Chapter 5: Continuous-time Fourier Transform Problem 5.1 (a)
The CTFT for x1(t) is given by ∞
∫
X 1 (ω) =
− jωt − jωt ∫ [ τt ] e dt + ∫ [1 − τt ] e dt
−∞
−τ
= 1+
[ ] t τ
− jωt
2 ω2 τ
=τ (b)
−
[]
e × − (− jω)
1 = − (− jω) =
τ
0
x1 (t )e − jωt dt = 1 +
[1τ ]×
(− jω)
ω2 τ
(ωτ / 2)
2
+ × 1− (− jω) 2 −τ e
1
2 cos(ωτ)
sin 2 (ωτ / 2)
1 τ
0
0
2
= 2×
+
− jωt
[1τ ]×
[ ] t τ
τ
e − jωt e − jωt × − − 1τ × (− jω) (− jω) 2 0
[ ]
e jωτ 1 e − jωτ 1 + × − − τ 2 2 (− jω) (− jω) (− jω)
[]
1 − cos(ωτ) ω2 τ
= 2×
[1τ ]×
(− jω) 1
2
2 sin 2 (ωτ / 2) ω2 τ
ωτ = τ sinc 2 . 2π
The CTFT for x2(t) is given by
X 2 (ω ) =
∞
∫
x2 (t )e− jω t dt =
−∞
= t 4
e− ( a+ jω ) t ( − ( a + jω ))
∞
∞
−∞
0
4 − at 4 − ( a + jω ) t − jω t dt ∫ t e u(t )e dt = ∫ t e
− ( a + jω ) t
+ 4t 3 (( −e( a + jω ))2 + 12t 2
e− ( a+ jω ) t ( − ( a + jω ))3
∞
+ 24t (( −e( a + jω ))4 + 24 ( −e( a + jω ))5 0 − ( a+ jω ) t
− ( a + jω ) t
= [0 + 0 + 0 + 0 + 0] − 0 + 0 + 0 + 0 + 24 ( − ( a +1 jω ))5 = ( a +24jω )5 . (c)
The CTFT for x3(t) is given by
X 3 (ω ) =
∞
∫ x (t )e 3
−∞
=
1 2
− jω t
∞
dt =
∫e
− at
−∞
cos(ω0t )u(t )e
− jω t
∞
dt =
1 2
∫e
− ( a + jω ) t
0
∞
∞
0
0
∞
e jω0t + e− jω0t dt ∞
0 0 − ( a + jω − jω0 ) t dt + 12 ∫ e− ( a + jω + jω0 )t dt = 12 ( −e( a + jω − jω0 )) + 12 ( −e( a + jω + jω0 )) ∫e 0 0 − ( a + jω − jω ) t
− ( a + jω + jω ) t
= 12 0 − ( − ( a + jω1 − jω0 )) + 12 0 − ( − ( a + jω1 + jω0 )) = 12 ( a + jω1− jω0 ) + ( a + jω1+ jω0 ) = ( a +ajω+ )jω2 +ω 2 . 0 (d)
The CTFT for x4(t) is given by
160
Chapter 5
X 4 (ω ) =
∞
∫
x4 (t )e− jω t dt =
−∞
∞
∫
2 2
∫
2
−∞
) = exp ( jωσ ⋅ 2σ 2
∞
exp( − 2tσ 2 )e− jω t dt =
∞
∫ exp −
( t + jωσ 2 )2 2σ 2
−∞
exp[ − t
2
+ 2 jωσ 2t 2σ
−∞
2
∞
]dt =
∫ exp −
t 2 + 2 jωσ 2t + ( jωσ 2 )2
−∞
2σ 2
exp ( jωσ 2 ) dt 2σ 2 2
dt = exp ( jωσ 2 ) 2πσ = 2πσ exp − ω 22σ 2 . 2σ 2 2
▌ Problem 5.2 (a)
By definition,
[ ]
π
∫
X 1 (ω) = 3e − jωt dt = 3 (e− jω) − jωt
0
3 e − jωπ / 2 jω
=−
π 0
=−
3 jω
[e
− jωπ
]
−1 = −
3 − jωπ / 2 e jω
[e
− jωπ / 2
By definition,
X 2 (ω ) =
0.5T
∫
0.5e
− jωt
1.5T
dt +
−0.5T
∫
0.5T
e− jωt dt = 0.5 (e− jω ) − jωt
− j 0.5ωT − e j 0.5ωT − = − 0.5 jω e = − 0.5 jω [ −2 j sin(0.5ωT ) ] −
1 jω
1 jω
0.5T − 0.5T
+ (e− jω ) − jωt
e− j1.5ωT − e− j 0.5ωT
e− jωT [ −2 j sin(0.5ωT )]
sin(0.5ωT ) sin(0.5ωT ) T T + 2e− jωT 0.5 = 0.5 0.5T × ω 0.5T × ω = 0.5Tsinc ( 0.5πωT ) + Te− jωT sinc ( 0.5πωT ) .
(c)
By definition, T
∫(
)
(
X 3 (ω) = 1 − Tt e − jωt dt = 1 − Tt 0
) (e− jω) − (− T1 ) (−e jω) − jωt
− jωt
2
T 0
( ) (−e jω) − (−1jω) − (− T1 ) (− j1ω)
= 0 − − T1 =−
1 ω2T
− jωT
e − jωT +
2
1 jω T
For ω = 0,
∫(
2
+
1 ω2T
)
=
1 jω
+
1 ω2T
(
(1 − e
X 3 (ω) = 1 − Tt dt = − T2 1 − Tt 0
]
/ 2) [− 2 j sin(ωπ / 2)] = 6e − jωπ / 2 [sin(ωπ ]= 6e − jωπ / 2 [21/ π × sin(ωπωπ/ 2/ 2) ] ω
= 3πe − jωπ / 2 sinc(ω / 2). (b)
− e jωπ / 2
− jωT
)2
T 0
).
= 0 + T2 = T2 .
1.5T 0.5T
Solutions (d)
By definition, 0
X 4 (ω) =
∫ (1 + )e t T
− jωt
T
∫(
−T
(
= 1 + Tt
(e)
2 ω2T
0
) (e− jω) − (T1 ) (−e jω) − jωt
− jωt
2
= ( − 1jω) − =
)
dt + 1 − Tt e − jωt dt
(T1 ) (− j1ω)
2
−0+
0 −T
(
+ 1 − Tt
) (e− jω) − (− T1 ) (−e jω) − jωt
− jωt
2
(T1 ) (−e jω) + 0 − (− T1 ) (−e jω) jωT
− jωT
2
2
T 0
( ) (− j1ω)
− ( − 1jω) + − T1
[1 − cos(ωT )] = 2×2 sinω (T0.5ωT ) = 1/(0.45 T ) × sin(0.(50ω.5Tω)T ) = Tsinc 2 ( 0.5πωT ) . 2
2
2
2
2
By definition, T
T
0
0
T
X 5 (ω ) = ∫ 1 − 0.5sin ( πTt ) e− jωt dt = ∫ e− jωt dt − 0.5∫ sin ( πTt ) e− jωt dt 0
=A
=B
We consider different cases for the above integral. Case I: ( ω = 0) −∞
X 5 (0) =
∫
∞
x(t )dt =
∞
=T
∫
−∞
+ π0.5 /T
T
T
0
0
1 − 0.5sin ( πTt ) dt = ∫ dt − 0.5∫ sin ( πTt ) dt T
cos( T ) 0 = T + πt
T 2π
[cos(π ) − cos(0)] = T − Tπ = T (1 − π1 )
Case II: ( ω ≠ 0, ω ≠ π/T): T
T A = ∫ e− jωt dt = −1jω e− jωt = −1jω e− jωT − 1 = j1ω 1 − e− jωT 0
[ω ≠ 0]
0
T
− jωt B = 0.5 πe2 2 {− jω sin ( πTt ) − πT cos ( πTt )} −ω 2 T 0
for ω ≠ 0, ± πT T
= π 20.5−ωT2T 2 e− jωt jω sin ( πTt ) + πT cos ( πTt ) =0 at t =0,T 0 2
= π 20.5−ωT2T 2 − πT e− jωT − πT = π 20.5−ωπ2TT 2 1 + e− jωT 2
Case III: ( ω = π/T):
2
161
162
Chapter 5
T
B = 0.5∫ sin ( πTt ) e− jωt dt = 0
T 0.5 2j
T
−j − j (ω − ) t − j (ω + ) t j − jωt ∫ e T − e T e dt = 0.52 j ∫ e T − e T dt πt
πt
0
0.5 − j 2Tπ t dt ω = π 2 j ∫ 1 − e T 0 = T 0.5 j 2Tπ t π − 2 j ∫ 1 − e dt ω = − T 0
π
π
0
T
± j 2Tπ t is periodic with period T , As e
T
∫e 0
± j 2Tπ t
dt = 0
0.5T = ± 0.5 2 j [ t ]0 = ± 2 j T
Combining, the above results, the CTFT can be expressed as T (1 − π1 ) X 5 (ω ) = j1ω 1 − e− jωT ∓ 0.52 jT 1 − jωT − π 20.5−ωπ2TT 2 1 + e− jωT jω 1 − e T (1 − π1 ) = ± 2jπT ∓ 4Tj 1 − jωT − π 20.5−ωπ2TT 2 1 + e− jωT jω 1 − e
ω =0 ω = ± πT otherwise
ω =0 ω = ± πT otherwise
▌ Problem P5.3
From magnitude and phase spectra shown in Fig. P5.3, the individual CTFT’s can be expressed as follows Fig. P5.3(b):
j 0.5ω X 1 (ω) = 1 × e 0
Fig. P5.3(c):
1× e− j 0.5ω X 2 (ω ) = 1× e j 0.5ω 0
Fig. P5.3(d):
1× e− jπ / 3 X 3 (ω ) = 1× e jπ / 3 0
−W ≤ ω≤W otherwise
−W ≤ ω ≤ 0 0 ≤ω ≤W otherwise −W ≤ ω ≤ 0 0 ≤ω ≤W otherwise
Using the CTFT synthesis Eq. (5.9), the function x1 (t ) is calculated as follows.
Solutions
x1 (t ) =
∞
1 2π
∫
X (ω )e jωt dω =
−∞
W
1 2π
∫
e j 0.5ω e jωt dω =
−W
W
1 2π
∫e
=
1 2 j sin [ (0.5 + t )W ] = j (0.5 + t ) 2π
1 e j (0.5+t )ω 1 e j (0.5+t )W − e− j (0.5+t )W = 2π j (0.5 + t ) −W 2π j (0.5 + t ) W
π
×
sin [ (0.5 + t )W ] (0.5 + t )W
=
W
dω
−W
W
=
j (0.5 + t )ω
sinc Wπ (t + 0.5) .
π
Using the CTFT synthesis Eq. (5.9), the function x2 (t ) is calculated as follows.
1 x2 (t ) = 2π
∞
0
1 ∫−∞ X (ω )e dω = 2π
∫e
jωt
j ( −0.5+ t )ω
−W
1 dω + 2π
0
W
∫e
j (0.5+ t )ω
dω
0
W
1 e j ( −0.5+t )ω 1 e j (0.5+t )ω 1 1 − e− j ( −0.5+t )W e j (0.5+t )W − 1 = + = + j (0.5 + t ) 2π j (−0.5 + t ) −W 2π j (0.5 + t ) 0 2π j (−0.5 + t ) =
2 jt sin(tW ) − cos(tW ) 1 1 + e j 0.5W 2 j (t 2 − 0.25) 2π j (t − 0.25)
=
1 1 + e j 0.5W ( 2 jt sin(tW ) − cos(tW ) ) . j 2π (t − 0.25) 2
Clearly, at (t = ±0.5), x2(t) is undefined in the above expression. Computing directly, we obtain W
At t = 0.5:
x2 (0.5) =
1 2π
j 0.5ω j 0.5ω ∫ e e dω =
W 1 2π
0
0
At t = −0.5: x2 ( −0.5) =
1 2π
∫
e − j 0.5ω e − j 0.5ω d ω =
−W
∫e
jω
dω =
1 2 jπ
0
0
1 2π
∫e
− jω
W
e jω = 0
dω =
1 −2 jπ
−W
1 2 jπ
e jW − 1 . 0
e − jω = −W
1 2 jπ
e jW − 1 .
