Bài giải mạch P18
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Chapter 18, Solution 1.
)2t()1t()1t()2t()t('f −δ+−δ−+δ−+δ=
2jjj2j
eeee)(Fj
ω−ω−ωω
+−−=ωω
ω−ω=
cos22cos2
F(ω) =
ω
ω−ω
j
]cos2[cos2
Chapter 18, Solution 2.
<<
=
otherwise,0
1t0,t
)t(f
-
δ(t-1)
δ(t)
f ”(t)
t
–δ’(t-1)
1
-δ(t-1)
0
1
f ‘(t)
t
f"(t) = δ(t) - δ(t - 1) - δ'(t - 1)
Taking the Fourier transform gives
-ω
2
F(ω) = 1 - e
-jω
- jωe
-jω
F(ω) =
2
j
1e)j1(
ω
−ω+
ω
or
F
∫
ω−
=ω
1
0
tj
dtet)(
But
∫
+−= c)1ax(
a
e
dxex
2
ax
ax
()
=−ω−
ω−
=ω
ω−
1
0
2
j
)1tj(
j
e
)(F
()
[ ]
1ej1
1
j
2
−ω+
ω
ω−
Chapter 18, Solution 3.
2t2,
2
1
)t('f,2t2,t
2
1
)t(f <<−=<<−=
∫
−
−
ω−
ω
−ω−
ω−
==ω
2
2
2
2
2
tj
tj
)1tj(
)j(2
e
dtet
2
1
)(F
[]
)12j(e)12j(e
2
1
2j2j
2
−ω−−ω−
ω
−=
ωω−
( )
[]
2j2j2j2j
2
eeee2j
2
1
ω−ωωω
−++ω−
ω
−=
()
ω+ωω−
ω
2sin2j2cos4j
2
1
2
−=
F(
ω
) =
)2cos22(sin
j
2
ωω−ω
ω
Chapter 18, Solution 4.
1
–2δ(t–1)
2δ
(t+1)
–1
g’
2
0
–2
t
–2δ’(t–1)
–2δ
(t+1)
1
–2δ(t–1)
2δ’
(t+1)
–1
g”
4δ(t)
0
–2
t
4sin4cos4
ej2e24ej2e2)(G)j(
)1t(2)1t(2)t(4)1t(2)1t(2g
jjjj2
+ωω−ω−=
ω−−+ω+−=ωω
−δ
′
−−δ−δ++δ
′
++δ−=
′′
ω−ω−ωω
)1sin(cos
4
)(G
2
−ωω+ω
ω
=ω
Chapter 18, Solution 5.
1
0
h’(t)
–1
–2δ(t)
1
t
δ(t+1)
–δ(t–1)
1
0
h”(t)
–1
–2δ’(t)
1
t
ω−ω=ω−−=ωω
δ
′
−−δ−+δ=
′′
ω−ω
j2sinj2j2ee)(H)j(
)t(2)1t()1t()t(h
jj2
H(ω) =
ω
ω
−
ω
sin
j2j2
2
Chapter 18, Solution 6.
dtetdte)1()(F
tj
0
1
1
0
tj ω−
−
ω−
∫∫
+−=ω
)1(cos
1
tsin
t
tcos
1
tsin
1
tdtcosttdtcos)(F Re
2
1
0
2
0
1
0
1
1
0
−ω
ω
=
ω
ω
+ω
ω
+ω
ω
−=
ω+ω−=ω
−
−
∫∫
Chapter 18, Solution 7.
(a) f
1
is similar to the function f(t) in Fig. 17.6.
)1t(f)t(f
1
−=
Since
ω
−ω
=
2
F ω
j
)1(cos
)(
=ω=ω
ω
)(Fe)(F
j
1
ω
−ω
ω−
j
)1(cose2
j
Alternatively,
)
)
2t()1t(2)t()t(f
'
1
−δ+−δ−δ=
e2e(eee21)(Fj
jjj2jj
1
ωωω−ω−ω−
+−=+−=ωω
)2cos2(e
j
−ω=
ω−
F
1
(ω) =
ω
−ω
ω−
j
)1e
j
(cos2
(b)
f
2
is similar to f(t) in Fig. 17.14.
f
2
(t) = 2f(t)
F
2
(ω) =
2
)cos1(4
ω
ω−
Chapter 18, Solution 8.
(a)
2
1
tj
2
2
1
tj
1
0
tj
2
1
tj
1
0
tj
)1tj(e
2
e
j
4
e
j
2
dte)t24(dte2)(F
−ω−
ω−
−
ω−
+
ω−
=
−+=ω
ω−ω−ω−
ω−ω−
∫∫
ω−ω−ω−
ω+
ω
−
ω
−
ω
+
ω
+
ω
=ω
2j
2
2jj
2
e)2j1(
2
e
j
4
j
2
e
j
22
)(F
(b) g(t) = 2[ u(t+2) – u(t-2) ] - [ u(t+1) – u(t-1) ]
ω
ω
−
ω
ω
=ω
sin22sin4
)(G
Chapter 18, Solution 9.
