Chapter 5: Continuous-time Fourier Transform Problem 5.1
(a)
The CTFT for x x1(t ) is given by ∞
∫ −∞
X 1 (ω) = x1 (t )e
0
∫−τ[ ]
− jωt
dt = 1 + τ e t
τ
− jωt
dt +
∫ [1 − τ ] e − ω dt j t
t
0
τ
0
t e − jωt 1 e − jωt t e − jωt e − jωt 1 = [1 + τ ]× − [τ ] × − [− τ ] × + [1 − τ ] × (− jω) ( − jω) ( − jω) 2 −τ ( − jω) 2 0 1 1 e jωτ 1 e − jωτ 1 1 1 1 1 [ ] [ ] [ ] = − [τ ]× + × + × − − × τ τ τ ( − jω) 2 (− jω) 2 (− jω) 2 ( − jω) (− jω) 2 (− jω) =
2
ω2 τ
=τ (b)
−
2 cos(ωτ)
ω2 τ
sin 2 (ωτ / 2) (ωτ / 2) 2
= 2×
1 − cos(ωτ)
ω2 τ
2 sin 2 (ωτ / 2)
= 2×
ω2 τ
ωτ = τ sinc 2 . 2π
The CTFT for x x2(t ) is given by ∞
X 2 (ω ) =
∫ x (t )e
∞
− jω t
2
dt
=
−∞
∫t e 4
∞
− at
u (t )e
−∞
− jω t
dt
= ∫ t 4e − ( a + jω )t dt 0
∞
− + − + − + − + − + = t 4 ( −e ( a + jω )) + 4t 3 (( −e( a + jω )) + 12t 2 ( −e( a+ jω )) + 24t (( −e( a+ jω )) + 24 ( −e( a+ jω)) 0 ( a jω ) t
( a jω ) t
( a jω ) t
2
( a jω ) t
3
(a ω j )t
4
= [0 + 0 + 0 + 0 + 0] − 0 + 0 + 0 + 0 + 24 ( −( a +1 jω )) = ( a+24jω ) 5
(c)
5
5
.
The CTFT for x x3(t ) is given by ∞
X 3 (ω ) =
∫ x (t )e 3
∞
− jω t
dt
=
−∞
∫e
∞
− at
−∞
∞
=
1 2
∫
cos(ω 0t )u (t )e
− jω t
e
dt
+
0
1 2
=
∫e
− ( a + jω )t
e jω t + e − jω t dt 0
0
0
∞
− ( a + jω − jω0 ) t
dt
1 2
∫
e −( a + jω + jω0 ) t d t =
0
1 2
∞
∞
( −e(−a ++jω−− jω )) + 21 ( −e(−a++jω++ jω )) 0 0 ( a jω jω0 ) t
( a jω
0
jω 0 ) t
0
= 12 0 − ( −( a + jω1 − jω )) + 21 0 − ( −( a+ jω1 + jω )) = 12 ( a+ jω1− jω ) + ( a+ jω1+ jω ) = ( a +ajω+ )jω +ω 0
(d)
The CTFT for x x4(t ) is given by
0
0
0
2
2 0
.
160
Chapter 5
∞
X 4 (ω ) =
∫ x (t )e
∞
− jωt
4
dt =
−∞
= exp
−∞ 2 2
( jωσ ) 2σ
∫ exp(−
2
∞
t 2 2σ 2
∞
)e
− jω t
dt
=
∫ exp[−
t 2 + 2 jωσ 2t 2σ 2
−∞
⋅ ∫ exp − (t + jωσ ) dt = exp ( jωσ ) −∞ 2σ 2σ 2 2
2 2
2
∞
]dt
∫ exp −
t 2 + 2 jωσ 2t + ( jωσ 2 ) 2 2σ 2
−∞
2π σ
2
=
=
exp ( jωσ ) dt 2σ 2 2
2
2π σ e xp − ω 2σ . 2
2
▌ Problem 5.2
(a)
By definition, π
[ ]π = − ω [e − ωπ − 1] = −
X 1 (ω) = 3e − jωt dt = 3 (e− jω)
∫
− jωt
0
j
3 j
0
= − j3ω e − jωπ / 2 [− 2 j sin(ωπ / 2)] = 6e − jωπ / 2
[
3 − jωπ / 2 e jω
sin(ωπ / 2)
ω
[e − ωπ j
] = 6e − ωπ [ j
= 3πe − jωπ / 2 sinc(ω / 2). (b)
By definition, 0.5T
∫
X 2 (ω ) =
1.5T
0.5e
− jωt
dt +
−0.5T
∫
e − jω t dt = 0.5 (e− jω ) − jωt
0.5T
0.5T
= − j0.5ω [ −2 j sin(0.5ωT )] −
1 jω
jω t
e − jω T [ −2 j sin(0.5ω T ) ]
T T = 0.5 × sin(0ω.5ωT ) + 2e− jω T 0.5 × sin(0ω .5ω T ) 0.5T 0.5T
= 0.5Tsinc ( 0.5πωT ) + Te− jω T sinc ( 0.5π ω T ) . (c)
By definition, T
X 3 (ω) =
∫ (1 − )e t T
0
T
− ω t e − ω 1) e ( dt = (1 − T ) − − ( − jω) T ( − jω) 0
− jωt
j t
j t
2
− ω = 0 − (− T 1 ) ( −e jω) − ( − j1ω) − (− T 1 ) ( − j1ω) j T
2
2
= − ω1T e − jωT + j1ω + ω1T = j1ω + ω1T (1 − e − jωT ). 2
2
T
For ω = 0,
2
T
t 2 X 3 (ω) = (1 − )dt = − T ( 1 − T ) = 0 + T 2 = T 2 . 2 0
∫ 0
t T
1.5T
(e−− jω ) + −0.5T 0.5T
= − j0.5ω e− j 0.5ωT − e j 0.5ωT − j1ω e− j1.5ωT − e − j 0.5ωT
− e jωπ / 2 ]
/2
/2
1 2/ π
× sin(ωπωπ/ 2/ 2)
]
160
Chapter 5
∞
X 4 (ω ) =
∫ x (t )e
∞
− jωt
4
dt =
−∞
= exp
−∞ 2 2
( jωσ ) 2σ
∫ exp(−
2
∞
t 2 2σ 2
∞
)e
− jω t
dt
=
∫ exp[−
t 2 + 2 jωσ 2t 2σ 2
−∞
⋅ ∫ exp − (t + jωσ ) dt = exp ( jωσ ) −∞ 2σ 2σ 2 2
2 2
2
∞
]dt
∫ exp −
t 2 + 2 jωσ 2t + ( jωσ 2 ) 2 2σ 2
−∞
2π σ
2
=
=
exp ( jωσ ) dt 2σ 2 2
2
2π σ e xp − ω 2σ . 2
2
▌ Problem 5.2
(a)
By definition, π
[ ]π = − ω [e − ωπ − 1] = −
X 1 (ω) = 3e − jωt dt = 3 (e− jω)
∫
− jωt
0
j
3 j
0
= − j3ω e − jωπ / 2 [− 2 j sin(ωπ / 2)] = 6e − jωπ / 2
[
3 − jωπ / 2 e jω
sin(ωπ / 2)
ω
[e − ωπ j
] = 6e − ωπ [ j
= 3πe − jωπ / 2 sinc(ω / 2). (b)
By definition, 0.5T
∫
X 2 (ω ) =
1.5T
0.5e
− jωt
dt +
−0.5T
∫
e − jω t dt = 0.5 (e− jω ) − jωt
0.5T
0.5T
= − j0.5ω [ −2 j sin(0.5ωT )] −
1 jω
jω t
e − jω T [ −2 j sin(0.5ω T ) ]
T T = 0.5 × sin(0ω.5ωT ) + 2e− jω T 0.5 × sin(0ω .5ω T ) 0.5T 0.5T
= 0.5Tsinc ( 0.5πωT ) + Te− jω T sinc ( 0.5π ω T ) . (c)
By definition, T
X 3 (ω) =
∫ (1 − )e t T
0
T
− ω t e − ω 1) e ( dt = (1 − T ) − − ( − jω) T ( − jω) 0
− jωt
j t
j t
2
− ω = 0 − (− T 1 ) ( −e jω) − ( − j1ω) − (− T 1 ) ( − j1ω) j T
2
2
= − ω1T e − jωT + j1ω + ω1T = j1ω + ω1T (1 − e − jωT ). 2
2
T
For ω = 0,
2
T
t 2 X 3 (ω) = (1 − )dt = − T ( 1 − T ) = 0 + T 2 = T 2 . 2 0
∫ 0
t T
1.5T
(e−− jω ) + −0.5T 0.5T
= − j0.5ω e− j 0.5ωT − e j 0.5ωT − j1ω e− j1.5ωT − e − j 0.5ωT
− e jωπ / 2 ]
/2
/2
1 2/ π
× sin(ωπωπ/ 2/ 2)
]
Solutions
(d)
By definition, 0
X 4 (ω) =
∫− (1 + )e t T
T
− jωt
dt +
T
∫ (1 − )e − ω dt j t
t T
0 0
T
− ω − ω − ω − ω = (1 + T t ) (e− jω) − (T 1 ) ( −e jω) + (1 − T t ) (e− jω) − (− T 1 ) ( −e jω) −T 0 j t
j t
j t
j t
2
2
= ( − j1ω) − (T 1 ) ( − j1ω) − 0 + (T 1 ) ( −e jω) jωT