Using the CTFT synthesis Eq. (5.9), the function x3 (t ) is calculated as follows.
x3 (t ) =
1 2π
∞
∫
X (ω )e jωt d ω =
−∞ 0
1 2π
0
∫
e − jπ / 3e jωt dω +
−W
1 2π
0
∫e
jπ / 3
e jωt dω
−W
W
jWt 1 − jπ / 3 e jωt 1 jπ / 3 e jωt 1 − jπ / 3 1 − e − jWt − 1 jπ / 3 e e e e = + = +e jt jt 2π jt −W 2π jt 0 2π
=
1 sin(Wt + π / 3) − sin(π / 3) [ 2 j sin(Wt + π / 3) − 2 j sin(π / 3)] = πt j 2π t
Clearly, at (t = 0), x3(t) is undefined in the above expression. Computing directly, we get
163
164
Chapter 5
x3 (0) =
1 4π
0 W (1 − j 3) ∫ d ω + (1 + j 3) ∫ d ω = −W 0
W 4π
1 − j 3 + 1 + j 3 =
W 2π
.
Although the functions x1 (t ) , x2 (t ) , and x3 (t ) have the same magnitude spectra, their phase spectra are different. As a result, the time domain representations of these functions are different.
abs(x2 (t))
x1 (t)
For the special case W = π, the three functions are plotted in Fig. S5.3. Since x2(t) is a complex function, its magnitude is plotted in Fig. S5.3. The Matlab code is also included below. ▌
Problem 5.3
1 0.5 0 -0.5 -1 -5
-4
-3
-2
-1
0 t
1
2
3
4
5
1 0.75 0.5 0.25 0 -5
-4
-3
-2
-1
0 t
1
2
3
4
5
-4
-3
-2
-1
0 t
1
2
3
4
5
x3 (t)
1 0.5 0 -0.5 -1 -5
Fig. S5.3. Plots of functions in Problem P5.3. % MATLAB code to plot the functions in Problem 5.3 del = 0.01; t = -5:del:5; W = pi ; x1 = (W/pi)*sinc((W/pi)*(t+0.5)) ; x2 = 1./(j*2*pi*(t.^2-0.25)).*(1+exp(j*0.5*W)*(2*j*t.*sin(t*W)-cos(t*W))); x2(t==0.5) = 1./(j*2*pi)*(exp(j*W)-1); x2(t==-0.5) = 1./(j*2*pi)*(exp(j*W)-1); x3 = (sin(W*t+pi/3)-sin(pi/3))./(pi*t); x3(t==0) = W/(2*pi) ; subplot(3,1,1), plot(t, x1), grid on title('Problem 5.3');
Solutions
xlabel('t') % Label of ylabel('x_1(t)') % Label of % subplot(3,1,2), plot(t, abs(x2)), grid xlabel('t') % Label of ylabel('abs(x_2(t))') % Label of
X-axis Y-axis on X-axis Y-axis
subplot(3,1,3), plot(t, x3), grid xlabel('t') % Label of X-axis ylabel('x_3(t)') % Label of Y-axis
Problem 5.4
(a)
The partial fraction expansion is given by X 1 (ω) =
(1 + jω) −1 2 ≡ + (2 + jω)(3 + jω) (2 + jω) (3 + jω)
Calculating the inverse CTFT, we obtain
x1 (t ) = − e −2t u (t ) + 2e −3t u (t ) . (b)
The partial fraction expansion is given by X 2 (ω) =
−1 1 0 .5 0.5 ≡ + + (1 + jω)(2 + jω)(3 + jω) (1 + jω) (2 + jω) (3 + jω)
Calculating the inverse CTFT, we obtain
x 2 (t ) = 0.5e −t u (t ) − e −2t u (t ) + 0.5e −3t u (t ) . (c)
The partial fraction expansion is given by
X 3 (ω) =
1 2
(1 + jω)(2 + jω) (3 + jω)
≡
−1 0 .5 0 − 0 .5 + + + (1 + jω) (2 + jω) (2 + jω) 2 (3 + jω)
Calculating the inverse CTFT, we obtain
x3 (t ) = 0.5e −t u (t ) − te −2t u (t ) + 0.5e −3t u (t ) . (d)
The partial fraction expansion is given by X 4 (ω) =
1 2
(1 + jω)(2 + 2 jω + ( jω) ) X 4 (ω) =
or,
≡
1 + jω 1 − (1 + jω) (2 + 2 jω + ( jω) 2 )
1 + jω 1 − (1 + jω) 1 + (1 + jω) 2
Calculating the inverse CTFT, we obtain
x 4 (t ) = e −t u (t ) − e −t cos t u (t ) . (e)
The partial fraction expansion is given by X 5 (ω) =
1 (1 + jω) 2 (2 + 2 jω + ( jω) 2 ) 2
≡
1 (1 + jω) 2
−
1.50 (2 + 2 jω + ( jω) 2 )
+
0.25(4 jω + ( jω) 2 ) (2 + 2 jω + ( jω) 2 ) 2
165
166
Chapter 5
or,
X 5 (ω) =
1 (1 + jω) 2
−
1.50 1 + (1 + jω) 2
+
0.25(4 jω + ( jω) 2 ) (1 + (1 + jω) 2 ) 2
Calculating the inverse CTFT, we obtain 0.25(4 jω + ( jω) x5 (t ) = te −t u (t ) − 1.50e −t sin t u (t ) + ℑ −1 (1 + (1 + jω) 2 ) 2
2
) .
▌
Problem 5.5
Consider an arbitrary function φ(t), and assume that ∞
p (t ) =
∫e
j ωt
dt .
−∞
Now, consider the integral ∞
∞
∞
∞
−∞
−∞
−∞
-∞
∞
−∞
jω ( t −T ) jω t − jωT ∫ φ (t ) p(t − T )dt = ∫ φ (t ) ∫ e dω dt = ∫ e ∫ φ (t )e dt dω = Changing the order of integration −∞
=
∫e
jω ′T
∞
∫e
Φ(ω ′)( −dω ′) =
jω ′T
∞
∫e
− jωT
Φ(−ω )dω
−∞
Φ(ω )=ℑ{φ ( t )}
Φ(ω ′)dω ′
−∞
∞
ω =−ω ′, dω =− dω ′ ∞
=
∫ Φ(ω )e
jωT
dω
................................................(1)
−∞
Note that the right hand side of Eq. (1) is the inverse CTFT of Φ(ω) computed at t = T, i.e., φ(T). Hence, ∞
∫
∞
φ(t ) p(t − T )dt =
−∞
∫ Φ(ω)e
jωT
dω = 2πφ(T ) .
−∞
The above equation is valid for any arbitrary φ(t) if and only if p(t) = 2πδ(t) as can be seen from the following property of the impulse response ∞
∫
2π φ(t )δ(t − T )dt = 2πφ(T ) . −∞
In other words, ∞
∫e
j ωt
dt = 2πδ(t ) .
−∞
Interchanging the variables, t and ω, we obtain the required identity ∞
∫e
−∞
j ωt
dω = 2πδ(ω) .
▌
Solutions
167
Alternate Proof:
Note that the above result can be proved directly from the CTFT pair CTFT 1 ← → 2πδ(ω) .
∞
∫
2πδ(ω) = 1 × e − jωt dt
By definition,
−∞ ∞
Since δ(−ω) = δ(ω),
∫e
2πδ(ω) = 2πδ(−ω) =
jωt
dt .
▌
−∞
Problem 5.6
Using Eq. (5.40), the CTFT for a real-valued even function x(t) can be expressed as ∞
∫
X (ω) =
∞
∫
x(t )e − jωt dt = 2 x(t ) cos(ωt )dt .
−∞
0
Since there is no complex value in the above equation, X(ω) is real valued, i.e., Im{X(ω)} = 0. ∞
∞
∫
∫
Also, X (−ω) = 2 x(t ) cos(−ωt )dt = 2 x(t ) cos(ωt )dt = X (ω) . 0
0
Therefore, X(ω) is also an even function with respect to ω. Since, X(ω) is real valued, Re{X(ω)} = Re{X(−ω)}. ▌ Problem 5.7
Using Eq. (5.40), the CTFT for a real-valued odd function x(t) can be expressed as ∞
X (ω) =
∫
∞
∫
x(t )e − jωt dt = − j 2 x(t ) sin(ωt )dt .
−∞
0
Since x(t) is real, the product x(t)sin(ωt) is also real and so is the integral. Therefore, X(ω) is pure imaginary, i.e., Re{X(ω)} = 0. ∞
Also,
∫
∞
∫
X (−ω) = 2 x(t ) sin( −ωt )dt = −2 x(t ) sin(ωt )dt = − X (ω) . 0
0
Therefore, X(ω) is also an odd function with respect to ω. Since, X(ω) is imaginary-valued, Im{X(ω)} = −Im{X(−ω)}. ▌ Problem 5.8
(a)
Since is not equal to
X 1 (− ω) =
5 2 + j ( − ω− 5 )
X 1∗ (ω) =
=
5 2 − j ( ω+ 5 )
5 2 − j (ω−5 )
,
168
Chapter 5 X1(ω) does not satisfy the Hermitian property. Its inverse CTFT x1(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x1(t) from the Hermitian property.
(b)
(
X 2 (− ω) = cos − 2ω +
Since
is not equal to
(
X 2∗ (ω) = cos 2ω +
π 6
π 6
)=
)=
3 2
3 2
cos(2ω) + 12 sin( 2ω)
cos(2ω) − 12 sin(2ω) ,
X2(ω) does not satisfy the Hermitian property. Its inverse CTFT x2(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x2(t) from the Hermitian property.
(c)
X 3 (− ω) =
Since
5 sin [4(− ω− π )] ( − ω− π )
X 3∗ (ω) =
is not equal to
=
5 sin [4(ω+ π )] ( ω+ π )
5 sin [4 (ω− π )] (ω − π )
[4ω] = 5 (sin ω+ π )
[4ω] , = 5 (sin ω− π )
X3(ω) does not satisfy the Hermitian property. Its inverse CTFT x3(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x3(t) from the Hermitian property.
(d)
Since X 4 (− ω) = (3 + 2 j ) δ(− ω − 10) + (1 − 2 j )δ(− ω + 10) = (3 + 2 j )δ(ω + 10) + (1 − 2 j )δ(ω − 10)
is not equal to
X 4∗ (ω) = (3 + 2 j ) δ(ω − 10) + (1 − 2 j )δ(ω + 10) ,
X4(ω) does not satisfy the Hermitian property. Its inverse CTFT x4(t) is not real valued and is complex. Nothing can be stated about the odd and even property of x4(t) from the Hermitian property. (e)
Since
X 5 (− ω) =
is equal to
X 5∗ (ω) =
1
(1− jω)(3− jω)2 (5+ ω2 ) 1
(1− jω)(3− jω)2 (5+ ω2 )
,
X4(ω) satisfies the Hermitian property. Its inverse CTFT x4(t) is real valued. Since X4(ω) is complex (neither pure real-valued or pure imaginary), x4(t) is neither even nor odd with respect to t. ▌ Problem 5.9
(a)
Applying the linearity property,
{
}
{
}
X 1 (ω) = ℑ 5 + 3 cos(10t ) − 7e −2t sin(3t )u (t ) = 5ℑ{1} + 3ℑ{cos(10t )} − 7 ℑ e −2t sin(3t )u (t ) . By selecting the appropriate CTFT pairs from Table 5.2, we get X 1 (ω) = 10δ(ω)ℑ{1} + 3πδ(ω − 10) + 3πδ(ω − 10) − (b)
Entry (8) of Table 5.2 provides the CTFT pair CTFT
sgn(t ) ← → Using the duality property,
2 jt
CTFT
2 jω
.
← → 2π sgn( −ω) ,
21 ( 2 + jω) 2 + 3 2
.
Solutions
1 πt
or, (c)
CTFT
← → − j sgn(ω) .
Entry (7) of Table 5.2 provides the CTFT pair
e Using the time shifting property, e
−4 t
−4 t −5
CTFT
← → 4+8jω .
