Fourier and Laplace Transforms 1 edition (Solution Manual, Solutions) R. J. Beerends, H. G. ter Mors

SOLUTION MANUAL

Answers to selected exercises for chapter 1

1.1

Apply cos(α + β) = cos α cos β − sin α sin β, then f1 (t) + f2 (t) = A1 cos ωt cos φ1 − A1 sin ωt sin φ1 + A2 cos ωt cos φ2 − A2 sin ωt sin φ2 = (A1 cos φ1 + A2 cos φ2 ) cos ωt − (A1 sin φ1 + A2 sin φ2 ) sin ωt = C1 cos ωt − C2 sin ωt, where C1 = A1 cos φ1 + A2 cos φ2 and C2 = A1 sin φ1 + A2 sin φ2 . Put A = p C12 + C22 and take φ such that cos φ = C1 /A and sin φ = C2 /A (this is possible since (C1 /A)2 +(C2 /A)2 = 1). Now f1 (t)+f2 (t) = A(cos ωt cos φ− sin ωt sin φ) = A cos(ωt + φ).

1.2

Put c1 = A1 eiφ1 and c2 = A2 eiφ2 , then f1 (t) + f2 (t) = (c1 + c2 )eiωt . Let c = c1 + c2 , then f1 (t) + f2 (t) = ceiωt . The signal f1 (t) + f2 (t) is again a time-harmonic signal with amplitude | c | and initial phase arg c.

1.5

The power P is given by Z π/ω Z ω A2 ω π/ω P = A2 cos2 (ωt + φ0 ) dt = (1 + cos(2ωt + 2φ0 )) dt 2π −π/ω 4π −π/ω A2 . = 2

1.6

The energy-content is E =

1.7

The power P is given by

R∞ 0

e−2t dt = 12 .

3

P =

1X | cos(nπ/2) |2 = 12 .

4 n=0 P∞

−2n , which is a geometric series with n=0 e

1.8

The energy-content is E = sum 1/(1 − e−2 ).

1.9

a If u(t) is real, then the integral, and so y(t), is also real. b Since ˛Z ˛ Z ˛ ˛ ˛ u(τ ) dτ ˛ ≤ | u(τ ) | dτ, ˛ ˛ it follows from the boundedness of u(t), so | u(τ ) | ≤ K for some constant K, that y(t) is also bounded. c The linearity follows immediately from the linearity of integration. The follows from the substitution ξ = τ − t0 in the integral Rtime-invariance t u(τ − t ) dτ representing the response to u(t − t0 ).

0 t−1 Rt d Calculating t−1 cos(ωτ ) dτ gives the following response: (sin(ωt) − sin(ωt − ω))/ω = R t2 sin(ω/2) cos(ωt − ω/2)/ω. e Calculating t−1 sin(ωτ ) dτ gives the following response: (− cos(ωt) + cos(ωt − ω))/ω = 2 sin(ω/2) sin(ωt − ω/2)/ω. f From the response to cos(ωt) in d it follows that the amplitude response is | 2 sin(ω/2)/ω |. g From the response to cos(ωt) in d it follows that the phase response is −ω/2 if 2 sin(ω/2)/ω ≥ 0 and −ω/2 + π if 2 sin(ω/2)/ω < 0. From

1

2

Answers to selected exercises for chapter 1

phase and amplitude response the frequency response follows: H(ω) = 2 sin(ω/2)e−iω/2 /ω.

1.11

1.12

a The frequency response of the cascade system is H1 (ω)H2 (ω), since the reponse to eiωt is first H1 (ω)eiωt and then H1 (ω)H2 (ω)eiωt . b The amplitude response is | H1 (ω)H2 (ω) | = A1 (ω)A2 (ω). c The phase response is arg(H1 (ω)H2 (ω)) = Φ1 (ω) + Φ2 (ω). ˛ ˛ √ a The amplitude response is | 1 + i | ˛ e−2iω ˛ = 2. b The input u[n] = 1 has frequency ω = 0, initial phase 0 and amplitude 1. Since eiωn 7→ H(eiω )eiωn , the response is H(e0 )1 = 1 + i for all n. c Since u[n] = (eiωn + e−iωn )/2 we can use eiωn 7→ H(eiω )eiωn to obtain that y[n] = (H(eiω )eiωn + H(e−iω )e−iωn )/2, so y[n] = (1 + i) cos(ω(n − 2)). d Since u[n] = (1 + cos 4ωn)/2, we can use the same method as in b and c to obtain y[n] = (1 + i)(1 + cos(4ω(n − 2)))/2.

1.13

a The power is the integral of f 2 (t) over [−π/ | ω | , π/ | ω |], times | ω | /2π.

Now cos2 (ωt + φ0 ) integrated over [−π/ | ω | , π/ | ω |] equals π/ | ω | and cos(ωt) cos(ωt + φ0 ) integrated over [−π/ | ω | , π/ | ω |] is (π/ | ω |) cos φ0 .

2 2 Hence, the power equals (A R 1 + 2AB cos(φ0 ) + B )/2. b The energy-content is 0 sin2 (πt) dt = 1/2.

1.14

The power is the integral of | f (t) |2 over [−π/ | ω | , π/ | ω |], times | ω | /2π, which in this case equals | c |2 .

1.16

a The amplitude response is | H(ω) | = 1/(1 + ω 2 ). The phase response is arg H(ω) = ω. b The input has frequency ω = 1, so it follows from eiωt 7→ H(ω)eiωt that the response is H(1)ieit = iei(t+1) /2.

1.17

a The signal is not periodic since sin(2N ) 6= 0 for all integer N . iω b The frequency response H(eiω ) equals A(eiω )eiΦe , hence, we obtain iω iω 2 that H(e ) = e /(1 + ω ). The response to u[n] = (e2in − e−2in )/2i is then y[n] = (e2i(n+1) − e−2i(n+1) )/(10i), so y[n] = (sin(2n + 2))/5. The amplitude is thus 1/5 and the initial phase 2 − π/2.

1.18

a If u(t) = 0 for t < 0, then the integral occurring in y(t) is equal to 0 for t < 0. For t0 ≥ 0 the expression u(t − t0 ) is also causal. Hence, the system is causal for t0 ≥ 0. b It follows from the boundedness of u(t), so | u(τ ) | ≤ K for some constant K, that y(t) is also bounded (use the triangle inequality and the inequality from exercise 1.9b). Hence, the system is stable. c If u(t) is real, then the integral is real and so y(t) is real. Hence, the system is real. d The response is Z t y(t) = sin(π(t − t0 )) + sin(πτ ) dτ = sin(π(t − t0 )) − 2(cos πt)/π. t−1

1.19

a If u[n] = 0 for n < 0, then y[n] is also equal to 0 for n < 0 whenever n0 ≥ 0. Hence, the system is causal for n0 ≥ 0. b It follows from the boundedness of u[n], so | u[n] | ≤ K for some constant K and all n, that y[n] is also bounded (use the triangle inequality): ˛ ˛ n n n ˛ X ˛ X X ˛ ˛ u[l] ˛ ≤ K + | u[l] | ≤ K + K, | y[n] | ≤ | u[n − n0 ] | + ˛ ˛ ˛ l=n−2

l=n−2

l=n−2

Answers to selected exercises for chapter 1

3

which equals 4K. Hence, the system is stable. c If u[n] is real, then u[n − n0 ] is real and also the sum in the expression for y[n] is real, hence, y[n] is real. This means that the system is real. d The response to u[n] = cos πn = (−1)n is y[n] = (−1)n−n0 +

n X

(−1)l = (−1)n−n0 + (−1)n (1 − 1 + 1)

l=n−2

= (−1)n (1 + (−1)n0 ).

Answers to selected exercises for chapter 2

2.1

2.2

p √ a The absolute values follow from x2 + y 2 and are given by 2, 2, 3, 2 respectively. The arguments follow from standard angles and are given by 3π/4, π/2, π, 4π/3 respectively. √ √ b Calculating modulus and argument gives 2 + 2i = 2 2eπi/4 , − 3 + i = 5πi/6 3πi/2 2e and −3i = 3e .

In the proof of theorem 2.1 it was shown that | Re z | ≤ | z |, which implies that − | z | ≤ ± | Re z | ≤ | z |. Hence, | z ± w |2 = (z ± w)(z ± w) = zz ± zw ± wz + ww = | z |2 ± 2Re(zw) + | w |2 ≥ | z |2 − 2 | z | | w | + | w |2 = (| z | − | w |)2 .

2.4

This shows that | z ± w |2 ≥ (| z | − | w |)2 . √ We have | z1 | = 4 2, | z√ 2 | = 4 and arg z1 = 7π/4, arg z2 = 2π/3. Hence, | z1 /z2 | = |√ z1 | / | z2 | = 2 and arg(z1 /z2 ) = arg(z1 ) − arg(z2 ) = 13π/12, so z1 /z2 = 2e13πi/12 . Similarly we obtain z12 z23 = 2048e3πi/2 and z12 /z23 = 1 3πi/2 e .

2

2.5

The solutions are given in a separate figure on the website.

2.6

a The four solutions ±1 ± i are obtained by using the standard technique to solve this binomial equation (as in example 2.3).√ b As part a; we now obtain the six solutions 6 2(cos(π/9 + kπ/3) + i sin(π/9 + kπ/3)) where k = 0, 1 . . . , 5. c By completing the square as in example 2.4 we obtain the two solutions −1/5 ± 7i/5.

2.7

4 4 4 Write z 5 − √z + z − 1 as (z − 1)(z + 1) and then solve z = −1 to find the roots 2(±1 ± i)/2. Combining linear factors √ with complex √ conjugate roots we obtain z 5 − z 4 + z − 1 = (z − 1)(z 2 + 2z + 1)(z 2 − 2z + 1).

2.8

Since 2i = 2eπi/2 the solutions are z = ln 2 + i(π/2 + 2kπ), where k ∈ Z.

2.9

Split F (z) as A/(z − 21 ) + B/(z − 2) and multiply by the denominator of F (z) to obtain the values A = −1/3 and B = 4/3 (as in example 2.6).

2.11

a Split F (z) as A/(z + 1) + B/(z + 1)2 + C/(z + 3) and multiply by the denominator of F (z) to obtain the values C = 9/4, B = 1/2 and, by comparing the coefficient of z 2 , A = −5/4 (as in example 2.8).

2.12

Trying the first few integers we find the zero z = 1 of the denominator. A long division gives as denominator (z − 1)(z 2 − 2z + 5). We then split F (z) as A/(z − 1) + (Bz + C)/(z 2 − 2z + 5). Multiplying by the denominator of F (z) and comparing the coefficients of z 0 = 1, z and z 2 we obtain that A = 2, B = 0 and C = −1.

2.13

a

2.14

Use integration by parts twice and the fact that a primitive of eiω0 t is eiω0 t /iω0 . The given integral then equals 4π(1 − πi)/ω03 , since e2πi = 1. ˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛ Since ˛ 1/(2 −˛ eit ) ˛ = 1/˛ ˛ 2 − eit ˛ and ˛ 2 − eit ˛ ≥ 2 − ˛ eit ˛ = 1, the result ˛R1 ˛ R1 follows from ˛ 0 u(t) dt ˛ ≤ 0 | u(t) | dt.

2.15

Using the chain rule we obtain f 0 (t) = −i(1 + it)−2 .

1

2

Answers to selected exercises for chapter 2

2.16

2.17

√ P 3 a Use that | an | = 1/ n6 + 1 ≤ 1/n3 and the fact that ∞ n=1 1/n converges (example 2.17).

P 2 b Use that | an | ≤ 1/n2˛ and the˛ fact that ∞ n=1 1/n converges. n ni n n ˛ ˛ c Use that | a | = 1/ ne e = 1/(ne ) ≤ 1/e and the fact that n P∞ n 1/e converges since it is a geometric series with ratio 1/e. n=1 a

Use the ratio test to conclude that the series is convergent: ˛ ˛ ˛ ˛ n!

1 ˛ = lim lim ˛˛ = 0. n→∞ (n + 1)! ˛ n→∞ n + 1 b The series is convergent; proceed as in part a: ˛ ˛ ˛ 2n+1 + 1 3n + n ˛ 2 + 1/2n 1 + n/3n 2 ˛ = lim = . lim ˛˛ n+1 ˛ n n n→∞ 3 + (n + 1)/3 1 + 1/2n n→∞ 3 +n+1 2 +1 3

2.19

Determine the radius of convergence as follows: ˛ n+1 2n+2 2 ˛ ˛ 2 ˛ ˛ ˛ ˛ 1 + 1/n2 z n + 1 ˛˛ ˛ lim ˛ = lim 2 ˛ z 2 ˛ = 2 ˛ z2 ˛ . ˛ 2 n 2n n→∞ (n + 1) + 1 2 z n→∞ 1 + 2/n + 2/n2 ˛ ˛ √ This is less than 1 if ˛ z 2 ˛ < 1/2, that is, if | z | < 2/2. Hence, the radius √ of convergence is 2/2.

2.20

This is a geometric series with ratio z −i and so it converges for | z − i | < 1; the sum is (1/(1 − i))(1/(1 − (z − i))), so 1/(2 − z(1 − i)).

2.23

b First solving w2 = −1 leads to z 2 = 0 or z 2 = −2i. The equation z 2 = −2i has solutions −1 + i and 1 − i and z 2 = 0 has solution 0 (with multiplicity 2). c One has P (z) = z(z 4 + 8z 2 + 16) = z(z 2 + 4)2 = z(z − 2i)2 (z + 2i)2 , so 0 is a simple zero and ±2i are two zeroes of multiplicity 2.

2.25

Split F (z) as (Az + B)/(z 2 − 4z + 5) + (Cz + D)/(z 2 − 4z + 5)2 and multiply by the denominator of F (z). Comparing the coefficient of z 0 , z 1 , z 2 and z 3 leads to the values A = 0, B = 1, C = −2 and D = 2.

R 2π Replace cos t by (eit + e−it )/2, then we have to calculate 0 (e2it + 1)/2 dt, which is π. √ a Using the ratio test we obtain as limit 5/3. This is less than 1 and so the series converges.

P b Since (n + in )/n2 = (1/n) + (in /n2 ) and the series ∞ n=1 1/n diverges, this series is divergent. ˛ ˛ P The series ∞ cn (z 2 )n converges for all z with ˛ z 2 ˛ < R, so it has radius n=0 √ of convergence R.

2.26 2.27

2.29 2.30

a

Determine the radius of convergence as follows: ˛ ˛ ˛ (1 + i)2n+2 z n+1 ˛ ˛ ˛ n+1 ˛ = lim | z | n + 1 ˛ (1 + i)2 ˛ = 2 | z | . lim ˛˛ n→∞ n+2 (1 + i)2n z n ˛ n→∞ n+2 This is less than 1 if | z | < 1/2, so the radius of convergence is 1/2. b Calculate f 0 (z) by termwise differentiation of the series and multiply this by z. It then follows that zf 0 (z) + f (z) =

∞ X n=0

(1 + i)2n z n =

∞ X

(2iz)n .

n=0

This is a geometric series with ratio 2iz and so it has sum 1/(1 − 2iz).

Answers to selected exercises for chapter 3

3.2

A trigonometric polynomial can be written as f (t) =

k X a0 + (am cos(mω0 t) + bm sin(mω0 t)).

2 m=1

Now substitute this for f (t) in the right-hand side of (3.4) and use the fact that all the integrals in the resulting expression are zero, except for R T /2 the integral −T /2 sin(mω0 t) sin(nω0 t) dt with m = n, which equals T /2.

Hence, one obtains bn .

3.4

The function g(t) = f (t) cos(nω0 t) has period T , so Z T /2 Z T Z T g(t)dt g(t)dt + g(t)dt = 0 T /2 0 Z T Z T /2 Z 0 Z T /2 g(t − T )dt + g(t)dt = g(τ )dτ + g(t)dt = T /2 0 −T /2 0 Z T /2 = g(t)dt. −T /2

Multiplying by 2/T gives an .

3.6

From a sketch of the periodic function with period 2π given by f (t) = | t | for t ∈ (−π, π) we obtain Z 0 Z π 1 1 cn = (−t)e−int dt + te−int dt.

2π −π 2π 0 As in example 3.2 these integrals can be calculated using integration by parts for n 6= 0. Calculating c0 separately (again as in example 3.2) we obtain c0 =

π , 2

cn =

(−1)n − 1 n2 π

Substituting these values of cn in (3.10) we obtain the Fourier series. One can also write this as a Fourier cosine series: ∞

π 4 X cos((2k + 1)t) − .

2 π (2k + 1)2 k=0

3.7

From the description of the function we obtain that Z 1 1 −(1+inπ)t cn = e dt.

2 0 This integral can be evaluated immediately and leads to cn =

´ inπ − 1 ` (−1)n e−1 − 1 .

2(n2 π 2 + 1)

The Fourier series follows from (3.10) by substituting cn .

3.9

The Fourier coefficients are calculated by splitting the integrals into a real and an imaginary part. For c0 this becomes:

3

4

Answers to selected exercises for chapter 3

c0 =

1 2

Z 1

t2 dt +

−1

i 2

Z 1 t dt = −1

1 .

3

For n 6= 0 we have that Z Z 1 1 2 −inπt i 1 −inπt cn = t e dt + te dt.

2 −1 2 −1 The second integral can be calculated using integration by parts. To calculate the first integral we apply integration by parts twice. Adding the results and simplifying somewhat we obtain the Fourier coefficients (and thus the Fourier series): cn =

(−1)n (2 − nπ) . n2 π 2

3.10

From the values of the coefficients cn calculated earlier in exercises 3.6, 3.7 and 3.9, one can immediately obtain the amplitude spectrum | cn | and the phase spectrum arg cn (note e.g. that arg cn = π if cn > 0, arg cn = −π if cn < 0, arg cn = π/2 if cn = iy with y > 0 and arg cn = −π/2 if cn = iy with y < 0). This results in three figures that are given separately on the website.

3.11

a

By substituting a = T /4 in (3.14) it follows that

cn =

sin(nπ/4) nπ

for n 6= 0,

c0 =

1 .

4

b As in a, but now a = T and we obtain c0 = 1,

for n 6= 0.

cn = 0

Hence, the Fourier series is 1 (!). This is no surprise, since the function is 1 for all t.

3.12

By substituting a = T /2 in (3.15) it follows that c0 =

1 , 2

cn = 0

for n 6= 0 even,

cn =

2 n2 π 2

for n odd.

3.14

We have that f (t) = 2p2,4 (t) − q1,4 (t) and so the Fourier coefficients follow by linearity from table 1: c0 = 3/4, cn = (2nπ sin(nπ/2) − 4 sin2 (nπ/4))/(n2 π 2 ) for n 6= 0.

3.15

Note that f (t) can be obtained from the sawtooth z(t) by multiplying the shifted version z(t − T /2) by the factor T /2 and then adding T /2, that is, f (t) = T2 z(t − T2 ) + T2 . Now use the Fourier coefficients of z(t) (table 1 e.g.) and the properties from table 2 to obtain that c0 =

T , 2

cn =

iT 2πn

for all n 6= 0.

3.17

Shifts over a period T (use the shift property and the fact that e−2πin = 1 for all n).

3.19

In order to determine the Fourier sine series we extend the function to an odd function of period 8. We calculate the coefficients bn as follows (the an are 0): Z Z 1 2 1 −2 (−2 sin(nπt/4)) dt + t sin(nπt/4) dt bn = 4 −4 4 −2

Answers to selected exercises for chapter 3

+

1 4

5

Z 4 2 sin(nπt/4) dt.

2

The second integral can be calculated by an integration by parts and one then obtains that 8 4 bn = 2 2 sin(nπ/2) − cos(nπ), n π nπ which gives the Fourier sine series. For the Fourier cosine series we extend the function to an even function of period 8. As above one can calculate the coefficients an and a0 (the bn are 0). The result is a0 = 3, 3.21

an =

8 (cos(nπ/2) − 1) n2 π 2

for all n 6= 0.

In order to determine the Fourier cosine series we extend the function to an even function of period 8. We calculate the coefficients an and a0 as follows (the bn are 0): Z 1 4 2 16 a0 = (x − 4x) dx = − , 2 0 3 while for n ≥ 1 we have Z Z 1 0 2 1 4 2 an = (x + 4x) cos(nπx/4) dx + (x − 4x) cos(nπx/4) dx 4 −4 4 0 Z 4 Z 4 1 = x2 cos(nπx/4) dx − 2 x cos(nπx/4) dx.

2 0 0 The first integral can be calculated by applying integration by parts twice; the second integral can be calculated by integration by parts. Combining the results one then obtains that an =

32((−1)n − 1) 32((−1)n + 1) 64(−1)n − = , 2 2 2 2 n π n π n2 π 2

which also gives the Fourier cosine series. One can write this series as −

∞ 8 16 X 1 + 2 cos(nπx/2).

3 π n=1 n2

For the Fourier sine series we extend the function to an odd function of period 8. As above one can calculate the coefficients bn (the an are 0). The result is bn =

64((−1)n − 1) n3 π 3

for all n ≥ 1.

3.24

If f is real and the cn are real, then it follows from (3.13) that bn = 0. A function whose Fourier coefficients bn are all 0 has a Fourier series containing cosine functions only. Hence, the Fourier series will be even. If, on the other hand, f is real and the cn are purely imaginary, then (3.13) shows that an = 0. The Fourier series then contains sine functions only and is thus odd.

3.25

Since sin(ω0 t) = (eiω0 t − e−iω0 t )/2i we have Z T /2 Z T /2 1 1 cn = ei(1−n)ω0 t dt − e−i(1+n)ω0 t dt.

2iT 0 2iT 0

6

Answers to selected exercises for chapter 3

The first integral equals T /2 for n = 1 while for n 6= 1 it equals i((−1)n + 1)/((1 − n)ω0 ). The second integral equals T /2 for n = −1 while for n 6= −1 it equals i((−1)n+1 − 1)/((1 + n)ω0 ). The Fourier coefficients are thus c1 = 1/(4i), c−1 = −1/(4i) and ((−1)n +1)/(2(1−n2 )π) for n 6= 1, −1; the Fourier series follows immediately from this.

3.27

b The even extension has period 2a, but it has period a as well. We can thus calculate the coefficients an and a0 as follows (the bn are 0): Z Z 2 a/2 2 0 a0 = 2bt/a dt − 2bt/a dt = b. a 0 a −a/2 while for n ≥ 0 we obtain from an integration by parts that Z Z 2 a/2 2 0 an = (2bt/a) cos(2nπt/a) dt − (2bt/a) cos(2nπt/a) dt a 0 a −a/2 n 2b((−1) − 1) = , n2 π 2 which gives the Fourier cosine series. It can also be determined using the result of exercise 3.6 by applying a multiplication and a scaling.

The odd extension has period 2a and the coefficients bn are given by (the an are 0): Z Z 1 a/2 2bt 1 −a/2 −2bt − 2b) sin(nπt/a) dt + sin(nπt/a) dt bn = ( a −a a a −a/2 a Z a 1 −2bt + ( + 2b) sin(nπt/a) dt a a/2 a 8b = 2 2 sin(nπ/2), n π where we used integration by parts.

Answers to selected exercises for chapter 4

4.1

a The periodic block function from section 3.4.1 is a continuous function on [−T /2, T /2], except at t = ±a/2. At these points f (t+) and f (t−) exist.

Also f 0 (t) = 0 for t 6= ±a/2, while f 0 (t+) = 0 for t = ±a/2 and t = −T /2 and f 0 (t−) = 0 for t = ±a/2 and t = T /2. Hence f 0 is piecewise continuous and so the periodic block function is piecewise smooth. Existence of the Fourier coefficients has already been shown in section 3.4.1. The periodic triangle function is treated analogously. b For the periodic block function we have ∞ X

| cn |2 ≤

n=−∞

∞ a2 8 X 1 + 2 2 2 T T ω0 n=1 n2

P P∞ 2 1 since sin2 (nω0 a/2) ≤ 1. The series ∞ n=−∞ | cn | n=1 n2 converges, so converges. The periodic triangle function is treated analogously.

4.2

This follows immediately from (3.11) (for part a) and (3.8) (for part b).

4.4

Take t = T /2 in the Fourier series of the sawtooth from example 4.2 and use that sin(nω0 T /2) = sin(nπ) = 0 for all n. Since (f (t+) + f (t−))/2 = 0, this agrees with the fundamental theorem.

4.6

a If we sketch the function, then we see that it is a shifted block function.

Using the shift property we obtain c0 =

1 , 2

cn = 0

even n 6= 0,

cn =

−i nπ

odd n.

The Fourier series follows by substituting the cn . One can write the series with sines only (split the sum in two pieces: one from n = 1 to ∞ and another from n = −1 to −∞; change from n to −n in the latter): ∞

1 2 X sin(2k + 1)t + .

2 π 2k + 1 k=0

b The function is piecewise smooth and it thus satisfies the conditions of the fundamental theorem. At t = π/2 the function f is continuous, so the series converges to f (π/2) = 1. Since sin((2k + 1)π/2) = (−1)k , formula (4.11) follows: ∞ X (−1)k k=0

4.7

2k + 1

=

π .

4

Rπ a We have that c0 = (2π)−1 0 t dt = π/4, while the Fourier coefficients for n 6= 0 follow from an integration by parts: Z π (−1)n i (−1)n − 1 1 cn = te−int dt = + .

2π 0 2n 2n2 π The Fourier series follows by substituting these cn : „ « ∞ X (−1)n i (−1)n − 1 π 1 + + eint .

4 2 n n2 π n=−∞,n6=0

12

Answers to selected exercises for chapter 4

13

b From the fundamental theorem it follows that the series will converge to 12 (f (π+) + f (π−)) = π/2 at t = π (note that at π there is a jump). If we substitute t = π into the Fourier series, take π4 to the other side of the =-sign, then multiply by 2, and finally split the sum into a sum from n = 1 to ∞ and a sum from n = −1 to −∞, then it follows that ∞ X (−1)n − 1 π =2 (−1)n 2π 2 n n=1

(the terms with (−1)n i/n cancel each other). For even n we have (−1)n − 1 = 0 while for odd n this will equal −2, so (4.10) results: ∞

X π2 1 = .

8 (2k − 1)2 k=1

4.9

a From f (0+) = 0 = f (0−) and f (1−) = 0 = f ((−1)+) it follows that f is continuous. We have that f 0 (t) = 2t+1 for −1 < t < 0 and f 0 (t) = −2t+1 for 0 < t < 1. Calculating the defining limits for f 0 from below and from above at t = 0 we see that f 0 (0) = 1 and since f 0 (0+) = 1 = f 0 (0−) it follows that f 0 is continuous at t = 0. Similarly it follows that f 0 is continuous at t = 1. Since f 00 (t) = 2 for −1 < t < 0 and f 00 (t) = −2 for 0 < t < 1 we see that f 00 is discontinuous. b The function f is the sum of g and h with period 2 defined for −1 < t ≤ 1 by g(t) = t and h(t) = t2 for −1 < t ≤ 0 and h(t) = −t2 for 0 < t ≤ 1.

Since g is a sawtooth, the Fourier coefficients are cn = (−1)n i/πn (see section 3.4.3). The function h is the odd extension of −t2 on (0, 1] and its Fourier coefficients have been determined in the first example of section 3.6. By linearity one obtains the Fourier coefficients of f . In terms of the an and bn they become an = 0 and bn = 4(1 − (−1)n )/π 3 n3 . Hence, they decrease as 1/n3 . c Use e.g. the fundamental theorem for odd functions to obtain ∞

f (t) =

8 X sin(2k + 1)πt . π3 (2k + 1)3 k=0

4.10

Now substitute t = 1/2 and use that f (1/2) = 1/4 and sin((2k + 1)π/2) = (−1)n to obtain the required result.

P 2 2 Use (3.8) to write the right-hand side of (4.14) as a20 /4 + 21 ∞ n=1 (an + bn ).

4.12

a The Fourier coefficients of f and g are (see table 1 or section 3.4.1), respectively, fn = (sin na)/nπ for n 6= 0 and f0 = a/π and gn = (sin nb)/nπ for n 6= 0 and g0 = b/π. Substitute into Parseval (4.13) and calculate the R a/2 integral (1/π) −a/2 1 dt (note that a ≤ b). Take all constants together and then again (as in exercise 4.7) split the sum into a sum from n = 1 to ∞ and a sum from n = −1 to −∞. The required result then follows. b Use that sin2 (nπ/2) = 1 for n odd and 0 for n even, then (4.10) follows.

