Contemporary Communication Systems 1st Edition Mesiya Solutions Manual Download:http://testbanklive.com/download/contemporary-communication-systems1st-edition-mesiya-solutions-manual/ Chapter 2 2.1 Consider the signals di splayed in Fi gure P2.1.Show that each of these signals can be expressed as the sum of r ectangular Π (t ) and/or triangular Λ (t ) pulses. Figure P2.1
Solution: ⎛t⎞⎛t⎞ a. x (t ) 1= Π ⎜ ⎟ + Π ⎜ ⎟ ⎝ ⎠2 ⎝4⎠ ⎛ t −3 ⎞ ⎛ t −3 ⎞ −Λ b. x (t2 )⎜ = ⎟2⎜ Λ ⎟ 6 2⎝⎠⎝⎠ 11 ⎞ 7⎞ 3⎞ 1⎞ 5⎞ 9 ⎛ ⎛ ⎛ ⎛ ⎛ ⎛ c. +Π t+ +Π t+ +Π t− +Π t− +Π t− ⎞ x (t ) = ... + Π t + + ...3 ⎜ ⎟ ⎜ 2 2⎟⎜ 2⎟⎜ 2⎟⎜ 2⎟⎜2⎟ ⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠ ∞
= ∑ Π ⎡⎣ t − ( 2 n + 0.5 ) n =−∞
1
⎛ t +4 ⎞ ⎛t⎞ ⎛ t −4 ⎞ + ... ⎟ ⎜+2Λ⎟ ⎜ 2 ⎟ + Λ ⎝⎠⎝⎠⎝⎠ ∞ ⎛ t −4 n ⎞ =∑ Λ⎜⎟ n =−∞ ⎝ 2 ⎠ 2.2 For the signal x2(t ) in Figure P2.1 (b) plot the following signals d. x (t4 )⎜ = ... + Λ 2
a. x2 (t − 3) b. x2 ( −t ) c. x (2t2 ) d. x2 (3 − 2t ) Solution:
x2(t − 3)
x2 ( − t )
1
1
0 1234
56789
−6 − 5 − 4 − 3 −
0t 123
2 −1
2
x (3 − 2t )
x2(2t )
1
1 0t 123456
2.3 Plot the following signals a. x1 (t ) = 2Π (t / 2) cos ( 6π t ) 11 b. x2(t ) = 2⎡ +⎤ s gn ( ⎢2 2 ⎣⎦ t) ⎥
− 2−1
0t 1234
2
c. x3 (t ) = x2 ( −t + 2) d. x (t4 ) = s i n c ( 2t ) Π ( t / 2 ) Solution:
2
x (t )
2
1. 5
x1 (t)
() 1t x
"
2
1 0. 5
0t
0
123456
-0. 5
-1 1 -1. 5 0. 8
-2
-1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0. 4 0. 6 0. 8 1 t
0. 6
t
x3 (t )
"
0. 4
x 4( t)
)()t)(t(t xx4x 4
4
0. 2
2 0
-0. 2
1234
0t
-0. 4
-1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0. 4 0. 6 0. 8 1 t
t
2.4 Determine whether the following signals are periodic. For periodic signals, determine the funda me ntal period. a. x1 (t ) = si n(π t ) + 5 cos( 4π t / 5) Solution: sin(π t ) is periodic with period 1T = 5)
2π
= 2 . cos( 4π t / is periodic with period
π
T 2π 5 1 = . x1 (t ) is peri odic if the ratio can be written as ratio of T2 = 4π / 5 2 T2 integers. In the present case, T1 2 × 2 4 == 55 T2
3
Therefore, x (t ) is periodic with fundame ntal period T such that 1o
T = 5T = 4T = 10 o12
b. x (t2 ) = e j 3 t + e j 9 t + cos (12t ) Solution: e j 3 t is periodic with period T T= 2
9
2π
1
=
2π j 9 t . Similarly, e is periodic with period 3
. cos(12t ) is periodic with period T =
3
To = 12L 3 C M ( T , T , T ) =
=
12 6
2π
funda me ntal period
2π
3
π
. x (t ) is periodic with
2
.
c. x3 (t ) = s i n( 2π t ) + cos(10t ) Solution: sin( 2π t ) is periodic with period T =1
T=
2π
=
π
2π = 1 . cos(10t is periodic with period 2π )
. x (t ) is peri odic if the ratio
2
10 5 3 In the present case,
T1 can be written as ratio of integers. T2
T1 1 × 5 5 == T2 π π Since π is an irrational number, the ratio is not rational. Therefore, x3(t ) is not periodic.
π⎞ ⎛ d. x (t ) = cos 2π t − + s i n(5π t ) 4⎜ 4⎟ ⎝⎠ Solution: 2π π⎞ ⎛ cos 2π t − is periodic with period = = 1 . sin(5π t ) is periodic with T 1 ⎜ ⎟ 2π 4 ⎝⎠ period T 2π 2
=
5Ď€
=
2 T . x (t ) 41 can be written as ratio T of 5 is 2 peri odic if the ratio
integers. In the present case,
4
T1 1 × 5 5 == 22 T2 Therefore, x (t ) is periodic with funda me ntal period T such that 4o
T = 2T = 5T = 2 o12
2.5 Classify the following signals as odd or even or neither. a. x (t ) = − 4t Solution: x ( − t ) = 4t = − ( −4t ) = − x x (t ) is odd. (t ) . So
b. x (t ) = e − t Solution: e −t = e −−t. So
x (t ) is even.
c. x (t ) = 5 cos(3t ) Solution: Since cos(t ) is even, 5 cos(3t ) is also even.
π d. x (t ) = s i n(3t − ) 2 Solution: x (t ) = si n(3t − 2
π
) = − cos(3t ) which is even.
e. x (t ) = u (t ) Solution: u (t ) is neither even nor odd; For example, u (1) = 1 but u ( −1) = 0 ≠ − u (1) .
5
f. x (t ) = si n( 2t ) + cos( 2t ) Solution: x (t ) is neither even nor odd; For example, x (π / 8) =
2 but x ( −π / 8) = 0 ≠ − x (π / 8) .
2.6 Determine whether the following signals are energy or power, or neither and calculate the corresponding ener gy or power in the signal: a. x1 (t ) = u (t ) Solution: The normalized average power of a signal x1 (t ) is defined as P = li m
1
x1 T →∞
T
T /2
∫ u1 (t )
T/2
2
− T /2 0
1T1 d t = li m ∫ 1 d t = li m = T →∞
T →∞
T
T22
Therefore, x1 (t ) is a power signal. b. x (t2 ) = 4 co s ( 2π t ) + 3 co s ( 4π t ) 2
Solution: 4 cos ( 2π is a power signal. 3 co s ( 4π t ) is also a power signal. t) Since
x (t ) is
sum of two power signals, it is a power signal. T /2
1 T →∞ T
Px = 2li m
∫ − T /2
= li m T →∞
Now li m
1
T /2
∫
1 T
T /2
4 cos ( 2π t ) + 3 cos ( 4π t )
2
T /2
1 T →∞ T
dt = lim
∫
4 cos( 2π t ) + 3 cos ( 4π t ) dt 2
− T /2
⎡16 cos 2 ( 2π t ) + 9 cos 2 ( 4π t ) + 24 cos( 2π t ) cos ( 4π t ) ⎤d t
∫ ⎣⎦
− T /2
16 ( 2π t )dt = li m 8 co2 s
T /2
∫
8 T
[+1
T →∞
co s( 4 π t) ]=dt li m = 8
T →∞
T − T /2
1 li m T →∞ T
T
T /2
∫ 9 cos
2
−T /2
T→∞
T
( 4π t )dt = 4.5
− T /2
6
24 li m
T →∞
T
T /2
∫ cos( 2π t ) cos ( 4π t )dtlim=
− T /2
T →∞
24 T
T /2
∫ ⎡⎣ co s ( 6π t ) + co s (⎦ 2π t ) ⎤dt = 0
− T /2
Therefore, Px2= 8 + 4.5 = 12.5 c. x (t3 ) = t
1
Solution: ⎛ 1 T /2 ⎞ 2 ⎛−⎞ E = li m 1 / t dt = li m = li m = 0 t −2 dt = li m − ⎝⎠ x3 ⎜ ⎟ ⎟ T→∞ ∫ T →∞ ∫ T →∞ ⎜ T →∞ T/2 ⎝ −tT /2 ⎠ − T /2 − T /2 T /2
T /2
2
P = li m x3
T →∞
1
1 / t dt = li m 2
T −∫ T /2
T /2
T/2
T /2
T →∞
T −∫T /
1⎛1 t −2 dt = li m − T→∞
T ⎜⎝
⎞12 ⎛−⎞ = li m = 0 ⎝⎠ ⎟ T→∞T ⎜ T / 2 ⎟ −tT /2 ⎠
2
