9 cos(0.2π)z −1 + (0.95) 2 z −2 1 − 1.5371z −1 + 0.9025z −2

The difference equation for implementing this system is therefore, y[n] = 1.5371y[n = 1] − 0.9025y[n − 2] + 0.3090x[n] − 0.7686x[n − 1]

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CHAPTER 10. IIR FILTERS P-10.29

DSP First 2e

(a) H3 (z) =

b0 −1 1 − a1 z − a2 z −2 − a3 z −3

(b) In general, the system function overall is 2b0 z −1 (1 − z −2 )(1 + 12 z −1 )

H (z) = H3 (z)H2 (z)H1 (z) = =

1 − a1 z −1 − a2 z −2 − a3 z 2b0 z −1 (1 + 21 z −1 − z −2 − 12 z −3 )

= −3

2b0 z −1 (1 − z −2 )(1 + 21 z −1 ) 1 − a1 z −1 − a2 z −2 − a3 z −3

1 − a1 z −1 − a2 z −2 − a3 z −3

For the given coefficients, we get H (z) =

14z −1 (1 − z −2 )(1 + 21 z −1 ) 1 − 21 z −1 − 12 z −2

(c) From the given information, we want H (z) = 8z −nd . Since we have a third-order polynomial in both the numerator and denominator, we just need to adjust the coefficients of the denominator so that the two polynomials are identical and therefore cancel. The required values are b0 = 4

a1 = − 12

a2 = 1

For these conditions, nd = 1.

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a3 = 21

CHAPTER 10. IIR FILTERS P-10.30

DSP First 2e

(a) yy=filter([0,0,1,0,1],1,xx);

ANS =1

(b) y[n] = 0.9y[n − 1] + 0.8y[n − 2] + x[n] − x[n − 3] (c) h[n] = u[n − 2] − u[n − 5] (d) y[n] =

3 √ X

k − 1 x[n − k]

ANS =4 ANS =5

k=1

(e) h[n] = 0.8n−2 u[n − 2] − 0.8n−3 u[n − 3]

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ANS =2

ANS =3

Appendix

Complex Numbers A-1

Problems

232 © 2022 Pearson Education.

A

APPENDIX A. COMPLEX NUMBERS P-A.1

DSP First 2e

(a) Cartesian: z = 0 + j2.

Polar: 2e jπ/2 = 2∠90◦

◦ (b) Cartesian: z = −3 − j4. Polar: 5e−j2.21 = 5e−j0.705π = 5 | ∠ − 126.87 {z } θ

(c) Cartesian: z = −1 + j; Polar:

√ j3π/4 2e = 2 |{z} ∠135◦ θ

(d) Cartesian: z = 0 − j; Polar: 1e−jπ/2 = 1 | ∠− 90}◦ {z θ

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APPENDIX A. COMPLEX NUMBERS P-A.2

(a)

DSP First 2e

√ j (3π/4) √ 2e = 2(cos(3π/4) + j sin(3π/4) ) = −1 + j | {z } | {z } √ =− 2/2

√ = 2/2

(b) z = 3e−j (π/2) = 3∠ − 90◦ ⇒ z = −3 j

(c) 1.6 ∠ (π/6) = 1.6e jπ/6 = 1.6(cos(π/6) + j sin(π/6) ) = 1.386 + j0.8 | {z } | {z } |{z} 30◦

√ =− 3/2

(d) 7∠ (7π) = 7∠π = 7e jπ = −7 + j0 |{z} 1260◦

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=0.5

APPENDIX A. COMPLEX NUMBERS P-A.3

DSP First 2e

(a) j 3 = j ( j) 2 = j (−1) = − j = 0 − j  n  +1 (b) j 2n = j 2 = (−1) n =   −1 

n even n odd

(n an integer)

(c) e j (π+2πm) = e jπ e j2πm = (−1)(1) = −1 + j0 (m an integer) 1/2

√ √ (this is one answer) = e jπ/4 = 2/2 + j 2/2 (d) j 1/2 = e jπ/2 1/2

√ √ j 1/2 = e jπ/2 e j2π = e jπ/4 |{z} e jπ = −e jπ/4 = − 2/2 − j 2/2 (this is the second answer) =−1

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APPENDIX A. COMPLEX NUMBERS P-A.4

DSP First 2e

√ Use this fact in (a,b,c): 2 − j2 = 2.449e−j0.304π (Angle is −0.955 rad, or −54.74◦ )

8 √ (a) ( 2 − j2) 8 = 2.449e−j0.304π = (2.449) 8 e−j8(.304π) = 1296e j0.433π (Angle is −1.36 rad, or −77.88◦ ) √ (b) ( 2 − j2) −1 = 1/2.449e+j0.304π = 0.4082e+j0.304π (Angle is +1.36 rad, or +77.88◦ )

1/2 √ (c) ( 2 − j2) 1/2 = 2.449e−j0.304π e j2π` = 1.565e−j0.152π e jπ` ` = 0, 1 There are two answers: 1.565e−j0.152π and 1.565e j0.848π

(d) Add the two complex numbers below and obtain −4.964 + j4.598 = 6.766e j0.762π (Angle is 2.394 rad, or 137.2◦ ) √ 3e j2π/3 = −3/2 + j3 3/2 √ −4e−jπ/6 = −4 3/2 + j2 ( ) ( ) ( ) (e) = je−jπ/3 = = e jπ/2 e−jπ/3 = = e jπ/6 = sin(π/6) = 0.5

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APPENDIX A. COMPLEX NUMBERS P-A.5

DSP First 2e

(a) z1∗ = (−4 + j3) ∗ = −4 − j3 = 5e−j0.795π = 5e j2.50

(b) z22 = (1 − j) 2 = 1 − 2 j + (− j) 2 = 1 − 2 j = 1 = −2 j 2 √ or, in polar form: z22 = 2e−jπ/4 = 2e−jπ/2 = − j2 (c) z1 + z2∗ = (−4 + j3) + (1 − j) ∗ = (−4 + j3) + (1 + j) = 3 + j4 = 5e j0.705π (Angle is 2.214 rad, or 126.872◦ ) (d) j z2 = j (1 − j) = j − j ( j) = j + 1 = 1 + j √ √ or, in polar form: e jπ/2 2e−jπ/4 = 2e jπ/4 (Angle is 0.785 rad, or 45◦ ) (e) Use polar form: z1−1 = 1/(5e j0.795π ) = 0.2e−j0.795π √ √ (f) Use polar form: z1 /z2 = (5e j0.795π )/( 2e−jπ/4 ) = (5 2/2)e j1.045π = 3.54e−j0.955π (Angle is −3.0 rad, or −171.8◦ ) (g) ez2 = e1−j = e1 e−j = e(cos(1) + j sin(1)) = 1.469 − j2.287 Polar form is e1 e−j = 2.7183e−j0.318π

(angle is −1 rad, or −57.3◦ )

(h) z1 z1∗ = (−4 + j3)(−4 + j3) ∗ = (−4 + j3)(−4 − j3) = 16 + 9 = 25 √ √ (i) Use polar form: z1 z2 = (5e j0.795π )/( 2e−jπ/4 ) = 5 2e j0.454π = 7.07e j1.713

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(98.13◦ )

APPENDIX A. COMPLEX NUMBERS P-A.6

DSP First 2e

z1 = e j9π/3 = e j3π = −1 + j0 z2 = e−j5π/8 = −0.383 − j0.924 z3 = e j13π/8 = e j13π/8 = 0.383 − j0.924z = −1 − j1.848 = 2.1e−j2.07

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APPENDIX A. COMPLEX NUMBERS P-A.7

DSP First 2e

(a) z = −3 + j4 = 5 e j0.705π 1/z = (1/5)e−j0.705π The angle is also 2.214 rad, or 126.87◦ . √ (b) z = −2 + j2 = 2 2 e j3π/4 √ 5 √ z 5 = 2 2 e j15π/4 = 128 2 e−jπ/4 because 15π/4 = 4π − π/4. (c) z = −5 + j13 |z| 2 = zz ∗ = (−5 + j13) (−5 − j13) = 25 + 169 = 194 = 194 e j0

(d) < (−2 + j5) |{z} e−jπ/2 = < (−2 + j5)(− j) = < − 2 j + 5 = 5 = 5e j0 =−j

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APPENDIX A. COMPLEX NUMBERS P-A.8

DSP First 2e

Recall that e j2π` = 1 when ` = integer. z 4 = j = e jπ/2 = e jπ/2 e j2π`

1/4 ⇒ z = e jπ/2 e j2π` = e jπ/8 e jπ`/2 ; ` = 0, 1, 2, 3

z ∈ e jπ/8, e j5π/8, |{z} e j9π/8, | e j13π/8 {z } e j π /8

e− j3π /8

There are four different answers shown as the four vectors below:

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APPENDIX A. COMPLEX NUMBERS P-A.9

DSP First 2e

z0 = e j2π/N ⇒ (z0 ) N −1 = e j2π/N

N −1

* 1−j2π/N

e j2π e z0N −1 = e j2π (N −1)/N = e j2π (N )/N e j2π (−1)/N = −1

= z0−1 = 1/z0 z0N −1 = e−j2π/N = e j2π/N

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APPENDIX A. COMPLEX NUMBERS P-A.10

DSP First 2e

Multiply by 1 = e j2π` which is true when ` = integer; this will generate all possible solutions. − j = 1e−jπ/2 = e j2π` e−jπ/2

1/2 ⇒ (− j) 1/2 = e j2π` e−jπ/2   e−jπ/4 = e jπ` e−jπ/4 =   e jπ e−jπ/4 = e j3π/4   1 Because e jπ` =   e jπ 

when ` is even when ` is odd

Thus, there are two solutions (shown below)

For each of the

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when ` is even when ` is odd

APPENDIX A. COMPLEX NUMBERS P-A.11

DSP First 2e

Use a calculator to check these √ 34 e−j1.03 z1 = √ = 0.707e−j1.275 68 e j0.245 z2 = 0.25 e jπ/2 z3 = e j5π/6 = e j2.618

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APPENDIX A. COMPLEX NUMBERS P-A.12

DSP First 2e

When N = 56, we can write N −1 X z=− e j2πk/N + e j2π (N −1)/N = e j2π(55)/56 = e−jπ/28 = e−j0.1122 k=0

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APPENDIX A. COMPLEX NUMBERS P-A.13

DSP First 2e

p = 32 and

√ q = 200 3 = 346.4102

Approach: Express the right-hand side (RHS) in Cartesian form using Euler’s formula. Then cross multiply by the LHS denominator to get a single complex-valued equation.

√

p = 200 + jq (0.08) −0.5 + j 3/2 The real and imaginary parts yield two (linear) equations in the two unknowns, p and q. One of these equations has only p, or only q, so solve it and then substitute into the other equation.

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APPENDIX A. COMPLEX NUMBERS P-A.14

DSP First 2e

Definite integral is

e jπt jπ

1.5 1

√ 2 −j3π/4 jθ e = 0.4502 e−j2.3562 . re = π Approach: The integral of an exponential is an exponential, but you end up with a j in the denominator because the exponent contains a j. After evaluating at the limits of the definite integral, the numerator contains a complex number. Finally, convert the complex numerator-denominator into polar form with a calculator.

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Appendix

Fourier Series C-1

Problem Solutions

247 © 2022 Pearson Education.

C

APPENDIX C. FOURIER SERIES P-C.1

DSP First 2e

(a) Sketch the periodic function x(t) over the time interval −2T0 < t < 2T0 for the case t c = T0 /4. x(t)

1

−T0

−2T0

1 (b) a0 = T0

T0 /2 Z −T0 /2

− 14 T0

1 x(t)dt = T0

Ztc /2

0

dt =

−tc /2

T0

1 4 T0

2T0

t

2t c T0

(c) Determine a formula for the Fourier series coefficients ak in the finite Fourier representation (??). Your final result should depend on t c and T0 .

1 ak = T0

T0 /2 Z

x(t)e

−j (2π/T0 )kt

−T0 /2

1 dt = T0

Ztc /2

e−j (2π/T0 )kt dt =

−tc /2

sin((2πt c /T0 )k) πk

k = ±1, ±2, . . .

(d) See plot below ak

tc = T0 /4

(d)

8f0

f

tc = T0 /10

(e)

8f0

f

1 4

–8f0

–4f0

0

4f0

ak 1 10

(e)

–8f0

–4f0

0

4f0

(f) We see by comparing the figures that the high-frequency components get bigger relative to the DC component as t c decreases. That is, very narrow pulses require much more bandwidth for an accurate representation by a finite Fourier series.

© 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.2

DSP First 2e

(a) Plot the periodic signal x(t) for −T0 ≤ t ≤ T0 . x(t)

1

− 21 T0

−T0

1 2 T0

T0

t

−1 Figure P-C.2

1 (b) a0 = T0

Z0 −T0 /2

1 x(t)dt + T0

T0 /2 T0 /2 T0 /2 Z Z Z 1 1 x(t)dt = x(τ + T0 /2)dτ + x(t)dt = 0 T0 T0 0

0

0

where we have made the substitution t = T0 /2 + τ in the first integral. (c) The same manipulation can be made on the general Fourier series coefficient integral.

1 ak = T0 1 = T0

1 = T0

Z0 x(t)e −T0 /2

−j (2π/T0 )kt

1 dt + T0

T0 /2 Z x(t)e−j (2π/T0 )kt dt 0

T0 /2 Z

T0 /2 Z x(t)e−j (2π/T0 )kt dt

0

0

1 x(τ + T0 /2)e−j (2π/T0 )k (τ+T0 /2) dτ + T0

 0  T0 /2  Z   T0 /2  Z 2 x(t)(1 − e−jπk )e−j (2π/T0 )kt dt =   x(t)e−j (2π/T0 )kt dt     T0 0 0 

k even k odd

Thus, when we have half-wave symmetry, the odd-indexed coefficients can be found by integrating over only half the period.

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APPENDIX C. FOURIER SERIES P-C.3

DSP First 2e

( (a) x(t) =

4+t 0

−4 ≤ t < 0 0≤t<4

(b) Here is a plot of x(t) over the range −8 ≤ t ≤ 8 s. x(t)

4

−8

−4

4

8

Figure P-C.3

(c) The DC value is just the average value of the waveform so a0 =

© 2022 Pearson Education.

area under curve 8 = = 1.

8 8

t

APPENDIX C. FOURIER SERIES P-C.4

DSP First 2e

(a) A sketch of the periodic function x(t) over the time interval −10 ≤ t ≤ 20. x(t)

2

−5

−10

5

10

Figure P-C.4

(b) The DC coefficient: a0 = 1 (c) a1 = 10

Z10

area under one period = (2 × 5)/10 = 1 10

2e−j (2π/10)t dt =

1 (e−j (2π/10)10 − e−j (2π/10)5 ) = j/π −10 j (2π/10)

5

(d) Adding a constant only changes the DC coefficient. b0 = 1 + a0 and b1 = a1 .

© 2022 Pearson Education.

t

APPENDIX C. FOURIER SERIES P-C.5

DSP First 2e

(a) Because each of the terms ak e jkt = t.

3 e jkt is paired with its complex conjugate ak∗ e−jkt giving a real function of 2 + jk

(b) Define ω0 = 1 and ak = 3/(2 + j k) then x(t) =

3 X

(− jak /2)e

j (k+7)t

+

k=−3

3 X

( jak /2)e

k=−3

− j1.5        2 + j (k − 7) j1.5 where bk =        2 + j (k + 7)  0

j (k−7)t

10 X

=

bk e jkt

k=−10

4 ≤ k ≤ 10 −10 ≤ k ≤ −4 otherwise

Note that b−k = b∗k as required for real x(t).

b−9

b b−8 −7 b−6

b−10

−10

−7

(c) See the explanation in part (b).

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b−5

b−4

−4

b4

0 Figure P-C.5

4

b5

b6

b7

7

b8

b9

b10

10

ω

APPENDIX C. FOURIER SERIES P-C.6

DSP First 2e

(a) Define ak =

j5 j5 . Each of the terms ak e j8πkt = e j8πkt is paired with its complex conjugate k + j1.5 k + j1.5

ak e j8πkt

∗

=

j5 − j5 −j8πkt e = e j8π (−k)t k − j1.5 (−k) + j1.5

for k = 1, 2 giving a real function of t for the sum. Multiplying by cos(3πt) is still real. (b) The frequencies in q(t) are ±5π, ±11π, ±13π, ±19π. Note that the frequencies ±3π have zero amplitude since a0 = 0.

The fundamental frequency is determined by the gcd of 5,11,13,19, which is 1, so ω0 = π. (c) The DC value is zero. There is no component at ω = 0. (d) Table listing all the nonzero Fourier coefficients of the signal q(t). ωk bk

−19π a−2 /2

−13π a−2 /2

−11π a−1 /2

−5π a−1 /2

5π a1 /2

11π a1 /2

13π a2 /2

19π a2 /2

(e) Spectrum plot showing only the positive frequency components. Note: ak =

0

© 2022 Pearson Education.

a1 /2

a1 /2

5π

11π Figure P-C.6

j5 . k + j1.5

a2 /2

a2 /2

13π

19π ω

APPENDIX C. FOURIER SERIES P-C.7

DSP First 2e

   0.5 (a) ak =  sin(πk/2)   πk 

k=0 k,0

(b) Sketch of the waveform y(t). y(t) = 2x(t − T0 /2)

2

−2T0

−T0

− 14 T0

0

1 4 T0

T0

2T0

t

Figure P-C.7

   1 bk = 2ak e−j (2π/T0 )kT0 /2 = 2ak e−jπk =  sin(πk/2)   2e−jπk πk 

k=0 k,0

Note that both ak and bk are zero for k a nonzero even integer. This is because both x(t) − 21 and y(t) − 1 have half-wave symmetry.

© 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.8

DSP First 2e

(a) Plot of x(t) over the interval −10 ≤ t ≤ 10 s with a period of 8 s. x(t)

7

3 −4

−8

(b) a0 =

4

8

t

area under one period 7 × 5 + 3 × 3 44 = = = 5.5 8 8 8

1 (c) ak = 8

Z5 7e

−j (2π/8)kt

1 dt + 8

0

Z8 3e−j (2π/8)kt dt 5

(d) 1 ak = 8

Z5 7e

−j (2π/8)kt

0

=

1 dt + 8

Z8

3e−j (2π/8)kt dt =

7(1 − e−j (π/4)k5 ) 3(e j (π/4)k5 − e−j (π/4)k8 ) + j (π/4)k8 j (π/4)k8

5

4(1 − e−j (π/4)k5 ) j2πk

= e−j5πk/8

4(e j5πk/8 − e−j5πk/8 ) 4 sin(5πk/8) = e−j5πk/8 j2πk πk

(e) Clearly, every eighth coefficient is zero since sin(5πk/8) = 0 for k a multiple of 8.

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APPENDIX C. FOURIER SERIES P-C.9

DSP First 2e

The Fourier series coefficients for the half-wave rectified sine wave are: 1      π(1 − k 2 )    ak =  ∓ j/4      0 

k even k = ±1 k odd and ≥ 3

The fundamental period is T0 = 0.02 s and therefore the fundamental frequency is f 0 = 1/0.02 = 50 Hz. Therefore the spectrum is 1 π

−j 4

−1 3π

0

50

100

−1 15π

−1 35π

−1 63π

−1 99π

200

300

400

500

f

The Fourier series coefficients for the full-wave rectified sine wave are: 2      π(1 − 4k 2 ) ak =      2  π

k even k=0

The fundamental period is T0 = T1 /2 = 0.01 s and therefore the fundamental frequency is f 0 = 1/0.01 = 100 Hz. The spectrum for the full-wave rectified sine wave is: 2 π

−2 3π

0

100

−2 15π

−2 35π

−2 63π

−2 99π

200

300

400

500

f

In both cases, the negative-frequency Fourier series coefficients are complex conjugates of the positive-frequency coefficients.

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APPENDIX C. FOURIER SERIES P-C.10

DSP First 2e

(a) Sketch of q(t) = p(t/5). The period is T0 = 5 × 0.1 = 0.5s and the pulse width is 5 × 0.04 = 0.2s. p(t)

1 −.1

−.05 −.02 0 .02

.05

.1

t

.5

t

q(t) = p(t/5)

1 −.25 −.1 0 .1 .25 Figure P-C.10

−.5

1 (b) ak = .1

Z.02

.02

e−j20πkt dt = 10

−.02

1 bk = .5

Z.1

sin(0.4πk) e−j20πkt = − j20πk −.02 πk .1

e−j4πkt dt = 2

−.1

sin(0.4πk) e−j4πkt = − j4πk −.1 πk

(c) Make the change of variable u = αt, and dt = du/α, in bk 1 bk = T0 /α

ZT0

1 x(u)e−j (2π/T0 )ku du/α = T0

0

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ZT0 0

x(u)e−j (2π/T0 )ku du = ak

APPENDIX C. FOURIER SERIES P-C.11

DSP First 2e

(a) One-sided spectrum for x(t). The negative frequency components are complex conjugates. j12

j6 j3

0

3

6

j2

12

18

f

Figure P-C.11

(b) Since 24 sin(6πt) = − j12e j6πt + j12e−j6πt it follows that the k = ±1 terms in x(t) are cancelled. Therefore, these components of the spectrum are missing in the spectrum of y(t). All other components are the same as in the spectrum of x(t). (c) The original 3 Hz fundamental is is not present. Now the fundamental is the gcd of 6, 12, 18 or f 0 = 6 Hz. (d) For x(t) with f 0 = 3 Hz and the Fourier coefficients are ak = {0, ± j12, ± j6, 0, ± j3, 0, ± j2} Thus y(t) has spectrum lines at {±6, ±12, ±18} hertz. The true fundamental frequency from the GCD is f 0 = 6 Hz. With respect to 6 Hz the formula for x(t) consists of the first, second and third harmonics. The formula for the relabeled Fourier coefficients is   j6/k bk = a2k =  0 

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for k = ±1, ±2, ±3 elsewhere

APPENDIX C. FOURIER SERIES P-C.12

DSP First 2e

(a) T0 = 4 s (b) f 0 = 1/4 = 0.25 Hz (c) a0 =

area under one period (2 × 1/2) + (3 × 1) − (2 × 1/2) = = 0.75 4 4

(d) The simplest expression would use symmetric limits: 1 ak = 4

Z−1

1 3e−jπkt/2 dt + 4

−2

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Z1 2te−jπkt/2 dt −1

APPENDIX C. FOURIER SERIES P-C.13

DSP First 2e

(a) T0 = 100 s (b) a3 = 5e−j0.644 , a−3 = 5e j0.644 , (c) b3 = 3e−jπ/2 , b−3 = 3e j1.571 ,

a2 = 0,

a1 = 2.236e−j0.464 , √ a−2 = 0, a−1 = 5e j0.464

a0 = 1,

b2 = 0,

b0 = 2.2e jπ ,

b1 = 2.236e−j0.464 , √ b−2 = 0, b−1 = 5e j0.464

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APPENDIX C. FOURIER SERIES P-C.14

DSP First 2e

(a) N-th harmonic is N = 3. (b) Since 3F0 = 2000, it follows that F0 = 2000/3 = 667 Hz (c) The signal x 2 (t) is the square wave since its Fourier series coefficients are ak =

sin(πk/2) πk

k = ±1, ±2, . . .

which are zero at the even harmonics.

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APPENDIX C. FOURIER SERIES P-C.15

DSP First 2e

(a) First, the fundamental frequenty of s(t) = | sin(2πt)| is F0 = 2 Hz. The FWRC can be expressed as x(t) = | cos(2πt)| = | sin(2π(t − .25))| = s(t − 0.25) so if ak denotes the Fourier coefficients of s(t), then the Fourier coefficients of x(t) are bk = ak e−j2π (2)(k)0.25 = ak e−jπk = (−1) k ak (b) The half-wave rectified sine is defined in (??) as ( sin((2π/T0 )t) 0 ≤ t < T0 /2 x(t) = 0 T0 /2 ≤ t < T0 where the fundamental period is T0 . Therefore, the FWRS can be expressed as s(t) = x(t) + x(t − T0 /2) i.e., we fill in the gaps in the HWRS by adding in a delayed version. Now if bk denotes the Fourier coefficients of the HWRS, the additive and delay properties give us ak = bk + bk e−j (2π/T0 )T0 /2 = (1 + e−jπk )bk Note that the fundamental frequency of the HWRS x(t) is 1/T0 but when we delay and add a copy the resulting FWRS has fundamental frequency 2/T0 . This is confirmed by observing that (1 + e−jπk )bk = 0 for k odd, which zeroes out the odd harmonics.

Since the odd harmonics are missing, the fundamental frequency is twice the fundamental of the HWRS. Thus, if we reindex the coefficients, we get ak = b2k (1 + e−j2πk ) = 2b2k . Recall the Fourier coefficients of the HWRS given in (??), which are 1       π(1 − k 2 ) bk =  ∓ j 41     0 

for k even for k = ±1 for k = ±3, ±5, . . .

Finally, we can write ak = b2k =

2 π(1 − 4k 2 )

(c) Since (−1) k = e−jπk = e−j (2π/T0 )kT0 /2 , multiplying the Fourier coefficients by (−1) k is equivalent to delaying the waveform by T0 /2. For the pulse wave with period 0.1s and pulse width 0.04s the plots are: p(t)

1 −.2

−.02 0 .02

−.2 −.07

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−.03

.1

t

q(t) = p(t − 0.05)

1 −.3

.05

0

.03 .05 .07

.1

.3

t

APPENDIX C. FOURIER SERIES P-C.16

DSP First 2e

d x(t). (a) Sketch of the triangle wave x(t), and its derivative y(t) = dt

x(t) 1 −T0

− T20

0 4 T0

−T0

− T20

T0 4 y(t) = dx(t) dt

0

T0 2

T0

t

T0 2

T0

t

Figure P-C.16

(b) The derivative property gives the result bk = ( j2πk/T0 )ak . (c) The Fourier coefficients of the square wave of amplitude 4/T0 are bk =

4 2 sin(πk/2) 8 sin(πk/2) = T0 πk πkT0

with b0 = 0

(d) Using the derivative property, we can write ( j2πk/T0 )ak =

8 sin(πk/2) πkT0

=⇒

ak = − j

4 sin(πk/2) π2 k 2

for k , 0. Note that a0 = 0, and thatt all even harmonics are zero since the triangle wave has half-wave symmetry.