2
+ 0 − (− 1 ) e− ω − 1 + (− 1 ) 1 T ( − jω) ( − jω) T ( − jω) j T
2
2
2
= ω2T [1 − cos(ωT )] = 2×2 sinω (T 0.5ωT ) = 1/(0.45 T ) × sin(0.(50ω.5T ω)T ) = T sinc 2 ( 0.5πωT ) . 2
2
(e)
2
2
2
2
By definition, T
X 5 (ω ) =
∫ 1 − 0.5sin (
T πt T
) e
− jωt
T
∫
dt = e
0
− jωt
∫
dt − 0.5 sin ( π T t ) e− jω t dt
0
= A
0
=B
We consider different cases for the above integral. Case I: ( ω = 0) −∞
X 5 (0) =
∞
∫ x(t )dt = ∫ 1 − 0.0.5 sin (
∞
πt T
−∞
T
T
0
0
) dt = ∫ dt − 0.5∫ sin ( π T t ) dt
T
π t T T 1 = T + π 0.5 cos( T ) 0 = T + 2π [cos(π ) − cos(0) ] = T − π = T (1 − π ) / T
Case II: (
ω ≠ 0, ω ≠ π/T ): ): T
T A = e − jωt dt = − j1ω e − jωt = −1jω e − j ωT
∫
0
− 1 =
1
jω
1 − e− j ωT
[ω ≠ 0]
0
T
− B = 0.5 e {− jω sin ( πTt ) − πT cos ( πTt )} −ω 0
for ω ≠ 0, ± πT
jω t
π 2 T 2
2
T
= π 0.5−ω T T 2
2
2
= π 0−.5ωT T
2
2
2
Case III: (
): ω = π/T ):
2
2
e− jω t jω sin ( π t ) + π cos ( π t ) T T T =0 at t =0,T 0 − Tπ e− jωT − πT = π 0−.5ωπ T T 1 + e− jω T 2
2 2
161
162
Chapter 5
T
T
∫
B = 0.5 sin ( π Tt ) e− jωt dt = 0
0.5 2j
T
−j − j (ω − j ∫0 e − e e− jω t dt = 02.5j ∫0 e πt T
0.5 T − j π 2 j ∫ 1 − e dt ω = T 0 = 0.5 T j π − 2 j ∫ 1 − e dt ω = − T 0 T = ± 02. j5 [t ]0 = ± 02.5jT
π t T
π T
)t
− e− j (ω + )t dt π T
2 π t T
2 π t T
± As e
j 2 π t T
T
is periodic with period T ,
∫ 0
e
± j 2T π t
dt = 0
Combining, the above results, the CTFT can be expressed as
T (1 − π 1 ) T X 5 (ω ) = j1ω 1 − e− jω T ∓ 0.5 2 1 − jωT 0.5π T jω 1 − e − π −ω T T (1 − π 1 ) T = ± j2T π ∓ 4 j 1 − jωT 0.5π T jω 1 − e − π −ω T 2
2
ω = 0 π ω = ± T
2 2
1 + e− jω T
otherwise ω = 0 ω = ± π T
2 2
1 + e− jω T
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):
Fig. P5.3(c):
Fig. P5.3(d):
j 0.5ω X 1 (ω) = 1 × e 0
− W ≤ ω ≤ W otherwise
1× e− j 0.5ω −W ≤ ω ≤ 0 X 2 (ω ) = 1× e j 0.5ω 0 ≤ ω ≤ W 0 otherwise 1× e− jπ /3 −W ≤ ω ≤ 0 X 3 (ω ) = 1× e jπ /3 0 ≤ ω ≤ W 0 otherwise
Using the CTFT synthesis Eq. (5.9), the function
1
(t ) is calculated as follows.
Solutions
∞
1
x1 (t ) =
∫ X (ω )e
jω t
2π
=
dω
W
1
∫e
2π
−∞
j 0.5ω
e
dω =
jωt
−W
W
1
e 2π ∫
j ( 0.5 + t ) ω
d ω
−W
W
e j ( 0.5+t )ω 1 e j ( 0.5+ t )W − e− j ( 0.5+ t )W 1 2 j sin [ (0.5 + t )W ] = = = 2π 2π j (0.5 + t ) −W 2π j ( 0 .5 + t ) j (0.5 + t ) 1
W
=
sin [ (0.5 + t )W ]
×
π
W
=
(0.5 + t )W
π
sinc W (t + 0.5). π
Using the CTFT synthesis Eq. (5.9), the function
x2 (t ) =
∞
1
∫ X (ω )e
jωt
2π
dω
−∞
=
0
1
∫e
2π
= =
j ( −0.5+ t ) ω
dω +
−W
0
e j ( −0.5+t )ω 1 = + 2π j (−0.5 + t ) −W 2π 1
(t ) is calculated as follows.
2
1
W
e 2π ∫
j ( 0.5 + t ) ω
d ω
0
W
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 ) + j (0.5 + t ) 0
1 j 0.5W 2 jt sin(tW ) − cos(tW ) + e 2π j (t 2 − 0.25) j (t 2 − 0.25) 1
1
1 + e j 0.5W ( 2 jt sin(tW ) − cos(tW ) ) . j 2π (t − 0.25) 2
Clearly, at (t (t = = ±0.5), x ±0.5), x2(t ) is undefined in the above expression. Computing directly, we obtain W
x2 (0.5) =
At t = t = 0.5:
1 2π
∫
W
e
j 0.5ω
e
j 0 .5 ω
dω
0
0
0
At t = t = −0.5: x2 ( −0.5) =
∫
1 2π
0
e
− j 0.5ω − j 0.5ω
e
dω
=
∫
1 2π
−W
x3 (t ) =
= =
2π 1 2π 1
∞
∫ X (ω)e
jω t
dω =
−∞
0
e
− jπ / 3
e− jω d ω =
1
−2 jπ
−W
Using the CTFT synthesis Eq. (5.9), the function
1
W
= 21π ∫ e jω d ω = 2 j1π e jω 0 = 2 1jπ e jW − 1 .
1 2π
3
(t ) is calculated as follows.
0
∫e
− jπ / 3
e
jωt
0
e− jω W = 2 1jπ e jW − 1 . −
dω +
−W
1
0
e 2π ∫
jπ / 3
e jωt d ω
−W
W
jWt e jωt − 1 1 jπ / 3 e jω t 1 − jπ / 3 1 − e− jWt jπ / 3 e + = + e e e jt jt jt jt −W 2π 0 2π
sin(Wt + π / 3) − sin(π / 3) π t
[ 2 j sin(Wt + π / 3) − 2 j sin(π / 3)] =
j 2π t
Clearly, at (t (t = = 0), x 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) ∫ dω + (1 + j 3) ∫ dω = 4Wπ 1 − j 3 +1 + j 3 = 2W π . −W 0
Although the functions x1 (t ) ,
2
(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. 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 ) t ( 1
x
) ) t ( 2
x ( s b a
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
1 ) t ( 3
x
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)
≡
( 2 + jω)(3 + jω)
−1 2 + (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 + jω)(2 + jω)(3 + jω)
0.5
≡
(1 + jω)
+
−1 0.5 + (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 (1 + jω)(2 + jω) 2 (3 + jω)
≡
0.5 (1 + jω)
+
0 (2 + jω)
+
−1 − 0.5 + (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 (1 + jω)(2 + 2 jω + ( jω) 2 ) X 4 (ω) =
or,
1 (1 + jω)
≡
−
1 (1 + jω)
−
1 + jω (2 + 2 jω + ( jω) 2 )
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 −t
x5 (t ) = te u (t ) − 1.50e
−t
sin t u (t ) + ℑ
0.25(4 jω + ( jω) ) −1 2
(1 + (1 + jω) 2 ) 2
▌
.