4.13

a The Fourier coefficients are (see table 1 or section 3.4.2 and use that sin2 (nπ/2) = 1 for n odd and 0 for n even): cn = 2/n2 π 2 for n odd, 0 for n 6= 0 even and c0 = 1/2. From Parseval for f = g, so from (4.14), it then follows that (calculate the integral occurring in this formula): ∞

1 1 8 X 1 = + 2 3 4 π (2k − 1)4 k=1

14

Answers to selected exercises for chapter 4

(again we split the sum in a part from n = 1 to ∞ and from n = −1 to −∞). Take all constants together and multiply by π 2 /8, then the required result follows. b Since S=

∞ X 1 n=1

n

= 4

∞ X k=1

∞

X 1 1 + , (2k)4 (2k + 1)4 k=0

it follows from part a that ∞

S=

1 π4 π4 1 X 1 = S+ . + 4 16 k 96 16 96 k=1

4.15

P 1 π4 Solving for S we obtain ∞ n=1 n4 = 90 .

Rb Rb Ra Since a f (t) dt = −T /2 f (t) dt − −T /2 f (t) dt, we can apply theorem 4.9 twice. Two of the infinite sums cancel out (the ones representing h0 in theorem 4.9), the other two can be taken together and lead to the desired result.

4.16

This follows from exercise 4.15 by using (3.8), so cn = (an − ibn )/2 and c−n = (an + ibn )/2 (n ∈ N).

4.17

a

The Fourier series is given by ∞ X

sin(2n + 1)t 4 . π n=0 2n + 1 b Since Z t cos(2n + 1)t 1 sin(2n + 1)τ dτ = − − , 2n + 1 2n +1 −π the integrated series becomes −

∞ ∞ 4X 1 4 X cos(2n + 1)t − . π n=0 (2n + 1)2 π n=0 (2n + 1)2

From (4.10) we see that the constant in this series R t equals −π/2. c The series in part b represents the function −π f (τ ) dτ (theorem 4.9 or better still, exercise 4.16). Calculating this integral we obtain the function g(t) with period 2π given for −π < t ≤ π by g(t) = | t | − π. d Subtracting π from the Fourier series of | t | in exercise 3.6 we obtain a Fourier series for g(t) which is in accordance with the result from part b.

4.19

This again follows as in exercise 4.16 from (3.8).

4.20

Since f 0 is piecewise smooth, f 00 is piecewise continuous and so the Fourier coefficients c00n of f 00 exist. Since f 0 is continuous, we can apply integration by parts, as in the proof of theorem 4.10. It then follows that c00n = inω0 c0n , where c0n are the Fourier coefficients of f 0 . But c0n = inω0 cn by theorem 4.10, so c00n = −n2 ω02 cn . Now apply the Riemann-Lebesgue lemma to c00n , then it follows that limn→±∞ n2 cn = 0.

4.22

a The Fourier coefficients have been determined in exercise 3.25: c1 = 1/(4i), c−1 = −1/(4i) and ((−1)n + 1)/(2(1 − n2 )π) for n 6= 1, −1. Taking positive and negative n in the series together, we obtain the following Fourier series:

Answers to selected exercises for chapter 4

15

∞

1 1 2X 1 sin t + + cos 2kt.

2 π π 1 − 4k2 k=1

b The derivative f 0 exists for all t 6= nπ (n ∈ Z) and is piecewise smooth.

According to theorem 4.10 we may thus differentiate f by differentiating its Fourier series for t 6= nπ: ∞

f 0 (t) =

1 4X k cos t − sin 2kt.

2 π 1 − 4k2 k=1

At t = nπ the differentiated series converges to (f 0 (t+) + f 0 (t−))/2, which equals 1/2 for t = 0, while it equals −1/2 for t = π. Hence, the differentiated series is a periodic function with period 2π which is given by 0 for −π < t < 0, 21 for t = 0, cos t for 0 < t < π, − 12 for t = π.

4.25

Write down the expression for Si(−x) and change from the variable t to −t, then it follows that Si(−x) = −Si(−x).

4.26

a From the definition of Si(x) it follows that Si0 (x) = sin x/x. So Si0 (x) = 0 if sin x/x = 0. For x > 0 we thus have Si0 (x) = 0 for x = kπ with k ∈ N. A candidate for the first maximum is thus x = π. Since sin x/x > 0 for 0 < x < π and sin x/x < 0 for π < x < 2π, it follows that Si(x) indeed has its first maximum at x = π. b The value at the first maximum is Si(π). Since Si(π) = 1.852 . . . and π/2 = 1.570 . . ., the overshoot is 0.281 . . .. The jump of f at x = 0 is π = 3.141 . . ., so the overshoot is 8.95 . . .%, so about 9%.

4.28

a The function f is continuous for t 6= (2k + 1)π (k ∈ Z) and it then converges to f (t), which is 2t/π for 0 ≤ | t | < π/2, 1 for π/2 ≤ t < π and −1 for −π < t ≤ −π/2. For t = (2k+1)π it converges to (f (t+)+f (t−))/2 = 0. b Since f is odd we have an = 0 for all n. The bn can be found using an integration by parts: Z π/2 Z 2(−1)n 4 2 π 4 . bn = 2 t sin nt dt + sin nt dt = 2 2 sin(nπ/2) − π 0 π π/2 n π nπ Since sin(nπ/2) = 0 if n even and (−1)k if n = 2k + 1, the Fourier series is −

∞ ∞ 4 X (−1)n 2 X (−1)n sin nt + 2 sin(2n + 1)t. π n=1 n π n=0 (2n + 1)2

Substituting t = 0 and t = π it is easy to verify the fundamental theorem for these values. c We cannot differentiate the series; the resulting series is divergent because limn→∞ (−1)n cos nt 6= 0. Note that theorem 4.10 doesn’t apply since f is not continuous. d We can integrate the series R t since theorem 4.9 can be applied (note that c0 = 0). If we put g(t) = −π f (τ ) dτ , then g is even, periodic with period 2π and given by (t2 /π)−(3π/4) for 0 ≤ t < π/2 and by t−π for π/2 ≤ t ≤ π.

4.29

Use table 1 to obtain the Fourier coefficients and then apply Parseval, that is, (4.13). Calculating the integral in Parseval’s identity will then give the first result; choosing a = π/2 gives the second result.

4.30

a The Fourier series has been determined in the last example of section 3.6. Since f is continuous (and piecewise smooth), the Fourier series converges to f (t) for all t ∈ R:

16

Answers to selected exercises for chapter 4

f (t) =

∞ 2 4X 1 − cos 2nt. π π n=1 4n2 − 1

b First substitute t = 0 in the Fourier series; since f (0) = 0 and cos 2nt = 1 for all n, the first result follows. Next substitute t = π/2 in the Fourier series; since f (π/2) = 1 and cos 2nt = (−1)n for all n, the second result follows. c One should recognize the squares of the Fourier coefficients here. Hence we have to apply Parseval’s identity (4.14), or the alternative form given in exercise 4.10. This leads to Z π ∞ 1 4 1 X 16 sin2 t dt = 2 + .

2 2π −π π 2π n=1 (4n2 − 1)2 Rπ Since −π sin2 t dt = π, the result follows.

4.31

a

Since f1 is odd it follows that Z 1 T /2 (f1 ∗ f2 )(−t) = − f1 (t + τ )f2 (τ ) dτ.

T −T /2

Now change the variable from τ to −τ and use that f2 is odd, then it follows that (f1 ∗ f2 )(−t) = (f1 ∗ f2 )(t). b The convolution product equals Z 1 1 (f ∗ f )(t) = τ f (t − τ ) dτ.

2 −1 Since f is odd, part a implies that f ∗ f is even. It is also periodic with period 2, so it is sufficient to calculate (f ∗ f )(t) for 0 ≤ t ≤ 1. First note that f is given by f (t) = t − 2 for 1 < t ≤ 2. Since −1 ≤ τ ≤ 1 and 0 ≤ t ≤ 1 we see that t − 1 ≤ t − τ ≤ t + 1. From 0 ≤ t ≤ 1 it follows that −1 ≤ t − 1 ≤ 0, and so close to τ = 1 the function f (t − τ ) is given by t − τ . Since 1 ≤ t + 1 ≤ 2, the function f (t − τ ) is given by t − τ − 2 close to τ = −1. Hence, we have to split the integral precisely at the point where t − τ gets larger than 1, because precisely then the function changes from t − τ to t − τ − 2. But t − τ ≥ 1 precisely when τ ≤ t − 1, and so we have to split the integral at t − 1: Z Z 1 1 1 t−1 τ (t − τ − 2) dτ + τ (t − τ ) dτ. (f ∗ f )(t) = 2 −1 2 t−1 It is now straightforward to calculate the convolution product. The result is (f ∗ f )(t) = −t2 /2 + t − 1/3. c From section 3.4.3 or table 1 we obtain the Fourier coefficients cn of the sawtooth f and applying the convolution theorem gives the Fourier coefficients of (f ∗ f )(t), namely c20 = 0 and c2n = −1/π 2 n2 (n 6= 0). d Take t = 0 R 1in part c; since f is odd and real-valued we can write (f ∗ f )(0) = 12 −1 | f (τ ) |2 dτ , and so we indeed obtain (4.13). e For −1 < t < 0 we have (f ∗ f )0 (t) = −t − 1, while for 0 < t < 1 we have (f ∗ f )0 (t) = −t + 1. Since f ∗ f is given by −t2 /2 + t − 1/3 for 0 < t < 2, (f ∗ f )0 (t) is continuous at t = 1. Only at t = 0 we have that f ∗ f is not differentiable. So theorem 4.10 implies that the differentiated series represents the function (f ∗ f )0 (t) on [−1, 1], except at t = 0. At t = 0 the differentiated series converges to ((f ∗ f )0 (0+) + (f ∗ f )0 (0−))/2 = (1 − 1)/2 = 0. f The zeroth Fourier coefficient of f ∗ f is given by

Answers to selected exercises for chapter 4

1 2

Z 1

Z 1 (f ∗ f )(t) dt =

−1

17

(−t2 /2 + t − 1/3) dt = 0.

0

This is in agreement with the result in part c since c20 = 0. Since this coefficient is 0, we can apply theorem 4.9. The function R t represented by the integrated series is given by the (periodic) function −1 (f ∗ f )(τ ) dτ . It is also odd, since f is even and for 0 ≤ t ≤ 1 it equals Z 0 Z t (−τ 2 /2 − τ − 1/3) dτ + (−τ 2 /2 + τ − 1/3) dτ = −t(t − 1)(t − 2)/6. −1

0

Answers to selected exercises for chapter 5

5.1

For a stable LTC-system the real parts of the zeroes of the characteristic polynomial are negative. Fundamental solutions of the homogeneous equations are of the form˛ x(t) ˛= tl est , where s is such a zero and l ≥ 0 some integer. Since ˛ tl est ˛ = | t |l e(Re s)t and Re s < 0 we have that limt→∞ x(t) = 0. Any homogeneous solution is a linear combination of the fundamental solutions.

5.2

The Fourier coefficients of u are u0 =

1 , 2

u2k = 0,

u2k+1 =

(−1)k (2k + 1)π

(u = pπ,2π , so use table 1 and the fact that sin(nπ/2) = (−1)k for n = 2k+1 odd and 0 for n even). Since H(ω) = 1/(iω + 1) and yn = H(nω0 )un = H(n)un it then follows that y0 =

1 , 2

y2k = 0,

y2k+1 =

(−1)k .

(1 + (2k + 1)i)(2k + 1)π

5.3

a The frequency response is not a rational function, so the system cannot be described by a differential equation (5.3). b Since H(nω0 ) = H(n) = 0 for | n | ≥ 4 (because 4 > π), we only need to consider the Fourier P coefficients of y with | n | ≤ 3. From Parseval it then follows that P = 3n=−3 | yn |2 with yn as calculated in exercise 5.2. This 20 sum is equal to P = 14 + 9π 2.

5.4

Note that u has period π and that the integral to be calculated is thus the zeroth Fourier coefficient of y. Since y0 = H(0ω0 )u0 = H(0)u0 Rand H(0) = π −1 (see example 5.6 for H(ω)), it follows that y0 = −u0 = − π1 0 u(t) dt = 2 −π.

5.5

a

According to (5.4) the frequency response is given by

H(ω) =

−ω 2 + 1 . −ω 2 + 4 + 2iω

Since H(ω) = 0 for ω = ±1, the frequencies blocked by the system are ω = ±1. b Write u(t) = e−4it /4 − e−it /2i + 1/2 + eit /2i + e4it /4. It thus follows that the Fourier coefficients unequal to 0 are given by u−4 = u4 = 1/4, u−1 = −1/2i, u1 = 1/2i and u0 = 1/2. Since yn = H(nω0 )un = H(n)un and H(1) = H(−1) = 0 we thus obtain that y(t) = y−4 e−4it + y−1 e−it + y0 + y1 eit + y4 e4it 15 1 1 1 15 1 = · e−4it + · + · e4it .

12 + 8i 4 4 2 12 − 8i 4 It is a good exercise to write this with real terms only: y(t) = 5.6

45 30 1 cos 4t + sin 4t + .

104 104 8

We have that 1 H(ω) = . −ω 2 + ω02

18

Answers to selected exercises for chapter 5

19

Since | ω0 | is not an integer, there are no homogeneous solutions having period 2π, while u does have period 2π. There is thus a uniquely determined periodic solution y corresponding to u. Since u(t) = πqπ,2π (t) the Fourier coefficients of u follow immediately from table 1: u0 =

π , 2

u2k = 0(k 6= 0),

u2k+1 =

2 .

(2k + 1)2 π 2

Since yn = H(nω0 )un = H(n)un = −n21+ω2 un , the line spectrum of y 0 follows.

5.7

For the thin rod the heat equation (5.8) holds on (0, L), with initial condition (5.9). This leads to the fundamental solutions (5.15), from which the superposition (5.16) is build. The initial condition leads to a Fourier series with coefficients Z Z 2 L/2 2 L An = x sin(nπx/L) dx + (L − x) sin(nπx/L) dx, L 0 L L/2 which can be calculated using an integration by parts. The result is: An = (4L/n2 π 2 ) sin(nπ/2) (which is 0 for n even). We thus obtain the (formal) solution ∞

u(x, t) = 5.9

a

4L X (−1)n −(2n+1)2 π2 kt/L2 e sin((2n + 1)πx/L). π 2 n=0 (2n + 1)2

The heat equation and initial conditions are as follows:

ut = kuxx ux (0, t) = 0, u(L, t) = 0 u(x, 0) = 7 cos(5πx/2L)

for 0 < x < L, t > 0, for t ≥ 0, for 0 ≤ x ≤ L.

b Separation of variables leads to (5.12) and (5.13). The function X(x) should satisfy X 00 (x) − cX(x) = 0 for 0 < x < L, X 0 (0) = 0 and X(L) = 0.

For c = 0 we obtain the trivial solution. For c 6= 0 the characteristic equation s2 − c = 0 has two distinct roots ±s1 . The general solution is then X(x) = αes1 x + βe−s1 x , so X 0 (x) = s1 αes1 x − s1 βe−s1 x . The first boundary condition X 0 (0) = 0 gives s1 (α − β) = 0, so β = α. Next we obtain from the second boundary condition X(L) = 0 the equation α(es1 L + e−s1 L ) = 0. For α = 0 we get the trivial solution. So we must have es1 L + e−s1 L = 0, implying that e2s1 L = −1. From this it follows that s1 = i(2n + 1)π/2L. This gives us eigenfunctions Xn (x) = cos((2n + 1)πx/2L) (n = 0, 1, 2, 3, . . .). Since Tn (t) remains as in the textbook (for other parameters), we have thus found the fundamental solutions 2

2

2

un (x, t) = e−(2n+1) π kt/4L cos((2n + 1)πx/2L).

Superposition gives u(x, t) =

∞ X

2

2

2

An e−(2n+1) π kt/4L cos((2n + 1)πx/2L).

n=0

Substituting t = 0 (and using the remaining initial condition) leads to u(x, 0) =

∞ X n=0

An cos((2n + 1)πx/2L) = 7 cos(5πx/2L).

20

Answers to selected exercises for chapter 5

Since the right-hand side consists of one harmonic only, it follows that A2 = 7 and An = 0 for all n 6= 2. The solution is thus u(x, t) = 2 2 7e−25π kt/4L cos(5πx/2L).

5.11

a

The equations are

ut = kuxx u(0, t) = 0, ux (L, t) = 0 u(x, 0) = f (x)

for 0 < x < L, t > 0, for t ≥ 0, for 0 ≤ x ≤ L.

b Going through the steps one obtains the same fundamental solutions as in exercise 5.9. The coefficients An cannot be determined explicitly here, since f (x) is not given explicitly.

5.12

The equations are given by (5.17) - (5.20), where we only need to substitute the given initial condition in (5.19), so u(x, 0) = 0.05 sin(4πx/L) for 0 ≤ x ≤ L. All steps to be taken are the same as in section 5.2.2 of the textbook and lead to the solution u(x, t) =

∞ X

An cos(nπat/L) sin(nπx/L).

n=1

Substituting t = 0 (and using the remaining initial condition) gives u(x, 0) =

∞ X

An sin(nπx/L) = 0.05 sin(4πx/L).

n=1

Since the right-hand side consists of one harmonic only, it follows that A4 = 0.05 and An = 0 for all n 6= 4. The solution is thus u(x, t) = 0.05 cos(4πat/L) sin(4πx/L).

5.15

Separation of variables leads to X 00 (x) − cX(x) = 0 for 0 < x < π, X 0 (0) = X 0 (π) = 0. For c = 0 we obtain the constant solution, so c = 0 is an eigenvalue with eigenfunction X(x) = 1. For c 6= 0 the characteristic equation s2 − c = 0 has two distinct roots ±s1 . The general solution is then X(x) = αes1 x + βe−s1 x , so X 0 (x) = s1 αes1 x − s1 βe−s1 x . The boundary condition X 0 (0) = 0 gives s1 (α − β) = 0, so β = α. From the boundary condition X 0 (π) = 0 we obtain s1 α(es1 π − e−s1 π ) = 0. For α = 0 we get the trivial solution. So we must have es1 π − e−s1 π = 0, implying that e2s1 π = 1. From this it follows that s1 = ni. This gives us eigenfunctions Xn (x) = cos(nx) (n = 0, 1, 2, 3, . . .). For T (t) we get the equation T 00 (t) + n2 a2 T (t) = 0. From the initial condition ut (x, 0) = 0 we obtain T 0 (0) = 0. The non-trivial solution are Tn (t) = cos(nat) (n = 0, 1, 2, 3, . . .) and we have thus found the fundamental solutions un (x, t) = cos(nat) cos(nx).

Superposition gives u(x, t) =

∞ X

An cos(nat) cos(nx).

n=0

Substituting t = 0 (and using the remaining initial condition) leads to u(x, 0) =

∞ X n=0

An cos(nx) = kx

for 0 < x < π.

Answers to selected exercises for chapter 5

21

Rπ Rπ We have A0 = (2/π) 0 kx dx = kπ and An = (2/π) 0 kx cos(nx) dx for n 6= 0, which can be calculated by an integration by parts: An = 0 for n even (n 6= 0) and An = −4k/n2 π for n odd. The solution is thus u(x, t) =

∞ kπ 4k X 1 − cos((2n + 1)at) cos((2n + 1)x).

2 π n=0 (2n + 1)2

5.16

a From H(−ω) = H(ω) and yn = H(nω0 )un follows that the response y(t) to a real signal u(t) is real: since u−n = un we also have y−n = yn . b Since we can write sin ω0 t = (eiω0 t − e−iω0 t )/2i, the response is equal to (H(ω0 )eiω0 t − H(−ω0 )e−iω0 t )/2i, which is ((1 − e−2iω0 )2 eiω0 t − (1 − e2iω0 )2 e−iω0 t )/2i. This can be rewritten as sin ω0 t − 2 sin(ω0 (t − 2)) + sin(ω0 (t − 4)).

P 2πint . c A signal with period 1 has Fourier series of the form ∞ n=−∞ un e P∞ 2πint , which is 0 since H(2πn) = 0 for The response is n=−∞ H(2πn)un e all n.

5.18

a The characteristic equation is s3 + s2 + 4s + 4 = (s2 + 4)(s + 1) = 0 and has zeroes s = −1 and s = ±2i. The zeroes on the imaginary axis correspond to periodic eigenfrequencies with period π and so the response to a periodic signal is not always uniquely determined. But see part b! b Since here the input has period 2π/3, we do have a unique response.

From Parseval and the relation yn = H(nω0 )un we obtain that the power is given by Z 2π/3 ∞ ∞ X X 3 P = | y(t) |2 dt = | yn |2 = | H(nω0 )un |2 .

2π 0 n=−∞ n=−∞ We have that H(ω) =

1 + iω .

4 − ω 2 + iω(4 − ω 2 )

Now use that only u3 = u−3 = 12 and that all other un are 0, then it follows that P = 1/50.

5.19

For the rod we have equations (5.8) - (5.10), where we have to take f (x) = u0 in (5.10). The solution is thus given by (5.16), where now the An are the Fourier coefficients of the function u0 on [0, L]. These are easy te determine (either by hand or using tables 1 and 2): An = 0 for n even, An = 4u0 /nπ for n odd. This gives u(x, t) =

∞ 2 2 2 4u0 X 1 e−(2n+1) π kt/L sin((2n + 1)πx/L).

2 π n=0 (2n + 1)

Substituting x = L/2 in the x-derivative and using the fact that cos((2n + 1)π/2) = 0 for all n leads to ux (L/2, t) = 0.

5.20

a As in the previous exercise the solution is given by (5.16). The An are R L/2 given by (2/L) 0 a sin(nπx/L) dx = 2a(1 − cos(nπ/2))/nπ, which gives the (formal) solution ∞

u(x, t) =

2 2 2 2a X 1 (1 − cos(nπ/2))e−n π kt/L sin(nπx/L). π n=1 n

22

Answers to selected exercises for chapter 5

b The two rods together form one rod and so part a can be applied with L = 40, k = 0.15 and a = 100. Substituting t = 600 in u(x, t) from part a then gives the temperature distribution. On the boundary between the rods we have x = 20, so we have to calculate u(20, 600); using only the contibution from the terms n = 1, 2, 3, 4 we obtain u(20, 600) ≈ 36.4. c Take k = 0.005, a = 100, L = 40, substitute x = 20 in u(x, t) from part a, and now use only the first two terms of the series to obtain the equation u(20, t) ≈ 63.662e−0.0000308t = 36 (terms of the series tend to 0 very rapidly, so two terms suffice). We then obtain 18509 seconds, which is approximately 5 hours.

Answers to selected exercises for chapter 6

6.1

6.2

6.4

R∞ We have to calculate (the improper integral) −∞ e−iωt dt. Proceed as in eaxample 6.1, but we now have to determine limB→∞ e−iωB . This limit does not exist.

R∞ a We have to calculate G(ω) = 0 e−(a+iω)t dt, which can be done precisely as in section 6.3.3 if we write a = α + iβ and use that e−(a+iω)R = e−αR e−i(β+ω)R . If we let R → ∞ then this tends to 0 since α > 0. b The imaginary of G(ω) is −ω/(a2 + ω 2 ) and applying the substituR part 2 tion rule gives ω/(a + ω 2 ) dω = 21 ln(a2 + ω 2 ), so this improper integral, which is the Fourier integral for t = 0, does not exits (limA→∞ ln(a2 + A2 ) does not exist e.g.). c We have lima→0 g(t) = lima→0 (t)e−at = (t), while for ω 6= 0 we have that lima→0 G(ω) = −i/ω.

To calculate the spectrum we split the integral at t = 0: Z 1 Z 0 G(ω) = te−iωt dt − te−iωt dt. −1

0

Changing Rfrom the variable t to −t in the second integral we obtain that 1 G(ω) = 2 0 t cos ωt dt, which can be calculated for ω 6= 0 using an integration by parts. The result is: G(ω) =

2(cos ω − 1) 2 sin ω + . ω ω2

R1 For ω = 0 we have that G(0) = 2 0 t dt = 1. Since limω→0 sin ω/ω = 1 and limω→0 (cos ω − 1)/ω 2 = − 21 (use e.g. De l’Hôpital’s rule), we obtain that limω→0 G(ω) = G(0), so G is continuous.

6.5

a

Calculating the integral we have that

F (ω) = 2i

cos(aω/2) − 1 ω

for ω 6= 0,

F (0) = 0.

b Using Taylor or De l’Hôpital it follows that limω→0 F (ω) = 0 = F (0), so F is continuous.

6.7

From the linearity and table 3 it follows that F (ω) =

6.8

Use (6.17) and table 3 for the spectrum of e−7| t | , then F (ω) =

6.9

sin2 (aω/2) 12 + 8i .

2 4+ω aω 2

7 7 + .

49 + (ω − π)2 49 + (ω + π)2

a From the shift property in the frequency domain (and linearity) it follows that the spectrum of f (t) sin at is F (ω − a)/2i − F (ω + a)/2i. b Write f (t) = p2π (t) sin t, obtain the spectrum of p2π (t) from table 3 and apply part a (and use the fact that sin(πω ± π) = − sin(ωπ)), then F (ω) =

2i sin(πω) . ω2 − 1

23

24

Answers to selected exercises for chapter 6

6.10

Use section 6.3.3 (or exercise 6.2) and the modulation theorem 6.17, and write the result as one fraction, then (F (t)e−at cos bt)(ω) =

a + iω . (a + iω)2 + b2

Similarly it follows from section 6.3.3 (or exercise 6.2) and exercise 6.9a that (F (t)e−at sin bt)(ω) = 6.12

b . (a + iω)2 + b2

Write Z ∞ F (ω) =

f (t)e−iωt dt +

0

Z 0

f (t)e−iωt dt

−∞

and Rchange from t to −t in the second integral, then it follows that F (ω) = ∞ −2i 0 f (t) sin ωt dt.

6.13

a We have F (−ω) = F (ω) and F (ω) is even, so F (ω) = F (ω), and thus F (ω) is real. b We have F (−ω) = F (ω) (by part a) and since | F (ω) | = (F (ω)F (ω))1/2 , it follows that | F (ω) | = | F (−ω) |.

6.14

Calculate the spectrum in a direct way using exactly the same techniques as in example 6.3.3 (or use (6.20) and twice an integration by parts): F (ω) =

−2iω .

1 + ω2

6.16

R a/2 The spectrum is given by −a/2 te−iωt dt, which can be calculated using an integration by parts. The result is indeed equal to the formula given in example 6.3.

6.17

a From the differentiation rule (and differentiating the Fourier transform √ √ 2 of the Gauss function, of course) it follows that −iω πe−ω /4a /(2a a) is the spectrum of tf (t). b If we divide the Fourier transform of −f 0 (t) by 2a, then we indeed obtain the same result as in part a.

6.18

Two examples are the constant function f (t) = 0 (k arbitrary), and the √ 2 Gauss function e−t /2 with k = 2π. Using exercise 6.17a we obtain the √ 2 function te−t /2 with k = −i 2π.

6.19

Use table 3 for (t)e−at and then apply the differentiation rule in the frequency domain, then the result follows: (a+iω)−2 . (Differentiate (a+iω)−1 just as one would differentiate a real function.)

6.20

The function e−a| t | is not differentiable at t = 0. The function t3 (1 + t2 )−1 e.g. is not bounded.

6.21

Use the fact that limx→∞ xa e−x = 0 for all a ∈ R and change to the variable 2 x = at2 in tk /eat (separate the cases t ≥ 0 and t < 0). Then part a follows and, hence, part b also follows since we have a finite sum of these terms.

6.22

Apply the product rule repeatedly to get an expression in terms of the derivatives of f and g (this involves the binomial coefficients and is sometimes called Leibniz rule). Since f and g belong to S, tn (f (t)g(t))(m) will be a sum of terms belonging to S, and so the result follows.

Answers to selected exercises for chapter 6

6.23

6.25 6.26

25

R∞ We have that ( ∗ )(t) = 0 (t − τ ) dτ . Now treat the cases t > 0 and t ≤ 0 separately, then it follows that ( ∗ )(t) = (t)t. (If t ≤ 0, then t − τ < 0 for τ > 0 and R t so (t − τ ) = 0; if t > 0 then (t − τ ) = 0 for τ > t and the integral 0 1 dτ = t remains.) Since (t)t is not absolutely integrable, the function ( ∗ )(t) is not absolutely integrable.