(a)
y(t) = u(t+2) – u(t-2) + 2[ u(t+1) – u(t-1) ]
ω
ω
+ω
ω
=ω sin
4
2sin
2
)(Y
(b)
)j1(
e22
)1tj(
e2
dte)t2()(
2
j
2
1
0
1
0
2
tj
tj
ω+
ω
−
ω
=−ω−
ω−
−
=−=ω
ω−ω−
ω−
∫
Z
Chapter 18, Solution 10.
(a) x(t) = e
2t
u(t)
X(ω) =
1/(2 + j
ω
)
(b)
<
>
=
−
−
0t,e
0t,e
e
t
t
)t(
∫∫∫
−−
ω−−ωω
+==ω
1
1
0
1
1
0
tjttjttj
dteedteedte)t(y)(Y
1
0
t)j1(
0
1
t)j1(
)j1(
e
j1
e
ω+−
+
ω−
=
ω+−
−
ω−
ω+
ω−ω
+
ω−
ω+ω
−
ω+
=
−
j1
sinjcos
j1
sinjcos
e
1
2
1
2
Y(ω) =
[ ]
)sin(cose1
1
2
1
2
ωω−ω−
ω+
−
Chapter 18, Solution 11.
f(t) = sin π t [u(t) - u(t - 2)]
( )
∫∫
ω−π−πω−
−=π=ω
2
0
2
0
tjtjtjtj
dteee
j2
1
dtetsin)(F
+
∫
π+ω−π+ω−+
2
0
t)(jt)(j
dt)ee(
j2
1
=
π+ω−
+
π−ω−
π+ω−
π−ω− 2
0
t)(j
2
0
t)(j
)(j
e
e
)(j
1
j2
1
=
ω+π
−
+
ω−π
−
ω−ω− 2j2j
e1e1
2
1
=
()
ω−
π+π
ω−π
2j
22
e22
)(2
1
=
F(ω) =
( )
1e
2j
22
−
π−ω
π
ω−
Chapter 18, Solution 12.
(a)
F =
dtedtee)(
0
2
0
t)j1(tjt
∫∫
∞
ω−ω−
=ω
=
ω−
=
ω− 2
0
t)j1(
e
j1
1
ω−
−
ω−
j1
1e
2j2
(b)
∫∫
−
ω−ω−
−+=ω
0
1
1
0
tjtj
dte)1(dte)(H
()( )
)cos22(
j
1
1e
j
1
e1
j
1
jj
ω+−
ω
=−
ω
+−
ω
ω−ω
−=
=
ω
ω−
=
j
2/sin4
2
2
2/
2/sin
j
ω
ω
ω
Chapter 18, Solution 13.
(a)
We know that
)]a()a([]at[cos +ωδ+−ωδπ=F
.
Using the time shifting property,
)a(e)a(e)]a()a([e)]a3/t(a[cos
3/j3/ja3/j
+ωδπ+−ωδπ=+ωδ+−ωδπ=π−
ππ−ωπ−
F
(b) sin tsinsintcoscostsin)1t(
π−=ππ+ππ=+π
g(t) = -u(t+1) sin (t+1)
Let x(t) = u(t)sin t, then
22
1
1
1)j(
1
)(X
ω−
=
+ω
=ω
Using the time shifting property,
1
e
e
1
1
)(G
2
j
j
2
−ω
=
ω−
−=ω
ω
ω
(c ) Let y(t) = 1 + Asin at, then
Y )]a()a([Aj)(2)( −ωδ−+ωδπ+ωπδ=ω
h(t) = y(t) cos bt
Using the modulation property,
)]b(Y)b(Y[
2
1
)(H −ω++ω=ω
[][]
)ba()ba()ba()ba(
2
Aj
)b()b()(H −−ωδ−−+ωδ++−ωδ−++ωδ
π
+−ωδ++ωδπ=ω
(d)
)14j(
e
j
e1
)1tj(
e
j
e
dte)t1()(
2
4j4j
2
4
0
2
tj
4
0
tj
tj
+ω
ω
−
ω
−
ω
=−ω−
ω−
−
ω−
=−=ω
ω−ω−ω−ω−
ω−
∫
I
Chapter 18, Solution 14.
