Chapter 18, Solution 1.
f ' ( t ) = δ( t + 2) − δ( t + 1) − δ( t − 1) + δ( t − 2)
jωF(ω) = e j2 ω − e jω − e − jω + e − jω2
= 2 cos 2ω − 2 cos ω
2[cos 2ω − cos ω]
F(ω) =
jω
Chapter 18, Solution 2.
t, 0 < t Chapter 18, Solution 3.
1 1
f (t) = t , − 2 < t < 2, f ' (t) = , − 2 < t < 2
2 2
2 1 jωt e − jωt
F(ω) = ∫ t e dt = (− jωt − 1) 2 2
−
−2 2 2(− jω) 2
=−
1
2ω 2
[
e − jω2 (− jω2 − 1) − e jω2 ( jω2 − 1) ]
=−
1
2ω 2
[
− jω2(e jω2 + e jω2 ) + e jω2 − e − jω2 ]
1
=− (− jω4 cos 2ω + j2 sin 2ω)
2ω 2
j
F(ω) = (sin 2ω − 2ω cos 2ω)
ω2
Chapter 18, Solution 4.
2δ(t+1)
g’
2
–1
1 t
0
–2
–2δ(t–1)
4δ(t)
2δ’(t+1)
g”
–1
1 t
0
–2δ(t+1) –2
–2δ(t–1)
–2δ’(t–1)
g ′′ = −2δ( t + 1) + 2δ′( t + 1) + 4δ( t ) − 2δ( t − 1) − 2δ′( t − 1)
( jω) 2 G (ω) = −2e jω + 2 jωe jω + 4 − 2e − jω − 2 jωe − jω
= −4 cos ω − 4ω sin ω + 4
4
G (ω) = (cos ω + ω sin ω − 1)
ω2
Chapter 18, Solution 5.
h’(t)
1
0 t
–1 1
–2δ(t)
h”(t)
δ(t+1) 1
1
0 t
–1
–δ(t–1)
–2δ’(t)
h ′′( t ) = δ( t + 1) − δ( t − 1) − 2δ′( t )
( jω) 2 H(ω) = e jω − e − jω − 2 jω = 2 j sin ω − 2 jω
2j 2j
H(ω) = − sin ω
ω ω2
Chapter 18, Solution 6.
0 1
− jωt
F(ω) = ∫ (−1)e dt + ∫ te − jωt dt
−1 0
0 1
Re F(ω) = − ∫ cos ωtdt + ∫ t cos ωtdt
−1 0
1 0 1 t 1 1
=− 2 cos ωt + ω sin ωt 0 = 2 (cos ω − 1)
sin ωt −1 +
ω ω ω
Chapter 18, Solution 7.
(a) f1 is similar to the function f(t) in Fig. 17.6.
f 1 ( t ) = f ( t − 1)
2(cos ω − 1)
Since F(ω) =
jω
2e − jω (cos ω − 1)
jω
F1 (ω) = e F(ω) =
jω
Alternatively,
f 1' ( t ) = δ( t ) − 2δ( t − 1) + δ( t − 2)
jωF1 (ω) = 1 − 2e − jω + e − j2 ω = e − jω (e jω − 2 + e jω )
= e − jω (2 cos ω − 2)
2e − jω (cos ω − 1)
F1(ω) =
jω
(b) f2 is similar to f(t) in Fig. 17.14.
f2(t) = 2f(t)
4(1 − cos ω)
F2(ω) =
ω2
Chapter 18, Solution 8.
1 2
− jωt
F(ω) = ∫ 2e dt + ∫ (4 − 2 t )e − jωt dt
(a) 0 1
2 − jωt 1 4 − jωt 2 2 − jωt 2
= e +
0 − jω
e 1
− e (− jωt − 1) 1
− jω −ω 2
2 2 − jω 2 4 − j2ω 2
F(ω) = + e + − e − (1 + j2ω)e − j2ω
ω 2 jω jω jω ω 2
(b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
4 sin 2ω 2 sin ω
G (ω) = −
ω ω
Chapter 18, Solution 9.