R∞ From the causality of f it follows that (f ∗ g)(t) = 0 f (τ )g(t − τ ) dτ . For Rt t < 0 this is 0. For t ≥ 0 it equals 0 f (τ )g(t − τ ) dτ . a

We use the definition of convolution and then split the integral at τ = 0: Z ∞ Z 0 (e−| v | ∗ e−| v | )(t) = e−τ e−| t−τ | dτ + eτ e−| t−τ | dτ. −∞

0

First we take t ≥ 0. Then − | t − τ | = τ −t for τ < 0. Furthermore we have for τ > t that − | t − τ | = t − τ and for 0 ≤ τ < t that − | t − τ | = τ − t.

Hence, Z t Z ∞ Z 0 (e−| v | ∗ e−| v | )(t) = e−t dτ + et−2τ dτ + e2τ −t dτ.

0

−∞

t

A straightforward calculation of these integrals gives (1 + t)e−t .

Next we take t < 0. Then − | t − τ | = t−τ for τ > 0. Furthermore we have for τ < t that − | t − τ | = τ − t and for t ≤ τ < 0 that − | t − τ | = t − τ .

Hence, Z ∞ Z 0 Z t (e−| v | ∗ e−| v | )(t) = et−2τ dτ + et dτ + e2τ −t dτ.

0

t

−∞

A straightforward calculation of these integrals gives (1 − t)et . b Use the result from section 6.3.3 and the convolution theorem to obtain the spectrum (2(1 + ω 2 )−1 )2 = 4/(1 + ω 2 )2 . c Since (1 + | t |)e−| t | = e−| t | + | t | e−| t | and the spectrum of e−| t | is 2(1 + ω 2 )−1 , we only need to determine the spectrum of f (t) = | t | e−| t | .

But f (t) = tg(t) with g(t) the function from exercise 6.14, whose spectrum we’ve already determined: G(ω) = −2iω(1 + ω 2 )−1 . Apply theorem 6.8 (differentiation rule in the frequency domain): the spectrum of f (t) is −G0 (ω)/i. Calculating this and taking the results together we obtain 4/(1 + ω 2 )2 , in agreement with part b.

6.28

a From the differentiation rule in the frequency domain we obtain that √ 2 the spectrum of tg(t) is iG0 (ω) = −iω 2πe−ω /2 . Since (F tg(t))(0) = 0, √ 2 we may apply the integration rule to obtain that F1 (ω) = − 2πe−ω /2 . b Apply the differentiation rule in the frequency domain with n = 2, then √ 2 F2 (ω) = 2π(1 − ω 2 )e−ω /2 . c Since f3 (t) = f2 (t − 1), it follows from the shift property that F3 (ω) = e−iω F2 (ω). √ 2 d From part a and exercise 6.9 it follows that F4 (ω) = (− 2πe−(ω−4) /2 + √ 2 2πe−(ω+4) /2 )/2i. e Use the scaling property from table 4 with c = 4, then F5 (ω) = G(ω/4)/4.

6.29

b Since p1 (τ ) = 0 for | τ | > 21 and 1 for | τ | < 21 , we have Z 1/2 (p1 ∗ p3 )(t) = p3 (t − τ ) dτ. −1/2

26

Answers to selected exercises for chapter 6

Here p3 (t − τ ) 6= 0 only if t − 3/2 ≤ τ ≤ t + 3/2. Moreover, we have that −1/2 ≤ τ ≤ 1/2, and so we have to separate the cases as indicated in the textbook: if t > 2, then (p1 ∗ p3 )(t) = 0; if t < −2, then also R 1/2 (p1 ∗ p3 )(t) = 0; if −1 ≤ t ≤ 1, then (p1 ∗ p3 )(t) = −1/2 1 dτ = 1; if R 1/2 1 < t ≤ 2, then (p1 ∗ p3 )(t) = t−3/2 1 dτ = 2 − t; finally, if −2 ≤ t < −1, R t+3/2 then (p1 ∗ p3 )(t) = −1/2 1 dτ = 2 + t. c Apply the convolution theorem to T (t) = (p1 ∗p3 )(t), then the spectrum of T (t) follows: 4 sin(ω/2) sin(3ω/2)/ω 2 .

Answers to selected exercises for chapter 7

7.1

From the spectra calculated in exerices 6.2 to 6.5 it follows immediately that the limits for ω → ±∞ are indeed 0: they are all fractions with a bounded numerator and a denominator that tends to ±∞. As an example we have from exercise 6.2 that limω→±∞ 1/(a + iω) = 0.

7.2

Use table 3 with a = 2A and substitute ω = s − t.

7.3

Take C > 0, then it follows by first changing from the variable Au to v and then applying (7.3) that Z C Z AC sin Au sin v π lim du = lim dv = .

A→∞ 0 A→∞ 0 u v 2

7.4

Split 1/(a+iω) into the real part 1/(1+ω 2 ) and the imaginary part −ω/(1+ ω 2 ). The limit of A → ∞ of the integrals over [−A, A] of these parts gives limA→∞ 2 arctan A = π for the real part and limA→∞ (ln(1 + A2 ) − ln(1 + (−A)2 )) = 0 for the imaginary part.

7.6

2 a In exercise 6.9b it was shown that F (ω) R ∞ = 2i sin(πω)/(ω R π − 1). The function f (t) is absolutely integrable since −∞ | f (t) | dt = −π | sin t | dt < ∞. Moreover, f (t) is piecewise smooth, so all conditions of the fundamental theorem are satisfied. We now show that the improper integral of F (ω) exists. First, F (ω) is continuous on R according to theorem R ∞ 6.10, so it is integrable over e.g. [−2, 2]. Secondly, the integrals 2 F (ω) dω and R −2 F (ω) dω both exist. For the former integral this can be shown as −∞ follows (the other integral can be treated similarly): ˛Z ∞ ˛ Z ∞ ˛ ˛ 2 ˛ ˛≤ F (ω) dω dω ˛ ˛ ω2 − 1 2 2

since | 2i sin(πω) | ≤ 2 (and ω 2 − 1 > 0 for ω > 2). The integral in the right-hand side is convergent. b Apply the fundamental theorem, then Z ∞ 2i sin(πω) iωt 1 f (t) = e dω 2π −∞ ω 2 − 1 for all t ∈ R (f is continuous). Now use that F (ω) is an odd function and that 2 sin πω sin ωt = cos(π − t)ω − cos(π + t)ω, then Z 1 ∞ cos(π − t)ω − cos(π + t)ω dω. f (t) = π 0 1 − ω2 7.8

a In exercise 6.15b it was shown that Fs (ω) = (1 − cos aω)/ω. This exercise used the odd extension to R. So f (t) is odd and using (7.12) we thus obtain Z 2 ∞ 1 − cos aω 1 sin ωt dω = (f (t+) + f (t−)). π 0 ω 2 Since f (t) is continuous for t > 0 and t 6= a we have for these values that Z 2 ∞ 1 − cos aω f (t) = sin ωt dω. π 0 ω

27

28

Answers to selected exercises for chapter 7

b At t = a the function is discontinuous, so we have convergence to 1 (f (a+) + f (a−)) = 12 .

2 7.10

If we take g(t) = 0 in theorem 7.4, then G(ω) = 0 and so we get the statement: if F (ω) = 0 on R, then f (t) = 0 at all points where f (t) is continuous. We now prove the converse. Take f (t) and g(t) as in theorem 7.4 with spectra F (ω) and G(ω) and assume that F (ω) = G(ω) on R. Because of the linearity of the Fourier transform, (F −G)(ω) is the spectrum of (f − g)(t); but (F − G)(ω) = F (ω) − G(ω) = 0. From our assumption it now follows that (f − g)(t) = 0 at all points where (f − g)(t) is continuous.

Hence f (t) = g(t) at all points where f (t) and g(t) are continuous, which is indeed theorem 7.4.

7.11

The spectrum of pa (t) is 2 sin(aω/2)/ω (table 3). From duality it then follows that the spectrum of sin(at/2)/t is πpa (ω) at the points where pa (t) is continuous; at ω = ±a/2 we should take the value π/2. (We can apply duality since the Fourier integral exists as improper integral; this is exercise 7.5b).

7.12

The spectrum of qa (t) is F (ω) = 4 sin2 (aω/2)/(aω 2 ) (see table 3). From duality it then follows that the spectrum of sin2 (at/2)/t2 is (aπ/2)qa (ω). (We can apply duality since qa is continuous, piecewise smooth, and absolutely integrable and since its Fourier integral exists as improper integral; this latter fact follows immediately if we use that F (ω) is even and continuous and that e.g. F (ω) ≤ 1/ω 2 for ω ≥ 1).

7.14

The function 1/(a + iω) is not integrable on R (see exercise 6.2), so duality cannot be applied.

7.15

These results follow immediately from duality (and calculating the right p 2 2 constants). For example: π/ae−t /4a ↔ 2πe−aω , now divide by 2π.

7.16

This is an important exercise: it teaches to recognize useful properties. a Complete the square, then one can apply a shift in time: f (t) = 1/(1 + (t − 1)2 ). Since the spectrum of 1/(1 + t2 ) is πe−| ω | , the result follows: πe−iω e−| ω | . b Here we have a shift from t to t − 3; from the spectrum of sin 2πt/t the result follows: πe−3iω p4π (ω) with value 21 at ω = ±2π. c We now have 1/(t2 − 4t + 7), multiplied by a sine function. The sine function is easy to deal with using exercise 6.9 (a variant of the modulation theorem). As in a we complete√the square and note that 1/(3 + (t − 2)2 ) √ has spectrum F (ω) = πe−2iω e− 3| ω | / 3. Hence, the result is now (F (ω − 4) − F (ω + 4))/2i. d We use that 3πq6 (ω) is the spectrum of sin2 (3t)/t2 and apply a shift in time from t to t − 1, then the result is 3πe−iω q6 (ω).

7.17

Again, this is an important exercise: it teaches to recognize useful properties for the inverse transform. a We immediately use table 3 to obtain that 1/(4 + ω 2 ) is the spectrum of f (t) = e−2| t | /4. b Apply a shift in the frequency domain to the spectrum πp2a (ω) of sin(at)/t, then it follows that f (t) = (eiω0 t + e−iω0 t ) sin(at)/(πt), so f (t) = 2 cos(ω0 t) sin(at)/(πt). c As in part b it follows that f (t) = 3e9it /(π(t2 + 9)).

7.19

From the convolution theorem it follows that F (Pa ∗ Pb )(ω) = (F Pa )(ω) ·

Answers to selected exercises for chapter 7

29

(F Pb )(ω) = e−(a+b)| ω | , where we also used table 3. But also (F Pa+b )(ω) = e−(a+b)| ω | , and since F is one-to-one (theorem 7.4) it then follows that Pa+b = Pa ∗ Pb .

7.21

a Use the result of exercise 6.14 (G(ω) = −2iω/(1+ω 2 )), the fundamental R∞ R∞ theorem and the fact that the spectrum is odd to change from −∞ to 0 .

It then follows that (use x instead of ω) Z ∞ x sin xt π dt = e−t .

2 1 + x 2 0 Since g is not continuous at t = 0, this result is not correct at t = 0. Here one should take the average of the jump, which is 0.

R∞ b We apply Parseval (formula (7.19)) and calculate −∞ | g(t) |2 dt = R0 R R ∞ ∞ e2t dt + 0 e−2t dt, which is 1. In −∞ | G(ω) |2 dω we can use the −∞ fact that the integrand is even. Writing x instead of ω, the result follows.

7.22

Parseval (7.18)R with f (t) = e−a| t | and g(t) = e−b| t | and calculate RUse ∞ ∞ f (t)g(t) dt = 2 0 e−(a+b)t dt = 2/(a + b). The spectra of f and g are −∞ 2 2 2a/(a + ω ) and 2b/(b2 + ω 2 ) (table 3).

7.23

a Since sin4 t/t4 is the square of sin2 t/t2 and (F sin2 t/t2 )(ω) = πq2 (ω) (table 3), it follows from the convolution theorem in the frequency domain that (F sin4 t/t4 )(ω) R ∞ = (π/2)(q2 ∗ q2 )(ω). b The integral −∞ sin4 t/t4 dt is the Fourier transform of sin4 x/x4 calR∞ culated at ω = 0, hence −∞ sin4 x/x4 dx = (π/2)(q2 ∗ q2 )(0). Using that q2 is an even function we obtain that Z ∞ Z 2 (q2 ∗ q2 )(0) = q2 (t)q2 (−t) dt = 2 (1 − t/2)2 dt. −∞

0

This integral equals 4/3 and so 7.24

R∞ −∞

sin4 x/x4 dx = 2π/3.

From table 3 we know that e−| t | /2 ↔ 1/(1 + ω 2 ). By the convolution theorem we then know that the spectrum of f (t) = (e−| v | /2∗e−| v | /2)(t) is 1/(1 + ω 2 )2 . Calculating this convolution product at t = 0 gives f (0) = 1/4 (or use exercise 6.26a, where it was shown that f (t) = (1 + | t |)e−| t | /4).

Now apply the fundamental theorem (formula (7.9)) at t = 0 and use that the integrand is even. We then obtain Z 1 ∞ 1 1 dω = f (0) = , π 0 (1 + ω 2 )2 4 which is indeed the case a = b = 1 from exercise 7.22.

7.26

∗

2

The Gauss function f (t) = e−at belongs to S and so we can apply Poisson’s p 2 summation formula. Since F (ω) = π/ae−ω /4a (see table 3), it follows from (7.23) with T = 1 that ∞ X n=−∞

2

e−an =

∞ X p 2 2 π/a e−π n /a . n=−∞

Replacing a by πx the result follows.

7.27∗

Take f (t) = a/(a2 +t2 ), then F (ω) = πe−a| ω | (see table 3); we can then use (7.22) with T = 1 (in example 7.8 the conditions were verified) to obtain

30

Answers to selected exercises for chapter 7

∞ X

a =π 2 + (t + n)2 a n=−∞

∞ ∞ X X 1+ e−2πn(a+it) + e−2πn(a−it) n=1

! .

n=1

Here we have also split a sum in terms with n = 0, n > 1 and n < −1, and then changed from n to −n in the sum with n < −1. The sums in the righthand side are geometric series with ratio r = e−2π(a+it) and r = e−2π(a−it) respectively. Note that | r | < 1 since a > 0. Using the formula for the sum of an infinite geometric series (example 2.16), then writing the result with a common denominator, and finally multiplying everything out and simplifying, it follows that ∞ a X 1 1 − e−4πa . = 2 2 −4πa π n=−∞ a + (t + n) 1+e − e−2πa (e2πit + e−2πit )

Multiplying numerator and denominator by e2πa the result follows.

7.28

a To determine the spectrum we write sin t = (eit − e−it )/2 and calculate the integral defining F (ω) in a direct way: „Z π « Z π 1 i(1−ω)t −i(1+ω)t F (ω) = e dt − e dt .

2i 0 0 Writing the result with a common denominator and using the fact that eπi = e−πi = −1 gives F (ω) = (1 + e−iωπ )/(1 − ω 2 ). From theorem 6.10 we know that F (ω) is continuous, so we do not have to calculate F (ω) at the exceptional points ω = ±1. b Apply the fundamental theorem, so (7.9), noting that f (t) is continuous on R. We then obtain Z ∞ 1 1 + e−iωπ iωt f (t) = e dω.

2π −∞ 1 − ω 2 Split the integral at t = 0 and change from ω to −ω in the integral over (−∞, 0]. Then Z ∞ iωt 1 e + e−iωt + eiω(t−π) + e−iω(t−π) f (t) = dω, 2π 0 1 − ω2 which leads to the required result. c Take t = π/2 in part b and use that f (π/2) = 1, thenRthe result follows. π d Apply Parseval’s identity (7.19) to f and use that 0 sin2 t dt = π/2, then it follows that Z ∞ 1 π | F (ω) |2 dω = .

2π −∞ 2 −iωπ/2 Since˛ F (ω) can cos(ωπ/2)/(1 − ω 2 ) and we have ˛ be rewritten as 2e ˛ −iωπ/2 ˛ 2 that ˛ e ˛ = 1, it follows that | F (ω) | = 4 cos2 (ωπ/2)/(1 − ω 2 )2 . This integrand being even, the result follows.

7.29

a We know from table 3 that p2a (t) ↔ 2 sin aω/ω and e−| t | ↔ 2/(ω 2 + 1).

From the convolution theorem it then follows that p2a (v)∗e−| v | ↔ 4f (ω) = G(ω). b We now determine g explicitly by calculating the convolution product (use the definition of p2a ): Z t+a Z a (p2a (v) ∗ e−| v | )(t) = e−| t−τ | dτ = e−| u | du −a

t−a

Answers to selected exercises for chapter 7

31

where we changed to the variable u = t − τ . Now if −a ≤ t ≤ a, then t − a ≤ 0 ≤ t + a and so Z t+a Z 0 u −| v | e−u du = 2 − 2e−a cosh t. e du + (p2a (v) ∗ e )(t) = t−a

0

If t > a, then t − a > 0 and so Z t+a e−u du = 2e−t sinh a. (p2a (v) ∗ e−| v | )(t) = t−a

Finally, if t < −a, then t + a < 0 and so Z t+a eu du = 2et sinh a. (p2a (v) ∗ e−| v | )(t) = t−a

c The function g from part b is continuous at t = a since limt↓a g(t) = 2e−a sinh a = e−a (ea − e−a ) = 1 − e−2a and g(a) = limt↑a g(t) = 2 − 2e−a cosh a = 2 − e−a (ea + e−a ) = 1 − e−2a . In the same way it follows that g(t) is continuous at t = −a. So g(t) is a piecewise smooth function which moreover is continuous. Also, g(t) is certainly absolutely integrable since e−| t | is absolutely integrable over | t | > a. Finally, the Fourier integral exists as improper ˛ Riemann˛ integral since G(ω) is even absolutely integrable: | G(ω) | ≤ 4/ ˛ ω(1 + ω 2 ) ˛. We can now apply the duality rule (theorem 7.5) and it then follows that G(−t) ↔ 2πg(ω), so f (−t) ↔ πg(ω)/2. Since f (−t) = f (t) we thus see that F (ω) = π − πe−a cosh ω for | ω | ≤ a and F (ω) = πe−| ω | sinh a for | ω | > a.

Answers to selected exercises for chapter 8

8.1

b For t 6= 0 we have that lim R ∞a↓0 Pa (t) = 0, while for t = 0 we have that lima↓0 Pa (t) = ∞. Since −∞ Pa (t) dt = 1, we see that Pa (t) fits the description of the delta function. c From table 3 it follows that Pa (t) ↔ e−a| ω | and lima↓0 e−a| ω | = 1.

Combining this with part b shows that it is reasonable to expect that the spectrum of δ(t) is 1.

8.2

a Since φ(a) ∈ C for all φ ∈ S, it follows from (8.10) that δ(t − a) is a mapping from S to C. For c ∈ C and φ ∈ S we have that hδ(t − a), cφi = (cφ)(a) = c hδ(t − a), φi , and for φ1 , φ2 ∈ S we have hδ(t − a), φ1 + φ2 i = (φ1 + φ2 )(a) = hδ(t − a), φ1 i + hδ(t − a), φ2 i .

So δ(t − a) is a linear mapping from S to C, hence a distribution. b Taking the limit inside the integral in (8.1) gives « Z ∞ „ 1 2 sin aω lim f (t − ω) dω = f (t) 2π a→∞ ω −∞ for any absolutely integrable and piecewise smooth function f (t) on R that is continuous at t. Using (8.3) this can symbolically be written as (take t = a) Z ∞ δ(ω)f (a − ω) dω = f (a) −∞

and by changing from ω to a − t we then obtain Z ∞ δ(a − t)f (t) dt = f (a). −∞

8.4

Using δ(a − t) = δ(t − a), which by (8.3) is reasonable to expect (see section 8.4 for a proof), this indeed leads to (8.11).

R∞ Since h1, φi = −∞ φ(t) dt ∈ C for all φ ∈ S, it follows that 1 is a mapping from S to C. The linearity of this mapping follows from the linearity of integration: for c ∈ C and φ ∈ S we have that Z ∞ Z ∞ h1, cφi = (cφ)(t) dt = c φ(t) dt = c h1, φi , −∞

−∞

and for φ1 , φ2 ∈ S we have Z ∞ Z ∞ Z ∞ h1, φ1 + φ2 i = (φ1 + φ2 )(t) dt = φ1 (t) dt + φ2 (t) dt, −∞

−∞

−∞

so h1, φ1 + φ2 i = h1, φ1 i + h1, φ2 i. This proves that 1 is a linear mapping from S to C, hence a distribution.

8.5

For φ ∈ S there exists a constant M > 0 such that (e.g.) (1+t2 ) | φ(t) | ≤ M for all t ∈ R. Hence, ˛Z ∞ ˛ Z ∞ Z ∞ ˛ ˛ 1 ˛ ˛≤ | φ(t) | dt ≤ M dt < ∞ φ(t) dt ˛ ˛ 1 + t2 0 0 0

32

Answers to selected exercises for chapter 8

8.7

33

R∞ (the latter integral equals [arctan]∞ 0 = π/2). The integral 0 φ(t) dt thus exists and one can now show that is indeed a distribution precisely as in exercise 8.4 (linearity of integration).

R∞ In example 8.4 it was already motivated why the integral −∞ | t | φ(t) dt exists: there exists a constant M > 0 such that (e.g.) (1 + t2 ) | tφ(t) | ≤ M for all t ∈ R. Hence, ˛Z ∞ ˛ Z ∞ Z ∞ ˛ ˛ 1 ˛ ˛≤ | tφ(t) | dt ≤ M | t | φ(t) dt dt < ∞. ˛ ˛ 2 −∞ 1 + t −∞ −∞ So h| t | , φi exists and one can now show that | t | is indeed a distribution precisely as in exercise 8.4 (linearity of integration).

8.9

a For the integral over [−1, 1] we have Z 1 Z 1 | t |−1/2 dt = 2 t−1/2 dt = [4t1/2 ]10 = 4, −1

0

√ R∞ hence, | t | is integrable over [−1, 1]. Since 0 t−1/2 dt = 2 limR→∞ R does not exist, | t |−1/2 is not integrable over R. R∞ b We first show that −∞ | t |−1/2 φ(t) dt exists for φ ∈ S. To do so, we split the integral in an integral over [−1, 1] and over | t | ≥ 1. For the first integral we note that | φ(t) | ≤ M1 for some constant M1 > 0. From part a we then get ˛Z 1 ˛ Z 1 Z 1 ˛ ˛ −1/2 −1/2 ˛ ˛≤ | t | φ(t) dt | t | | φ(t) | dt ≤ M | t |−1/2 dt < ∞.

1 ˛ ˛ −1/2

−1

−1

−1

−1/2

For the second integral we use that | t | ≤ 1 for | t | ≥ 1. Hence, ˛Z ˛ Z Z ∞ ˛ ˛ ˛ ˛ | t |−1/2 φ(t) dt ˛ ≤ | φ(t) | dt ≤ | φ(t) | dt. ˛ ˛ | t |≥1 ˛ | t |≥1 −∞ In example 8.1 itEhas been shown that the latter integral exists. This shows D −1/2 that | t | , φ exists and one can now show that | t |−1/2 is indeed a distribution (linearity of integration; see e.g. exercise 8.4).

8.10

a For φ ∈ S there exists a constant M > 0 such that (e.g.) t2 ) | tφ(t) | ≤ M for all t ∈ R. Hence, ˛Z ∞ ˛ Z ∞ Z ∞ ˛ ˛ 1 ˛ dt < ∞. tφ(t) dt ˛˛ ≤ | tφ(t) | dt ≤ M ˛ 1 + t2 −∞ −∞ −∞

(1 +

As in exercise 8.3 this shows that t defines a distribution.

8.12

a From the linearity for immediately that the com√ distributions itRfollows ∞ plex number 2φ(0) + i 3φ0 (0) + (1 + i) 0 ((φ(t) − φ(−t))dt is assigned. b This defines a distribution if 1, t and t2 are distributions. The first one is known from example 8.1, the other two from exercise 8.10. From definition 8.15 it thus follows that f (t) defines a distribution as well.

8.14

a From (8.17) it follows that the complex number −φ(3) (0) is assigned. b This number −φ(3) (0) is meaningfull for all functions that are 3 times continuously differentiable.

8.15

a First apply example 8.3, then definition 8.4, and finally integration by parts, then

34

Answers to selected exercises for chapter 8

˙

¸ (sgn t)0 , φ =

Z ∞

(φ0 (−t) − φ0 (t)) dt = [φ(t)]0−∞ − [φ(t)]∞ 0 = 2φ(0),

0

hence, (sgn t)0 = 2δ(t). b Since sgn t = 2 (t) − 1 (verify this), it follows from the linearity of differentiation that (sgn t)0 = 2 0 (t) = 2δ(t). Here we used that 10 = 0 and that 0 (t) = δ(t) (see (8.18)). c Since | t |0 = sgn t it follows from part a that | t |00 = (sgn t)0 = 2δ(t).

8.16

Since the function | t | from example 8.9 is continuously differentiable outside t = 0, it follows from the jump formula that | t |0 = sgn t (at t = 0 there is no jump and outside t = 0 this equality holds for the ordinary derivatives). The function from example 8.10 has a jump of magnitude 1 at t = 0, while for t < 0 the derivative is 0 and for t > 0 the derivative is − sin t. Hence, the jump formula implies that ( (t) cos t)0 = δ(t) − (t) sin t.

8.17

a The function pa has a jump of magnitude 1 at t = −a/2 and of magnitude −1 at t = a/2. Outside t = 0 the ordinary derivative is 0, so it follows from the jump formula that pa (t)0 = δ(t + a/2) − δ(t − a/2). b The function (t) sin t has no jump at t = 0, for t < 0 the ordinary derivative is 0 and for t > 0 the ordinary derivative is cos t, so it follows from the jump formula that ( (t) sin t)0 = (t) cos t.

8.18

a This is entirely analagous to exercises 8.10 and 8.12b. b The function is differentiable outside t = 1 and the ordinary derivative is 1 for t < 1 and 2t − 2 for t > 1. We denote this derivative as the distribution Tf 0 . At t = 1 the jump is 2, so according to the jump formula the derivative is Tf 0 + 2δ(t − 1).

8.19

From the jump formula it follows that the derivative as distribution is given by a (t)eat + δ(t), so f 0 (t) − af (t) = δ(t) as distributions.

8.22

Subsequently apply definition 8.6 and the definition of δ(t − a) in (8.10), then hp(t)δ(t − a), φi = (pφ)(a) = p(a)φ(a) = p(a) hδ(t − a), φi .

Now use definition 8.5, then hp(t)δ(t − a), φi = hp(a)δ(t − a), φi, which shows that p(t)δ(t − a) = p(a)δ(t − a).

8.23

a The definition becomes: hf (t)δ 0 (t), φi = hδ 0 (t), f φi. The product f φ of two continuously differentiable functions is continuously differentiable, so this definition is correct and it gives a mapping from S to C. The linearity follows immediately from the linearity of δ 0 (t), so f (t)δ 0 (t) is a distribution. b According to part a we have hf (t)δ 0 (t), φi = −(f φ)0 (0), where we also applied δ 0 (t) to f φ. Now apply the product rule for differentiation and write the result as −f 0 (0) hδ(t), φi + f (0) hδ 0 (t), φi = hf (0)δ 0 (t) − f 0 (0)δ(t), φi. c If f (t) = t, then f (0) = 0 and f 0 (0) = 1, so tδ 0 (t) = −δ(t); if f (t) = t2 then f (0) = 0 and f 0 (0) = 0, so t2 δ 0 (t) = 0.

8.25

First apply definition 8.6 and then the definition of pv(1/t) from example 8.5 to obtain Z Z tφ(t) ht · pv(1/t), φi = lim dt = lim φ(t) dt. α↓0 | t |≥α α↓0 | t |≥α t Since φ ∈R S is certainly integrable over R, the limit exists and it will be ∞ equal to −∞ φ(t) dt. Hence, t · pv(1/t) = 1.

8.26

Let T be an even distribution, then T (−t) = T (t) (definiton 8.8), so

Answers to selected exercises for chapter 8

35

hT (t), φ(t)i = hT (t), φ(−t)i for all φ ∈ S, where we used definition 8.7.

Similarly for odd T .

8.27

a From the definition of sgn t in example 8.3 it follows that hsgn t, φ(t)i = − hsgn t, φ(−t)i for all φ ∈ S. This shows that sgn t is odd according to exercise 8.26. Similarly for pv(1/t) (change from t to −t in the integrals defining pv(1/t)). b From the definition of | t | in example 8.4 it follows that h| t | , φ(t)i = h| t | , φ(−t)i for all φ ∈ S (change from t to −t in the integral defining | t |).

This shows that | t | is even according to exercise 8.26.