Problem 5.5
Consider an arbitrary function φ(t ), and assume that ∞
p(t ) =
∫−∞e ω dt . j t
Now, consider the integral ∞
∞
∞ ∞ ∞ ∞ jω (t −T ) − jωT jωt ∫−∞ φ (t ) p(t − T )dt = −∞∫ φ(t) −∞∫ e dω dt = -∫∞ e −∞∫ φ( t) e dt dω = −∞∫ e− jωT Φ(−ω ) dω Changing the order of integrat ion
−∞
=∫e
∞
jω ′T
Φ(ω ′)(−d ω ′) =
∞ ω =−ω ′,
d ω =− d ω ′
∫e
Φ (ω )=ℑ{φ ( t )}
jω ′T
Φ(ω ′)d ω ′
−∞
∞
= ∫ Φ(ω )e jω T d ω
................................................(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 ω
j T
The above equation is valid for any arbitrary following property of the impulse response
φ(t )
d ω = 2πφ(T ) .
if and only if p(t ) = 2πδ(t ) as can be seen from the
∞
2π
∫−∞φ(t )δ(t − T )dt = 2πφ(T ) .
In other words, ∞
∫−∞e ω dt = 2πδ(t ) . j t
Interchanging the variables, t and ω, we obtain the required identity ∞
∫−∞e ω d ω = 2πδ(ω) . j t
▌
Solutions
167
Alternate Proof:
Note that the above result can be proved directly from the CTFT pair CTFT
1 ← → 2πδ(ω) . ∞
2πδ(ω) =
By definition,
∫−∞1 × e − ω dt j t
∞
Since δ(−ω) = δ(ω),
2πδ(ω) = 2πδ(−ω) =
∫−∞e ω dt . j t
▌
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 Re{ X (−ω)}.
ω.
Since, X (ω) is real valued, 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 −Im{ X (−ω)}.
ω. Since, X (ω) is imaginary-valued, Im{ X (ω)} =
▌
Problem 5.8
(a)
Since is not equal to
X 1 (− ω) = 2+ j (−5ω−5)
= 2− j (5ω+5)
X 1∗ (ω) = 2− j (5ω−5 ) ,
168
Chapter 5
X 1(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 + π6
Since is not equal to
(
X 2∗ (ω) = cos 2ω + π6
)=
3 2
3 2
cos(2ω) + 12 sin( 2ω)
cos( 2ω) − 12 sin(2ω) ,
X 2(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
[4ω] = 5 sin(ω[4+(ωπ+) π )] = 5(sin ω+ π )
5 sin [4 (ω− π )]
(ω− π )
[4ω] = 5 (sin ω− π ) ,
X 3(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) ,
X 4(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 )
,
X 4(ω) satisfies the Hermitian property. Its inverse CTFT x4(t ) is real valued. Since X 4(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,
{
}= 5ℑ{1} + 3ℑ{cos(10t )} − 7ℑ{e −
X 1 (ω) = ℑ 5 + 3 cos(10t ) − 7e −2t sin(3t )u (t )
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 ) ← → j2ω . Using the duality property,
2 jt
CTFT ← → 2π sgn( −ω) ,
21 ( 2 + jω) 2
+ 32
.
Solutions
1
or, (c)
πt
169
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+8 jω .
CTFT ← → 4+8 jω e − j5ω .
Using the frequency differentiation property, t 2 e or, (d)
t 2 e
−4 t −5
−4 t −5
CTFT ← →( j ) 2
CTFT ← → 200e − j 5ω
d 2 d ω2 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
and
3 sinc(3t ) = 3
sin(3πt ) 3πt
CTFT ← → rect(6ωπ )
5 sinc(5t ) = 5
sin( 5 πt ) 5 πt
CTFT ← → rect (10ωπ )
Using the multiplication property CTFT π 2 × sin(π3t πt ) × sin(π5t πt ) ← → 2ππ [rect(6ωπ ) ∗ rect(10ωπ )] 2
sin( 3πt ) sin( 5 πt )
or, or,
2
t
5
sin(3π t )sin(5π t ) 2
t
CTFT ← → π2 [rect ( 6ωπ ) ∗ rect (10ωπ )] ,
CTFT ← → 52π rect ( 6ωπ ) ∗ rect ( 10ω π ) ,
where * is the convolution operation. (e)
Entry (17) of Table 5.2 provides the CTFT pair
and
3 sinc(3t ) = 3
sin(3πt ) 3πt
CTFT ← → rect(6ωπ )
4 sinc( 4t ) = 4
sin( 4 πt ) 4 πt
CTFT ← → rect (8ωπ ).
Using the time differentiation property, 1 d sin( 4 πt ) π dt t
CTFT ← →( jω) rect(8ωπ ).
Using the convolution property CTFT d sin( 4 πt ) π 2 × sin(π3t πt ) ∗ 1π dt ← → 2ππ [rect(6ωπ ) × jωrect(8ωπ )] t 2
or,
sin( 3πt ) t
or,
4
CTFT d sin( 4 πt ) ∗ dt ← → π2 [rect(6ωπ ) × jωrect(8ωπ )], t
sin( 3πt ) t
CTFT d sin( 4 πt ) ∗ dt ← → j 2π rect(6ωπ ) . t
▌
170
Chapter 5
Problem 5.10
Using the linearity property,
{
X (ω ) = ℑ
6 13
}
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+1 jω − 136 π2 δ (ω − 3) + π 2 δ (ω + 3) − 136 9 j−ωω + 134 −2jπ δ (ω − 3) + j2π δ (ω + 3) + 134 9−3ω = 136 2+1 jω − 9 j−ω ω + 9−2ω − 26π [6δ (ω − 3) + 6δ (ω + 3) + 4 jδ (ω − 3) − 4 jδ (ω + 3)] 2
2
2
2
− jω )(2+ jω ) π = 136 (9−ω(2)++ j(2ω )(9 ] −ω ) − 26 [(6 + j 4)δ (ω − 3) + (6 − j 4)δ (ω + 3) 2
2
6 = (2+ jω )(9 − π (3 + j 2)δ (ω − 3) + (3 − j 2)δ (ω + 3) ] −ω ) 13 [ 2
▌
which is the required result.
Problem 5.11 ∞
∫
F { x(at )} = x( at )e − jωt dt .
From the definition of CTFT,
−∞
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 )} = x (r )e Case 2:
− j ( ωa ) r
∞
dr a
=
1 a
∫ −∞
x (r )e
− j ( ωa ) r
( )
dr = 1a X ω . a
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
Combining the two cases, yields F { x(at )} =
1 a
− j (ωa ) r
dr = −
∞
1 a
∫ −∞
x( r )e
( )
X ωa .
− j ( ωa ) r
( )
dr = − 1a 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] ,
( ) − 4sinc 2 (ωπ ) .
H (ω) = 16sinc 4πω
▌
Solutions
171
Problem 5.13
Using the definition of CTFT, we obtain
{
F e
∞
jω0t
}= ∫ e
x(t )
∞ jω0t
x(t )e
∫
− jωt
dt = x(t )e
−∞
− j ( ω− ω0 )t
dt = X (ω − ω 0 ) .
▌
−∞
Problem 5.14
Using the convolution property,
[
CTFT
]
x(t ) ∗ u (t ) ← → X (ω) 2πδ(ω) + j1ω , ∞
∫ 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)
CTFT
Using the time scaling property, x(2t ) ← →
1 2
( )
X ω2 .
Using the frequency shifting property, e − j 5t x(2t ) ← → CTFT
1 2
X ( ω2+ 5 ) .