(a) y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
2 4
Y(ω) = sin 2ω + sin ω
ω ω
1
− 2e − jωt 2 2e − j ω
(b) Z(ω) = ∫ (−2 t )e − jωt dt =
1
(− jωt − 1) 0 = − (1 + jω)
0 − ω2 ω 2 2
ω
Chapter 18, Solution 10.
(a) x(t) = e2tu(t)
X(ω) = 1/(2 + jω)
e − t , t > 0
(b) e −( t ) = t
e , t < 0
1 0 1
Y(ω) = ∫ y( t )e jωt dt = ∫ e t e jωt dt + ∫ e − t e − jωt dt
−1 −1 0
e (1− jω) t e − (1+ jω) t
= 0
−1 + 1
0
1 − jω − (1 + jω)
2 cos ω + jsin ω cos ω − jsin ω
= − e −1 +
1+ ω2
1 − jω 1 + jω
Y(ω) =
2
1+ ω 2
[
1 − e −1 (cos ω − ω sin ω) ]
Chapter 18, Solution 11.
f(t) = sin π t [u(t) - u(t - 2)]
0
2
F(ω) = ∫ sin πt e − jωt dt =
1 2 j πt
2j
(
∫0 e − e e dt
− j πt − j ωt
)
1 2 + j( − ω + π ) t
+ e − j( ω + π ) t )dt
2 j ∫0
= (e
1 1 e − j( ω+ π ) t 2
= e − j ( ω− π ) t 0 +
2
0
2 j − j(ω − π) − j(ω + π)
1 1 − e − j2 ω 1 − e − j2 ω
= +
2 π−ω
π+ω
=
1
2π + 2πe − j2 ω ( )
2(π − ω )
2 2
F(ω) =
π
(
e − jω 2 − 1 )
ω −π22
Chapter 18, Solution 12.
∞ 2
(a) F(ω) = ∫ e t e − jωt dt = ∫ e (1− jω) t dt
0 0
1 e 2− jω 2 − 1
= e (1− jω) t 2
0 =
1 − jω 1 − jω
0 1
(b) H(ω) = ∫ e − jωt dt + ∫ (−1)e − jωt dt
−1 0
=−
1
jω
( )
1 − e jω +
1 − jω
jω
e −1 = (1
jω
)
(−2 + 2 cos ω)
2
− 4 sin 2 ω / 2 sin ω / 2
= = jω
jω ω/ 2
Chapter 18, Solution 13.
(a) We know that F[cos at ] = π[δ(ω − a ) + δ(ω + a )] .
Using the time shifting property,
F[cos a ( t − π / 3a )] = πe − jωπ / 3a [δ(ω − a ) + δ(ω + a )] = πe − jπ / 3δ(ω − a ) + πe jπ / 3δ(ω + a )
(b) sin π( t + 1) = sin πt cos π + cos πt sin π = − sin πt
g(t) = -u(t+1) sin (t+1)
1 1
Let x(t) = u(t)sin t, then X(ω) = =
( jω) 2 + 1 1 − ω2
Using the time shifting property,
1 e jω
G (ω) = − e jω =
1 − ω2 ω2 − 1
(c ) Let y(t) = 1 + Asin at, then Y(ω) = 2πδ(ω) + jπA[δ(ω + a ) − δ(ω − a )]
h(t) = y(t) cos bt
Using the modulation property,
1
H(ω) = [Y(ω + b) + Y(ω − b)]
2
jπA
H(ω) = π[δ(ω + b) + δ(ω − b)] + [δ(ω + a + b) − δ(ω − a + b) + δ(ω + a − b) − δ(ω − a − b)]
2
4
e − j ωt e − j ωt 1 e − j4ω e − j4ω
(d) I(ω) = ∫ (1 − t )e − jωt dt =
4
− (− jωt − 1) 0 = − − ( j4ω + 1)
− jω − ω 2 ω2 jω ω2
0
Chapter 18, Solution 14.