8.29

a

Applying (8.12) to f (t) gives Z 0 Z ∞ hTf , φi = 2tφ(t) dt + t2 φ(t) dt −∞

0

and in e.g. exercises 8.10, 8.12b and 8.18a we have seen that such integrals are well-defined for φ ∈ S. This gives a mapping from S to C and the linearity of this mapping follows precisely as in e.g. exercise 8.3 or 8.4.

Hence, f indeed defines a distribution Tf . b Apply the jump formula (8.21): outside t = 0 the function f is continuously differentiable with derivative f 0 (t) = 2t for t > 0 and f 0 (t) = 2 for t < 0. Note that f 0 again defines a distribution Tf 0 . At t = 0 the function has no jump, hence (8.21) implies that Tf0 = Tf 0 . c Again we have that the function f 0 is continuously differentiable outside t = 0 and f 00 (t) = 2 for t > 0 and f 00 (t) = 0 for t < 0. Let Tf 00 be the distribution defined by f 00 . At t = 0 the function f 0 has a jump f 0 (0+) − f 0 (0−) = 0 − 2 = −2, and according to (8.21) (applied to Tf 0 and using that Tf0 = Tf 0 and so Tf00 = Tf0 0 ) we have that Tf00 = Tf0 0 = Tf 00 + (f 0 (0+) − f 0 (0−))δ(t) = Tf 00 − 2δ(t).

The second derivative of f considered as distribution is the same as the second derivative of f outside t = 0, minus the distribution 2δ(t) at t = 0.

8.30

a Since δ 00 (t) can be defined for all twice continuously differentiable functions, the product f (t)δ 00 (t) can also be defined for all twice continuously differentiable functions f (t) by hf (t)δ 00 (t), φ(t)i = hδ 00 (t), f (t)φ(t)i. This is because it follows from the product rule that the product f (t)φ(t) is again twice continuously differentiable. b From part b and the definition of the second derivative of a distribution (formula (8.17) for k = 2) we obtain hf (t)δ 00 (t), φ(t)i = hδ(t), (f (t)φ(t))00 i.

Since (f (t)φ(t))00 = f 00 (t)φ(t) + 2f 0 (t)φ0 (t) + f (t)φ00 (t) we thus obtain that hf (t)δ 00 (t), φ(t)i = f 00 (0)φ(0) + 2f 0 (0)φ0 (0) + f (0)φ00 (0), which equals hf 00 (0)δ(t) − 2f 0 (0)δ 0 (t) + f (0)δ 00 (t), φ(t)i (φ ∈ S). This proves the identity. c Apply part b to the function f (t) = t2 and use that f (0) = f 0 (0) = 0 and f 00 (0) = 2, then t2 δ 00 (t) = 2δ(t). Next apply b to f (t) = t3 and use that f (0) = f 0 (0) = f 00 (0) = 0, then it follows that t3 δ 00 (t) = 0. d According to definition 8.7 we have that ˙ 00 ¸ ˙ ¸ δ (at), φ(t) = | a |−1 δ 00 (t), φ(a−1 t) .

Now put ψ(t) = φ(a−1 t), then the right-hand side equals | a |−1 ψ 00 (0).

Next we use the chain rule twice to obtain that ψ 00 (0) = a−2 φ00 (0). Hence hδ 00 (at), φ(t)i = | a |−1 a−2 hδ 00 , φi.

Answers to selected exercises for chapter 9

9.1

Let φ ∈ S. From theorem 6.12 it follows that the spectrum Φ belongs to S. Since T is a distribution, we then have that hT, Φi ∈ C, and so hFT, φi = hT, Φi ∈ C as well. So F T is a mapping from S to C. The linearity of F T follows from the linearity of T and F ; we will only give the necessary steps for hFT, cφi, since the rule for hFT, φ1 + φ2 i follows similarly. hFT, cφi = hT, F (cφ)i = hT, cΦi = c hT, Φi = c hFT, φi .

9.3

a Use table 5 to obtain that δ(t − 4) ↔ e−4iω . b Again use table 5 to obtain that e3it ↔ 2πδ(ω − 3). c First write the sine function as combination of exponentials, so sin at = (eiat − e−iat )/2i. From linearity and table 5 it then follows that sin at ↔ −πi(δ(ω − a) − δ(ω + a)). d First determine the spectrum of pv(1/t) and 4 cos 2t = 2e2it + 2e−2it using table 5 and then (again) apply linearity to obtain the spectrum 4π(δ(ω − 2) + δ(ω + 2)) + 2πsgn ω.

9.4

a From example 9.1 (or table 5) we obtain the result e−5it /2π. b See example 9.2: 2 cos 2t. c The spectrum of pv(1/t) is −πisgn ω (table 5). Note that 2 cos ω = eiω + e−iω and that the spectrum of δ(t − a) is e−iaω (table 5). Hence the answer is iπ −1 pv(1/t) + δ(t − 1) + δ(t + 1).

9.5

Let T be an even distribution with spectrum U . We have to show that U (−ω) = U (ω), so hU, φ(t)i = hU, φ(−t)i for all φ ∈ S (see exercise 8.26). But hU, φ(−t)i = hT, F φ(−t)i and from table 4 we know that (F φ(−t))(ω) = Φ(−ω) if Φ is the spectrum of φ. Since T is even, we have that hT, Φ(−ω)i = hT, Φ(ω)i. From these observations it follows that hU, φ(−t)i = hT, Φ(ω)i = hU, φ(t)i, which shows that U is even. Similarly for odd T .

9.7

It is obvious that (t) = (1 + sgn t)/2 by looking at the cases t > 0 and t < 0. Since 2πδ(ω) is the spectrum of 1 and −2ipv(1/ω) is the spectrum of (t), it follows that (t) has spectrum πδ(ω) − ipv(1/ω).

9.8

a Let φ ∈ S have spectrum Φ. From definition 9.1 and the action of δ 00 it follows that hFδ 00 , φi = hδ 00 , Φi = hδ, Φ00 i = Φ00 (0). From the differentiation rule in the frequency domain (table 4) with k = 2 we see that Φ00 (ω) = F ((−it)2 φ(t))(ω), and hence δ 00 ↔ −ω 2 is proven as follows: Z ∞ Z ∞ ˙ 00 ¸ F δ , φ = F (−t2 φ(t))(0) = − t2 φ(t) dt = − ω 2 φ(ω) dω, −∞ 00

−∞

2

¸ so hFδ , φi = −ω , φ for all φ ∈ S, proving the required result.

Parts b and c can be proven using similar steps.

9.9

˙

a Subsequently apply definitions 9.1 and 8.7: hFT (at), φi = hT (at), Φi = ˙ ¸ | a |−1 T, Φ(a−1 ω) (φ ∈ S having spectrum Φ). From table 4 we see that Φ(a−1 ω) = | a | (F φ(at))(ω), so it follows that hFT (at), φi = hT, F φ(at)i = hU, φ(at)i, where we again used definition 9.1 in Now again ˙ the final step. ¸ apply definition 8.7, then hFT (at), φi = | a |−1 U (a−1 ω), φ , which proves T (at) ↔ | a |−1 U (a−1 ω).

36

Answers to selected exercises for chapter 9

37

b We have that δ(4t+3) is the distribution δ(t+3) scaled by 4. According to the shift rule in the time domain (see table 6) it follows from δ(t) ↔ 1 that δ(t+3) ↔ e3iω . From part a it then follows that δ(4t+3) ↔ 4−1 U (ω/4) with U (ω) = e3iω . Hence, δ(4t + 3) ↔ 4−1 e3iω/4 . (This can also be solved by considering δ(4t + 3) as the distribution δ(4t) shifted over −3/4.)

9.11

From table 5 it follows that δ 0 (t) ↔ iω. Using (9.12) we then obtain that −itδ 0 (t) ↔ (iω)0 = i, so tδ 0 (t) ↔ −1. Exercise 8.23c gives: tδ 0 (t) = −δ(t) and since δ(t) ↔ 1 we indeed get tδ 0 (t) ↔ −1 again. Similarly we get tδ 00 (t) ↔ −2iω using (9.12) or using exercise 8.30b: tδ 00 (t) = −2δ 0 (t).

9.12

From iωT = 1 we may not conclude that T = 1/(iω) since there exist distributions S 6= 0 such that ωS = 0 (e.g. δ(ω)).

9.13

The linearity follows as in definition 8.6. The main point is that one has to show that eiat φ(t) ∈ S whenever φ ∈ S. So we˛ have to show that for ˛ ˛ ˛ any m, n ∈ Z+ there exists an M > 0 such that ˛ tn (eiat φ(t))(m) ˛ < M .

From the product rule for differentiation it follows that (eiat φ(t))(m) is a iat (k) + sum of ˛ terms of the˛ form ce φ (t) (k ∈ Z ). It is now sufficient to show ˛ ˛ that ˛ tn eiat φ(k) (t) ˛ < M for some M > 0 and all k, n ∈ Z+ . But since ˛ ˛ ˛ iat ˛ ˛ e ˛ = 1 this means that we have to show that ˛˛ tn φ(k) (t) ˛˛ < M for some M > 0 and all k, n ∈ Z+ , which indeed holds precisely because φ ∈ S.

9.15

From˙ definition¸9.1 ˙and the definition of e¸iat T (see exercise 9.13) it follows ¸ ˙ iat iat iat that F e T, φ = e T, Φ = T, e Φ (φ ∈ S having spectrum Φ). According to the shift property in the frequency domain (table 4) we have that eiat Φ(t) = F (φ(ω+a))(t) (note that for˙ convenience ¸ we’ve interchanged the role of the variables ω and t). Hence, F eiat T, φ = hT, F (φ(ω + a))(t)i = hU, φ(ω + a)i = hU (ω − a), φi, where we used definition 9.2 in the last step.

So we indeed have eiat T ↔ U (ω − a).

9.16

a Use table 5 for (t) and apply a shift in the time domain, then it follows that (t − 1) ↔ e−iω (πδ(ω) − ipv(1/ω)). b Use table 5 for (t) and apply a shift in the frequency domain, then it follows that eiat (t) ↔ πδ(ω − a) − ipv(1/(ω − a)). c We have (t) ↔ πδ(ω) − ipv(1/ω) and if we now write the cosine as a combination of exponentials, then we can use a shift in the frequency domain (as in part b) to obtain that (t) cos at ↔ 12 (πδ(ω − a) − ipv(1/(ω − a)) + πδ(ω + a) − ipv(1/(ω + a))). d Use that 1 ↔ 2πδ(ω) and δ 0 (t) ↔ iω (table 5), so 3i ↔ 6iπδ(ω) and (apply a shift) δ 0 (t − 4) ↔ e−4iω iω; the sum of these gives the answer. e First note that (t)sgn t = (t) and the spectrum of this is known; furthermore we have that t3 ↔ 2πi3 δ (3) (ω) (table 5), so the result is 2π 2 δ (3) (ω) + πδ(ω) − ipv(1/ω).

9.17

a Use table 5 for the sign function and apply a shift: 21 ieit sgn t. b Write sin t as a combination of exponentials and apply a shift to 21 isgn t, then we obtain the result 14 (sgn(t + 3) − sgn(t − 3)). c Apply reciprocity to (t), then we obtain (πδ(−t) − ipv(−1/t))/2π ↔ (ω). Now δ(−t) = δ(t) and pv(−1/t) = −pv(1/t), hence, the result is: 1 iπ −1 pv(1/t) + 12 δ(t).

2 d Apply the scaling property (table 6) to 1 ↔ 2πδ(ω) to obtain 1 ↔ 6πδ(3ω). Next we apply a shift in the frequency domain (table 6), which

38

Answers to selected exercises for chapter 9

results in e2it/3 ↔ 6πδ(3ω − 2). Using the differentiation rule in the frequency domain (table 6) we obtain from 1/2π ↔ δ(ω) that (−it)2 /2π ↔ δ 00 (ω). From linearity it then follows that (3−1 e2it/3 − t2 )/2π ↔ δ(3ω − 2) + δ 00 (ω).

9.19

We know that δ 0 ∗ T = T 0 , so δ 0 ∗ | t | = | t |0 = sgn t (by example 8.9).

9.20∗

According to definition 9.3 we have that hT (t) ∗ δ(t − a), φi = hT (τ ), hδ(t − a), φ(t + τ )ii .

Since hδ(t − a), φ(t + τ )i = φ(a + τ ), the function τ → hδ(t − a), φ(t + τ )i belongs to S, so T (t) ∗ δ(t − a) exists and hT (t) ∗ δ(t − a), φi = hT (τ ), φ(a + τ )i = hT (τ − a), φ(τ )i (the last step uses definition 9.2). This proves that T (t)∗δ(t−a) = T (t−a).

∗

Use exercise 9.20 with T (t) = δ(t − b). The convolution theorem leads to the obvious e−ibω e−iaω = e−i(a+b)ω .

9.24

a Use table 5: δ(t − 3) ↔ e−3iω . b Since δ(t + 4) ↔ e4iω (as in part a) and cos t = (eit + e−it )/2 we apply a shift in the frequency domain: cos tδ(t + 4) ↔ (e4i(ω−1) + e4i(ω+1) )/2. c From table 5 we have (t) ↔ πδ(ω)−ipv(1/ω). Apply the differentiation rule in the time domain (with n = 2), then we obtain t2 (t) ↔ −πδ 00 (ω) + ipv(1/ω)00 . d Apply the differentiation rule in the time domain to the result obtained in exercise 9.16c, then it follows that (2 (t) cos t)0 ↔ iω(πδ(ω − 1) + πδ(ω + 1) − ipv(1/(ω − 1)) − ipv(1/(ω + 1))). e Since δ(t) ↔ 1 it follows from first the scaling property and then a shift in the time domain that δ(7(t − 1/7)) ↔ e−iω/7 7. Finally apply the differentiation rule in the time domain to obtain the result: (δ(7t − 1))0 ↔ iωe−iω/7 7. f This is a convergent Fourier series and so we can determine the spectrum term-by-term. Since 1 ↔ 2πδ(ω) and e(2k+1)it ↔ 2πδ(ω − (2k + 1)) we obtain the following result:

9.21

π 2 δ(ω) − 4

∞ X

(2k + 1)−2 δ(ω − (2k + 1)).

k=−∞

9.25

a From table 5 we know that eit ↔ 2πδ(ω − 1) and similarly for e−it .

Hence, iπ −1 sin t ↔ δ(ω − 1) − δ(ω + 1). b Apply the differentiation rule in the time domain (with n = 2) to δ(t) ↔ 1, then −δ 00 (t) ↔ ω 2 . c From table 5 we obtain that δ(t + 12 )/4 ↔ eiω/2 /4. d From table 5 (and linearity) we obtain that (δ(t + 1) − δ(t − 1))/2i ↔ (eiω −e−iω )/2i, which is sin ω. Now apply differentiation in the time domain (with n = 3), then we obtain that (δ (3) (t + 1) − δ (3) (t − 1))/2 ↔ ω 3 sin ω. e From exercise 9.4c and a shift in the frequency domain it follows that e4it (δ(t + 1) + δ(t − 1))/2 ↔ cos(ω − 4).

9.26

a In exercise 9.25b it was shown that −δ(t)00 ↔ ω 2 . Applying a shift in the frequency domain leads to −eit δ 00 (t) ↔ (ω − 1)2 . b From the differentiation rule in the time domain and table 5 it follows as in exercise 9.25b that δ 0 (t) ↔ iω and δ 00 (t) ↔ −ω 2 , so −δ 00 (t) + 2iδ 0 (t) + δ(t) ↔ ω 2 − 2ω + 1.

Answers to selected exercises for chapter 9

39

c Since (ω − 1)2 = ω 2 − 2ω + 1, the results in part a and b should be the same. Using exercise 8.30b with f (t) = eit we indeed obtain that eit δ 00 (t) = δ 00 (t) − 2iδ 0 (t) − δ(t).

9.27

a From exercise 9.25b it follows that δ 00 (t) ↔ −ω 2 . The convolution theorem then implies that T ∗ δ 00 (t) ↔ −ω 2 U where U is the spectrum of T . This also follows by applying the differentiation rule in the time domain to T 00 , which equals T ∗ δ 00 (t) by (9.21). b As noted in part a we have that δ 00 ∗ | t | = | t |00 . In exercise 8.15c it was shown that | t |00 = 2δ, so we indeed get δ 00 ∗ | t | = 2δ. Now let V be the spectrum of | t |. Since δ 00 ↔ −ω 2 and δ ↔ 1 it then follows as in part a from the convolution theorem that ω 2 V = −2.

Answers to selected exercises for chapter 10

10.1

a When the system is causal, then the response to the causal signal δ(t) is again causal, so h(t) is causal. On R t the other hand, if h(t) is causal, then it follows that y(t) = (u ∗ h)(t) = −∞ h(t − τ )u(τ ) dτ and if we now have a causal input u, then the integral will be 0 for t < 0 and so y(t) is causal as well, proving that the system is causal. b When the system is real, then the response to the real signal δ(t) is again real, so h(t) is real. On the other hand, if h(t) is real, then it follows from the integral for y(t) = (u ∗ h)(t) that if u(t) is real, then y(t) is also real, proving that the system is real.

10.2

a If we substitute u(t) = δ(t) then it follows that h(t) = δ(t−1)+ (t)e−2t , so h(t) is causal and real and according to exercise 10.1 the system is then causal and real. b We R t substitute u(t) = (t), then it follows for t ≥ 0 that a(t) = (t − 1) + 0 e−2(t−τ ) dτ = (t − 1) + 21 (1 − e−2t ), while for t < 0 the integral is 0 and so a(t) = (t − 1). This result can be written for all t as a(t) = (t − 1) + 12 (t)(1 − e−2t ).

10.3

We can express p2 (t) as p2 (t) = (t + 1) − (t − 1). Since a(t) is (by definition) the response to (t) and we have a linear system, the response to p2 (t) = (t + 1) − (t − 1) is a(t + 1) − a(t − 1).

10.5

a We differentiate a(t) in distribution sense, which results in h(t) = δ(t) − (t)e−3t (2 sin 2t + 3 cos 2t), since a(t) has a jump at t = 0 of magnitude 1 and we can differentiate in ordinary sense outside t = 0. b We use theorem 10.1, which implies that we may ignore the delta component and only have to Rshow that (t)e˛−3t (2 sin 2t +3 ˛ cos 2t) is absolutely ˛ R∞ ∞˛ integrable. Since both 0 R˛ e−3t sin 2t ˛ dt and 0 ˛ e−3t cos 2t ˛ dt exist ∞ (e.g., both are smaller than 0 e−3t dt), this is indeed the case and hence the system is stable.

10.6

a The impulse response is h1 ∗ h2 . b If the input for the first system is bounded, then the output is bounded since the system is stable. This output is then used as input for the second system, which is again stable. So the output of the second system is again bounded. This means that the cascade system itself is stable: the response to a bounded input is bounded.

10.7

a The spectrum of h is H(ω) = 1+1/(1+iω)2 since δ ↔ 1 and te−t (t) ↔ 1/(1 + iω)2 . b Since eiωt 7→ H(ω)eiωt it follows that the response is given by eiωt (1 + 1/(1 + iω)2 ).

10.8

a We need to determine the inverse Fourier transform of the function H(ω) = cos ω/(ω 2 + 1). First note that 21 e−| t | ↔ 1/(ω 2 + 1). Writing the cosine as a combination of exponentials we obtain from the shift rule that h(t) = 21 (e−| t+1 | + e−| t−1 | ). b It now suffices to use the time-invariance of the system: the reponse to δ(t) is h(t), so the response to δ(t − 1) is h(t − 1).

10.9

a We need to determine the inverse Fourier transform of the function from figure 10.3, which is H(ω) = qωc (ω). This is h(t) = 2 sin2 (ωc t/2)/(πωc t2 ).

40

Answers to selected exercises for chapter 10

41

b The function u has a Fourier series with terms cn eint (note that ω0 = 1).

But the response to eint is H(n)eint and H(n) = 0 for n > 1 and n < −1.

So we only have to determine c0 , c1 and c−1 . These can easily be calculated from the defining integrals: c0 = 21 and c−1 = c1 = −1/π. Hence, the 2 response follows: y(t) = H(0)c0 +H(1)c1 eit +H(−1)c−1 e−it = 12 − 3π cos t.

10.11

a Put s = iω and apply partial fraction expansion to the system function (s + 1)(s − 2)/(s − 1)(s + 2). A long division results in 1 − 2s/(s − 1)(s + 2) and a partial fraction expansion then gives (s + 1)(s − 2) 2 1 4 1 2 1 4 1 =1− − =1− − . (s − 1)(s + 2) 3s−1 3s+2 3 iω − 1 3 iω + 2 Now δ(t) ↔ 1 and (t)e−2t ↔ 1/(iω + 2) (table 3, no. 7) and from time 1 = reversal (scaling with a = −1 from table 4, no. 5) it follows for iω−1 t t −2t −1 1 2 4 that − (−t)e ↔ .

Hence, h(t) = δ(t) + (−t)e − (t)e . i(−ω)+1 iω−1 3 3 b The impulse reponse h(t) is not causal, so the system is not causal. c The modulus of H(ω) is 1, so it is an all-pass system and from Parseval it then follows that the energy-content of the input is equal to the energycontent of the output (if necessary, see the textbook, just above example 10.7).

10.13

a From the differential equation we immediately obtain the frequency response: H(ω) =

ω 2 − ω02 √ . ω 2 − i 2ω0 ω − ω02

b Write the cosine as a combination of exponentials, then it follows from eiωt 7→ H(ω)eiωt that y(t) = (H(ω0 )eiω0 t + H(−ω0 )e−iω0 t )/2. However H(±ω0 ) = 0, so y(t) = 0 for all t. c Note that we cannot use the method from part b. Instead we use (10.6) to determine the spectrum of the response y(t). From table 5 we obtain that (t) ↔ pv(1/iω) + πδ(ω). Write the cosine as a combination of exponentials, then it follows from the shift rule that the spectrum of u(t) = cos(ω0 t) (t) is given by U (ω) = pv(1/(2i(ω − ω0 )) + pv(1/(2i(ω + ω0 )) + (π/2)δ(ω − ω0 ) + (π/2)δ(ω + ω0 ). To determine Y (ω) = H(ω)U (ω) we use that H(ω)δ(ω ± ω0 ) = H(±ω0 )δ(ω ± ω0 ) = 0. Hence, writing √ everything with a common denominator, Y (ω) = ω/i(ω 2 − i 2ω0 ω − ω02 ).

Put s = iω and apply√partial fraction expansion √ to obtain that 2Y (ω) = (1 + i)/(s + ω0 (1 − i)/ 2) + (1 − i)/(s + ω0 (1 + i)/ 2). The inverse Fourier √ √ √ transform of this equals e−ω0 t/ 2 (cos(ω0 t/ 2) − sin(ω0 t/ 2)) (t).

10.14

a From the differential equation we immediately obtain the frequency response: H(ω) =

1 + α/2 1 iω + 1 + α = + , 2iω + α 2 2iω + α

where we also used a long division. Using the tables it then follows that h(t) = 12 (δ(t) + (1 + α/2) (t)e−tα/2 ). b We have to interpret the differential equation ‘the other way around’, so with input and output interchanged. This means that the system function is now 1/H(ω), so 2iω + α 2+α =2− iω + 1 + α iω + 1 + α

42

Answers to selected exercises for chapter 10

where we also used a long division. Using the tables it then follows that h1 (t) = 2δ(t) − (2 + α) (t)e−t(1+α) . c Note that the spectrum of (h ∗ h1 )(t) is the function H(ω) · (1/H(ω)), which is 1. Since δ(t) ↔ 1, it follows that (h ∗ h1 )(t) = δ(t).

10.16

We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into uyy + uxx = 0. This gives for some arbitrary constant c (the separation constant) that X 00 + cX = 0, Y 00 − cY = 0. In order to satisfy the linear homogeneous condition as well, X(x)Y (y) has to be bounded, and this implies that both X(x) and Y (y) have to be bounded functions. Solving the differential equations from the boundedness condition that √ we obtain √ i cx −i cx X(x) = 1 if c = 0 and e , e if c > 0. Similarly Y (y) = 1 if c =√0 and √ √ √ e cy , e− cy if c > 0. But e cy is not bounded for y > 0 so Y (y) = e− cy for c ≥ 0. We put c = s2 , then it follows that the class of functions satisfying the differential equation and being bounded, can be described by X(x)Y (y) = eisx e−| s |y ,

where s ∈ R.

By superposition we now try a solution u(x, y) of the form Z ∞ u(x, y) = e−| s |y F (s)eisx ds. −∞

If we substitute y = 0 in this integral representation, then we obtain that Z ∞ 1 u(x, 0) = = F (s)eisx ds.

1 + x2 −∞ Since 12 e−| t | ↔ 1/(1 + ω 2 ) this means that Z 1 1 ∞ −| t | −iωt = e e dt.

1 + ω2 2 −∞ The (formal) solution is thus given by Z ∞ Z ∞ u(x, y) = 21 e−| s |(1+y) eisx ds = e−| s |(1+y) cos(sx) ds. −∞

10.18

0

a

We substitute u(t) = δ(t),then Z t Z ∞ h(t) = e−(t−τ ) δ(τ ) dτ = ( (τ + 1 − t) − (τ − t))e−(t−τ ) δ(τ ) dτ, −∞

t−1 −t

so h(t) = ( (t) − (t − 1))e . Applying table 3, no. 7 and a shift in the time domain gives H(ω) = (1 − e−(1+iω) )/(1 + iω). b The impulse response is causal, so the system is causal. c It is straightforward to verify that the impulse response is absolutely integrable, so the system is stable. d The response y(t) to the block function p2 (t) is equal to the convolution of h(t) with p2 (t), which equals Z 1 Z t+1 y(t) = h(t − τ ) dτ = h(τ ) dτ. −1

t−1

This is 0 for t < −1 or for t > 2. For −1 < t < 0 it equals 1 − e−(t+1) , for 0 ≤ t < 1 it equals 1 − e−1 , and for 1 ≤ t < 2 it equals e−(t−1) − e−1 .

10.19

a The impulse response is the derivative in the sense of distributions of a(t) = e−t (t), which is δ(t) − e−t (t).

Answers to selected exercises for chapter 10

43

b The frequency response is H(ω) = iω/(1 + iω). (Apply e.g. the differentiation rule to h(t) = a0 (t).) c We have Y (ω) = H(ω)U (ω) (using obvious notations), so Y (ω) = iω/(1 + iω)2 = 1/(1 + iω) − 1/(1 + iω)2 . The response is the inverse Fourier transform of Y (ω): y(t) = (1 − t)e−t (t).

10.21

a The frequency response H(ω) is the triangle function qωc (ω). The inverse Fourier transform follows from table 3: h(t) = 2 sin2 (ωc t/2)/(πωc t2 ). b Since a0 (t) = h(t) and h(t) ≥ 0, the function a(t) is a monotone increasing function.

10.22

The frequency response follows immediately from the differential equation: H(ω) =

1 + iω . −ω 2 + 2iω + 2

Applying partial fraction expansion (use s = iω) we obtain H(ω) =

1 1 + .

2(iω + 1 − i) 2(iω + 1 + i)

The inverse Fourier transform is then h(t) = (e−t cos t) (t). Integrating this over (−∞, t] gives the step response a(t) = 12 (1 + e−t (sin t − cos t)) (t). b We have Y (ω) = H(ω)U (ω) (using obvious notations), so Y (ω) = 1/(−ω 2 + 2iω + 2). Applying partial fraction expansion we obtain Y (ω) =

10.23

1 1 − .

2i(iω + 1 − i) 2i(iω + 1 + i)

The response is the inverse Fourier transform of Y (ω), which gives as repsonse the function y(t) = (e−t sin t) (t). ˛ ˛ a Since | iω − 1 − i | = | iω + 1 − i | and ˛ e−iωt0 ˛ = 1 we have | H(ω) | = 1 and so L is an all-pass system. b First write (iω − 1 − i)/(iω + 1 − i) as 1 − 2/(iω + 1 − i) and then use the inverse Fourier transform (the tables) to obtain δ(t) − 2e−(1−i)t (t) ↔ 1 − 2/(iω + 1 − i). From the shift property in the time domain we obtain h(t) = δ(t − t0 ) − 2e−(1−i)(t−t0 ) (t − t0 ). c Ignoring˛ the delta ˛ function it is easy to verify that h(t) is absolutely ˛ ˛ integrable (˛ ei(t−t0 ) ˛ = 1), hence the system is stable. d Write u(t) = 1 + eit + e−it and use that eiωt 7→ H(ω)eiωt , then y(t) = H(0)+H(1)eit +H(−1)e−it , which equals −i−e−it0 eit +(4i−3)eit0 e−it /5.