← → 4+8jω e − j 5ω . CTFT
Using the frequency differentiation property,
t2 e t2 e
or, (d)
169
−4 t − 5
−4 t −5
CTFT
← →( j ) 2
← → 200e − j 5ω CTFT
d2 dω 2 1 4+ jω
{e
− j 5ω
8 4 + jω
+ 16e − j 5ω
}
1 ( 4 + jω) 3
.
Entry (17) of Table 5.2 provides the CTFT pair
(6ωπ ) sin(5 πt ) CTFT 5 sinc(5t ) = 5 5πt ← → rect (10ωπ ) sin(3πt ) 3πt
3 sinc(3t ) = 3
and
CTFT
← → rect
Using the multiplication property
[ (6ωπ )∗ rect(10ωπ )] sin( 3πt ) sin( 5 πt ) CTFT ← → π2 [rect (6ωπ ) ∗ rect (10ωπ )] , t
π2 × or, or,
sin( 3πt ) πt
×
sin( 5 πt ) πt
CTFT
← → 2ππ rect 2
2
πt) CTFT ← → 52π rect ( 6ωπ ) ∗ rect ( 10ωπ ) , 5 sin(3π t t)sin(5 2
where * is the convolution operation. (e)
Entry (17) of Table 5.2 provides the CTFT pair
(6ωπ ) sin( 4 πt ) CTFT 4 sinc(4t ) = 4 4 πt ← → rect (8ωπ ). sin(3πt ) 3πt
3 sinc(3t ) = 3
and
CTFT
← → rect
Using the time differentiation property, 1 d sin( 4 πt ) π dt t
CTFT
← →( jω) rect
(8ωπ ).
Using the convolution property
[ (6ωπ )× jωrect(8ωπ )] sin( 3πt ) sin( 4 πt ) CTFT ∗ dtd ← → π2 [rect (6ωπ ) × jωrect (8ωπ )], t t sin( 3πt ) sin( 4 πt ) CTFT 4 t ∗ dtd ← → j 2π rect (6ωπ ). t
π2 × or, or,
sin(3πt ) πt
∗ π1
d sin( 4 πt ) dt t
CTFT
← → 2ππ rect 2
▌
170
Chapter 5
Problem 5.10
Using the linearity property,
{
}
X (ω ) = ℑ 136 e−2t − 136 cos(3t ) + 134 sin(3t ) u (t )
{
}
= 136 ℑ e−2t u (t ) − 136 ℑ{cos(3t )u (t )} + 134 ℑ{sin(3t )u (t )} = 136 2+1jω − 136 π2 δ (ω − 3) + π2 δ (ω + 3) − 136 9−jωω 2 + 134 − 2jπ δ (ω − 3) +
jπ 2
δ (ω + 3) + 134 9−3ω
2
π = 136 2+1jω − 9−jωω 2 + 9−2ω 2 − 26 [6δ (ω − 3) + 6δ (ω + 3) + 4 jδ (ω − 3) − 4 jδ (ω + 3)] π = 136 (9−ω(2)++j(2ω −)(9jω−ω)(22 )+ jω ) − 26 [(6 + j 4)δ (ω − 3) + (6 − j 4)δ (ω + 3)] 6 = (2+ jω )(9 − 13π [ (3 + j 2)δ (ω − 3) + (3 − j 2)δ (ω + 3)] −ω 2 ) 2
which is the required result.
▌
Problem 5.11 ∞
F {x(at )} =
From the definition of CTFT,
∫ x(at )e
− jω t
dt .
−∞
We consider two different cases (a > 0) and (a < 0) Case 1:
Assume a > 0. Substitute r = at in the above expression. The upper and lower limits of integration stay the same and dr = a dt. The final result is F {x(at )}
Case 2:
∞
∞
− j (ω ) r = x(r )e a dra −∞
∫
=
1 a
∫ x ( r )e
− j ( ωa ) r
−∞
dr = 1a X
(ωa ) .
Assume a < 0. Substitute r = at in the above expression. The upper limit of integration is r → −∞ and the lower limit of integration is r → ∞ , and dr = a dt. The final result is F {x(at )} =
∞
∫
x ( r )e
−∞
− j ( ωa ) r dr a
−∞
=
1 a
∫
x ( r )e
− j ( ωa ) r
∞
dr = −
∞
Combining the two cases, yields F {x(at )} =
1 a
X
1 a
∫ x ( r )e
−∞
(ωa ) .
− j ( ωa ) r
dr = −
1 a
X
(ωa ) . ▌
Problem 5.12
Comparing with Fig. 5.9(a), we observe that h(t ) = x1
H (ω) = 2 X 1 (2ω)
Using the scaling property, or, which simplifies to
(2t ) .
H (ω) =
2 ω2
[2ω sin( 4ω) + cos(2ω) − 1] ,
H (ω) = 16sinc
(4πω ) − 4sinc 2 (ωπ ) .
▌
Solutions
171
Problem 5.13
Using the definition of CTFT, we obtain
{
Fe
jω 0 t
∞
} ∫
x(t ) = e
jω 0 t
x(t )e
− jω t
−∞
∞
dt =
∫ x(t )e
− j ( ω − ω0 ) t
dt = X (ω − ω 0 ) .
▌
−∞
Problem 5.14
Using the convolution property,
[
x(t ) ∗ u (t ) ← → X (ω) 2πδ(ω) + CTFT
1 jω
],
∞
∫ x(τ )u(t − τ )dτ ←→ 2π X (0)δ (ω ) + CTFT
or,
X (ω ) jω
,
−∞
∞
∫ x(τ )u(−(τ − t ))dτ ←→ 2π X (0)δ (ω ) + CTFT
or,
X (ω ) jω
,
−∞
t
∫ x(τ )dτ ←→ 2π X (0)δ (ω ) + CTFT
or,
X (ω ) jω
.
−∞
Problem 5.15
(a)
→ Using the time scaling property, x(2t ) ← CTFT
1 2
X
(ω2 ).
CTFT → Using the frequency shifting property, e − j 5t x(2t ) ←
1 2
X ( ω2+ 5 ) .
Substituting the value of X(ω), we obtain
1 − 3 ℑ{e − j 5t x(2t )} = 12 0 ω12+11 −ω = 112 0
ω +5
(b)
ω +5 ≤ 3 elsewhere −11 ≤ ω ≤ −5 −5 ≤ ω ≤ −1 elsewhere.
Using the frequency differentiation property, CTFT ( jt ) 2 x(t ) ← →
d2X dω2
,
CTFT t 2 x(t ) ← → − ddωX2 . 2
or, The CTFT of t2 x(t) is given by
{
}
F t 2 x(t ) = −
d2 dω 2
[∆(ω3 )] = − ddω [rect(ω3 )] = −[δ(ω + 3) − δ(ω − 3)] = [δ(ω − 3) − δ(ω + 3)] .
▌
172
(c)
Chapter 5 =t Express (t + 5) dx dt
dx dt
+ 5 dx . dt
Using the time differentiation property, the CTFT of dx dt
dx dt
is given by
CTFT ← → jωX (ω) .
Applying the frequency differentiation property to the above CTFT pair, gives
t
dx dt
CTFT ← → j ddω [ jωX (ω)] = − X (ω) − ω dX dω .
The CTFT of ( t + 5 ) dx dt is given by
{
}
ℑ (t + 5) dx = − X (ω) − ω dX + 5 jωX (ω) . dt dω
Substituting the value of X(ω), we obtain
{
ℑ (t + 5)
(d)
dx dt
}
( (
) ( ) (
j 5ω 1 − ω3 − 1 − = j 5ω 1 + ω3 − 1 + 0
2ω 3 2ω 3
) )
0≤ω≤3 −3≤ω≤0 elsewhere.
Using the time multiplication property, x(t ) ⋅ x(t ) ← → 21π [ X (ω) ∗ X (ω)] , CTFT
which implies that F {x(t ) ⋅ x(t )} =
(e)
1 2π
[∆(ω3 )∗ ∆(ω3 )].
Using the time convolution property, CTFT x(t ) * x(t ) ← → X (ω) ⋅ X (ω) ,
which reduces to 2 ω 1 − 3 F {x(t ) ∗ x(t )} = 0
(f)
ω ≤3
1 + = elsewhere
ω2 9
x(t ) ⋅ cos ω 0 t ← → 21π X (ω) ∗ πδ(ω − ω 0 ) +
1 2π
−
2ω 3
0
ω ≤3 elsewhere.
Using the time multiplication property, CTFT
or,
x(t ) ⋅ cos ω 0 t ← → 12 X (ω − ω 0 ) + CTFT
1 2
X (ω) ∗ πδ(ω + ω 0 ) , X (ω + ω 0 )
Case I: For ω0 = 3/2, we obtain
(
x(t ) ⋅ cos(3t / 2) ← → 12 X ω − CTFT
3 2
) + 12 X (ω + 32 ).
The two replicas overlap over (−3/2 ≤ ω < 3/2), therefore,
Solutions
12 + ω+63 / 2 1 F {x(t ) cos(3t / 2)} = 1 ω−3 / 2 2 + 6 0
173
− 92 ≤ ω ≤ − 32 − 32 ≤ ω ≤ 32 3 ≤ω≤ 9 2 2 elsewhere.
Case II: For ω0 = 3, we obtain x(t ) ⋅ cos 3t ← → 12 X (ω − 3) + CTFT
1 2
X (ω + 3) .
Since there is no overlap between the two shifted replicas, 1 − ω+3 3 ω−3 1 F {x(t ) cos 3t} = 2 1 − 3 0
or,
1 − ω+ 3 6 2 1 ω− 3 F {x(t ) cos 3t} = 2 − 6 0
ω+3 ≤3 ω−3 ≤3 elsewhere. −6≤ω<0 0≤ω<6 elsewhere.
Case III: For ω0 = 6, we obtain x(t ) ⋅ cos 6t ← → 12 X (ω − 6 ) + CTFT
1 2
X (ω + 6 ) .
Since there is no overlap between the two shifted replicas, 1 − ω+ 6 3 ω− 6 1 F {x(t ) cos 3t} = 2 1 − 3 0
or,
1 − ω+ 6 6 2 1 ω− 6 F {x(t ) cos 3t} = 2 − 6 0
ω+6 ≤3 ω−6 ≤3 elsewhere. − 9 ≤ ω < −3 3≤ω<9 elsewhere.
Problem 5.16
(a)
From Table 5.2,
5 2 + jω
inverse CTFT ← → 5e −2t u (t ) .
Using the frequency shifting property, 5 2 + j ( ω− 5 )
implying that (b)
From Table 5.2,
[
]
← → 5e − 2t u (t ) × e j 5t inverse CTFT
x1 (t ) = 5e ( −2+ j 5)t u (t ) . CTFT cos(2t ) ← → π[δ(ω − 2) + δ(ω + 2)]. .
▌
174
Chapter 5 Using the duality property, CTFT π[δ(t − 2) + δ(t + 2)] ← → 2π cos(− 2ω) = 2π cos(2ω) .
Using the frequency shifting property, π[δ(t − 2) + δ(t + 2)] e
− j 12π t
(
)
CTFT π , ← → 2π cos 2(ω + 12 )
implying that x 2 (t ) =
or, (c)
1 2
[δ(t − 2) + δ(t + 2)] e − j
π 12
t
−j πt −jπt = 12 δ(t − 2)e 12 + δ(t + 2)e 12
jπ j 2π − jπ x 2 (t ) = 12 δ(t − 2)e 6 + δ(t + 2)e 6 = 14 ( 3 − j )δ(t − 2) + ( 3 + j )δ(t + 2)e 3 .
From Table 5.2,
4 πω / 2 π ) CTFT rect ( 4t )← → 4sinc( 42ωπ ) = 4 sin( = 2 sin(ω2ω) . ( 4 πω / 2 π )
Using the time scaling property, CTFT rect ( 4t⋅2 ) ← → 2 ⋅ 2 sin(2 ω4 ω) = 2 sin(ω4 ω) .
Using the frequency shifting property, rect
sin( 4 ( ω− π )) (8t )e jπt ←CTFT → 2 ( ω− π) ,
implying that x3 (t ) = 52 rect
(d)
(8t )e jπt .