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CHAPTER 3. SPECTRUM P-3.1

DSP First 2e

(a) x(t) = 11 + 14 cos(100πt − π/3) + 8 cos(350πt − π/2) (b) Since the gcd of 50 and 175 is 25, x(t) is periodic with period T0 = 1/25 = 0.04 s. (c) Negative frequencies are implicit in the cosine terms. They are needed to give a real signal when combined with their corresponding positive-frequency terms.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.2

DSP First 2e

(a) A single plot labeled with complex amplitudes is sufficient. The spectrum consists of the lines {(400, 5e jπ/4 ), (−400, 5e− jπ/4 ), (600, 3.5e− jπ/3 ), (−600, 3.5e jπ/3 ), (800, 1.5), (−800, 1.5)} where the frequencies are in Hz. 5e− jπ/4

5e jπ/4

3.5e jπ/3

3.5e− jπ/3

1.5 −800

1.5 −600

−400

0

400

600

800

f

(b) The signal x(t) is periodic with fundamental frequency 200 Hz or period 1/200 = 0.005 s since the gcd of {400, 600, 800} is 200. (c) The spectrum has the added components {(500, 2.5e jπ/2 ), (500, 2.5e− jπ/2 )}. Now we seek the gcd of {400, 500, 600, 800} so the fundamental frequency changes to 100 Hz and the period is 0.01 s.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.3

DSP First 2e

(a) x(t) = 2 cos(2t) + 4 cos(4t − π/3) + 2 cos(6t + π/4) (b) The spectrum is {(2, 1), (−2, 1), (4, 2e− jπ/3 ), (−4, 2e jπ/3 ), (6, e jπ/4 ), (−6, e− jπ/4 )} The frequencies are all in rad/s. e− jπ/4 −6

2e jπ/3

−4

1 −2

1 0

2

2e− jπ/3 4

e jπ/4 ω

6

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.4

DSP First 2e

(a) Determine a formula for x(t) as the real part of a sum of complex exponentials.

Use Euler’s formula for the sine function obtaining !3 e j27πt − e− j27πt sin (27πt) = 2j

1 j27πt3 e − 3e j27πt2 e− j27πt + 3e j27πt e− j27πt2 − e− j27πt3 = −8 j 3 1 = sin(27πt) − sin(81πt) 4 4 3

(b) What is the fundamental period for x(t)?

The fundamental frequency is 27/2 so the fundamental period is 2/27. (c) Plot the spectrum for x(t).

The spectrum is {(27π, − j3/8), (−27π, j3/8), (81π, j/8), (−81π, − j/8)}, where the frequencies are in rad/s.

j3/8

− j3/8

− j/8 −81π

j/8 −27π

0

27π

ω

81π

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CHAPTER 3. SPECTRUM P-3.5

DSP First 2e

There are seven spectral components: {(−3, 1/(1 − j3)), (−2, 1/(1 − j2)), (−1, 1/(1 − j)), (0, 1), (1, 1/(1 + j)), (2, 1/(1 + j2)), (3, 1/(1 + j3))}, where the frequencies are all in rad/s.

Putting all the complex numbers in polar form gives the following plot: 1 0.707e j0.25π

0.707e− j0.25π

0.447e j0.3524π 0.316e j0.398π

−3

−2

0.447e− j0.3524π 0.316e− j0.398π

−1

0

1

2

3

ω

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CHAPTER 3. SPECTRUM P-3.6

DSP First 2e

(a) In this case we need to find the gcd of 36 and 84, which is 12. Thus, the fundamental frequency is ω0 = 1.2π rad/s. (b) The fundamental period is T0 = 2π/ω0 = 1/0.6 = 5/3 s. (c) The DC value is −7. (d) The ak coefficients are nonzero for k = 0, ±3, ±7.

Here is the list of the nonzero Fourier series coefficients in a table. k ak

−7 3e− jπ/4

−3 4e jπ/3

0 7e jπ

3 4e− jπ/3

7 3e jπ/4

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CHAPTER 3. SPECTRUM P-3.7

DSP First 2e

(a) The phasor representation is z(t) = Ae j2π( f c − f∆ )t + Be j2π ( f c + f∆ )t (b) z(t) = e j2π f c t ( Ae− j2π f∆ t + Be j2π f∆ t ) = e j2π f c t ( A cos(2π f ∆ t) − j A sin(2π f ∆ t) + B cos(2π f ∆ t) − j B sin(2π f ∆ t)) = e j2π f c t [( A + B) cos(2π f ∆ t) − j ( A − B) sin(2π f ∆ t)] Therefore, the real part is x(t) = <{z(t)} = ( A + B) cos(2π f ∆ t) cos(2π f c t) + ( A − B) sin(2π f ∆ t) sin(2π f c t) so C = A + B and D = A − B. If A = B = 1, C = 2 and D = 0, so using the trigonometric identity cos α cos β = 1 1 2 cos(α − β) + 2 cos(α + β), it follows that x(t) = 2 cos(2π f ∆ t) cos(2π f c t) = 2[ 21 cos(2π( f c − f ∆ )t) + 12 cos(2π( f c + f ∆ )t)] (c) The values are A = 1 and B = −1. In this case, ! j2π f c t ! e j2π f∆ t − e− j2π f∆ t e − e− j2π f c t x(t) = 2 2j 2j

j2π( f c + f ∆ )t j2π( f c − f ∆ )t 1 = −2 e −e − e− j2π( f c − f∆ )t + e− j2π( f c + f∆ )t The spectrum is {(− f c − f ∆, −0.5), (− f c + f ∆, 0.5), ( f c − f ∆, 0.5), ( f c + f ∆, −0.5)}, and the plot is 0.5

−0.5

−( f c + f ∆ )

− fc

−( f c − f ∆ )

0.5

0

( fc − f∆)

−0.5

fc

( fc + f∆)

July 29, 2015 © 2022 Pearson Education.

f

CHAPTER 3. SPECTRUM P-3.8

DSP First 2e

(a) Using Euler’s relation we get x(t) = 10 + 10e jπ/4 e j2π(100)t + 10e− jπ/4 e− j2π(100)t + 5e j2π(250)t + 5e− j2π (250)t The gcd of 100 and 250 is 50 so f 0 = 50 and therefore N = 5. The nonzero Fourier coefficients are, therefore, a−5 = 5, a−2 = 10e− jπ/4 , a0 = 10, a2 = 10e jπ/4 , and a5 = 5. (b) The signal is periodic because all the frequencies are multiples of 50 Hz. Therefore, the fundamental period is T0 = 1/50 = 0.02s. (c) Here is the spectrum plot of this signal versus f in Hz. 10e− jπ/4

10

10e jπ/4

5

−250

5

−100

0

100

f

250

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.9

DSP First 2e

(a) Use phasors to show that x(t) can be expressed in the form x(t) = A1 cos(ω1 t + ϕ1 ) + A2 cos(ω2 t + ϕ2 ) + A3 cos(ω3 t + ϕ3 ) where ω1 < ω2 < ω3 ; i.e., find values of the parameters A1, A2, A3, ϕ1, ϕ2, ϕ3, ω1, ω2, ω3 .

Using Euler’s relation we get f g

x(t) = 12 − 3.5 je− jπ/3 e jπt + 3.5 je jπ/3 e− jπt 0.5e j13πt + .5e− j13πt = 12 cos(13πt) − 1.75 je− jπ/3 (e j14πt + e− j12πt ) + 1.75 je jπ/3 (e j12πt + e− j14πt ) = 12 cos(13πt) + 1.75e− jπ/2 e− jπ/3 (e j14πt + e− j12πt ) + 1.75e jπ/2 e jπ/3 (e j12πt + e− j14πt ) = 12 cos(13πt) + 1.75e− j5π/6 e j14πt + 1.75e− j5π/6 e− j12πt + 1.75e j5π/6 e j12πt + 1.75e j5π/6 e− j14πt = 12 cos(13πt) + 3.5 cos(14πt − 5π/6) + 3.5 cos(12πt + 5π/6) The requested parameters are easily picked off from this equation. (b) Sketch the two-sided spectrum of this signal on a frequency axis. Be sure to label important features of the plot.

Label your plot in terms of the numerical values of Ai , ϕi , and ωi .

12

3.5e j5π/6 −14π

12

3.5e− j5π/6 −13π

−12π

3.5e j5π/6 0

12π

3.5e− j5π/6 13π

14π

July 29, 2015 © 2022 Pearson Education.

ω

CHAPTER 3. SPECTRUM P-3.10

DSP First 2e

(a) Assume without limitation that ω2 − ω1 > 0. For periodicity with period T0 we require that ω0 = 2π/T0 . This means that k 1 ω0 = ω2 − ω1 and k 2 ω0 = ω2 + ω1 , where k1 and k 2 are integers and k 2 > k1 . (b) Part (a) gives two equations for ω1 and ω2 . If we solve them in terms of ω0 we get ω1 = (k2 − k1 )ω0 /2 and ω2 = (k2 + k 1 )ω0 /2, so the main condition is that both ω1 and ω2 are integer multiples of ω0 /2. This is the most general condition.

Therefore, the relationship between ω2 and ω1 is ω2 =

k2 + k1 ω1 k2 − k1

if x(t + T0 ) = x(t). Thus, ω2 could be an integer multiple of ω1 if k2 − k1 divides into k 2 + k1 with no remainder, but that is not necessary for periodicity of x(t).

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.11

DSP First 2e

(a) The gcd of 40, 60, 120 is 20 so ω0 = 20π and the fundamental period is T0 = 2π/ω0 = 0.1s. The finite Fourier series has components indexed by 0, ±2, ±3, ±6 so N = 6. The coefficients are a0 = 2, a±2 = 2e∓ jπ/5 , a±3 = 1.5e∓ jπ/2 , a±6 = 2e∓π/3 (b) The spectrum is {(−120π, 2e jπ/3 ), (−60π, 1.5e jπ/2 ), (−40π, 2e jπ/5 ), . . .

(0, 2), (40π, 2e− jπ/5 ), (60π, 1.5e− jπ/2 ), (120π, 2e− jπ/3 )} 2e jπ/3

2e jπ/5

2

2e− jπ/5

1.5e jπ/2

−120π

−60π

2e− jπ/3 1.5e− jπ/2

−40π

0

40π

60π

120π

(c) Now the fundamental frequency is 10π rad/s because the gcd of 20, 40, 50, and 120 is 10. Therefore, the period is T0 = 2π/10π = 1/5 = 0.2s. The spectrum is the same as in part (b) except there are two additional components at ±50π rad/s: (−50π, 5e jπ/6 ) and (50π, 5e− jπ/6 ).

July 29, 2015 © 2022 Pearson Education.

ω

CHAPTER 3. SPECTRUM P-3.12

DSP First 2e

(a) Make a table of the frequencies of the tones of the octave beginning with middle C, assuming that the A above middle C is tuned to 440 Hz. Note name Note number Frequency Note name Note number Frequency

C 40 262 F# 46 370

C# 41 277 G 47 392

D 42 294 G# 48 415

E[ 43 311 A 49 440

E 44 330 B[ 50 466

F 45 349 B 51 494

F# 46 370 C 52 523

(b) The formula for the frequency f as a function of note number n is f = 440 · 2(n−49)/12 (c) The spectrum would have the form: {(−440, a3∗ ), (−370, a2∗ ), (−294, a1∗ ), (294, a1 ), (370, a2 ), (440, a3 )} To sound like a musical chord, the coefficients should have similar magnitudes, but the phases could be arbitrarily chosen. A chord from a real instrument would have overtones (higher harmonics) of each individual note.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.13

DSP First 2e

(a) The frequency of the DC component is by definition 0. The waveform is periodic with period 25 ms so the frequency is 1/0.025 = 40 Hz. (b) The DC level is (20 − 10)/2 = 5, the amplitude of the cosine is (20 + 10)/2 = 15, and the cosine is delayed by 0.005 s, so x(t) = 5 + 15 cos(2π(40)(t − .005)) = 5 + 15 cos(80πt − 0.4π) (c) x(t) = 5 + 7.5e j (80πt−0.4π) + 7.5e− j (80πt−0.4π) = 7.5e j0.4π e− j80πt + 5 + 7.5e− j0.4π e j80πt (d) Plot of the two-sided spectrum of the signal x(t).

7.5e j0.4π

7.5e− j0.4π 5

−80π

0

80π

ω

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.14

(a) Using symmetry we obtain √ √ X−1 = 2 − j 2 = 2e− jπ/4

DSP First 2e

X2 = 8e jπ/3

ω1 = 70π

ω2 = −100π

(b) x(t) = 20 + 4 cos(70πt + π/4) + 16 cos(100πt + π/3) (c) The gcd of 70 and 100 is 10, so the fundamental frequency of the signal is f 0 = 5 Hz and the fundamental period is T0 = 1/5 = 0.2 s. (d) Note that the −20 ≤ 4 cos(70πt+π/4)+16 cos(100πt+π/3) ≤ 20 since the individual terms satisfy −4 ≤ 4 cos(70πt+ π/4) ≤ 4 and −16 ≤ 16 cos(100πt + π/3) ≤ 16. The value ±20 would be attained only if the phases of the two cosines are such that 4 cos(70π(t − t 0 )) + 16 cos(100π(t − t 0 )).

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.15

DSP First 2e

(a) x(t) = cos2 (7πt − 0.1π) We need to express x(t) in terms of complex exponentials

2 x(t) = 0.5e j (7πt−0.1π) + 0.5e− j (7πt−0.1π) = 0.25e j0.2π e− j14πt + 0.5 + 0.25e− j0.2π e j14πt 0.5 0.25e j0.2π

−14π

0.25e− j0.2π

0

ω

14π

(b) y(t) = cos2 (7πt − 0.1π) cos(77πt + 0.1π)

x(t) = 0.25e j0.2π e− j14πt + 0.5 + 0.25e− j0.2π e j14πt 0.5e j0.1π e j77πt + 0.5e− j0.1π e− j77πt = 0.125e j0.1π e− j91πt + 0.25e− j0.1π e− j77πt + 0.125e− j0.3π e− j63πt + 0.125e j0.3π e j63πt + 0.25e j0.1π e j77πt + 0.125e− j0.1π e j91πt

0.25e− j0.1π 0.125e j0.1π

0.25e j0.1π

0.125e− j0.3π

−91π −77π −63π ω axis is not to scale.

0.125e j0.3π

0

63π

0.25e− j0.1π

77π

91π

ω Note:

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.16

DSP First 2e

(a) Plotting spectrum of v(t) versus f in kHz, 0.5e− j0.3π

−3

0.5e j0.3π

0

f (kHz)

3

(b) The spectrum for x(t) versus f in kHz. 0.75

0.25e− j0.3π

0.75

0.25e j0.3pi

−683 −680 −677

0.25e− j0.3π 0

677

0.25e j0.3π 680

683

f (kHz)

Note: f axis is not to scale.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.17

DSP First 2e

(a) The gcd of 105 and 180 is 15, so the given frequencies are the 7th and 12th harmonics of f 0 = 15 Hz. (b) x(t) = 22 cos(2π(105)t − 0.4π) + 14 cos(2π(180)t − 0.6π) (c) Simplify the numerical values for the complex amplitudes, i.e., phases should be in [−π, π]. x 2 (t) = 22 cos(2π(105)(t − 0.05) − 0.4π) + 14 cos(2π(180)(t − 0.05) − 0.6π) = 22 cos(2π(105)t − 10.5π − 0.4π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 10π − 0.9π) + 14 cos(2π(180)t − 18π − 0.6π) = 22 cos(2π(105)t − 0.9π) + 14 cos(2π(180)t − 0.6π) Note that even multiples of 2π rad can be dropped from the equation. Thus, the spectrum is: {(−180, 7e j0.6π ), (−105, 11e j0.9π ), (105, 11e− j0.9π ), (180, 7e− j0.6π )} where the frequencies are in hertz. Therefore the plot of the spectrum looks just like Fig. P-3.17 except the phase is different at frequencies ±105. (d) The effect of this operation is simply to increase all the frequencies by 105 Hz, or, in other words, to shift the spectrum of x(t) to the right by 105 Hz. Therefore the spectrum line at −105 Hz will move to f = 0. The DC component is therefore, the value of the spectrum at f = −105, or 11e j0.9π .

July 29, 2015 © 2022 Pearson Education.

CHAPTER 3. SPECTRUM P-3.18

DSP First 2e

(a) The spectrum of y(t) is the spectrum of x(t) with an added DC component of size 8.

11e j0.4π

11e− j0.4π 8

7e j0.6π

−180

7e− j0.6π

105

0

−105

f

180

(b) The spectrum of z(t) is the same as that of x(t) with the addition of components of size 9e± j0.8π at frequencies ±40 Hz. 11e j0.4π 7e j0.6π

−180

−105

9e− j0.8π

−40

11e− j0.4π

9e j0.8π

0

7e− j0.6π

40

105

f

180

(c) The fundamental frequency is the gcd of 40, 105, 180, which is 5 Hz. (d) The derivative operation multiplies each spectrum component by j2π f , where f is the frequency of the complex exponential component. So we get − j2π(105)11e j0.4π

j2π(105)11e− j0.4π

− j2π(180)7e j0.6π

−180

j2π(180)7e− j0.6π

−105

0

105

180

f

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CHAPTER 3. SPECTRUM P-3.19

DSP First 2e

(a) X1 = 8e− jπ/3

and ω1 = −10π

(b) Here is the plot of the spectrum of x(t).

8e jπ/3

−10π

8e− jπ/3

0

ω

10π

(c) The symmetry implies that ωb = 20π and B = +4 j. Furthermore, symmetry requires that ωa = 0. To find A, ωc , and ϕ we can write y(t) as y(t) = 0.5x(t)e jϕ e jω c t + 0.5x(t)e− jϕ e− jω c t , which shows that the spectrum of y(t) will consist of the sum of scaled copies of the spectrum of x(t) shifted right (up) by ωc and left (down) by ωc . In order to have only three components we must choose ωc = 10π so that two of the shifted spectrum lines over lap at ω = 0.

4e jπ/3 e jϕ + 4e− jπ/3 e− jϕ 4e− j π /3 e jϕ

4e jπ/3 e− jϕ

−20π

0

20π

ω

− jπ/2 Now, 4e− jπ/3 e jϕ √ = 4e √ so ϕ = −π/6. Finally, note that the DC value can be written as A = 8 cos(π/3 + ϕ) = 8 cos(π/6) = 8 3/2 = 4 3.

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CHAPTER 3. SPECTRUM P-3.20

DSP First 2e

(a) The gcd of 40 and 90 is 10, so f 0 = 10 Hz. (b) The fundamental period is T0 = 1/ f 0 = 1/10 = 0.1 s. (c) From the plot, the DC value is 0.5. (d) With f 0 = 10, the harmonics are k = 0, ±4, ±9. k

−9

−4

ak

0.4e− j2

0.6e j1.4

0

4

9

0.5

0.6e− j1.4

0.4e j2

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CHAPTER 3. SPECTRUM P-3.21

DSP First 2e

(a) The instantaneous frequency is ωi (t) = dψ dt = 2αt + β, so ω1 = ω i (0) = β and ω2 = ω i (T2 ) = 2αT2 + β. (b) The instantaneous frequency versus time is ωi (t) = 80t + 27 (c) Here is the plot of the instantaneous frequency (in Hz) versus time over the range 0 ≤ t ≤ 1 sec. i (t ) f i (t) = ω2π 6

107 2π

27 2π

0

1

t

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CHAPTER 3. SPECTRUM P-3.22

DSP First 2e

(a) The general form for the chirp signal is x(t) = cos(αt 2 + βt + ϕ). The instantaneous frequency of this signal is ωi (t) = 2αt + β. From this we observe that ω1 = 2π f 1 = 2π(4800) = ωi (0) = β. To obtain α, we note that ω2 = 2π(800) = ωi (2) = 2α(2) + β = 4α + 9600π so α = −2000π. Therefore, the signal is x(t) = cos(−2000πt 2 + 9600πt + ϕ) where ϕ is an arbitrary phase constant. (b) The instantaneous frequency is ωi = 800πt + 500π, so ω1 = ωi (0) = 500π and ω2 = ωi (3) = 800π(3) + 500π = 2900π rad/s.

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CHAPTER 3. SPECTRUM P-3.23

DSP First 2e

(a) The instantaneous frequency is ωi (t) = 2αt + β. Substituting the given parameters gives α = 4π and β = 2π, so the signal with the given parameters is x(t) = cos(4πt 2 + 2πt + ϕ).

frequency in Hz

(b–f) The solution to this problem is given in the following figure. Note that the times at which f i (t) is equal to 4 Hz and 8 Hz are indicated with dashed lines. Careful scrutiny of the plots confirms that the waveform of the chirp signal does match the waveforms of the 4 Hz and 8 Hz constant-frequency sinusoids at the corresponding two times.

Instantaneous Frequency of Chirp Signal

10 5 0

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

1.2

1.4

1.6

1.8

2

4 Hz Signal

1 0.5 0 -0.5 -1

0

0.2

0.4

0.6

0.8

1

Chirp Signal

1 0 -1

0

0.2

0.4

0.6

0.8

1

8 Hz Signal

1 0.5 0 -0.5 -1

0

0.2

0.4

0.6

0.8

1

time in seconds

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CHAPTER 3. SPECTRUM P-3.24

DSP First 2e

(a) f 1 (t) =

1 √

(c) f 3 (t) = e2t /π

frequency in Hz

frequency in Hz

frequency in Hz

frequency in Hz

2π (t) (b) f 2 (t) = t/π

(d) f 4 (t) = − sin(2πt) Solution to Problem 3.24

0.8 0.6 0.4 0.2 0

0

0.5

1

1.5

2

2.5

3

3.5

4

1 0.5 0 -0.5 -1 -3

-2

-1

0

1

2

3

0

0.5

1

1.5

2

2.5

3

0

0.5

1

1.5

2

2.5

3

150 100 50 0 1 0.5 0 -0.5 -1

time in seconds

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CHAPTER 3. SPECTRUM P-3.25

DSP First 2e

(a) Let x(t) be given by the Fourier series x(t) = x(0) =

∞ X

∞ :1 X

(2π/T )k (0) 0 e j = ak . ak

k=−∞

∞ X

ak e j (2π/T0 )k t . Then it follows that

k=−∞

k=−∞

(b) Let f 3 = − f 2 = f 0 and f 4 = − f 1 = 3 f 0 so that from the spectrum we can write x(t) = 12 cos(2π f 0 t + π/4) + 4 cos(6π f 0 t + 3π/4) √ √ √ Therefore x(0) = 12 cos(π/4) + 4 cos(3π/4) = 6 2 − 2 2 = 4 2. Now if we add the coefficients of the Fourier series we get √ a1 + a2 + a3 + a4 = 2e− j3π/4 + 6e− jπ/4 + 6e jπ/4 + 2e j3π/4 = 12 cos(π/4) + 4 cos(3π/4) = 4 2

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CHAPTER 3. SPECTRUM P-3.26

DSP First 2e

The equations corresponding to the spectra are: The matches are

(a) (b) (c) (d) (e)

(3) (1) (2) (5) (4)

x 1 (t) = 4 cos(4πt + π) + 4 cos(6πt + π/2) x 2 (t) = 2 cos(4πt + π/4) + 4 cos(6πt − 0.333π) x 3 (t) = −3 + 2 cos(4πt + π/4) x 4 (t) = −2 + 4 cos(4πt + π) x 5 (t) = 4 cos(2πt + π) + 4 cos(4πt + π)

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CHAPTER 3. SPECTRUM P-3.27

(a) x(t) = cos(−250πt 2 ) Spectrogram (2) (b) x(t) = cos(100πt − π/4) + cos(400πt) Spectrogram (5) (c) x(t) = cos(1000πt − 250πt 2 ) Spectrogram (4)

DSP First 2e

(d) x(t) = cos(100πt) cos(400πt) Spectrogram (1) (e) x(t) = cos(200πt 2 ) Spectrogram (6) (f) x(t) = cos(30e2t ) Spectrogram (3)

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CHAPTER 4. SAMPLING AND ALIASING P-4.1

DSP First 2e

(a) A = 11, ϕ = −π/3, and ω/ f s = 7π/9 = ω̂1 . (b) A = 11, ϕ = +π/3, and ω/ f s = 7π/6 → ω̂1 = 5π/6. (c) A = 11, ϕ = −π/3, and ω/ f s = 7π/3 → ω̂1 = π/3.

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CHAPTER 4. SAMPLING AND ALIASING P-4.2

DSP First 2e

x 1 [n] = x 3 [n] and x 2 [n] = x 4 [n] = x 5 [n]

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CHAPTER 4. SAMPLING AND ALIASING P-4.3

DSP First 2e

(a) y(t) = 8 cos(2160πt − π/3) (b) y(t) = 4 cos(2880πt + π/4) (c) y(t) = 2 cos(1440πt − π/5)

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CHAPTER 4. SAMPLING AND ALIASING P-4.4

DSP First 2e

(a) Sketch will have four spectral lines at ω̂ = ±4000π, ±12000π. The complex amplitude of each spectral line will be 0.25.

6 0.25 0.25 0.25 0.25

−12000π

−4000π

0

4000π

12000π

ω

(b) The Nyquist rate for x(t) is 12,000 samples/s. (c) The Nyquist rate for r (t) is 15 × 106 samples/s. (d) The Nyquist rate for v(t) is 7 × 106 samples/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.5

DSP First 2e

(a) The Nyquist rate for x(t) is 50 samples/s. (b) The Nyquist rate for r (t) is 72 samples/s. (c) The Nyquist rate for v(t) is 45 samples/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.6

DSP First 2e

(a) Sketch will have four spectral lines: at ω̂ = ±0.2π with complex amplitudes equal to 13e± jπ/4 , and at ±0.5π with complex amplitudes of 7e± j3π/4 .

There is no aliasing when f s = 2000 samples/s.

13e− jπ/4

6 13e jπ/4

7e− j3π/4

−π

−0.5π

7e j3π/4

−0.2π

0

0.2π

0.5π

π

ω̂

(b) The reconstructed output when the D-to-C rate is f s = 1600 samples/s will be y(t) = 26 cos(320πt + π/4) + 14 cos(800πt + 3π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.7

DSP First 2e

(a) No, because there is aliasing. (b) Sketch will have four spectral lines: at ω̂ = ±4π/9 with complex amplitudes equal to 13e± jπ/4 , and at ±2π/9 with complex amplitudes of 7e± j3π/4 .

When f s = 450 samples/s, there will be an alias for the lines at ±500 Hz, i.e., the spectral line first maps to ω̂ = ±20π/9 and then aliases to ±2π/9.

13e− jπ/4 7e− j3π/4

−π

−4π/9

13e jπ/4

6

−2π/9

7e j3π/4

0

2π/9

4π/9

π

ω̂

(c) The reconstructed output when f s = 450 samples/s is y(t) = 26 cos(400πt + π/4) + 14 cos(100πt + 3π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.8

DSP First 2e

(a) Yes. f Nyquist = 1000 Hz. (b) Sketch will have two spectral lines at ω̂ = ±4π/7 with complex amplitudes equal to 6e± jπ/4 .

When f s = 700 samples/s, there will be a folded alias for the lines at ±500 Hz, i.e., the spectral line first maps to ω̂ = ±10π/7 and then aliases (via folding) to ∓4π/7. The complex amplitudes of the coincident spectral lines must be summed.

−π

6e− jπ/4

6

6e jπ/4 = 13e jπ/4 + 7e− j3π/4

−4π/7

0

4π/7

π

ω̂

(c) The reconstructed output when f s = 700 samples/s is y(t) = 12 cos(400πt + π/4).

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CHAPTER 4. SAMPLING AND ALIASING P-4.9

DSP First 2e

(a) ω = 2π(15200) rad/s. (b) f s = 120 samples/s. (c) f s = 75 samples/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.10

DSP First 2e

(a) ω1 = 18π rad/s. (b) ω̂ = 0.6π rad. (c) ω̂1 = π/7 and ω̂2 = −π/7, but b[n] involves a folded alias so ω̂2 = π/7. (d) Unequal because the phases are different: a[n] = cos(πn/7 + 0.1π) and b[n] = cos(πn/7 − 0.1π).