Substituting the value of X (ω), we obtain
ℑ{e
(b)
1 − ω 3+5 ω + 5 ≤ 3 x(2t )} = elsewhere 0 −11 ≤ ω ≤ −5 ω 12+11 1−ω = 12 −5 ≤ ω ≤ −1 0 elsewhere.
− j 5t
1 2
Using the frequency differentiation property, ( jt ) x(t ) ← → 2
CTFT
d 2 X d ω2
,
2
t x(t ) ← → − d X 2 . 2
or,
CTFT
d ω
The CTFT of t 2 x(t ) is given by
{
}= − d d ω [∆(ω3 )] = − d d ω [rect (ω3 )] = −[δ(ω + 3) − δ(ω − 3)] = [δ(ω − 3) − δ(ω + 3)] .
F t 2 x(t )
2
2
▌
172
(c)
Chapter 5
Express (t + 5) dx = t dxdt + 5 dxdt . 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 CTFT
← → j d d ω [ jω X (ω)] = − X (ω) − ω dX t dx . dt d ω The CTFT of ( t + 5 ) dx is given by dt
} = − X (ω) − ω dX ℑ{(t + 5) dx + 5 jωX (ω) . dt d ω Substituting the value of X (ω), we obtain
j 5ω(1 − ω3 ) − (1 − 23ω ) 0 ≤ ω ≤ 3 } = j 5ω(1 + ω3 ) − (1 + 23ω ) − 3 ≤ ω ≤ 0 ℑ{(t + 5) dx dt 0 elsewhere. (d)
Using the time multiplication property, CTFT
x(t ) ⋅ x (t ) ← → 21π [ X (ω ) ∗ X (ω )] , which implies that F { x (t ) ⋅ x(t )} = 21π
(e)
[∆(ω3 ) ∗ ∆(ω3 )].
Using the time convolution property, CTFT
x(t ) * x (t ) ← → X (ω) ⋅ X (ω ) , which reduces to
ω 2 1 − 3 F { x(t ) ∗ x (t )} = 0 (f)
ω ≤3
1 + = elsewhere
ω2 9
−
2ω 3
0
ω ≤3 elsewhere.
Using the time multiplication property, CTFT
x(t ) ⋅ cos ω0 t ← → 21π X (ω) ∗ πδ(ω − ω 0 ) + 21π X (ω ) ∗ πδ(ω + ω 0 ) , or,
CTFT
x(t ) ⋅ cos ω0 t ← → 12 X (ω − ω0 ) +
1 2
X (ω + ω 0 )
Case I: For ω0 = 3/2, we obtain CTFT
(
x(t ) ⋅ cos(3t / 2) ← → 12 X ω −
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 ≤ 92 2 elsewhere.
Case II: For ω0 = 3, we obtain CTFT
x(t ) ⋅ cos 3t ← → 12 X (ω − 3) +
1 2
X (ω + 3) .
Since there is no overlap between the two shifted replicas,
1 − ω+3 3 ω −3 F { x(t ) cos 3t } = 12 1 − 3 0
or,
ω+3 ≤3 ω−3 ≤3 elsewhere.
1 − ω+ 3 6 2 ω −3 F { x(t ) cos 3t } = 12 − 6 0
−6≤ω<0 0≤ ω< 6 elsewhere.
Case III: For ω0 = 6, we obtain CTFT
x(t ) ⋅ cos 6t ← → 12 X (ω − 6 ) +
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,
ω+ 6 ≤3 ω−6 ≤3 elsewhere.
1 − ω+ 6 6 2 ω −6 F { x(t ) cos 3t } = 12 − 6 0
− 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,
inverse CTFT ← →[5e − 2t u (t )] × e j 5t
x1 (t ) = 5e (
−2 + j 5)t
u (t ) .
cos( 2t ) ← → π[δ(ω − 2) + δ(ω + 2)]. . CTFT
▌
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
π 12
t
CTFT ← → 2π cos(2(ω + 12π ) ) ,
implying that x 2 (t ) =
or, x 2 (t ) = (c)
1 2
1 2
π π π [δ(t − 2) + δ(t + 2)] e − j t = 12 δ(t − 2)e − j t + δ(t + 2)e − j t 12
12
δ(t − 2)e − j π + δ(t + 2)e j π = 1 ( 4 6
From Table 5.2,
3
6
rect ( 4t ) ← → 4sinc ( 42ωπ ) = 4 CTFT
12
− j )δ(t − 2) + ( sin( 4 πω / 2 π ) ( 4 πω / 2 π )
3
π + j )δ(t + 2)e j . 2 3
= 2 sin(ω2ω) .
Using the time scaling property, rect ( 4t ⋅2 ) ← → 2 ⋅ 2 CTFT
sin( 4 ω) 2ω
= 2 sin(ω4ω) .
Using the frequency shifting property, rect
(8t )e jπt ← CTFT → 2 sin((4ω(−ωπ−)π )) ,
implying that x 3 (t ) =
(d)
5 2
rect
(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+2π2 j ) e j10t + (1−2π 2 j ) e − j10 t . Expanding the exponential terms using the Euler’s formula, we obtain x 4 (t ) =
(cos 10t + j sin 10t ) +
x 4 (t ) = π2 cos10t −
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 )
≡ (1+ A jω) +
B (3+ jω)
+
A = 0.0625, B = 0.25, C = 0.125, D = −0.3125,
C (3+ jω) 2
and
ω+ E + jD (5+ ω ) 2
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
1 ) t ( 1 x
0.5
0 -5
-4
-3
-2
-1
0 t
1
2
3
4
5
0.4 1 ) t ( 2 x
0.3 ) t ( 3 0.2 x
0.5 0 -0.5 -5
0.1
-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
0.5 ) t ( 4 x
) t ( 5 x
0
0
-0.5
-50
-1 -100 -1
1000 2000 3000 4000 5000 6000 7000 8000 9000 10000 t
-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
0.5 ) t ( 6 x
) t ( 7 x
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
− a t
∞
0
dt =
−∞
∫
∫
e dt + e− at dt = 1a eat at
−∞
0
∞
0
+ ( −1a ) e− at 0 = 1a [1 − 0] − 1a [0 − 1] = 2a < ∞. −∞
Since the condition in Eq. (5.59) is satisfied, the CTFT for x1(t ) exists. ∞
(b)
∫
−∞
∞
x 2(t ) dt =
∫
∞
e− at cos(ω 0 t )u(t ) dt
−∞
= 12 ∫ e− at e jω t + e− jω t dt ≤ 12 0
0
0
∞
∫
e
− ( a − jω 0 )t
∞
dt
∞
I =
∫
e
−( a − jω0 )t
0
while Integral II is given by
dt
0
dt
I
II
∞
0
− ( a + jω0 )t
0
= 12 [e−( a− jω ) ]0 = 12 [0 − −( a −1 jω ) ] = 12 [( a −1jω ) ] , − ( a − jω0 ) t
∫e
0
Integral I is given by 1 2
+ 12
0
Solutions
∞
II =
1 2
∫
−( a + jω0 ) t
e
= 12
dt
[
e
− ( a + jω 0
) t
−( a + jω0 )
]
∞ 0
177
= 12 [0 − −( a+1 jω ) ] = 12 [( a +1jω ) ]. 0
0
0
Therefore, ∞
∫ x (t ) dt = I + II = 4
−∞
1 1 2 ( a − jω0 )
+ 12
1 ( 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 =
−∞
∫
∞
4
− at
t e u (t ) dt
−∞
∞
− − − − − = ∫ t 4e− at dt = t 4 (e− a ) + 4t 3 ( −e a ) + 12t 2 ( e−a) + 24t ( e−a) + 24 ( e−a) 0 0 at
at
at
2
at
3
at
4
5
= [ 0 + 0 + 0 + 0 + 0] − 0 + 0 + 0 + 0 + 24 ( −1a ) = a24 < ∞. 5
5
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, ∞
∞
−∞
0
∫ x4(t) dt = ∫ sin(ln(t)) dt →∞
Therefore, the CTFT for x4(t ) does not exist. ∞
(e)
∞
∫ x5(t ) dt = ∫
−∞
∞ 1 t
dt = 2
−∞
∫
1 t
∞
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)
∞
∫ x6(t ) dt = ∫ cos (
−∞
π / 2 t
→ ∞. ) dt
−∞
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( −
−∞
−∞
∞ t2 2σ 2
) dt
=
∫ exp( −
−∞
t 2 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
[ ]dt = ∫
exp
t 2 2σ 2
∞
1 2 πσ
[
exp
( t − m ) 2 2σ 2
]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
Dn
with fundamental frequency
3 2 = 0 3 jnπ
n=0 even n, n ≠ 0 odd n,
ω0 = 1 radians/s. Therefore, the CTFT is given by ∞
∑ =−∞
X 1(ω) = 2π
Dn δ(ω − nω 0 ) =3πδ(ω) − j 6
n
(b)
with fundamental frequency X 2(ω) = 2π
n odd n
∞
3 = 0.5 4 − nπ sin(0.5nπ)
n=0 n≠0
ω0 = π/T radians/s. Therefore, the CTFT is given by
∑ =−∞
Dn δ(ω − nω 0 ) =1.5πδ(ω) −
n
∞
sin(0.5nπ)δ(ω − ∑ =−∞ 1 n
nπ ) T
.
n n ≠0
From the solution of Problem P4.11(c), the CTFS coefficients Dn of the rectangular pulse train are obtained as Dn with fundamental frequency
12 , = 1 j 2 nπ ,
n=0 n ≠ 0.