Using the linearity property, we obtain
x4 (t ) = ℑ−1 {(3 + 2 j ) δ (ω − 10) + (1 − 2 j )δ (ω + 10)} = (3 + 2 j ) ℑ−1 {δ (ω − 10)} + (1 − 2 j )ℑ−1 {δ (ω + 10)} =
(3+ 2 j ) 2π
e j10t + (1−2π2 j ) e − j10t .
Expanding the exponential terms using the Euler’s formula, we obtain x 4 (t ) =
(cos 10t + j sin 10t ) +
x 4 (t ) = π2 cos 10t −
or, (e)
( 3+ 2 j ) 2π
(1− 2 j ) 2π
( 2− j ) π
(cos 10t − j sin 10t )
sin 10t .
Taking the partial fraction expansion X 5 (ω) = where
1 (1+ jω) (3+ jω) 2 (5+ ω2 )
≡
A (1+ jω)
+
B (3+ jω)
+
C (3+ jω) 2
+
jDω+ E ( 5 + ω2 )
A = 0.0625, B = 0.25, C = 0.125, D = −0.3125, and E = 0.6876 .
Calculating the inverse CTFT transform yields
(
)
x5 (t ) ≅ Ae −t u (t ) + Be −3t u (t ) + Cte −3t u (t ) + D cos( 5t )u (t ) + E / 5 sin( 5t )u (t ) .
▌
Solutions
175
Problem 5.17
(i) The functions are plotted in Fig. S5.17. The MATLAB code used to generate the plots is given below. % MATLAB code to plot the functions in Problem 5.17 t = -10:0.01:10 ; t4 = 0:0.001:10000 ; % for plotting x4(t) t = t + eps; t4 = t4 + eps; %for plotting x4(t) % x1 = exp(-2*abs(t)); % a=2 x2 = exp(-2*t).*(cos(5*t)).*(t>=0); % a=2, w=5 x3 = (t.^4).*exp(-2*t).*(t>=0); % a=2 x4 = sin(log(t4)); x5 = 1./t ; x6 = cos(pi./(2*t)) ; x7 = exp(-(t.^2)./(2*3*3)) ; % sigma=3 % subplot(4,2,1), plot(t, x1), grid xlabel('t') % Label of X-axis ylabel('x1(t)') % Label of Y-axis axis([-5 5 0 1.3]) % subplot(4,2,3), plot(t, x2), grid xlabel('t') % Label of X-axis ylabel('x2(t)') % Label of Y-axis % axis([-5 5 -0.5 1.3]) % subplot(4,2,4), plot(t, x3), grid xlabel('t') % Label of X-axis ylabel('x3(t)') % Label of Y-axis % axis([-4 4 0 0.4]) % subplot(4,2,5), plot(t4, x4), grid xlabel('t') % Label of X-axis ylabel('x4(t)') % Label of Y-axis % axis([0.001 10000 -1.3 1.3]) % subplot(4,2,6), plot(t, x5), grid xlabel('t') % Label of X-axis ylabel('x5(t)') % Label of Y-axis axis([-1 1 -100 100]) % subplot(4,2,7), plot(t, x6), grid xlabel('t') % Label of X-axis ylabel('x6(t)') % Label of Y-axis % axis([-5 5 -1.3 1.3]) % subplot(4,2,8), plot(t, x7), grid xlabel('t') % Label of X-axis ylabel('x7(t)') % Label of Y-axis % axis([-5 5 0 1.3])
176
Chapter 5
x1(t)
1
0.5
0 -5
-4
-3
-2
-1
0 t
1
2
3
4
5
0.4 1
0.3 x3(t)
x2(t)
0.5 0
0.2 0.1
-0.5 -5
-4
-3
-2
-1
0 t
1
2
3
4
0 -4
5
-3
-2
-1
0 t
1
2
3
4
100 1 50 x5(t)
x4(t)
0.5 0
0
-0.5
-50
-1 1000
2000
3000
4000
5000 t
6000
7000
8000
-100 -1
9000 10000
-0.8
-0.6
-0.4
-0.2
0 t
0.2
0.4
0.6
0.8
1
-4
-3
-2
-1
0 t
1
2
3
4
5
1 1 x7(t)
x6(t)
0.5 0 -0.5
0.5
-1 -5
-4
-3
-2
-1
0 t
1
2
3
4
0 -5
5
Fig. S5.17: Time-domain Waveforms for Problem 5.17 (ii) ∞
∫
(a)
∞
x1(t ) dt =
−∞
∫
e
0
−a t
dt =
−∞
∫
−∞
∞
eat dt + ∫ e− at dt = a1 eat 0
0 −∞
+ ( −1a ) e− at
∞ 0
=
1 a
[1 − 0] − a1 [0 − 1] = a2 < ∞.
Since the condition in Eq. (5.59) is satisfied, the CTFT for x1(t) exists. (b)
∞
∫
−∞
∞
x 2(t ) dt =
∫
e− at cos(ω0t )u(t ) dt =
−∞
∞
1 2
∫ 0
e− at e jω0t + e− jω0t dt ≤
∞
1 2
∞
− ( a − jω ) t − ( a + jω ) t ∫ e 0 dt + 12 ∫ e 0 dt 0
0
I
II
Integral I is given by ∞
I=
1 2
∫e
−( a − jω0 ) t
0
while Integral II is given by
dt = 12
[
]
∞ e − ( a − jω 0 ) t −( a − jω0 ) 0
[
= 12 0 −
1 − ( a − jω0 )
]= [
1 1 2 ( a − jω0 )
],
Solutions ∞
II =
1 2
∫
e −( a + jω0 )t dt = 12
[
0
] = [0 −
− ( a + jω 0 ) t ∞ e − ( a + jω0 ) 0
1 2
1 − ( a + jω0 )
]= [
1 1 2 ( a + jω0 )
177
].
Therefore, ∞
∫x
4 (t ) dt
= I + II = 12
1 ( a − jω0 )
−∞
+
=
1 1 2 ( a + jω0 )
1 a 2 + ω02
< ∞.
Since the condition in Eq. (5.59) is satisfied, the CTFT for x2(t) exists. ∞
(c)
∫
∞
x3(t ) dt =
−∞
∫
∞
t e u (t ) dt = ∫ t 4e− at dt = t 4 4 − at
−∞
e− at (−a)
− at
+ 4t 3 ( e− a )2 + 12t 2
e− at ( − a )3
0
= [ 0 + 0 + 0 + 0 + 0] − 0 + 0 + 0 + 0 + 24 ( −1a )5 =
24 a5
∞
+ 24t (e− a )4 + 24 (e− a )5 0 − at
− at
< ∞.
Since the condition in Eq. (5.59) is satisfied, the CTFT for x3(t) exists. (d)
The function x 4(t ) = sin(ln(t ))u(t )
is plotted in Fig. S5.17. Note that the horizontal axis uses a logarithmic scale. It is observed that the function oscillates like a sine wave (although not with a constant period). Therefore, the function has an infinite number of maximas and minimas. In addition, ∞
∫
−∞
∞
x4(t ) dt = ∫ sin(ln(t )) dt →∞ 0
Therefore, the CTFT for x4(t) does not exist. ∞
(e)
∫
∞
x5(t ) dt =
−∞
∫
−∞
∞
1 t
dt = 2∫ 1t dt = 2 [ ln(t )]0 →∞ ∞
0
Since the condition in Eq. (5.59) is not satisfied, the CTFT for x5(t) does not exist. ∞
(f)
∫
∞
x 6(t ) dt =
−∞
∫ cos ( π ) dt → ∞. /2 t
−∞
Clearly the area enclosed by the cosine term would be infinite. Since the condition in Eq. (5.59) is not satisfied, the CTFT for x6(t) does not exist. Also, it can be checked that x6(t) has an infinite number of maximas and minimas, which is a second violation of the existence of the CTFT. ∞
(g)
∫
−∞
∞
x7(t ) dt =
∫
−∞
∞
exp( − 2tσ 2 ) dt = 2
∫ exp(−
−∞
t2 2σ 2
)dt = 2π σ < ∞ .
In evaluating the above result, we used the fact that the area enclosed by a bell curve is 1. Mathematically, this implies that ∞
∫
∞
1 2 πσ
[ ] 2
exp 2tσ 2 dt =
∞
∫
∞
1 2 πσ
[ ] 2
exp (t2−σm2) dt = 1 ,
where m is a constant. Since the condition in Eq. (5.59) is satisfied, the CTFT for x7(t) exists.
▌
178
Chapter 5
Problem 5.18
(a)
From the solution of Problem P4.11(a), the CTFS coefficients Dn of the rectangular pulse train are obtained as
3 2 Dn = 0 3 jnπ
n=0 even n, n ≠ 0 odd n,
with fundamental frequency ω0 = 1 radians/s. Therefore, the CTFT is given by
X 1(ω) = 2π
(b)
∞
∞
n = −∞
n = −∞ odd n
∑ Dn δ(ω − nω0 ) =3πδ(ω) − j6 ∑ δ(ω − n) .
From the solution of Problem P4.11(b), the CTFS coefficients Dn of the rectangular pulse train are obtained as Dn = −
3 4
n=0 n≠0
0.5 sin(0.5nπ) nπ
with fundamental frequency ω0 = π/T radians/s. Therefore, the CTFT is given by X 2(ω) = 2π
∞
∑
Dn δ(ω − nω 0 ) =1.5πδ(ω) −
n = −∞
(c)
∞
∑ 1n sin(0.5nπ)δ(ω − nTπ ) .
n = −∞ n≠0
From the solution of Problem P4.11(c), the CTFS coefficients Dn of the rectangular pulse train are obtained as 1 , n=0 Dn = 12 j 2 nπ , n ≠ 0. with fundamental frequency ω0 = 2π/T radians/s. Therefore, the CTFT is given by X 3(ω) = 2π
∞
∑
n = −∞
(d)
∞
Dn δ(ω − nω 0 ) =πδ(ω) − j
∑ 1n δ(ω − 2Tnπ ) .
n = −∞ n≠0
From the solution of Problem P4.11(d), the CTFS coefficients Dn of the rectangular pulse train are obtained as 1, 2 Dn = 0, 2 ( nπ) 2
n=0 even n, n ≠ 0 odd n, n ≠ 0.
with fundamental frequency ω0 = π/T radians/s. Therefore, the CTFT is given by
Solutions
X 4(ω) = 2π
(e)
∞
∞
n = −∞
n = −∞ n≠0 odd n
179
∑ Dn δ(ω − nω0 ) =πδ(ω) + π4 ∑ n1 δ(ω − nTπ ) . 2
From the solution of Problem P4.11(e), the CTFS coefficients Dn of the rectangular pulse train are obtained as 12 (1 − π1 ) ∓ j π1 − 18 Dn = 1 2π ( n2 −1) 1 jnπ
(
n=0
0.3408 n = ±1 ∓ j 0.1933 = 0 ≠ n = even 0.1592 n2 −1 − j 0.3183 ± 1 ≠ n = odd n n=0
)
n = ±1 0 ≠ n = even ± 1 ≠ n = odd
with fundamental frequency ω0 = π/T radians/s. Therefore, the CTFT is given by X 5(ω) = 2π
∞
∑ Dnδ(ω − nω0 ) = 0.6816
n = −∞
= 0.6816πδ(ω) + j 0.3866πδ(ω + Tπ ) − j 0.3866πδ(ω − Tπ ) +
∞
∞
∑ n 1−1 δ(ω − nTπ ) − j 2 ∑ 1n δ(ω − nTπ )
.
2
n = −∞ n ≠ 0,1 even n
n = −∞ n ≠ 0,1 odd n
▌ Problem 5.19
(a)
From the solution of Problem P5.2(a), the CTFT of the aperiodic signal is given by
3π ω=0 X 1 (ω) = 3πe − jωπ / 2 sinc(ω / 2) = 3 − jωπ ) ω ≠ 0. jω (1 − e The signal shown in Fig. P4.6(a) is a periodic signal with a fundamental period T0 = 2π with one period matching the function shown in Fig. P5.2(a). The fundamental frequency ω0 = 1 and the exponential CTFS coefficients are given by
Dn =
1 T0
X 1 (ω)
ω= nω0
=
1 2π
3 3π n = 0 n=0 2 = 3 − jnω0 π ) n ≠ 0 j 23πn (1 − e − jnπ ) n ≠ 0 jnω0 (1 − e
which simplifies to
3 2 Dn = = jn3π 0 (b)
n=0 odd n even n, n ≠ 0.