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CHAPTER 4. SAMPLING AND ALIASING P-4.11

DSP First 2e

(a) Average is 9 samples per period. (b) ω = 8000π rad/s. (c) Average is 0.9 samples per period, or 9 samples taken every 10 periods.

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CHAPTER 4. SAMPLING AND ALIASING P-4.12

DSP First 2e

Frequencies will be: 1000 Hz, 4000 Hz (but x(t) will have a different phase), and 6000 Hz. x 1 (t) = 9.1 cos(2000πt + 0.4π) x 2 (t) = 9.1 cos(8000πt − 0.4π) x 3 (t) = 9.1 cos(6000πt + 0.4π)

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CHAPTER 4. SAMPLING AND ALIASING P-4.13

DSP First 2e

(a) f = 1600 Hz. Phase will be −2.03 rad. (b) fsamp = 9600 Hz. (c) f = 2200 Hz. Duration = 12.8 s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.14

DSP First 2e

(a) Sketch will have six spectral lines at ω = ±2000π, ±(2000π ± 6000) rad/s.

In hertz, these frequencies are f ≈ ±47.07, ±1000, ±1954.93 Hz. The complex amplitudes of the spectral lines are 14 e± jπ/2, 1.5e± j0, 14 e∓ jπ/2 .

1.5

1.5 6 − 4j

j 4

−1955

−1000

−47 0

− 4j

j 4

47

1000

1955

f (Hz)

(b) Not periodic. (c) f s > 3909.859 Hz.

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CHAPTER 4. SAMPLING AND ALIASING P-4.15

DSP First 2e

(a) x 1 (t) = cos(2π(2800)t + 0.2π) and x 2 (t) = cos(2π(5200)t − 0.2π) (b) y(t) = cos(2π(1750)t + 0.2π)

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CHAPTER 4. SAMPLING AND ALIASING P-4.16

DSP First 2e

(a) x 1 (t) = cos(2π(8700)t + 0.2π) and x 2 (t) = cos(2π(7300)t − 0.2π) (b) x(t) = cos(2π(7300)t − 0.2π) (c) f = 8000 − f s

ω̂ Hz, and ϕ = −0.2π rad.

2π

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CHAPTER 4. SAMPLING AND ALIASING P-4.17

DSP First 2e

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CHAPTER 4. SAMPLING AND ALIASING P-4.18

DSP First 2e

(a) ω1 = 12π rad/s. (b) ω̂ = 0.8π rad. (c) ω1 = ω2 = 200π rad/s. ) ( (d) The three possible answers are 2π(14 72 ), 2π(28 47 ), 2π(33 31 ) rad/s.

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CHAPTER 4. SAMPLING AND ALIASING P-4.19

DSP First 2e

(a) Run the code to make the stem plot.

(b) Period of x[n] in the plot is 5 samples. (c) ω̂0 = 0.4π rad, and x[n] = 2 cos(0.4πn − 0.6π). (d) 2π(380/100) = 7.6π > 2π and 7.6π = 4(2π) − 0.4π.

Over the total duration of 0.3 s, a 380 Hz continuous-time signal would have 114 periods, but x[n] only has 6 complete periods.

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CHAPTER 4. SAMPLING AND ALIASING P-4.20

DSP First 2e

(a) Expect 34 zero crossings in 17 periods. Number of positive peaks = number of negative peaks = 17. (b) 11 zero crossings, 6 positive peaks and 6 negative peaks. (c) T = 0.2 s, so f = 5 Hz. Sampling a 15-Hz sinusoid at 20 Hz (Ts = 0.05) s. involves aliasing, and a 5-Hz sinusoid is a folded version of the 15-Hz sinusoid. (d) x[n] = 1.005 cos(0.4πn − 0.1). (e) To obey the Sampling Theorem, Ts < (1/30) s, but to make a smooth plot Ts < 0.05 s would be needed.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 4. SAMPLING AND ALIASING P-4.21

DSP First 2e

x(t) = (3 − j0.5e jπt + j0.5e− jπt )(0.5e j13πt+ jπ/2 + 0.5e− j13πt− jπ/2 ) = 1.5e j13πt+ jπ/2 + 1.5e− j13πt− jπ/2 + 0.25e j14πt + 0.25e− j14πt + 0.25e j12πt+ jπ + 0.25e− j12πt− jπ = 3 cos(13πt + π/2) + 0.5 cos(14πt) + 0.5 cos(12πt + π) (a) ω1 = 12π rad/s, A1 = 0.5, ϕ1 = π; ω2 = 13π rad/s, A2 = 3, ϕ2 = π/2; ω3 = 14π, A3 = 0.5, ϕ3 = 0; (b) Use the second line above to plot six spectrum lines. −3 j 6 0.5

0.5

−14π −13π −12π

3j

0.5 0

12π

0.5 13π

14π

ω (rad/s)

(c) Nyquist rate is 28π rad/s, or 14 Hz.

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CHAPTER 4. SAMPLING AND ALIASING P-4.22

DSP First 2e

(a) Sampling rate is greater than 2(300π) rad/s, or 300 Hz. (b) x[n] = x(t) t=n/250 = 2 cos(2π(50)n/250 + π/2) + cos(2π(150)n/250) = 2 cos(2πn/5 + π/2) + cos(6πn/5) = 2 cos(0.4πn + 0.5π) + cos(0.8πn) (c) Plot four lines; two at ±0.4π with complex amps equal to 1e± j0.5π ; and two at ±0.8π with complex amps equal to 10.5. Note: amplitudes are not to scale.

10.5

e− jπ/2 −π

−0.8π

10.5

6

−0.4π

e jπ/2 0

0.4π

0.8π

π

ω̂

(d) f s = 300π rad/s, or 150 Hz.

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CHAPTER 4. SAMPLING AND ALIASING P-4.23

DSP First 2e

(a) The discrete-time exponential when flashing at 15 flashes/s is p[n] = e+ j2π(13)n/15+ jπ/2 = e j2π(−2)n/15+ jπ/2 . Thus, at each flash the spot moves −4π/15 rad, which is 48◦ clockwise. (b) p(t) = p[n]

n=15t

= e j2π(−2)(15t )/15+π/2

appears to rotate through an angle of −4π every second. This is 2 rev/s clockwise. (c) Thirteen flashes in one second; disk moving CCW which is negative. The disk moves 2π(−15/360) + 2π` rad per flash. Thus the disk moves 2π(13)(−1/24 + `) in one second. When ` = 0, the rotation speed is 13/24 = 0.54 rev/s CCW. When ` < 0, the rotation speed is |13` − 13/24| rev/s CCW. When ` > 0, the rotation speed is |13` − 13/24| rev/s CW. For example, when ` = 1, the speed is 12.46 rev/s CW. Complex Exponential Derivation of Rotation Speeds (r) in rev/s: The discrete-time exponential after flashing at f s flashes/s is p[n] = e+ j2π (r )n/ f s . Thus, to get the movement of 15◦ CCW, which is −2π/24 rad/s CCW, we must determine whether or not there is is an integer ` for which the following is true:

* 1e j2π` e− j2π(15/360)

p[1] = e+ j2π (r )(1)/ f s = p[0]

e+ j2π (r )/ f s = e− jπ/12 e j2π`

⇒ 2π(r)/ f s = 2π` − 2π/24

⇒ r = f s (` − (1/24))

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CHAPTER 4. SAMPLING AND ALIASING P-4.24

DSP First 2e

(a) p(t) = e+ j2π(12)t rotates through an angle of 24π every second. (b) The discrete-time exponential after flashing at n flashes/s is p[m] = e+ j2π(12)m/n . Thus, given n, we must determine whether or not there is is an integer ` for which the following is true: p[1] = p[0] e

+ j2π (12)(1)/n

= e j2π`

⇒ 24π/n = 2π`

⇒ 12/n = `

So, n must be a divisor of 12, which gives flashing rates of n = 1, 2, 3, 4, 6, 12 flashes/s. (c) When n = 13, p[1] = e+ j2π(12)/13 = e− j2π(1)/13 , so the disk rotates clockwise by 2π/13 rad, or 360/13 = 27.69◦ , for every flash. The observed rotation is 1 rev/s clockwise.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 4. SAMPLING AND ALIASING P-4.25

DSP First 2e

(a) Plot 8 unit-length vectors (with zvect). The angles will be {0.2π, 0.5π, 0.8π, 0, 0.3π, −0.5π, 0.5π, −0.5π}. (b) Period is 20. (c) Plot 5 unit-length vectors (with zvect). The angles will be {0, π/15, 4π/15, −14π/15, −14π/15}. (d) Shortest period is 30.

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CHAPTER 4. SAMPLING AND ALIASING P-4.26

DSP First 2e

(a) θ[n] = 0.07π, 1.75π, 7π are the angles of the vectors. When reduced to the −π to +π range, these angles are θ[n] = 0.07π, −0.25π, π. (b) The instantaneous frequency of the signal below is 2π(44.8 × 103 )t Hz. The plot runs from t = 0 to t = 200/8000 = 1/40 s, so it starts at f = 0 and ends at (1/40)44.8 × 103 = 1120 Hz. x(t) = x[n]

n=8000t

= cos(π(0.7 × 10−3 × 64 × 106 )t 2 ) = cos(π(44.8 × 103 )t 2 )

(c) 2800 Hz

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CHAPTER 4. SAMPLING AND ALIASING P-4.27

DSP First 2e

ALPHA = 437.5, BETA = 2000, fs = 8000

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CHAPTER 5. FIR FILTERS P-5.1

DSP First 2e

y[n] = 7x[n] + x[n − 3] − 5x[n − 4]

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CHAPTER 5. FIR FILTERS P-5.2

G = 9e− j0.5π(3) − 9e− j0.5π(4)

DSP First 2e

√ e− j0.5πn = 9 2 e+ j3π/4

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CHAPTER 5. FIR FILTERS P-5.3

DSP First 2e

(a) First of all, y[n] = 0 for n = −2, −1. The system is causal. Then y[n] = n + 1 for n = 0, 1, 2, 3; and y[n] = 4 for n ≥ 4. (b) Plot with Matlab.

(c)   0     y[n] =  n + 1   L 

n<0 0≤n<L n≥L

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CHAPTER 5. FIR FILTERS P-5.4

DSP First 2e

(a) Use numerical convolution table: n n<0 0 1 2 3 4 2 0

7 n>7

1

1

1

0 0

1 2 2 −3

h[0]x[n] h[1]x[n−1] h[2]x[n−2]

0 0 0

2 4 6 4 2 2 2 2 2 0 −3 −6 −9 −6 −3 −3 −3 −3 0 0 2 4 6 4 2 2 2

y[n]

0

2

2 −1

1

6

x[n] h[n]

1

3 2

5

2

3

1

1

1

1

(b) (c) h[n] = 2δ[n] − 3δ[n − 1] + 2δ[n − 2]

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CHAPTER 5. FIR FILTERS P-5.5

DSP First 2e

Here are the figures: 2 2

xŒn

xŒn

v2 Œn Unit Delay xŒn 1

Unit Delay

3 3

v1 Œn Unit Delay xŒn 2

Unit Delay

2

2

yŒn

yŒn

(a)

(b)

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CHAPTER 5. FIR FILTERS P-5.6

DSP First 2e

(a) P = N + M (b) Input signal is time-shifted by N1 . Thus, the output starts at N3 = N1 , and the last nonzero value in the output is at N4 = N2 + M. The length of y[n] is N4 − N3 + 1 = N2 − N1 + M + 1.

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CHAPTER 5. FIR FILTERS P-5.7

DSP First 2e

(a) Make a plot in Matlab.

(b) First of all, y[n] = 0 for n = −2, −1. The system is causal. Then y[n] = 1 for n = 0; and y[n] = (−0.5) n + (−0.5) n−1 for n = 1, 2, 3, 4, 5, 6, 7, 8. The nonzero values starting at n = 0 are 1.0, 0.5, −0.25, 0.125, −0.0625, 0.03125, −0.015625, 0.0078125, −0.00390625 (c) Causality implies that y[n] = 0 for n < 0.

Running onto the signal (partial overlap) region which is 0 ≤ n < L−1: sum (−0.5) k for k = 0 to k = n n X

(−0.5) k =

k=0

(−0.5) 0 − (−0.5) n+1 1 − (−0.5) n+1 = 1 − (−0.5) 1.5

Complete overlap region: sum (−0.5) k for k = n − L + 1 to k = n which is n ≥ L−1: n X

(−0.5) k =

k=n−L+1

(−0.5) −L − 1 (−0.5) n−L+1 − (−0.5) n+1 = (−0.5) n+1 1 − (−0.5) 1.5

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CHAPTER 5. FIR FILTERS P-5.8

DSP First 2e

Filter order is M = 4, and bk = {3, 7, 13, 9, 5}

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CHAPTER 5. FIR FILTERS P-5.9

DSP First 2e

Impulse response is h[n] = 13δ[n]−13δ[n−1]+13δ[n−2]. The output can be obtained via convolution: y[n] = h[n]∗ x[n].

Assume that the input is nonzero for −∞ < n < ∞, which implies that x[n − 2] is equal to x[n]. Also, x[n − 1] will be equal to 0 for n odd, and equal to 1 for n even.   26 y[n] = 13x[n] − 13x[n − 1] + 13x[n − 2] = 26x[n] − 13x[n − 1] =   −13 

n odd n even

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CHAPTER 5. FIR FILTERS P-5.10

DSP First 2e

(a) y[n] = x[n] cos(0.2πn)

(1) linear YES, (2) time-invariant NO, and (3) causal YES.

(b) y[n] = x[n] − x[n − 1]

(1) linear YES, (2) time-invariant YES, and (3) causal YES.

(c) y[n] = |x[n]|

(1) linear NO, (2) time-invariant YES, and (3) causal YES.

(d) y[n] = Ax[n] + B, where A and B are nonzero constants.

(1) linear NO, (2) time-invariant YES, and (3) causal YES.

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CHAPTER 5. FIR FILTERS P-5.11

DSP First 2e

(a) Plot of y[n]

(b) Use LTI properties with x[n] = 7 (δ[n] − δ[n − 1]) +7 (δ[n − 1] − δ[n − 2]). | {z } first x[n] in table

Thus, y[n] = 7 (δ[n] − δ[n − 1] + 2δ[n − 3]) +7 (δ[n − 1] − δ[n − 2] + 2δ[n − 4]) | {z } first x[n] in table

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CHAPTER 5. FIR FILTERS P-5.12

DSP First 2e

Note that the Delay by 2 block in (b) is the cascade of two unit delays.

1 xŒn

3

v3 Œn

xŒn

Unit Delay xŒn 1

Delay by 2

4 2

v2 Œn Unit Delay xŒn 2

Unit Delay

2 4

v1 Œn Delay by 2 xŒn 4

Unit Delay

3

1

yŒn

yŒn

(a)

(b)

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CHAPTER 5. FIR FILTERS P-5.13

DSP First 2e

(a) Use the LTI properties. The output will be y2 [n] = y1 [n] − 2y1 [n − 2]. y2 [n] = δ[n] − 3δ[n − 1] − 3δ[n − 2] − 2δ[n − 2] + 6δ[n − 3] + 6δ[n − 4] = δ[n] − 3δ[n − 1] − 5δ[n − 2] + 6δ[n − 3] + 6δ[n − 4] (b) A = 1 and B = −1. (c) Using LTI properties, h[n] = ( A/3)y1 [n] + (B/3)y1 [n − 1] = (1/3) (δ[n] − 3δ[n − 1] − 3δ[n − 2]) − (1/3) (δ[n − 1] − 3δ[n − 2] − 3δ[n − 3]) = (1/3)δ[n] − (4/3)δ[n − 1] + δ[n − 3]

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CHAPTER 5. FIR FILTERS P-5.14

DSP First 2e

Do the convolutions with h[n] = δ[n] − aδ[n − 1] (a) x[n] ∗ h[n] = a n u[n] ∗ (δ[n] − aδ[n − 1]) = a n u[n] − aa n−1 u[n − 1] = a n (u[n] − u[n − 1]) = δ[n] | {z } δ[n]

(b) Use the result from the previous part. h[n] ∗ a n (u[n] − u[n − 8]) = h[n] ∗ a n u[n]) − h[n] ∗ a8 a n−8 u[n − 8] {z } | {z } | δ[n]

a 8 δ[n−8]

(c) Sketch the plot which has only two nonzero stems, 1 at n = 0, and a8 at n = 8.

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CHAPTER 5. FIR FILTERS P-5.15

DSP First 2e

Use trial and error, and just do the convolutions (a) x[n] ∗ δ[n − 3] = x[n − 3], so we want x[n − 3] = 5u[n − 2] − u[n − 6]. Unshift to get x[n] = 5u[n + 1] − u[n − 3]. (b) y[n] = u[n − 2] − u[n − 3] which simplifies to y[n] = δ[n − 2] (c) x[n] − x[n − 1] = y[n] = δ[n − 2] − δ[n − 7].

Rewrite y[n] as y[n] = u[n − 2] − u[n − 3] − u[n − 7] − u[n − 8]. {z } | {z } | δ[n−2]

δ[n−7]

Rewrite again as y[n] = u[n − 2] − u[n − 7] − (u[n − 3] − u[n − 8]) . | {z } | {z } x[n]

x[n−1]

(d) x[n] + x[n − 1] = y[n] = −8δ[n] − 8δ[n − 3].

Mimic the previous part to obtain x[n] = −8u[n] + 8u[n − 3]

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CHAPTER 5. FIR FILTERS P-5.16

DSP First 2e

It is easy to verify these; just do the convolutions (a) h[n] = δ[n − 1] (b) h[n] = δ[n] − δ[n − 1] (c) h[n] = δ[n − 1] − 0.5δ[n − 2] because ( 21 ) n−1 u[n − 1] − 21 ( 21 ) n−2 u[n − 2] is zero for n ≥ 2.

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CHAPTER 5. FIR FILTERS P-5.17

DSP First 2e

If an FIR filter with impulse response h[n] is used to deconvolve, then y[n] = h[n] ∗ (δ[n] + δ[n − 1]) = h[n] + h[n − 1] 0 >

We want y[0] = δ[0] = 1 ⇒ 1 = y[0] = h[0] + h[−1] ⇒ h[0] = 1

1

We want y[1] = δ[1] = 0 ⇒ 0 = y[1] = h[1] + h[0] ⇒ h[1] = 1 At this point we know that the order of the filter is at least M = 1. For an M th order FIR filter the coefficient of z −M must be nonzero, i.e., b M , 0. With that fact, the following shows that y[M + 1] will also be nonzero, so y[n] cannot be equal to δ[n]. h[n] =

M X

bk δ[n − k ]

where b M , 0

k=0

*0

⇒ y[M + 1] = h[M + 1] + h[M] = h[M] = b M , 0

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CHAPTER 5. FIR FILTERS P-5.18

DSP First 2e

(a) Let x 1 [n] = δ[n], x 2 [n] = δ[n], and x 3 [n] = δ[n], S1 : S2 : S3 :

h1 [n] = δ[n] + δ[n − 1] h2 [n] = δ[n] − δ[n − 2] h3 [n] = δ[n − 1] − δ[n − 2]

(b) Carry out the convolutions: h[n] = h1 [n] ∗ h2 [n] ∗ h3 [n] h[n] = (δ[n] + δ[n − 1) ∗ (δ[n] − δ[n − 2]) ∗ (δ[n − 1] − δ[n − 2]) | {z } h[n] = (δ[n] + δ[n − 1) ∗ (δ[n − 1] − 2δ[n − 2] + δ[n − 3]) h[n] = δ[n − 1] − δ[n − 2] − δ[n − 3] + δ[n − 4] (c) Get the filter coefficients from h[n], and then y[n] = x[n − 1] − x[n − 2] − x[n − 3] + x[n − 4]

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CHAPTER 5. FIR FILTERS P-5.19

DSP First 2e

In this case, the three systems are as follows: S1 : S2 : S3 :

y1 [n] = x[n] + 2x[n − 1] + 3x[n − 2] + 4x[n − 3] y2 [n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3] y[n] = w[n] − w[n − 1]

h1 [n] = δ[n] + 2δ[n − 1] + 3δ[n − 2] + 4δ[n − 3] h2 [n] = δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3] h3 [n] = δ[n] − δ[n − 1]

where w[n] = y1 [n] + y2 [n]. The impulse responses are also given above (a) We need three blocks and one adder. Systems S 1 and S 2 are in parallel; their outputs are added and the sum is processed by S 3 . (b) We can get h[n] from the combination of the individual systems, and the write the difference equation from the filter coefficients. h[n] = (h1 [n] + h2 [n]) ∗ h3 [n] h[n] = (2δ[n] + 3δ[n − 1] + 4δ[n − 2] + 5δ[n − 3]) ∗ (δ[n] − δ[n − 1]) h[n] = 2δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3] − 5δ[n − 4] y[n] = 2x[n] + x[n − 1] + x[n − 2] + x[n − 3] − 5x[n − 4] If we had the z-transform at this point we would write

H (z) = 1 + 2z −1 + 3z −2 + 4z −3 + 1 + z −1 + z −2 + z −3 1 − z −1

H (z) = 2 + 3z −1 + 4z −2 + 5z −3 1 − z −1 = 2 + z −1 + z −2 + z −3 − 5z −4

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CHAPTER 6. FREQUENCY RESPONSE P-6.1

DSP First 2e

y[n] = x[n] − x[n − 1] = e j (0.4πn−0.5π) − e j (0.4π (n−1)−0.5π) = (1 − e− j0.4π )e j (0.4πn−0.5π) = H (e j0.4π )e j (0.4πn−0.5π) H (e j0.4π ) = 1 − e− j0.4π = 1.1756e j0.3π Therefore, A = 1.1756, ϕ = 0.3π − 0.5π = −0.2π, and ω̂0 = 0.4π.

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CHAPTER 6. FREQUENCY RESPONSE P-6.2

DSP First 2e

(a) y[n] = x[n] + x[n − 1] = (e j0.1πn + e j0.4πn ) + (e j0.1π (n−1) + e j0.4π (n−1) ) = (1 + e− j0.1π )e j0.1πn + (1 + e− j0.4π )e j0.4πn = H (e j0.1π )e j0.1πn + H (e j0.4π )e j0.4πn where H (e j ω̂ ) = 1 + e− j ω̂ . (b) The system is LTI, so superposition applies. Each complex exponential is modified by the frequency response evaluated at its frequency. (c) H (e j0.1π ) = 1 + e− j0.1π = A1 e jϕ1 = 1.9754e− j0.05π H (e j0.1π ) = 1 + e− j0.4π = A2 e jϕ2 = 1.6180e− j0.2π

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CHAPTER 6. FREQUENCY RESPONSE P-6.3

DSP First 2e

2

(a) y[n] = e j0.1πn + e j0.4πn = e j0.2πn + 2e j0.5πn + e j0.8πn (b) Clearly, the squaring operation produces three complex exponential signals each with a different frequency that is not one of the original two frequencies. (c) The system is clearly not linear since the output cannot be written as the sum of two terms involving the two input frequencies. The principle of superposition is not satisfied, therefore the system is NOT linear.

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CHAPTER 6. FREQUENCY RESPONSE P-6.4

DSP First 2e

(a) H (e j ω̂ = 2 − 2e− j ω̂ + 2e− j ω̂2 = 2e− j ω̂ (e j ω̂ − 1 + e− j ω̂ ) = 2e− j ω̂ (2 cos(ω̂) − 1) Thus, R(e j ω̂ ) = 4 cos(ω̂) − 2 and n0 = 1. (b) The period = 2π

5 0 −π π

−π/2

0

π/2

−π/2

0

π/2

π

3π/2

2π

5π/2

3π

π

3π/2

2π

5π/2

3π

0

6

H(ej ω̂ )

|H(ej ω̂ )|

10

(c)

−π −π

ω̂

(d) The output is zero whenever the magnitude of the frequency response is zero; i.e., when |2 cos(ω̂) − 1| = 0 or when cos(ω̂) = 0.5. The solution is therefore ω̂ = π/3 rad. (e) We know that for a cosine input with frequency ω̂0 = 0.5, the output is y[n] = |H (e j0.5π )| cos(0.5πn + ∠H (e j0.5π )) where H (e j0.5π ) = 2e− j0.5π (2 cos(0.5π) − 1) = −2e− j0.5π = 2e j0.5π ⇒ y[n] = 2 cos(0.5πn + 0.5π) = 2 cos(0.5π(n + 1)).

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CHAPTER 6. FREQUENCY RESPONSE P-6.5

DSP First 2e

(a) H (e j ω̂ ) = 3 + 2e− j ω̂ + e− j ω̂2 (b) y[n] = H (e j0 ) + 2|H (e j0.5π )| cos(0.5πn + 0.1π + ∠H (e j0.5π )) √ where H (e j0 ) = 6 and H (e j0.5π ) = 3 − 2e− j0.5π + e− jπ = 2e− j0.25π . √ √ Thus, y[n] = 6 + 2 2 cos(0.5πn + 0.1π − 0.25π) = 6 + 2 2 cos(0.5πn − 0.15π) (c) 0      3 y[n] = 3u[n] + 2u[n − 1] + u[n − 2] =   5     6

n<0 n=0 n=1 n≥2

(d) The unit step input is identical to a suddenly applied complex exponential of zero frequency. Thus, for an FIR filter like this, the steady state response is simply H (e j0 ) times 1. The steady state is reached at n = 2 because the filter order is 2.