(a)
)t3cos()0(t3sin)1(t3cossint3sincost3cos)t3cos( −=−−=π−π=π+
(f
)t(ut3cose)t
t
−
−=
F(ω) =
( )
()
9j1
j1
2
+ω+
ω+−
(b)
[]
)1t(u)1t(utcos)t('g −−−ππ=
g(t)
t
1
-1
-1
1
-
π
-1 1
g’(t)
t
π
)1t()1t()t(g)t("g
2
−πδ++πδ−π−=
ω−ω
π+π−ωπ−=ωω−
jj22
ee)(G)(G
( )
ωπ−=−π−=ωω−π
ω−ω
sinj2)ee()(G
jj22
G(ω) =
22
sinj2
π−ω
ωπ
Alternatively, we compare this with Prob. 17.7
f(t) = g(t - 1)
F(ω) = G(ω)e
-jω
()
( )
ωω−ω
−
π−ω
π
=ω=ω
jj
22
j
eee)(FG
22
sin2j
π−ω
ωπ−
=
G(ω) =
22
sinj2
ω−π
ωπ
(c)
tcos)0(tsin)1(tcossintsincostcos)1t(cos π−=π+−π=ππ+ππ=−π
Let ex
=
)t(he)1t(u)1t(cos)t(
2)1t(2
−=−−π
−−
and
)t(u)tcos(e)t(y
t2
π=
−
22
)j2(
j2
)(Y
π+ω+
ω+
=ω
)1t(x)t(y −=
ω−
ω=ω
j
e)(X)(Y
()
()
2
2
j
j2
ej2
)(X
π+ω+
ω+
=ω
ω
)(He)(X
2
ω−=ω
)(Xe)(H
2
ω−=ω
−
=
()
()
2
2
2j
j2
ej2
π+ω+
ω+−
−ω
(d) Let x
)t(y)t(u)t4sin(e)t(
t2
−=−−=
−
)t(x)t(p −=
where )t(ut4sine)t(y
t2
=
()
2
2
4j2
j2
)(Y
+ω+
ω+
=ω
()
16j2
j2
)(Y)(X
2
+ω−
ω−
=ω−=ω
=ω−=ω )(X)(p
()
162j
2j
2
+−ω
−ω
(e)
2j2j
e
j
1
)(23e
j
8
)(Q
ω−ω−
ω
+ωπδ−+
ω
=ω
Q(ω) =
2j2j
e)(23e
j
6
ω−ω
ωπδ−+
ω
Chapter 18, Solution 15.
(a) F =−=ω
ω−ω
3j3j
ee)( ω
3sinj2
(b)
Let g
ω−
=ω−δ=
j
e2)(G),1t(2)t(
=ω)(F
F
∫
∞−
t
dt)t(g
)()0(F
j
)(G
ωδπ+
ω
ω
=
)()1(2
j
e2
j
ωδ−πδ+
ω
=
ω−
=
ω
ω−
j
e
j
2
(c)
F
[]
1
2
1
)t2( ⋅=δ
=ω−⋅=ω j
2
1
1
3
1
)(F
2
j
3
1
ω
−
Chapter 18, Solution 16.
(a) Using duality properly
2
2
t
ω
−
→
ωπ→
−
2
t
2
2
or
ωπ−→ 4
t
4
2
F(
ω
) =
F
=
2
t
4
ωπ4−
(b)
ta
e
−
22
a
a2
ω+
22
ta
a2
+
ω−
π
a
e2
22
ta
8
+
ω−
π
2
e4
G(ω) =
F
=
+
2
t4
8
ω−
π
2
e4
Chapter 18, Solution 17.
(a) Since H(ω) =
F
()()(
[]
000
FF
2
1
)t(ftcos ω−ω+ω+ω=ω
)
where F(ω) =
F
()
[]
()
2,
j
1
tu
0
=ω
ω
+ωπδ=
() ()
()
()
()
−ω
+−ωπδ+
+ω
++ωπδ=ω
2j
1
2
2j
1
2
2
1
H
()()
[]
()()
−ω+ω
−ω++ω
−−ωδ++ωδ
π
=
22
22
2
j
22
2
H(ω) =
()()
[]
4
j
22
2
2
−ω
ω
−−ωδ++ωδ
π
(b)
G(ω) =
F
[]
()(
[]
000
FF
2
j
)t(ftsin ω−ω−ω+ω=ω
)
where F(ω) =
F
()
[]
()
ω
+ωπδ=
j
1
tu
() ()
()
()
()
−ω
−−ωπδ−
+ω
++ωπδ=ω
10j
1
10
10j
1
10
2
j
G
()()
[]
+ω
−
−ω
+−ωδ−+ωδ
π
=
10
j
10
j
2
j
1010
2
j
=
()()
[]
100
10
1010
2
2
j
−ω
−−ωδ−+ωδ
π
Chapter 18, Solution 18.