ω0 = 2π/T radians/s. Therefore, the CTFT is given by
X 3(ω) = 2π
∞
∞
D δ(ω − nω ) =πδ(ω) − j ∑ ∑ = −∞ = −∞ n
1 n
0
n
(d)
δ(ω − n) . ∑ = −∞
From the solution of Problem P4.11(b), the CTFS coefficients Dn of the rectangular pulse train are obtained as Dn
(c)
∞
δ(ω − 2T nπ ) .
n n≠0
From the solution of Problem P4.11(d), the CTFS coefficients Dn of the rectangular pulse train are obtained as
Dn
with fundamental frequency
1, 2 = 0, 2 ( nπ)
n=0 even n, n ≠ 0 2
odd n, n ≠ 0.
ω0 = π/T radians/s. Therefore, the CTFT is given by
Solutions
X 4(ω) = 2π
∞
∞
D δ(ω − nω ) =πδ(ω) + π ∑ ∑ = −∞ = −∞ n
4
0
n
(e)
n n≠0 odd n
179
δ(ω − nT π ) .
1 n2
From the solution of Problem P4.11(e), the CTFS coefficients Dn of the rectangular pulse train are obtained as n=0
X 5(ω) = 2π
n = ±1 0 ≠ n = even
2
2
with fundamental frequency
n=0
0.3408 n = ±1 ∓ j 0.1933 = 0.1592 0 ≠ n = even n −1 ± 1 ≠ n = odd − j 0.3183 n
12 (1 − π 1 ) ∓ j ( π 1 − 18 ) Dn = 1 2π (n −1) 1 jnπ
± 1 ≠ n = odd
ω0 = π/T radians/s. Therefore, the CTFT is given by
∞
D δ(ω − nω ) = 0.6816 ∑ = −∞ n
0
n
π ) − j 0.3866 πδ(ω − π ) +
= 0.6816πδ(ω) + j 0.3866 πδ(ω + T
T
∞
∑
δ(ω − −1
1
n n = −∞ n ≠ 0,1 even n
2
nπ ) T
− j 2
∞
∑
1 n
n = −∞ n ≠ 0,1 odd n
δ(ω − nT π )
.
▌ Problem 5.19
(a)
From the solution of Problem P5.2(a), the CTFT of the aperiodic signal is given by
3π ω=0 − jωπ 3 ) ω ≠ 0. jω (1 − e
X 1 (ω) = 3πe − jωπ / 2 sinc(ω / 2) =
The signal shown in Fig. P4.6(a) is a periodic signal with a fundamental period T 0 = 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 X 1 ( T 0
ω)
ω=nω0
=
1 2π
3 jnω
3π jn (1 − e − ω0 π )
0
3 2 = 3 jn n ≠ 0 j 2πn (1 − e − π )
n=0
n=0 n≠0
which simplifies to
Dn
(b)
3 23 = = jnπ 0
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) = 0.5T sinc
ω
(1 + 2e − ω ) j T
ω=0 ω ≠ 0.
180
Chapter 5
The signal shown in Fig. P4.6(b) is a periodic signal with a fundamental period T 0 = 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
=
1 X 2 ( T 0
ω)
ω=nω0
=
1 2T
3 1.5T ω = 0 = sin(nπ / 2) 4 − jnπ sin(nω T / 2) − jnω T (1 + 2e ) ω ≠ 0 2nπ (1 − e ) nω 0
0
n=0 n≠0
0
which simplifies to
Dn
(c)
34 1 − = = 21nπ 2nπ 0
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
0.5T ω=0 X 3 (ω) = − jωT 1 1 ) ω ≠ 0. jω + ω T (1 − e 2
The signal shown in Fig. P4.6(c) is a periodic signal with a fundamental period T 0 = T with one period matching the function shown in Fig. P5.2(c). The fundamental frequency ω0 = 2 π/T and the exponential CTFS coefficients are given by
Dn
=
1 X 3 ( T 0
ω)
ω=nω0
0.5T ω = 0 0.5 = 1 = − jnω T 1 ) ω ≠ 0 j 21nπ − 4n1π (1 − e − j 2nπ ) jnω − n ω T (1 − e 1 T
0
0
2
2
2 0
2
n=0 n≠0
which simplifies to
Dn (d)
0.5 = 1 j 2nπ
n=0 n ≠ 0.
From the solution of Problem P5.2(d), the CTFT of the aperiodic signal is given by
T X 3 (ω) = T sinc 2 ( 0.5πωT ) = 2 T sinc ( 0.5πωT )
ω=0 ω ≠ 0.
The signal shown in Fig. P4.6(d) is a periodic signal with a fundamental period T 0 = 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
= T 1 X 3 (ω) 0
ω=nω0
which simplifies to
T ω=0 0.5 = 21T = 2 ω 0 . 5 n T 2 T sinc ( π ) ω ≠ 0. 0.5sinc (0.5n) 0
Dn
0.5 = 0 2 ( nπ)
2
n=0 even n, n ≠ 0 odd n, n ≠ 0.
n=0 n≠0
Solutions
181
(e) From the solution of Problem P5.2(e), the CTFT of the aperiodic signal is obtained as
T (1 − π 1 ) T X 5 (ω ) = ± 2 jT π ∓ 4 1 − jωT − jω T 0.5π T jω 1 − e − π −ω T 1 + e 2
2
ω = 0 ω = ± π T otherwise
2
The signal shown in Fig. P4.6(e) is a periodic signal with a fundamental period T 0 = 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
= T1
0
X (nω0 ) =
1 2T
X (nω 0 )
T (1 − π 1 ) n=0 1 1 − e jω T ∓ 0.5T n = ±1 = 21T × ± jω 2j 1 π T 1 − e− jnω T − π −0.5 1 + e− jnω T otherwise ( nω ) T jnω T (1 − π 1 ) n =0 = 21T × ± jT π 1 − e jπ ∓ 0.52 jT n = ±1 T − jnπ − jnπ 0.5π T otherwise jnπ 1 − e − π −n π 1 + e 12 − 21π n=0 = ± j1π ∓ 0.52 jT n = ±1 1 n n 1 otherwise j 2nπ 1 − (−1) + 4π (n −1) 1 + ( −1) 12 (1 − π 1 ) n=0 n = ±1 ∓ j ( π 1 − 18 ) = 1 0 ≠ n = even 2π (n −1) 1 ± 1 ≠ n = odd jnπ ∓
0
0
0
0
2
0
2
2
0
∓
2
2
2
2
2
▌ Problem 5.20
(a)
Calculating the CTFT of both sides and applying the time differentiation property, yields
( jω )
3
2
Y ( ω ) + 6 ( jω ) Y ( ω ) + 11( jω ) Y ( ω ) + 6Y ( ω )
or,
( ( jω )
or,
H ( ω ) =
3
= X ( ω ) ,
+ 6 ( jω ) +11( jω ) + 6 ) Y ( ω ) = X ( ω ) , 2
Y ( ω ) X ( ω )
=
1
(
jω )
3
2
+ 6 ( jω ) + 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 + jω)(2 + jω)(3 + jω)
≡
0.5 (1 + jω)
+
−1 0.5 + (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 ), ( jω)2 + 3( jω) + 2)Y (ω) = X (ω) ,
or, or,
H (ω) =
Y (ω) X (ω)
=
1
( jω)2 + 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 2
( jω) + 3( jω) + 2
≡
1 (1 + jω)
−
1 ( 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) =
+ 1( jω) + 1)Y (ω) = X (ω) ,
2
Y (ω) X (ω)
=
1
( jω)2 + 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 − u (t ) . 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) ,
( jω) + 4 1 = . X (ω) ( jω)2 + 6( jω) + 8 2 + jω Y (ω)
=
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)
or,
H (ω) =
3
+ 8( jω)2 + 19( jω) + 12 )Y (ω) = X (ω) , Y (ω) X (ω)
=
1 3
2
( jω) + 8( jω) + 19( jω) + 12
.