From the solution of Problem P5.2(b), the CTFT of the aperiodic signal is given by
(0.5πωT ) + Te − jωT sinc(0.5πωT ) = sin(ωT / 2)1.5T
X 2 (ω) = 0.5Tsinc
ω
(1 + 2e
− jωT
)
ω=0 ω ≠ 0.
180
Chapter 5 The signal shown in Fig. P4.6(b) is a periodic signal with a fundamental period T0 = 2T with one period matching the function shown in Fig. P5.2(b). The fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by
Dn = T1 X 2 (ω) 0
ω= nω0
=
1 2T
1.5T sin(nω0T / 2) 1 + 2e − jnω0T nω0
(
)
3 ω = 0 n=0 = sin(nπ / 2) 4 − jnπ ) n≠0 ω ≠ 0 2nπ (1 − e
which simplifies to
34 1 − Dn = = 21nπ 2nπ 0 (c)
n=0 n = 4k + 1 n = 4k + 3 even n, n ≠ 0.
From the solution of Problem P5.2(c), the CTFT of the aperiodic signal is given by
X 3 (ω) = 1 jω +
0.5T 1 ω2T
(1 − e
− jωT
)
ω=0 ω ≠ 0.
The signal shown in Fig. P4.6(c) is a periodic signal with a fundamental period T0 = T with one period matching the function shown in Fig. P5.2(c). The fundamental frequency ω0 = 2π/T and the exponential CTFS coefficients are given by
Dn =
1 T0
X 3 (ω)
ω= nω0
= 1 jnω0 − 1 T
0.5T 1 n 2ω02T
(1 − e
− jnω0T
)
ω = 0 = ω ≠ 0 j 21nπ −
0.5 1 4 n 2 π2
(1 − e
− j 2 nπ
)
n=0 n≠0
which simplifies to
0.5 Dn = 1 j 2nπ (d)
n=0 n ≠ 0.
From the solution of Problem P5.2(d), the CTFT of the aperiodic signal is given by
ω=0 T X 3 (ω) = Tsinc 2 ( 0.5πωT ) = 2 0.5ωT Tsinc ( π ) ω ≠ 0. The signal shown in Fig. P4.6(d) is a periodic signal with a fundamental period T0 = 2T with one period matching the function shown in Fig. P5.2(d). The fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by
Dn = T1 X 3 (ω) 0
ω= nω0
which simplifies to
=
1 2T
ω=0 T 0.5 n=0 = 2 2 0.5nω0T Tsinc ( π ) ω ≠ 0. 0.5sinc (0.5n) n ≠ 0 0.5 Dn = 0 2 ( nπ) 2
n=0 even n, n ≠ 0 odd n, n ≠ 0.
Solutions
181
(e) From the solution of Problem P5.2(e), the CTFT of the aperiodic signal is obtained as T (1 − π1 ) X 5 (ω ) = ± 2jπT ∓ 4Tj 1 − jωT − π 20.5−ωπ2TT 2 1 + e− jωT jω 1 − e
ω=0 ω = ± πT otherwise
The signal shown in Fig. P4.6(e) is a periodic signal with a fundamental period T0 = 2T and whose one period is identical to the function shown in Fig. P5.2(e). Therefore, the fundamental frequency ω0 = π/T and the exponential CTFS coefficients are given by Dn = T10 X (nω0 ) =
1 2T
X (nω0 )
T (1 − 1 ) π 1 1 = 2T × ± jω0 1 − e∓ jω0T ∓ 0.52 jT 1 − jnω0T πT − π 2 −0.5 1 + e− jnω0T jnω0 1 − e ( nω0 )2 T 2 T (1 − π1 ) = 21T × ± Tjπ 1 − e∓ jπ ∓ 0.52 jT T − jnπ − jnπ 0.5π T jnπ 1 − e − π 2 −n2π 2 1 + e 12 − 21π = ± j1π ∓ 0.52jT 1 n n 1 j 2nπ 1 − (−1) + 4π ( n2 −1) 1 + (−1) 12 (1 − π1 ) ∓ j π1 − 18 = 1 2π ( n2 −1) 1 jnπ
(
n=0 n = ±1 otherwise n=0 n = ±1 otherwise n=0 n = ±1 otherwise n=0
)
n = ±1 0 ≠ n = even ± 1 ≠ n = odd
▌ Problem 5.20
(a)
Calculating the CTFT of both sides and applying the time differentiation property, yields
( jω ) Y (ω ) + 6 ( jω ) Y (ω ) + 11( jω ) Y (ω ) + 6Y (ω ) = X (ω ) , 3
2
or,
(( jω ) + 6 ( jω )
or,
H (ω ) =
3
2
)
+ 11( jω ) + 6 Y (ω ) = X (ω ) ,
Y (ω ) 1 = . 3 X (ω ) ( jω ) + 6 ( jω )2 + 11( jω ) + 6
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as H (ω) =
−1 1 0.5 0 .5 ≡ + + (1 + jω)(2 + jω)(3 + jω) (1 + jω) (2 + jω) (3 + jω)
182
Chapter 5 Calculating the inverse CTFT, we obtain h(t ) = 0.5e −t u (t ) − e −2t u (t ) + 0.5e −3t u (t ) .
(b)
Calculating the CTFT of both sides and applying the time differential property, yields
( jω)2 Y (ω) + 3( jω)Y (ω) + 2Y (ω) = X (ω) , or,
(( jω)
or,
H (ω) =
2
)
+ 3( jω) + 2 Y (ω) = X (ω) , Y (ω) 1 . = 2 X (ω) ( jω) + 3( jω) + 2
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as H (ω) =
1
( jω )
2
+ 3( jω) + 2
≡
1 1 − (1 + jω) (2 + jω)
Calculating the inverse CTFT, we obtain
h(t ) = e −t u (t ) − e −2t u (t ) . (c)
Calculating the CTFT of both sides and applying the time differentiation property, yields
( jω)2 Y (ω) + 2( jω)Y (ω) + Y (ω) = X (ω) , or,
(( jω)
or,
H (ω) =
2
)
+ 1( jω) + 1 Y (ω) = X (ω) , Y (ω) 1 . = 2 X (ω) ( jω) + 2( jω) + 1
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as 1 H (ω) = (1 + jω)2 Calculating the inverse CTFT, we obtain h(t ) = te −t u (t ) . (d)
Calculating the CTFT of both sides and applying the time differentiation property, yields
( jω)2 Y (ω) + 6( jω)Y (ω) + 8Y (ω) = ( jω)X (ω) + 4 X (ω) , or,
(( jω)
or,
H (ω) =
2
)
+ 6( jω) + 8 Y (ω) = (( jω) + 4)X (ω) ,
( j ω) + 4 1 Y (ω) = . = 2 X (ω) ( jω) + 6( jω) + 8 2 + jω
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which is given by
Solutions
183
h(t ) = e −2t u (t ) . (e)
Calculating the CTFT of both sides and applying the time differential property, yields
( jω)3 Y (ω) + 8( jω)2 Y (ω) + 19( jω)Y (ω) + 12Y (ω) = X (ω) , or,
(( jω)
+ 8( jω)2 + 19( jω) + 12 Y (ω) = X (ω) ,
or,
H (ω) =
Y (ω) 1 . = 3 X (ω) ( jω) + 8( jω)2 + 19( jω) + 12
3
)
The impulse response h(t) can be obtained by calculating the inverse CTFT of H(ω), which can be expressed as H (ω) =
1
( j ω)
3
+ 8( jω) + 19( jω) + 12 2
≡
1/ 6 − 1/ 2 1/ 3 + + (1 + jω) (3 + jω) ( 4 + jω)
Calculating the inverse CTFT, we obtain h(t ) = 16 e −t u (t ) − 12 e −3t u (t ) + 13 e −4t u (t ) .
▌
Problem 5.21
(a)
Calculating the CTFT of the input and output signals, we obtain X (ω) =
1 2 + jω
and
Y (ω) =
5 . 2 + jω
The transfer function is given by H (ω) =
Y (ω) = 5. X (ω)
Calculating the inverse CTFT, the impulse response is given by h(t ) = 5 δ(t ) . In the frequency domain, the input-output relationship is given by
Y (ω) = 5 X (ω) y (t ) = 5 x(t ) .
or, in the time domain, (b)
Calculating the CTFT of the input and output signals, we obtain X (ω) =
1 2 + jω
and
Y (ω) =
3 e − j 4ω . 2 + jω
The transfer function is given by H (ω) =
Y (ω) = 3e − j 4ω . X (ω)
Calculating the inverse CTFT, the impulse response is given by h(t ) = 3 δ(t − 4) .
184
Chapter 5 In the frequency domain, the input-output relationship is given by Y (ω) = 3e − j 4ω X (ω)
y (t ) = 3x(t − 4) .
or, in the time domain, (c)
Calculating the CTFT of the input and output signals, we obtain X (ω) =
1 2 + jω
and
Y (ω) =
6 ( 2 + jω) 4
.
The transfer function is given by H (ω) =
Y (ω) 6 . = X (ω) (2 + jω) 3
Taking the inverse CTFT, the impulse response is given by
h(t ) = 3 t 2 e −2t u (t ) . In the frequency domain, the input-output relationship is given by (2 + jω) 3 Y (ω) = 6 X (ω) , or,
[(8 + 12( jω) + 6( jω)
2
]
+ ( jω) 3 Y (ω) = 6 X (ω) .
Calculating the inverse CTFT, the resulting differential equation is obtained as
d3y dt (d)
3
+6
d2y dt
2
+ 12
dy + 8 y (t ) = 6 x(t ) . dt
Calculating the CTFT of the input and output signals, we get X (ω) =
1 2 + jω
and
Y (ω) =
1 1 + . (1 + jω) (3 + jω)
The transfer function is given by H (ω) =
Y (ω) (4 + 2 jω)(2 + jω) 2(2 + jω) 2 = = . (1 + jω)(3 + jω) (1 + jω)(3 + jω) X (ω)
Using partial fraction expansion, the transfer function is given by H (ω) =
2(2 + jω) 2 1 1 ≡2+ − (1 + jω) (3 + jω) (1 + jω)(3 + jω)
Taking the inverse CTFT, the impulse response is given by h(t ) = 2δ(t ) + e −t u (t ) + e −3t u (t ) . In the frequency domain, the input-output relationship is given by (1 + jω)(3 + jω)Y (ω) = 2(2 + jω) 2 X (ω) , or,
[(3 + 4( jω) + ( jω) ] Y (ω) = 2[(4 + 4( jω) + ( jω) ] X (ω) . 2
2
Solutions Calculating the inverse CTFT, the resulting differential equation is obtained as d2y dy d 2x dx + 4 + 3 y ( t ) = 2 + 8 + 8 x(t ) . 2 2 dt dt dt dt
185
▌
Problem 5.22
The transfer function of the RC series circuit is given by H (ω) =
1 /( jωC ) 1 1 1 = = × . R + 1 /( jωC ) (1 + jωCR ) CR 1 /(CR ) + jω
Calculating the inverse CTFT, the impulse response of the system is obtained as h(t ) =
1 × e −t /(CR ) . CR
The output response is calculated by convolving the input signal with the impulse response h(t) in the time domain. Figure S5.22 shows the convolution using graphical approach. We consider the three cases separately: Case I (t <= −T/2):
Since there is no overlap between h(τ) and v(t − τ), the output y(t) is 0.
Case II (−T/2 < t <= T/2):
y (t ) = Case III (t > T/2):
1 CR
The output y(t) is given by t +T / 2
∫
e−τ /(CR ) dt =
0
t +T / 2 1 (−CR) e−τ /(CR ) = 1 − e−T /(2CR ) e− t /(CR ) . 0 CR
The output y(t) is given by
1 y (t ) = CR
t +T / 2 − τ /( CR )
∫e
t −T / 2
dt =
1 (−CR )e − τ /(CR ) CR
t +T / 2 t −T / 2
[
]
= e T /( 2CR ) − e −T /( 2CR ) e −t /(CR ) .