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CHAPTER 6. FREQUENCY RESPONSE P-6.6

DSP First 2e

(a) H (e j ω̂ ) = 1 − e− j ω̂4 = e− j ω̂2 (e j ω̂2 − e− j ω̂2 ) = 2 je− j ω̂2 sin(2ω̂) (b) y[n] = 4H (e j0 ) + |H (e j0.25π )| cos(0.25πn − 0.25π + ∠H (e j0.25π )) where H (e j0 ) = 0 and H (e j0.25π ) = 2 j (− j) sin(π/2) = 2. Thus, y[n] = 2 cos(0.25πn − 0.25π) (c) Since the impulse response has length 5 samples, the output reaches the steady state at n = 4 so the output due to the suddenly applied signal is identical to the output in (b) for n ≥ 4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.7

DSP First 2e

(a) h[n] = 3δ[n] − δ[n − 2] (b) We need to expand H (e j ω̂ ) in terms of complex exponentials. ! j2ω̂ + e− j2ω̂ j ω̂ − j3ω̂ e = 2.5e− j ω̂ + 2.5e− j5ω̂ H (e ) = 5e 2 Therefore h[n] = 2.5δ[n − 1] + 2.5δ[n − 5]. (c) We should recognize this as the form of an L sample running sum whose frequency response is H (e j ω̂ ) = e− j ω̂(L−1)/2

L−1 sin(ω̂L/2) X − j ω̂n e = sin(ω̂/2) n=0

Thus, it follows that L = 10 and the impulse response is ( 1 0≤n≤9 h[n] = u[n] − u[n − 10] = 0 otherwise

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CHAPTER 6. FREQUENCY RESPONSE P-6.8

DSP First 2e

(a) To get the difference equation we need to expand the factors of H (e j ω̂ ) so that we get powers of e− j ω̂ . √ H (e j ω̂ ) = (1 − e− j ω̂ )(1 − 2 cos(3π/4)e− j ω̂ + e− j2ω̂ ) = (1 − e− j ω̂ )(1 − 2e− j ω̂ + e− j2ω̂ ) √ √ = 1 + ( 2 − 1)e− j ω̂ − ( 2 − 1)e− j2ω̂ − e− j3ω̂ = 1 + 0.4142e− j ω̂ − 0.4142e− j2ω̂ − e− j3ω̂ Therefore, the difference equation is y[n] = x[n] + 0.4142x[n − 1] − 0.4142x[n − 2] − x[n − 3] (b) 0        1  1.4142 y[n] = u[n] + 0.4142u[n − 1] − 0.4142u[n − 2] − u[n − 3] =      1    0 

n<0 n=0 n=1 n=2 n≥3

(c) We need to find the values of ω̂ such that H (e j ω̂ ) = 0. These are easily found from the given factored form as ω̂ = 0 and ω̂ = ±3π/4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.9

DSP First 2e

(a) We need to multiply out the factors of H (e j ω̂ ) to get: H (e j ω̂ ) = 1 + 0.5e− j ω̂ − 0.25e− j2ω̂ + 0.25e− j3ω̂ Therefore, the difference equation is y[n] = x[n] + 0.5x[n − 1] − 0.25x[n − 2] + 0.25x[n − 3] (b) By definition, the impulse response is: h[n] = δ[n] + 0.5δ[n − 1] − 0.25δ[n − 2] + 0.25δ[n − 3] (c) We need to find the values of ω̂ such that H (e j ω̂ ) = 0. The factored form shows that this can be true for values of ω̂ such that 1 + e− j2ω̂ = 0 or ω̂ = ±π/2 rad.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 6. FREQUENCY RESPONSE P-6.10

DSP First 2e

(a) Plot {3, 0, −2, 1} for 0 ≤ n ≤ 3. (b) The first entry in the table says that the impulse response is h[n] = δ[n] − δ[n − 3]. Therefore, y[n] = x[n] − x[n − 3].

For the given input the output is y[n] = 3δ[n] − 2δ[n − 2] + δ[n − 3] − 3δ[n − 3] + 2δ[n − 5] − δ[n − 6] = 3δ[n] − 2δ[n − 2] − 2δ[n − 3] + 2δ[n − 5] − δ[n − 6] Plot {3, 0, −2, −2, 0, 2, −1} for 0 ≤ n ≤ 6. (c) In this case we can use the third result in the table, which says that the frequency response at ω̂ = π/3 is H (e jπ/3 ) = 2; i.e., the phase shift is zero. Therefore, the output for this input is y[n] = 2 cos(π(n − 3)/3) (d) It is FALSE because the impulse response is δ[n] − δ[n − 3], so the frequency response is H (e j ω̂ ) = 1 − e− j3ω̂ which is nonzero at ω̂ = π/2.

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CHAPTER 6. FREQUENCY RESPONSE P-6.11

DSP First 2e

(a) The zeros marked with o are at multiples of 2π/8 with gaps at 0, ±2π, ±4π, . . ..

D8 (ω̂)

1 0

−1 −3π

−2π

−π

0 Figure P-6.11

π

2π

3π

(b) The period is 4π. (c) The maximum value occurs at ω̂ = 0. Using the small angle approximation for the sine functions we get D L (0) ≈

(Lω̂/2) =1 L(ω̂/2)

We get the same result using L’ Hospital’s Rule.

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CHAPTER 6. FREQUENCY RESPONSE P-6.12

DSP First 2e

(a) From the difference equation, we can write down H (e j ω̂ ) as H (e j ω̂ ) = 1 − 3e− j ω̂ + 3e− j2ω̂ − e− j3ω̂ = (1 − e− j ω̂ ) 3 = e− j ω̂3/2 (e j ω̂/2 − e− j ω̂/2 ) 3 = e− j ω̂3/2 (2 j) 3 (sin(ω̂/2)) 3 = 8e− j ω̂3/2 (− j)(sin(ω̂/2)) 3 = 8 sin3 (ω̂/2)e j (−π/2−3ω̂/2)

|H(ej ω̂ )|

10 5 0 −2π

−3π/2

−π

−π/2

0

π/2

π

3π/2

2π

Figure P-6.12 (b) (c) We need to determine the frequency response at 0 and 0.5π. √ √ These are H (e j0 ) = 0 and H (e j0.5π ) = 8(1/ 2) 3 e j (−π/2−3π/4) = 2 2e− j7π/4 . Therefore, y[n] = 5H (e j0 ) + 6|H (e j0.5π )| cos(0.5π(n − 1) + ∠H (e j0.5π )) √ = 12 2 cos(0.5π(n − 1) − 7π/4) (d) h[n] = δ[n] − 3δ[n − 1] + 3δ[n − 2] − δ[n − 3] (e) y[n] = 5H (e j0 ) + 6|H (e j0.5π )| cos(0.5πn + ∠H (e j0.5π )) + 9h[n − 3] √ = 12 2 cos(0.5πn − 7π/4) + 9h[n − 3] where h[n] is as determined in part (d).

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CHAPTER 6. FREQUENCY RESPONSE P-6.13

DSP First 2e

(a) S1 :

H1 (e j ω̂ ) = 1 − e− j ω̂

S2 :

H2 (e j ω̂ ) = 1 + e− j2ω̂

S3 :

H3 (e j ω̂ ) = e− j ω̂ + e− j2ω̂

(b) H (e j ω̂ ) = (1 − e− j ω̂ )(1 + e− j2ω̂ )(e− j ω̂ + e− j2ω̂ ) (c) Multiply out all the factors in H (e j ω̂ ) and then pick off the coefficients to get the difference equation.

H (e j ω̂ ) = (1 − e− j ω̂ + e− j2ω̂ − e− j3ω̂ )(e− j ω̂ + e− j2ω̂ ) = e− j ω̂ − e− j5ω̂ Therefore, y[n] = x[n − 1] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.14

DSP First 2e

(a) H (e j ω̂ ) =

4 X

e− j ω̂n =

n=0

(b) H (e

j ω̂

sin(ω̂5/2) 1 − e− j5ω̂ = e− j ω̂2 − j ω̂ sin(ω̂/2) 1−e

) = 0 for ω̂ = ±0.4π, ±0.8π

(c) We need the frequency response at ω̂ = 0, 0.25π, and 0.4π.

H (e j0 ) = 5 H (e j0.25π ) = e− j0.5π

sin(5π/8) = 2.4142e− j0.5π sin(π/8)

H (e j0.4π ) = 0 Therefore, y[n] = 35 + 8(2.4142) cos(0.25πn − π/2) (d) Since the length of the impulse response is 5, the steady-state output is attained for n ≥ 4.

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CHAPTER 6. FREQUENCY RESPONSE P-6.15

DSP First 2e

The sampled signal is x[n] = x(n/ f s ) = 7 + 8 cos(1000πn/4000) + 9 cos(1600πn/4000) = 7 + 8 cos(0.25πn) + 9 cos(0.4πn + 0.7π) This is the same input signal as in Problem P-6.14 and the discrete-time system is the same as in that problem. Therefore, from the solution to Problem P-6.14(c), the output sequence is y[n] = 35 + 8(2.4142) cos(0.25πn − π/2) The output of the D-to-C converter is therefore y(t) = 35 + 8(2.4142) cos(1000πn − π/2)

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CHAPTER 6. FREQUENCY RESPONSE P-6.16

DSP First 2e

(a) f s > 1000 Hz. Since h[n] = δ[n], y[n] = x[n]. In other words, the system is a C-to-D converter followed by a D-to-C converter, so the issue is sampling above the Nyquist rate. The highest frequency in the input signal is 500 Hz, so the Nyquist rate is 2(500) = 1000 Hz. (b) A delay of 10 samples must correspond to 0.001 s. After sampling x(t) at a rate of f s x[n] = 10 + 20 cos(ω0 (n/ f s ) + π/3) y[n] = x[n − 10] = 10 + 20 cos(ω0 (n − 10)/ f s + π/3) Reconstructing y(t) from y[n] with an ideal D-to-C converter gives y(t) = 10 + 20 cos(ω0 ( f s t − 10)/ f s + π/3) = 10 + 20 cos(ω0 t − 10ω0 / f s + π/3) We want y(t) = x(t − 0.001) = 10 + 20 cos(ω0 (t − 0.001) + π/3) which requires −10ω0 / f s = ω0 (−0.001)

⇒

f s = 10, 000 Hz

In addition, the output frequency must be the same as the input frequency, so there must be no aliasing, i.e., the relationship between ω0 and f s = 10, 000 Hz must obey the Nyquist condition.

2ω0 < 2π f s

⇒

ω0 < 2π(5000) rad/s

(c) To get y(t) = A, we need y[n] = constant. Since x[n] = 10 + 20 cos(ω0 (n/ f s ) + π/3), the filter must null out the cosine term; or the cosine term must alias to DC. sin(5ω̂/2) − j ω̂2 e are at ω̂ = 2π/5, −2π/5, 4π/5, −4π/5.

Nulling: The zeros of H (e j ω̂ ) = 5 sin(ω̂/2) For ω0 and f s = 2000 Hz, these frequencies are 2π(±400) and 2π(±800) rad/s.

The negative frequencies must be used when we include aliases. The complete list of input frequencies that alias to the zeros of H (e j ω̂ ) is 2π(400 + 2000`) = 2π(400), 2π(2400), 2π(4400), . . .

2π(800 + 2000`) = 2π(800), 2π(2800), 2π(4800), . . .

2π(−400 + 2000`) = 2π(−400), 2π(1600), 2π(3600), . . .

2π(−800 + 2000`) = 2π(−800), 2π(1200), 2π(3200), . . .

For these input frequencies, the cosine term will be filtered out, and only the DC remains. Since H (e j0 ) = 1 and the DC value of x(t) = 10, the output will be y(t) = A = 10.

Cosine aliases to DC: When ω0 is a frequency that aliases to DC, then ω̂0 = (ω0 + 2π` f s )/ f s = 0, and x[n] = 10 + 20 cos(0n + π/3) = 10 + 20 cos(π/3) = 20.

The frequencies for which this happens are 2π(0 + 2000`) = 2π(0), 2π(2000), 2π(4000), . . .

In this case, the DC value of x(t) = 20 and H (e j0 ) = 1, so the output will be y(t) = A = 20.

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CHAPTER 6. FREQUENCY RESPONSE P-6.17

DSP First 2e

(a) The constant component passes through the first system as a constant. Then the first difference system gives a zero output for a constant input. By superposition, the output is thus equal to y1 [n], the output due to the other input component x 1 [n]. (b) The overall frequency response function is the product of the frequency responses. Therefore, H (e

j ω̂

!

4 X 1 − e− j5ω̂ − j ω̂n + − j ω̂ * (1 − e− j ω̂ ) = 1 − e− j5ω̂ )= e (1 − e )= − j ω̂ 1 − e , n=0

(c) The frequency response is zero at the 5th roots of unity or ω̂ = 2πk/5 for k = 0, 1, 2, 3, 4. (d) From the frequency response, the overall difference equation is y[n] = x[n] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.18

DSP First 2e

(a) Need a plot of h[n] = −δ[n − 2] + δ[n − 12]

(b) H (e j ω̂ ) = −e− j2ω̂ +e− j12ω̂ = e− j7ω̂ (−e j5ω̂ +e− j5ω̂ ) = e− j7ω̂ (−2 j) R(e j ω̂ ) = −2 sin(5ω̂) and n0 = −7.

e j5ω̂ − e− j5ω̂ = j (−2 sin(ω̂5))e− j7ω̂ Therefore, 2j

1 0 −π 5

−π/2

0

π/2

π

0

π/2

π

0

6

H(ej ω̂ )

|H(ej ω̂ )|

(c,d) Plots of |H (e j ω̂ )| and the principal value of ∠H (e j ω̂ ) for −π < ω̂ ≤ π.

2

-5 −π

−π/2

ω̂

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CHAPTER 6. FREQUENCY RESPONSE P-6.19

DSP First 2e

(a) H (e j ω̂ ) = H1 (e j ω̂ )H2 (e j ω̂ ) = (1 + 2e− j ω̂ + e− j ω̂2 )(1 − e− j ω̂ + e− j ω̂2 − e− j ω̂3 ) = 1 + e− j ω̂ − e− j ω̂4 − e− j ω̂5 (b) h[n] = δ[n] + δ[n − 1] − δ[n − 4] − δ[n − 5]

(c) y[n] = x[n] + x[n − 1] − x[n − 4] − x[n − 5]

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CHAPTER 6. FREQUENCY RESPONSE P-6.20

DSP First 2e

(a) We need to have zeros of the frequency response at ω̂ = 0 and 0.2π. Thus, H (e j ω̂ ) = (1 − e− j ω̂ )(1 − e− j0.2π e− j ω̂ )(1 − e j0.2π e− j ω̂ ) = (1 − e− j ω̂ )(1 − 2 cos(0.2π)e− j ω̂ + e− j ω̂2 ) = 1 − 2.6180e− j ω̂ + 2.6180e− j ω̂2 − e− j ω̂3 Therefore, the difference equation is y[n] = x[n] − 2.6180x[n − 1 + 2.6180x[n − 2] − x[n − 3] (b) We need to null the frequencies ω̂ = 0, ±0.2π, ±0.4π, ±0.6π, ±0.8π, π. A system whose frequency response is zero at all these frequencies is H (e j ω̂ ) = 1 − e− j ω̂10 .

If we set H (e j ω̂ ) = 0 we get the equation 1 − e− j ω̂10 = 0 or e− j ω̂10 = 1 or, equivalently, e j ω̂10 = 1 = e j2πk where k = 0, 1, . . . , 9. If we then equate the exponents of the last equation we get ω̂10 = 2πk, or ω̂ = 2πk/10 = 0.2πk.

The difference equation for this system is y[n] = x[n] − x[n − 10].

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CHAPTER 6. FREQUENCY RESPONSE P-6.21

DSP First 2e

(a) The two normalized frequencies are 120π/ f s = 0.12π and 240π/ f s = 0.24π. (b) We need to place zeros at ω̂ = ±0.12π and ±0.24π. Therefore, choose H (e j ω̂ ) = (1 − e− j0.12π e− j ω̂ )(1 − e j0.12π e− j ω̂ )(1 − e− j0.24π e− j ω̂ )(1 − e j0.24π e− j ω̂ ) = 1 − 3.3175e− j ω̂ + 4.7111e− j ω̂2 − 3.3175e− j ω̂3 + e− j ω̂4 The corresponding difference equation is therefore: y[n] = x[n] − 3.3175x[n − 1] + 4.7111x[n − 2] − 3.3175x[n − 3] + x[n − 4]

|H(ej ω̂ )|

(c) Here is the Matlab plot of the frequency response (magnitude) versus ω̂, and also versus the analog frequency f in hertz. The frequencies 60 Hz and 120 Hz are marked and are snown to be nulled out.

Note: the plot of |H (e j ω̂ )| versus ω̂ is zoomed vertically so as to see the detail of the zeros.

1

|H(ej2πf /fs )|

0.5 0 −π

−π/2

0 normalized frequency ω̂

π/2

π

20 10 0 -500

-120-60 0 60 120 analog frequency f

500

Figure P-6.21

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CHAPTER 6. FREQUENCY RESPONSE P-6.22

DSP First 2e

(a) To find y[n] we need the frequency response evaluated at frequencies 0 and 0.5π which are from the graph: H (e j0 ) = 0

H (e j0.5π ) = 1.4e− jπ/4

It is easy to get these analytically as well since H (e j ω̂ ) = 1 − e− j ω̂ = je− j ω̂/2 2 sin(ω̂/2) = 2 sin(ω̂/2)e j (π/2−ω̂/2) Thus, the output is y[n] = 10 √1 cos(0.5πn − π/4) = 7.0711 cos(0.5πn − π/4).

2

(b) The discontinuity of size π radians is due to the fact that H (e j ω̂ ) is zero at ω̂ = 0 so there is a sign change from one side of 0 to the other. This is evident in H (e j ω̂ ) equation above since the function sin(ω̂/2) is negative for −π < ω̂ < 0 and positive for 0 < ω̂ < π.

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CHAPTER 6. FREQUENCY RESPONSE P-6.23

DSP First 2e

(a) When the input is x[n] = 10 + 10 cos(0.2πn) + 10 cos(0.5πn), we need to determine the frequency response at frequencies 0, 0.2π and 0.5π. From the curves we get H (e j0 ) = 1

H (e j0.2π ) ≈ 0.9e− j0.4π

H (e j0.5π ) = 0

Therefore, y[n] = 10 + 9 cos(0.2πn − 0.4π). (b) There are two kinds of discontinuities in the phase. At ω̂ = 2π(0.17), the discontinuity of size 2π is due to taking the principal value. At ω̂ = 0.5π the frequency response is zero and the discontinuity is a phase jump of ±π due to the sign change as the frequency response crosses zero.

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CHAPTER 7. DTFT P-7.1

DSP First 2e

This solution requires mostly table lookup and applying the delay property for the DTFT. DTFT

(a) x 1 [n] = 2δ[n − 3] ←→ X1 (e j ω̂ ) = 2e− j ω̂3 DTFT

(b) x 2 [n] = 3δ[n − 2] − δ[n − 3] + 3δ[n − 4] ←→ X2 (e j ω̂ ) = 3e− j ω̂2 − e− j ω̂3 + 3e− j ω̂4  7 1≤n≤8 (c) x 3 [n] = 7u[n − 1] − 7u[n − 9] =  0 otherwise  Note how the step sequence is used to create a finite-length sequence. Plug into the DTFT definition and set n = m + 1 in the sum. Then apply the formula for the sum of terms of a geometric series.

X3 (e j ω̂ ) =

8 X

7e− j ω̂n = 7e− j ω̂

n=1

(d) x 4 [n] =

sin(0.25πn) DTFT ←→ X4 (e j ω̂ ) = 9πn

7 X m=0

( 1 9

0

e− j ω̂m = 7e− j ω̂

1 − e− j ω̂8 sin(4ω̂) − j ω̂9/2 =7 e sin(ω̂/2) 1 − e− j ω̂

|ω̂| < 0.25π 0.25π < |ω̂| ≤ π

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CHAPTER 7. DTFT P-7.2

DSP First 2e

This solution requires table lookup and use of DTFT properties DTFT

(a) Y1 (e j ω̂ ) = 2π ←→ y1 [n] = 2πδ[n] DTFT

(b) Y2 (e j ω̂ ) = 5e− j3ω̂ ←→ y2 [n] = 5δ[n − 3] DTFT

(c) Y3 (e j ω̂ ) = 6 cos(3ω̂) = 3e j3ω̂ + 3e− j3ω̂ ←→ y3 [n] = 3δ[n + 3] + 3δ[n − 3] DTFT

(d) Y4 (e j ω̂ ) = j sin(7ω̂) = 12 e j7ω̂ − 21 e− j7ω̂ ←→ y4 [n] = 12 δ[n + 7] − 21 δ[n − 7]

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CHAPTER 7. DTFT P-7.3

DSP First 2e

This solution involves either evaluating the definitions of the direct and inverse DTFT or using table lookup and the properties of the DTFT. 1 (a) v1 [n] = 2π (b) V2 (e j ω̂ ) =

0.3π Z

e j ω̂n d ω̂ =

e j0.3πn − e− j0.3πn sin(0.3πn) = j2πn πn

−0.3π

9 X

1 − e− j ω̂10 e− j ω̂5 (e j ω̂5 − e− j ω̂5 ) sin(ω̂5) − j ω̂4.5 = = e − j ω̂ − j ω̂/2 j ω̂/2 − j ω̂/2 1−e e (e −e ) sin(ω̂/2)

e− j ω̂n =

n=0

(c) Note that V3 (e j ω̂ ) = 1 − V1 (e j ω̂ ). Thus, v3 [n] = δ[n] − v1 [n] = δ[n] −

sin(0.3πn) πn

Alternatively, you can plug into the inverse DTFT integral as in part (a) obtaining 1 v3 [n] = 2π

(d) V4 (e j ω̂ ) =

9 X

−0.3π Z

e

j ω̂n

−π

e jπn e− j ω̂n =

n=0

1 d ω̂ + 2π

9 X n=0

Zπ 0.3π

e j ω̂n d ω̂ =

sin(πn) sin(0.3πn) sin(0.3πn) − = δ[n] − πn πn πn

e− j (ω̂−π)n = V2 (e j (ω̂−π) ) =

sin((ω̂ − π)5) − j (ω̂−π)4.5 e sin((ω̂ − π)/2)

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CHAPTER 7. DTFT P-7.4

DSP First 2e

This solution requires using the linearity and delay properties of the DTFT and the following basic DTFT pair: ( sin(ω̂0 n) DTFT 1 |ω̂| < ω̂0 j ω̂ h[n] = ←→ H (e ) = 0 ω ˆ 0 < |ω̂| ≤ π πn It might be helpful to sketch the different bandlimited DTFTs that are involved. ( ( sin(0.25πn) DTFT 1 |ω̂| < 0.25π 0 j ω̂ ←→ H1 (e ) = 5 − 5 (a) h1 [n] = 5δ[n] − = 0 0.25π < |ω̂| ≤ π 5 0.2πn (b) h2 [n] =

sin(0.4πn) sin(0.1πn) − 0.1πn 0.1πn

 1 ←→ H2 (e j ω̂ ) = 10  0  DTFT

(c) h3 [n] =

|ω̂| < 0.25π 0.25π < |ω̂| ≤ π

  0 0 ≤ |ω̂| ≤ 0.1π      0 ≤ |ω̂| ≤ 0.4π 1 0 ≤ | ω̂| ≤ 0.1π   − 10  = 10  1 0.1π < |ω̂| ≤ 0.4π  0.4π < |ω̂| ≤ π   0 0.4π < |ω̂| ≤ π  0 0.1π < |ω̂| ≤ π 

sin(0.4π(n − 8)) sin(0.1π(n − 8)) − = 0.1h2 [n − 8] π(n − 8) π(n − 8)

DTFT

←→ H3 (e

j ω̂

) = 0.1H2 (e

j ω̂

)e

− j ω̂8

=e

    0 0 ≤ |ω̂| ≤ 0.1π 1 0.1π < |ω̂| ≤ 0.4π     0 0.4π < |ω̂| ≤ π 

− j ω̂8  

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CHAPTER 7. DTFT P-7.5

DSP First 2e

Write down the system function directly from the difference equation obtaining H (e j ω̂ ) = 3

6 X

e− j ω̂k .

k=2

With a change of index in the sum we can apply the formula for the sum of terms in a geometric series.

H (e j ω̂ ) = 3

sin(ω̂5/2) − j ω̂4 e sin(ω̂/2)

Thus, the answers as requested are: L = 5, α = 3, and β = 4.

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CHAPTER 7. DTFT P-7.6

DSP First 2e

The exponential factors suggest a delay, but we cannot use the DTFT delay property directly because the delay would be a fraction of a sample. This means that we should use the inverse DTFT if we can evaluate it, and, of course, we can. (a) Inverse DTFT of G1 (e j ω̂ ) 1 ←→ g1 [n] = 2π DTFT

0.4π Z

5e

− j0.3ω̂ j ω̂n

e

−0.4π

1 d ω̂ = 2π

0.4π Z

5eω̂(n−0.3) d ω̂ = 5

−0.4π

e j0.4π(n−0.3) − e− j0.4π(n−0.3) 2π(n − 0.3)

sin 0.4π(n − 0.3) . π(n − 0.3) This looks like it was obtained using the delay property, but remember that the delay property only holds in general for integer shifts.

so g1 [n] = 5

(b) Using the approach of (a), the solution is g2 [n] = 4

sin 0.7π(n − 0.1) sin π(n − 0.1) −4 π(n − 0.1) π(n − 0.1)

If the delay had been an integer, the first function would reduce to an impulse, but in this case it does not. (c) Again using the direct evaluation of the inverse DTFT, the solution is g3 [n] = 9

sin 0.4π(n − 0.2) sin 0.7π(n − 0.2) −9 π(n − 0.2) π(n − 0.2)

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CHAPTER 7. DTFT P-7.7

(a)

DSP First 2e

sin(0.25πn) sin(0.44πn) sin(0.25πn) ∗ = 9πn 4πn 36πn The DTFT of this signal is the product of two ideal lowpass filter frequency responses. Therefore the result of the convolution is a sinc function corresponding to the lowest cutoff frequency and scaled by the product of the gains.

(b)

sin(0.25πn) sin(0.25πn) sin(0.25πn) sin(0.25πn) ∗ ∗ = 9πn 8πn 7πn (9 × 8 × 7)πn In this case we have the convolution of three sinc functions all with the same cutoff frequency so the result is the same signal but with the product of the gains.

(c) Considerable simplification is possible.

(0.3) n u[n] ∗ (10δ[n − 2] − 3δ[n − 3]) = 10(0.3) n−2 u[n − 2] − 3(0.3) n−3 u[n − 3] | {z } break into 2 terms

= 10δ[n − 2] + 10(0.3)(0.3) n−3 u[n − 3] − 3(0.3) n−3 u[n − 3] = 10δ[n − 2]

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CHAPTER 7. DTFT P-7.8

DSP First 2e

We will need the impulse response of the system to do parts (a) and (c). We can get it by expanding H (e j ω̂ ) using the inverse Euler relation; i.e., H (e j ω̂ ) = 21 e j ω̂3 + 21 e− j ω̂3 , so h[n] = 21 δ[n + 3] + 12 δ[n − 3]. (a) By definition, y1 [n] = h[n − 3] = 21 δ[n] + 21 δ[n − 6]. (b) Here we can use the frequency response directly if we express x 2 [n] as x 2 [n] = 1.5e j0.25πn + 1.5e− j0.25πn . Then y2 [n] = 1.5H (e j0.25π )e j0.25πn + 1.5H (e− j0.25π )e− j0.25πn

√ = 1.5 cos(0.75π)e j0.25πn + 1.5 cos(−0.75π)e− j0.25πn = −1.5 2 cos(0.25πn)

 1 0 ≤ n ≤ 8 (c) This signal is a length-9 pulse: x 3 [n] = u[n] − u[n − 9] =   0 otherwise  This one should be done by convolution because the impulse response is so simple and a frequency-domain solution would not be feasible.

y3 [n] = x 3 [n] ∗ h[n] = x 3 [n] ∗ 21 δ[n + 3] + 21 δ[n − 3] 1   2     1 = 21 (u[n + 3] − u[n − 6] + u[n − 3] − u[n − 12]) =  1    2    0

−3 ≤ n ≤ 2 3≤n≤5 6 ≤ n ≤ 11 otherwise

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CHAPTER 7. DTFT P-7.9

DSP First 2e

(a) Expand cosine with the inverse Euler relation: x[n] = a n cos(ω̂0 n)u[n] = a n DTFT

(b) Using the DTFT pair a n u[n] ←→ of x[n] as X (e j ω̂ ) =

1

=

1 and the frequency-shift property of the DTFT we can write the DTFT 1 − ae− j ω̂

1 1 2 2 + 1 − ae− j (ω̂−ω̂0 ) 1 − ae− j (ω̂+ω̂0 )

= 2

=

! e j ω̂0 n + e− j ω̂0 n u[n] = 12 e j ω̂0 n a n u[n] + 12 e− j ω̂0 n a n u[n] 2

− 21 ae− j ω̂0 e− j ω̂ + 12 − 12 ae j ω̂0 e− j ω̂

(1 − ae j ω̂0 e− j ω̂ )(1 − ae− j ω̂0 e− j ω̂ ) 1 − 21 a(e j ω̂0 + e− j ω̂0 )e− j ω̂

1 − a(e j ω̂0 + e− j ω̂0 )e− j ω̂ + a2 e− j2ω̂ 1 − a cos(ω̂0 )e− j ω̂ 1 − 2a cos(ω̂0 )e− j ω̂ + a2 e− j ω̂2

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CHAPTER 7. DTFT P-7.10

DSP First 2e

(a) The plot of |V (e j ω̂ )| 2 for −π < ω̂ ≤ π is below.