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)3 + 8( jω)2 + 19( jω) + 12
≡
1/ 6
+
(1 + jω)
− 1/ 2 1/ 3 + (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
Y (ω)
and
5
=
2 + jω
.
The transfer function is given by H (ω) =
Y (ω) X (ω)
= 5.
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 2 + jω
e − j 4ω .
The transfer function is given by H (ω) =
Y (ω) X (ω)
= 3e − j 4ω .
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 ) = 3 x(t − 4) .
or, in the time domain, (c)
Calculating the CTFT of the input and output signals, we obtain 1
X (ω) =
2 + jω
6
Y (ω) =
and
.
( 2 + jω) 4
The transfer function is given by H (ω) =
Y (ω) X (ω)
=
6
.
(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 d 3 y 3
dt (d)
+6
d 2 y 2
dt
+ 12
dy dt
+ 8 y (t ) = 6 x(t ) .
Calculating the CTFT of the input and output signals, we get X (ω) =
1
and
2 + jω
Y (ω) =
1 (1 + jω)
+
1 (3 + jω)
.
The transfer function is given by H (ω) =
Y (ω) X (ω)
=
(4 + 2 jω)(2 + jω) (1 + jω)(3 + jω)
=
2( 2 + jω) 2 (1 + jω)(3 + jω)
.
Using partial fraction expansion, the transfer function is given by H (ω) =
2(2 + jω) 2 (1 + jω)(3 + jω)
≡2+
1 (1 + jω)
−
1 (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
185
Calculating the inverse CTFT, the resulting differential equation is obtained as d 2 y 2
dt
+4
dy dt
+ 3 y (t ) = 2
d 2 x 2
dt
+8
dx dt
+ 8 x(t ) .
▌
Problem 5.22
The transfer function of the RC series circuit is given by H (ω) =
1 /( jωC ) R + 1 /( jωC )
=
1 (1 + jωCR)
=
1 CR
×
1 1 /(CR) + jω
.
Calculating the inverse CTFT, the impulse response of the system is obtained as h(t ) =
1 CR
× e −t /(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): y (t ) =
1 CR
The output y(t ) is given by t +T / 2
∫
e−τ /( CR ) dt =
0
1 CR
( −CR) e−τ /(CR )
t +T / 2 0
= 1 − e−T /(2CR ) e−t /(CR ) .
The output y(t ) is given by 1
t +T / 2
CR ∫ −
e − τ /(CR ) dt =
t T / 2
1 CR
(−CR)e − τ /(CR )
t +T / 2 t −T / 2
= [e T /( 2CR ) − e −T /( 2CR ) ] e −t /(CR ) .
Combining the three cases, we obtain
0 t ≤ −T / 2 y (t ) = 1 − e−T /(2CR ) e−t /(CR ) −T / 2 < t ≤ T / 2 eT /( 2CR ) − e−T /(2CR ) e−t /(CR ) t >T /2 For T
= 2,
R = 1M Ω, C
= 1µ F ,
0 t ≤ −1 − ( t +1) y (t ) = 1 − e −1 < t ≤ 1 − (e − 1/ e) e t t > 1 ≈2.3504
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 CR
h(τ)
e − τ / CR τ
0 v(τ)
1
v(τ) τ
0
− T 2
T 2
v(t − τ)
1
v(t − τ) τ
t −
T 2
t +
0
T 2
h(τ) v(t −
τ
1 CR
e
1 CR
Case I: (t < −T /2)
) − τ / CR
τ
t − T 2
0
t + T 2
h(τ) v(t − 1 CR
Case II: (−T /2 < t < T /2)
)
τ
1 − τ / CR e CR
τ
t −
0 t + T 2
T 2
h(τ) v(t −
Case III: (t > T /2)
1 CR
1 CR
)
τ
e − τ / CR τ
0
y (t ) ( for T = 2, R = 1 MΩ, C = 1 µ F)
t − T 2
t + T 2
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 CR
×
1 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
×
CR
1 1 /(CR) + jω
δ(ω − ω0 ) +
π CR
×
1 1 /(CR ) + jω
δ(ω + ω 0 )
which reduces to
π
Y (ω) =
or,
Y (ω) =
CR
π CR
× ×
1 1 /(CR ) + jω0 1 /(CR) − jω 0 1 /(CR)
Y (ω) = or,
+
π CR
2
×
π CR
+ω
2 0
δ( ω − ω 0 ) + δ(ω − ω 0 ) +
1 /(CR) 1 /(CR) 2 + ω 02 − jω 0
×
1 /(CR) 2
+ ω 20
π CR
π CR
× ×
1 1 /(CR) − jω0 1 /(CR) + jω0 1 /(CR) 2
+ ω02
δ(ω + ω 0 ) , δ(ω + ω0 ) ,
[δ(ω − ω 0 ) + δ(ω + ω0 )] [δ(ω − ω0 ) − δ(ω + ω0 )] .
Calculating the inverse CTFT, we obtain Y (ω) =
1 CR
×
1 /(CR) 1 /(CR ) 2
+ ω 20
cos(ω0 t ) +
1 CR
×
− jω0 j sin( ω 0 t ) , 1 /(CR) 2 + ω02
or, y (t ) = which can be expressed as
1 1 + C 2 R 2 ω 20
[cos(ω 0 t ) + ω 0 CR sin(ω0 t )] ,
187
188
Chapter 5
1
y (t ) = (b)
1 + C 2 R 2 ω02
cos
[ω t − tan − (ω CR)]. 1
0
0
As determined in Problem 5.22, the transfer function of the RC series circuit is given by H (ω) =
1 CR
×
1 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 jCR
×
1 1 /(CR) + jω
δ( ω − ω 0 ) −
π jCR
×
1 1 /(CR) + jω
δ( ω + ω 0 )
which reduces to Y (ω) = Y (ω) =
or,
π
×
jCR
π
×
jCR
1 1 /(CR) + jω0 1 /(CR ) − jω0 1 /(CR )
Y (ω) = or,
π jCR
+
2
+ω
×
π jCR
2 0
δ(ω − ω 0 ) − δ(ω − ω0 ) −
1 /(CR) 2
2
1 /(CR) + ω 0 − jω 0
×
1 /(CR ) 2
+ ω 02
π jCR
π jCR
× ×
1 1 /(CR) − jω0 1 /(CR) + jω0 1 /(CR)
2
+ω
2 0
δ( ω + ω 0 ) , δ( ω + ω 0 ) ,
[δ(ω − ω0 ) − δ(ω + ω 0 )] [δ(ω − ω 0 ) + δ(ω + ω 0 )] .