x3(t ) is neither an energy nor a power signal. d. x (t4 ) = e − α t u (t ) Solution: T /2
E = li m
x
T→∞
4
⎜⎟
−αt
e m
T /2
2
u (t ) dt = li
∫
T →∞
∫ − T /2 0
Thus x4(t ) is an energy signal. e. x5 (t ) = Π (t / 3) + Π (t ) Solution:
⎛ e −2α t T /2 ⎞ 1 = li m e −2α t dt = li m − ⎜ T →∞
2α
⎝ 0⎠
⎟
2α T → ∞
− e −αT +1
( ) 2α
=
1
x (t ) 5
2
1 − 3/ 2
3/ 2 t
7
−1/ 2 1/ 2
T /2
Π (t / 3) + Π (t ) dt T →∞ ∫ = − T /2
E x = 5li m
2
−1/ 2
∫ 1d t ∫4+ d t ∫ 1d t +
−1/ 2
− 3 /2
−3 / 2
=1+4+1=6 Thus x5(t ) is an energy signal. f. x (t6) = 5e ( −2 t + j 10 π t ) u (t ) Solution: T /2
E x = 6li m
∫
T →∞
T /2
2
5e ( −2 t + j 10 π t ) u (t ) dt = li m
∫
T/2
4t 5e −2 tej 10π )t ∫ dt = li m 2 5 e − dt 2
T →∞
T→∞
− T /2 0 0
⎛ e −4 tT⎞/2 25 25 ⎟= = 25 li m ⎜ li m ( − e −2 T + 1) = − T →∞ 4 T →∞ ⎜ 40 ⎠⎟ 4 ⎝ Thus x6(t ) is an energy signal. ∞
g. x (t7 )∑= ⎣Λ ⎡ ( t − 4 n ) / 2 n =−∞
Solution: x7(to ) is a periodic signal with period T = 4 . 1 P=⎡Λ x7
T
To / 2 1 1
22 t / 2 ⎤ dt =
∫ ⎣( ) ⎦
T∫
o − To / 2
1⎛ = ⎜ t − t 2+ 232⎜ ⎝ ⎠ 0⎝ ⎠
1− t
()
o00
t 3⎞
2
dt =
1
1 − 2t + t 2 dt
2∫
()
1
111
⎛ ⎞ ⎟= 1−1+ = 3⎟6
2.7 Evaluate the following expressions by us ing the properties of the delta function: a. x1 (t ) = δ (4t ) s i n(2t )
Solution: 1 δ (4t ) = δ (t ) 4 11 x1 (t ) = δ (t ) s i n ( 2t ) = δ (t ) s i n ( 0) = 0 44
8
π⎞ ⎛ b. x (t2 )⎜ = ⎟δ (t ) cos 30π t + 4 ⎝⎠ Solution:
π⎞ π⎞ ⎛ ⎛ ⎛π⎞ x2(t⎜ ) = δ (t ) cos 30π t + = δ (t ) cos ⎟0⎜ + ⎟ = cos δ (t ) ⎟ ⎜ 4 4 4 ⎝⎠⎝⎠⎝⎠ c. x3 (t ) = δ (t )s i nc (t + 1) Solution: x3 (t ) = δ (t )s i nc (t + 1) = δ ( t )s i nc ( 0 + 1) = δ ( t )s i nc (1) = δ ( t ) × 0 = 0 d. x (t4 ) = δ (t − 2) e − t s i n ( 2.5π t ) Solution: x4(t ) = δ (t − 2 ) e − t s i n ( 2 . 5π t ) = δ (t − 2) e −2 si n ( 2.5π × 2) = δ (t − 2) e −2 sin (π ) = 0 ∞
e. x5 (t ) =
∫ δ ( 2t )s i nc (t ) d t −∞
Solution:
5
1
δ ( 2t )si n c (t )dt
x (t )
= ∫∞ ∞
=
−∞ −∞
1 δ (t )sinc ( 0) d t =
2∫
∞∞
−∞ −∞
f. x6 (t ) =
δ (t )si n c (t ) d t
2∫
11 δ (t ) d t = ∫ = 2 2
∞
∫ δ (t − 3) cos (t ) d t −∞
Solution: x6 (t ) =
∞∞
∫ δ (t − 3) cos (t ) dt = cos (3) ∫ −∞ − ∞
δ (t − 3) dt = cos (3)
9
∞
g. x (t ) 7
δ 2(
=∫
−
t)
−∞
1
dt
1−t3
Solution: ∞
∞
1 1 x7(t ) = δ ( 2 − t ) d t = ∫ δ (t − 2 ) 3d t 3 ∫ 1t 1 t −∞ − −∞ −
δ (t 2) =∞ ∞ ∫− h. x8 (t ) =
1
1
dt
=−
∫
1−2 3
δ (t 2) d t −=−
7
1 7
−∞ −∞ ∞
∫ δ (3t − 4) e
−3 t
dt
−∞
Solution: ∞∞
1 ⎛ 4 ⎞ 3t x8(t ) = ∫ δ (3t − 4 ) e −3 t dt = ∫ δ ⎜ t − ⎟ e − dt 3 ⎝ 3 −∞ ⎠ −∞ 1
∞
∞
e −4 e −4 ⎛ 4 ⎞ −3 ×3 ⎛ 4⎞ e dt = =δt− δ t − dt = 4
∫ ⎜⎟ 3 −∞ ⎝ 3−∞ ⎠
∫ ⎜⎟ 3⎝3⎠3
i. x9 (t ) = δ '(t ) ⊗ Π (t ) Solution: ∞
τdτ)δ '(τ ) x (t ) (t = Π − = − Π− 9 ∫ −∞ d
( 1)
d
τ
(t τ ) τ=0
= Π ' (t ) = δ ( t + 0.5 ) − δ ( t − 0.5 ) 2.8 For each of the following continuous-time systems, determine whether or not the system is (1) linear, (2) time-invariant, (3) me moryless, and (4) casual. a. y (t ) = x (t − 1) Solution:
The system is linear, time-invariant, causal, and has me mory. The system has me mory because current value of the output depends onthepreviousvalueof theinput. The system is causal because current value of the output does not depend on future inputs. To prove linearity, let x (t ) = ι x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is
10
y (t ) = x (t − 1) = α x1 (t − 1) + β x2 ( t − 1) = α y1 (t ) + β y 2 (t ) where x1 (t1 2)2 ⎯T⎯ → y (t ) and x (t ) ⎯T⎯→ y (t ) . To show that the syst em is time-invari ant, let input. The corresponding output is
x1 (t ) = x (t − t o ) be the system
y1 (t ) = x1 (t − 1) = x (t − o ) = y (t − t o ) 1−t b. y (t ) = 3 x (t ) − 2 Solution: The system is nonlinear, time-invariant, causal, and me moryless. The system is me moryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is y (t ) = 3 x (t ) − 2 = 3 [α x1 (t ) + β x2 (t ) ] − 2
≠ α y1 (t ) + β y 2 (t ) = α [ 3 x1 (t ) − 2 ] + β [ 3 x2 ( t ) − 2 ]
To show that the syst em is time-invari ant, let input. The corresponding output is
x1 (t ) = x (t − t o ) be the system
y1 (t ) = 3 x1 (t ) − 2 = 3 x o ) − 2 = y (t − t o ) (t − t c. y (t ) = x (t )
Solution: The system is nonlinear, time-invariant, causal, and me moryless. The system is me moryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is
11
y (t ) = α x1 (t ) + β x2 (t ) ≤ α x1 (t ) + β x2 (t ) Because α x (t ) + β x (t ) = α
x (t ) + β x (t ) ≠ α y (t ) + β y (t ) fo r all α an d β ,
121212
the system is nonlinear. To show that the syst em is time-invari ant, let input. The corresponding output is
x1 (t ) = x (t − t o ) be the system
y (t ) = x (t ) = x (t − ) = y (t − t ) t 11oo
d. y (t ) = [ co s ( 2t ) ] x (t ) Solution: The system is linear, time-variant, causal, and memoryless. The system is me moryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is y (t ) = [ cos ( 2t ) ] x (t ) = [ cos ( 2t ) ] [α x1 (t ) + β x2 (t ) ]
= α [ cos ( 2t ) ] x1 (t ) + β [ cos ( 2t ) ] x2 ( t ) = α y1 ( t ) + β y 2 (t )
To prove time-variance, let x1 (t ) = x (t − t o ) be the system input. The corresponding output is y1 (t1 o)o = [ cos ( 2t ) ] x (t ) = [ cos ( 2t ) ] x ( t − t ) ≠ y ( 2 ( t − t x ( t − t o ) t − t ) = cos o)
e. y (t ) = e x ( t ) Solution: The system is nonlinear, time-invariant, causal, and me moryless. The system is me moryless because current value of the output de pends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs.