1 1 so |V (e j ω̂ )| 2 = .

From the table, V (e j ω̂ ) = 1 − ae− j ω̂ 1 + a2 − 2a cos ω̂ A sketch of this for a = 0.95 would show a maximum at ω̂ = 0 of size monotonically with increasing ω̂ to a value of

1 = 400 and it would fall off (1 − a) 2

1 = 0.2630 at ω̂ = ±π.

(1 + a) 2

(b) If V (e j ω̂ ) were a frequency response, is the filter lowpass, highpass or bandpass? LOWPASS (c) Using the frequency shift property of the DTFT we obtain W (e j ω̂ ) = 21 V (e j (ω̂−0.3π) ) + 21 V (e j (ω̂+0.3π) ). Therefore, a plot of |W (e− j ω̂ )| 2 will look like two shifted versions of the plot in part (a) with everything scaled by 41 . This will place peaks at approximately ω̂ = ±0.3π with heights of approximately 100.

Note: the plots use a semilog scale to show details.

(d) If W (e j ω̂ ) were a frequency response, is the filter lowpass, highpass or bandpass? BANDPASS

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CHAPTER 7. DTFT P-7.11

DSP First 2e

(a) Using the definition of the ideal lowpass filter we can write     1 |ω̂| ≤ ω̂co    1 |ω̂| ≤ ω̂co  0 |ω̂| ≤ ω̂co Hhp (e j ω̂ ) =   1 ω̂co < |ω̂| ≤ π −  0 ω̂co < |ω̂| ≤ π =  1 ω̂co < |ω̂| ≤ π    which is the frequency response of an ideal highpass filter.

Note: A plot of the function 1 − Hlp (e j ω̂ ) might make this easier to visualize. (b) The cutoff frequency of the ideal HPF is ω̂co . (c) From part (a) it follows that hhp [n] = δ[n] − hlp [n] so hhp [n] = δ[n] −

sin(ω̂co n) πn

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.12

DSP First 2e

(a) To get a bandpass filter, we need to subtract the low frequency band from a wider lowpass filter. Thus, we can write

Hbp (e

j ω̂

 1 )= 0 

  0 |ω̂| < 0.3π      |ω̂| ≤ 0.6π 1 |ω̂| ≤ 0.3π   − =  1 0.3π ≤ |ω̂| ≤ 0.6π  0.6π < |ω̂| ≤ π   0 0.6π < |ω̂| ≤ π  0 0.3π < |ω̂| ≤ π  

Therefore, the cutoff frequencies of the two lowpass filters are ω̂lp1 = ω̂co2 = 0.6π, and ω̂lp2 = ω̂co1 = 0.3π. (b) hbp [n] =

sin(0.6πn) sin(0.3πn) − πn πn

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.13

DSP First 2e

(a) A sketch would show a negative exponential decaying to the left for n ≤ −1. (b) The DTFT exists for values of b such that |x[n]| is absolutely summable; i.e.,

−1 X

| − bn | < ∞.

−∞

This will be true for |b| > 1. (c) By definition, the DTFT of x[n] = −bn u[−n − 1] when |b| > 1 is X (e j ω̂ ) =

−1 X

−bn e− j ω̂n = −

n=−∞

=1− =

∞ X

b−n e j ω̂n = 1 −

n=1

∞ X

b−n e j ω̂n

n=0

−b−1 e j ω̂ 1 = 1 − b−1 e j ω̂ 1 − b−1 e j ω̂

−b−1 e j ω̂ 1 = −b−1 e j ω̂ (1 − be− j ω̂ ) 1 − be− j ω̂

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.14

DSP First 2e

0.25π Z ∞ 2 X 1 1 1 sin(0.25πn) d ω̂ = (a) = 3πn 2π 3 12 n=−∞ −0.25π

since

(b) 4

Zπ −π

 1/3 sin(0.25πn) DTFT  ←→  0 3πn 

0 ≤ |ω̂| ≤ 0.25π 0.25π < |ω̂| ≤ π

2

sin(ω̂5) − j ω̂9/2 20 e d ω̂= sin(ω̂/2) π since

9 X k=0

DTFT

2δ[n − k] ←→ 2

sin(ω̂5) − j ω̂9/2 e sin(ω̂/2)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.15

DSP First 2e

(a) This one is easy with the DTFT since X1 (e j ω̂ ) = e− j ω̂4 and Cx1 x1 (e j ω̂ ) = |X1 (e j ω̂ )| 2 = 1. Therefore, cx1 x1 [n] = δ[n]. In the index domain, it is also easy: cx1 x1 [n] = x 1 [n] ∗ x 1 [−n] = δ[n − 4] ∗ δ[−n − 4] = δ[n] (b) This one can be done easily by discrete convolution, but let’s use the DTFT. Since X2 (e j ω̂ ) = e− j ω̂4 − e− j ω̂6 , it follows that Cx2 x2 (e j ω̂ ) = |X2 (e j ω̂ )| 2 = (e− j ω̂4 − e− j ω̂6 )(e j ω̂4 − e j ω̂6 ) = −e j ω̂2 + 2 − e− j ω̂2 so cx2 x2 [n] = −δ[n + 2] + δ[n] − δ[n − 2]. (c) Let’s try this with the DTFT. First note that x 3 [n] is identical to the impulse response of a 10-point running sum system with additional delay of 6 samples. Therefore, !2 sin(ω̂5) − j ω̂4.5 − j ω̂6 sin(ω̂5) e e and Cx3 x3 (e j ω̂ ) = |X3 (e j ω̂ )| 2 = X3 (e j ω̂ ) = . sin(ω̂/2) sin(ω̂/2) (Note that the exponential factors corresponding to delay disappear when we take the magnitude.)

To determine cx3 x3 [n] we have to determine the inverse DTFT. We cannot evaluate the integral with what we know and we have no entry in our table of transforms for this function. Therefore, we should try convolution, which is not too difficult. The result is   10 − |n| cx3 x3 [n] =  0 

−9 ≤ n ≤ 9 otherwise

This might be a good transform pair to know in a general case. In particular, you can show the following for any sequence x[n] (including an autocorrelation function):   L − |n| x[n] =  0 

|n| ≤ (L − 1) otherwise

DTFT

←→

X (e

j ω̂

sin(ω̂L/2) )= sin(ω̂/2)

!2

  1/9 |ω̂| < 0.25π sin(0.25πn) DTFT ←→ X4 (e j ω̂ ) =  0 9πn 0.25π < |ω̂| ≤ π  This one is a clear candidate for using the DTFT since it would be impossible to evaluate the discrete-time autocorrelation in the time domain.

We should recognize that x 4 [n] is identical to the impulse response of an ideal lowpass filter with cutoff frequency 0.25π and gain 1/9. If we take the squared-magnitude of X4 (e j ω̂ ), we get the same cutoff but the gain is squared. sin(0.25πn) Therefore, it follows that cx4 x4 [n] = 81πn

(d) Recall the DTFT pair: x 4 [n] =

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.16

DSP First 2e

(a) This is accomplished by evaluating the discrete-time convolution cx x [n] = x[n] ∗ x[−n] and the result is cx x [n] =    5 − |n| |n| ≤ 4 . 0 otherwise  (b) To do this, we don’t try to find the DTFT of cx x [n] in part (a) directly, but instead use the fact that Cx x (e j ω̂ ) = |X (e j ω̂ )| 2 . We can look up the DTFT of x[n] in our table of DTFT pairs with L = 5. Thus Cx x (e

j ω̂

2

sin(ω̂5/2) sin(ω̂5/2) − j ω̂(L−1)/2 )= e = sin(ω̂/2) sin(ω̂/2)

!2

Cx x (e j ω̂ ) is clearly real and positive because it is a squared-magnitude of the complex DTFT X (e j ω̂ ). (c) Since Cy y (e j ω̂ ) = |Y (e j ω̂ )| 2 = |X (e j ω̂ )e− j ω̂2 | 2 = |X (e j ω̂ )| 2 , it follows that cy y [n] = cx x [n]. In fact the auto correlation function for a delayed signal is always the same as the autocorrelation function for the signal itself.

Delay always cancels out when you evaluate a correlation.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.17

DSP First 2e

(a) Plot shows details of the passband and stopband magnitudes.

Magnitude of Frequency Response of Lowpass Filter M = 32

1.5 1 0.5 0

6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0021663 ! s = 0.4932:

#10 -3

4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0029736 ! p = 0.7034:

1.004 1.002 1 0.998 0.996 0.994 0.6

0.65

0.7

0.75

0.8

!/:

0.85

0.9

0.95

1

Figure 7-1: Solution for Problem 7-17(a).

(b) From the plots the measured parameters are ω̂ p = 0.7034π, ω̂ s = 0.4932π, δ p = 0.003 and δ s = 0.0022. (c) A good rule of thumb is that with a Hamming window, the transition region satisfies ∆ω̂ M = 8π/M, where M is the order of the filter. For M = 32, this formula gives ∆ω̂32 = 0.25π, whereas the measured value is ∆ω̂32 = (0.7034 − 0.4932)π = 0.2102π, so the formula over-estimates the transition width in this case. (d) However, the formula gives us a rationale for estimating the effect of going from M = 32 to M = 80. We should have ∆ω̂80 ≈ ∆ω̂32 32/80 = 0.2102π(32/80) = .0841π. The formula would give a value for ∆ω̂80 of 0.1π. The measured cutoff frequencies for M = 80 are ω̂ p = (.6 + .0841/2)π = 0.642π and ω̂ s = (0.6 − .0841/2)π = 0.558π.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT

Magnitude of Frequency Response of Lowpass Filter M = 80 1 0.8 0.6 0.4 0.2 0

5

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0020404 ! s = 0.5584:

#10 -3

4 3 2 1 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0018343 ! p = 0.6418:

1.005 1 0.995 0.99 0.985 0.98 0.6

0.65

0.7

0.75

0.8

!/:

0.85

0.9

0.95

1

Figure 7-2: Solution for Problem 7-17d. The estimates of the cutoff frequencies from part (c) are seen to be quite

accurate.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.18

DSP First 2e

To solve this problem use Matlab to design the filter using filterdesign. Save the impulse response to the work space as h24 for use. The solution to each of the following parts requires that we evaluate the frequency response at a given frequency. The frequencies are 0, 0.5π and 0.75π. This can be done in many ways, but the most straightforward way is to use Matlab’s freqz( ) function as follows: >> H24 = freqz(h24,1,[0,0.5*pi,0.75*pi]) H24 = Columns 1 through 2 0.000134504777373 + 0.000000000000000i Column 3 -0.996401749465568 - 0.000000000000001i

0.041265977941552 + 0.000000000000000i

We will need the magnitude and angle at these three frequencies so we do the following: >> zprint(H24) Z = X + jY 0.0001345 0 0.04127 7.164e-17 -0.9964 -1.142e-15

Magnitude 0.0001345 0.04127 0.9964

Phase 0.000 0.000 -3.142

Ph/pi 0.000 0.000 -1.000

Ph(deg) 0.00 0.00 -180.00

(a) If the input is x 1 [n] = 10 for all n, the output is y1 [n] = 10H24 (e j0 ), so we use Matlab as above to find H24 (e j0 ).H32=sum(h32)=0.0001345, so y1 [n] = 0.001345. This is a highpass filter so the DC is greatly attenuated. (b) When the input is a zero-phase cosine, the output is y2 [n] = 10|H24 (e j0.5π )| cos(0.5πn + ∠H24 (e j0.5π )), so from the Matlab analysis, y2 [n] = 0.4127 cos(0.5πn + 0). This frequency is in the transition zone near the stopband so its amplitude is much reduced. (c) As in part (b) the output in this case is y3 [n] = 10|H24 (e j0.75π )| cos(0.75πn + ∠H24 (e j0.75π )), so from the Matlab analysis, y3 [n] = 9.964 cos(0.75πn − π). This frequency is in the passband so the amplitude is barely reduced.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.19

Matlab program to make the requested plots is as follows:

DSP First 2e

A

% solution to problem 7-19 % assumes that h37 has been saved in workspace by filterdesign GUI % h37 = fir1(37,0.3,ones(1,38); n=0:99; x1=10*ones(1,100); x2=10*cos(0.5*pi*n); y1=filter(h37,1,x1); y2=filter(h37,1,x2); sn=x1+x2; snOutput=filter(h37,1,sn); % subplot(411) stem(n,y1,’filled’);hold on; plot([37,37],[-9,14.5],’r--’);hold off title(’Output y_1[n] due to x_1[n] = 10’) text(20,1,5,’Transient region’);hold off subplot(412) stem(n,y2,’filled’);hold on; plot([37,37],[-10,15],’r--’);hold off title(’Output y_2[n] due to x_2[n] = 10cos(0.5 \pi n)’) subplot(413) stem(n,y1+y2,’filled’);hold on; plot([37,37],[-10,15],’r--’);hold off title(’Sum of Outputs’) subplot(414) stem(n,snOutput,’filled’) title(’Output Due to Sum of Inputs’) xlabel(’time index n’) hold on han=stem(n,snOutput-y1-y2);hold on; plot([37,37],[-10,15],’r--’);hold off The resulting plot is shown in the figure above. (a) When the input x 1 [n] is DC with a DC level of 10, the output level y1 [n] converges to 10H (e j0 ) = 10(1) = 10 after a brief transient of less than 37 samples. (b) When the input is the sinusoid x 2 [n] = 10 cos(0.5πn) we must evaluate H (e j ω̂ ) at ω̂ = 0.5π rad. We obtain H (e j0.5π ) = 0.02814e j0.75π , so the output would be y2 [n] = 0.2814 cos(0.5πn + 0.75π). (c) Note that the bottom two figures are identical, thereby showing that superposition holds. This is also shown by the open circles, which show the difference between y1+y2 and snoutput.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT

Output y [n] due to x [n] = 10 1

1

10 0 −10 0

Transient region is n=0 to 37 10

20

30 40 50 60 70 Output y [n] due to x [n] = 10cos(0.5 π n) 2

2 1 0 −1 0

10

20

30

80

90

100

70

80

90

100

70

80

90

100

70

80

90

100

2

40

50

60

Sum of Outputs 15 10 5 0 0

10

20

30

40

50

60

Output Due to Sum of Inputs 15 10 5 0 0

10

20

30

40 50 60 Time Index (n)

Figure 7-3: Solution for Problem 7-19.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 7. DTFT P-7.20

DSP First 2e

(a) Use the GUI to find this frequency to be approximately 0.712π rad. The unwrapped phase at this frequency is ϕ ≈ −10.5 rad. The principal value of the phase is −10.5 + 2π ≈ 2.01 rad. (b) Here again, you can use the GUI to find the frequencies where |H (e j ω̂ )| = 1.

These are approximately {0.71π, 0.788π, 0.854π, 0.914π, 0.972π} (c) Use the GUI to find the frequencies where |H (e j ω̂ )| = 0.

These are approximately {0.03π, 0.091π, 0.151π, 0.212π, 0.272π, 0.334π, 0.398π, 0.486π}.

The following figure shows the places where the frequency response is one and zero more accurately than is possible with the GUI. This plot can be constructed by saving the impulse response to the work space and then using freqz to compute the frequency response.

Magnitude of Frequency Response of Lowpass Filter M = 32

1.5 1 0.5 0

6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Detail of Stopband: d1 = 0.0021663 ! s = 0.4932:

#10 -3

4 2 0

0

0.1

0.2

0.3

0.4

0.5

0.6

Detail of Passband: d2 = 0.0029736 ! p = 0.7034: 1.002 1 0.998 0.996 0.994 0.6

0.65

0.7

0.75

0.8

!/:

0.85

0.9

0.95

1

Figure 7-4: Solution for Problem 7-20.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.1

DSP First 2e

(a) The DTFT of x 0 [n] is X0 (e j ω̂ ) = 1 for all ω̂. Therefore, X0 [k] = X0 (e j (2πk /10) ) = 1 for k = 0, 1, . . . , 9. In Matlab >> X0=fft([1,zeros(1,9)]) X0 = 1 1 1 1

1 9 X

(b) The DTFT of x 1 [n] is X1 (e j ω̂ ) =

1

e− j ω̂n =

n=0

1

1

1

1 − e− j ω̂10 . Therefore, 1 − e− j ω̂

  1 − e− j2πk  10 = − j (2π/10)k 0 1−e 

X1 [k] = X1 (e j (2π/10)k ) =

1

k=0 k = 1, 2, . . . , 9

k = 0, 1, . . . , 9

1 − e− j2πk is indeterminant when we try to evaluate for k = 0. L’ Hospital’s 1 − e− j (2π/10)k rule doesn’t work here because k is an integer variable. Therefore we need to go back to the sum form where we 9 X see that X1 [0] = X1 (e j (2π/10)0 ) = 1 = 10.

Note that the closed form X1 [k] =

n=0

Matlab verification: >> X1 = fft(ones(1,10)) X1 = 10 0 0

0

0

0

0

0

(c) The DTFT of x 2 [n] is X2 (e j ω̂ ) = e− j ω̂4 so X2 [k] = e− j (2π/10)k4 Matlab verification: >> X2 = fft([zeros(1,4),1,zeros(1,5)]) X2 = 1.0000 + 0.0000i -0.8090 - 0.5878i 0.3090 - 0.9511i -0.8090 + 0.5878i -0.8090 - 0.5878i 0.3090 + 0.9511i -0.8090 + 0.5878i (d) The DTFT of x 3 [n] is X3 (e j ω̂ ) =

9 X n=0

e j2πn/5 e− j ω̂n =

9 X n=0

0

0

k = 0, 1, . . . , 9.

0.3090 + 0.9511i 1.0000 + 0.0000i 0.3090 - 0.9511i

e− j (ω̂−2π/5)n =

1 − e− j (ω̂−2π/5)10 .

1 − e− j (ω̂−2π/5)

Therefore, X3 [k] = X3 (e

j (2π/10)k

  0    1 − e− j ((2π/10)k−2π/5)10 1 − e− j2π(k−2)  )= = = 10  1 − e− j ((2π/10)k−2π/5) 1 − e− j (2π/10)(k−2)   0 

k = 0, 1 k=2 k = 3, 4, . . . , 9

As in part (b), we have to be careful in evaluating this expression for k = 2.

Matlab verification: >> X3 = fft(exp(j*2*pi*(0:9)/5)) X3 = -0.0000 + 0.0000i -0.0000 + 0.0000i 10.0000 - 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 + 0.0000i

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.2

DSP First 2e

(a) Substituting into the inverse DFT definition we get, 9

x a [n] =

1 X δ[k]e j (2π/10)k n = 0.1 10 k=0

n = 0, 1, . . . , 9

Matlab verification: >> xa = ifft([1,zeros(1,9)]) xa = 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000 0.1000

0.1000

0.1000

(b) This one is the first entry in Table 8-1, so we can write the answer down by inspection: x b [n] = δ[n]. Alternatively, you can plug into the inverse DFT definition and get 9 1 1 X j (2π/10)k n 1 − e j (2π/10)10n  x b [n] = = e = 0 10 k=0 1 − e j (2π/10)n 

n=0 = δ[n] n = 1, 2, . . . , 9

Matlab verification: >> xb = ifft(ones(1,10)) xb = 1 0 0 0

0

0

0

0

0

0

(c) Since there are only two nonzero terms, we can write down the answer directly and manipulate it if possible. x c [n] = 0.1(e j (2π/10)3n + e j (2π/10)7n ) = 0.1(e j (2π/10)3n + e j (2π/10)(10−3)n ) = 0.1(e j (2π/10)3n + e j (2π/10)10n e− j (2π/10)3n ) = 0.1(e j (2π/10)3n + e− j (2π/10)3n ) = 0.2 cos((2π/10)3n) Matlab verification: >> xc = ifft([0,0,0,1,0,0,0,1,0,0]) xc = 0.2000 -0.0618 -0.1618 0.1618 0.0618 0.1618 -0.1618 -0.0618

0.0618

-0.2000

(d) We can observe that X d [k] = cos(2πk/5)X b [k] = 21 e j (2π/10)2k X b [k] + 21 e− j (2π/10)2k X b [k]

= 21 e j (2π/10)2k X b [k] + 21 e j (2π/10)(10−2)k X b [k]

= 21 e j (2π/10)2k X b [k] + 21 e j (2π/10)8k X b [k] so by the time-shift property of Table 8-2,

x d [n] = 21 x b [n − 2] + 21 x b [n − 8] = 12 δ[n − 2] + 21 δ[n − 8] Matlab verification: >> xd = ifft(cos(2*pi*(0:9)/5)) xd = -0.0000 + 0.0000i -0.0000 + 0.0000i 0.0000 - 0.0000i -0.0000 - 0.0000i -0.0000 + 0.0000i 0.0000 + 0.0000i -0.0000 - 0.0000i

0.5000 + 0.0000i 0.0000 + 0.0000i 0.5000 - 0.0000i

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.3

DSP First 2e

(a) Express (−1) n as e jπn = e j (2π/12)6n and substitute into the DFT definition.

Y0 [k] = 3

11 X

e j (2π/12)6n e− j (2π/12)k n = 3

n=0

  0    = 36    0 

11 X

e− j (2π/12)(k−6)n = 3

n=0

1 − e− j2π(k−6) 1 − e− j (2π/12)(k−6)

k = 0, 1, . . . , 5 k=6 k = 7, 8, . . . , 11

Note that the numerator of the closed form is zero for all integer k, but when k = 6 we have an indeterminant form that evaluates to 3 × 12 = 36.

Matlab verification: >> Y0 = 3*fft((-1).^(0:11)) Y0 = 0 0 0 0

0

0

36

0

0

0

0

0

(b) This can be looked up in Table 8-1. It is the third entry with L = 4 and N = 12 so Y1 [k] =

sin(πk/3) − jπk /4 e sin(πk/12)

Matlab verification: >> Y1 = fft([ones(1,4),zeros(1,8)]) Y1 = 4.0000 + 0.0000i 2.3660 - 2.3660i 0.0000 + 0.0000i 1.0000 + 0.0000i 0.0000 + 0.0000i 0.6340 + 0.6340i 0.0000 + 0.0000i 0.0000 + 1.7321i

0.0000 - 1.7321i 0.6340 - 0.6340i 1.0000 + 0.0000i 2.3660 + 2.3660i

(c) In this problem, the even samples are 1 and the odd indexed samples are zero. This solution is facilitated if we 11

X j (2π/12)6n 1 write y2 [n] as y2 [n] = 2 1 + e δ[n − m]. Therefore, it follows from the linearity and frequency-shift m=0

properties in Table 8-2 and the third entry in Table 8-1 with L = N = 12 that Y2 [k] = 21

sin(πk) − jπk11/12 1 sin(π(k − 6)) − jπ(k−6)11/12 +2 e e sin(πk/12) sin(π(k − 6)/12)

The first term is zero except at k = 0 where its value is Y2 [0] = 6. The second term is zero except when k = 6 where its value is Y2 [6] = 6. Thus  6 Y2 =  0 

k = 0, 6 k = 1, 2, . . . , 5, 7, 8, . . . , 11

We can get another closed form expression by summing the DFT expression only over the even indices using a index m = 2n, n = 0, 1, . . . , 5. This gives us Y2 [k] =

5 X m=0

e− j (2π/12)2k m =

1 − e− j (2π/12)6k 1 − e− jπk = 1 − e− j (2π/12)2k 1 − e− j (2π/6)k

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CHAPTER 8. DFT

In this case, the expression is indeterminant for k = 0, 6. The value at these DFT indices is Y2 [0] =

11 X

y2 [n]e− j (2π/12)6n =

n=0

11 X

y2 [n] = 6,

n=0

and because the odd-indexed samples are zero, Y2 [6] =

11 X

y2 [n]e− jπn = 6

n=0

Matlab verification: >> Y2 = fft([1,0,1,0,1,0,1,0,1,0,1,0]) Y2 = 6 0 0 0 0 0

6

0

0

0

0

0

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.4

DSP First 2e

(a) The DFT of y[n] is Y [k] =

5 X

7e− j (2π/6)k n − 3W [k] = 7

n=0

1 − e− j2πk − 3W [k] = 7(6)δ[k] − 3W [k] 1 − e− j (2π/6)k

Therefore, Y [k] = {39, 0, −3 j, 0, +3 j, 0} (b) If we write v[n] = 4(−1) n w[n] as v[n] = 4e j (2π/6)3n w[n], we see that the new DFT is V [k] = 4W [k − 3] where the shift is periodic with period 6.

Therefore, V [k] = {0, −4 j, 0, 4, 0, 4 j}

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CHAPTER 8. DFT P-8.5

DSP First 2e

(a) X[0] =

255 X

x[n]e− j2π (0)n/256 =

255 X

n=0

X[128] =

x[n] is real since all x[n] are real-valued.

n=0

255 X

x[n]e− j2π(128)n/256 =

(b) X[k] = 3 − e

x[n](−1) n is real since all x[n] are real-valued.

n=0

n=0 − j (2π/4)k

255 X

+ 3e

− j (2π/4)2k

− e− j (2π/4)3k = {4, 0, 8, 0}

3

(c) v[2] =

(d) Y [0] =

1 X V [k]e j (2π/4)(2)k = 41 (3 + (5 j)(−1) + 0 + (−5 j)(−1)) = 43 N k=0 9 X

y[n] = 0

n=0

Y [5] =

9 X

7(−1) n e− j (2π/10)5n = 70

n=0

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CHAPTER 8. DFT P-8.6

DSP First 2e

(a) Sketch should show x[n] = {0, 1, 2, 3, 4, 0, 0, 0} starting at n = 0. (b) Sketch should show y[n] = {2, 3, 4, 0, 0, 0, 0, 1} because y[n] = x[n − 6] evaluated with a period of 8.

Problem 8.6(a), x[n] 4 3 2 1 0 0

1

2

3

4

5

6

7

5

6

7

Problem 8.6(b), y[n] 4 3 2 1 0 0

1

2

3 4 Time Index (n)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.7

DSP First 2e

(a) Plot y[n] = {2, 1, 0, 0, −2, −1} starting at n = 0.