Calculating the inverse CTFT, we obtain Y (ω) =
1 CR
×
1 /(CR ) 1 /(CR) 2
+ ω 02
sin(ω0 t ) −
1 CR
×
ω0 1 /(CR) 2
+ ω02
cos(ω 0 t ) ,
or, y (t ) =
1 2
1 + C R 2 ω 20
[sin( ω 0 t ) − ω0 CR cos(ω0 t )] ,
which can be expressed as y (t ) =
1 2
1 + C R
2
ω
2 0
sin
[ω t − tan − (ω CR)]. 1
0
Problem 5.24
The transfer functions obtained in Problem 5.20 are as follows: (a) H ( ω ) =
(b) H ( ω ) =
1 3
2 ( jω ) + 6 ( jω ) + 11( jω ) + 6
1 2
( jω ) + 3 ( jω ) + 2
0
▌
Solutions
1
(c) H ( ω ) =
2
(e) H ( ω ) =
3
( jω ) + 2 ( jω ) + 1 ( jω ) + 4 1 (d) H ( ω ) = . = 2 2 j ω + j ω 6 j ω 8 + + ( ) ( ) 1 2 ( jω ) + 8 ( jω ) + 19 ( jω ) + 12
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.
| ) ω (
0.15
H | : ) a ( 0 2 . 5 P
0. 1
) ω (
1
1
H < : ) a ( 0 2 . 5 P
0.05 -5
-4
-3
-2
-1
ω | ) ω (
2
H | : ) b ( 0 2 . 5 P
-4
-3
-2
-1
0 1 (radians/s)
2
3
4
5
-2
-1
0
1
2
3
1
2
3
1
2
3
1
2
3
1
2
3
2 0 -2 -5
) ω (
-4
-3
-2
-1
0
2
3
H < : ) c ( 0 2 . 5 P
0. 2 -4
-3
-2
-1
ω
0 1 (radians/s)
2
3
4
5
0 -2 -5
-4
-3
-2
-1
0
ω (radians/s)
0. 5
) ω (
1
4
0. 4
H < : ) d ( 0 2 . 5 P
0. 3
-4
-3
-2
-1
ω
0 1 (radians/s)
2
3
4
5
5
H < : ) e ( 0 2 . 5 P
0.04 0.02 -3
-2
-1
ω
0 1 (radians/s)
-1 -5
) ω (
0.06
-4
0
-4
-3
-2
-1
0
ω (radians/s)
0.08
-5
-3
ω (radians/s)
0. 4
0. 2 -5
-4
2
0. 6
4
-2 -5
H < : ) b ( 0 2 . 5 P
0. 8
-5
0
ω (radians/s)
0. 1
H | : ) c ( 0 2 . 5 P
H | : ) e ( 0 2 . 5 P
5
0. 2
3
5
4
0. 3
| ) ω (
| ) ω (
3
) ω (
ω
H | : ) d ( 0 2 . 5 P
2
0. 4
-5
| ) ω (
0 1 (radians/s)
2
2
3
4
5
2 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 (radians/s)
Solutions
(a) H (ω) = (b) H (ω) = (c) H (ω) =
(d) H (ω) =
Y (ω) X (ω) Y (ω)
X (ω) Y (ω) X (ω) Y (ω) X (ω)
= 5. = 3e − j 4ω . = =
6 ( 2 + jω) 3
.
(4 + 2 jω)(2 + jω) (1 + jω)(3 + jω)
=
2( 2 + jω)
2
(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. 1 | ) ω (
4
) ω (
H | : ) a ( 1 2 . 5 P
2
H < : ) a ( 1 2 . 5 P
0 -5
0.5
1
1
-4
-3
-2
-1
0 1 ω (radians/s)
2
3
4
0 -0.5 -1 -5
5
-4
-3
-2
-1 ω
0 1 (radians/s)
2
3
4
5
0 1 (radians/s)
2
3
4
5
0 1 (radians/s)
2
3
4
5
0 1 (radians/s)
2
3
4
5
3 | ) ω (
2
H | : ) b ( 1 2 . 5 P
) ω ( 2 H < : ) b ( 1 2 . 5 P
2
1
0 -5
-4
-3
-2
-1 ω
| ) ω (
0 1 (radians/s)
2
3
4
5
H | : ) d ( 1 2 . 5 P
-4
-3
-2
-1
2
3
H < : ) c ( 1 2 . 5 P
0.4 0.2
-4
-3
-2
-1 ω
4
-2 -5
) ω (
0.6
-5
| ) ω (
0
ω
3
H | : ) c ( 1 2 . 5 P
2
0 1 (radians/s)
2
3
4
5
0
-2 -5
-4
-3
-2
-1 ω
2.6
) ω (
0.1
4
2.4
H < : ) d ( 1 2 . 5 P
2.2
0 -0.1
2 -5
-4
-3
-2
-1 ω
0 1 (radians/s)
2
3
4
5
-5
-4
-3
-2
-1 ω
Fig. S5.25: Gain and phase responses for Problem 5.25.
Problem 5.26 Case II with input signal given by sin(
0t ):
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 ( −ω0 )δ(ω + ω 0 ) − H (ω 0 )δ(ω − ω0 )] .
or,
Expressing H (ω0) and H (−ω0) in terms of their real and imaginary components, we obtain Y (ω) = jπ A(ω 0 )[δ(ω + ω0 ) − δ(ω − ω0 )] + πB (ω0 )[δ(ω + ω 0 ) + δ(ω − ω0 )] . Calculating the inverse CTFT, the output y(t ) is obtained as
y (t ) = A(ω0 ) sin(ω0 t) + B(ω0 ) cos(ω0 t )
= ( A(ω0 )) 2 + ( B(ω0 )) 2 sin {ω0t + tan −1( B(ω0 ) / A(ω0 ))} . = H (ω0 ) sin {ω0t + < H (ω 0 )} where
H (ω0 )
=
( A(ω0 )) 2
+ ( B(ω0 )) 2
Case I with input signal given by cos(
< H (ω0 ) = tan −1 ( B(ω0 ) / A(ω0 )) .
and
0t ):
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 ( −ω0 )δ(ω + ω0 ) + H (ω0 )δ(ω − ω0 )] .
or,
Expressing H (ω0) and H (−ω0) in terms of their real and imaginary components, we get
Y (ω) = π A(ω 0 )[δ(ω + ω0 ) + δ(ω − ω0 )] − jπB(ω0 )[δ(ω + ω 0 ) − δ(ω − ω 0 )] . Taking the inverse CTFT, the output y(t ) is given by
y (t ) = A(ω0 ) cos(ω0 t) − B(ω0 )sin(ω0 t )
= ( A(ω0 )) 2 + ( B( ω0 )) 2 cos {ω0t + tan −1 ( B( ω0 ) / A(ω0 ))} . = H (ω0 ) cos {ω0t + < H (ω 0 )} where H (ω0 )
=
( A(ω0 )) 2
+ ( B(ω0 )) 2
and
< H (ω 0 ) = tan −1 ( B(ω 0 ) / A(ω0 )) .
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(ω0 t + < H (ω0 ) ) where
H (ω) =
1 1 + jωCR
.
Substituting the value of the magnitude and phase of H (ω0) at the fundamental frequency the output is given by
ω = ω0,
194
Chapter 5
y (t ) =
(ii)
1 1 + (ω0 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 (ω0 ) sin (ω 0 t + < H (ω 0 ) ) 1
H (ω) =
where
1 + jωCR
.
Substituting the value of the magnitude and phase of H (ω0) at the fundamental frequency the output is given by y (t ) =
1 1 + (ω0 CR ) 2
sin
(ω0 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,
→ H (3) sin (3t + < H (3) ) . sin (3t ) 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 + 0.32
= 0.9578
and
< H (3) = −16.700 .
The output is given by
(
0
)
y1 (t ) = 0.9578 sin 3t − 16.70 . (ii)
Based on the solution of Problem 5.26,
→ H (3) cos(3t + < H (3) ) cos(3t ) 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 + 0.32
= 0.9578
ω = ω0,
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
= 0.8575
1 + 0.6 2
and
195
< H (6) = −30.960 .
Using the linearity property, the output is given by
(
y1 (t ) = 0.9578 cos 3t − 16.700 (iii)
)− 4.2875 sin(6t − 0.96 ). 0
Based on the solution of Problem 5.26,
→ H (2) cos(2t + < H (2) ) cos(2t ) and
→ H (2000) sin (2000 t + < H (2000) sin ( 2000t ) ) At ω = 2 radians/s, the magnitude and phase of the RC circuit is given by H (2)
=
1 1 + 0.3
= 0.9806
2
and
< H (3) = −11.310 .
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 (iv)
)+ 0.0050 sin (2000t − 89.71 ) . 0
Based on Eq. (5.75)
→ H (3) exp( j3t + j < H (3) ) exp( j3t ) 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
= 0.9578
1 + 0.32
and
< H (3) = −16.700 .
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 j3t − 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+1 jω ,
H (ω) = 2 +1 jω .
and
The CTFT of the output is then given by
Y (ω ) = H (ω ) X (ω ) =
1
1
(1+ jω ) ( 2 + jω )
= 1+1 jω − 2+1jω .