(a) cos(3t + π) = cos 3t cos π − sin 3t sin π = cos 3t (−1) − sin 3t (0) = − cos(3t )
f ( t ) = −e − t cos 3t u ( t )
− (1 + jω )
F(ω) =
(1 + jω)2 + 9
(b)
g(t)
1
-1
1 t
-1
g’(t)
π
-1 1
t
-π
g ' ( t ) = π cos πt[u ( t − 1) − u ( t − 1)]
g" ( t ) = −π 2 g( t ) − πδ( t + 1) + πδ( t − 1)
− ω 2 G (ω) = − π 2 G (ω) − πe jω + πe − jω
(π 2 − ω2 )G(ω) = −π(e jω − e − jω ) = −2 jπ sin ω
2 jπ sin ω
G(ω) =
ω2 − π 2
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F(ω) = G(ω)e-jω
π
G (ω) = F(ω)e jω = (e − jω − e jω )
ω −π
2 2
− j2π sin ω
=
ω2 − π 2
2 jπ sin ω
G(ω) =
π 2 − ω2
(c) cos π( t − 1) = cos πt cos π + sin πt sin π = cos πt (−1) + sin πt (0) = − cos πt
Let x ( t ) = e −2( t −1) cos π( t − 1)u ( t − 1) = −e 2 h ( t )
and y( t ) = e −2 t cos(πt )u ( t )
2 + jω
Y(ω) =
(2 + jω) 2 + π 2
y( t ) = x ( t − 1)
Y(ω) = X(ω)e − jω
X(ω) =
(2 + jω)e jω
(2 + jω)2 + π 2
X(ω) = −e 2 H(ω)
H(ω) = −e −2 X(ω)
− (2 + jω)e jω− 2
=
(2 + jω)2 + π 2
(d) Let x ( t ) = e −2 t sin( −4t )u (− t ) = y(− t )
p( t ) = − x ( t )
where y( t ) = e 2 t sin 4t u ( t )
2 + jω
Y (ω) =
(2 + jω)2 + 4 2
2 − jω
X(ω) = Y(−ω) =
(2 − jω)2 + 16
jω − 2
p(ω) = −X(ω) =
(jω − 2 )2 + 16
8 − jω 2 1
(e) Q(ω) = e + 3 − 2 πδ(ω) + e − jω2
jω jω
6 jω 2
Q(ω) = e + 3 − 2πδ(ω)e − jω 2
jω
Chapter 18, Solution 15.
(a) F(ω) = e j3ω − e − jω3 = 2 j sin 3ω
(b) Let g( t ) = 2δ( t − 1), G (ω) = 2e − jω
F(ω) = F ∫ g ( t ) dt
t
−∞
G (ω)
= + πF(0)δ(ω)
jω
2e − j ω
= + 2πδ(−1)δ(ω)
jω
2e − jω
=
jω
1
(c) F [δ(2t )] = ⋅1
2
1 1 1 jω
F(ω) = ⋅ 1 − jω = −
3 2 3 2
Chapter 18, Solution 16.
(a) Using duality properly
−2
t →
ω2
−2
→ 2π ω
t2
4
or → − 4π ω
t2
4
F(ω) = F 2 = − 4π ω
t
−at 2a
(b) e
a + ω2
2
2a −a ω
2π e
a + t2
2
8 −2 ω
4π e
a + t2
2
8 −2 ω
G(ω) = F 2
= 4π e
4+t
Chapter 18, Solution 17.