10.24

We use separation of variables, so we substitute u(x, y) = X(x)Y (y) into uxx − 2uy = 0. This gives for some arbitrary constant c (the separation constant) that X 00 +cX = 0, 2Y 0 +cY = 0. Here X(x) and Y (y) have to be bounded functions. Solving the differential equations,√we obtain from the √ boundedness condition that X(x) = 1 if c = 0 and ei cx , e−i cx if c > 0.

For c ≥ 0 we obtain for Y (y) the solution e−cy/2 . We put c = s2 with s ∈ R. It then follows that the class of functions satisfying the differential equation and being bounded, can be described by 2

X(x)Y (y) = eisx e−s y/2 ,

where s ∈ R.

By superposition we now try a solution u(x, y) of the form Z ∞ 2 u(x, y) = e−s y/2 F (s)eisx ds. −∞

44

Answers to selected exercises for chapter 10

If we substitute y = 0 in this integral representation, then we obtain that Z ∞ u(x, 0) = xe−x (x) = F (s)eisx ds. −∞ −t

1 Since te (t) ↔ 1/(1+iω)2 this means that F (s) = 2π(1+is) 2 . The (formal) solution is thus given by Z ∞ 2 1 1 u(x, y) = e−s y/2 eisx ds 2π −∞ (1 + is)2 Z ∞ (1 − s2 ) cos sx + 2s sin sx −s2 y/2 1 = e ds.

2π −∞ (1 + s2 )2

Answers to selected exercises for chapter 11

11.1

In parts a, b and c the domain is C and the range is C as well. In part d the domain is C \ {−3} and the range is C \ {0}.

11.2

a If we write z = x + iy, then z = x − iy, so the real part is x and the imaginary part is −y. b Expanding z 3 = (x + iy)3 we see that the real part is x3 − 3xy 2 and that the imaginary part is 3x2 y − y 3 . c The real part is x − 4 and the imaginary part is −y − 1. d The real part is (3y − 2x − 6)/((x + 3)2 + y 2 ) and the imaginary part is (2y + 3x + 9)/((x + 3)2 + y 2 ). To see this, write z = x + iy, then z + 3 = x + 3 + iy and so f (z) =

3y − 2x − 6 + i(2y + 3x + 9) (3i − 2)(x + 3 − iy) = . (x + 3 + iy)(x + 3 − iy) (x + 3)2 + y 2

11.3

Apply definition 11.3 and expand the squares; several of the exponentials cancel and only 1/2+1/2 remains, so sin2 z +cos2 z = 1. Similarly it follows by substitution that 2 sin z cos z = sin 2z.

11.5

a From definition 11.3 it follows that sin(iy) = (e−y − ey )/2i = i sinh y and cos(iy) = (e−y + ey )/2 = cosh y. b Write z = x + iy; from exercise 11.4 with z = x and w = iy it follows that sin(x + iy) = sin x cos(iy) + cos x sin(iy). Now apply part a, then sin(x + iy) = sin x cosh y + i cos x sinh y, so the real part is sin x cosh y and the imaginary part is cos x sinh y.

11.6

The proofs can be copied from the real case; this is a straightforward matter. The same applies to exercise 11.7.

11.8

This rational function is continuous for all z ∈ C for which the denominator is unequal to 0. But the denominator is 0 for z = 1, z = −i or z = 2i. So g(z) is continuous on G = C \ {1, −i, 2i}.

11.9

The proof can be copied from the real case; this is a straightforward matter: f (w) − f (z) w2 − z 2 = lim = lim (w + z) = 2z. w→z w − z w→z w→z w−z lim

11.11

a The derivative is 4(z − 1)3 ; the function is differentiable on C, so it is analytic on C. b The derivative is 1 − 1/z 2 ; since the function is not differentiable at z = 0, it is analytic on C \ {0}. c The derivative is ((z 3 + 1)(2z − 3) − 3z 2 (z 2 − 3z + 2))/(z 3 + 1)2 ; the function is not differentiable when z 3 = −1. √ Solving this equation one obtains that it is analytic on C \ {−1, 12 ± 12 i 3}.

2

d The derivative is 2zez ; the function is analytic on C. 11.12 11.13 11.14∗

Using definition 11.3 and the chain rule it follows that (cos z)0 = (ieiz − ie−iz )/2 = (−eiz + e−iz )/2i = − sin z. √ √ √ The step wz = w z cannot be applied for non-real numbers (so e.g. for w = −1, z = −1).

The real and imaginary part are u(x, y) = x2 − y 2 and v(x, y) = 2xy

45

46

Answers to selected exercises for chapter 11

(example 11.4). So ∂u/∂x = 2x, which equals ∂v/∂y, and ∂u/∂y = −2y which equals −∂v/∂x. Hence, the Cauchy-Riemann equations are satisfied on the whole of R2 .

11.16

a This is not true; take e.g. z = 2i, then cos 2i = (e−2 + e2 )/2 > 3. b Use definition 11.3; expand the exponentials in the resulting expression for cos z cos w − sin z sin w. c Write z = x + iy and use part b, then cos(x + iy) = cos x cos iy − sin x sin iy. But cos iy = cosh y and sin iy = i sinh y (see exercise 11.5a), so cos(x + iy) = cos x cosh y − i sin x sinh y, which gives the real and imaginary parts. d Since ez is analytic on C, it follows from theorem 11.5 that cos z = (eiz + e−iz )/2 is also analytic on C.

11.17

a This function is analytic on C \ {1}. From the quotient rule it follows that the derivative is given by (2z 3 − 3z 2 − 1)/(z − 1)2 . √ √ b This function is analytic when z 4 6= −16, so on C \ {± 2 ± i 2}. The 3 4 11 derivative is given by −40z /(z + 16) . √ c This function is analytic when z 2 6= −3, so on C \ {±i 3}. The derivative is given by ez (z 2 − 2z + 3)/(z 2 + 3)2 (use the quotient rule). d This function is analytic on C since both ez and sin w are analytic on C. The derivative is given by ez cos ez (use the chain rule).

Answers to selected exercises for chapter 12

12.1

a The complex number e−iωR lies on the unit circle for all R and so the limit R → ∞ does not exist. ˛ ˛ b Since limR→∞ e−σR = 0 only for σ > 0, and ˛ e−iωR ˛ = 1, it follows that the limit exists precisely for σ > 0.

12.2

The integrand equals e(a−s)t and a primitive of this is given by e(a−s)t /(a − s). The lower limit 0 leads to 1/(s − a), while the upper limit results in the limit limR→∞ e(a−s)R (note that 1/(a−s) does not influence the outcome of this limit). Write s = σ + iω, a = α + iβ and use that limR→∞ e(α−σ)R = 0 only if α−σ < 0, so if σ > α. Hence, the Laplace transform equals 1/(s−a) for Re s > Re a.

12.3

b The function equals 1 for 0 ≤ t < b and is 0 elsewhere. It is easy to calculate the Laplace transform using the definition and an integration by parts and it is given by (1 − e−bs )/s for all (!) s 6= 0, while it equals b for s = 0.

12.4

One can use the method of example 12.9 and apply an integration by parts for s 6= 0. It then follows that Z 1 2 ∞ −st F (s) = − lim R2 e−sR + te dt. s R→∞ s 0 The remaining integral is the Laplace transform of t for Re s > 0, which is 1/s2 (example 12.9). Since limR→∞ R2 e−σR = 0 for σ > 0, it follows as before that limR→∞ R2 e−sR = 0 for Re s > 0. This shows that F (s) = s23 .

12.5

a From examples 12.2 and 12.8 it follows that 1/(s + 2) is the Laplace transform and that σa = σc = −2. b From example 12.7 it follows that e−4s /s is the Laplace transform and that σa = σc = 0. c From exercise 12.2 it follows that 1/(s − 2 − 3i) is the Laplace transform and that σa = σc = 2.

12.6

a b c d

12.8

a Using the method of example 12.10 (write cos t as 12 eit + 12 e−it ), we obtain the Laplace transform „ « 1 1 1 + 2 s−i s+i

From example 12.1 it follows that f (t) = 1 (that is, (t)).

From example 12.7 it follows that f (t) = (t − 3).

From example 12.2 it follows that f (t) = e7t (that is, (t)e7t ).

From exercise 12.4 it follows that f (t) = t2 /2 (that is, (t)t2 /2).

and this is equal to s/(s2 + 1) (for Re s > 0). b As in part a we now obtain the Laplace transform s/(s2 + a2 ). c Using the same method as in parts a and b it follows that the Laplace transform of cosh at = (eat + e−at )/2 is given by „ « 1 1 s 1 + = 2 .

2 s−a s+a s − a2 12.9

Follow the hint, e.g. write cos(at + b) = cos at cos b − sin at sin b, then apply

47

48

Answers to selected exercises for chapter 12

linearity and the fact that we know the transforms of cos at and sin at. In table 7, lines 8 and 9 the answers are given.

12.10

In all these exercises we have to use linearity and/or table 7. a (20/s3 ) − (5/s2 ) + ((8i − 3)/s) b 4/(s2 + 16) c s/(s2 − 25) d (1/s2 ) + (1/s) − (s/(s2 + 1)) e (1/(s − 2)) + (1/(s + 3)) f (1/2s) − (s/2(s2 + 4)) g (cos 2 − s sin 2)/(s2 + 1) h 1/(s − ln 3) (use that 3t = et ln 3 ).

12.11

a The function is 1 for 0 ≤ t < 1, 3 − 2t for 1 ≤ t < 2 and 1 − t for t ≥ 2. b The Laplace transform of 1 equals 1/s. Also, (t − 1)(2t − 2) = 2 (t − 1)(t − 1) and so we can apply a shift in the time-domain, which gives as Laplace transform 2e−s /s2 . The same applies to (t − 2)(t − 2), which gives as Laplace transform e−2s /s2 . Add these three results.

12.12

The function equals cos t for 0 ≤ t < 2π and is 0 elsewhere since cos(t − 2π) = cos t. The Laplace transform of cos t is s/(s2 + 1). A shift in the time-domain shows that se−2πs /(s2 + 1) is the Laplace transform of (t − 2π) cos(t − 2π). Adding these results leads to s(1 − e−2πs )/(s2 + 1).

12.15

Apply the scaling property to f (t), then G(s) = (s2 −2s+4)/(4(s+1)(s−2)).

12.16

For all these exercises one first needs to recognize the basic form of the function, then apply table 7, in combination with a shift in the time- or s-domain. a 1/(s − 2)2 ; use a shift in the s-domain. b 2e−s /s3 ; use a shift in the time-domain. c 5/((s + 3)2 + 25); use a shift in the s-domain. d (s − b)/((s − b)2 + a2 ); use a shift in the s-domain. e se−3s /(s2 − 1); use a shift in the time-domain. f 2e−3 /(s − 1)3 ; use a shift in the s-domain.

12.17

a The function is 0 for 0 ≤ t < 1 and t − 1 for t ≥ 1. From table 7 and a shift in the time-domain the Laplace transform e−s /s2 follows. b The function equals t − 1 for t ≥ 0. From table 7 (and linearity) the Laplace transform follows: (1/s2 ) − (1/s). c The function equals 0 for 0 ≤ t < 1 and t for t ≥ 1 (so there is a jump at t = 1 of magnitude 1). In order to apply a shift in the time-domain we write f (t) = (t − 1)(t − 1) + (t − 1). Using part a and table 7 (and linearity of course), the Laplace transform follows: (e−s /s2 ) + (e−s /s).

12.18

Write f (t) = (t)t − (t − 1)t, then it follows from exercise 12.17c and table 7 that the Laplace transform is given by (1 − (1 + s)e−s )/s2 .

12.19

For this exercise one again first has to recognize the basic form of the Laplace transform, e.g. using table 7, and then, if necessary, combine it with properties like a shift in the time- or s-domain. a 2e3t b 3 sin t c 4 cos 2t d (sinh 2t)/2 e (t − 2)(t − 2); use a shift in the time-domain. f (t − 3) cos(t − 3); use a shift in the time-domain.

Answers to selected exercises for chapter 12

49

g (et sin 4t)/4; use a shift in the s-domain. h e−t (3 cos t − sin t); use a shift in the s-domain; in order to do so we first have to write F (s) as 3

s+1 1 − , (s + 1)2 + 1 (s + 1)2 + 1

which can be obtained by noting that 3s + 2 = 3(s + 1) − 1. i −3e3t t2 ; use a shift in the s-domain. j e2t cos 2t; use a shift in the s-domain (first complete the square in the denominator, writing it as (s − 2)2 + 4). k 41 (t − 1) cos(3(t − 1)/2); use a shift in the time-domain (first write the function F (s) as 14 e−s s/(s2 + (9/4)).

12.21

Apply De l’Hôpital’s rule repeatedly (n times) to the limit t → ∞ of tn /eαt and use that limt→∞ e−αt = 0 for any α > 0. Theorem 12.3 implies that the Laplace transform of tn exists for Re s > 0 (it is easy to show that tn = (t)tn is of exponential order for α > 0 arbitrary).

12.23

We know that (L1)(s) = 1/s. According to (12.13) we then have for n ∈ N (and Re s > 0) that dn 1 = (−1)n (Ltn )(s), dsn s hence, (−1)n n!/sn+1 = (−1)n (Ltn )(s), that is, (Ltn )(s) = n!/sn+1 .

12.25

a We know (e.g. from table 7) that (Ltn )(s) = n!/sn+1 and using the shift rule in the s-domain we then obtain that F (s) = n!/(s − a)n+1 . b We know (e.g. from table 7) that (Leat )(s) = 1/(s − a) and using the differentiation rule in the s-domain we then obtain that F (s) = (−1)n

dn 1 n! = . dsn s − a (s − a)n+1

12.27

The causal function sinh at is continuous on R, so it follows from the integration rule (table 8) that „ Z t « 1 L sinh aτ dτ (s) = F (s) s 0 Rt with F (s) = (L sinh at)(s) = a/(s2 − a2 ). Hence f (t) = 0 sinh aτ dτ = (cosh at − 1)/a.

12.28

a Use the differentiation rule in the s-domain and table 7 for the Laplace transform of cos at. Then f has Laplace transform 2s(s2 − 3a2 ) d2 s = . ds2 s2 + a2 (s2 + a2 )3 b Using the differentiation rule in the s-domain and table 7 for the Laplace transform of sinh 3t one obtains that (Lt sinh 3t)(s) = −

3 6s d = 2 ds s2 − 9 (s − 9)2

and (Lt2 sinh 3t)(s) =

18(s2 + 3) 3 d2 . = ds2 s2 − 9 (s2 − 9)3

50

Answers to selected exercises for chapter 12

Finally apply linearity, then one obtains that F (s) = 12.30

18(s2 + 3) 18s 6 − 2 + 2 . (s2 − 9)3 (s − 9)2 s −9

a The function equals 2 (t)t for t < 1 and since t = 2t − t it follows that f (t) = 2 (t)t − (t − 1)t = 2 (t)t − (t − 1)(t − 1) − (t − 1). b From part a, table 7 and the shift rule in the time domain it follows that 2 e−s e−s . − 2 − 2 s s s c The function is not differentiable at t = 0 and t = 1. Apart from these two points the derivative equals 0 for t < 0, 2 for 0 < t < 1 and 1 for t > 1. Hence, f 0 (t) = 2 (t) − (t − 1) for t 6= 0, 1. Since the Laplace transform does not depend on the value at these points, we have that (Lf 0 )(s) = (2/s) − (e−s /s). d According to the differentiation rule in the time domain one should have (Lf 0 )(s) = s(Lf )(s), so

F (s) =

2 − e−s (s + 1) 2 − e−s =s . s s2 This is not correct since we cannot apply the differentiation rule in the present situation: f has a jump at t = 1 and so it isn’t differentiable on R. 12.31

a From the shift rule in the s-domain it follows that (Leibt f (t))(s) = F (s− ib) and since sin at = (eiat − e−iat )/2i it follows that (Lf (t) sin at)(s) = (F (s − ia) − F (s + ia))/2i. b First apply the scaling property and then the shift rule in the s-domain, then (Le−2t f (3t))(s) = e−(s+2)/3 /(s + 2). c We can apply the integration here (t3 f (t) is continuous on R).

R t rule 3 Hence the Laplace transform of 0 τ f (τ ) dτ is given by G(s)/s with G(s) = (Lt3 f (t))(s). Now apply the differentiation rule in the s-domain, then it Rt follows that the Laplace transform of 0 τ 3 f (τ ) dτ is given by −

12.32

1 d3 F (s). s ds3

a Since 3et−2 = 3e−2 et it follows that (L3et−2 )(s) = 3e−2 /(s − 1) (see table 7). We also have that (L (t − 2))(s) = e−2s /s so (for Re s > 1) F (s) =

3e−2 s + e−2s (s − 1) . s(s − 1)

b Apply linearity (and table 7) to (t − 1)2 = t2 − 2t + 1, then we obtain that F (s) =

s2 − 2s + 2 . s3

c From (L (t − 4))(s) = e−4s /s and the shift property in the s-domain (table 8) it follows that F (s) = e−4(s−2) /(s − 2). d Note that f (t) = e2it e−t = e(−1+2i)t and applying table 7 we thus obtain that F (s) = 1/(s + 1 − 2i). (One can also use the Laplace transforms of sin 2t and cos 2t and a shift in the s-domain.) e The Laplace transform of sin t is 1/(s2 +1). Applying a shift in the timedomain we obtain that (L (t − 2) sin(t − 2))(s) = e−2s /(s2 + 1). Finally we note that et+3 = e3 et and so we apply a shift in the s-domain:

Answers to selected exercises for chapter 12

F (s) =

51

e3 e−2(s−1) . (s − 1)2 + 1

f The Laplace transform of cos 2t is s/(s2 + 4). Since 3t = et ln 3 we apply a shift in the s-domain: F (s) =

s − ln 3 . (s − ln 3)2 + 4

g One can write f (t) as the following combination of shifted unit step functions: f (t) = (t) − (t − 1) + (t − 2). From table 7 we then obtain that F (s) = 12.33

1 − e−s + e−2s − e−3s . s

a Since (L1)(s) = 1/s and (L (t − 1))(s) = e−s /s (table 7) it follows that f (t) = 1 − (t − 1). b Since F (s) = 1/s + 3/s4 we obtain from table 7 that f (t) = 1 + t3 /2. c Since (Lte−t )(s) = 1/(s + 1)2 and (L sinh 2t)(s) = 2/(s2 − 4) (table 7) it follows that te−t + 12 sinh 2t has Laplace transform 1/(s + 1)2 + 1/(s2 − 4).

Furthermore we have that (L sin t)(s) = 1/(s2 + 1) (table 7) and from a shift in the time-domain it then follows that (L (t − π) sin(t − π))(s) = e−πs /(s2 + 1). Hence f (t) = te−t + 21 sinh 2t + sin t + (t − π) sin(t − π). d The denominator equals (s − 2)2 + 16 and if we now use that 3s − 2 = 3(s − 2) + 4, then it follows that F (s) = 3

s−2 4 + . (s − 2)2 + 16 (s − 2)2 + 16

From table 7 and a shift in the s-domain we then obtain that f (t) = 3e2t cos 4t + e2t sin 4t. e The denominator equals (s+4)2 and if we now use that s+3 = (s+4)−1, then it follows that 1 1 F (s) = − . s+4 (s + 4)2 From table 7 and a shift in the s-domain we then obtain that f (t) = e−4t (1 − t). f Apply a shift in the time domain to (Lt2 e2t )(s) = 2/(s − 2)3 (table 7), then we obtain that f (t) = 21 (t − 4)e2t−8 (t − 4)2 . g Applying the integration rule to (L sin 3t)(s) = 3/(sR2 + 9) (the causal t function sin t is continuous on R) we obtain that (L 0 sin 3τ dτ )(s) = 2 3/(s(s + 9)). But the integral equals (1 − cos 3t)/3, so (L(1 − cos 3t))(s) = 9/(s(s2 + 9)). From a shift in the time domain it then follows that f (t) = (t − 1)(1 − cos 3(t − 1))/9.

Answers to selected exercises for chapter 13

13.1

The integral defining the convolution can be calculated by using the formula for the product of two cosines. The convolution then equals 12 t cos t + 1 sin t. Using the convolution theorem we obtain s2 /(s2 +1)2 as the Laplace 2 transform. On the other hand we obtain from the Laplace transforms of cos t and sin t and the differentiation rule in the s-domain the Laplace transform (s2 − 1)/(2(s2 + 1)2 ) + (1/(2(s2 + 1)), which agrees with the result obtained from the convolution theorem.

13.2

a Table 7 gives f (t) = eat . b The convolution theorem implies that g(t) = eav ∗ ebv ; to determine g(t) explicitly, we need toR calculate this convolution. From the definition t it follows that g(t) = ebt 0 eτ (a−b) dτ . If a = b, then g(t) = teat . If a 6= b at bt then g(t) = (e − e )/(a − b). Next we can verify the convolution theorem.

Write G(s) = (Lg)(s). If a = b, then G(s) = 1/(s − a)2 (table 7, no. 10).

If a 6= b then G(s) = (1/(s − a) − 1/(s − b))/(a − b) (table 7, no. 2), which equals 1/(s − a)(s − b).

13.4

a Consider this as the product of the Laplace transforms of t and e−t , which gives t ∗ e−t as result. b Similarly we now obtain e−2t ∗ cos 2t. c sinh t ∗ cosh t.

1 d 16 sinh 4t ∗ sinh 4t.

13.5

This is not possible, since lims→∞ sn does not exist, contradicting theorem theorem 13.2.

13.6

a From table 7 we obtain that F (s) = s/(s2 − 9). We indeed have f (0+) = 1 = lims→∞ sF (s). b From table 7 we obtain that (L sin t)(s) = 1/(s2 + 1) and (L1)(s) = 1/s. Applying the differentiation rule in the s-domain it follows that (Lt sin t)(s) = 2s/(s2 + 1)2 and so F (s) = (2/s) + (2s/(s2 + 1)2 ). We indeed have f (0+) = 2 = lims→∞ sF (s). c From the integration rule (table 8) we obtain that F (s) = (Lg)(s)/s and so sF (s) = G(s) with G(s) the Laplace transform of g(t). Applying theorem 13.2 to g(t) we obtain that lims→∞ sF (s) = lims→∞ G(s) = 0 and for f we indeed have that f (0+) = f (0) = 0.

13.8

a From table 7 we obtain that F (s) = 1/(s + 3). Since f (∞) exists, we may apply the final value theorem and we indeed have f (∞) = 0 = lims→0 sF (s). b From table 7 we obtain that (L sin 2t)(s) = 2/(s2 + 4). Applying the shift property in the s-domain we obtain that F (s) = 2/((s + 1)2 + 4).

Since f (∞) exists, we may apply the final value theorem and we indeed have f (∞) = 0 = lims→0 sF (s). c From table 7 we obtain that F (s) = (1/s) − (e−s /s). Since f (∞) exists, we may apply the final value theorem and we indeed have f (∞) = 0 = lims→0 (1 − e−s ).

13.9

For the functions cos t and sinh t the value f (∞) does not exist and so the final value theorem cannot be applied.

13.11

a For a periodic function f (∞) will in general not exist and so the final value theorem cannot be applied. b Theorem 13.5 implies that 52

Answers to selected exercises for chapter 13

sF (s) =

s 1 − e−sT

Z T

53

f (t)e−st dt.

0

Taking the limit s → 0 gives (note that the integral is over a bounded interRT RT val) lims→0 0 f (t)e−st dt = 0 f (t) dt. From the definition of derivative (definition 11.7) it follows that (e−zT )0 (0) = lims→0 (e−sT − 1)/s and since (e−zT )0 = −T e−zT we thus obtain that lim

s

s→0 1 − e−sT

=

1 , T

RT which shows that lims→0 sF (s) = T1 0 f (t) dt. c In example 13.4 we have sF (s) = 1/(1 + e−s ) and so lims→0 sF (s) = 12 .

R2 The function has period 2 and so the integral is 12 0 f (t) dt = 12 , which verifies the result of part b for f .

13.12

a For t < a we have φ(t) = (t); for t < 2a we then have φ(t) = (t) − 2 (t − a); finally, for all t we have φ(t) = (t) − 2 (t − a) + (t − 2a). If Φ(s) is the Laplace transform of φ(t), then it follows from table 7 that Φ(s) =

1 2e−as e−2as − + . s s s

Since f has period 2a we then obtain from theorem 13.5 (or table 8) that F (s) =

1 − 2e−as + e−2as . s(1 − e−2as )

b Multiply numerator and denominator by eas , write the denominator as s(eas/2 + e−as/2 )(eas/2 − e−as/2 ), and use the definitions of the hyperbolic sine and cosine functions, then it follows that F (s) = tanh(as/2)/s.

13.13

b Let φ(t) denote the restriction of f (t) to one period, then φ(t) = t (t) − (t − 2) (t − 2) − 2 (t − 1). Since (Lt)(s) = 1/s2 , we obtain from a shift in the time-domain that (L(t − 2) (t − 2))(s) = e−2s /s2 . Also (L (t − 1))(s) = e−s /s and so Φ(s) =

e−2s 2e−s 1 − 2 − .

2 s s s

Theorem 13.5 (or table 8) then implies that F (s) = 13.16

1 2e−s 1 1 − = 2 − . s2 s(1 − e−2s ) s s sinh s

Apply (13.7) (or definition 13.2) and the definition of distribution derivative, then it follows that D E D E (Lδ (n) (t − a))(s) = δ (n) (t − a), e−st = (−1)n δ(t − a), (e−st )(n) .

Since (e−st )(n) = (−s)n e−st , we then obtain from the definition of the shifted delta function that (Lδ (n) (t − a))(s) = sn e−as .

13.18

This follows immediately from table 9, no 2 and (9.21) together with (13.9).

13.19

Consult tables 7 and 9 for this exercise and use linearity, and for part d the convolution theorem. a 1 + (1/(s2 + 1)) = (s2 + 2)/(s2 + 1), b s + 3s2 ,

54

Answers to selected exercises for chapter 13

c 1/s + e−2s + 2is2 e−4s , d s3 e−as .

13.20

Consult tables 7 and 9 for this exercise, but now in the opposite direction, and use mainly linearity. a δ 0 (t) + 3δ(t) − δ(t − 2), b δ 00 (t) − 4δ 0 (t) + 4δ(t) + e2t (first write (s − 2)2 as s2 − 4s + 4), c (t − 2) sin(t − 2) + δ (3) (t − 2) (apply a shift in the time-domain to the Laplace transform of sin t), d δ(t) − sin t (first write s2 /(s2 + 1) as 1 − 1/(s2 + 1)).

13.22

This is an important exercise: it shows how to use partial fraction expansions. a Factorise the denominator as (s + 2)(s + 3) and apply partial fraction expansion to obtain 3/(s + 3) − 2/(s + 2). From table 7 it then follows that f (t) = 3e−3t − 2e−2t . b Completing the square in the denominator gives (s + 3)2 + 1 and from table 7 it then follows that f (t) = e−3t sin t. c Apply partial fraction expansion to obtain − 43 (1/(s + 3)) + 23 (1/(s + 1 2)) + 12 (1/(s − 1)). From table 7 it then follows that f (t) = −3e−3t /4 + 2e−2t /3 + et /12. d First put y = s2 and then apply partial fraction expansion in the variable y, which gives 35 (1/(y − 4)) − 23 (1/(y − 1)). With y = s2 we obtain from table 7 the inverse Laplace transform f (t) = (5 sinh 2t − 4 sinh t)/6. e A partial fraction expansion leads to −1/(s + 1) + 2/(s − 1) − 4/(s + 1)2 .

From table 7 it then follows that f (t) = −e−t + 2et − 4te−t . f Since s2 − 1 = (s − 1)(s + 1) one could solve this using partial fraction expansions. However, it is easier to note that d s 1 2 =− 2 − 2 ds s2 − 1 s −1 (s − 1)2 and so 1 1 d s 1 1 =− − . (s2 − 1)2 2 ds s2 − 1 2 s2 − 1 We have (L sinh t)(s) = 1/(s2 − 1) and (L cosh t)(s) = s/(s2 − 1) and the differentiation rule in the s-domain then implies that (Lt cosh t)(s) = −

d s . ds s2 − 1

Hence, f (t) = (t cosh t − sinh t)/2. g Since the degree of the numerator equals the degree of the denominator we first perform a long division: s3 + 4 s2 − 4s + 8 = 1 + . (s2 + 4)(s − 1) (s2 + 4)(s − 1) Applying a partial fraction expansion to the second term leads to 1/(s−1)− 4/(s2 +4). From tables 7 and 9 it then follows that f (t) = δ(t)+et −2 sin 2t. h The factor e−2s can be dealt with afterwards by applying a shift in the time-domain. Since the degree of the numerator is greater than the degree of the denominator we first perform a long division: s6 + s2 − 1 2s2 − 1 = s2 + 1 + 2 2 . s2 (s2 − 1) s (s − 1)

Answers to selected exercises for chapter 13

55

To the rational part we apply partial fraction expansion (again we first put y = s2 ), which leads to (1/s2 ) + (1/(s2 − 1)). From table 7, a shift in the time-domain and table 9 it then follows that f (t) = δ 00 (t − 2) + δ(t − 2) + (t − 2)(t − 2 + sinh(t − 2)).