12
To prove nonlinearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is y (t ) = eα x1 ( t ) + β x2 ( t ) = eα x1 ( t ) e β x2 ( t )1≠2 α y (t ) + β y (t ) = α e x1 ( t ) + β e x2 ( t ) for all α and β To show that the syst em is time-invari ant, let input. The correspondi ng output is
x1 (t ) = x (t − t o ) be the system
y1 (t ) = e x1 ( t ) = e x ( t − to ) = y (to − t ) f. y (t ) = t x (t ) Solution: The system is linear, time-variant, causal, and me moryless. The system is me moryless because current value of the output de pends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is y (t ) = t [α x1 (t ) + β x2 ( t ) ] = α t x1 ( t ) + β t x2 (t ) = α y1 (t ) + β y 2 ( t ) To prove time-variance, let x1 (t ) = x (t − t o ) be the system input. The corresponding output is y1 (t1 o)o = o o t x (t ) = t x ( t − t ) ≠ y (t − t ) = ( t − t ) x ( t − t ) t
g. y (t ) = ∫ x ( 2τ ) dτ −∞
Solution: The system is linear, time-invariant, causal, and has me mory. The system has me mory because current value of the output depends only on the past values of the input. The system is causal because current value of the output does not depend on future inputs. To prove linearity, let x (t ) = α x1 ( t ) + β x2 (t ) . The response of the system to x (t ) is
13
ttt
y (t ) = ∫ [α x1 ( 2τ ) + β x2 ( 2τ ) ] dτ = α ∫ x1 ( 2τ ) dτ + β ∫ x2 ( 2τ ) dτ = α y1 (t ) + β y 2 (t ) −∞ −∞ −∞
To check for time-variance, let x1 (t ) = x (t − t o ) be the system input. The corresponding output is tt
y1(t∫) 1=∫x (o 2τ ) dτ = x [ 2(τ − t ) ] dτ =
t−t o
x 2α d α = y ( t − t ) ∫ ( ) o
−∞ −∞ −∞
2.9 Calculate the output y (t ) of the LTI system for the followi ng cases: a. x (t ) = e −2 t u (t ) and h (t ) = u (t − 2) − u (t − 4) Solution: h (t ) = u (t − 2) − u (t − 4) =⎜ Π⎟ ⎝⎠2 y (t )
⎛ t −3 ⎞
∞∞
=−=
τ−3
e 2( t −τ ) u (t τ )
h (τ ) x τ ) (t dτ
−Π
−
dτ
⎛⎞
∫ ∫ ⎜2⎟
⎝⎠
−∞ −∞
For t < 2 , there is no overlap and y (t ) = 0. tt−2
For 2 ≤ t < 4 y (t ) = ,
∫
−2τ
t−
e 2 − −2τ e −2( t −τ ) dτ e dτ = = −2 2
1 e −2 ( t − 2 )
=∫
200
For t ≥ 4 , 44
e 2τ
y (t−2) = e −2( t −τ ) dτ = e −2 t ∫ e 2τ dτ =e t ∫ 222
4
e4 8 − e = e −2 t = e −2 ( t − 2 )
222
(e
4
− 1)
⎧1 − e −2( t − 2) ,2≤t≤4 ⎪ ⎪2 4 ⎪ ( e − 1) y (t ) = ⎨ e −2( t − 2) , t>4 ⎪2 ⎪ 0, otherwise ⎪ ⎪⎩ b. x (t ) = e − t u (t ) and h ( t ) = e −2 t u (t )
14
Solution: ∞t
y (t ) = ∫ e −τ u (τ ) e −2( t −τ ) u (t − τ ) dτ = ∫ e −τ e − 2 ( t −τ ) dτ −∞ 0
For t < 0 , there is no overlap and y (t ) = 0. t −tτ
y (t ) = e
2
−tτ
∫ e dτ =e
2
0
t
−tt
(
e =e e 0 2
−t−t
2 − e 1) = − e , t ≥ 0
c. x (t ) = u ( − t ) an d h (t ) = δ ( t ) − 3e −2 t u (t ) Solution: y (t ) =
∞∞
h (τ )dxτ (t=− τ ) ∫ ∫ ⎣⎦
⎡δ (τ ) − 3e −2τ u (τ ) ⎤ u (τ − t ) dτ
−∞ −∞
Now ∞
∫ δ (τ )u (τ − t ) dτ = u ( −t ) −∞
∞
∞
−2τ 3e −2τ 3 3e dτ = = For t < 0 , −2 2 ∫ 00
∞∞
∞
−2τ 3e −2τ 3 −2τ For t ≥ 0 , ∫ 3e u (τ )u (τ − t ) 3e dτ = = −2 2 dτ = ∫ −∞
⎧3
5 +1= , ⎪22 y (t )⎨ = ⎪3e −2 t ,
e −2 t
tt
t<0 t≥0
2 d. x (t ) = δ ( t − 2) + 3e 3 t u ( −t ) and h (t ) = u (t ) − u (t − 1) Solution:
15
∞∞
y (t ) =
h (τ )dxτ (t=− τ ) ∫ ∫ ⎣⎦
⎡δ (t − τ − 2 ) + 3e 3( t −τ ) u ( − t + τ ) ⎤ [ u (τ ) − u (τ − 1) ] dτ
−∞ −∞ ∞∞
= ∫ δ (t − τ − 2) [ u (τ ) − u (τ − 1) ] dτ + ∫ 3e 3( t −τ ) u ( − t + τ ) [ u (τ ) − u (τ − 1) ] dτ −∞ −∞
Now ∞
∫ δ (t − τ − 2) [ u (τ ) − u (τ − 1) ] dτ = u (t − 2) − u ( t − 2 − 1) = Π ( t − 2.5 )
−∞
For t < 0 , ∞11
∫ 3e
3( t −τ )
−∞ 0 0
u ( − t + τ ) [ u (τ ) − u (τ − 1) ] dτ = ∫ 3e 3( t −τ ) dτ = 3e 3 t ∫ e − 3τ dτ = 3e 3t
For 0 < t ≤ 1 , ∞11
∫ 3e
3( t −τ )
−∞ t t
3t
1⎞ ⎛ e 3t 1 −
1
⎛ 1⎞ = e 3⎜t 1 ⎟ e3 ⎠ −3 0 − ⎝
u ( − t + τ ) [ u (τ ) − u (τ − 1) ] dτ = ∫ 3e 3( t −τ ) dτ = 3e 3 t ∫ e − 3τ dτ = 3e
⎧
e −3τ
e −3τ
1
1⎞ ⎛ = e 3 ⎜t e −3 t ⎟ e3 ⎠ −3 t − ⎝
t≤0 , ⎪ ⎜ e 3⎟ ⎪⎝⎠ ⎪ 1⎞ ⎛ ,0<t≤1 y (t ) =⎨ ⎜ e 33 ⎟t e −3 t − ⎪ ⎝ ⎠e ⎪1, 2 ≤ t ≤ 3 ⎪ ⎩ 0, ot herwi s e 2.10 The impulse response function of a continuous-time LTI is displayed in Fi gure P2.2(b). Assumi ng the input x (t ) to the system is waveform illustrated in Figure P2.2(a), determine the syst em output wavefor m y (t ) and sketch it. Solution:
For t < 1 , there is no overlap and
y (t ) = 0.
16
Figure P2.2 x (τ )
(a)
1 0123
τ
4567
(b)
h (t − τ ) 1
1<t<2
τ
01234
h (t − τ )
(c)
5<t<6
1
τ
01234 t -1
As shown in Figure (b), y (t ) = ∫ dτ = t for 1 ≤ t < 2 −1
1
0
For 2 ≤ t < 3 y (t ) = ∫ dτ , =1 0 1
For 3 ≤ t < 4 y (t ) , =
∫ dτ =1 − t + 3 = 4 − t t−3 t
Referring to Figure (c), y (t ) = ∫ dτ t5− = t 5
For 7 ≤ t < 8 y (t ) = ,
∫ dτ = t − t + 2 = 2 t−2 7
For 8 ≤ t < 1 0 ,
y (t ) = y
(
t
) 2
for 5 ≤ t < 7
∫ dτ = 7 − t + 3 = 10 − t
t−3
1
1234
t
5 6 7 8 9 10
17
−0.5 ( t − 2 ) u (t − 2) . 2.11 An LTI system has the impulse response h (t ) = e
a. Is the system casual? Solution: Yes. The system is casual because h (t ) = 0 for t < 0 . b. Is the system stable? Solution: ∞ ∞ −0.5 x ∞
The system is stable because
∫ h (t ) dt = e ∫
−0.5 x
e dx = = 2 −0.5
−∞ 0 0
c. Repeat parts (a) and (b) for
h (t ) = e −0.5 ( t + 2 ) u (t + 2) .
Solution: The system is not causal but stable. 2.12 Wr ite down the exponential Fourier series coe fficients of the signal x (t ) = 5 s in( 40π t ) + 7 cos ( 80π t − π / 2 ) − cos (160π t + π / 4 ) Solution: Applying the Euler’s formula, we get the following terms: ⎡ e j 40 π t − e − j 40 π t ⎤ ⎡ e j 80 π t e − jπ / 2 + e − j 80 π t e jπ / 2 ⎤ ⎡ e j 160 π t e jπ / 4 + e − j 160 π t e − jπ / 4 ⎤ x (t ) = 5 ⎢ ⎥ + 7 ⎢ ⎥ − ⎢ ⎥ 2⎦⎣2⎦ ⎣2j⎦⎣ = − j 2.5e j 40 π t + j 2.5e − j 40 π t − j 3.5e j 80 π t + j 3.5e − j 80π t − 0.5e jπ / 4 e j 160 π t − 0.5e − jπ / 4 e − j 160 π t The Fourier coefficients are C1 = − j 2.5, C −1 j 2.5 = C2 = − j 3.5, C −2 j 3.5 =
− jπ /4 C3 = 0.5e jπ / 4 , C−3 = −0.5e −
a. Is x (t ) period ic? If so, what is its peri od? Solution:
18
Yes. ω = 40π ⇒ f = 20, T ooo 20 20
1 1 = . The period is = 0.05 sec .