1
(a) Since H(ω) = F (cos ω0 t f ( t ) ) = [F(ω + ω0 ) + F(ω − ω0 )]
2
1
where F(ω) = F [u (t )] = πδ(ω) + , ω0 = 2
jω
1 1 1
H(ω) = πδ (ω + 2) + ( + πδ(ω − 2) +
2 j ω + 2) j (ω − 2)
=
π
[δ(ω + 2) + δ(ω − 2)] − j ω + 2 + ω − 2
2 2 (ω + 2)(ω − 2)
π
H(ω) = [δ(ω + 2) + δ(ω − 2 )] − 2jω
2 ω −4
j
(b) G(ω) = F [sin ω0 t f ( t )] = [F(ω + ω0 ) − F(ω − ω0 )]
2
1
where F(ω) = F [u (t )] = πδ (ω) +
jω
j 1 1
G (ω) = πδ(ω + 10) + j(ω + 10) − πδ(ω − 10) − j(ω − 10 )
2
jπ
= [δ(ω + 10) − δ (ω − 10)] + j j − j
2 2 ω − 10 ω + 10
jπ
= [δ(ω + 10) − δ(ω − 10 )] − 2 10
2 ω − 100
Chapter 18, Solution 18.
1
Let f (t ) = e − t u (t ) F(ω) =
j + jω
1
f (t ) cos t [F(ω − 1) + F(ω + 1)]
2
1 1 1
Hence Y(ω) = +
2 1 + j (ω − 1) 1 + j (ω + 1)
1 1 + jω + j + 1 + jω − j
=
2 [1 + j(ω − 1)][1 + j (ω + 1)]
1 + jω
=
1 + jω + j + jω − j − ω 2 + 1
1 + jω
=
2 jω − ω 2 + 2
Chapter 18, Solution 19.
∫0 (e + e )e dt
∞ 1 1 j2 πt
F(ω) = ∫ f ( t )e jωt dt = − j2 πt − jωt
−∞ 2
F(ω) =
2 ∫0 e [
1 1 − j( ω + 2 π ) t
]
+ e − j(ω− 2 π )t dt
1
1 1 1
= e − j( ω + 2 π ) t + e − j( ω − 2 π ) t
2 − j (ω + 2π ) − j(ω − 2π ) 0
1 e − j( ω+ 2 π ) − 1 e − j( ω− 2 π ) − 1
=− +
2 j (ω + 2π) j(ω − 2π )
But e j2 π = cos 2π + j sin 2π = 1 = e − j2 π
1 e − jω − 1 1 1
F(ω) = −
ω + 2π + ω − 2π
2 j
=
jω
(
e − jω − 1 )
ω − 4π
2 2
Chapter 18, Solution 20.
(a) F (cn) = cnδ(ω)
( )
F c n e jnωo t = c n δ(ω − nωo )
∞ ∞
F ∑ c n e jnωo t =
n = −∞
∑ c δ(ω − nω )
n = −∞
n o
2π
(b) T = 2π ωo = =1
T
1 T 1 π
∫0 f (t ) e o dt = 2π ∫0 1⋅ e dt + 0
− jnω t − jnt
cn =
T
1 1 jnt
2πn (e − 1)
π j − jnπ
= − e =
2π jn
0
But e − jnπ = cos nπ + j sin nπ = cos nπ = (−1) n
cn =
j
2nπ
[
(− 1)n − 1 = 0−,j ,] n = even
n = odd , n ≠ 0
nπ
for n = 0
1 π 1
cn =
2π ∫0 1dt = 2
Hence
∞
1 j jnt
f (t) = − ∑ e
2 n = −∞ nπ
n ≠0
n = odd
∞
1 j
F(ω) = δω − ∑ δ(ω − n )
2 n = −∞ nπ
n≠0
n = odd
Chapter 18, Solution 21.
Using Parseval’s theorem,
∞ 2 1 ∞ 2
∫− ∞ f ( t )dt =
2π ∫− ∞ | F(ω) | dω
If f(t) = u(t+a) – u(t+a), then
2
∞ a 1 ∞ 2 sin aω
∫−∞ f 2 ( t )dt = ∫ (1) 2 dt = 2a = ∫−∞ 4a aω dω
−a 2π
or
2
∞ sin aω 4πa π
∫− ∞ aω dω = 4a 2 = a as required.