Combining the three cases, we obtain 0 −T /(2 CR ) − t /( CR ) y (t ) = 1− e e T /(2CR ) −T /(2CR ) − t /( CR ) e −e e
For T = 2, R = 1M Ω, C = 1µ F ,
t ≤ −T / 2 −T / 2 < t ≤ T / 2 t >T /2
0 y (t ) = 1 − e − (t +1) (e − 1/ e) e − t ≈ 2.3504
t ≤ −1 −1 < t ≤ 1 t >1
The above output y (t ) is plotted in the last row of Fig. S5.22. The output response matches our expectation from our circuit theory knowledge. At t = −T / 2 , the input voltage becomes 1 volt, and the capacitor starts charging resulting in an increase in the output voltage. The increase continues until t = T / 2 at which the input becomes zero. After t = T / 2 , the capacitor starts discharging resulting in an ▌ exponential decrease of the output voltage. The output voltage becomes zero at t = ∞ .
186
Chapter 5
h(τ) 1 CR
1 e − τ / CR CR
h(τ)
τ
0
1
v(τ)
v(τ) 0
− T2
τ
T 2
v(t − τ)
1
v(t − τ) t − T2
τ
0
t + T2
h(τ) v(t − τ) 1 CR
Case I: (t < −T/2) t − T2
1 e − τ / CR CR
τ
0
t + T2
h(τ) v(t − τ) 1 CR
Case II: (−T/2 < t < T/2)
1 e − τ / CR CR
τ
0 t+T 2
t − T2
h(τ) v(t − τ)
Case III: (t > T/2)
1 CR
1 e − τ / CR CR
0
y (t ) ( for T = 2, R = 1 MΩ, C = 1 µ F)
t − T2
t + T2
τ
Solutions Figure S5.22: Convolution of the input signal v(t) with the impulse response h(t) in Problem 5.22.
Program: The MATLAB code to plot y(t) in Problem 5.22 t = -2:0.001:3; % P5.20(a) y = 0*(t<=-1)+(1-exp(-t-1)).*(t>-1).*(t<=1)+(exp(1)-exp(-1))*exp(-t).*(t>1); plot(t,y); grid on; xlabel('t'); ylabel('y(t)');
Problem 5.23 (i)
As determined in Problem 5.22, the transfer function of the RC series circuit is given by H (ω) =
1 1 × . CR 1 /(CR ) + jω
Calculating the CTFT of the input, we obtain
X 1 (ω) = π[δ(ω − ω 0 ) + δ(ω + ω 0 )] . Using the modulation property, the CTFT of the output signal is obtained as Y (ω) = H (ω) X 1 (ω) =
π π 1 1 × δ(ω − ω 0 ) + × δ( ω + ω 0 ) CR 1 /(CR ) + jω CR 1 /(CR ) + jω
which reduces to
or,
Y (ω) =
π 1 π 1 × δ( ω − ω 0 ) + × δ( ω + ω 0 ) , CR 1 /(CR ) + jω 0 CR 1 /(CR ) − jω 0
Y (ω) =
π 1 /(CR ) − jω 0 π 1 /(CR ) + jω 0 × δ( ω − ω 0 ) + × δ( ω + ω 0 ) , 2 2 CR 1 /(CR ) + ω 0 CR 1 /(CR ) 2 + ω 02 1 /(CR ) π [δ(ω − ω0 ) + δ(ω + ω0 )] × CR 1 /(CR ) 2 + ω 02 − jω 0 π [δ(ω − ω0 ) − δ(ω + ω0 )] . + × CR 1 /(CR ) 2 + ω 02
Y (ω) =
or,
Calculating the inverse CTFT, we obtain Y (ω) =
− jω 0 1 /(CR ) 1 1 × cos( ω t ) + × j sin(ω 0 t ) , 0 CR 1 /(CR ) 2 + ω 02 CR 1 /(CR ) 2 + ω 02
or, y (t ) = which can be expressed as
1 2
1 + C R 2 ω 02
[cos(ω0 t ) + ω0 CR sin(ω0 t )] ,
187
188
Chapter 5
y (t ) = (b)
1 2
1+ C R
2
ω 02
[
]
cos ω 0 t − tan −1 (ω 0 CR ) .
As determined in Problem 5.22, the transfer function of the RC series circuit is given by H (ω) =
1 1 × . CR 1 /(CR ) + jω
Calculating the CTFT of the input, we obtain X 1 (ω) = jπ[δ(ω − ω 0 ) − δ(ω + ω 0 )] . Using the modulation property, the CTFT of the output signal is given by Y (ω) = H (ω) X 1 (ω) =
π 1 π 1 × δ( ω − ω 0 ) − × δ( ω + ω 0 ) jCR 1 /(CR ) + jω jCR 1 /(CR ) + jω
which reduces to
or,
Y (ω) =
π π 1 1 × δ( ω − ω 0 ) − × δ( ω + ω 0 ) , jCR 1 /(CR ) + jω 0 jCR 1 /(CR ) − jω 0
Y (ω) =
1 /(CR ) + jω 0 1 /(CR ) − jω 0 π π δ( ω − ω 0 ) − × δ( ω + ω 0 ) , × 2 2 jCR 1 /(CR ) + ω 0 jCR 1 /(CR ) 2 + ω 02 1 /(CR ) π [δ(ω − ω0 ) − δ(ω + ω0 )] × jCR 1 /(CR ) 2 + ω 02 − jω 0 π [δ(ω − ω0 ) + δ(ω + ω0 )] . + × jCR 1 /(CR ) 2 + ω 02
Y (ω) = or,
Calculating the inverse CTFT, we obtain Y (ω) =
ω0 1 /(CR ) 1 1 sin( ) × ω t − × cos(ω 0 t ) , 0 CR 1 /(CR ) 2 + ω 02 CR 1 /(CR ) 2 + ω 02
or, y (t ) =
1 2
1 + C R 2 ω 02
[sin(ω0 t ) − ω0 CR cos(ω0 t )] ,
which can be expressed as y (t ) =
1 1 + C 2 R 2 ω 02
[
Problem 5.24
The transfer functions obtained in Problem 5.20 are as follows: (a) H (ω ) = (b) H (ω ) =
1
( jω )
3
+ 6 ( jω ) + 11( jω ) + 6 2
1
( jω )
2
+ 3 ( jω ) + 2
]
sin ω 0 t − tan −1 (ω 0 CR ) .
▌
Solutions
1
(c) H (ω ) = (d)
( jω ) + 2 ( jω ) + 1 ( jω ) + 4 1 = . H (ω ) = 2 ( jω ) + 6 ( jω ) + 8 2 + jω
(e) H (ω ) =
2
1
( jω )
3
+ 8 ( jω ) + 19 ( jω ) + 12 2
The MATLAB code to plot the magnitude and phase spectra is given below: w = -5:0.001:5; % % P5.20(a) H = 1./((j*w).^3+6*(j*w).^2+11*(j*w)+6); subplot(5,2,1) plot(w,abs(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.20(a): |H_1(\omega)|'); axis tight subplot(5,2,2) plot(w,angle(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.20(a):
189
190
Chapter 5
subplot(5,2,8) plot(w,angle(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.20(d):
The spectra are plotted in Fig. S5.24.
▌
1
P5.20(a):
1
P5.20(a): |H (ω)|
0.1 5 0. 1 0.0 5 -5
-4
-3
-2
-1
0
1
2
3
4
5
2 0 -2 -5
-4
-3
-2
-1
0. 2 0. 1 -4
-3
-2
-1
0 1 ω (radia ns /s )
2
3
4
5
0. 6 0. 4 0. 2 -4
-3
-2
-1
0
1
2
3
4
3
1
2
3
1
2
3
1
2
3
-2 -4
-3
-2
-1
0
2 0 -2 -5
-4
-3
-2
-1
0
ω (rad ians /s )
0. 5
1
4
P5.20(d):
4
P5.20(d): |H (ω)|
0. 4 0. 3
-4
-3
-2
-1
0 1 ω (radia ns /s )
2
3
4
5
0 -1 -5
-4
-3
-2
-1
0
ω (rad ians /s ) P5.20(e):
0.0 6
2
5
5
2
0
-5
5
0.0 8 P5.20(e): |H (ω)|
1
2
ω (radia ns /s )
0.0 4 0.0 2 -5
3
ω (rad ians /s )
0. 8
0. 2 -5
2
2
0. 3
-5
1
3
3
P5.20(c): |H (ω)|
P5.20(b):
0. 4
-5
0
ω (rad ians /s )
P5.20(c):
2
P5.20(b): |H (ω)|
ω (radia ns /s )
-4
-3
-2
-1
0 1 ω (radia ns /s )
2
3
4
5
0 -2 -5
-4
-3
-2
Fig. S5.24: Gain and Phase responses for Problem 5.24. Problem 5.25
The transfer functions obtained in Problem 5.21 are as follows:
-1
0
ω (rad ians /s )
Solutions
(a) H (ω) =
Y (ω) = 5. X (ω)
(b) H (ω) =
Y (ω) = 3e − j 4ω . X (ω)
(c) H (ω) =
Y (ω) 6 . = X (ω) (2 + jω) 3
(d) H (ω) =
Y (ω) (4 + 2 jω)(2 + jω) 2(2 + jω) 2 = = . X (ω) (1 + jω)(3 + jω) (1 + jω)(3 + jω)
The MATLAB code to plot the magnitude and phase spectra is given below: w = -5:0.001:5; % % P5.21(a) H = 5*ones(size(w)); subplot(4,2,1) plot(w,abs(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.21(a): |H_1(\omega)|'); axis([-5 5 0 5.25]); subplot(4,2,2) plot(w,angle(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.21(a):
191
192
Chapter 5
xlabel('\omega (radians/s)'); ylabel('P5.21(d): |H_4(\omega)|'); axis tight subplot(4,2,8) plot(w,angle(H)); grid on; xlabel('\omega (radians/s)'); ylabel('P5.21(d):
The gain and phase spectra are plotted in Fig. S5.25.
▌ P5.21(a):
P5.21(a): |H1(ω)|
1 4
2
0 -5
-4
-3
-2
-1
0
1
2
3
4
0.5 0 -0.5 -1 -5
5
-4
-3
-2
-1
ω (radians/s)
0
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
1
2
3
4
5
ω (radians/s)
P5.21(b):
P5.21(b): |H2(ω)|
3 2 1 0 -5
-4
-3
-2
-1
0
1
2
3
4
5
2 0 -2 -5
-4
-3
-2
-1
0.6 0.4 0.2 -5
-4
-3
-2
-1
0
ω (radians/s)
P5.21(c):
P5.21(c): |H3(ω)|
ω (radians/s)
0 1 ω (radians/s)
2
3
4
5
2 0 -2 -5
-4
-3
-2
-1
0
ω (radians/s)
P5.21(d):
P5.21(d): |H4(ω)|
2.6 2.4 2.2
0.1 0 -0.1
2 -5
-4
-3
-2
-1
0
1
2
3
4
5
-5
-4
-3
-2
-1
ω (radians/s)
0
ω (radians/s)
Fig. S5.25: Gain and phase responses for Problem 5.25. Problem 5.26 Case II with input signal given by sin(ω0t):
Let us assume that the transfer function H(ω0) = A(ω0) + jB(ω0) at the fundamental frequency ω0 of the sine wave. From the Hermitian property, we note that the real component of A(ω0) of H(ω0) is even, while the imaginary component B(ω0) of H(ω0) is odd. Therefore, H(−ω0) = A(ω0) − jB(ω0). Using the modulation property, the CTFT of the output of the system is given by Y (ω) = H (ω) × jπ[δ(ω + ω 0 ) − δ(ω − ω 0 )] ,
Solutions
193
Y (ω) = jπ[H (−ω 0 )δ(ω + ω 0 ) − H (ω 0 )δ(ω − ω 0 )] .
or,
Expressing H(ω0) and H(−ω0) in terms of their real and imaginary components, we obtain Y (ω) = jπA(ω 0 )[δ(ω + ω 0 ) − δ(ω − ω 0 )] + πB (ω 0 )[δ(ω + ω 0 ) + δ(ω − ω 0 )] . Calculating the inverse CTFT, the output y(t) is obtained as
y (t ) = A(ω0 ) sin(ω0t ) + B(ω0 ) cos(ω0t ) = ( A(ω0 )) 2 + ( B(ω0 )) 2 sin {ω0t + tan −1 ( B(ω0 ) / A(ω0 ))} . = H (ω0 ) sin {ω0t + < H (ω0 )} where
H (ω 0 ) = ( A(ω 0 )) 2 + ( B(ω 0 )) 2
and < H (ω 0 ) = tan −1 ( B(ω 0 ) / A(ω 0 )) .