Problem 8.7, y[n] 2 1 0 −1 −2 0

1

2

3

4

5

Using the method of synthetic polynomial multiplication to evaluate the convolution, we get 1 2 2

1 -1 1

1 -1 0

1 0

-2

-1

so y[n] = 2δ[n] + δ[n − 1] − 2δ[n − 4] − δ[n − 5]

(b) Applying the definition of the DFT and writing out the individual terms gives X[k] = 1 + e− j (2π/N )k + e− j (2π/N )k (2) + e− j (2π/N )k3 + e− j (2π/N )k (4) H[k] = 2 − e− j (2π/N )k − e− j (2π/N )k (2) (c) Forming the product Y [k] = H[k]X[k] is the same as doing the polynomial multiplication as in part (a) so we get Y [k] = 2 + e− j (2π/N )k − 2e− j (2π/N )k (4) − e− j (2π/N )k (5) . (d) For N = 6, the DFT is Y [k] = 2 + e− j (2π/6)k − 2e− j (2π/6)k (4) − e− j (2π/6)k (5) . All we need to do is pick off the coeficients of Y [k] to obtain y[n] = {2, 1, 0, 0, −2, −1} (e) We need to adjust the exponents of W [k] so that any power of e− j (2π/4)k (n+4) is replaced by e− j (2π/4)k n .

Thus, Y [k] = 2 + e− j (2π/4)k − 2e− j (2π/4)k (4) − e− j (2π/4)k (5) is equivalent to Y [k] = 2 + e− j (2π/4)k − 2e− j (2π/4)k (0) − e− j (2π/4)k = 0. (f) Since the length of the convolution is L + M − 1, we need to be able to represent the sequence of this length by the DFT. Therefore, N ≥ L + M − 1 is required.

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CHAPTER 8. DFT P-8.8

DSP First 2e

In general, the 100-point DFT would be X[k] = X (e j (2π/100)k ) = 7e− j (2π/100)18k

sin(12.5(2π/100)k) sin(πk/100)

If this is evaluated for k = 13 we get X[13] = X (e j (2π/100)13 ) = 12.46 e j0.32π = 12.46 e j1.005

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CHAPTER 8. DFT P-8.9

DSP First 2e

(a) From Q(e j ω̂ ) it is clear that the peaks will occur around DFT indices corresponding to the frequencies ω̂0 and 2π − ω̂0 ; i.e. when (2π/32)k1 = ω̂0 and (2π/32)(32 − k1 ) = 2π − ω̂0 . Therefore, ω̂0 = (2π/32)10 = 0.625π. (b) The second peak occurs when (2π/32)(32 − k1 ) = (2π/32)k2 or when k2 = 22. The third nonzero value occurs at k 3 = 0, corresponding to the DC component of 0.1 in s[n]. The DFT of that DC component is 0.1N δ[k], which is zero for all k except k = 0. The Dirichlet components are also zero except at k1 = 10 for the first one and k 2 = 22 for the second one. (c) The peak magnitude at k = 10 is 12 ( A)(32) = 50 so A = 100/32 = 3.125.

Matlab verification: >> x=0.1+(100/32)*cos(2*pi*10*(0:31)/32); >> X=fft(x); >> abs(X) ans = 3.2000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.0000 50.0000 0.0000 50.0000 0.0000

0.0000 0.0000 0.0000 0.0000 0.0000

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CHAPTER 8. DFT P-8.10

DSP First 2e

(a) N = 32 + 48 = 80. (b) k peaks = 8, and −8 → (80 − 8 = 72) because (2π/80)k peaks = ±0.2π The peaks will occur around DFT indices corresponding to ω̂ = ±0.2π, or when (2π/80)k = 0.2π or k = 8 and when (2π/80)(80 − 8) = 2π − 0.2π or k = 72. (c) max{|R[k]|} = 118.4553. Evaluate in Matlab via k

r=1+7*cos(0.2*pi*(0:31)); R=fft(r,80); [mx,ix]=max(abs(R))

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CHAPTER 8. DFT P-8.11

DSP First 2e

The impulse response h[n] =   1/5 H (e j ω̂ ) =  0 

sin(0.15πn) corresponds to a frequency response 5πn

|ω̂| ≤ 0.15π 0.15π < |ω̂| ≤ π

The input signal has fundamental frequency ω̂0 = 0.08π. Thus, only DC and the first harmonic are below the cutoff of the filter. Since the gain is 1/5, we get three terms in the output

y[n] = (1/5) 1 + 2e j0.08πn + 2e− j0.08πn = 0.2 + 0.4 cos(0.08πn)

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CHAPTER 8. DFT P-8.12

DSP First 2e

(a) In order to show the spectrum only for positive frequencies, we need to convert the frequencies by adding 2πn to the exponent of each of the components having negative frequency. Therefore, x[n] = 3 + 2e j0.2πn + 2e j (2π−0.2π)n − 7 je j0.7πn + 7 je j (2π−0.7π)n = 3 + 2e j0.2πn + 2e j1.8πn − 7 je j0.7πn + 7 je j1.3πn Therefore, the spectrum plot is as follows: 7j

2 −π

7j

−7 j

3

−.2π 0

−.7π

2

2 .2π

.7π

π

1.3π

3

1.8π 2π

ω̂

The dashed lines show negative frequencies that differ by 2π from frequencies in the range 0 ≤ ω̂ < 2π. (b) If we carry out the multiplication and adjust any resulting negative frequencies as we did in part (a), we get x 1 [n] = e j0.4πn x[n] = 3e j0.4πn + 2e j0.6πn + 2e0.2πn − 7 je j1.1πn + 7 je j (−0.3π)n = 3e j0.4πn + 2e j0.6πn + 2e0.2πn − 7 je j1.1πn + 7 je j1.7πn This is an example of the frequency shift property since by adding 0.4π to each frequency we shift the spectrum to the right by that amount. The resulting spectrum is −7 j

7j

2 −.9π

−.3π

0

3

−7 j

7j

π 1.1π

1.7π

2

.2π .4π .6π

2π

ω̂

(c) Now we can write x 2 [n] = (−1) n x[n] = e jπn x[n] and we see that by the same process as in part (b), each of the original frequencies will be increased by π radians. If we choose the alias frequencies that are in the range 0 ≤ ω̂ < 2π, we can write x[n] = 3e jπn + 2e j1.2πn + 2e j0.8πn − 7 je j1.7πn + 7 je j0.3πn Therefore the spectrum plot is that of part (a) shifted to the right by π radians.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT 7j

−7 j

3

2

2

−π −.8π

−7 j

−.3π

0

.3π

.8π

3

2

π 1.2π

1.7π

2π

ω̂

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.13

DSP First 2e

(a) The sampled signal is x[n] = x(n/60) = 6 cos(42πn/60) + 4 cos(18πn/60 − 0.5π) = 6 cos(2π(7/20)n) + 4 cos(2π(3/20)n − 0.5π) = 6 cos(0.7πn) + 4 cos(0.3πn − 0.5π) Since 7 and 3 have no common factors, the period is N = 20 and the fundamental frequency is ω̂0 = 2π/20. (b) x[n] = 2e− j0.5π e j (2π/20)3n + 2e j0.5π e− j (2π/20)3n + 3e j (2π/20)7n + 3e− j (2π/20)7n (c) From the answer to part (b), the DFS coefficients are c3 = 2e− j0.5π , c−3 = 2e j0.5π , c7 = 3, c−7 = 3. (d) Here is the spectrum plot of the DFS: 3

3 2j

−π

−.7π

−.3π

3 2j

−2 j

0

.3π

3

π

1.3π

1.7π

−2 j

2π

2.3π

2.7π

3π

ω̂

Dashed lines are aliases of the frequencies in the range −π < ω̂ ≤ π.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.14

DSP First 2e

sin(4ω̂) − j ω̂7/2 e . The zeros of this function are at frequencies sin(ω̂/2) where 4ω̂ = πr where r is an integer; i.e., where ω̂ = 0.25π, 0.5π, 0.75π, π. The equivalent values of DFT index k are when (2π/N )k = 0.25πr or k = Nr/8. For N = 512, these values are multiples of 64.

(a) The frequency response of this filter is H (e j ω̂ ) =

Problem 8.14 Magnitude

8 6 4 2 0 0

32

64

32

64

96

128

160

192

224

256

96 128 160 192 DFT Frequency Index (k)

224

256

Phase (rad)

pi/2 0 −pi/2 −pi 0

(b) Note that the given frequencies are all multiples of 0.25π so all the frequencies except DC fall at the zeros of H (e j ω̂ ). Therefore, these frequencies do not appear in the output. The gain of the filter at ω̂ = 0 is H (e j0 ) = 8, so the output is y[n] = 3(8) = 24 for all n. You can pick all these values off the plot, or evaluate H (e j ω̂ ) at individual frequencies with freqz.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.15

DSP First 2e

(a) We have two constraints. To avoid aliasing we need 2(1000) < f s . To meet the frequency spacing condition, we need f s /N ≤ 5, where N = 2ν .

The combined constraints are given by 2000 < f s ≤ 5 · 2ν , or equivalently 400 < f s /5 ≤ ·2ν .

Thus we require N = 512 = 29 . If we fix N at 512, the condition on f s is 2000 < f s ≤ 2560 Hz. (b) As discussed in Section 8-7.4.1, the width of the main lobe of the Hann and Hamming windows is approximately ∆ω̂ = 8π/L where L is the window length. Therefore, for L = N/2 = 256, we should expect a main lobe width of about 8π/256 = π/32 rad. This is equivalent to analog radian frequency of ∆ω = (π/32) f s or using analog frequency in hertz, ∆ f = f s /64.

Using the constraints on f s obtained in part (a), we get 2000/64 < ∆ f ≤ 2560/64, or 31.25 < ∆ f ≤ 40.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.16

DSP First 2e

(a) Here’s a Matlab plot of the ideal spectrogram. A sketch should look like this.

Problem 8-16a: Ideal Spectrogram 2000 1800 1600

frequency

1400 1200 1000 800 600 400 200 0

0

0.2

0.4

0.6

0.8

1

1.2

time axis

(b) The width of the main lobe of the DTFT of the Hann window is approximately ∆ω̂ = 8π/L, where L is the window length. Therefore the normalized main lobe width for the L = 200 window is ∆ω̂ = 0.04π. In terms of analog cyclic frequency, this would be equivalent to ∆ f = ∆ω̂ f s /(2π) = 0.04π8000/(2π) = 160 Hz. Thus, the spectrogram sketch should show bars about 160 Hz wide and there should be blurred regions of approximately 200 samples (or equivalently, 200/8000 = 0.025 s) duration at the beginning and end of each sinusoidal component.

The following is a Matlab spectrogram of the signal.

Problem 8-16b: Spectrogram with 200-point Window

2000 1800 1600

frequency

1400 1200 1000 800 600 400 200 0

0

0.2

0.4

0.6

0.8

1

1.2

time axis

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.17

DSP First 2e

Note that there are 8 equal duration notes in (approximately) 2 seconds of time. Therefore, each note has duration 2/8 s or 250 ms.

Spectrogram of C Major scale, L D 500; fs D 4 kHz

600 550

Frequency (Hz)

500 450 400 350 300 250 200 0

0.2

0.4

0.6

0.8

1 1.2 Time (s)

1.4

1.6

1.8

2

Figure P-8.17

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.18

DSP First 2e

The ratio of the window lengths is 256/100 ≈ 2.5.

The sketch will look similar to the spectrogram in Fig. P-8.17, but it will have bars that are 2.5 times as wide in the frequency dimension. The frequency width of the horizontal bars in Fig. P-8.17 is approximately 16 Hz, so it will change to 40 Hz in the new figure.

In the time dimension, the fuzzy regions at the time of frequency change will be shorter—their duration will be 40% of the duration of those in Fig. P-8.17 which are approximately 0.03 s. Thus, in the new figure the duration of the fuzzy regions will be approximately 0.01 s.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.19

DSP First 2e

(a) The Hamming window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. For the given parameters, this would give a main lobe width equivalent to ∆ f = 4 f s /L = 4(8000)/256 = 125 Hz. Thus, two sinusoids whose frequencies are farther apart than 125 Hz will show up as two distinct peaks, but if the two frequencies are closer that 125 Hz they will tend to blend together in the spectrum. (b) The window length is equivalent to 256/8000 =32 ms so if there are abrupt changes in the waveform, the spectral characteristics will look blurred over approximately 32 ms around the abrupt change.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.20

DSP First 2e

(a) The Hamming window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. In terms of analog cyclic frequency this corresponds to ∆ f = 4 f s /L, where L is the window length. For the given parameters, we want 40000/L = 250 or L = 160. (b) Here is the plot of the DTFT magnitude:

DTFT of Windowed Sinusoids at Frequencies 3000 and 3250 Hz

45 40 35 30 25 20 15 10 5 0

0

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

analog frequency in Hz

Figure 8-1: Solution for Problem 8-20.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.21

DSP First 2e

(a) The notes are in the octave above A-440, so the frequencies are given by the equation f = 440(2n/12 ) with n = 0, 1, . . . , 12 covering one octave from A-440 to A-880. Here are the notes and their theoretical frequencies.

(Subscript denotes octave number.) note A4 B4 [ B4 C5 C5 ] D5 D5 ] E5 F5 F5 ] G5 G5 ] A5 f 440 466 494 523 554 587 622 659 698 740 784 831 880 Now the first 9 notes of Für Elise are { E5 , D5 ], E5 , D5 ], E5 , B4 , D5 , C5 , A4 }. Therefore, the theoretical frequencies for the first nine notes are { 659, 622, 659, 622, 659, 494, 587, 523, 440 }.

Spectrogram of Fur Elise (9 notes), L D 1000; fs D 8 kHz

800 750 700

E

650

Frequency (Hz)

600

D

550 500

C B

450

A

400 350

E

300 250 200

6

6.25

6.5

6.75

7

7.25

7.5 7.75 Time (s)

8

8.25

8.5

8.75

9

(b) The spectrogram in the figure above is marked with the true frequencies. Observe that the passage of interest starts at t = 6.4 s (approximately). It is difficult to obtain very accurate estimates of the frequencies by measuring on the spectrogram because the spectral peaks have a width of about 10 Hz. However, careful examination of the figure above confirms that the theoretical note frequencies are close to the center of the spectral peaks. (c) Observe that the passage of interest starts at t = 6.4 s (approximately). From the score in Fig. 3-22, the first eight notes are 16th notes while note nine is an eighth note, which should have twice the duration.

The first five notes are easy to distinguish, but overlap in time due to the sliding window of the spectrogram. Their total duration is 7.7 − 6.4 = 1.3 s, so each 16th note has a duration of approximately 0.26 s. The sixth note (B4 ) appears shorter (≈ 0.15 s), but might be softer so its second half is hard to see; likewise, for note eight (C5 ). The seventh note (D5 ) is barely visible, but there is evidence of a spectral peak in the 0.2 s gap between B4 and C5 . The ninth note (A4 ) is easy to distinguish, and its duration seems a bit longer (≈ 0.3 s), but not twice as long.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 8. DFT P-8.22

DSP First 2e

(a) From the spectrogram it appears that ω̂0 = 0.7π and ω̂1 = 0.75π. To estimate the chirp parameter α, recall that the instantaneous frequency is ω̂i [n] = 2αn. From the spectrogram we see that ω̂i [n] starts at 0 and increases linearly to ω̂i [2000] = 0.5π = 2α(2000). Therefore, α = π/8000. (b) The Hann window has a main lobe width of approximately ∆ω̂ = 8π/L, where L is the window length. Thus, if we multiply the window length by 4, we divide the main lobe width by 4. Thus, the spectrogram with L = 400 will like Fig. P-8.22 except that the bars will be much narrower—they will have 1/4 the width.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.1

X1 (z) = 7,

DSP First 2e

X2 (z) = z −1,

X3 (z) = 5z −4,

X4 (z) = 7 − z −1 + 5z −4

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.2

DSP First 2e

Y (z) = X (z) − z −1 X (z) = (1 − z −1 )X (z) = H (z)X (z) where H (z) = 1 − z −1 is the z-transform of h[n] = δ[n] − δ[n − 1], the impulse response of the system.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.3

DSP First 2e

(a) y[n] = x[n] + 5x[n − 2] − 3x[n − 3] + 2x[n − 5] + 4x[n − 7], (b) h[n] = δ[n] + 5δ[n − 2] − 3δ[n − 3] + 2δ[n − 5] + 4δ[n − 7], Plot {1, 0, 5, −3, 0, 2, 0, 4} for 0 ≤ n ≤ 8.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.4

DSP First 2e

(a) H (z) = 0.1(1 − z −1 + z −2 ) (b) The roots of H (z) are e±π/3 so we can write H (z) = 0.1(1 − e jπ/3 z −1 )(1 − e− jπ/3 z −1 ) = 0.1

(z − e jπ/3 )(z − e− jπ/3 ) z2

Imaginary Part

1 0.5 2

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) H (e j ω̂ ) = 0.1(1 − e− j ω̂ + e− j ω̂2 ) 0.3 0.2π

0.1

H(ej ω̂ )

0 −π 5

−0.8π −0.6π −0.4π −0.2π

0

0.2π

0.4π

0.6π

0.8π

π

0

0.2π

0.4π

0.6π

0.8π

π

6

0

(d)

-5 −π

−0.8π −0.6π −0.4π −0.2π

ω̂

(e) For this we need the frequency response evaluated at ω̂ = 0, π/4, 2π/3.

H (e j0 = 1

H (e jπ/4 ) = 0.4142e− jπ/4

H (e j2π/3 ) = 2e jπ/3

Therefore, the output is y[n] = 9 − 8(0.4142) cos[0.25π(n − 1) − π/4] + 7(2) cos[(2π/3)n + π/3] = 9 − 3.3136 cos[0.25π(n − 2)] + 14 cos[(2π/3)n + π/3]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.5

DSP First 2e

(a) We have to multiply out the factors to get a polynomial in z −1 .

H (z) = 1 + 2z −1 + 2.64z −2 + 2.28z −3 + 1.28z −4 + 0.64z −5 Then the difference equation is: y[n] = x[n] + 2x[n − 1] + 2.64x[n − 2] + 2.28x[n − 3] + 1.28x[n − 4] + 0.64x[n − 5] (b) There are five poles at z = 0 and the five zeros are at z = −1, ± j0.9, and e± j2π/3 .

Imaginary Part

1 0.5 5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) We need to find the frequencies where H (e j ω̂ ) = 0.

These frequencies are the angles of the zeros that are on the unit circle; i.e., (1 + z −1 ) = 0, (1 − e j2π/3 z −1 ) = 0 and (1 − e− j2π/3 z −1 ) = 0, or ω̂ = π, ±2π/3.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.6

DSP First 2e

(a) H (z) = H2 (z)H1 (z) = 14 (1 + z −1 + z −2 + z −3 ) 31 (1 + z −1 + z −2 )

1 = 12 (1 + 2z −1 + 3z −2 + 3z −3 + 2z −4 + z −5 )

(b) Y2 (z) = X (z)H2 (z) = (1 − z −1 ) 31 (1 + z −1 + z −2 )

= 31 (1 − z −3 )

W (z) = Y2 (z)H1 (z) = 31 (1 − z −3 ) 41 (1 + z −1 + z −2 + z −3 )

1 = 12 (1 + z −1 + z −2 − z −4 − z −5 − 3z −3 )

1 (c) y[n] = 12 (x[n] + 2x[n − 1] + 3x[n − 2] + 3x[n − 3] + 2x[n − 4] + x[n − 5]) Clearly, the samples are not weighted equally although the sum of the weights (coefficients) is equal to one.

(d) There are five poles at z = 0, and the five zeros are at z = −1, ± j, and e± j2π/3

Imaginary Part

1 0.5 5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.7

DSP First 2e

>> z=roots([1,-.5,.5,-1]) z = 1.0000 -0.2500 + 0.9682i -0.2500 - 0.9682i >> magz=abs(z) magz = 1.0000 1.0000 1.0000 >> angz=angle(z)/pi angz = 0 0.5804 -0.5804 Thus, the zeros are at z = 1, e± j0.5804π and there is one pole at z = 0 for each of the zeros.

Imaginary Part

1 0.5 3

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.8

DSP First 2e

(a) Impulse response: h[n] = δ[n] − δ[n − 1] + δ[n − 2] − δ[n − 3] + δ[n − 4] =

4 X

(−1) k δ[n − k]

k=0

(b) H (z) =

4 X

(−1) k z −k =

k=0

4 X

(−1z −1 ) k =

k=0

z5 + 1 1 − (−z −1 ) 5 1 + z −5 = 4 = −1 −1 1 − (−z ) 1+z z (z + 1)

(c) Plot the poles and zeros of H (z) in the complex z-plane.

Imaginary Part

1 0.5 4

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

cos(ω̂2.5) 1 + e− j ω̂5 = e− j ω̂2 cos(ω̂/2) 1 + e− j ω̂ (e) Sketch of the frequency response (magnitude and phase) , or Matlab plot.

(d) H (e j ω̂ ) =

|H(ej ω̂ )|

6 4 2

H(ej ω̂ )

0 −π 5

−0.6π

−0.2π

0

0.2π

0.6π

π

−0.6π

−0.2π

0

0.2π

0.6π

π

6

0

-5 −π

ω̂ Figure P-9.8

(f) We need the frequency response evaluated at ω̂ = 0, 0.5π, 0.6π. These values are H (e jω̂ ) = 1, 1, 0.

Therefore, y[n] = 5 + 4 cos(0.5πn).

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.9

DSP First 2e

(a) H (z) = (1 − z −1 ) 5 (b) Plot of the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 5

5

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c)

5 H (e j ω̂ ) = (1 − e− j ω̂ ) 5 = e− j ω̂5/2 = 2e jπ/2 e− j ω̂/2 sin(ω̂/2) = 32e (π−ω̂)5/2 sin5 (ω̂/2) (d) H (z) = (1 − z −1 ) 5 = 1 − 5z −1 + 10z −2 − 10z −3 + 5z −4 − z −5 h[n] = δ[n] − 5δ[n − 1] + 10δ[n − 2] − 10δ[n − 3] + 5δ[n − 4] − δ[n − 5]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.10

DSP First 2e

(a) We can find N1 as the lowest power of z −1 of the product X (z)H (z). That will be N1 = 0 + 6 = 6. Similarly, N2 is the highest power of z −1 if the product X (z)H (z). That is N2 = 9 + 9 = 18. (b) Again, we can use the polynomials of X (z) and H (z) to get the answer. y[6] is the product of the coefficients of the two lowest order terms or y[6] = 2 × 1 = 2. Similarly, y[18] is the product of the coefficients of the highest order terms, or y[18] = −1 × 3 = −3. (c) y[n] = 2δ[n − 6] − 4δ[n − 7] + 8δ[n − 8] + δ[n − 9] − 6δ[n − 10] + 12δ[n − 11] − 11δ[n − 12] +16δ[n − 13] − 32δ[n − 14] + 11δ[n − 15] − 6δ[n − 16] + 12δ[n − 17] − 3δ[n − 18]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.11

DSP First 2e

(a) H (z) = 1 + z −1 + z −2 = (1 − e j2π/3 z −1 )(1 − e− j2π/3 z −1 ), so there are zeros at z = e± j2π/3 . (b) To solve this part we need the frequency response evaluated at ω̂ = π/6. That is, H (e j ω̂ ) = 1 + e− j ω̂ + e− j ω̂2 = e− j ω̂

sin(ω̂3/2) sin(ω̂/2)

so H (e jπ/6 ) = 2.7321e− jπ/6 and, therefore, the formula for the steady-state output is y[n] = 2.7321 cos((π/6)(n − 1) − π/4) (c) omega = 2*pi/3 gives zero for the steady-state output after n ≥ 2. (Until the last two samples.)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.12

DSP First 2e

(a) y[n] = x[n] − x[n − 4] (b) H (e j ω̂ ) = 1 − e− j ω̂4 = 2e j (π/2−ω̂2) sin(ω̂2) (c) |H (e

j ω̂

)| = 2| sin(ω̂2)|

and

∠H (e

j ω̂

( )=

π/2 − 2ω̂ π/2 − 2ω̂ + 2π

0 < ω̂ < π/2 π/2 < ω̂ < π

(d) The frequencies nulled are the angles of the zeros of H (z); ω̂0 = 0, ±π/2, π. (e) y[n] = |H (e jπ/3 )| cos(πn/3 + ∠H (e jπ/3 )) = 1.7321 cos(πn/3 − π/6)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.13

DSP First 2e

We should find the output as the sum of the response to DC plus the response to an impulse of size -70 plus the response to the sinusoid of frequency 0.5π. We need the impulse response and the frequency response function to do this.

From H (z) in expanded form, h[n] = δ[n] − 5δ[n − 2] + 4δ[n − 4] and H (e j ω̂ ) = (1 − e− j ω̂2 )(1 − 4e− j ω̂2 ) = 1 − 5e− j ω̂2 + 4e− j ω̂4 so H (e j0 ) = 0 and H (e j0.5π ) = 1 − 5e− j2π/2 + 4e− j4π/2 = 1 − 5e−π + 4e− j4π/2 = 10. Therefore, y[n] = −70δ[n] + 350δ[n − 2] − 280δ[n − 4] + 300 cos(0.5πn + π/4)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.14

DSP First 2e

After sampling, we get the sequence x[n] = 4 + cos(1000πn/2000 − π/4) − 3 cos(500πn/2000). Thus, we need to evaluate the frequency response at ω̂ = 0, π/2, π/4 to find the outputs due to each component of the input. These values are H (e j0 ) = 0, H (e jπ/2 ) = 0, and H (e jπ/4 ) = 1.0824e jπ/8 . Therefore, y[n] = −3(1.0824 cos(πn/4 + π/8) = 3.2472 cos(πn/4 − π + π/8) = 3.2472 cos(πn/4 − 2.7489) and therefore the output of the D-to-C converter with f s = 2000 is y(t) = 3.2472 cos(π2000t/4 − 2.7489) = 3.2472 cos(500πt − 2.7489)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.15

DSP First 2e

(a) α = 4, β = 10 From H (z) we get H (e j ω̂ ) = 5(1 + .8e− j ω̂ + e− j ω̂2 . By factoring e− j ω̂ out and using the inverse Euler relation we get H (e j ω̂ ) = e− j ω̂ (4 + 10 cos(ω̂)) (b) In the plot the DC value is +14, H (e jπ ) = −6, and H (e j ω̂ ) = 0 at ω̂ = 0.631π = 1.9823.

|H(ej ω̂ )|

15 10 5

H(ej ω̂ )

0 −π 2

0.2π

0.4π

0.6π

0.8π

π

0.2π 0 ω̂ Figure P-9.15

0.4π

0.6π

0.8π

π

−0.8π −0.6π −0.4π −0.2π

0

6

0

-2 −π

−0.8π −0.6π −0.4π −0.2π

(c) We want the system function to have the form H (z) = (1−e j ω̂0 z −1 )(1−e− j ω̂0 z −1 ) = 1−2 cos ω̂0 z −1 +z −2 . Therefore, we must set b1 = −2 cos ω̂0 . The normalized frequency that we want to null is ω̂0 = 2π f 0 / f s = 2π(1600/8000) = 0.4π. Thus, we need to set b1 to b1 = −0.618.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.16

DSP First 2e

(a) Factoring the two second-order factors gives H (z)(1 − z −1 )(1 + z −1 )(1 − 0.8e jπ/4 z −1 )(1 − 0.8e− jπ/4 z −1 )(1 + e jπ/2 z −1 )(1 + e− jπ/2 z −1 )

Imaginary Part

1 0.5 6

0

-0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-9.16 (b) To satisfy the condition, we must group the zeros at ±1 and e± jπ/2 together to form H1 (z) as in H1 (z) = (1 − z −2 )(1 + z −2 ) = (1 − z −4 ) . That leaves the remainder for H2 (z) as H2 (z) = (1 − 0.8e jπ/4 z −1 )(1 − 0.8e− jπ/4 z −1 ) = 1 − 1.6 cos(π/4)z −1 + 0.64z −2 . (c) y[n] is the output of H2 (z) when the input is y1 [n] so from H2 (z) we get y[n] = y1 [n] − 1.6 cos(π/4)y1 [n − 1] + 0.64y1 [n − 2]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.17

DSP First 2e

(a) h[n] = 7δ[n] − 14.35δ[n − 1] + 22.4175δ[n − 2] − 14.35δ[n − 3] + 7δ[n − 4] We need to determine the expanded version of H (z) with an additional gain factor.