2.13 A signal has the two-sided spectrum representation shown in Figure P2.3. Figure P2.3
a. Wr ite the equation for x(t ). Solution:
π⎞ π⎞ ⎛ ⎛ x (t ) = 14 cos + 10 + 8 cos 300π t − ⎜ 3 ⎟100 ⎜ 2 π⎟ t − ⎝⎠⎝⎠ b. Is the signal periodic? If so, what is its period? Solution: 1 Yes. It is periodic with fundame ntal period T = = 0.02 . o 50 c. Does the signal have energy at DC? Solution: Yes as indicated by the presence of DC term C0 = 10 . 2.14 Wr ite down the complex exponential Fourier seri es for each of the periodic signals shown in Figure P2.4. Use odd or even symme t ry whenever possible. Solution: a. T = 3, f = 1 / 3 oo
19
1 ⎡ 23⎤ 1 C0= ⎢2d∫t − 1d⎥ t = ( 4 − 1) = 1 3 ⎣ 0∫ 3 2⎦ 1⎡
2
2e C= n⎢ ∫ ∫ ⎥ 3 ⎣ 02⎦
2 πnt −j 3
j3⎡
= ⎢ 2e 3 3 × 2π n ⎢ ⎣
3 2 πnt −j 3
dt − e
−j
2 πnt 2
dt
−j
⎤
2 πnt 3
⎤
− e 3⎥
⎥⎦
02
j = ⎡ 2e −⎣ j⎦4 π n / 3 − 2 − e − j 2 π n + e − j 4 π n / 3 ⎤ 2π n j 3 3e − j 2 π n / 3 ⎛ e + j 2 π n / 3 − e − j 2 π n / 3 ⎞ = ( e − j 4 π n / 3 − 1) = ⎜ ⎟ πn⎝2j⎠ 2π n 3e −j 2 πn / 3
⎛ 2πn ⎞
= sin , πn⎝3⎠ ⎜
⎟
n = ± 1, ±2, .....
b. T = 2, f = 1 / 2 oo
1 ⎡ 1 ⎤ 1 t 21 C0= t dt⎥ = ⎢∫ 222
1
=
4
(1 − 1) = 0
⎣ −1 ⎦ −1 1 ⎡ ⎤ 1 −⎡ jπ⎤ nt d t = − C t d⎢(⎥ e − jπ n t ) ⎢ ⎥∫ t e n = ∫ 2 j 2π n ⎣ −1 ⎦ ⎣ −1 ⎦ 11
⎤ j ⎡1 e − jπ n t d t = te − jπ n t − ⎢ −1 ∫ ⎥ 2π n ⎣ −1⎦ j ⎡ 1 1⎤ = e − jπ ⎢n + e + jπ n + e − jπ n t −1 ⎥ 2π n jπ n ⎣⎦ j ⎡ e − jπ n e + jπ n e − jπ n e jπ n ⎤ = ⎢ + + 2 πn π n − jπ 2 n 2 jπ 2 n 2 ⎥ 1
⎣⎦ je − jπ n j ( − 1) , n = ± 1, ± 2, ..... == πn πn n
c. T = 2, f = 1 / 2 oo
2A⎡
1
⎤ ⎛ t 2⎞
1
⎛1⎞A
C0= ⎢ ∫ (1 − t )dt ⎥ = A ⎜ t − = A ⎜ 1 − ⎟ = 2 2 2 2 ⎟ ⎣ 0⎦ ⎝ ⎠ 0⎝ ⎠
20
Since x3(t ) is an even function of time, ⎤ 11 A⎡2A1 Cn= n = ⎢ (1 − t ) cos(π nt )dt ⎥ = A ∫ cos(π nt )dt − A ∫ t cos(π nt )dt 22 ∫ ⎣ 0⎦ 00 A A1 1⎡ 1 ⎤ = 0 − ∫ td ( si n(π nt ) )dt = ⎢ − t si n(π nt ) 0 + ∫ sin(π nt )dt ⎥ π n 0π⎣ 0n⎦ 1 A ⎡ co s (π nt ) ⎤ A A =⎢0− ⎥ = − 2 2 [ cos(π n ) − 1] = 2 2 [1 − cos(π n ) ] π n π n π n ⎢⎣ π n 0⎥ ⎦ ⎧ 0, ⎪ = ⎨ 2A
n even
, π 2n 2
n odd
d. T = 3, f = 1 / 3 oo
2A ⎡1 3 / 2 2 t dt + C= 0⎢ ∫ ∫ ⎥ 3 ⎣ 01⎦
1
⎤ 2A ⎡ t 3/ 2 1 ⎡ 1 1 ⎤ 2A 1dt = + t ⎤= + = ⎢ ⎥ 1 3 ⎢⎣20 ⎥3⎦ 2 2 3
Since x4(t ) is an even function of time, ⎤ 2 A ⎡1 3 / 2 A⎡2 A 1.5 C= = x (t ) cos( 2π nt / 3)dt t cos( 2π nt / 3)dt = + ⎢ ⎥ ⎢∫ ∫ ⎥ n 4 2 3∫ 3 ⎣ 0⎦ ⎣ 01⎦ n
cos( 2π nt / 3)dt
⎤
Now 3 ⎡ ⎤ 3 ⎡ 1⎤ t cos( 2 π nt / 3)dt = t d ( sin( 2π nt / 3) ) ⎥ = ⎢ ⎢ t si n( 2π nt / 3) 0 − ∫ ∫ 2π n ∫ 2π n sin( 2π nt / 3)dt ⎥ 0⎣ 0⎦ ⎣ 0⎦ ⎡ 1⎤ 3 3 = sin( 2π n / 3) + cos( 2π nt / 3) 2π n 2π n 0 111
⎡⎤ 3 3 = sin( 2π n / 3) + [ cos( 2π n / 3) − 1] 2π n 2π n
3/ 2
2π nt / 3)dt = sin( 2π nt / 3) = − si n( 2π n / 3) ∫2πcos( n 2π n 3
3/ 2
3
1
1
Substituting yields
21
2A ⎛ 3 ⎞ ⎡ ⎛ 2πn ⎞ ⎤ 3 A ⎡ ⎛ 2πn ⎞ ⎤ C=⎜⎟ ⎢ cos ⎜ ⎟ − 1 = ⎦⎥ ⎢ cos ⎜ ⎟ −⎥ 1 ⎦ n 2 2 3 2π n ⎣ 3 ⎝ 2π n ⎠ ⎣ ⎝ ⎠ ⎝3⎠ 2
∞
e. x 5(t ) =
∑ p( t−8n)
n =−∞
where p ( t ) = Π ⎡⎣ ( t − 2 + Π ⎣⎡ ( t − 2 ) − Π ⎡⎣ ( t − 6 ) − Π ⎡⎣ ( t − 6 ) / 4
) /2
/4
/2
The FT of the pulse shape p ( t ) over [ 0,
is given by
To ] To
P ( f ) = ∫ p (t ) e − j 2 π ft d t 0
The FS coefficients of a periodic signal with basic pulse shape p ( t ) are gi ven by T
1 o C n = ∫ p (t ) e − j 2 π nf o t d t T o0 Comparing yields 1 Cn = P ( f ) f=nfo To Now ℑ Π ⎡⎣ ( t − 2 ) / ←⎯ → 2s i nc ( 2 ℑ ←⎯ → 4s i nc ( Π ⎡⎣ ( t − 2 ) / ℑ ←⎯ → 2s i nc ( 4Π ⎤⎦ ⎡⎣ ( t − 6 ℑ ←⎯ → 4s i nc ( / 2 ⎤ Π ⎡ t − ) ⎦ ⎣(
6 ) / 4 ⎤⎦
2 f ) e −j4πf 4 f ) e −j4πf 2 f ) e − j 12 π f 4 f ) e − j 12 π f
Therefore, P ( f ) = 2si nc ( 2 f ) e − j⎣4⎦π ⎣f ⎡⎦1 − e − j 8π f ⎤ + 4si nc ( 4 f ) e − j 4 π f ⎡1 − e − j 8π f ⎤ The FS coefficients of a periodic signal x5(t ) are now obtained as
22
{
1 Cn = 2si nc ( 2 nf o To
) e − j 4 π nf⎣ ⎦⎡1 − e − j 8π nf ( 4 nf o
o
}
⎤ + 4si nco ) e − j 4 π nf⎣ o⎦⎡1 − e − j 8 π nf o ⎤
{
}
1 = sinc n / 4 ) e − jπ n /2 ⎡1 − e − jπ n ⎤ + 2sinc ( n / 2 ) e − jπ n /2 ⎡1 − e − jπ n ⎤ ( ⎣ ⎦ ⎣ ⎦ 4
2.15 For the rectangular pulse train in Figure 2.23, compute the Fourier coefficients of the new periodic signal y( t) given by a. y (t ) = x (t − 0.5To ) Solution: ⎡ ( t −nTo ) ⎤
∞
The FS expansion of a periodic pulse train x (t )
of
=∑ Π⎢⎥ n =−∞ ⎣ τ ⎦ rectangular pul ses is given by x (t ) =
∞
∑C
x n
e j 2 π nf o t
n =−∞
where the exponential FS coefficients are τ C x = sinc ( nf τ ) no To Let the FS expansion of a periodic pulse train y (t ) = x (t − 0.5To ) be expressed as y (t ) =
∞
∑C
y n
e j 2 π nf o t
n =−∞
where
C y= n
1 T∫
y (t1) e − j 2 π n f o t dt x (t − 0.5T )e − j 2 π n f o t dt = o T∫
o To
o To
1 1 = ∫j 2=πxnf(ν(ν +)e0 .−5To o) d−νj 2eπ n f o ( 0 .5 To ) T oTTo
∫ x (ν )e oT
o
C nx
= e − jπ n C nx
dν
j 2− π onf ν
Time Shifting introduces a linear phase shift in the FS coeffi cients; their magnitudes are not changed.