Case I with input signal given by cos(ω0t):
Using the modulation property, the CTFT of the output of the system is given by Y (ω) = H (ω) × π[δ(ω + ω 0 ) + δ(ω − ω 0 )] ,
Y (ω) = π[H (−ω 0 )δ(ω + ω 0 ) + H (ω 0 )δ(ω − ω 0 )] .
or,
Expressing H(ω0) and H(−ω0) in terms of their real and imaginary components, we get
Y (ω) = πA(ω 0 )[δ(ω + ω 0 ) + δ(ω − ω 0 )] − jπB(ω 0 )[δ(ω + ω 0 ) − δ(ω − ω 0 )] . Taking the inverse CTFT, the output y(t) is given by
y (t ) = A(ω0 ) cos(ω0t ) − B(ω0 ) sin(ω0t ) = ( A(ω0 )) 2 + ( B (ω0 ))2 cos {ω0t + tan −1 ( B (ω0 ) / A(ω0 ))} . = H (ω0 ) cos {ω0t + < H (ω0 )} where H (ω 0 ) = ( A(ω 0 )) 2 + ( B(ω 0 )) 2
and < H (ω 0 ) = tan −1 ( B(ω 0 ) / A(ω 0 )) .
The above result states that the output of an LTI system with real-valued impulse response and a sinusoidal signal at the input is another sinusoidal signal of the same fundamental frequency as the input. Only the magnitude and phase of the sinusoidal signal are modified. ▌ Problem 5.27
(i)
With x1(t) = cos(ω0t), the output of the RC circuit shown in Fig. P5.22 is given by y (t ) = H (ω 0 ) cos(ω 0 t + < H (ω 0 ) ) where
H (ω) =
1 . 1 + jωCR
Substituting the value of the magnitude and phase of H(ω0) at the fundamental frequency ω = ω0, the output is given by
194
Chapter 5
y (t ) = (ii)
(
1 1 + (ω 0 CR )
)
cos ω 0 t − tan −1 (ωCR ) .
2
As in part (i), the output of the RC circuit shown in Fig. P5.22 for x2(t) = sin(ω0t), is given by y (t ) = H (ω 0 ) sin (ω 0 t + < H (ω 0 ) ) 1 . 1 + jωCR
H (ω) =
where
Substituting the value of the magnitude and phase of H(ω0) at the fundamental frequency ω = ω0, the output is given by y (t ) =
1 1 + (ω 0 CR )
2
(
)
sin ω 0 t − tan −1 (ωCR ) .
The answers obtained above match with those obtained in Problem 5.23. Problem 5.28
(i)
Based on the solution of Problem 5.26, sin (3t ) → H (3) sin (3t + < H (3) ) .
For R = 1MΩ, and C = 0.1µF, the transfer function is given by
H (ω) =
1 . 1 + j 0.1ω
At ω = 3 radians/s, the magnitude and phase of the RC circuit is given by H (3) =
1 1 + 0.3
2
= 0.9578 and < H (3) = −16.700 .
The output is given by
(
)
y1 (t ) = 0.9578 sin 3t − 16.700 . (ii)
Based on the solution of Problem 5.26, cos(3t ) → H (3) cos(3t + < H (3) ) and
sin ( 6t + 300 ) → H (6) sin ( 6t + 300 + < H (6) ) At ω = 3 radians/s, the magnitude and phase of the RC circuit is given by H (3) =
1 1 + 0.3
2
= 0.9578 and < H (3) = −16.700 .
Similarly, at ω = 6 radians/s, the magnitude and phase of the RC circuit is given by
▌
Solutions
H (6) =
1 1 + 0.6
195
= 0.8575 and < H (6) = −30.960 .
2
Using the linearity property, the output is given by
(
)
(
)
y1 (t ) = 0.9578 cos 3t − 16.700 − 4.2875 sin 6t − 0.960 . (iii) Based on the solution of Problem 5.26, cos(2t ) → H ( 2) cos(2t + < H (2) ) and sin ( 2000t ) → H (2000) sin ( 2000t + < H (2000) )
At ω = 2 radians/s, the magnitude and phase of the RC circuit is given by H ( 2) =
1 1 + 0.3
= 0.9806 and < H (3) = −11.310 .
2
Similarly, at ω = 2000 radians/s, the magnitude and phase of the RC circuit is given by H (2000) =
1 1 + 2000
2
= 0.0050 and < H ( 2000) = −89.710 .
Using the linearity property, the output is given by
(
)
(
)
y1 (t ) = 0.9806 cos 2t − 11.3`0 + 0.0050 sin 2000t − 89.710 . (iv)
Based on Eq. (5.75) exp( j 3t ) → H (3) exp( j 3t + j < H (3) ) and exp( j 2000t ) → H (2000) exp( j 3t + j < H (2000) )
At ω = 3 radians/s, the magnitude and phase of the RC circuit is given by H (3) =
1 1 + 0.3
= 0.9578 and < H (3) = −16.700 .
2
Similarly, at ω = 2000 radians/s, the magnitude and phase of the RC circuit is given by H (2000) =
1 1 + 2000
2
= 0.0050 and < H (2000) = −89.710 .
Using the linearity property, the output is given by
(
)
(
)
y1 (t ) = 0.9578 exp j 3t − j16.700 + 0.0050 exp j 2000t − 89.710 .
▌
196
Chapter 5
Problem 5.29
(a)
In Example 3.6, it was shown that
[
]
y (t ) = e − t u (t ) ∗ e − 2t u (t ) = e − t − e − 2t u (t ).
(b)
From Table 5.2, the CTFT of x(t ) and h(t ) are obtained as
X (ω) = 1+1jω , and
H (ω) =
1 2 + jω
.
The CTFT of the output is then given by
Y (ω ) = H (ω ) X (ω ) = (1+1jω ) ( 2+1jω ) = 1+1jω − 2+1jω . Calculating the inverse CTFT results in the output signal
[
]
y (t ) = e − t − e − 2t u (t ). (c)
As H (ω) =
Y ( ω) X ( ω)
=
1 2 + jω
, the Fourier-domain input-output relationship can be expressed as jωY (ω) + 2Y (ω) = X (ω) .
Calculating the inverse CTFT of both sides results in the following differential equation dy + 2 y (t ) = x (t ) . dt
The output can be obtained by solving the differential equation with input x(t ) = e − t u (t ) and zero initial conditions y(0−) = 0. Zero-input Response: Due to zero initial condition, the zero-input response is yzi(t) = 0. Zero-state Response: The characteristics equation is given by (s + 2) = 0 resulting in a single pole at s = −2. The homogenous component of the zero-state response is given by h y zs (t ) = Ae −2t .
Since the input x(t) = exp(−t) u(t), the particular solution is of the form
y zsp (t ) = Ke − t for t ≥ 0 . Inserting the particular solution in the differential equation results in K = 1. Therefore,
y zsp (t ) = e − t u (t ) . The overall zero-state response is, therefore, given by y zs (t ) = Ae −2t + e − t
for t ≥ 0. To determine the value of A, we insert the initial condition y(0−) = 0 giving
A + 1 = 0 ⇒ A = −1 or, A = −1. The zero state response is given by
y zs (t ) = ( e − t − e −2t ) u (t )
.
Solutions
197
Total Response: By adding the zero-input and zero-state responses, the overall output is given by
y (t ) = y zi (t ) + y zs (t ) = ( e − t − e −2t ) u (t ). =0
It is observed that Methods (a) – (c) yield the same result.
▌
Problem 5.30
(a)
For part (a), we assume that T = 1 in H1 (ω ) and H 2 (ω ) . From the solution of Problem P5.19(a), the CTFT of Fig. P4.6(a) is given by X 1(ω) = 3πδ(ω) − j 6
∞
1 δ( ω − n ) . n n = −∞
∑
odd n
Output for H1(ω): Since H1(ω) eliminates all frequency components outside the range |ω| ≤ 4 (as T = 1), the output is given by Y 1(ω) = j 2δ(ω + 3) + j 6δ(ω + 1) + 3πδ(ω) − j 6δ(ω − 1) − j 2δ(ω − 3) . Output for H2(ω): Since H2(ω) eliminates all frequency components outside the range 4 ≤ |ω| ≤ 8, the output is given by Y 1(ω) = j 76 δ(ω + 7) + j 65 δ(ω + 5) − j 56 δ(ω − 1) − j 76 δ(ω − 3) . (b)
From the solution of Problem P5.19(b), the CTFT of Fig. P4.6(b) is given by
X 2(ω ) = 1.5πδ (ω ) −
∞
∑
n =−∞ n≠0
1 n
sin(0.5nπ )δ (ω − nTπ ) .
Output for H1(ω): Since H1(ω) eliminates all frequency components outside the range |ω| ≤ 4/T, the output is given by Y 2(ω) = −δ(ω + Tπ ) + 1.5πδ(ω) − δ(ω − Tπ ) . Output for H2(ω): Since H2(ω) eliminates all frequency components outside the range 4/T ≤ |ω| ≤ 8/T, the output is given by Y 2(ω) = 0 . (c)
From the solution of Problem P5.19(c), the CTFT of Fig. P4.6(c) is given by X 3(ω) = πδ(ω) − j
∞
∑ 1n δ(ω − 2Tnπ ) .
n = −∞ n≠0
Output for H1(ω): Since H1(ω) eliminates all frequency components outside the range |ω| ≤ 4/T, the output is given by Y 3(ω) = πδ(ω) . Output for H2(ω): Since H2(ω) eliminates all frequency components outside the range 4/T ≤ |ω| ≤ 8/T, the output is given by
198
Chapter 5
Y 3(ω) = jδ(ω + (d)
2π )− T
jδ(ω − 2Tπ ) .
From the solution of Problem P5.19(d), the CTFT of Fig. P4.6(d) is given by X 4(ω) = 2π
∞
∞
n = −∞
n = −∞ n≠0 odd n
∑ Dnδ(ω − nω0 ) =πδ(ω) + π4 ∑ n1 δ(ω − nTπ ) . 2
Output for H1(ω): Since H1(ω) eliminates all frequency components outside the range |ω| ≤ 4/T, the output is given by Y 4(ω) = π4 δ(ω + Tπ ) + πδ(ω) + π4 δ(ω − Tπ ) . Output for H2(ω): Since H2(ω) eliminates all frequency components outside the range 4/T ≤ |ω| ≤ 8/T, the output is given by Y 4(ω) = 0 . (e)
From the solution of Problem P5.19(e), the CTFT of Fig. P4.6(e) is given by
X 5(ω ) = 2π
∞
∑ D δ (ω − nω )
n =−∞
n
0
= 0.6816πδ (ω ) + j 0.3866πδ (ω + πT ) − j 0.3866πδ (ω − πT ) +
∞
∑
n =−∞ n≠0 even n
1 n 2 −1
δ (ω − nTπ )
∞
− j 2 ∑ 1n δ (ω − nTπ ). n =−∞ n ≠ 0,1 odd n
Output for H1(ω): Since H1(ω) eliminates all frequency components outside the range |ω| ≤ 4/T, the output is given by Y 5(ω) = j 0.3866πδ(ω + Tπ ) + 0.6816πδ(ω) − j 0.3866πδ(ω − Tπ ) . Output for H2(ω): Since H2(ω) eliminates all frequency components outside the range 4/T ≤ |ω| ≤ 8/T, the output is given by Y 5(ω) = 13 δ(ω +
2π ) + 13 δ(ω − 2Tπ ) . T
▌
Problem 5.31
(a)
The magnitude spectra of the two systems are calculated below
H1 (ω) = 1 H 2 (ω) = 0
400 + ω 2 400 + ω 2
=1
ω ≥ 20 elsewhere.
The magnitude spectra are plotted in Fig. S5.31. From Fig. S5.31(a), we observe that the magnitude |H1(ω)| is 1 at all frequencies. Therefore, System H1(ω) is an all pass filter.