H (z) = A(1 − 0.8e jπ/3 z −1 )(1 − 0.8e− jπ/3 z −1 )(1 − 1.25e jπ/3 z −1 )(1 − 1.25e− jπ/3 z −1 ) = (1 − 0.8z −1 + 0.64z −2 )(1 − 1.25z −1 + 1.5625z −2 ) Clearly, when expanded out, the H (z) polynomial will be of the form H (z) = A + . . . + A(0.8) 2 (1.25) 2 z −4 , so if h[0] = 7 then A = 7. (b) H (z) = 1.81406 − 3.71882z −1 + 5.80952z −2 − 3.71882z −3 + 1.81406z −4 If we expand H (z) into a polynomial with unknown gain we get H (z) = A(1 − 2.05z −1 + 3.2025z −2 − 2.05z −3 + z −4 ) The DC value of the frequency response is H (e j0 ) = A(1 − 2.05 + 3.2025 − 2.05 + 1) = 1.1025A = 2 so A = 1.8141 and H (z) = 1.8141(1 − 2.05z −1 + 3.2025z −2 − 2.05z −3 + z −4 )

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.18

DSP First 2e

(a) H (1) = b0 + b1 − b1 − b0 = 0 (b) H (e j ω̂ ) = b0 + b1 e− j ω̂ − b1 e− j ω̂3 − b0 e− j ω̂4 = e− j ω̂2 (b0 e j ω̂2 + b1 e j ω̂ − b1 e− j ω̂ − b0 e− j ω̂2 ) = [2b0 sin(2ω̂) + 2b1 sin(ω̂)]e j (π/2− j ω̂

(from the inverse Euler formula)

(c) H (z −1 ) = b0 + b1 z − b1 z 3 − b0 z 4 = z 4 (b0 z −4 + b1 z −3 − b1 z −1 − b0 ) = −z 4 H (z) (d) In the general case with bk = −b M−k , the condition H (1/z) = −z M H (z) holds for M even or odd.

The frequency response has the form H (e j ω̂ ) = 2e j (π/2−ω̂ M /2)

M /2 X

b M /2−k sin(ω̂k)

M an even integer

k=1

and H (e j ω̂ ) = 2e j (π/2−ω̂ M /2)

(M+1)/2 X

b(M+1)/2−k sin(ω̂(2k − 1)/2)

M an odd integer

k=1

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM P-9.19

DSP First 2e

(a) H1 (z) = H2 (z) = 1 + z −1 + z −2 + z −3 (b) H (z) = H1 (z)H2 (z) = (1 + z −1 + z −2 + z −3 ) 2 (c) H (z) = 1 + 2z −1 + 3z −2 + 4z −3 + 3z −4 + 2z −5 + z −6 so h[n] = δ[n] + 2δ[n − 1] + 3δ[n − 2] + 4δ[n − 3] + 3δ[n − 4] + 2δ[n − 5] + δ[n − 6] (d) y[n] = x[n] + 2x[n − 1] + 3x[n − 2] + 4x[n − 3] + 3x[n − 4] + 2x[n − 5] + x[n − 6] (e) We can use the formula for the sum of N terms of a geometric series to write H1 (z), or H2 (z) as H1 (z) = H2 (z) = 1 + z −1 + z −2 + z −3 =

1 − z −4 1 − z −1

(f) From part (e) we see that (1 + z −1 + z −2 + z −3 ) 2 so all the zeros at {−1, ± j} are double zeros, and there are six poles at z = 0.

2

Imaginary Part

1 0.5 0

2

6

-0.5 2

-1 -1

-0.5

0 0.5 Real Part

1

Figure P-9.19 (g) From the result in (e), we can write H1 (e j ω̂ ) = H2 (e j ω̂ ) =

e− j ω̂2 (e j ω̂2 − e− j ω̂2 ) 1 − e− j ω̂4 sin(2ω̂) − j3ω̂/2 = e = 1 − e− j ω̂ e− j ω̂/2 (e j ω̂/2 − e− j ω̂/2 ) sin(ω̂/2)

(h) Note that H (e j ω̂ ) = H2 (e j ω̂ )H1 (e j ω̂ ) = [H1 (e j ω̂ )]2 = [H2 (e j ω̂ )]2 .

H (e j ω̂ ) = H1 (e j ω̂ )H2 (e j ω̂ ) =

sin(2ω̂) sin(ω̂/2)

!2

e− j3ω̂

Since the squared-term is nonnegative, it is the magnitude, and the system has linear phase with time delay equal to 3 samples.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 9. z-TRANSFORM

|H(ej ω̂ )|

20 10

H(ej ω̂ )

0 −π 5

0.2π

0.4π

0.6π

0.8π

π

0.2π 0 ω̂ Figure P-9.19

0.4π

0.6π

0.8π

π

−0.8π −0.6π −0.4π −0.2π

0

6

0

-5 −π

−0.8π −0.6π −0.4π −0.2π

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.1

DSP First 2e

(a) The system function is H (z) =

1 1 − z −1 − z −2

=

1 (1 − p1 z −1 )(1 − p2 z −1 )

√ where, by the quadratic formula, p1,2 = (1 ± 5)/2 = 1.618, −0.618. The partial fraction expansion of H (z) is √ √ 5+1 5−1 p2 p1 √ √ p1 − p2 p2 − p2 2 5 2 5 + = + H (z) = 1 − p1 z −1 1 − p2 z −1 1 − p1 z −1 1 − p2 z −1 Therefore, the impulse response is √ n √ n √ √ 5 + 1+ *1 + 5+ 5 − 1+ *1 − 5+ * * h[n] = u[n] + u[n] √ √ , 2 5 -, 2 , 2 5 -, 2 Try a few terms of this closed form expression to see if it generates the Fibonacci sequence, 1, 1, 2, 3, 5, 8, 13, ... √ (b) The poles determined in part (a) are p1,2 = (1 ± 5)/2 = 1.618, −0.618. Since one of the poles is outside the unit circle, this is an unstable system.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.2

DSP First 2e

First, find the system function by inspection as !

1 z −1 = z −1 = G(z)z −1 H (z) = 1 − 0.9z −1 + 0.9z −2 1 − 0.9z −1 + 0.9z −2 The z −1 factored out suggests that the impulse response will be zero prior to n = 1. Then we can find the partial fraction expansion of the rational function part and write G(z) as G(z) =

0.568e− j0.4942 0.568e j0.4942 + j1.0766 −1 1 − 0.9487e z 1 − 0.9487e− j1.0766 z −1

and

n

n g[n] = 0.568e− j0.4942 0.9487e j1.0766 u[n] + 0.568e j0.4942 0.9487e− j1.0766 u[n] = 1.136(0.9487) n cos(1.0766n − 0.4942)u[n] Therefore, the impulse response is h[n] = g[n − 1] = 1.136(0.9487) n−1 cos(1.0766(n − 1) − 0.4942)u[n − 1] which, by virtue of u[n − 1], is the desired formula showing h[n] being zero for n < 1 and being a decaying sinusoid for n ≥ 1.

It should be noted that we could have made a partial fraction expansion of H (z) directly. The resulting inverse transform would be a formula that is valid from n = 0 since the complex exponentials would be multiplied by u[n].

However, when the formula is evaluated at n = 0, the result would be zero.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.3

DSP First 2e

The system function is H (z) =

5z −7 1 − 0.5z −1

Us the z-transform table and then the delay property, h[n] = 5(0.5) n−7 u[n − 7] Note that h[n] = 0 for n < 7. It decays exponentially as (0.5) n for n ≥ 7.

h[n]

6 4 2 0 0

5

10

15 20 time index n Figure P-10.3

25

30

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.4

DSP First 2e

The impulse response is h[n] = 5(0.5) n u[n], so the output for the given input is y[n] = 3h[n − 1] − 2h[n − 2] + 3h[n − 4] − h[n − 5] = 15(0.5) n−1 u[n − 1] − 10(0.5) n−2 u[n − 2] + 15(0.5) n−4 u[n − 4] − 5(0.5) n−5 u[n − 5] Evaluate with Matlab in order to make a plot.

y[n]

20 10 0 -10 0

5

10

15 20 time index n Figure P-10.4

25

30

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.5

DSP First 2e

The easiest way to work this is to write the output in terms of shifted impulse responses as y[n] = h[n] + βh[n − 2] + h[n − 4] √ and then examine y[n] for n ≥ 4. Define a = 1/ 2 so h[n] = 10a n u[n]. Then for n ≥ 4, the output is y[n] = 10a n + 10 βa n−2 + 10a n−4 = 10a n (1 + βa−2 + a−4 ) To make y[n] = 0 for n ≥ 4, we need 1 + βa−2 + a−4 = 0. Solving this equation for β gives β = −(1 + a4 )/a2 , or for the given value of a, β = −2.5. If you use this value of β, the output looks like the following

y[n]

5 0 -5 -10 0

5

10

15 20 time index n Figure P-10.5

25

30

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.6

DSP First 2e

(a) The system function is H (z) =

−1 + 3z −1 − 2z −2

1 − 31 z −1 − 21 z −2 The zeros are at z = 1, 2, and the poles at z = 0.8931, −0.5598

Imaginary Part

2 1 0 -1 -2 -2

-1

0 Real Part

(b) The system function is H (z) =

1

2

−1 + 3z −1 + 2z −2 1 − 31 z −1 + 21 z −2

The zeros are now at z = 3.5616, −0.5616, and the poles at z = 0.7071e± j1.3329 .

Imaginary Part

4

2

0

-2

-4 -4

-2

0 Real Part

2

4

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.7

DSP First 2e

(a) y[n] = 13 y[n − 1] − 21 y[n − 2] − x[n] H (z) =

−1 1 − 31 z −1 + 12 z −2

=

−z 2 z 2 − 13 z + 21

=⇒Poles: z = 0.7071e± j1.3329 Zeros: z = 0, 0 (b) y[n] = 31 y[n − 1] − 21 y[n − 2] − x[n − 2] H (z) =

−z −2 1 − 31 z −1 + 12 z −2

=

−1 z 2 − 13 z + 21

=⇒Poles: z = 0.7071e± j1.3329 Zeros: z = ∞, ∞ (c) y[n] = 13 y[n − 1] − 21 y[n − 2] − x[n − 4] H (z) =

−z −4 1 − 31 z −1 + 12 z −2

=

−1 z 2 (z 2 − 31 z + 12 )

=⇒Poles: z = 0.7071e± j1.3329, 0, 0 Zeros: z = ∞, ∞, ∞, ∞ In (c) there are four zeros at infinity balanced by two additional poles at the origin.

(b) Two zeros at z = ∞

(a) Two zeros at origin

0.5 (2)

0 -0.5 -1

1 Imaginary Part

1 Imaginary Part

Imaginary Part

1

(c) Four zeros at z = ∞

0.5 0 -0.5 -1

-1

0 Real Part

1

0.5 (2)

0 -0.5 -1

-1

0 Real Part

1

-1

0 Real Part

1

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.8

DSP First 2e

(a) By inspection, H (z) =

1 1 − 21 z −1

=

z z − 21

H (z) has a pole at z = 21 and a zero at z = 0. The corresponding impulse response is h[n] = (0.5) n u[n]. (b) Write the input as x[n] = δ[n] + δ[n − 2] + δ[n − 5]. Therefore, it follows that for this input the output is y[n] = h[n] + h[n − 2] + h[n − 5]. If we substitute for h[n] we get y[n] = (0.5) n u[n] + (0.5) n−2 u[n − 2] + (0.5) n−5 u[n − 5] Taking the step sequence into account we get 0      (0.5) n y[n] =   5(0.5) n    n  37(0.5)

n<0 n = 0, 1 n = 2, 3, 4 n≥5

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.9

DSP First 2e

−15 + 20z −1 −15(z − 4/3) = z − 0.75 1 − 0.75z −1 (b) Need plot. Pole at z = 0.75, zero at z = 4/3. (a) H (z) =

(c) H (e j ω̂ ) =

−15 + 20e− j ω̂ 1 − 0.75e− j ω̂

(d) |H (e j ω̂ )| 2 = 400 conjugate

}| {! !z −15 + 20e− j ω̂ −15 + 20e j ω̂ j ω̂ 2 j ω̂ ∗ j ω̂ |H (e )| = H (e )H (e ) = 1 − 0.75e− j ω̂ 1 − 0.75e j ω̂ =

625 − 600 cos(ω̂) 1.5625 − 1.5 cos(ω̂)

=

400(1.5625 − 1.5 cos(ω̂)) = 400 1.5625 − 1.5 cos(ω̂)

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.10

DSP First 2e

(a) By inspection, H (z) =

1 1 = √ √ −2 1 + 0.8z (1 − 0.8e jπ/2 z −1 )(1 − 0.8e− jπ/2 z −1 )

=

√

z2

√ (z − 0.8e jπ/2 )(z − 0.8e− jπ/2 ) √ Zeros at z = 0, 0 and poles at z = 0.8e± jπ/2 . (b) To find the output for the given input, determine the impulse response.

H (z) =

0.5 0.5 1 = + √ √ 1 + 0.8z −2 1 − 0.8e jπ/2 z −1 1 − 0.8e− jπ/2 z −1

so √ √ √ h[n] = 12 ( 0.8e jπ/2 ) n u[n] + 21 ( 0.8e− jπ/2 ) n u[n] = 0.8 cos( π2 n)u[n] Note: h[n] = 0 for n even, which could have been predicted from the fact that the initial H (z) has only z −2 in the denominator—an even power.

Therefore, y[n] = 5h[n] + 2.5h[n − 1] + 4h[n − 2] + 2h[n − 3] √ √ = 5 0.8 cos( π2 n)u[n] + 2.5 0.8 cos( π2 (n − 1))u[n − 1] √ √ + 4 0.8 cos( π2 (n − 2))u[n − 2] + 2 0.8 cos( π2 (n − 3))u[n − 3]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.11

DSP First 2e

(a) By inspection, H (z) =

z5 1 = 1 − 0.97z −5 z 5 − 0.97

(b) There are 5 zeros at z = 0, and the poles are at the 5th roots of 0.97, z k = (0.97) 0.2 e j2πk /5 = (0.9939)e j2πk /5 .

Imaginary Part

1 0.5 5

0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-10.11 (c) The z-transform of the output is Y (z) = H (z)X (z) =

1 (1 + βz −p ) 1 − 0.97z −5

If we choose β = −0.97 and p = 5, we get Y (z) = 1 or y[n] = δ[n], which satisfies the condition y[n] = 0 for n ≥ 5.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.12

DSP First 2e

(a) By using the delay and linearity properties, ha [n] = (−0.77) n u[n] − (−0.77) n−1 u[n − 1] or, by making the partial fraction expansion, Ha (z) = −1/.77 +

1.77/0.77 ⇐⇒ x a [n] = −1/0.77 + (1.77/0.77) (−0.77) n u[n] | {z } | {z } 1 + 0.77z −1 −1.2987

2.2987

(b) By using the delay and linearity properties, hb [n] = (0.9) n u[n] + 0.8(0.9) n−1 u[n − 1] or, by making the partial fraction expansion, Ha (z) = −0.8/0.9 +

1.7/0.9 ⇐⇒ x a [n] = −0.8/0.9 + (1.7/0.9) (0.9) n u[n] | {z } | {z } 1 − 0.9z −1 −0.8889

1.8889

(c) Using the delay property, hc [n] = (0.9) n−2 u[n − 2] (d) Using the transform of an impulse and the delay property, hd [n] = δ[n] − δ[n − 1] + 2δ[n − 3] − 3δ[n − 4]

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CHAPTER 10. IIR FILTERS P-10.13

DSP First 2e

(a) Using residuez or hand calculation, determine the partial fraction expansion of H (z) as X a (z) =

1 − z −1 (1 − 21 z −1 )(1 + 31 z −1 )

=

−0.6 1 − 21 z −1

+

1.6 1 + 31 z −1

Therefore, x a [n] = −0.6( 21 ) n u[n] + 1.6(− 31 ) n u[n] (b) Determine the partial fraction expansion as 1 + z −2 (1 − 0.9e j2π/3 z −1 )(1 − 0.9e− j2π/3 z −1 ) 0.6556e j1.7506 0.6556e− j1.7506 = 1.2346 + + 1 − 0.9e j2π/3 z −1 1 − 0.9e− j2π/3 z −1

X b (z) =

Now the inverse z-transform is by table look up, x b [n] = 1.2346 δ[n] + 0.6556e j1.7506 (0.9e j2π/3 ) n u[n] + 0.6556e− j1.7506 (0.9e− j2π/3 ) n u[n] = 1.2346 δ[n] + 1.3112(0.9) n cos((2π/3)n + 1.7506)u[n] (c) The partial fraction expansion is X c (z) =

1 + z −1 1.1176 −0.1176 = + (1 − 0.9z −1 )(1 + 0.8z −1 ) 1 − 0.9z −1 1 + 0.8z −1

so the inverse transform is x c [n] = 1.1176(0.9) n u[n] − 0.1176(−0.8) n u[n]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.14

S1 : S2 :

DSP First 2e

y[n] = 0.9y[n − 1] + 21 x[n] + 21 x[n − 1] #2 y[n] = −0.9y[n − 1] + 9x[n] + 10x[n − 1] #3

S4 :

1 (1 − z −1 ) H (z) = 2 #6 1 + 0.9z −1 y[n] = 41 x[n] + x[n − 1] + 23 x[n − 2] + x[n − 3] + 41 x[n − 4] not used

S5 :

H (z) = 1 − z −1 + z −2 − z −3 + z −4

S3 :

S6 :

y[n] =

2 X

#5

x[n − k] #4

k=0

S7 :

y[n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4] + x[n − 5] #1

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CHAPTER 10. IIR FILTERS P-10.15

DSP First 2e

(a) By inspection, H (z) =

1 1 − 21 z −1

. Therefore,

! −1 1 2 1 * + = Y (z) = H (z)X (z) = + −1 1 −1 1 −1 1 − z −1 1 − 2z ,1 − 2 z - 1 − z Therefore, y[n] = 2u[n] − ( 12 ) n u[n] = (2 − ( 21 ) n )u[n]. (b) Use a partial fraction expansion: 1 1 1 − jπ/4 1 − 2e jπ/4 = 1 − 2e + 1 −1 1 − e jπ/4 z −1 1 − 2z

1 1 + Y (z) = H (z)X (z) = * jπ/4 1 −1 z −1 ,1 − 2 z - 1 − e !

1 1 + e j (π/4)n u[n] Therefore, y[n] = ( 21 ) n u[n] + * jπ/4 1 − jπ/4 1 − 2e e 1 − , 2 | {z }!

H (e j π/4 )

(c) Note that H (e j ω̂ ) = y[n] =

1 1 − 21 e− j ω̂

so the second term in y[n] in part (b) is the steady-state component.

!

1 ( 1 ) n u[n] + H (e jπ/4 )e j (π/4)n u[n] | {z } 1 − 2e jπ/4 2 steady state

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CHAPTER 10. IIR FILTERS P-10.16

DSP First 2e

(a) Multiply out the factors in numerator and denominator: H (z) = Therefore, the difference equation is

1 − 0.1756z −1 − 0.1756z −2 + z −3 1 + 0.4944z −1 + 0.64z −2

y[n] = −0.4944y[n − 1] − 0.64y[n − 2] + x[n] − 0.1756x[n − 1] − 0.1756x[n − 2] + x[n − 3] (b) Get the poles and zeros from the given factored form of H (z); then plot in the complex z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-10.16 (c) H (z) has zeros on the unit circle at ω̂ = 0 and ω̂ = ±0.3π. Therefore, any complex exponential with one of these frequencies will have zero output.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.17

S1 : S2 :

DSP First 2e

y[n] = 0.9y[n − 1] + 21 x[n] + 21 x[n − 1] #1 y[n] = −0.9y[n − 1] + 9x[n] + 10x[n − 1] #4

S4 :

1 (1 − z −1 ) H (z) = 2 #2 1 + 0.9z −1 y[n] = 41 x[n] + x[n − 1] + 23 x[n − 2] + x[n − 3] + 41 x[n − 4] #6

S5 :

H (z) = 1 − z −1 + z −2 − z −3 + z −4

S3 :

S6 :

y[n] =

2 X

#5

x[n − k] #3

k=0

S7 :

y[n] = x[n] + x[n − 1] + x[n − 2] + x[n − 3] + x[n − 4] + x[n − 5]

not used

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CHAPTER 10. IIR FILTERS P-10.18

DSP First 2e

(A) #2 (B) #4 (C) #1 (D) #6 (E) #3

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CHAPTER 10. IIR FILTERS P-10.19

DSP First 2e

(a) Expand the numerator and denominator to get H (z) =

1 − z −1 + z −2 1 − 2 cos(π/3)z −1 + z −2 = 1 − 2r cos(π/3)z −1 + r 2 z −2 1 − r z −1 + r 2 z −2

Therefore, the difference equation is y[n] = r y[n − 1] − r 2 y[n − 2] + x[n] − x[n − 1] + x[n − 2]. (b) Plot the poles and zeros of H (z) for the case r = 0.9. Angles to poles and zeros are ±π/3.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

|H(ej ω̂ )|

(c) Magnitude of the frequency response plotted for the case r = 0.9. Notches are at ±π/3.

1.5

1

0.5 0 −π

−0.8π −0.6π −0.4π −0.2π

0

0.2π

0.4π

0.6π

0.8π

π

ω̂

(d) The notches get sharper and the magnitude at other frequencies approaches 1. (e) Frequency = Angle. Change the angle of the poles and zeros from π/3 to whatever frequency is desired.

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CHAPTER 10. IIR FILTERS P-10.20

DSP First 2e

(J) #6 (K) #5 (L) #4 (M) #2 (N) #3

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CHAPTER 10. IIR FILTERS P-10.21

DSP First 2e

−0.8 + z −1 z − 1.25 = (−0.8) z − 0.8 1 − 0.8z −1 (b) The pole is at 0.8 and the zero is at 1.25 = 1/0.8. (a) H (z) =

(c) |H (e j ω̂ )| 2 = H (e j ω̂ )H ∗ (e j ω̂ ) =

(−0.8 + e− j ω̂ )(−0.8 + e j ω̂ ) 1.64 − 1.6 cos(ω̂) =1 = 1.64 − 1.6 cos(ω̂) (1 − 0.8e− j ω̂ )(1 − 0.8e j ω̂ )

(d) This results because the magnitude response is unity for all ω̂. Only the phases will change. (e) By evaluating H (e j ω̂ ) at the two frequencies (with Matlab), we get ϕ1 = −2.4826 and ϕ2 = −3.0285.

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CHAPTER 10. IIR FILTERS P-10.22

DSP First 2e

If we sample with f s = 4000, the input to the LTI system is x[n] = 9 + 8 cos(0.3πn) + 7 cos(0.15πn − 0.7π) so we need to evaluate the frequency response at ω̂ = 0, 0.3π, 0.15π. The frequency response is obtained by substituting z = e j ω̂ into the expression for H (z) and then evaluating at the required frequencies.

Since H (z) has zeros on the unit circle at ω̂ = ±0.3π, the component at that frequency is nulled out. Therefore, the output of the LTI system is y[n] = 9H (e j0 ) + 7|H (e j0.15π )| cos(0.15πn − 0.7π + ∠H (e j0.15π )) y[n] = 9(0.7725) + 7(0.6010) cos(0.15πn − 0.7π − 0.3190)) y[n] = 6.9525 + 4.2070 cos(0.15πn − 0.7π − 0.3190)) Now use the sampling frequency to write the output of the D-to-C converter as y(t) = 6.9525 + 4.2070 cos(600πt − 0.7π − 0.3190))

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CHAPTER 10. IIR FILTERS P-10.23

DSP First 2e

(a) Plot the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

(b) The frequency response is constant for all ω̂ at (1.25) 2 = 1.5625. The vector lengths should cancel out to a constant. |H(ej ω̂ )|

1.5625 1.5625 1.5625 −π

0

-5

5

π

ω̂

(c) The zeros of an all-pass filter are at reciprocals of the poles, so the angle stays the same but the radius of the zeros must be decreased to 1/0.9 =1.1111.

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CHAPTER 10. IIR FILTERS P-10.24

DSP First 2e

(a) Expand H (z) to H (z) =

(1 + z −2 ) (1 + z −2 ) = (1 − 0.9z −1 )(1 + 0.5z −1 ) 1 − 0.4z −1 − 0.45z −2

Therefore, the difference equation is y[n] = 0.4y[n − 1] + 0.45y[n − 2] + x[n] + x[n − 2]. (b) Plot the poles and zeros of H (z) in the z-plane.

Imaginary Part

1 0.5 0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

(c) The pole at z = 0.9 dominates around ω̂ = 0 and the zeros pull down the frequency response around ω̂ = ±π/2.

The pole at z = −0.5 is symmetrical with the other pole, but much farther away from the unit circle. Therefore the maximum value occurs at ω̂ = 0.

The estimate of the maximum value is obtained as the ratio of the product of the lengths of the vectors from the zeros √ to z = 1 divided by the product of the lengths of the vectors from the poles to z = 1. The maximum value is ( 2) 2 /(0.1 × 1.5) = 13.33. (d) At ω̂ = ±π/2, |H (e j ω̂ )| = 0.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.25

DSP First 2e

(a) This system has a pole at z = −0.8 and a zero at z = 0.9. (b) |H (e j ω̂ )| 2 = H (e j ω̂ )H ∗ (e j ω̂ ) =

(1 − 0.9e− j ω̂ )(1 − 0.9e j ω̂ ) 1.81 − 1.8 cos(ω̂) = (1 + 0.8e− j ω̂ )(1 − 0.8e j ω̂ ) 1.64 + 1.6 cos(ω̂)

(c) This is a highpass filter since it has a zero near z = 1 or ω̂ = 0 and a pole near z = −1 or ω̂ = π.

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CHAPTER 10. IIR FILTERS P-10.26

DSP First 2e

! !

15 10 + 5z −1 3 15z = = z − 0.7 2 + z −1 1 − 0.7z −1 1 − 0.7z −1 Therefore, H (z) has a zero at z = 0 and a single pole at z = 0.7.