Calculating the inverse CTFT results in the output signal
[
]
y (t ) = e − t − e − 2t u (t ). (c)
Y ( ω)
As H (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 dt
+ 2 y (t ) = x(t ) . − t
The output can be obtained by solving the differential equation with input x (t ) = e 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 p y zs (t ) = Ke −t for t ≥ 0 .
Inserting the particular solution in the differential equation results in K = 1. Therefore, p y zs (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 H 1 (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 H 1(ω): Since H 1(ω) eliminates all frequency components outside the range | ω| = 1), the output is given by
≤ 4 (as T
Y 1(ω) = j 2δ(ω + 3) + j 6δ(ω + 1) + 3πδ(ω) − j 6δ(ω − 1) − j 2δ(ω − 3) . Output for H 2(ω): Since H 2(ω) eliminates all frequency components outside the range 4 the output is given by
≤ |ω| ≤ 8,
Y 1(ω) = j 76 δ(ω + 7) + j 65 δ(ω + 5) − j 65 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 − nT π ) .
Output for H 1(ω): Since H 1(ω) eliminates all frequency components outside the range | ω| ≤ 4/T , the output is given by π ) + 1.5πδ(ω) − δ(ω − π ) . Y 2(ω) = −δ(ω + T T
Output for H 2(ω): Since H 2(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
∞
δ( ω − ∑ = −∞ 1 n
2 nπ ). T
n n≠0
Output for H 1(ω): Since H 1(ω) eliminates all frequency components outside the range | ω| ≤ 4/T , the output is given by Y 3(ω) = πδ(ω) . Output for H 2(ω): Since H 2(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 + 2T π ) − jδ(ω − 2T π ) . (d)
From the solution of Problem P5.19(d), the CTFT of Fig. P4.6(d) is given by X 4(ω) = 2π
∞
∑ = −∞
Dn δ(ω − nω0 ) =πδ(ω) + π4
n
∞
∑ = −∞
n n≠0 odd n
1 n2
δ(ω − nT π ) .
Output for H 1(ω): Since H 1(ω) eliminates all frequency components outside the range | ω| ≤ 4/T , the output is given by π ) + πδ(ω) + 4 δ(ω − π ) . Y 4(ω) = π4 δ(ω + T T π
Output for H 2(ω): Since H 2(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
0
n =−∞
= 0.6816πδ (ω ) + j 0.3866πδ (ω + ∞
− j 2 ∑
n =−∞ n ≠ 0,1 odd n
1 n
π T
) − j 0.3866πδ (ω − ) + π T
∞
∑
n =−∞ n≠0 even n
1 n 2 −1
δ (ω − nT π )
δ (ω − nT π ).
Output for H 1(ω): Since H 1(ω) eliminates all frequency components outside the range | ω| ≤ 4/T , the output is given by π ) + 0.6816 πδ(ω) − j 0.3866 πδ(ω − π ) . Y 5(ω) = j 0.3866πδ(ω + T T
Output for H 2(ω): Since H 2(ω) eliminates all frequency components outside the range 4/ T ≤ |ω| 8/T , the output is given by Y 5(ω) =
1 3
δ(ω + 2T π ) + 13 δ(ω − 2T π ) .
≤
▌
Problem 5.31
(a)
The magnitude spectra of the two systems are calculated below H 1 (ω) H 2 (ω)
=
1 = 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 |H 1(ω)| is 1 at all frequencies. Therefore, System H 1(ω) is an all pass filter.
Solutions
199
From Fig. S5.31(b), we observe that the magnitude | H 2(ω)| is zero at frequencies below 20 radians/s. At frequencies above 20 radians/s, the magnitude is 1. Therefore, System H 2(ω) is a highpass filter.
1
H 1 (ω)
1
H 2 (ω)
ω
− ωc
ω
ωc
0
− 20
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 h2 (t ) = ℑ−1 { 1 − rect( 40 ) } = ℑ−1 { 1 } − ℑ −1 { rect( 40 ) } = δ (t ) −
20 π
sinc( 20π t ) .
▌
Problem 5.32
The transfer functions for the three LTIC systems are given by System (a):
System (b):
System (c):
H 1 (ω) =
2
H 2 (ω) = πδ(ω) + H 3 (ω) = −2 +
.
(1 + jω) 2
5 2 + jω
1 jω
=
.
1 − j 2ω 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 0 ) B d (
-20
e d u t i n -40 g a M
-60 -80 0
) g e d ( e s a h P
-45 -90 -135 -180 -1
0
10
1
10
2
10
10
Frequency (rad/sec)
(a) Bode Plot for System-2
Bode Plot for System-3 10
40
) B d (
e d u t i n g a M
20
) B d (
e d u t i n g a M
0
-20
e s a h P
0
-5
-10 360
-40 -89
) g e d (
5
-89.5
) g e d (
-90
e s a h P
-90.5
315 270 225 180
-91 -1
0
10
10
Frequency (rad/sec)
1
10
2
10
-1
0
10
10
Frequency (rad/sec)
(b)
(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
Y 1 (ω ) =
2 (1 + jω )
2
× π [δ (ω − 1) + δ (ω + 1) ] = 2π ( (1+ j11)
2
δ (ω −1) + (1−1j1)2 δ (ω + 1)
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 × π [δ (ω − 1) + δ (ω + 1)] = π j1 δ (ω − 1) + −1j δ (ω + 1) jω = − jπ [δ (ω − 1) − δ (ω + 1)] .
Y 2 (ω ) = πδ (ω ) +
Calculating the inverse CTFT, we obtain
y2 (t ) = sin t . System (c): Using the convolution property, the output of system (c) is given by
Y 2 (ω ) =
1 − j 2ω 2 + jω
× π [δ (ω − 1) + δ (ω + 1) ] = π 12−+ j j2 δ (ω −1) + 12+−j j2 δ (ω +1)
= − 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 Hermitian Symmetric H ( ω)
cos(t ) → H (1) cos(ω0t + < H (1) ) . In other words, the output for x(t ) = cos(t ) depends only on the magnitude and phase of the system at 1. For the three systems, we note that H 1 (1)
= H 2 (1) =
H 3 (1)
=1
ω=
and
< H 1 (1) = H 2 (1) = H 3 (1) = − π2 . Since the magnitudes and phases of the three system transfer functions at systems produce the same output y (t ) = sin t for x(t ) = cos(t ).
ω = 1 are identical, the three
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 1 | ) t ( 1 x |
2
) t (
1
x <
1
0
-1 0 -60
-40
-20
0 20 ω (radians/s)
40
60
-60
x 2(t) = sin(8π t)+sin(20π t): Magnitude spectrum
-40
-20
0 20 ω (radians/s)
40
60
x 2(t) = sin(8π t)+sin(20π t): Phase spectrum
3 1 | ) t (
2
) t ( 2 x <
2
x |
1
0 -1
0
-100
-50
0 (radians/s) ω
50
100
-100
-50
0 (radians/s) ω
50
100
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 0.15 ) t ( y t u p t u o
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) 0
) B -50 d ( e d u t i n g a -100 M
-150 0
) g e d ( e s a h P
-90
-180
-270 -2
10
-1
0
10
10 ω
1
10
2
10
(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 -20 ) B d ( e d u t i n g a M
-40 -60 -80 -100 0
) g e d ( e s a h P
-45 -90 -135 -180 -2
10
-1
0
10
10 ω
1
10
2
10
(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 -20 ) B d ( e d u t i n g a M
-40 -60 -80 -100 0
) g e d (
e s a h P
-45 -90 -135 -180 -2
10
-1
0
10
10 ω
1
10
2
10
(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 -10 ) B d ( -20 e d u t i n -30 g a M
-40 -50 0
) g e d ( e s a h P
-45
-90 -2
10
-1
0
10
10 ω
1
10
2
10
(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 -40 ) B -60 d ( e d -80 u t i n g a -100 M
-120 -140 0
) g e d ( e s a h P
-90
-180
-270 -2
10
-1
0
10
10 ω
1
10
2
10
(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 -20 ) B d ( e d u t i n g a M
-40 -60 -80 -100 -120 0
) g e d (
e s a h P
-90
-180
-270 -2
10
-1
0
10
10 ω
1
10
2
10
(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.