Chapter 18, Solution 22.
F [f ( t ) sin ωo t ] = ∫ f ( t )
∞ (e jω o t
)
− e − j ω o t − j ωt
e dt
−∞ 2j
1 ∞
f ( t )e − j(ω− ωo )t dt − ∫ e − j(ω+ ωo )t dt
∞
∫− ∞
=
2j −∞
1
= [F(ω − ω o ) − F(ω + ωo )]
2j
Chapter 18, Solution 23.
1 10 30
(a) f(3t) leads to ⋅ =
3 (2 + jω / 3)(5 + jω / 3) (6 + jω)(15 + jω)
30
F [f (− 3t )] =
(6 − jω)(15 − jω)
1 10 20
(b) f(2t) ⋅ =
2 (2 + jω / 2)(15 + jω / 2) (4 + jω)(10 + jω)
20e − jω / 2
f(2t-1) = f [2(t-1/2)]
(4 + jω)(10 + jω)
1 1
(c) f(t) cos 2t F(ω + 2) + F(ω + 2 )
2 2
5 5
= +
[2 + j(ω + 2)][5 + j(ω + 2)] [2 + j(ω − 2 )[5 + j(ω − 2)]]
jω10
(d) F [f ' (t )] = jω F(ω) =
(2 + jω)(5 + jω)
F(ω)
∫ f (t ) dt + πF(0 )δ(ω)
t
(e)
−∞ j(ω)
10 x10
= + πδ(ω)
jω(2 + jω)(5 + jω) 2x5
10
= + πδ(ω)
jω(2 + jω)(5 + jω)
Chapter 18, Solution 24.
(a) X (ω) = F(ω) + F [3]
= 6πδ(ω) +
ω
(
j − jω
e −1 )
(b) y(t ) = f (t − 2 )
je − j2ω − jω
Y(ω) = e − jω2 F(ω) =
ω
e −1 ( )
(c) If h(t) = f '(t)
H(ω) = jωF(ω) = jω
ω
(
j − jω
)
e − 1 = 1 − e − jω
2 5 3 3 3 3
(d) g(t ) = 4f t + 10f t , G (ω) = 4 x F ω + 10x F ω
3 3 2 2 5 5
= 6⋅
3
j
(e − j3ω / 2
−1 +) 3
(
6 j − j3ω / 5
e )
−1
ω ω
2 5
=
ω
e(
j4 − j3ω / 2
−1 +
ω
e )
j10 − j3ω / 5
−1 ( )
Chapter 18, Solution 25.
10 A B
(a) F(s ) = = + , s = jω
s(s + 2) s s + 2
10 10
A= = 5, B = = −5
2 −2
5 5
F(ω) = −
jω jω + 2
5
f(t) = sgn(t ) − 5e −2 t u(t )
2
jω − 4 A B
(b) F(ω) = = +
( jω + 1)( jω + 2) jω + 1 jω + 2
s−4 A B
F(s ) = = + , s = jω
(s + 1)(s + 2) s + 1 s + 2
A = 5, B = 6
−5 6
F(ω) = +
1 + jω 2 + jω
(
f(t) = − 5e − t + 6e −2 t u(t ) )
Chapter 18, Solution 26.
(a) f ( t ) = e −( t −2) u ( t )
(b) h ( t ) = te −4 t u ( t )
sin ω
(c) If x ( t ) = u ( t + 1) − u ( t − 1)
→ X(ω) = 2
ω
By using duality property,
2 sin t
G (ω) = 2u (ω + 1) − 2u (ω − 1)
→ g( t ) =
πt
Chapter 18, Solution 27.