13.24

a

First write F (s) as

F (s) =

1 s 2 .

2 s2 + 4 s2 + 4

From table 7 and the convolution theorem (or table 8) it then follows that (g ∗ h)(t) = (cos 2v ∗ 21 sin 2v)(t).

Rt b The definition of convolution gives f (t) = 12 0 cos 2τ sin(2t−2τ ) dτ and using the trigonometricRformula 2 cos a sin b = sin(a+b)−sin(a−b) this can t be written as f (t) = 14 0 (sin 2t − sin(4τ − 2t)) dτ . Caculating this integral 1 gives f (t) = 4 t sin 2t. c Applying the differentiation rule in the time-domain to (L sin 2t)(s) = 2/(s2 + 4), we obtain that (Lt sin 2t)(s) = 4s/(s2 + 4)2 , so f (t) = 14 t sin 2t.

13.25

a The function f is piecesewise smooth and f (∞) = 2, so the final value theorem implies that lims→0 sF (s) = 2. Since f (0+) = 0, the inital value theorem implies that lims→∞ sF (s) = 0. b Apply a shift in the time-domain to (Lt)(s) = 1/s2 , then (L (t − 2)(t − 2))(s) = e−2s /s2 , so F (s) = (1 − e−2s )/s2 . Hence, lims→0 sF (s) = lims→0 (1 − e−2s )/s. But according to definition 11.7 this limit equals −(e−2z )0 (0), which is −2, and so lims→0 sF (s) = 2. This agrees with part a. Finally, applying De l’Hôpital’s rule we obtain that lims→∞ sF (s) = lims→∞ (1 − e−2s )/s = lims→∞ 2e−2s = 0, also in agreement with part a.

13.26

a Note that lims→0 sF (s) = 1. We now determine the inverse Laplace transform of F (s) using partial fraction expansion. We have F (s) =

s+1 1 − s (s + 1)2 + 4

and from table 7 and a shift in the s-domain it then follows that f (t) = 1 − e−t cos 2t. We see that f (∞) = 1 = lims→0 sF (s), which verifies the final value theorem. b Note that lims→0 sF (s) = 0. We now determine the inverse Laplace transform of F (s). Since F (s) is a function of y = s2 , we use partial fraction expansion for y/(y − 1)(y + 4), which gives F (s) =

2 2 1 1 + .

5 s2 − 1 5 s2 + 4

From table 7 we then obtain that f (t) = (sinh t + 2 sinh 2t)/5. Since f (∞) does not exits, the final value theorem cannot be applied.

13.27

a For 0 ≤ t < 1 we have f (t) = (t)t2 ; for 0 ≤ t < 2 we then have f (t) = (t)t2 − (t − 1)t2 . b Let φ(t) = ( (t) − (t − 1))t2 for all t, and let Φ(s) be the Laplace transform of φ(t), then it follows from theorem 13.5 (or table 8) that F (s) = Φ(s)/(1−e−2s ). If we write φ(t) as φ(t) = (t)t2 − (t−1)(t−1)2 − (t−1)− 2 (t − 1)(t − 1), then it follows from table 7 and a shift in the time-domain (to get the Laplace transforms of (t − 1)(t − 1)2 and (t − 1)(t − 1)) that Φ(s) = −

2e−s 2 2e−s e−s − 2 + 3 − 3 , s s s s

56

Answers to selected exercises for chapter 13

which then also gives F (s) = Φ(s)/(1 − e−2s ).

13.28

a A partial fraction expansion of G(s) = 1/s(s + 1) gives G(s) = (1/s) − (1/(s + 1)), which has inverse Laplace transform g(t) = 1 − e−t . Combining this with a shift in the time-domain we obtain that f (t) = 1 − e−t − (t − 3)(1 − e−t+3 ). b A partial fraction expansion gives F (s) =

2 1 2 2 + 2 − + . s s s−1 (s − 1)2

From table 7 we then obtain that f (t) = 2 + t − 2et + 2tet . c Write F (s) = G(s) + H(s) where G(s) =

1 s2 (s2 + 4)

and

H(s) = e−πs

s5 − 4s4 − 8s + 64 . s2 (s2 + 4)

We first determine the inverse Laplace transform g(t) of G(s). Since G(s) is a function of y = s2 , we apply a partial fraction expansion to 1/y(y + 4), which gives G(s) = (1/4s2 ) − (1/4(s2 + 4)) and so (by table 7) g(t) = (2t − sin 2t)/8. Next we determine the inverse Laplace transform h(t) of H(s). Since the degree of the numerator is larger than the degree of the denominator, we first perform a long division: s5 − 4s4 − 8s + 64 4s3 − 16s2 + 8s − 64 =s−4− . s2 (s2 + 4) s2 (s2 + 4) Partial fraction expansion of the rational function gives 4s3 − 16s2 + 8s − 64 2 16 2s = − 2 + 2 . s2 (s2 + 4) s s s +4 Hence, H(s) = se−πs − 4e−πs − e−πs

„

16 2s 2 − 2 + 2 s s s +4

« .

From table 7 and the shift property in the time-domain it then follows that h(t) = δ 0 (t − π) − 4δ(t − π) − (t − π) (2 − 16(t − π) + 2 cos 2(t − π)) and then f (t) = g(t) + h(t).

Answers to selected exercises for chapter 14

14.1

a We have to determine the Laplace transform H(s) of h(t). From tables 7 (for te−t ) and 9 (for δ) it follows immediately that H(s) = 1 + 1/(s + 1)2 . b Using that H(s) exists for Re s > −1, we can substitute s = iω into H(s) to obtain the frequency response 1 + 1/(iω + 1)2 . c We first determine the response y(t) by calculating the convolution product y(t) = (h ∗ u)(t). Since δ ∗ u = u we only need to calculate te−t ∗ u, that is Z t Z t u(τ )h(t − τ ) dτ = e−τ sin τ e−(t−τ ) (t − τ ) dτ.

0

0

Here e−τ and eτ cancel each other and the integral that remains can be calculated by an integration by parts. This gives te−t − sin te−t and so y(t) = u(t) + te−t − sin te−t = te−t .

Next we determine y(t) using the Laplace transform: Y (s) = H(s)U (s) and since U (s) = 1/((s+1)2 +1) (table 7 and a shift in the s-domain), it follows that Y (s) = 1/(s + 1)2 and so (inverse Laplace transform) y(t) = te−t .

14.2

Again use the important formula Y (s) = H(s)U (s), so H(s) = Y (s)/U (s).

In this case U (s) = 1/s2 and Y (s) = 1/s2 − s/(s2 + 4), so H(s) = 1 − s3 /(s2 + 4) = 1 − s + 4s/(s2 + 4) (divide s3 by s2 + 4). The inverse Laplace transform gives h(t) = δ(t) − δ 0 (t) + 4 cos 2t.

14.3

a It is clear that y(0) = 0. The derivative at t = 0 can be obtained from the definition of the derivative (and e.g. using De l’Hôpital’s rule) and we indeed obtain that y 0 (0) = 0. For t > 0 it is straightforward to check that y 00 − y = 2t. So in ordinary sense y(t) does indeed satisfy the differential equation.

We now differentiate in distribution sense. Since y has no jump at t = 0 and y 0 also has no jump at t = 0, the jump formula (8.21) will give the same result (outside t = 0 the derivative can be taken in ordinary sense). b Since h has no jump at t = 0 we have that h0 (t) = 2e2t − et (in the sense of distributions). Now h0 has a jump 1 at t = 0, so h00 (t) = 4e2t − et + δ(t).

Hence, we indeed have that h00 − 3h0 + 2h = δ(t). c Similar as in part b: there is no jump at t = 1 and so y 0 (t) = 2 (t − 1)(2e2t−2 −et−1 ). Now y 0 has a jump 2 at t = 1, so y 00 (t) = 2 (t−1)(4e2t−2 − et−1 ) + 2δ(t − 1). Thus, y 00 − 3y 0 + 2y = 2δ(t − 1).

14.4

Taking Laplace transforms of the left- and right-hand side gives (note the condition of initial rest) Y (s) = 48/((s2 + 4)(s2 + 16)). A partial fraction expansion (in the variable y = s2 ) results in 4/(s2 + 4) − 4/(s2 + 16) and from table 7 it then follows that y(t) = 2 sin 2t − sin 4t.

14.5

a Using (14.9) we obtain that H(s) = 1/(s2 − 5s + 4) = 1/((s − 1)(s − 4)) and its zeroes do not satisfy Re s < 0, so the system is not stable. b Partial fraction expansion of H(s) shows that H(s) = (1/3(s − 4)) − (1/3(s − 1)) and from table 7 we then obtain that h(t) = (e4t − et )/3; this function is not absolutely integrable, which is in agreement with the fact that the system is not stable. c Integrating the impulse response over [0, t] gives the step response a(t) = (e4t − 4et + 3)/12. d Either calculate the convolution product (h ∗ u)(t) or use Laplace transforms; in general the latter is preferable and we will apply it here. Since 57

58

Answers to selected exercises for chapter 14

Y (s) = U (s)H(s) = 1/((s − 1)(s − 4)(s − 2)) we use partial fraction expansion: 1 1 1 Y (s) = − + .

3(s − 1) 2(s − 2) 6(s − 4) From table 7 we then obtain that y(t) = (e4t + 2et − 3e2t )/6. e Use time-invariance (and linearity): 3δ(t − 1) has response 3h(t − 1) = (t − 1)(e4t−4 − et−1 ).

14.7

a From the differential equation we get H(s) = 1/(R(s + 1/RC)) (use (14.9)) and so H(s) has only one zero s = −1/RC. Since this zero lies in Re s < 0 (because RC > 0), the system is stable. b Let Q and V be the Laplace transforms of q and v, then Q(s) = V (s)H(s) = E/(Rs(s+1/RC)) and partial fraction expansion gives Q(s) = EC(1/s − 1/(s + 1/RC)). The inverse transform of this is q(t) = EC(1 − e−t/RC ) and since i(t) = q 0 (t) it follows that i(t) = Ee−t/RC /R. c As in part b it now follows that Q(s) = aE/(R(s + 1/RC)(s2 + a2 )), and partial fraction expansion results in Q(s) = p(1/(s + 1/RC) − s/(s2 + a2 ) + 1/(RC(s2 + a2 ))), where p = aE/(R(a2 + 1/R2 C 2 )). The inverse transform of this is q(t) = p(e−t/RC − cos at + (1/aRC) sin at). d Use time-invariance (and linearity): Eδ(t − 3) has as response Eh(t − 3) = E (t − 3)e−(t−3)/RC /R (h(t) follows from part a).

14.8

a From the differential equation we get H(s) = 1/(L(s + R/L)) (use (14.9)) and so H(s) has only one zero s = −R/L. Since this zero lies in Re s < 0 (because R/L > 0), the system is stable. The inverse transform gives h(t) = e−Rt/L /L. b Integrating the impulse response over [0, t] gives the step response a(t) = (1 − e−Rt/L )/R. c Let V be the Laplace transform of v, then it follows that Y (s) = V (s)H(s) = e−as /(Ls(s + R/L)). Partial fraction expansion results in Y (s) = (e−as /R)(1/s − 1/(s + R/L)). The inverse transform of this is y(t) = (t − a)(1 − e−(t−a)R/L )/R (use a shift in the time-domain). However, it is much simpler here to use time-invariance: the response to (t − a) is a(t − a) = (t − a)(1 − e−R(t−a)/L )/R.

14.10

Let Y (s) be the Laplace transform of y(t), then we obtain from the differential equation (and tables 7 and 8) that (s2 Y (s)−sy(0)−y 0 (0))+Y (s) = 1/s2 .

Since y(0) = 0 and y 0 (0) = 1 it follows that s2 Y (s) − 1 + Y (s) = 1/s2 , so (s2 + 1)Y (s) = 1 + 1/s2 = (s2 + 1)/s2 and thus Y (s) = 1/s2 . The inverse Laplace transform of Y (s) is y(t) = t.

14.12

Let Y (s) be the Laplace transform of y(t), then we obtain from the differential equation and the initial conditions y(0) = 3 and y 0 (0) = 1 (and tables 7 and 8) that (s2 Y − 3s − 1) − 4(sY − 3) − 5Y = 3/(s − 1). Solving for Y gives (s2 − 4s − 5)Y (s) + 11 − 3s = 3/(s − 1), hence, Y (s) =

3 (s − 1)(s2 − 4s − 5)

+

3s − 11 s2 − 4s − 5

=

3s2 − 14s + 14 . (s − 1)(s2 − 4s − 5)

Partial fraction expansion gives Y (s) =

31 3 19 − + .

12(s + 1) 8(s − 1) 24(s − 5)

From table 7 we obtain the inverse Laplace transform y(t) = (62e−t − 9et + 19e5t )/24.

Answers to selected exercises for chapter 14

14.13

59

Note that u(t) = t − (t − 2)(t − 2), so (shift in time-domain) U (s) = (1 − e−2s )/s2 . From the differential equation and the initial conditions we obtain that s2 Y − s + Y = (1 − e−2s )/s2 , hence (apply partial fraction expansion in s2 to the first term), „ « 1 − e−2s 1 s 1 s Y (s) = 2 2 + 2 = (1 − e−2s ) . − + 2 s (s + 1) s +1 s2 s2 + 1 s +1 From tables 7 and 8 we obtain the inverse Laplace transform y(t) = t − sin t + cos t − (t − 2)(t − 2 − sin(t − 2)).

14.14

From tables 7 and 9 we know that the Laplace transforms of 1 and δ(t − 2) are given by 1/s and e−2s . From the differential equation and the initial conditions we obtain that (s2 Y − 2s + 2) + 2(sY − 2) + 5Y = 2e−2s + 1/s, and thus (s2 + 2s + 5)Y (s) = 2s + 2 + 2e−2s + 1/s. Hence, Y (s) =

2(s + 1) 2e−2s 1 + + . (s + 1)2 + 4 (s + 1)2 + 4 s(s2 + 2s + 5)

Using table 7 and a shift in the s-domain it is easy to get the inverse Laplace transform of the first two terms since (Le−t cos 2t)(s) = (s+1)/((s+1)2 +4) and (Le−t sin 2t)(s) = 2/((s + 1)2 + 4). For the third term we use partial fraction expansion: (s + 2) 1 1 = − . s(s2 + 2s + 5) 5s 5(s2 + 2s + 5) We write the second term as (s + 1)/5(s2 + 2s + 5) + 1/5(s2 + 2s + 5) and taking everything together now we get Y (s) =

9(s + 1) 2 1 1 + − + e−2s .

5s 5((s + 1)2 + 4) 5((s + 1)2 + 4) (s + 1)2 + 4

From our previous remarks and a shift in the time-domain it then follows that y(t) = (2 + 18e−t cos 2t − e−t sin 2t)/10 + (t − 2)e−(t−2) sin 2(t − 2).

14.16

Let X(s) and Y (s) be the Laplace transforms of x(t) and y(t). From table 7 we know that (L cos 2t)(s) = s/(s2 + 4) and (L sin 2t)(s) = 2/(s2 + 4).

Applying the Laplace transform to the system and substituting the initial conditions x(0) = −1 and y(0) = 0 we obtain the algebraic system  sX + Y = −1 + 2s/(s2 + 4), X + sY = 2/(s2 + 4).

Next we solve this system of two linear equations in the unknowns X = X(s) and Y = Y (s). We can find Y (s) by multiplying the second equation by −s and adding it to the first equation; we then obtain Y − s2 Y =

2s 2s −1− 2 = −1 s2 + 4 s +4

and so Y (s) = 1/(s2 − 1). One similarly obtains −s2 X + X =

−2s2 2 +s+ 2 s2 + 4 s +4

and so X(s) = 2/(s2 + 4) − s/(s2 − 1). The inverse Laplace transform gives the solution x(t) = sin 2t − cosh t, y(t) = sinh t.

14.17

Let X(s) and Y (s) be the Laplace transforms of x(t) and y(t). Applying the Laplace transform to the system and substituting the initial conditions

60

Answers to selected exercises for chapter 14

x(0) = 0, x0 (0) = 2, y(0) = −1 and y 0 (0) = 0 we obtain the algebraic system  2 s X + sY = 1, X − sY = 1.

We can find X(s) by adding these two equations; this gives (s2 + 1)X = 2 and thus X(s) = 2/(s2 + 1). Since sY = X − 1 this gives Y =

1 1 2 1 1 s X− = − = −2 2 , s s s(s2 + 1) s s s +1

where we also applied partial fraction expansion to 2/(s(s2 + 1)). The inverse Laplace transform gives the solution x(t) = 2 sin t, y(t) = 1 − 2 cos t.

14.18

As in exercise 14.16 we obtain the algebraic system  (2s + 1)Y + (5s − 2)X = 3 + 2/(s + 1), −sY + (1 − 2s)X = −1 + 1/(s2 + 1).

Next we solve this system of two linear equations in the unknowns X = X(s) and Y = Y (s). We can find X by multiplying the first equation by s, and adding it to the second equation multiplied by 2s + 1; we then obtain, after some simplification, X(s) =

2s 2s + 1 1 + 2 + . (s − 1)2 (s + 1) (s + 1)(s − 1)2 s−1

One similarly obtains Y (s) =

2 − 4s 5s − 2 1 − 2 − . (s + 1)(s − 1)2 (s + 1)(s − 1)2 s−1

Four terms in X and Y need a partial fraction expansion, which leads to X(s) = − Y (s) =

1 5 s 1 1 + + + − 2 , 2(s + 1) s−1 2(s − 1)2 2(s2 + 1) s +1

3 7 5 s 5 − − + 2 + .

2(s + 1) 2(s − 1) 2(s − 1)2 s +1 2(s2 + 1)

The inverse Laplace transform gives the solution x(t) = (−e−t +2et +5tet + cos t − 2 sin t)/2, y(t) = (3e−t − 7et − 5tet + 2 cos t + 5 sin t)/2.

14.20

Apply Laplace transform with respect to t to the partial differential equation and substitute the initial conditions u(x, 0) = 2 sin 2πx and ut (x, 0) = 0, then it follows as in example 14.16 that s2 U (x, s) − (2 sin 2πx)s = 4Uxx , where U (x, s) is the Laplace transform of u(x, t). Hence, we have obtained the ordinary differential equation U 00 −

s2 1 U = − s sin 2πx.

4 2

The general solution of the homogeneous equation is U (x, s) = Aesx/2 + Be−sx/2 , where A and B can still be functions of s. A particular solution can be found using the ’classical method’, so by trying U (x, s) = a sin 2πx. It then follows that a(4π 2 + s2 /4) = s/2 and so the general solution follows:

Answers to selected exercises for chapter 14

U (x, s) = Aesx/2 + Be−sx/2 +

61

2s sin 2πx. s2 + (4π)2

To determine A and B, we translate the remaining boundary conditions to the s-domain by Laplace transforming them. From the conditions u(0, t) = u(1, t) = 0 it follows that U (0, s) = U (1, s) = 0. Using U (0, s) = 0 we obtain that A + B = 0 and using U (1, s) = 0 we obtain that A(es/2 − e−s/2 ) = 0. Hence, A = B = 0 and so U (x, s) =

2s sin 2πx. s2 + (4π)2

The inverse Laplace transform gives the solution u(x, t) = 2 cos 4πt sin 2πx.

14.22

As in exercise 4.20, for example, one obtains the ordinary differential equation U 00 − (s + 6)U = − cos(x/2).

The general solution of the homogeneous equation is √

√

U (x, s) = Aex s+6 + Be−x s+6 , where A and B can still be functions of s. A particular solution can be found by trying U (x, s) = a cos(x/2). The general solution then follows: √

√

U (x, s) = Aex s+6 + Be−x s+6 +

4 cos(x/2).

4s + 25

To determine A and B, we translate the remaining boundary conditions to the s-domain by Laplace transforming them, which results in U (π, s) = 0 and U 0 (0, s) = 0. This leads to A = B = 0 and so U (x, s) =

1 cos(x/2). s + 25/4

The inverse Laplace transform gives the solution u(x, t) = e−25t/4 cos(x/2).

14.23

a The impulse response is the derivative (in distribution sense if necessary) of the step response. Since the step response has no jump at t = 0, it follows that h(t) = a0 (t) = 2 sinh 2t + 2 sin t − e−t . The transfer function is the Laplace transform of h(t) and from table 7 we obtain that H(s) =

2 1 4 + 2 − . s2 − 4 s +1 s+1

b Since (Lδ(t − 1))(s) = e−s , it follows that Y (s) = 2e−s H(s), where Y (s) = (Ly)(s). From part a and the shift rule in the time domain it follows that y(t) = 2 (t − 1)(2 sinh(2t − 2) + 2 sin(t − 1) − e−t+1 ). One can also use the time-invariance of the system. c Since (Lt)(s) = 1/s2 it follows that Y (s) =

4 2 1 + 2 2 − 2 . s2 (s2 − 4) s (s + 1) s (s + 1)

A partial fraction expansion of these terms leads to Y (s) =

2 1 1 1 − 2 + − . s2 − 4 s +1 s s+1

The inverse Laplace transform gives y(t) = 12 sinh 2t − 2 sin t + 1 − e−t .

14.24

a

From the differential equation it immediately follows that

62

Answers to selected exercises for chapter 14

H(s) =

1 1 , L s2 + ω02

where ω0 = (LC)−1/2 . From table 7 we obtain: h(t) = (sin ω0 t)/(Lω0 ). b The system is not stable since h(t) is not absolutely integrable (or: since the poles iω0 and −iω0 do not lie in the half-plane Re s < 0). c Let Q and V be the Laplace transforms of q and v, where v(t) = e−at (a > 0). Then Q(s) = V (s)H(s) =

1 1 .

L (s + a)(s2 + ω02 )

A partial fraction expansion gives L · Q(s) =

1 1 1 s a 1 − 2 . + 2 a2 + ω02 s + a a + ω02 s2 + ω02 a + ω02 s2 + ω02

The inverse Laplace transform gives „ « 1 a −at q(t) = e + sin ω t − cos ω t .

0 0 L(a2 + ω02 ) ω0 d Note that ω0 6= a. Since V (s) = s/(s2 + a2 ) it follows as in part c that „ « s 1 s s 1 = − Q(s) = , L (s2 + a2 )(s2 + ω02 ) L(ω02 − a2 ) s2 + a2 s2 + ω02 where we also applied a partial fraction expansion. The inverse Laplace transform gives q(t) = e

1 (cos at − cos ω0 t) .

L(ω02 − a2 )

Since V (s) = 2(L cos ω0 t)(s) = 2s/(s2 + ω02 ) it follows as in part c that

Q(s) =

1 2s .

L (s2 + ω02 )2

Since (L sin ω0 t)(s) = ω0 /(s2 + ω02 ), it follows from the differentiation rule in the s-domain that (Lt sin ω0 t)(s) = −

d ω0 2sω0 = 2 . ds s2 + ω02 (s + ω02 )2

Hence, q(t) = (t sin ω0 t)/(Lω0 ). This keeps increasing (in an oscillating way) as t increases.

14.25

a One can write u(t) = cos t + (t − π) cos(t − π). Taking the Laplace transform gives U (s) = (Lu)(s) = (1 + e−πs )s/(s2 + 1). Applying the Laplace transform to the differential equation and substituting the initial conditions gives (s2 +s−2)Y −s−2 = U (s). Since s2 +s−2 = (s−1)(s+2) we thus obtain, after a little simplifying, that Y (s) =

` ´ s 1 + 1 + e−πs . s−1 (s + 2)(s − 1)(s2 + 1)

For the second term we use a partial fraction expansion, which eventually leads to ` ´ 1 Y (s) = + 1 + e−πs s−1 „ « 2 1 3s 1 · + − + .

15(s + 2) 6(s − 1) 10(s2 + 1) 10(s2 + 1)

Answers to selected exercises for chapter 14

63

From table 7 and a shift in the time domain it follows by the inverse Laplace transform that y(t) = et + g(t)/30 + (t − π)g(t − π)/30 with g(t) = 4e−2t + 5et − 9 cos t + 3 sin t. b We have (Lδ(t−2))(s) = e−2s and (Lδ 0 (t−3))(s) = se−3s (table 9). Applying the Laplace transform to the differential equation and substituting the initial conditions gives as in part a: Y (s) =

1 3e−2s + 6se−3s + . s−1 (s + 2)(s − 1)

Partial fraction expansion leads to „ „ « « 1 1 1 1 2 −2s −3s Y (s) = +e − + 2e + . s−1 s−1 s+2 s−1 s+2 From a shift in the time domain it follows by the inverse Laplace transform that y(t) = et + (t − 2)(et−2 − e−2t+4 ) + 2 (t − 3)(et−3 + 2e−2t+6 ).

14.26

a From table 7 we know that (LE)(s) = E/s. Applying the Laplace transform to the system of differential equations and substituting the initial conditions gives  2RI2 + (3R + sL)I1 = E/s, 2(R + sL)I2 − (R + sL)I1 = 0, where I1 and I2 are the Laplace transforms of i1 and i2 . From the second equation we get I1 = 2I2 . Substituting this into the first equation we obtain I2 =

E 1 .

2L s(s + 4R/L)

After a partial fraction expansion we obtain from the inverse Laplace transform the solution ” E “ 1 − e−4Rt/L . i1 (t) = 2i2 (t) = 4R b Since (L sin 2t)(s) = 2/(s2 + 4) we obtain as in part a that I2 =

1 1 .

L (s2 + 4)(s + 4R/L)

After a partial fraction expansion we obtain from the inverse Laplace transform the solution „ « L 2R −4Rt/L i1 (t) = i2 (t) = e − cos 2t + sin 2t .

8R2 + 2L2 L 14.27

Apply the Laplace transform with respect to t to the partial differential equation and substitute the initial conditions u(x, 0) = 0 and ut (x, 0) = 2 sin πx, then it follows that s2 U (x, s) − 2 sin πx = 4Uxx , where U (x, s) is the Laplace transform of u(x, t). Hence, U (x, s) satisfies the ordinary differential equation U 00 −

1 s2 U = − sin πx.

4 2

64

Answers to selected exercises for chapter 14

The general solution of the homogeneous equation is U (x, s) = Aesx/2 + Be−sx/2 , where A and B can still be functions of s. A particular solution can be found by trying U (x, s) = a sin πx. It then follows that a(π 2 + s2 /4) = 1/2 and so the general solution follows: U (x, s) = Aesx/2 + Be−sx/2 +

2 sin πx. s2 + 4π 2

To determine A and B, we translate the remaining boundary conditions to the s-domain by Laplace transforming them. From the conditions u(0, t) = u(2, t) = 0 it follows that U (0, s) = U (2, s) = 0. Using U (0, s) = 0 we obtain that A+B = 0 and using U (2, s) = 0 we obtain that A(es −e−s ) = 0.

Hence A = B = 0 and so 2 U (x, s) = 2 sin πx. s + 4π 2 The inverse Laplace transform gives u(x, t) = π −1 sin 2πt sin πx as solution.

Answers to selected exercises for chapter 15

15.1

For n ≥ 4 we have f [n] = 1 − 1 = 0, for n < −1 we have f [n] = 0 − 0 = 0, and for −1 ≤ n < 4 we have f [n] = 0 − 1 = −1. Hence, f [n] = −(δ[n + 1] + δ[n] + δ[n − 1] + δ[n − 2] + δ[n − 3]).

15.2

Since f [n] has period 5 we obtain that f [n] = δ5 [n] + δ5 [n − 2] + δ5 [n − 3].

15.3

Write z = eiπ/6 , then z 12 = 1, so f [n] = z n has period 12.

15.4

Since [k] = 0 for k < 0 and [k] = 1 for k ≥ 0 we can write the sum in the given right-hand side as ∞ X

∞ X

δ[n − k] =

k=0

[k]δ[n − k].

k=−∞

From theorem 15.1 it then follows that the sum equals [n].