23
b. y (t ) = x (t ) e j 2 π t / To Solution: 1
C y= n
TT
∫ y (t ) e
1
j 2 π−nof t
dt =
o To
∫ x (t ) e
j 2 π tof − oj 2 π nf t
e dt
o To
1 = x (t∫)e − o dtj−12 )π tf ( n To To = C−1x n c. y (t ) = x (α t ) Solution: 1 C = y
n
TT
∫
1 ( ) − πj 2o n f t yte dt =
o To
1
αT∫
j2nft
dt
o To − j2π n
=
x( α ∫ t )e
−πo
x ( )e ν ⎝⎠ν
⎛ f ⎞o ⎜α⎟
ν
d
o α To
=C
x n
Time Scaling does not change FS coefficients but the FS itself has changed as the f 2f 3f harmonic components are now at the frequencies ± o , ± o , ± o , … ααα 2.16 Draw the one-sided power spectrum for the square wave in Figure P2.5 with duty cycle 50%. Figure P2.5
Solution:
The FS expansion for the square wave is given by
24
∞
∑Ce
x (t ) =
n
j 2π nf o t
n =−∞
where Cn =
1 T
∫ x (t ) e
− j 2π n f ot
dt
o To T /2 T ⎫⎪ 1 ⎧⎪ o o − j 2 π nf o t dt − A ∫ e − j 2 π nf o t dt ⎬ = ⎨A ∫ e T o 0 To / 2
A =
e −j2πnft
{
T
To / 2
T ( − j 2π nf )
e −j2πnft o
−o
o
To / 2
}
0
oo
jA − jπ n − j 2 π n − jπ n = e − 1( − e+e ) 2π n Therefore, ⎧ j2 A − , ⎪ Cn ⎨= πn ⎪⎩ 0, ot herwise
n odd
Averag e power in the frequency component at = nf equals C f on
displays the one-sided power spectrum for the square wave.
One-s i ded P ower S pec t rum of S quare wave 0. 9
0. 8
One-s i ded P ower S pec t rum
0. 7 0. 6 0. 5 0. 4 0. 3 0. 2 0. 1 0
0 5 10 15 20 25 F requenc y (x fo )
a. Calculate the normalized average power.
2
. Figure
Solution:
25
The norma lized average power for a periodic signal x (t ) is given by 1 To /2 2 AT 2 P=∫ x (t ) dt = o = A 2 x
T T o − To /2 o
b. Determine the 98% power bandwidth of the pulse train. Solution:
n
1 3 5 7 9 11 13 15 17 19 21
Total Power including current Fourier coefficient 0.8106 0.9006 0.9331 0.9496 0.9596 0.9663 0.9711 0.9747 0.9775 0.9798 0.9816
98% Power bandwidt h = 21 × f o
2.17 Determine the Fourier transforms of the signals shown in Figure P2.6. Figure P2.6
26
Solution (a) The pulse x1 (t ) can be expressed as x1 (t ⎣ ) = A ⎡ Π ( t + 0.5 ) − Π ( t − 0.5 ) Now ℑ Π ( t ) ←⎯ → sinc ( f )
Applying the time-shifting property of the FT, we obtain ℑ Π ( t + 0.5 ) ←⎯ → sinc ( f ) e jπ f ℑ Π ( t − 0.5 ) ←⎯ → sinc ( f ) e − jπ f
Adding − jπ f jπ f
jπ f
X 1( f ) = A ⎣⎡ s i nc ( f )e
− sinc ( f ) e
− jπ f
= Asinc ( f ) ⎡⎣ e − e
= Aj 2sinc ( f ) s in ( π f j 2π f A⎣ ⎦⎡ si nc 2 ( f ) ⎤
)=
Solution (b) The pulse x2(t ) can be expressed as x2(t ) = Π ( 2t ) co s ( 2π t ) Now 1 ℑ Π (2t ) ←⎯ → s i nc ( f / 2) 2 1 ℑ cos( 2π t ) ←⎯ → ⎣ ⎡δ ( f − 1) + δ ( f + 1) 2 1 1 X 2( f ) = ℑ {Π (2t )} ⊗ ℑ {cos (2π t )} = s i nc ( f / 2) ⊗ ⎡δ ( f − 1) + δ ( f + 1) ⎣ 22 1
=
⎡ si nc ⎡⎣ 0.5 ( f − 1) 4⎣
+ si nc ⎡⎣ 0.5 ( f + 1) ⎤⎦ Solution (c) The pulse x3(t ) can be expressed as
27
x3(t ) = p ( t ) + p ( − t ) where p ( t ) = e − t ⎡⎣ u ( t ) − u ( t − 1) From Table 2.2, we have 1 ℑ e − t u (t ) ←⎯ → 1 + j 2π f Using the time-shifting property of the FT, we obtain eℑ− j 2 π f − 1 e − (1+ j 2 π f ) e − t u ( t − 1) = e −1e − ( t −1) u ( t − 1) ←⎯ →e = 1 + j 2π f 1 + j 2π f Combining
1 e − (1+ j 2 π f ) 1 ℑ p (t ) ←⎯ →P( f) =−= 1 + j 2π f 1 + j 2π f
⎡1 − e − (1+ j 2 π f ) ⎤ 1 + j 2π f ⎣ ⎦
By time-reversal pr operty, 1 − (1− j 2 π f ) ℑ ⎡1 − ⎤ p ( −t ) ←⎯ →P( −f ⎣ ⎦e 1 − j 2π )= f Xf=Pf
+P−f =
1
()()() 3
1 + j 2π f ⎣ ⎦
1− e
+
1
1−e
⎡ − (1+ j 2 π f ) ⎤ ⎡
− (1− j 2 π f )
1 − j 2π f ⎣ ⎦
{
}
1 = (1 − j 2π2 f ) ⎡1 − e − (1+⎣j ⎦2 π⎣ f⎦) ⎤ + (1 + j 2π f ) ⎡1 − e − (1− j 2 π f ) ⎤ 1 + ( 2π f ) 2 −1 = cos ( 2π f ) + 2π f e −1 sin ( 2π f ) ⎤ 2 ⎡1 − e 1 + ( 2π f ) ⎣ ⎦ Solution (d) The pulse x4(t ) can be expressed as
⎤
x4(t ) = Π ( t / 2 ) + Π ( t / 4 ) Now
28
ℑ Π ( t / 2 ) ←⎯ → 2s i nc ( 2 f ) ℑ Π ( t / 4 ) ←⎯ → 4s i nc ( 4 f )
Adding X 4( f ) = 2s i n c ( 2 f ) + 4s i n c ( 4 f ) 2.18 Use properties of the Fourier transform to compute the Four ier transform of following signals. a. sinc 2 (Wt ) Solution:
τ ⎛ fτ ⎞ ℑ Λ (t / τ ) ←⎯ → ⎜ s i⎟nc 2 2 2 ⎝⎠ Using the duality property, we obtain ℑ Wsinc 2 (Wt ) ←⎯ → Λ ( f / 2W )
1 ℑ sinc 2 (Wt ) ←⎯ → Λ ( f / 2W ) W Thus the Fourier transform of a sinc 2 pulse is a triangular function in frequency. b. Π (t / T ) cos( 2π f c t ) Solution: ℑ Π ( t / T ) ←⎯ → T s i nc ( f T )
1 ℑ cos( 2π fc t )⎣ ←⎯ → ⎡δ ( f − f ) + δ ( f + f ) cc 2 1 X ( f ) = ℑ Π t / T ⊗ ℑ co s ( 2π f t = T s i f T ⊗ ⎡δ f−f +δf+f⎤ nc ) } ( ) ( { ( )} { c) ⎦ c c) ( 2⎣ =
T 2
{sin c ⎣⎡T⎦ ⎣( ⎦f − f
c. ( e − t cos 10π t ) u (t ) Solution:
c
) ⎤ + si nc ⎡T ( f + f c ) ⎤}
29
From Table 2.2, 1 ℑ e − t u (t ) ←⎯ → 1 + j 2π f Now ℑ
{( e
−t
u (t ) ) cos 10π t =
} = ℑ {( e
−t
}
u (t ) ) ⊗ ℑ {cos 10π t }
11 ⊗ ⎡⎣δ ( f − 5 ) + δ ( f + 5 ) 1 + j 2π f 2
1⎡11⎤ =⎢+⎥ 2 ⎣ 1 + j 2π ( f − 5 ) 1 + j 2π ( f + 5 ) ⎦ =
1 ⎡ 1 + j 2π ( f + 5 ) + 1 + j 2π ( f − 5 ) ⎤ ⎥ ⎢ 22 2 ⎢⎣ (1 + j 2π f ) − ( j 2π 5 ) ⎥⎦ ⎡ 1 + j 2π f ⎤ =⎢ ⎥ (1 + j 2π f 2)2 + 100π
⎡ 1 + j 2π f ⎤ = ⎢ ⎥ 22 2 + π⎣j − ⎢ ( + π 1) 100 ⎦4 fπ4 f ⎥ d. te − t u (t )
Solution: Applying the differentiation in frequency domain property in (2.85), we obtain j d − −tℑ t te u (t ) ←⎯→ ℑ {e u (t )} 2π df That is, 1− d ⎡ (1 + j 2π f ) ⎤ j 1 j ℑ {te −ut (t )⎣} ⎦= =− 2π df 2π 1 =2