100 A B
(a) Let F(s ) = = + , s = jω
s (s + 10) s s + 10
100 100
A= = 10, B = = −10
10 − 10
10 10
F(ω) = −
jω jω + 10
f(t) = 5 sgn(t ) − 10e −10 t u(t )
10s A B
(b) G (s ) = = + , s = jω
(2 − s )(3 + s ) 2 − s s + 3
20 − 30
A= = 4, B = = −6
5 5
4 6
G (ω) = −
= − jω + 2 jω + 3
g(t) = 4e 2 t u(− t ) − 6e −3 t u(t )
60 60
(c) H (ω) = =
( j ω) 2
+ j40ω + 1300 ( jω + 20)2 + 900
h(t) = 2e −20 t sin( 30t ) u(t )
1 ∞ δ(ω)e jωt dω 1 1 1
y (t ) = ∫−∞ (2 + jω)( jω + 1) = 2 π ⋅ 2 = 4 π
2π
Chapter 18, Solution 28.
1 ∞ 1 ∞ πδ(ω) e jωt
2π ∫−∞ 2π ∫−∞ (5 + jω)(2 + jω)
(a) f (t) = F(ω)e jωt dω = dω
1 1 1
= = = 0.05
2 (5)(2) 20
1 ∞ 10δ(ω + 2) jωt 10 e − j2 t
2π ∫−∞ jω( jω + 1)
(b) f (t) = e dω =
2π (− j2)(− j2 + 1)
j5 e − j2 t ( −2 + j)e − j2 t
= =
2π 1 − j2 2π
1 ∞ 20δ(ω − 1)e jωt 20 e jt
2π ∫−∞ (2 + jω)(3 + 5ω)
(c) f (t) = dω =
2π (2 + j)(3 + j)
20e jt (1 − j)e jt
= =
2π(5 + 5 j) π
5πδ(ω) 5
(d) Let F(ω) = + = F1 (ω) + F2 (ω)
(5 + jω) jω(5 + jω)
1 ∞ 5πδ(ω) jωt 5π 1
f1 ( t ) =
2π ∫−∞ 5 + jω e dω = 2π ⋅ 5 = 0.5
5 A B
F2 (s) = = + , A = 1, B = −1
s(5 + s) s s + 5
1 1
F2 (ω) = −
jω jω + 5
1 1
f 2 (t) = sgn( t ) − e −5 t = − + u ( t ) − e 5 t
2 2
f ( t ) = f 1 ( t ) + f 2 ( t ) = u( t ) − e − 5 t
Chapter 18, Solution 29.
(a) f(t) = F -1 [δ(ω)] + F -1 [4δ(ω + 3) + 4δ(ω − 3)]
1 4 cos 3t 1
= + = (1 + 8 cos 3t )
2π π 2π
(b) If h ( t ) = u ( t + 2) − u ( t − 2)
2 sin 2ω
H(ω) =
ω
1 8 sin 2 t
G (ω) = 4H(ω) g( t ) = ⋅
2π t
4 sin 2t
g(t) =
πt
(c) Since
cos(at) ↔πδ(ω + a ) + πδ(ω − a )
Using the reversal property,
2π cos 2ω ↔ πδ( t + 2) + πδ( t − 2)
or F -1 [6 cos 2ω] = 3δ(t + 2) + 3δ(t − 2)
Chapter 18, Solution 30.
2 1
(a) y( t ) = sgn( t )
→ Y(ω) = , X(ω) =
jω a + jω
Y(ω) 2(a + jω) 2a
H(ω) = = = 2+
→ h ( t ) = 2δ( t ) + a[u ( t ) − u (− t )]
X(ω) jω jω
1 1
(b) X(ω) = , Y(ω) =
1 + jω 2 + jω
1 + jω 1
H(ω) = = 1−
→ h ( t ) = δ( t ) − e − 2 t u ( t )
2 + jω 2 + jω
(c ) In this case, by definition, h ( t ) = y( t ) = e −at sin bt u ( t )
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