15.7

Let Φ(ω) be the spectrum of φ(t), then Φ(ω) = 0 for | ω | > ω0 for some ω0 . According to the convolution theorem the spectrum of the convolution is equal to the product of the spectra. This product is then also equal to 0 for | ω | > ω0 , hence the convolution is also band-limited.

15.9

Using Fourier series one can determine P (ω) explicitly as follows. We have P (ω) =

∞ X

cn einω0 ω = eiω0 ω + e−iω0 ω ,

n=−∞

where ω0 = 2. Hence, F (ω) = (eiω0 ω + e−iω0 ω )pωs (ω). From table 3 we know that p(t) = sin(ωs t/2)/(πt) ↔ pωs (ω) and applying the shift rule we then obtain that f (t) = p(t + 2) + p(t − 2).

15.10

We have to determine the Nyquist frequency of the convolution. By definition of convolution we have Z ∞ Z π Z ω+π F (ω) = pπ (ω − u)p2π (u) du = pπ (ω − u) du = pπ (u) du. −∞

−π

ω−π

Since pπ (ω) is 0 outside [−π/2, π/2] it follows that F (ω) = 0 outside [−3π/2, 3π/2]. The Nyquist frequency is thus equal to 3π and the sampling frequency should satisfy 2π/T > 3π, hence T < 2/3.

15.11

We write f (t) = (ei(φ0 +ω0 t) + e−i(φ0 +ω0 t) )/2. Then the spectrum F (ω) = π(eiφ0 δ(ω − ω0 ) + e−iφ0 δ(ω + ω0 ). In the proof of the sampling theorem we see that the spectrum Fr (ω) of the reconstructed signal fr (t) equals Fr (ω) =

∞ X

F (ω − kωs )pωs (ω).

k=−∞

Hence, Fr (ω) = πeiφ0

∞ X

δ(ω − ω0 − kωs )pωs (ω)

k=−∞

+ πe−iφ0

∞ X

δ(ω + ω0 − kωs )pωs (ω).

k=−∞

65

66

Answers to selected exercises for chapter 15

Since f (t)δ(t − t0 ) = f (t0 )δ(t − t0 ) we thus have Fr (ω) = πeiφ0

∞ X

δ(ω − ω0 − kωs )pωs (ω0 + kωs )

k=−∞

+ πe−iφ0

∞ X

δ(ω + ω0 − kωs )pωs (kωs − ω0 ).

k=−∞

Since ω0 = 2ωs /3 we have pωs (kωs ± ω0 ) = pωs ((k ± 23 )ωs ). This means that only the terms k = −1 and k = 1 contribute to the sum. Hence, Fr (ω) = πeiφ0 δ(ω + 12 ω0 ) + πe−iφ0 δ(ω − 21 ω0 ).

Applying the inverse Fourier transform leads to the reconstructed signal fr (t) = cos(φ0 − 12 ω0 t).

15.12

a From T = 4/3 it follows that ωs = 3π/2. The sampling condition is satisfied, so fr (t) = f (t). b From T = 2 it follows that ωs = π. Although the sampling condition is not satisfied, it is easy to see that in this case Fs (ω) = 1 for all ω and so Fs (ω) = pπ (ω) = F (ω). Again, fr (t) = f (t). c From T = 8/3 it follows that ωs = 3π/4. The sampling condition is not satisfied. We have Fr (ω) = 2pωs (ω) − pα (ω) with α = ωs /3 = π/4. Using table 3 fr (t) follows: fr (t) =

15.13

a

2 sin(3πt/8) sin(πt/8) − . πt πt

The impulse response follows from table 3:

h(t) =

2 sin2 (πt/2) . π 2 t2

b Let u(t) be an input with spectrum U (ω), then the spectrum Y (ω) of the output y(t) is given by Y (ω) = U (ω)H(ω). The Nyquist frequency of u(t) is π, so U (ω) = 0 outside [−π/2, π/2]. Hence, also Y (ω) = 0 outside [−π/2, π/2]. c For T = 1 the sampling frequency equals 2π. This is greater than the Nyquist frequency of the input and so the sampling condition is satisfied.

From the sampling theorem it then follows that 1

u(t) =

1 X sin(π(t − nT )) . π n=−1 t − nT

To determine y(t) it suffices to know the response to the signal (sin πt)/t since the system is linear and time-invariant. The spectrum of (sin πt)/(πt) is p2π (ω) and the spectrum of the corresponding output is then given by p2π (ω)qπ (ω) = qπ (ω). The response to (sin πt)/(πt) is thus the inverse Fourier transform of qπ (ω), which is 2 sin2 (πt/2)/(π 2 t2 ) (table 3). Hence, 2 sin2 (π(t − 1)/2) 2 sin2 (π(t + 1)/2) 2 sin2 (πt/2) + + .

2 2 2 2 π t π (t − 1) π 2 (t + 1)2 P a We know that f [n] = 3k=0 f [k]δ4 [n − k] holds for all periodic discretetime signals with period 4. Hence, we only have to show that the sampling has period 4, which is easy because nπ π f [n] = f ( + 2π) = f ((n + 4) ) = f [n + 4].

2 2 y(t) =

15.15

Answers to selected exercises for chapter 15

67

b The sampling frequency is 2π/T = 4 > 3, so the sampling condition is satisfied. The signal can thus be reconstructed completely from the sampling.

P∞ ikω0 t be the Fourier series of f , where ω0 = c Let f (t) = k=−∞ ck e 2π/T = 1. Since eiω0 t ↔ 2πδ(ω − ω0 ) we can write the spectrum as F (ω) = 2π

∞ X

ck δ(ω − kω0 ).

k=−∞

This is a line spectrum with lines at kω0 . The periodic signal is band-limited with Nyquist frequency 3. Hence, the spectrum is 0 outside [−3/2, 3/2].

This implies that cn = 0 for | n | ≥ 2. d From part c follows that f (t) = c0 +c1 eit +c−1 e−it . Next substitute the values t = 0, −π/2, π/2, π and use the given sampling from part b. Then we obtain the three equations c0 + c1 + c−1 = 0, c0 + ic1 − ic−1 = 1 and c0 − ic1 + ic−1 = 1. Hence, c0 = 1.

15.16

a The Nyquist frequency is 2π. The sampling frequency is 2π/T = 3π, so the sampling condition is satisfied. The signal can thus be reconstructed completely from the sampling. b We know that (use table 3, no 2 and table 4, shift in the time domain, twice) f (t) =

sin π(t + 1) sin π(t − 1) t sin πt + = .

2π(t + 1) 2π(t − 1) π(1 − t2 )

c Using the sampling theorem we obtain that Z ∞ Z ∞ X 1 ∞ sin π(t − nT )/T f [n] dt. f (t) dt = T π −∞ t − nT −∞ n=−∞ R∞ P The integral in the right-hand side is π, so −∞ f (t) dt = T ∞ n=−∞ f [n]. d According to Parseval’s identity we have Z ∞ Z ∞ Z π 1 1 1 E= | f (t) |2 dt = | F (ω) |2 dω = cos2 ω dω = .

2π 2π 2 −∞ −∞ −π

Answers to selected exercises for chapter 16

16.1

The formula for the 2-point DFT is 1 X

F [k] =

f [n]e−2πink/2 .

n=0 −πi

Since e F [k] =

1 X

= −1 we have f [n](−1)nk = f [0] + (−1)k f [1].

n=0

16.2

The signal f [n] = (−1)n has period 2 and, hence, period 4 as well. The 2-point DFT F2 [k] is given by 1 − (−1)k (see 16.1). The 4-point DFT F4 [k] is by definition given by F4 [k] =

3 X

f [n]e−2πink/4 = 1 − e−πik/2 + e−πik − e−3πik/2

n=0

= 1 − (−i)k + (−1)k − ik = (1 + (−1)k )(1 − ik ).

16.3

The Fourier coefficients are given by Z 1 1 T /2 −ikω0 t (1 − e−ikω0 T /2 ). e dt = ck = T 0 ikω0 T Since ω0 T = 2π we obtain ck = (1 − (−1)k )/2πik for k 6= 0 and c0 = 1/2.

The value F [0] follows immediately from the formula for F [k]: F [0] =

N −1 X n=0

f [n] =

N −1 X

f (nT /N ).

n=0

When N is even then F [0] 1 1 = (f (0) + f (T /N ) + · · · + f (T /2)) = = c0 .

N N 2 When N is odd then F [0] N −1 1 1 1 = = − = c0 − .

N 2N 2 2N 2N Hence, we do not have c0 = F [0]/N for all N .

16.5

It is better to define the function value at the jumps as the average value of the left-hand and right-hand limit. Using the DFT we then find a better approximation of the Fourier coefficients. (In exercise 16.4 we do have c0 = F [0]/N for all N .)

16.7

Apply the inverse 4-point DFT to F [k], then “ ” f [n] = F [0] + F [1]eπin/2 + F [2]eπin + F [3]e3πin/2 /4 = (1 + (−i)n ) /4.

16.8

Since F [k] = | F [k] | ei arg(F [k]) we obtain from the given amplitude spectrum and phase spectrum that F [k] = 2eπik/2 and so F [0] = 2, F [1] = 2i, F [2] = −2, F [3] = −2i. Next apply the inverse DFT: “ ” f [n] = 2 + 2ieπin/2 − 2eπin − 2ie3πin/2 /4

68

Answers to selected exercises for chapter 16

69

and hence, f [0] = 0, f [1] = 0, f [2] = 0, f [3] = 2, that is, f [n] = 2δ4 [n − 3].

16.10

a

On [0, 2π] the periodic function f is given by ( t for 0 ≤ t ≤ π, π |t − π| f (t) = 1 − = π 2 − t for π ≤ t ≤ 2π. π

Since f (t + π) = 1 − t/π for 0 < t ≤ π and f (t + π) = f (t − π) = t/π − 1 for π ≤ t ≤ 2π we obtain that f (t) + f (t + π) = 1 for all t. b Since f [n] = f (2πn/N ) we obtain from part a that f [n]+f [n+N/2] = 1. c Apply the N -point DFT to f [n] + f [n + N/2] = 1, using the shift rule in the n-domain. Since 1 ↔ N δN [k] (table 11), we the obtain that F [k] + e2πiN k/2N F [k] = N δN [k], hence (1 + eπik )F [k] = N δN [k]. When k is even and not a multiple of N then 2F [k] = N δN [k] = 0, so F [k] = 0.

When k is a multiple of N then δN [k] = 1, hence, F [k] = N/2.

16.12

The signal is real, so F [−k] = F [k], which gives F [3] = F [−1] = F [1] = −i.

Applying the inverse DFT leads to f [n] = (1 + in+1 + (−i)n+1 )/4.

16.13

Apply the definition of the cyclical convolution and use theorem 15.2, then one obtains that (f ∗f )[n] = f [n]+f [n−1] = δN [n]+2δN [n−1]+δN [n−2].

16.14

Use the convolution theorem: first determine the functions f1 and f2 with f1 ↔ cos(2πk/N ) = (e2πik/N + e−2πik/N )/2 and f2 ↔ sin(4πk/N ) = (e4πik/N − e−4πik/N )/2i and then calculate f [n] = (f1 ∗ f2 )[n]. Since δN [n − m] ↔ e−2πimk/N (table 11 and table 12, shift in the n-domain), we have f1 [n] = (δN [n − 1] + δN [n + 1])/2 and f2 [n] = (δN [n + 2] − δN [n − 2])/2i. From the convolution theorem and theorem 15.2 it then follows that f [n] = (f1 ∗ f2 )[n] = (f2 [n + 1] + f2 [n − 1])/2, which equals (δN [n + 3] + δN [n + 1] − δN [n − 1] − δN [n − 3])/4i.

16.16

From table 11 we have that δN [n] ↔ 1 and so (table 12, shift in the ndomain) δN [n − l] ↔ e−2πilk/N . Since cos2 (πk/N ) = (1 + cos(2πk/N ))/2 = (2 + eπik/N + e−πik/N )/4 it follows that f [n] = (2δN [n] + δN [n − 1] + δN [n + 1])/4. The power equals N −1

1 X | f [n] |2 .

N n=0 Now | f [n] |2 = (4δN [n] + δN [n − 1] + δN [n + 1])/16 since δN [n]δN [n + 1] = δN [n]δN [n − 1] = δN [n − 1]δN [n + 1] = 0. Also note that δN [n + 1] = δN [n − (N − 1)] and hence, | f [0] |2 = 1/4, | f [1] |2 = 1/16, | f [N − 1] |2 = 1/16, while all other values are 0. This means that „ « 1 1 3 1 1 + + = .

P = N 4 16 16 8N 16.18

a We calculate G[k] from the expression for the 5-point DFT. Note that g[3] = g[−2] = c−2 = 1 and g[4] = g[−1] = c−1 = 2. Hence, G[k] = g[0] + g[1]e−2πik/5 + g[2]e−4πik/5 + g[3]e−6πik/5 + g[4]e−8πik/5 = 1 + 2e−2πik/5 + e−4πik/5 + e−6πik/5 + 2e−8πik/5 = 1 + 4 cos(2πk/5) + 2 cos(4πk/5). b Since the Fourier coefficients of f are known, we can express f as a Fourier series: f (t) = c0 + c1 eiω0 t + c−1 e−iω0 t + c2 e2iω0 t + c−2 e−2iω0 t .

Hence, f (2πm/5ω0 )

70

Answers to selected exercises for chapter 16

= g[0] + g[1]e2πim/5 + g[2]e4πim/5 + g[3]e6πim/5 + g[4]e8πim/5 = G[−m]. c

First use Parseval for Fourier series: Z T ∞ X 1 | f (t) |2 dt = | cn |2 .

T 0 n=−∞

P2 P 2 2 Since ck = 0 for | k | ≥ 3 we obtain that ∞ n=−2 | g[n] | = n=−∞ | cn | = P4 2 n=0 | g[n] | . Applying Parseval for the DFT one obtains the result.

16.19

a From table 11 we have that δN [n] ↔ 1 and so (table 12, shift in the n-domain) f [n] ↔ e2πik/N − 1 + e−2πik/N = 2 cos(2πk/N ) − 1. b Calculate the convolution using (16.14) and theorem 15.2, then (f ∗ g)[n] =

N −1 X

f [l]g[n − l] = g[n + 1] − g[n] + g[n − 1].

l=0

c First write g[n] as complex exponentials and then determine the N point DFT using table 11 and the shift rule in the k-domain: G[k] =

N (δN [k − 2] + δN [k + 2]).

2

Finally apply (16.19) to calculate the power: «2 N −1 N −1 „ 1 1 X 1 X N | δN [k − 2] + δN [k + 2] | = . | g[n] |2 = 2 N n=0 N 2 2 k=0

16.20

a Since f (t) is real and even, the sampling f [n] is real and even since f [−n] = f (−nT /5) = f (nT /5) = f [n]. b Apply the inverse DFT, where the values of F [3] and F [4] are calculated using the fact that F has period 5 and is even. Hence, f [n] = (1 + 2e2πin/5 + 2e8πin/5 + e4πin/5 + e6πin/5 )/5, which equals (1 + 4 cos(2πn/5) + 2 cos(4πn/5))/5. c The function f is band-limited with band-width 10π/T . The Fourier coefficients ck of f contribute to the frequencies kω0 = 2πk/T . Hence, these are 0 for | k | ≥ 3. This means that f (t) is equal to the Fourier series c0 + c1 eiω0 t + c−1 e−iω0 t + c2 e2iω0 t + c−2 e−2iω0 t . Substituting t = nT /5 (and rearranging) we obtain that f [n] = c0 + c1 e2πin/5 + c2 e4πin/5 + c−2 e6πin/5 + c−1 e8πin/5 . But this is precisely the expression for the inverse DFT, which implies that c0 = F [0]/5, c1 = F [1]/5, c2 = F [2]/5, c−2 = F [3]/5 = F [−2]/5, c−1 = F [4]/5 = F [−1]/5. Hence ck = F [k]/5 for | k | ≤ 2. Since F [k] has period 5 and ck = 0 for | k | > 2 we do not have ck = F [k]/5 for | k | > 2.

Answers to selected exercises for chapter 17

17.1

We can write (17.2) for N = 5 as follows: F [k] = f [0] + f [1]w5−k + · · · + f [4]w5−4k where w5 = e2πi/5 = w. Hence, f [0] + f [1] + · · · + f [4] = F [0], f [0] + f [1]w−1 + · · · + f [4]w−4 = F [1], f [0] + f [1]w−2 + · · · + f [4]w−8 = F [2], f [0] + f [1]w−3 + · · · + f [4]w−12 = F [3], f [0] + f [1]w−4 + · · · + f [4]w−16 = F [4].

Using that w−5 = 1 we then obtain a system that is equal to the system arising from the matrix representation.

17.2

As in exercise 17.1 we can write the formula for the inverse DFT in matrix form: 01 1 1 1 1 1 0 F [0] 1 0 f [0] 1 B1 1B B1 5@ 1 1

w w2 w3 w4

w2 w4 w w3

w3 w w4 w2

w4 C B F [1] C B f [1] C CB C B C w3 C B F [2] C = B f [2] C .

A @ A 2 A@ F [3] f [3] w w F [4] f [4]

The matrix in the left-hand side is thus the inverse of the matrix in exercise 17.1.

17.4

Take N1 = 2 and N2 = 2. Note that f [3] = f [−1]. The matrix Mf now looks as follows: „ « „ « f [0] f [2] 2 2 Mf = = . f [1] f [3] 0 1 The 2-point DFT of the rows of this matrix gives „ « 4 0 C= .

1 −1 Multiplying this by the twiddle factors w4−µν with w4 = eπi/2 = i gives „ « 4 0 Ct = .

1 i Now calculate the 2-point DFT of the columns of this matrix to get the 4-point DFT: „ « 5 i MF = .

3 −i Hence, F [0] = 5, F [1] = i, F [2] = 3, F [3] = −i.

17.6

The matrix Mf looks as follows: „ « f [0] f [2] . . . f [N − 2] Mf = .

1 1 ...

1 The N/2-point DFT of the first row of this matrix is A[k], while the N/2point DFT of the second row follows from table 11. This gives the matrix C:

71

72

Answers to selected exercises for chapter 17

„ C=

A[0] N/2

A[1] 0

A[N/2 − 1] 0

... ...

« .

Multiplying this by the twiddle factors will not change this matrix because of the zeroes in the second row of C and hence Ct = C. Now calculate the 2-point DFT of the columns of Ct = C to get the matrix MF and, hence, the required N -point DFT of f [n]: „ « A[0] + N/2 A[1] . . . A[N/2 − 1] MF = „ A[0] − N/2 A[1] . . . A[N/2 − 1] « F [0] F [1] . . . F [N/2 − 1] = .

F [N/2] F [N/2 + 1] . . . F [N − 1] 17.7

Let A1 [k] be the 2N -point DFT of f [2n] and B1 [k] the 2N -point DFT of f [2n + 1]. Then we have for the 4N -point DFT of f [n] (see (17.14)): −ν F [ν] = A1 [ν] + w4N B1 [ν], −ν F [2N + ν] = A1 [ν] − w4N B1 [ν],

where ν = 0, 1, . . . , 2N − 1. The 2N -point DFT of f [2n] follows analogously from the N -point DFT of f [4n] and f [4n + 1]: −ν A1 [ν] = A[ν] + w2N C[ν], −ν C[ν], A1 [N + ν] = A[ν] − w2N

where ν = 0, 1, . . . , N − 1. Also, the 2N -point DFT of f [2n + 1] follows from the N -point DFT of f [4n + 1] and f [4n + 3]: −ν D[ν], B1 [ν] = B[ν] + w2N −ν D[ν], B1 [N + ν] = B[ν] − w2N

where ν = 0, 1, . . . , N − 1. Combining these results leads to the 4N -point DFT of f [n]: −3ν −ν −ν D[ν], B[ν] + w4N C[ν] + w4N F [ν] = A[ν] + w2N −3ν −ν −ν D[ν], B[ν] − w4N C[ν] + w4N F [N + ν] = A[ν] − w2N −ν −ν −3ν F [2N + ν] = A[ν] + w2N C[ν] − w4N B[ν] − w4N D[ν], −3ν −ν −ν D[ν], B[ν] + w4N F [3N + ν] = A[ν] − w2N C[ν] − w4N

where ν = 0, 1, . . . , N − 1.

17.8

RT Since f (t) is causal we have FT (ω) = 0 f (t)e−iωt dt. Apply the trapezium rule to the integral and substitute ω = (2k + 1)π/T , then it follows that N −1

FT ((2k + 1)π/T ) ≈

T X −πin/N e f [n]e−2πin/N .

N n=0

This shows that the spectrum at the frequencies ω = (2k + 1)π/T can be approximated by the N -point DFT of e−πin/N f [n].

17.9

Let F (ω) be the Fourier transform of f (t). Applying the trapezium rule to RT RT f (t)e−iωt dt and to 0 f (t)eiωt dt leads to (T /N )F [k] and (T /N )F [−k] 0 respectively, where F [k] denotes the N -point DFT of f [n]. Adding this gives F(

T 2πk ) ≈ (F [k] + F [−k]) T N

for | k | < N/2.

Answers to selected exercises for chapter 17

73

This means that we can efficiently approximate the spectrum of f at the frequencies 2πk/T with | k | < N/2 by using an N -point DFT. 17.11

First we determine the DFT’s of the signals using table 11. Since f1 [n] = δN [n] + δN [n − 1] ↔ F1 [k] = 1 + e−2πik/N and f2 [n] = δN [n] + δN [n + 1] ↔ F2 [k] = 1 + e2πik/N we obtain the DFT of the cross-correlation as follows: ρ12 ↔ F1 [k]F2 [k] = (1 + e2πik/N )2 = 1 + 2e2πik/N + e4πik/N .

17.13

17.14

Applying the inverse transform gives the cross-correlation ρ12 = δN [n] + 2δN [n + 1] + δN [n + 2]. √ Let P (z) = f [0] + f [1]z + f [2]z 2 and w3 = e2πi/3 = (−1 + i 3)/2 = 1/w. −k k k 2k Then F [k] = P (w3 ) = P (w ) = f [0]+f [1]w +f [2]w = f [0]+wk (f [1]+ wk f [2]).

Let Mf be the 3 × N -matrix given by 0 1 f [0] f [3] . . . f [3N − 3] Mf = @ f [1] f [4] . . . f [3N − 2] A . f [2] f [5] . . . f [3N − 1] The N -point DFT of the rows of this matrix are given by A[k], B[k] and C[k] respectively. The matrix C is then given by 1 0 A[0] A[1] . . . A[N − 1] C = @ B[0] B[1] . . . B[N − 1] A .

C[0] C[1] . . . C[N − 1] −νµ Multiplying this by the twiddle factors w3N gives the matrix Ct. The 3-point DFT of the columns of Ct will give us the matrix MF : 0 1 F [0] F [1] . . . F [N − 1] MF = @ F [N ] F [N + 1] . . . F [2N − 1] A .

F [2N ] F [2N + 1] . . . F [3N − 1]

Since the 3-point DFT of g[n] is given by G[k] = g[0] + g[1]w3−k + g[2]w3−2k we conclude that −(ν+N µ)

F [µN + ν] = A[ν] + w3N 17.15

−2(ν+N µ)

B[ν] + w3N

C[ν].

m−1

Take N1 = 3 and N2 = 3 and consider the N1 × N2 -matrix 0 1 f [0] f [3] . . . f [N − 3] Mf = @ f [1] f [4] . . . f [N − 2] A . f [2] f [5] . . . f [N − 1] Let the N2 -point DFT of the rows f [3n], f [3n + 1] and f [3n + 2] be given

by A[k], B[k] and C[k], then the matrix C is given by 0 1 A[0] A[1] . . . A[N2 − 1] C = @ B[0] B[1] . . . B[N2 − 1] A .

C[0] C[1] . . . C[N2 − 1] Multiplying this by the twiddle factors and then applying the 3-point DFT of the columns gives the matrix MF containing the N -point DFT of f [n]: 0 1 F [0] F [1] . . . F [N2 − 1] MF = @ F [N2 ] F [N2 + 1] . . . F [2N2 − 1] A .

F [2N2 ] F [2N2 + 1] . . . F [N − 1]

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Answers to selected exercises for chapter 17

17.16

To calculate the 3-point DFT we need 6 additions and 4 multiplications, hence 10 elementary operations. The N2 -point DFT of f [3n], of f [3n + 1], and of f [3n + 2] can be determined by repeatedly splitting this into (multiples of 3), (multiples of 3) + 1, and (multiples of 3) + 2. We thus only have to determine 3-point DFT’s.

P Let h[n] = N l=0 f [l]g[n − l]. Since g[n] is causal and g[n] = 0 for n > N we have that h[n] = 0 for n < 0 or n > 2N . Hence, it suffices to calculate h[n] for n = 0, 1, . . . , 2N . Now let fp [n] and gp [n] be two periodic signals with period 2N + 1 and such that fp [n] = f [n] and gp [n] = g[n] for n = 0, 1, . . . , 2N . Then h[n] =

2N X

fp [l]gp [n − l] = (fp ∗ gp )[n]

for n = 0, 1, . . . , 2N .

l=0

Now let fp [n] ↔ Fp [k] and gp [n] ↔ Gp [k]. Then we have 2N

h[n] =

X 1 Fp [k]Gp [k]e2πink/(2N +1) 2N + 1 k=0

for n = 0, 1, . . . , 2N . Although these are really 2N + 1-point DFT’s, we continue for convenience our argument with 2N . If we assume that calculating a 2N -point DFT using the FFT requires 2N (2 log N ) elementary operations, then the number of elementary operations to calculate h[n] will (approximately) equal 2N (2 log N )+2N (2 log N )+(2N +1)+2N (2 log N ) = 6N (2 log N ) + 2N + 1. A direct calculation would require in the order N 2 operations, which for large N is much less efficient.

17.17

Let P (z) = f [0] + f [1]z + f [2]z 2 + f [3]z 3 and w = e−2πi/4 = −i. Then F [k] = P (wk ), hence, F [0] = P (1), F [1] = P (−i), F [2] = P (−1), F [3] = P (i). Now f [n] is even and so f [3] = f [−1] = f [1]. Then F [k] is also even, hence F [3] = F [1]. This gives F [0] = f [0] + f [1] + f [2] + f [3] = f [0] + 2f [1] + f [2], F [1] = f [0] − if [1] − f [2] + if [3] = f [0] − f [2], and F [2] = f [0] − f [1] + f [2] − f [3] = f [0] − 2f [1] + f [2].

Answers to selected exercises for chapter 18

18.1

When f [n] = 0 for | n | > N for some N > 0, then ∞ X

F (z) =

f [n]z −n =

n=−∞

−1 X

f [n]z −n +

n=−N

N X

f [n]z −n .

n=0

The anti-causal part converges for all z, while the causal part converges for all z 6= 0. The region of convergence is thus given by 0 < | z | < ∞. If f [n] = 0 for n ≥ 1 then the z-transform converges for all z ∈ C. 18.2

The anti-causal part converges for all z. The causal part can be written as «n „ «n « ∞ „„ X 1 1 + .

2z 3z n=0 Hence, the z-transform converges for | 2z | > 1 and | 3z | > 1, that is, for | z | > 1/2.

18.3

a

A direct calculation of F (z) gives

∞ ““ ” X z n n=0

2

+

“ z ”n ” 3

.

This series converges for | z/2 | < 1 and | z/3 | < 1, hence, for | z | < 2. b The z-transform is given by F (z) =

∞ X

cos(πn/2)z −n .

n=0

P −n Since | cos(πn/2) | ≤ 1 for all n and since ∞ converges n=0 | z | P∞ for | z | > 1, the z-transform also converges for | z | > 1. Moreover, n=0 cos(πn/2) diverges since limn→∞ cos(πn/2) 6= 0. We conclude that the z-transform converges for | z | > 1. c From parts a and b it follows immediately that the region of convergence is the ring 1 < | z | < 2.

18.4

a We rewrite f [n] in order to apply a shift in the n-domain: f [n] = 2(n − 2) [n − 2] + 4 [n − 2]. Since [n] ↔ z/(z − 1) for | z | > 1, we obtain from the shift rule that 4 [n − 2] ↔ 4z −1 /(z − 1). It also follows from [n] ↔ z/(z − 1) and the differentiation rule that n [n] ↔ z/(z − 1)2 and applying the shift rule we then obtain that (n − 2) [n − 2] ↔ z −1 /(z − 1)2 .