(1 + j 2π f ) e. e− π t
2
j 2π
(1 + j 2π f ) 2
Solution:
30
∞∞
2π
t + j 2 ft
− ( 2) − π t − j 2 π ft X ( f ) = e= e dt e dt ∫ ∫ −∞ −∞
Multiplying the ri ght hand side by −eπef
∞∞ f −π
X ( f ) e 2 fe2 =∫ =∫
−π
2
πf2
yields
( t 2 + j 2 f t − f 2 ) dt e − π e − π ( t +2jf ) dt
−∞ −∞
Substituting t + j f = ν , we obtain X(f)=e− πf
∞ 22
∫e
− πν
dν
− ∞ 1
f. 4si nc (t ) cos (100π t ) 2
Solution: ⎛f ⎞ ℑ 4si nc 2 ( t ) ←⎯ → 4⎜Λ⎟ ⎝2⎠ 1 ℑ cos (100π t ) ←⎯ → ⎣ ⎡⎦δ ( f − 50 ) + δ ( f + 50 ) ⎤ 2 ⎛ f ⎞ 1 X ( f ) = ℑ { 4sinc 2 ( t )} ⊗ ℑ {cos (100π t )} = 4 Λ ⊗ ⎡δ ( f − 50 ) + δ ( f + 50 )⎤ 2⎜⎝ ⎟⎠ 2 ⎣ ⎦ =2
{ Λ ⎡⎣ 0.5 ( f −
}
+ Λ ⎡⎣ 0.5 ( f − 50 ) ⎤⎦
50 ) 2.19 Find the following convolutions: a. sinc (Wt ) ⊗ si nc ( 2Wt ) Solution: We use the convolution property of Fourier transfor m in (2.79). x (t ) ⊗ y (t ) ←⎯ℑ → X ( f )Y ( f ) Now
1 ℑ sinc ( 2Wt ) ←⎯ → Π ( f / 2W ) 2W
31
Therefore, 1 1 ℑ sinc (Wt ) ⊗ s i nc ( 2Wt ) ←⎯ → Π ( f / W ) × Π ( f / 2W ) W 2W 1 = Π(f/W) 2W 2 b. sinc 2 (Wt ) ⊗ sinc ( 2Wt ) Solution: Again using the convolution property of Fourier transform ℑ sinc 2 (Wt ) ⊗ si nc ( 2Wt ) ←⎯ →
1
Λ ( f / 2W ) × W 2W
1
Π ( f / 2W )
1 Λ ( f / 2W ) 2W 2 2.20 The FT of a signal x (t ) is described by =
X(f)=
1 5 + j 2π f
Determine the FT V ( f ) of the following signals: a. v (t ) = x (5t − 1) Solution: ⎡ ⎛ 1⎞⎤ v (t ) = x (5t − 1) = x⎢ ⎜ 5 ⎟ ⎥t − ⎣ ⎝5⎠ ⎦ Let 1 f ⎛ ⎞1 1 ℑ y (t ) = x (5t ) ←⎯ → Y ( f ) = X⎜ = 5 ⎝ ⎟⎠55 ( j 2π f / 5 ) + 5 =
1 j 2π f + 25
−j 2 π f ⎛ 1⎞ ℑ ←⎯→ V ( f ) = Y ( f ) 5 = v (t ) = y t − e ⎜ ⎟ ⎝⎠ 5
1
−j 2 π f
e5
j 2π f + 2 5
32
b. v (t ) = x (t ) cos(100π t ) Solution: πt ⎡+ x (t⎤ ) e j 100 x (t ) e − j 100 π t v (t ) = x (t ) cos(100π t ) = ⎣ ⎦ 2
Now πt −j ⎡+ x (t⎤ ) e j 100 100 π tx (t ) e ⎣⎦ 2
1 ℑ ←⎯ → V ( f ) = X ( f − 50 ) + X ( f + 50 ) ⎤ ⎦ 2 =
11 + j 2π ( f − 50 ) + 5 j 2π ( f + 50 ) + 5
After simplification, we get V(f)=
j 2π f +5 j 20π f + π 2 1 0 4 + 25 − 4π 2 f 2
(
)
c. v (t ) = x (t ) e j 10 t Solution: 5⎞ 1 ⎛ ℑ v (t ) = x (t ) e j 10 t ←⎯ →V(f)=X f− = ⎜⎟ ⎛ ⎝ π ⎞⎠ j 2π ⎜
5⎟ f−+5 ⎝⎠ π
d. v (t ) =
dx (t ) dt
Solution: Using the differentiation property of FT j 2π f V ( f ) = j 2π f X ( f ) = 5 + j 2π f
d x (t ) ←⎯ℑ → j 2πfX ( f ) , we obtain dt
e . v (t ) = x (t ) ⊗ u (t ) Solution: Using the convolution property of FT x (t ) ⊗ y (t ) ←⎯ℑ → X ( f )Y ( f ) , we can write
33
1 ⎛11 ⎞ V ( f ) = X ( f )U ( f ) = δ ( f ) +⎜ 5 + j 2π f 2 j 2π f ⎟ ⎝⎠ 1 1 1 = δ(f) + 2 5 + j 2π f j 2π f ( 5 + j 2π f ) 1 1 =δ(f)+ 10 j 2π f ( 5 + j 2π f ) 2.21 Consider the delay eleme nt y (t ) = x (t − 3) . a. What is the impulse response h( t)? Solution: Si nc e y (t ) = h (t ) ⊗ x (t ) = x (t − 3) , the impulse response of the delay eleme nt is given from ( 2.16) as h (t ) = δ (t − 3) b. What is the ma gnitude and phase response function of the system? Solution: H ( f ) = ℑ {δ (t − 3)} = e − j 6 π f H(f)=1 (H ( f ) = − 6π f 2.22 The periodic input x (t ) to an LTI system is displaye d in Figure P2.7. The frequency response function of the system is given by H(f)= Figure P2.7
2 2 + j 2π f
34
a. Wr ite the complex exponential FS of input x (t ) . Solution: The input x (t ) is rectangular pulse train in Example 2.24 shifted by / 4 and o T duty cycle
τ = 0.5 . That is, T o
x (t ) g
( tT = To − o
⎡ ( t −nTo − To /4) ⎤
∞
/4)
=∑ Π⎢
⎥ 0.5T ⎣ o⎦ The comple x exponential FS of gT(t ) from Exa mple 2.24 with o To = o 2 ( f = 0.5 ) and
n =−∞
τ
= 0.5 is given by To
∞
gT(t ) = 0.5 ∑ s i nc ( 0.5 n )e jπ nt o n =−∞
In Exercise 2.15(a), we showed that time shifting introduces a linear phase shift in the FS coefficients; their ma gnitudes are not changed. The phase shift is equal x (t ) are to e − j 2 π nf o u for a time shift of u. The exponential FS coefficients Cn = 0.5s i nc ( 0.5n ) e − j 2 π nf o ( 0. 2 5To ) = e − jπ n /2 0.5s inc ( 0.5n ) ∞
− jπ n / 2 s i nc ( 0.5 n ) ⎤e jπ nt x (t ) = 0.5 ⎡ e ∑ ⎣⎦ n =−∞
b. Plot the ma gnitude and phase response functions for H ( f ) . Solution: 2 1 H(f)= = 2 + j 2π f 1 + jπ f
H ( nf o ) = jπ f
11 = 1+ 1 + jπ nf o
35
00 -1 0 -5 -2 0 -3 0
M agni t ude Res pons e(dB )
-1 0
P has e Res pons e(degrees )
-4 0
-1 5
-5 0
-2 0
-6 0
-7 0 -2 5 -8 0 -3 0
0
246 F r equenc y (f)
8
-9 0 0
2
4 6 F r eq uenc y ( f)
8
c. Compute the complex exponential FS of the output y (t ) . Solution: Using (2.114), the output of an LTI system to the input x (t ) =
∞
∑
n =−∞
⎡ 0.5e − jπ n / 2 s i nc ( 0.5 n ) ⎤e jπ nt ⎦ Cn
is given by ⎢ 0.5e y (t ) = ∑ C n H ( 0.5n ) e ∞∞
⎥
⎡⎤
− jπ n / 2 jπ n t
=∑ s inc⎢ ⎥( 0.5n ) e jπ n t n = −∞ 1 ⎢ ⎥+ j 05π n
n =−∞ FS co effi cien t o f y ( t)
2.23 The frequency response of an ideal LP filter is given by ⎧⎪5e − j 0.0025π f , H(f)=⎨
f < 1000 Hz
0, f
> 1000 Hz
Determine the output signal in each of the followi ng cases:
a. x (t ) = 5 sin ( 400π t ) + 2 cos (1200π t − π / 2 ) − cos ( 2200π t + π / 4 )
36
Solution: < 1000 Hz
5, f H(f)=⎨ 0, ⎪⎩ ⎪⎧ −0.0025π f ,
f > 1000 Hz f < 1000 Hz
(H ( f ) = ⎨ ⎪⎩ 0,
f > 1000 Hz