Solutions
199
From Fig. S5.31(b), we observe that the magnitude |H2(ω)| is zero at frequencies below 20 radians/s. At frequencies above 20 radians/s, the magnitude is 1. Therefore, System H2(ω) is a highpass filter. H1 (ω)
1
− ωc
ω
ωc
0
H 2 (ω)
1
− 20
ω
0
(a)
20
(b) Fig. S5.31: Magnitude Spectra for Problem 5.31.
(b)
Calculating the inverse CTFT, the impulse response of the two systems is given by h1 (t ) = ℑ−1
{
40 − 20 − jω 20 + jω
}= ℑ { −1
40 20 + jω
}− ℑ
−1
{1 } = 40e − 20t u (t ) − δ(t ) .
ω ω 20 t h2 (t ) = ℑ−1 { 1 − rect( 40 ) } = ℑ−1 { 1 } − ℑ−1 { rect( 40 ) } = δ (t ) − 20 π sinc( π ) .
▌
Problem 5.32
The transfer functions for the three LTIC systems are given by System (a): System (b): System (c):
H1 (ω) =
2 . (1 + jω) 2
H 2 (ω) = πδ(ω) +
H 3 (ω) = −2 +
1 . jω
5 1 − j 2ω = 2 + jω 2 + jω
The following Matlab code generates the magnitude and phase spectra of the three LTIC systems. %MATLAB Program for Problem P5.32 %System (a) clear; num_coeff = [2]; denom_coeff = [1 2 1]; sys = tf(num_coeff,denom_coeff); figure(1) bode(sys,{0.02,100}); grid; title('Bode Plot for System-1') %System (b) clear; num_coeff = [1]; denom_coeff = [1 0]; sys = tf(num_coeff,denom_coeff); figure(2) bode(sys,{0.02,100}); grid; title('Bode Plot for System-2') %System (vc)
% % % %
clear the MATLAB environment NUM coeffs. in decreasing powers of s DEN coeffs. in decreasing powers of s specify the transfer function
% sketch the Bode plots % % % %
clear the MATLAB environment NUM coeffs. in decreasing powers of s DEN coeffs. in decreasing powers of s specify the transfer function
% sketch the Bode plots
200
Chapter 5
clear; num_coeff = [-2 1]; denom_coeff = [1 2]; sys = tf(num_coeff,denom_coeff); figure(3) bode(sys,{0.02,100}); grid; title('Bode Plot for System-3')
% % % %
clear the MATLAB environment NUM coeffs. in decreasing powers of s DEN coeffs. in decreasing powers of s specify the transfer function
% sketch the Bode plots
The resulting Bode plots are shown in Fig. S5.32.
Bode Plot for System-1 20
Magnitude (dB)
0 -20 -40 -60 -80 0
Phase (deg)
-45 -90 -135 -180 -1
0
10
10
10
1
2
10
Frequency (rad/sec)
(a) Bode Plot for System-3
20
5 Magnitude (dB)
10
0
-20
0
-5
-40 -89
-10 360
-89.5
315 Phase (deg)
Phase (deg)
Magnitude (dB)
Bode Plot for System-2 40
-90 -90.5
270 225 180
-91 -1
0
10
10
Frequency (rad/sec)
(b)
10
1
2
10
-1
0
10
10
Frequency (rad/sec)
(c)
Figure S5.32. Magnitude and phase spectra for systems in Problem 5.32. Calculating Output:
System (a): Using the convolution property, the output of system (a) is given by
1
10
2
10
Solutions
Y1 (ω ) =
2 × π [δ (ω − 1) + δ (ω + 1) ] = 2π (1 + jω ) 2
(
1 (1+ j1) 2
δ (ω − 1) + (1−1j1) δ (ω + 1) 2
201
)
= − jπ [δ (ω − 1) − δ (ω + 1) ] . Calculating the inverse CTFT, we obtain y1 (t ) = sin t . System (b): Using the convolution property, the output of system (b) is given by
1 Y2 (ω ) = πδ (ω ) + × π [δ (ω − 1) + δ (ω + 1) ] = π 1j δ (ω − 1) + −1j δ (ω + 1) jω = − jπ [δ (ω − 1) − δ (ω + 1) ] . Calculating the inverse CTFT, we obtain
y2 (t ) = sin t . System (c): Using the convolution property, the output of system (c) is given by
Y2 (ω ) =
1 − j 2ω × π [δ (ω − 1) + δ (ω + 1) ] = π 12−+j j2 δ (ω − 1) + 12+−j j2 δ (ω + 1) 2 + jω
= − jπ [δ (ω − 1) − δ (ω + 1) ] . Calculating the inverse CTFT, we obtain
y3 (t ) = sin t . To explain why the three systems produce the same output for input x(t) = cost, consider Eq. (5.77), which for ω0 = 1 is given by cos(t ) → H (1) cos(ω0t + < H (1) ) . Hermitian Symmetric H ( ω)
In other words, the output for x(t) = cos(t) depends only on the magnitude and phase of the system at ω = 1. For the three systems, we note that H1 (1) = H 2 (1) = H 3 (1) = 1
and
< H1 (1) = H 2 (1) = H 3 (1) = − π2 . Since the magnitudes and phases of the three system transfer functions at ω = 1 are identical, the three ▌ systems produce the same output y (t ) = sin t for x(t) = cos(t). Problem 5.33
The MATLAB code for calculating the CTFTs is listed below. % Problem 5_33(i) ws = 200*pi; % sampling rate Ts = 2*pi/ws; % sampling interval tmin = -2; tmax = 2; t = tmin:Ts:tmax; % define time instants x = sin(5*pi*t); y = fft(x); % fft computes CTFT z = (2*pi*Ts/(tmax-tmin))*y;% scale the magnitude of y z = fftshift(z); % centre CTFT about w = 0
202
Chapter 5
w = -ws/2:ws/length(z):ws/2-ws/length(z); subplot(221); plot(w,abs(z)); % CTFT plot of cos(w0*t) axis([-20*pi 20*pi 0 max(abs(z))]); xlabel('\omega (radians/s)'); ylabel('|x_1(t)|'); title('x_1(t) = sin(5\pi t): Magnitude spectrum') grid on subplot(222); plot(w,angle(z).*abs(z)/max(abs(z))); axis([-20*pi 20*pi -0.5*pi 0.5*pi]); xlabel('\omega (radians/s)'); ylabel('
The magnitude and phase spectra are shown in Fig. S5.33.
▌
Solutions
x 1(t) = sin(5π t): Magnitude spectrum
203
x 1(t) = sin(5π t): Phase spectrum
3
|x 1(t)|
1 2 1
0 -1
0 -60
-40
-20
0
20
40
60
-60
ω (radians/s) x 2(t) = sin(8π t)+sin(20π t): Magnitude spectrum
-40
-20
0
20
40
60
ω (radians/s) x 2(t) = sin(8π t)+sin(20π t): Phase spectrum
3 2
|x 2(t)|
1
1
0 -1
0
-100
-50
0
50
100
ω (radians/s)
-100
-50
0
50
100
ω (radians/s)
Figure S5.33. Magnitude and phase spectra for the sinusoidal signals in Problem 5.33. Problem 5.35
The MATLAB code for calculating the output is listed below. % Problem 5.35 t = -5:0.001:5; % time waveforms with N samples each x = exp(-t).*(t>=0); h = exp(-2*t).*(t>=0); % CTFT calculated for (2N-1) samples Xfreq = fft(x,length(x)+length(h)-1); Hfreq = fft(h,length(x)+length(h)-1); % Scale the ffts Xfreq = (2*pi*0.001/10) * Xfreq; Hfreq = (2*pi*0.001/10) * Hfreq; % Output Yfreq = Xfreq .* Hfreq; y = ifft(Yfreq); y = (10/(2*pi*0.001))*y; % plot plot([-10:0.001:10],real(y)); xlabel('time (t)'); ylabel('output y(t)'); title('Problem 5.35');
The magnitude and phase spectra are shown in Fig. S5.35.
▌
204
Chapter 5 0.2
output y (t)
0.15 0.1 0.05 0 -0.05 -10
-8
-6
-4
-2
0 time (t)
2
4
6
8
10
Figure S5.35: Output waveform for Problem 5.35. Problem 5.36 (i) Bode plots for the LTIC systems specified in Problem 5.20:
The MATLAB code for calculating the Bode plots for the LTIC systems specified in Problem 5.20 is listed below. % Problem 5.36 % Bode plot for Problem 5.20(a) figure(1) w = 0.01:0.01:100; num = [1]; den = [1 6 11 6]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(a)'); grid on % % Bode plot for Problem 5.20(b) figure(2) w = 0.01:0.01:100; num = [1]; den = [1 3 2]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(b)'); grid on % % Bode plot for Problem 5.20(c) figure(3) w = 0.01:0.01:100; num = [1]; den = [1 2 1]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(c)'); grid on
Solutions
205
% % Bode plot for Problem 5.20(d) figure(4) w = 0.01:0.01:100; num = [1 4]; den = [1 6 8]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(d)'); grid on % % Bode plot for Problem 5.20(e) figure(5) w = 0.01:0.01:100; num = [1]; den = [1 8 19 12]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(e)'); grid on
The Bode plots are shown in Fig. S5.36.1 to Fig. S5.36.5 below.
P5.20(a)
Magnitude (dB)
0
-50
-100
Phase (deg)
-150 0
-90
-180
-270 10
-2
10
-1
0
10
1
10
10
2
ω (radians/s) (rad/sec)
Figure S5.36.1: Bode plot for LTI system specified in Problem 5.20(a) as required in Problem 5.36.
206
Chapter 5
P5.20(b) 0
Magnitude (dB)
-20 -40 -60 -80 -100 0
Phase (deg)
-45 -90 -135 -180
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.2: Bode plot for LTI system specified in Problem 5.20(b) as required in Problem 5.36.
P5.20(c) 0
Magnitude (dB)
-20 -40 -60 -80 -100 0
Phase (deg)
-45 -90 -135 -180
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.3: Bode plot for LTI system specified in Problem 5.20(c) as required in Problem 5.36.
Solutions
207
P5.20(d) 0
Magnitude (dB)
-10 -20 -30 -40
Phase (deg)
-50 0
-45
-90
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.4: Bode plot for LTI system specified in Problem 5.20(d) as required in Problem 5.36. P5.20(e) -20
Magnitude (dB)
-40 -60 -80 -100 -120
Phase (deg)
-140 0
-90
-180
-270
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.5: Bode plot for LTI system specified in Problem 5.20(e) as required in Problem 5.36. The Bode plots have a one to one correspondence with the magnitude and phase spectra plotted in Problem 5.20. (ii) Bode plots for the LTIC systems specified in Problem 5.21:
Since the transfer function H(ω) = 5 in part (a), the magnitude plot for part (a) is constant at 5 for all frequencies. The phase is 0. The transfer function H(ω) = 3e−j4ω in part (b). Therefore, the magnitude plot for part (b) is constant at 3 for all frequencies. The phase is −4ω represented by a straight line with a slope of −4. These two plots are not plotted.
208
Chapter 5
The MATLAB code for calculating the Bode plots for the LTIC systems specified in parts (c) and (d) of Problem 5.21 is listed below. The plots are shown in Fig. S5.36.6 and S5.36.7. ▌ % Problem 5.36 % Bode plot for Problem 5.21(c) figure(1) w = 0.01:0.01:100; num = [6]; den = [1 6 12 8]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(c)'); grid on % Bode plot for Problem 5.21(c) figure(2) w = 0.01:0.01:100; num = [2 8 8]; den = [1 4 3]; sys = tf(num,den); bode(sys,{0.01,100}); xlabel('\omega (radians/s)'); title('P5.20(d)'); grid on
P5.20(c) 0
Magnitude (dB)
-20 -40 -60 -80 -100
Phase (deg)
-120 0
-90
-180
-270
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.6: Bode plot for LTI system specified in Problem 5.21(c) as required in Problem 5.36.
Solutions
209
P5.20(d) 9
Magnitude (dB)
8
7
6
Phase (deg)
5 0
-5
-10
-15
-2
10
-1
10
0
10
1
10
2
10
ω (radians/s) (rad/sec)
Figure S5.36.7: Bode plot for LTI system specified in Problem 5.21(d) as required in Problem 5.36.