Combining these results gives the z-transform F (z) of f [n]: F (z) =

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

for | z | > 1.

b We rewrite f [n] in order to apply a shift in the n-domain: f [n] = 2(n + 2) [−(n + 2)] − 4 [−(n + 2)]. From [n] ↔ z/(z − 1) we obtain from time reversal that [−n] ↔ (1/z)/((1/z) − 1) = 1/(1 − z) for | z | > 1. From the shift rule it follows that 4 [−(n + 2)] ↔ 4z 2 /(1 − z). It also follows from [−n] ↔ 1/(1 − z) and the differentiation rule that n [−n] ↔ −z/(1 − z)2 and applying the shift rule we then obtain that 2(n + 2) [−(n + 2)] ↔ −2z 3 /(1 − z)2 . Combining these results gives

75

76

Answers to selected exercises for chapter 18

F (z) = c

−2z 3 4z 2 2z 3 − 4z 2 − = .

2 (1 − z) 1−z (1 − z)2

We can, for example, calculate this z-transform in a direct way:

f [n] = (−1)n [−n] ↔

∞ X

(−1)n z n =

n=0

1 1+z

for | z | < 1.

d Again we can, for example, calculate this z-transform in a direct way: f [n] = [4 − n] ↔

∞ X

zn =

n=−4

z −4 1−z

for | z | < 1.

One can also use [−n] ↔ 1/(1 − z) and apply a shift rule. e From example 18.6 it follows that (n2 − n) [n] ↔ 2z/(z − 1)3 and n [n] ↔ z/(z − 1)2 . Adding these results gives n2 [n] ↔

z(z + 1) (z − 1)3

for | z | > 1.

From example 18.2 we obtain that 4n [n] ↔ z/(z − 4) for | z | > 4. The differentiation rule implies that n4n [n] ↔ 4z/(z−4)2 for | z | > 4. Together these results give F (z) = 18.5

a

z(z + 1) 4z + (z − 1)3 (z − 4)2

for | z | > 4.

A direct calculation of the z-transform gives (for | z | > 1):

F (z) =

∞ X

cos(nπ/2)z −n = 1 − z −2 + z −4 − · · · =

n=0

z2 1 = . −2 1+z 1 + z2

b A direct calculation of the z-transform gives (for | z | > 1): F (z) =

∞ X

sin(nπ/2)z −n = z −1 − z −3 + · · · =

n=0

z −1 z = .

1 + z −2 1 + z2

A direct calculation of the z-transform of einφ [n] gives (for | z | > 1):

c

∞ X einφ z −n = 1 + eiφ z −1 + e2iφ z −2 + · · · = n=0

1 z = .

1 − eiφ /z z − eiφ

A direct calculation of the z-transform of 2n einφ [−n] gives (for | z | < 2): 0 X

2n einφ z −n =

n=−∞

∞ X

2−n e−inφ z n =

n=0

1 .

1 − e−iφ z/2

Hence we get for 1 < | z | < 2: F (z) = 18.8

z 1 + . z − eiφ 1 − e−iφ z/2

One could apply a partial fraction expansion here (see e.g. exercise 18.10).

However, in this case it is easy to obtain the z-transform in a direct way by developing F (z) in a series expansion. Since the z-transform has to converge for | z | > 2 (there are poles at z = ±2i and f [n] has a finite switch-on time) we develop F (z) as follows: „ « 1 1 4 16 F (z) = 2 = 2 1 − 2 + 4 + ··· . z +4 z z z

Answers to selected exercises for chapter 18

77

From the series we now obtain that f [n] = 0 for n ≤ 0 and that f [2n] = (−1)n−1 22n−2 , f [2n + 1] = 0, for n > 0.

18.9

As in the previous exercise we obtain the z-transform in a direct way by developing F (z) in a series expansion. Since the unit circle | z | = 1 has to belong to the region of convergence (f [n] has to be absolute convergent) we develop F (z) for | z | < 2 as follows: « „ 1 1 z4 z2 F (z) = = + + · · · .

1 − 4(1 + z 2 /4) 4 4 16 From the series we obtain that f [n] = 0 for n ≥ 1 and f [2n] = (−1)n 22n−2 , f [2n − 1] = 0, for n ≤ 0.

18.10

The poles are at z = −1/2 and z = −3; the signal f [n] must have a finite switch-on time, hence, the z-transform has to converge for | z | > 3. A partial fraction expansion of F (z)/z gives F (z) 1/10 18/5 =1+ − . z z + 1/2 z+3 Applying (18.14) to the expansion of F (z) gives (−1/2)n [n] ↔

z z for | z | > 1/2, (−3)n [n] ↔ for | z | > 3. z + 1/2 z+3

Combining this gives f [n] = δ[n + 1] + ((−1/2)n [n] − 36(−3)n [n])/10.

18.11

In exercise 18.10 we obtained the partial fraction expansion. Now the signal f [n] has to be absolutely convergent, which means that | z | = 1 has to belong to the region of convergence. Applying (18.14) to z/(z + 1/2) gives z (−1/2)n [n] ↔ for | z | > 1/2, z + 1/2 while applying (18.15) to z/(z + 3) gives (−3)n [−n − 1] ↔

−z for | z | < 3. z+3

Combining this gives f [n] = δ[n+1]+((−1/2)n [n]+36(−3)n [−n−1])/10.

18.12

A direct application of the definition of the convolution product gives (f ∗ g)[n] =

M1 X

f [l]g[n − l]

l=N1

= f [N1 ]g[n − N1 ] + f [N1 + 1]g[n − N1 − 1] + · · · + f [M1 − 1]g[n − M1 + 1] + f [M1 ]g[n − M1 ].

Since g[n − N1 ] = 0 for n < N1 + N2 and, hence, also g[n − N1 − 1] = 0, g[n − N1 − 2] = 0, etc., we have that (f ∗ g)[n] = 0 for n < N1 + N2 . This means that the switch-on time is N1 + N2 .

Since g[n − M1 ] = 0 for n > M1 + M2 and, hence, also g[n − M1 + 1] = 0, g[n − M1 + 2] = 0, etc., we have that (f ∗ g)[n] = 0 for n > M1 + M2 . This means that the switch-off time is M1 + M2 .

18.13

a The signal is causal and F (z) has poles at z = ±i. The z-transform converges for | z | > 1. Write F (z) as the sum of a geometric series (or use a partial fraction expansion):

78

Answers to selected exercises for chapter 18

F (z) =

z 1 = z2 + 1 z

« „ 1 1 1 − 2 + 4 + ··· . z z

From the series we obtain that f [n] = 0 for n < 0 and f [2n + 1] = (−1)n , f [2n] = 0, for n ≥ 0. b The convolution theorem gives (f ∗ f )[n] ↔ F 2 (z). Hence, ! ∞ n X X h[n] = (f ∗ f )[n] = f [l]f [n − l] = f [l]f [n − l] [n]. l=−∞

l=0

From this we obtain that h[n] = 0 for n < 0 while for m ≥ 0 we have: h[2m + 1] = f [0]f [2m + 1] + f [1]f [2m] + · · · + f [2m + 1]f [0] = 0, h[2m] = f [0]f [2m] + f [1]f [2m − 1] + · · · + f [2m]f [0] = 0 + (−1)0 (−1)m−1 + 0 + · · · + (−1)m−1 (−1)0 + 0 = m(−1)m−1 .

18.14

Apply the definition of the convolution product and use that [n − l] = 0 for l > n and [n − l] = 1 for l ≤ n.

18.15

Use that f [n] ↔ F (z) and δ[n] ↔ 1 and apply the convolution theorem (assuming that the intersection of the regions of convergence is non-empty) then (f ∗ δ)[n] ↔ F (z) · 1 = F (z). Hence, f [n] = (f ∗ δ)[n].

P l−n Define a discrete-time signal h[n] by h[n] = n f [l], which is the l=−∞ 2 −n −n convolution product of f [n] with 2 [n]. Now 2 [n] ↔ z/(z − 1/2) for | z | > 1/2 and f [n] ↔ F (z). Hence, h[n] ↔ zF (z)/(z − 1/2). Assuming that | z | = 1 belongs to the region of convergence of the z-transform of f [n] we get

18.17

∞ X eiω F (eiω ) = h[n]e−inω . eiω − 1/2 n=−∞

Using (18.22) to determine h[n] gives Z 1 π eiω F (eiω )einω dω. h[n] = π −π 2eiω − 1 18.19

Applying (18.31) gives Z π ˛ Z π ∞ ˛2 X 1 1 1 ˛ iω ˛ | f [n] |2 = F (e ) dω = cos2 ω dω = . ˛ ˛ 2π 2π 2 −π −π n=−∞ (One can also determine f [n] first and then calculate ectly.)

P∞

2 n=−∞ | f [n] | dir-

18.20

We have that ρ[n] = (g ∗ f )[n] with g[l] = f [−l]. The convolution theorem and property (18.25) (or table 15, entry 2) imply that the spectrum of ρ[n] is given by G(eiω )F (eiω ). But G(eiω ) = F (eiω ) (combine table 15, entries ˛ ˛2 2 and 5) and hence ρ[n] ↔ F (eiω )F (eiω ) = ˛ F (eiω ) ˛ .

18.22

a The signal f [n] is causal. The region of convergence is the exterior of a circle. Applying the shift rule to 2−n [n] ↔ z/(z − 1/2) for | z | > 1/2 gives 2−(n+2) [n + 2] ↔ z 3 /(z − 1/2) for | z | > 1/2. Hence, F (z) = 4z 3 /(z − 1/2) for | z | > 1/2. b One could write g[n] = (f ∗ )[n], apply the convolution theorem and then a partial fraction However, it is easier to do a direct P expansion. −l calculation: g[n] = n l=−∞ 2 [l + 2] and hence, g[n] = 0 for n < −2 while P −l g[n] = n = 8(1 − 2−n−3 ) for n ≥ −2 (it is a geometric series). We l=−2 2

Answers to selected exercises for chapter 18

79

thus obtain that g[n] = (8 − 2−n ) [n + 2]. c The z-transform of f [n] converges for | z | > 1/2, which contains the unit circle | z | = 1. Hence, the Fourier transform of f [n] equals F (eiω ) = 4e3iω /(eiω − 1/2).

18.23

a The signal f [n] is absolutely summable. The z-transform has one pole at z = −2 and therefore the region of convergence is | z | < 2, since it must contain | z | = 1. Since F (z)/z = 1 − 2/(z + 2) we have F (z) = z − 2z/(z + 2) for | z | < 2. The inverse transform of this gives f [n] = δ[n + 1] − (−2)n+1 [−n − 1]. b The Fourier transform of f [n] equals F (eiω ) = e2iω /(eiω + 2). c We have f [n] ↔ F (z) for | z | < 2. The scaling property (table 14, entry 5) gives 2n f [n] ↔ F (z/2) for | z | < 4. The spectrum of 2n f [n] is thus equal to F (eiω /2) = e2iω /(2eiω + 8).

18.24

a The signal f [n] is causal. The poles of F (z) are at ±i/2 and at 0. The region of convergence is the exterior of the circle | z | = 1/2. This contains the unit circle and so the signal is absolutely summable. The spectrum of f [n] equals F (eiω ) = 1/(eiω (4e2iω + 1)). b First apply a partial fraction expansion to F (z)/z (the denominator equals z 2 (2z + i)(2z − i)): F (z) 1 i i = 2 + − . z z z − i/2 z + i/2 Applying (18.14) to F (z) gives f [n] = δ[n − 1] + (i(i/2)n − i(−i/2)n ) [n]. c If F (eiω ) = F (e−iω ), then the signal is real. Since F (eiφ ) = 1/(e−iφ (4e−2iφ + 1)) = F (e−iφ ),

18.25

the signal is indeed real. (One can also write (i(i/2)n − i(−i/2)n ) = in+1 2−n (1 − (−1)n ), which equals 0 for n = 2k and 21−n (−1)k+1 for n = 2k + 1, showing clearly that it is real.)

P For n < 0 we have that n l=0 g[l]g[n − l] = 0 since g[n] is causal. Thus f [n] is also causal. Since g[l] = 0 for l < 0 and g[n − l] = 0 for l > n we can write f [n] =

∞ X

g[l]g[n − l] = (g ∗ g)[n].

l=−∞

This implies that F (z) = G(z)2 , so we need to determine G(z). To do so, we write G(eiω ) = 1/(4 + cos 2ω) as a function of eiω : 1 1 2e2iω = = 4iω .

1 2iω −2iω 4 + cos 2ω e + 8e2iω + 1 4 + 2 (e +e ) Taking z = eiω we find that G(z) = 2z 2 /(z 4 + 8z 2 + 1), which gives F (z) = G(z)2 .

Answers to selected exercises for chapter 19

19.2

a

Substituting δ[n] for u[n] we find that

h[n] =

n−1 X

2l−n δ[l] = 2−n [n − 1].

l=−∞

Use the definition of δ[n] and [n] to verify (19.3): y[n] =

n−1 X

∞ X

2l−n u[l] =

l=−∞

2l−n [n − 1 − l]u[l] =

l=−∞

∞ X

h[n − l]u[l].

l=−∞

b As in part a we obtain that h[n] = (δ[n + 1] + δ[n − 1])/2 and ∞ 1 X u[l] (δ[n − l + 1] + δ[n − l − 1]) 2 l=−∞ l=−∞ 1 = (u[n + 1] + u[n − 1]) = y[n].

2 P l−n c We now have h[n] = ∞ δ[l] = 2−n [−n] and l=n 2 ∞ X

h[n − l]u[l] =

∞ X l=−∞

19.3

∞ X

h[n − l]u[l] =

u[l]2l−n [l − n] =

l=−∞

∞ X

u[l]2l−n = y[n].

l=n

a An LTD-system is causal if and only if the impulse response is a causal signal. So this system is causal. The system is stable P∞if and only if the impulse response is absolutely summable. Since n=−∞ | h[n] | = P∞ −n 2 = 1 < ∞, this system is stable. n=1 b This system is not causal, but it is stable since ∞ X

| h[n] | =

n=−∞

∞ X

(δ[n + 1] + δ[n − 1])/2 = 1 < ∞.

n=−∞

c P∞Thisn system is not causal, and it is not stable since n=0 2 = ∞.

P∞

n=−∞ | h[n] | =

19.4

a The step response is the P response to [n] and can be calculated using ∞ (19.3): a[n] = (h ∗ )[n] = l=−∞ (δ[l] − 2δ[l − 1] + δ[l − 2]) [n − l] = [n] − 2 [n − 1] + [n − 2]. b The response to an arbitrary input also follows from (19.3): y[n] = (h ∗ u)[n] = u[n] − 2u[n − 1] + u[n − 2].

19.5

Since δ[n] = [n] − [n − 1], it follows from linearity and time-invariance that h[n] = a[n] − a[n − 1]. Now use (19.3) to calculate the response y[n] to u[n] = 4−n [n]: y[n] = (h ∗ u)[n] =

∞ X

a[l]u[n − l] −

l=−∞

=

∞ X l=−∞

80

a[l]u[n − l] −

∞ X

a[l − 1]u[n − l]

l=−∞ ∞ X l=−∞

a[l]u[n − 1 − l].

Answers to selected exercises for chapter 19

81

We now calculate the first sum; the second one then follows by replacing n by n − 1. ∞ X

a[l]u[n − l] =

l=−∞

∞ “ ” X 2−l [l] − 3−l [l − 1] u[n − l] l=−∞ ∞ X −l l−n

∞ X [n − l] − 3−l 4l−n [n − l] l=0 l=1 ! n n X X −n l −n = 4 2 [n] − 4 (4/3)l [n − 1]

=

2 4

l=0

l=1

= 4−n (2n+1 − 1) [n] − 3 · 4−n ((4/3)n+1 − (4/3)) [n − 1] = (21−n − 4−n ) [n] − 4(3−n − 4−n ) [n − 1].

Hence, replacing n by n − 1 and then taking terms together in the sum, y[n] = (21−n −4−n ) [n]−4(3−n +2−n −2·4−n ) [n−1]−4(31−n −41−n ) [n−2].

19.8

The transfer function follows from an [n] ↔ z/(z − a) for | z | > | a |; this is because we can write cos nφ = (einφ + e−inφ )/2, and hence „ « z z 1 + h[n] ↔ 2 z − eiφ /2 z − e−iφ /2 ˛ iφ ˛ for | z | > ˛ e /2 ˛ = 1/2. We thus obtain: H(z) =

z(z − cos φ/2) z 2 − cos φz + 1/4

for | z | > 1/2. The poles are at eiφ /2 and e−iφ /2, which is inside the unit circle. Therefore the system is stable.

19.9

a The impulse response h[n] can be determined by a partial fraction expansion of H(z)/z. Since „ « 1/3 H(z) 1 1 = − z 9 z + 1/3 (z + 1/3)2 we have H(z) =

1 9

„

z/3 z − z + 1/3 (z + 1/3)2

« .

The system is stable, so the region of convergence must contain | z | = 1.

This region is thus given by | z | > 1/3. Using table 13 we find that h[n] = ((−1/3)n + n(−1/3)n ) [n]/9. b Write u[n] = (einπ/2 − e−inπ/2 )/2i and use (19.10): e±inπ/2 has response H(e±iπ/2 )e±inπ/2 , so the response to u[n] is (H(eiπ/2 )einπ/2 − H(e−iπ/2 )e−inπ/2 )/2i, which is of the form (w − w)/2i = Im (w). Hence the response is „ « „ « −1 8 + 6i Im einπ/2 = Im (cos(nπ/2) + i sin(nπ/2)) , −9 + 6i + 1 100 which is (3 cos(nπ/2) + 4 sin(nπ/2))/50.

19.11

a The frequency be written as H(eiω ) = 1 + e2iω + e−2iω .

P∞response can iω −inω Since H(e ) = n=−∞ h[n]e it follows that h[0] = 1, h[2] = h[−2] = 1 and h[n] = 0 for all other n. Hence, the impulse response is h[n] = δ[n] + δ[n − 2] + δ[n + 2]. The input is u[n] = δ[n − 2]. Since δ[n] 7→ h[n], we have

82

Answers to selected exercises for chapter 19

u[n] 7→ h[n − 2], so the response to u[n] is y[n] = δ[n − 2] + δ[n − 4] + δ[n]. b The impulse response is not causal, so the system is not causal.

19.13

From (19.15) and the fact that H(eiω ) is even we obtain that Z π Z 1 1 ωb h[n] = cos(nω) dω H(eiω ) cos(nω) dω = 2π −π π ωa 1 = (sin nωb − sin nωa ), nπ which can be written as 2(sin 21 n(ωb − ωa ) cos 12 n(ωb + ωa ))/(nπ) for n 6= 0 and Z 1 ωb ωb − ωa h[0] = dω = . π ωa π

19.14

The spectrum Y (eiω ) of y[n] is a periodic function with period 2π. Apply Parseval for periodic functions and substitute Y (eiω ) = H(eiω )U (eiω ) to get the desired result.

19.15

a Apply the z-transform to the difference equation, using the shift rule in the n-domain. We then obtain that (1 + 21 z −1 )Y (z) = U (z). Since H(z) = Y (z)/U (z) it follows that H(z) =

z 1 = .

1 + z −1 /2 z + 1/2

From table 13 we get the impulse response h[n] = (−1/2)n [n]. b We could use z-transforms here: from Y (z) = H(z)U (z) we get Y (z) = z 2 /((z + 1/2)(z − 1/2)) (use table 13) and applying a partial fraction expansion to Y (z)/z then leads to y[n] (again use table 13). However, in this case it is easier to follow the direct way: (h ∗ u)[n] =

∞ X

(−1/2)l [l](1/2)n−l [n − l].

l=−∞

P l Now if n < 0 then this is 0, while if n ≥ 0 then it equals (1/2)n n l=0 (−1) = n+1 n (1/2) (1 + (−1) ). c The transfer function H(z) has one pole at z = −1/2, which is inside the unit circle, so the system is stable.

19.17

a From the difference equation we obtain that (1 − z −2 /4)Y (z) = (1 + z −1 )U (z) and hence H(z) =

z(z + 1) 1 + z −1 = 2 .

1 − z −2 /4 z − 1/4

A partial fraction expansion of H(z)/z gives −1/2 3/2 H(z) = + . z z + 1/2 z − 1/2 Hence, H(z) = (−z/2)/(z + 1/2) + (3z/2)/(z − 1/2). Using table 13 we find that h[n] = ((−1/2)n+1 + 3(1/2)n+1 ) [n]. b The (rational) transfer function H(z) has poles at z = ±1/2, which lie inside the unit circle, so the system is stable. c The z-transform of [n] is z/(z − 1) and therefore the z-transform A(z) of the step response a[n] is given by A(z) = H(z)U (z) = z 2 (z + 1)/((z − 1)(z + 1/2)(z − 1/2)). A partial fraction expansion of A(z)/z gives

Answers to selected exercises for chapter 19

83

A(z) 8/3 3/2 1/6 = − − . z z−1 z − 1/2 z + 1/2 Multiply this by z and use table 13 to obtain a[n] = ((8/3) − 3(1/2)n+1 − (1/3)(−1/2)n+1 ) [n]. d Since u[n] = [n] + [n − 2] and [n] 7→ a[n] we get u[n] 7→ a[n] + a[n − 2].

19.18

a To find the impulse response we first apply a partial fraction expansion to H(z)/z, which gives H(z) 1/4 1 1 = − − . z z z + 1/2 (z + 1/2)2 Multiply this by z and use table 13 to obtain that h[n] = δ[n] − ((−1/2)n − (1/2)n(−1/2)n ) [n]. b According to (19.9) we have that z n 7→ H(z)z n . Substituting z = −1 we get the response to the input (−1)n . Since H(−1) = 0, the response is the null-signal. c The (rational) transfer function H(z) has one pole at z = −1/2, which is inside the unit circle, so the system is stable. d The impulse response is real, so the system is real. e Let u[n] 7→P y[n], then Y (eiω ) = H(eiω )U (eiω ). Furthermore we have −inω that U (eiω ) = ∞ . Comparing this with U (eiω ) = cos 2ω = n=−∞ u[n]e 2iω −2iω (e +e )/2 we may conclude that u[2] = u[−2] = 1/2 and u[n] = 0 for n 6= ±2, hence, u[n] = (δ[n − 2] + δ[n + 2])/2. By superposition it then follows from δ[n] 7→ h[n] that y[n] = (h[n − 2] + h[n + 2])/2.

19.19

a The response to u[n + N ] is y[n + N ] (time-invariance), but also u[n] = u[n + N ] for n ∈ Z, so y[n] = y[n + N ] for n ∈ Z. b In (16.7) (with f replaced by u) the input u[n] is written as superposition of the signals e2πink/N . Since z n 7→ H(z)z n , we have e2πink/N 7→ H(e2πik/N )e2πink/N . Hence we obtain from (16.7) that N −1

y[n] =

1 X U [k]H(e2πik/N )e2πink/N .

N k=0

On the other hand we have from the inverse DFT for y[n] that N −1

y[n] =

1 X Y [k]e2πink/N , N k=0

where Y [k] is the N -point DFT of y[n]. Hence, Y [k] = U [k]H(e2πik/N ).

19.20

a Since H(eiω ) = cos 2ω = (e2iω + e−2iω )/2 we see from definition (19.11) of the frequency response that h[2] = h[−2] = 1/2 and h[n] = 0 for n 6= ±2.

Hence, h[n] = (δ[n + 2] + δ[n − 2])/2. b Using the inverse DFT we can recover u[n] from the DFT F [k] of u[n]: 3

u[n] =

1X F [k]e2πink/4 , 4 k=0

where F [k] is the 4-point DFT of u[n]. We have F [0] = 1, F [1] = −1, F [2] = 0, F [3] = 1 and since e2πink/4 7→ H(e2πik/4 )e2πink/4 , it follow by superposition that y[n] = (1 + 2i sin(nπ/2))/4.

19.21

a Since δ[n] = [n] − [n − 1] we have for the responses that h[n] = a[n] − a[n − 1] and hence h[n] = n2 (1/2)n [n] − (n − 1)2 (1/2)n−1 [n −

84

Answers to selected exercises for chapter 19

1]. For the z-transforms we have (use the shift rule in the n-domain) H(z) = A(z) − A(z)/z = A(z)(z − 1)/z. From table 13` it´ follows that n 2 n (1/2)n [n] ↔ z/(z −1/2), n(1/2)n [n] ` ´ ↔ (z/2)/(z −1/2) , 2 (1/2) [n] ↔ (z/4)/(z − 1/2)3 and since n2 = 2 n2 + n it then follows that z/2 z/2 1 z(z + 1/2) + = . (z − 1/2)3 (z − 1/2)2 2 (z − 1/2)3

A(z) =

This means that the transfer function H(z) equals H(z) =

1 (z − 1)(z + 1/2) .

2 (z − 1/2)3

b The impulse response is causal, so the system is causal. c From einω 7→ H(eiω )einω it follows that y[n] = 19.22

1 (eiω − 1)(eiω + 1/2) inω e .

2 (eiω − 1/2)3

a From the difference equation we obtain that (1 − z −1 /2)Y (z) = (z −1 + z −2 )U (z) and hence H(z) =

z+1 z −1 + z −2 = 2 .

1 − z −1 /2 z − z/2

Since [n] ↔ z/(z − 1) = U (z) we have a[n] ↔ A(z) = H(z)U (z) =

z+1 . (z − 1/2)(z − 1)

A partial fraction expansion of A(z)/z gives A(z) 2 4 6 = + − . z z z−1 z − 1/2 Multiply this by z and use table 13 to obtain that a[n] = 2δ[n] + (4 − 6(1/2)n ) [n]. b The (rational) transfer function H(z) has poles at z = 0 and z = 1/2, which lie inside the unit circle, so the system is stable.

19.23

a From the difference equation we obtain that (6 − 5z −1 + z −2 )Y (z) = (6 − 6z −2 )U (z) and hence H(z) =

6(z 2 − 1) 6 − 6z −2 = 2 . −1 −2 6 − 5z + z 6z − 5z + 1

A partial fraction expansion of H(z)/z (note that the denominator equals z(2z − 1)(3z − 1)) gives H(z) 6 16 9 =− + − . z z z − 1/3 z − 1/2 Multiply this by z and use table 13 to obtain that h[n] = −6δ[n] + (16(1/3)n − 9(1/2)n ) [n]. b The frequency response H(eiω ) is obtained from the transfer function H(z) by substituting z = eiω , so H(eiω ) = 6(e2iω − 1)/(6e2iω − 5eiω + 1). c From (19.10) we know that Y (eiω ) = H(eiω )U (eiω ). Since y[n] is identically 0, we know that Y (eiω ) = 0 for all frequencies ω. Since H(eiω ) 6= 0 for e2iω 6= 1 we have U (eiω ) = 0 for e2iω 6= 1. The frequencies ω = 0 or ω = π may still occur in the input. These frequencies correspond to

Answers to selected exercises for chapter 19

85

the time-harmonic signals ei0n = 1 and eiπn = (−1)n respectively. Hence, the input consists of a linear combination of 1 and (−1)n , which means that the solution equals u[n] = A + B(−1)n , where A and B are complex constants.

19.24

a The frequency response can be written as H(eiω ) =P 1+2 cos ω+cos 2ω = −inω it 1 + eiω + e−iω + e2iω /2 + e−2iω /2. Since H(eiω ) = ∞ n=−∞ h[n]e follows that h[n] = δ[n] + δ[n + 1] + δ[n − 1] + δ[n + 2]/2 + δ[n − 2]/2. b If we write U (eiω ) = 1+sin ω+sin 2ω as complex exponentials, as in part a, then it follows that u[0] = 1, u[1] = u[2] = i/2, u[−1] = u[−2] = −i/2 and so u[n] = δ[n] P + i(δ[n − 1]2+ δ[n − 2] − δ[n + 1] − δ[n + 2])/2. We have to calculate E = ∞ n=−∞ | y[n] | . In this case it is easiest to do this directly (and so not using Parseval). From the expression for u[n] it follows (by linearity and time-invariance) that y[n] = h[n] + i(h[n − 1] + h[n − 2] − h[n + 1] − h[n + 2])/2. Substituting h[n] from part a we get y[n] as a linear combination of δ[n − 4], δ[n − 3], . . . , δ[n + 3], δ[n + 4]. The coefficients in this combination are not hard to determine. In fact, they are −i/4, −3i/4, (1/2) − i, 1 − 3i/4, 1, 1 + 3i/4, (1/2) + i, 3i/4, i/4. Their contribution to E is 1/16, 9/16, 20/16, 25/16, 1, 25/16, 20/16, 9/16, 1/16. The sum of these contributions is 126/16, hence, E = 7 87 .