Since H ( 200) = 5 and (H ( 200) = −π / 2 , the output of the system for an input 5sin(400πt ) can now be expressed using (2.117) as 25sin(400πt − π/2). Next H ( 600) = 5 and (H ( 600) = −3π / 2 = π / 2 , the output of the system for an input 2 cos (1200π t − π / 2 ) can now be expressed using (2.117) as 10 cos (1200π t − π / 2 + π / 2 ) = 10 cos (1200π t ) . Now H (1100) = 0. So the LP filter doesn’t pass cos ( 2200π t + π / 4 ) . The output of the LP filter is therefore given by y (t ) = 25 sin ( 400π t − π / 2 ) + 10 cos (1200π t ) sin ( 2200π t b. x (t ) = 2 si n( 400π t ) +
)
πt
Solution: Since H ( 200) = 5 and (H ( 200) = −π / 2 , the output of the system for an input 2sin(400πt ) is 10sin(400πt − π/2). sin ( 2200π t ) To calculate the response to , we note that πt sin ( 2200π t ) ⎛ f ⎞ ℑ = 2200 × sinc ( 2200t ) ←⎯ →Π π t ⎜ 2200 ⎟ ⎝⎠ In fr equency domain, the output of LP filter to
⎛ f ⎞ − j 0.0025 π f ⎛ f ⎞ ⎛ f ⎞ − j 0.0025π f Π⎜ 5e⎟ ⎜ Π = 5Π e ⎟ ⎜ ⎟ ⎝ 2200 ⎠ ⎝ ⎠ ⎝ ⎠ 2000 2000
sin ( 2200π t ) is πt
37
Now ⎛f ⎞ ℑ Π ⎜ ⎟ e − j 0.0025 π f ←⎯ → 2000sinc ⎝ 2000 ⎠
2000 ( t − 0.00125 )
The output of the LP filter is therefore given by y (t ) = 10 sin ( 400π t − π / 2 ) + 10000si nc
2000 ( t − 0.00125 )
sin (1000π t ) c. x (t ) = cos( 400π t ) + πt Solution: Since H ( 200) = 5 and (H ( 200) = −π / 2 , the output of the system for an input cos( 400π t ) is 5 cos ( 400π t − π / 2 ) . To calculate the response to
sin (1000πt ) , we note that πt
sin (1000π t ) ⎛ f ⎞ ℑ = 1000 × si nc (1000t ) ←⎯ →Π π t ⎜ 1000 ⎟ ⎝⎠ In fr equency domain, the output of LP filter to
5e − j 0.0025π f Π⎜
sin (1000πt ) is πt
⎛f ⎞ ⎛ f ⎞ − j 0.0025π f =⎟ 5Π ⎜ ⎟ e 1000 ⎝ ⎠1000 ⎝⎠
Now
⎛f ⎞ πf ℑ Π ⎜ ⎟ e − j⎣0.0025 ←⎯ → 1000sinc ⎡1000 ( t − 0.00125 ) ⎤ ⎦ ⎝ 1000 ⎠ The output of the LP filter is therefore given by y (t ) = 5 cos ( 400π t − π / 2 ) + 5000si ⎣ ⎦ nc ⎡1000 ( t − 0.00125 ) ⎤ d. x (t ) = 5 cos(800π t ) + 2δ (t ) Solution:
38
Since H ( 400) = 5 and (H ( 4 00) = −π , the output of the system for an input 5 cos ( 800π t ) is 25 cos ( 800π t − π ) . The response of the system to δ (t ) is h (t ) . The impulse response of the ideal LP filter is 10000sinc 2000 ( t − 0.00125 ) ⎤⎦ . Co mbining y (t ) = 25 cos ( 800π t − π ) + 2 h (t ) 2000 ( t − 0.00125 ) = 25 cos ( 800π t − π ) + 20 × 103 sinc
2.24 The frequency response of an ideal HP filter is given by ⎧⎪4, H ( f ) =⎨ ⎪⎩ 0,
f > 20 Hz , f < 20 Hz
Determine the output signal y (t ) for the input a. x (t ) = 5 + 2 cos ( 5 0π t − π / 2 ) − cos ( 75π t + π / 4 ) Solution: y (t ) = 8 cos ( 50π t − π / 2 ) − 4 cos ( 75π t + π / 4 ) b. x (t ) = cos ( 2 0π t − 3π / 4 ) + 3 cos (100π t + π / 4 ) Solution: y (t ) = 12 cos (100π t + π / 4 ) 2.25 The frequency response of an ideal BP filter is gi ven by ⎧⎪2e − j 0.0005 π f , 900< f H(f)=⎨ ⎪⎩ 0, otherwise
< 1000 Hz ,
Determine the output signal y (t ) for the input
a. x (t ) = 2 cos (1850π t − π / 2 ) − cos (1900π t + π / 4 ) Solution: H (925) = 2 and H (950) = 2 .
39
(H ( f ) = − 0.0005π f . Therefore, (H (925) = − 0.0005π = −0.0005π × 925 = − 0.462π f = − 0.0005π × 950 = − 0.475π (H (950) = −0.0005π f y (t ) = 4 cos (1850π t − π / 2 − 0.462π ) − 2 cos (1900π t + π / 4 − 0.475π ) b. x (t ) = si nc ( 60t ) cos (1900π t ) Solution: X(f)=
⎛f ⎞1 Π⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 60 ⎝ ⎠ 60 2 950) ]
1
1 ⎡ ⎛ f −950 ⎞ ⎛ f +950 ⎞ ⎤ = Π + Π⎢ ⎜ 120 60 ⎟ ⎜ 60 ⎟ ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ Y( f) =H( f) X( f) − ⎞ ⎛ f+ ⎞950 ⎤ 1 f⎡ ⎛ 950 = 2 e − j 0. 0005 π f × × Π + Π ⎜ ⎟ ⎢ 120 60 ⎜ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦
60 ⎟ ⎥
Now 1 f 950 ⎛−⎞ × Π ←⎯→ s i nc 60t e ⎜⎟
ℑ
60 ⎝
()
j 1900 π t
60 ⎠
1 f 950 ⎛−⎞ e × Π ←⎯→ si nc ⎡ 60 − j 0.0005π f ℑ
60 ⎝
⎜⎟ 60 ⎠ ⎣ ⎦
t − 0.00025 ⎤ e
( ) j 1900 π ( t − 0.00025 )
Similarly 1 f 950 ⎛+⎞ × Π ←⎯→ s i nc 60t e ⎜⎟ 60 ⎝
ℑ
() 60 ⎠
− j 1900 π t
1 f 950 ⎛+⎞ e × Π ←⎯→ si nc ⎡ 60 − j 0.0005π f ℑ
60 ⎝
⎜⎟ 60 ⎠ ⎣ ⎦
t − 0.00025 ⎤ e
( ) − j 1900 π ( t − 0.00025 )
Therefore, j 1900 π ( t − 0.00025 ) y (t ) = sinc⎣ ⎦⎡ ⎣60 + e − j 1900 π ( t − 0.00025 ) ⎤ ⎦ ( t − 0.00025 ) ⎤ ⎡ e
= 2s inc ⎡⎣ 60 ( t −
cos ⎣⎡1900π ( t − 0.00025 )
0.00025 )
40
c. x (t ) = si nc 2 ( 30t ) cos (1900π t ) Solution: 1⎛f⎞1 X(f)= Λ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 30 ⎝ ⎠ 60 2 950) ] 1 ⎡ ⎛ f −950 ⎞ ⎛ f +950 ⎞ ⎤ = Λ + ⎢Λ⎜ 60 60 ⎟ ⎜ 60 ⎟ ⎥ ⎣⎝⎠⎝⎠⎦ Y( f) =H( f) X( f) ⎡ ⎛ 1− ⎞ ⎛ f+ ⎞ ⎤950 f 950 = 2 e − j 0. 0005π f × Λ + Λ 60 ⎢ ⎜ 60 ⎟ ⎜ 60 ⎟ ⎥ ⎣⎝⎠⎝⎠⎦ Now 1
⎛ f −950 ⎞ ×Λ ⎜ ⎟ ℑ 30 ⎝ 60
()
⎛ f −950 ⎞
30 ⎝
⎡ 30 t − 0.00025 ⎤ e
←⎯→ sinc
×Λ ⎜ ⎟ ℑ
− j 0.0005π f
j 1900 π t
⎠
1
e
←⎯→ sinc 30t e 2
2
()
j 1900 π ( t − 0.00025 )
60 ⎠ ⎣ ⎦
Similarly 1 f 950 ⎛+⎞ × Π ←⎯→ s i nc 30t e ⎜⎟
ℑ2
30 ⎝
()
− j 1 90 0 π t
60 ⎠
1 f 950 ⎛+⎞ e × Π ←⎯→ si nc − j 0.0005π f
⎜⎟ 30 ⎝
ℑ2
⎡ 30 t − 0.00025 ⎤ e
()
− j 1900 π ( t − 0.00025 )
60 ⎠ ⎣ ⎦
Therefore, ⎡ j 1900 π ( t − 0. 00025 ) + e − j 1900 π ( t − 0. 00025 ) ⎤ y (t ) = sinc⎣2⎦⎡ ⎣30 ⎦ ( t − 0.00025 ) ⎤ e
= 2s inc 2 ⎡⎣ 30 ( t −
cos ⎣⎡1900π ( t − 0.00025 )