(a) H (z) = H1 (z)H2 (z) =

(b) Since H1 (z) =

H (z) =

1 + 21 z −1

1 − 21 z −1

, it follows that

3z −1 (1 + 21 z −1 )

(1 − 21 z −1 )(1 − 56 z −1 + 61 z −2 )

=

3z −1 + 23 z −2

7 −2 1 −3 z − 12 z 1 − 34 z −1 + 12

7 1 (c) y[n] = 34 y[n − 1] − 12 y[n − 2] + 12 y[n − 3] + 3x[n] + 32 x[n − 2 − 1]

(d) For the systems in (b), plot the poles and zeros of H (z) in the complex z-plane.

Imaginary Part

1 0.5 2

0 -0.5 -1 -1

-0.5

0 0.5 Real Part

1

Figure P-10.26

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CHAPTER 10. IIR FILTERS P-10.27

DSP First 2e

(a) Since we require H (z) = H1 (z)H2 (z) = 1 for the output to equal the input, then H2 (z) = 1/H1 (z). (b) Since H1 (z) = 1 − 34 z −1 , H2 (z) =

1 1 = H1 (z) 1 − 43 z −1

(c) Since the poles of H2 (z) must be the zeros of H1 (z), both the poles and the zeros of H1 (z) must be inside the unit circle of the z-plane.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.28

DSP First 2e

(a) Here is a stem plot of y[n] versus n, over the range 0 ≤ n ≤ 10.

y[n]

1 0 -1 0

5

10

15 20 time index n Figure P-10.28

25

30

(b) We need to find a system whose impulse response is h[n] = (0.95) n cos(0.2πn + 0.4π)u[n]. Do this by finding the corresponding system function, writing it as a ratio of polynomials in z −1 , and picking off the coefficients of the corresponding difference equation.

First write h[n] as h[n] = 21 e j0.4π (.95e j0.2π ) n u[n] + 21 e− j0.4π (.95e− j0.2π ) n u[n] The corresponding z-transform is H (z) =

1 − j0.4π 1 j0.4π 2e 2e + 1 − 0.95e j0.2π z −1 1 − 0.95e− j0.2π z −1

If we put this over a common denominator and collect terms using the inverse Euler formula, we get H (z) =

cos(0.4π) − 0.95 cos(0.2π)z −1 0.3090 − 0.7686z −1 = −1 2 −2 1 − 1.9 cos(0.2π)z + (0.95) z 1 − 1.5371z −1 + 0.9025z −2

The difference equation for implementing this system is therefore, y[n] = 1.5371y[n = 1] − 0.9025y[n − 2] + 0.3090x[n] − 0.7686x[n − 1]

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.29

DSP First 2e

(a) H3 (z) =

b0 −1 1 − a1 z − a2 z −2 − a3 z −3

(b) In general, the system function overall is 2b0 z −1 (1 − z −2 )(1 + 21 z −1 )

H (z) = H3 (z)H2 (z)H1 (z) = =

1 − a1 z −1 − a2 z −2 − a3 z 2b0 z −1 (1 + 12 z −1 − z −2 − 21 z −3 )

= −3

2b0 z −1 (1 − z −2 )(1 + 12 z −1 ) 1 − a1 z −1 − a2 z −2 − a3 z −3

1 − a1 z −1 − a2 z −2 − a3 z −3

For the given coefficients, we get H (z) =

14z −1 (1 − z −2 )(1 + 21 z −1 ) 1 − 21 z −1 − 21 z −2

(c) From the given information, we want H (z) = 8z −n d . Since we have a third-order polynomial in both the numerator and denominator, we just need to adjust the coefficients of the denominator so that the two polynomials are identical and therefore cancel. The required values are b0 = 4

a1 = − 21

a2 = 1

a3 = 12

For these conditions, nd = 1.

July 29, 2015 © 2022 Pearson Education.

CHAPTER 10. IIR FILTERS P-10.30

DSP First 2e

(a) yy=filter([0,0,1,0,1],1,xx);

ANS =1

(b) y[n] = 0.9y[n − 1] + 0.8y[n − 2] + x[n] − x[n − 3] (c) h[n] = u[n − 2] − u[n − 5] (d) y[n] =

3 √ X

k − 1 x[n − k]

ANS =3

ANS =4 ANS =5

k=1

(e) h[n] = 0.8n−2 u[n − 2] − 0.8n−3 u[n − 3]

ANS =2

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.1

DSP First 2e

(a) Cartesian: z = 0 + j2.

Polar: 2e jπ/2 = 2∠90◦

◦ (b) Cartesian: z = −3 − j4. Polar: 5e− j2.21 = 5e− j0.705π = 5 | ∠ − 126.87 {z } θ

(c) Cartesian: z = −1 + j; Polar:

√

2e j3π/4 = 2 |{z} ∠135◦ θ

(d) Cartesian: z = 0 − j; Polar: 1e− jπ/2 = 1 | ∠− 90}◦ {z θ

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.2

(a)

DSP First 2e

√

√ 2 e j (3π/4) = 2(cos(3π/4) + j sin(3π/4) ) = −1 + j | {z } | {z } √ =− 2/2

√ = 2/2

(b) z = 3e− j (π/2) = 3∠ − 90◦ ⇒ z = −3 j

(c) 1.6 ∠ (π/6) = 1.6e jπ/6 = 1.6(cos(π/6) + j sin(π/6) ) = 1.386 + j0.8 |{z} | {z } | {z } 30◦

√ =− 3/2

=0.5

(d) 7∠ (7π) = 7∠π = 7e jπ = −7 + j0 |{z} 1260◦

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.3

DSP First 2e

(a) j 3 = j ( j) 2 = j (−1) = − j = 0 − j  +1 n  (b) j 2n = j 2 = (−1) n =   −1 

n even n odd

(n an integer)

(c) e j (π+2πm) = e jπ e j2πm = (−1)(1) = −1 + j 0 (m an integer)

1/2 √ √ (this is one answer) (d) j 1/2 = e jπ/2 = e jπ/4 = 2/2 + j 2/2 1/2

√ √ = e jπ/4 |{z} e jπ = −e jπ/4 = − 2/2 − j 2/2 (this is the second answer) j 1/2 = e jπ/2 e j2π =−1

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.4

DSP First 2e

√

Use this fact in (a,b,c): 2 − j2 = 2.449e− j0.304π (Angle is −0.955 rad, or −54.74◦ ) 8

√ (a) ( 2 − j2) 8 = 2.449e− j0.304π = (2.449) 8 e− j8(.304π) = 1296e j0.433π (Angle is −1.36 rad, or −77.88◦ ) √ (b) ( 2 − j2) −1 = 1/2.449e+ j0.304π = 0.4082e+ j0.304π (Angle is +1.36 rad, or +77.88◦ )

1/2 √ (c) ( 2 − j2) 1/2 = 2.449e− j0.304π e j2π` = 1.565e− j0.152π e jπ` ` = 0, 1 There are two answers: 1.565e− j0.152π and 1.565e j0.848π (d) Add the two complex numbers below and obtain −4.964 + j4.598 = 6.766e j0.762π (Angle is 2.394 rad, or 137.2◦ ) √ 3e j2π/3 = −3/2 + j3 3/2 √ −4e− jπ/6 = −4 3/2 + j2 ( ) ( ) ( ) (e) = je− jπ/3 = = e jπ/2 e− jπ/3 = = e jπ/6 = sin(π/6) = 0.5

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APPENDIX A. COMPLEX NUMBERS P-A.5

DSP First 2e

(a) z1∗ = (−4 + j3) ∗ = −4 − j3 = 5e− j0.795π = 5e j2.50

(b) z22 = (1 − j) 2 = 1 − 2 j + (− j) 2 = 1 − 2 j = 1 = −2 j 2 √ or, in polar form: z22 = 2e− jπ/4 = 2e− jπ/2 = − j2 (c) z1 + z2∗ = (−4 + j3) + (1 − j) ∗ = (−4 + j3) + (1 + j) = 3 + j4 = 5e j0.705π (Angle is 2.214 rad, or 126.872◦ ) (d) j z2 = j (1 − j) = j − j ( j) = j + 1 = 1 + j √ √ or, in polar form: e jπ/2 2e− jπ/4 = 2e jπ/4 (Angle is 0.785 rad, or 45◦ ) (e) Use polar form: z1−1 = 1/(5e j0.795π ) = 0.2e− j0.795π √ √ (f) Use polar form: z1 /z2 = (5e j0.795π )/( 2e− jπ/4 ) = (5 2/2)e j1.045π = 3.54e− j0.955π (Angle is −3.0 rad, or −171.8◦ ) (g) e z 2 = e1− j = e1 e− j = e(cos(1) + j sin(1)) = 1.469 − j2.287 Polar form is e1 e− j = 2.7183e− j0.318π

(angle is −1 rad, or −57.3◦ )

(h) z1 z1∗ = (−4 + j3)(−4 + j3) ∗ = (−4 + j3)(−4 − j3) = 16 + 9 = 25 √ √ (i) Use polar form: z1 z2 = (5e j0.795π )/( 2e− jπ/4 ) = 5 2e j0.454π = 7.07e j1.713

(98.13◦ )

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APPENDIX A. COMPLEX NUMBERS P-A.6

DSP First 2e

z1 = e j9π/3 = e j3π = −1 + j0 z2 = e− j5π/8 = −0.383 − j0.924 z3 = e j13π/8 = e j13π/8 = 0.383 − j0.924z = −1 − j1.848 = 2.1e− j2.07

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APPENDIX A. COMPLEX NUMBERS P-A.7

DSP First 2e

(a) z = −3 + j4 = 5 e j0.705π 1/z = (1/5)e− j0.705π The angle is also 2.214 rad, or 126.87◦ . √ (b) z = −2 + j2 = 2 2 e j3π/4 √ 5 √ z 5 = 2 2 e j15π/4 = 128 2 e− jπ/4 because 15π/4 = 4π − π/4. (c) z = −5 + j13 |z| 2 = zz ∗ = (−5 + j13) (−5 − j13) = 25 + 169 = 194 = 194 e j0

(d) < (−2 + j5) |{z} e− jπ/2 = < (−2 + j5)(− j) = < − 2 j + 5 = 5 = 5e j0 =− j

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APPENDIX A. COMPLEX NUMBERS P-A.8

DSP First 2e

Recall that e j2π` = 1 when ` = integer. z 4 = j = e jπ/2 = e jπ/2 e j2π` 1/4

= e jπ/8 e jπ`/2 ; ` = 0, 1, 2, 3 ⇒ z = e jπ/2 e j2π`

z ∈ e jπ/8, e j5π/8, |{z} e j9π/8, | e j13π/8 {z } e j π/8

e − j3π/8

There are four different answers shown as the four vectors below:

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.9

DSP First 2e

z0 = e j2π/N ⇒ (z0 ) N −1 = e j2π/N

N −1

* 1− j2π/N

z0N −1 = e j2π(N −1)/N = e j2π(N )/N e j2π (−1)/N = e j2π e

−1 z0N −1 = e− j2π/N = e j2π/N = z0−1 = 1/z0

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APPENDIX A. COMPLEX NUMBERS P-A.10

DSP First 2e

Multiply by 1 = e j2π` which is true when ` = integer; this will generate all possible solutions. − j = 1e− jπ/2 = e j2π` e− jπ/2 1/2

⇒ (− j) 1/2 = e j2π` e− jπ/2   e− jπ/4 = e jπ` e− jπ/4 =   e jπ e− jπ/4 = e j3π/4   1 Because e jπ` =   e jπ 

when ` is even when ` is odd

when ` is even when ` is odd

Thus, there are two solutions (shown below)

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APPENDIX A. COMPLEX NUMBERS P-A.11

DSP First 2e

Use a calculator to check these √ 34 e− j1.03 = 0.707e− j1.275 z1 = √ 68 e j0.245 z2 = 0.25 e jπ/2 z3 = e j5π/6 = e j2.618

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.12

DSP First 2e

When N = 56, we can write N −1 X e j2πk /N + e j2π(N −1)/N = e j2π(55)/56 = e− jπ/28 = e− j0.1122 z=− k=0

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.13

p = 32

DSP First 2e

and

√ q = 200 3 = 346.4102

Approach: Express the right-hand side (RHS) in Cartesian form using Euler’s formula. Then cross multiply by the LHS denominator to get a single complex-valued equation.

√

p = 200 + jq (0.08) −0.5 + j 3/2 The real and imaginary parts yield two (linear) equations in the two unknowns, p and q. One of these equations has only p, or only q, so solve it and then substitute into the other equation.

July 29, 2015 © 2022 Pearson Education.

APPENDIX A. COMPLEX NUMBERS P-A.14

DSP First 2e

Definite integral is √ re jθ =

e jπt jπ

1.5 1

2 − j3π/4 e = 0.4502 e− j2.3562 . π

Approach: The integral of an exponential is an exponential, but you end up with a j in the denominator because the exponent contains a j. After evaluating at the limits of the definite integral, the numerator contains a complex number.

Finally, convert the complex numerator-denominator into polar form with a calculator.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.1

DSP First 2e

(a) Sketch the periodic function x(t) over the time interval −2T0 < t < 2T0 for the case t c = T0 /4. x(t)

1

−T0

−2T0 1 (b) a0 = T0

T0 /2 Z −T0 /2

1 x(t)dt = T0

− 14 T0 Zt c /2

0

T0

1 4 T0

2T0

t

2t c T0

dt =

−t c /2

(c) Determine a formula for the Fourier series coefficients ak in the finite Fourier representation (??). Your final result should depend on t c and T0 .

1 ak = T0

T0 /2 Z

x(t)e −T0 /2

− j (2π/T0 )k t

1 dt = T0

Zt c /2

e− j (2π/T0 )k t dt =

−t c /2

sin((2πt c /T0 )k) πk

k = ±1, ±2, . . .

(d) See plot below ak

tc = T0 /4

(d)

8f0

f

tc = T0 /10

(e)

8f0

f

1 4

–8f0

–4f0

0

4f0

ak 1 10

(e)

–8f0

–4f0

0

4f0

(f) We see by comparing the figures that the high-frequency components get bigger relative to the DC component as t c decreases. That is, very narrow pulses require much more bandwidth for an accurate representation by a finite Fourier series.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.2

DSP First 2e

(a) Plot the periodic signal x(t) for −T0 ≤ t ≤ T0 . x(t)

1

− 21 T0

−T0

1 2 T0

T0

t

−1 Figure P-C.2

1 (b) a0 = T0

Z0 −T0 /2

1 x(t)dt + T0

T0 /2 T0 /2 T0 /2 Z Z Z 1 1 x(t)dt = x(τ + T0 /2)dτ + x(t)dt = 0 T0 T0 0

0

0

where we have made the substitution t = T0 /2 + τ in the first integral. (c) The same manipulation can be made on the general Fourier series coefficient integral.

1 ak = T0 1 = T0

1 = T0

Z0 x(t)e

− j (2π/T0 )k t

−T0 /2 T0 /2 Z

x(τ + T0 /2)e

0

1 dt + T0

T0 /2 Z x(t)e− j (2π/T0 )k t dt 0

− j (2π/T0 )k (τ+T0 /2)

1 dτ + T0

T0 /2 Z x(t)e− j (2π/T0 )k t dt 0

 0  T0 /2  Z  T0 /2  Z  − jπk − j (2π/T0 )k t 2 x(t)(1 − e )e dt =   x(t)e− j (2π/T0 )k t dt     T0 0 0 

k even k odd

Thus, when we have half-wave symmetry, the odd-indexed coefficients can be found by integrating over only half the period.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.3

DSP First 2e

( (a) x(t) =

4+t 0

−4 ≤ t < 0 0≤t<4

(b) Here is a plot of x(t) over the range −8 ≤ t ≤ 8 s. x(t)

4

−8

−4

4

8

t

Figure P-C.3

(c) The DC value is just the average value of the waveform so a0 =

area under curve 8 = = 1.

8 8

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.4

DSP First 2e

(a) A sketch of the periodic function x(t) over the time interval −10 ≤ t ≤ 20. x(t)

2

−5

−10

5

10

t

Figure P-C.4 (b) The DC coefficient: a0 = 1 (c) a1 = 10

Z10

area under one period = (2 × 5)/10 = 1 10

2e− j (2π/10)t dt =

5

1 (e− j (2π/10)10 − e− j (2π/10)5 ) = j/π −10 j (2π/10)

(d) Adding a constant only changes the DC coefficient. b0 = 1 + a0 and b1 = a1 .

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.5

DSP First 2e

(a) Because each of the terms ak e j k t = of t.

3 e j k t is paired with its complex conjugate ak∗ e− j k t giving a real function 2 + jk

(b) Define ω0 = 1 and ak = 3/(2 + j k) then x(t) =

3 X

(− jak /2)e j (k+7)t +

k=−3

3 X

( jak /2)e j (k−7)t =

k=−3

10 X

bk e j k t

k=−10

− j1.5   4 ≤ k ≤ 10    2 + j (k − 7)   j1.5 where bk =   −10 ≤ k ≤ −4    2 + j (k + 7)   otherwise  0 Note that b−k = b∗k as required for real x(t).

b−9

b b−8 −7 b−6

b−10

−10

b−5

−7

b−4

−4

b4

0 Figure P-C.5

4

b5

b6

b7

7

b8

b9

b10

10

ω

(c) See the explanation in part (b).

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.6

DSP First 2e

(a) Define ak =

j5 j5 . Each of the terms ak e j8πk t = e j8πk t is paired with its complex conjugate k + j1.5 k + j1.5

ak e j8πk t

∗

=

j5 − j5 − j8πk t e = e j8π(−k )t k − j1.5 (−k) + j1.5

for k = 1, 2 giving a real function of t for the sum. Multiplying by cos(3πt) is still real. (b) The frequencies in q(t) are ±5π, ±11π, ±13π, ±19π. Note that the frequencies ±3π have zero amplitude since a0 = 0. The fundamental frequency is determined by the gcd of 5,11,13,19, which is 1, so ω0 = π. (c) The DC value is zero. There is no component at ω = 0. (d) Table listing all the nonzero Fourier coefficients of the signal q(t). ωk −19π −13π −11π −5π 5π 11π 13π 19π bk a−2 /2 a−2 /2 a−1 /2 a−1 /2 a1 /2 a1 /2 a2 /2 a2 /2 (e) Spectrum plot showing only the positive frequency components. Note: ak =

0

a1 /2

a1 /2

5π

11π Figure P-C.6

j5 . k + j1.5

a2 /2

a2 /2

13π

19π ω

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APPENDIX C. FOURIER SERIES P-C.7

DSP First 2e

 k=0   0.5 sin(πk/2) (a) ak =   k,0  πk  (b) Sketch of the waveform y(t). y(t) = 2x(t − T0 /2)

2

−2T0

−T0

− 14 T0

0

1 4 T0

T0

2T0

t

Figure P-C.7

   1 bk = 2ak e− j (2π/T0 )kT0 /2 = 2ak e− jπk =   2e− jπk sin(πk/2)  πk 

k=0 k,0

Note that both ak and bk are zero for k a nonzero even integer. This is because both x(t) − 12 and y(t) − 1 have half-wave symmetry.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.8

DSP First 2e

(a) Plot of x(t) over the interval −10 ≤ t ≤ 10 s with a period of 8 s. x(t)

7

3 −4

−8 (b) a0 =

4

8

t

area under one period 7 × 5 + 3 × 3 44 = = = 5.5 8 8 8

1 (c) ak = 8

Z5

7e

− j (2π/8)k t

0

1 dt + 8

Z8

3e− j (2π/8)k t dt

5

(d) 1 ak = 8

Z5

7e

0

− j (2π/8)k t

1 dt + 8

Z8 5

3e− j (2π/8)k t dt =

7(1 − e− j (π/4)k5 ) 3(e j (π/4)k5 − e− j (π/4)k8 ) + j (π/4)k8 j (π/4)k8

=

4(1 − e− j (π/4)k5 ) = e− j5π j2πk

(e) Clearly, every eighth coefficient is zero since sin(5πk/8) = 0 for k a multiple of 8.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.9

DSP First 2e

The Fourier series coefficients for the half-wave rectified sine wave are: 1        π(1 − k 2 ) ak =   ∓ j/4      0 

k even k = ±1 k odd and ≥ 3

The fundamental period is T0 = 0.02 s and therefore the fundamental frequency is f 0 = 1/0.02 = 50 Hz. Therefore the spectrum is 1 π

−j 4

−1 3π

0

50

100

−1 15π

−1 35π

−1 63π

−1 99π

200

300

400

500

f

The Fourier series coefficients for the full-wave rectified sine wave are: 2      π(1 − 4k 2 ) ak =      2  π

k even k=0

The fundamental period is T0 = T1 /2 = 0.01 s and therefore the fundamental frequency is f 0 = 1/0.01 = 100 Hz. The spectrum for the full-wave rectified sine wave is: 2 π

−2 3π

0

100

−2 15π

−2 35π

−2 63π

−2 99π

200

300

400

500

f

In both cases, the negative-frequency Fourier series coefficients are complex conjugates of the positive-frequency coefficients.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.10

DSP First 2e

(a) Sketch of q(t) = p(t/5). The period is T0 = 5 × 0.1 = 0.5s and the pulse width is 5 × 0.04 = 0.2s. p(t)

1 −.1

−.05 −.02 0 .02

.05

.1

t

.5

t

q(t) = p(t/5)

1 −.25 −.1 0 .1 .25 Figure P-C.10

−.5

1 (b) ak = .1

Z.02

.02

e− j20πk t dt = 10

−.02

1 bk = .5

Z.1

sin(0.4πk) e− j20πk t = − j20πk −.02 πk .1

e− j4πk t dt = 2

−.1

sin(0.4πk) e− j4πk t = − j4πk −.1 πk

(c) Make the change of variable u = αt, and dt = du/α, in bk 1 bk = T0 /α

ZT0 x(u)e 0

− j (2π/T0 )k u

1 du/α = T0

ZT0

x(u)e− j (2π/T0 )k u du = ak

0

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.11

DSP First 2e

(a) One-sided spectrum for x(t). The negative frequency components are complex conjugates.

j12

j6 j3

0

3

6

j2

12

18

f

Figure P-C.11 (b) Since 24 sin(6πt) = − j12e j6πt + j12e− j6πt it follows that the k = ±1 terms in x(t) are cancelled. Therefore, these components of the spectrum are missing in the spectrum of y(t). All other components are the same as in the spectrum of x(t). (c) The original 3 Hz fundamental is is not present. Now the fundamental is the gcd of 6, 12, 18 or f 0 = 6 Hz. (d) For x(t) with f 0 = 3 Hz and the Fourier coefficients are ak = {0, ± j12, ± j6, 0, ± j3, 0, ± j2} Thus y(t) has spectrum lines at {±6, ±12, ±18} hertz. The true fundamental frequency from the GCD is f 0 = 6 Hz. With respect to 6 Hz the formula for x(t) consists of the first, second and third harmonics. The formula for the relabeled Fourier coefficients is   j6/k bk = a2k =  0 

for k = ±1, ±2, ±3 elsewhere

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.12

DSP First 2e

(a) T0 = 4 s (b) f 0 = 1/4 = 0.25 Hz area under one period (2 × 1/2) + (3 × 1) − (2 × 1/2) = = 0.75 4 4 (d) The simplest expression would use symmetric limits: (c) a0 =

1 ak = 4

Z−1

3e

−2

− jπk t /2

1 dt + 4

Z1

2te− jπk t /2 dt

−1

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.13

DSP First 2e

(a) T0 = 100 s (b) a3 = 5e− j0.644 , a−3

= 5e j0.644 ,

(c) b3 = 3e− jπ/2 , b−3 = 3e j1.571 ,

a2 = 0,

a1 = 2.236e− j0.464 , √ a−2 = 0, a−1 = 5e j0.464

a0 = 1,

b2 = 0,

b0 = 2.2e jπ ,

b1 = 2.236e− j0.464 , √ b−2 = 0, b−1 = 5e j0.464

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.14

DSP First 2e

(a) N-th harmonic is N = 3. (b) Since 3F0 = 2000, it follows that F0 = 2000/3 = 667 Hz (c) The signal x 2 (t) is the square wave since its Fourier series coefficients are ak =

sin(πk/2) πk

k = ±1, ±2, . . .

which are zero at the even harmonics.

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.15

DSP First 2e

(a) First, the fundamental frequenty of s(t) = | sin(2πt)| is F0 = 2 Hz. The FWRC can be expressed as x(t) = | cos(2πt)| = | sin(2π(t − .25))| = s(t − 0.25) so if ak denotes the Fourier coefficients of s(t), then the Fourier coefficients of x(t) are bk = ak e− j2π(2)(k )0.25 = ak e− jπk = (−1) k ak (b) The half-wave rectified sine is defined in (??) as ( sin((2π/T0 )t) 0 ≤ t < T0 /2 x(t) = 0 T0 /2 ≤ t < T0 where the fundamental period is T0 . Therefore, the FWRS can be expressed as s(t) = x(t) + x(t − T0 /2) i.e., we fill in the gaps in the HWRS by adding in a delayed version. Now if bk denotes the Fourier coefficients of the HWRS, the additive and delay properties give us ak = bk + bk e− j (2π/T0 )T0 /2 = (1 + e− jπk )bk Note that the fundamental frequency of the HWRS x(t) is 1/T0 but when we delay and add a copy the resulting FWRS has fundamental frequency 2/T0 . This is confirmed by observing that (1 + e− jπk )bk = 0 for k odd, which zeroes out the odd harmonics.

Since the odd harmonics are missing, the fundamental frequency is twice the fundamental of the HWRS. Thus, if we reindex the coefficients, we get ak = b2k (1 + e− j2πk ) = 2b2k . Recall the Fourier coefficients of the HWRS given in (??), which are 1      π(1 − k2)  bk =  1 ∓ j   4   0 

for k even for k = ±1 for k = ±3, ±5, . . .

Finally, we can write ak = b2k =

2 π(1 − 4k 2 )

(c) Since (−1) k = e− jπk = e− j (2π/T0 )kT0 /2 , multiplying the Fourier coefficients by (−1) k is equivalent to delaying the waveform by T0 /2. For the pulse wave with period 0.1s and pulse width 0.04s the plots are: p(t)

1 −.2

−.02 0 .02

−.2 −.07

−.03

.1

t

q(t) = p(t − 0.05)

1 −.3

.05

0

.03 .05 .07

.1

.3

t

July 29, 2015 © 2022 Pearson Education.

APPENDIX C. FOURIER SERIES P-C.16

DSP First 2e

d x(t). (a) Sketch of the triangle wave x(t), and its derivative y(t) = dt

x(t) 1 −T0

− T20

0 4 T0

−T0

− T20

T0 4 y(t) = dx(t) dt

0

T0 2

T0

t

T0 2

T0

t

Figure P-C.16 (b) The derivative property gives the result bk = ( j2πk/T0 )ak . (c) The Fourier coefficients of the square wave of amplitude 4/T0 are bk =

4 2 sin(πk/2) 8 sin(πk/2) = T0 πk πkT0

with b0 = 0

(d) Using the derivative property, we can write ( j2πk/T0 )ak =

8 sin(πk/2) πkT0

=⇒

ak = − j

4 sin(πk/2) π2 k 2

for k , 0. Note that a0 = 0, and thatt all even harmonics are zero since the triangle wave has half-wave symmetry.

July 29, 2015 © 2022 Pearson Education.