0.00025 ) 2.26 The signal 2e −2 t u (t ) is input to an ideal LP filter with passband edge frequency equal to 5 Hz . Find the energy density spectrum of the output of the filter. Calculat e the energy of the input signal and the output signal. Solution:
41
∞
∞
2
−4 t dt = E = 2e −2 t u (t ) dt = 4e 4 x∫ ∫ −4 −∞ 0 0
e −4 t
∞
=1
2 ℑ x (t ) = 2e −2 t u (t ) ←⎯ → 2 + j 2π f 2
The energy density spectrum of the output y( t ), Y ( f ) , is related to the energy density spectrum of the input x (t ) 2
2 ⎛f ⎞ Y ( f ) = H ( f ) X ( f ) = Π⎜ ⎟ ⎝ ⎠10 2 + j 2π f 2
2
∞∞∞22 2
E = y (t ) dt = y
Y df ( f )
∫∫∫
2
=Π
−∞ −∞ −∞
=∫
2
2
⎛f ⎞
2
df
⎜⎝ 10 ⎟⎠ 2 + j 2π f
2
5
5 1 1 df df = 2 1 + jπ 1 + π 2f 2 ∫ f −5 0
1 Making change of variables π f = u ⇒ d = d u , we obtain π f 2 5π1 2 du = tan yE = ∫ π 0 1 + u 2π
−1
(
2 u ) = (1.507 ) = 0.9594 0 π 5π
Thus output of the LP filter cont ains 96% of the input signal energy. 2.27 Calculate and sketch the power spectral density of the following signals. Calculate the normalized average power of the signal in each case. a. x (t ) = 2 cos (1000π t − π / 2 ) − cos (1850π t + π / 4 ) x1 ( t )
x2 ( t )
Solution:
1 R (τ ) = li m x
T →∞
T
T /2
∫
− T /2
x (tR) x (t − τ ) d t R (τ ) + = x1 x2
(τ ) − R
(τ ) − R x1 x2
(τ ) x2 x1
Now
42
T /2
R (τ ) = li m x1
T →∞
1
x (t ) x (t − τ ) dt
T
− T /2
∫
11
π⎞⎡π⎤ ⎛ = li m 2 cos ⎜ 1000π t − ⎟ 2 cos 1000π ( t − τ ) − d t T ∫ 2 2 ⎢⎥ T →∞ − T /2 ⎝ ⎠ ⎣ ⎦ T /2
1
1 T →∞ T
= 2 lim
∫ {cos (1000πτ ) + cos⎣ ⎦⎡1000π ( 2t − τ ) − π ⎤}
T /2
dt
− T /2
= 2 cos (1000πτ ) The second term is zero because it integrates a sinusoidal func tion over a period. Similarly, Rx (τ ) = li m 2T
T /2
1 T
→∞
∫ x (t2) x (t − τ ) dt =
1
22
cos (1850πτ )
− T /2
The cross-correlation term Rx x (τ ) = li m
1 2T
→∞
1
T /2
T
x (t ) x (t − τ ) = lim 1
∫ − T /2
= li m
T →∞
12
T
1
T /2
T →∞
T /2
2/ 4cos ⎤dt(1000π t − π / 2 ) cos ⎡1850π ( t − τ ) + π
∫ ⎣⎦ − T /2
{cos ( 850π t − 1850πτ + 3π / 4 ) + s in ( 2850π t − 1850πτ − π / T −∫ T / 24 ) dt }
is zero because it integrates a si nusoi dal function over a period in each case. Similarly, it can be shown that all ot he r cross-correlation terms are zero. Therefore, 1 Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) = 2 cos (1000πτ ) +
2
cos (1850πτ
)
1 Gx ( f ) = ℑ { Rx (τ )} = ⎧ℑ ⎨ ⎫2 cos (1000πτ ) + cos (1850πτ ) ⎬ 2 ⎩⎭ 1 = [δ ( f − 500) + δ ( f + 500) ] + [δ ( f − 925) + δ ( f + 925) ] 4 The norma lized average power is obtained by using (2.172) as
43
∞∞ 1∞ Px = G ( f ∫ x )df
= ∫ [δ ( f − 500) + δ ( f + 500) ] df +
] df
4∫
−∞ −∞ −∞
=1+1+
115 + = 4 4 2
P ower S pec t ral Dens ity
1
0. 8
0. 6
0. 4
0. 2
0
-1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0. 4 0. 6 0. 8 1 F requenc y (k Hz )
b. x (t ) = [1 + sin( 200π t ) ] cos( 2000π t ) Solution: x (t ) = [1 + s in( 200π t ) ] cos( 2000π t ) = cos( 2000π t ) + s in( 200π t ) cos( 2000π t ) = cos( 2000π t ) + x1 ( t )
2
11 sin ( 2200π t ) − x2 ( t )
2
sin (1800π t )
x3 ( t )
Now 1 R (τ ) = li m x
T →∞
T
T /2
∫
x (tR) x (t − τ ) d t R (τ ) + =
− T /2
where 1 cos ( 2000πτ ) 2 1 Rx2 (τ ) = cos ( 2200πτ ) 8 1 Rx1 (τ ) =
x1 x2 x3
(τ ) + R
(τ )
[δ ( f − 925) + δ ( f + 925)
Rx3 (τ ) =
8
cos (1800πτ )
44
Therefore, 111 cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) 2 8 8 ⎧111⎫ Gx ( f ) = ℑ { Rx (τ )} = ℑ ⎨ cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) ⎬ 8 8 ⎩⎭ 2 11 = [δ ( f − 1000) + δ ( f + 1000) ] + [δ ( f − 1100) + δ ( f + 1100) ] 4 16 1 + [δ ( f − 900) + δ ( f + 900) ] 16 Rx (τ ) =
The normalized average power is obtained by using (2.172) as ∞
1 ∞1 ∞ Px = ∫ Gx ( f )df = 4 ∫ [δ ( f − 1000) + δ ( f + 1000) ] df + 16 ∫ [δ ( f − 1100) + δ ( f + 1100) ] df −∞ −∞ −∞
1 ∞1 1 1 3 [δ ( f − 900) + δ ( f + 16 ∫ 900) ] d f +
−∞
P ower S pec t ral Dens ity
0. 25
0. 2
0. 15
0. 1
0. 05
0
-1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0. 4 0. 6 0. 8 1 Frequenc y (k Hz )
c. x (t ) = cos 2 ( 200π t ) s i n (1800π t ) Solution:
=
2
+
8
+
8
=
4
45
1 1 1 x (t ) = ⎣ ⎡1 + cos ( 400π t ) ⎤ sin (1800π t ) = si n (1800π t ) + sin (1800π t ) cos ( ⎦ 2 22 400π t ) =
111 sin (1800π t ) + s i n (1400π t ) + si n ( 2200π t ) 2 4 4
Now Rx (τ ) = R1x (τ ) + Rx
2
(τ ) + Rx
3
(τ )
where Rx1 (τ ) =
1 cos (1800πτ ) 8
Rx2 (τ ) =
1 cos (1400πτ ) 32
Rx3 (τ ) =
1 cos ( 2200πτ ) 32
Therefore, Rx (τ ) =
111 cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ 8 32 32
)
⎧111⎫ Gx ( f ) = ℑ { Rx (τ )} = ℑ ⎨ cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) ⎬ 32 32 ⎩⎭ 8 11 = [δ ( f − 900) + δ ( f + 900) ] + [δ ( f − 700) + δ ( f + 700) ] 16 64 1 + [δ ( f − 1100) + δ ( f + 1100) ] 64 The normalized average power is obtained by using (2.172) as ∞ 1 ∞1 ∞ Px = G ∫ x ( f )df = 16 ∫ [δ ( f − 900) + δ ( f + 900) ]df + 64 ∫ [δ ( f − 700) + δ ( f + 700) ] df −∞ −∞ −∞
1 ∞1 1 1 3 [δ ( f − 1100) + δ ( f + 64 ∫ 1100) ] df +
−∞
=
8
+
32
+
32
=
16
46
0. 1 0. 09 0. 08
P ower S pec t ral Dens i ty
0. 07 0. 06 0. 05 0. 04 0. 03 0. 02 0. 01 0 -1 -0. 8 -0. 6 -0. 4 -0. 2 0 0. 2 0. 4 0. 6 0. 8 1
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