Let f
() ()
tuet
t
−
=
()
ω+
=ω
jj
1
F
()
tcostf
()(
[]
1F1F
2
1
+ω+−ω
)
Hence
()
() ()
+ω+
+
−ω+
=ω
1j1
1
1j1
1
2
1
Y
()
[]
()
[]
+ω+−ω+
−ω+++ω+
=
1j11j1
jj1jj1
2
1
1jjjj1
j1
2
+ω−−ω++ω+
ω+
=
=
2j2
j1
2
+ω−ω
ω+
Chapter 18, Solution 19.
()
( )
∫∫
∞
∞−
ω−π−πω
+==ω dteee
2
1
dte)t(fF
tj
1
0
t2jt2jtj
()
() ()
[]
∫
π−ω−π+ω−
+=ω
1
0
t2jt2j
dtee
2
1
F
()
()
()
()
1
0
t2jt2j
e
2j
1
e
2j
1
2
1
π−ω−
+
π+ω−
=
π−ω−π+ω−
()
()
( )
()
π−ω
−
+
π+ω
−
−=
π−ω−π+ω−
2j
1e
2j
1e
2
1
2j2j
But
π−π
==π+π=
2j2j
e12sinj2cose
()
π−ω
+
π+ω
−
−=ω
ω−
2
1
2
1
j
1e
2
1
F
j
=
( )
1e
4
j
j
22
−
π−ω
ω
ω−
Chapter 18, Solution 20.
(a)
F
(c
n
) = c
n
δ(ω)
F
( )
( )
on
tjn
n
ncec
o
ω−ωδ
ω
=
F
=
∑
∞
−∞=
ω
n
tjn
n
o
ec
()
∑
∞
−∞=
ω−ωδ
n
on
nc
(b)
π= 2T
1
T
2
o
=
π
=ω
()
∫∫
+⋅
π
==
π
−
ω−
T
00
jnt
tjn
n
0dte1
2
1
dtetf
T
1
c
o
()
1e
n2
j
e
jn
1
2
1
jn
0
jnt
−
π
=
−
π
=
π−π
But
e
njn
)1(ncosnsinjncos −=π=π+π=
π−
()
[]
=−−
π
=
=
≠=
π
−
evenn,0
0n,oddn,
n
j
n
n
11
n2
j
c
for n = 0
∫
π
=
π
=
0
n
2
1
dt1
2
1
c
Hence
∑
∞
=
≠
−∞=
π
−=
oddn
0n
n
jnt
e
n
j
2
1
)t(f
F(ω) =
()
∑
∞
=
≠
−∞=
−ωδ
π
−δω
oddn
0n
n
n
n
j
2
1
Chapter 18, Solution 21.
Using Parseval’s theorem,
∫∫
∞
∞−
∞
∞−
ωω
π
=
d|)(F|
2
1
dt)t(f
22
If f(t) = u(t+a) – u(t+a), then
∫∫∫
∞
∞−
∞
∞−−
ω
ω
ω
π
=== d
a
asin
a4
2
1
a2dt)1(dt)t(f
2
2
a
a
22
or
a
a4
a4
d
a
asin
2
2
π
=
π
=ω
ω
ω
∫
∞
∞−
as required.
Chapter 18, Solution 22.
F
[]
( )
∫
∞
∞−
ω−
ω−ω
−
=ω dte
j2
ee
)t(ftsin)t(f
tj
tjtj
o
oo
() ()
−=
∫∫
∞
∞−
∞
∞−
ω+ω−ω−ω−
dtedte)t(f
j2
1
tjtj
oo
=
()(
[]
oo
FF
j2
1
ω+ω−ω−ω
)
Chapter 18, Solution 23.
(a) f(3t) leads to
()()()( )
ω+ω+
=
ω+ω+
⋅
j15j6
30
3/j53/j2
10
3
1
F
[]
()
=− t3f
()( )
ω−ω− j15j6
30
(b) f(2t)
()( )()( )
ω+ω+
=
ω+ω+
⋅
j10j4
20
2/j152/j2
10
2
1
f(2t-1) = f [2(t-1/2)]
()( )
ω+ω+
ω−
j10j4
e20
2/j
(c) f(t) cos 2t
()()
2F
2
1
2F
2
1
+ω++ω
=
[]
()()
[]
()()
[][]
2j52j2
5
2j52j2
−ω+−ω+
+
+ω++ω+
5
(d)
F
[]
() ()
()()
ω+ω+
ω
=ωω=
j5j2
10j
Fjt'f
(e)
()
∫
f
∞−
t
dtt
( )
()
()( )
ωδπ+
ω
ω
0F
j
F
()()
()
5x2
10x
j5j2j
ωπδ+
ω+ω+ω
10
=
=
()()
()
ωπδ+
ω+ω+ω
j5j2j
10
Chapter 18, Solution 24.
(a)
() ()
ω=ω FX
+
F
[3]
=
()
( )
1e
j
ω
+ωπδ
ω−
j
6 −
(b)
() ( )
2tfty −=
() ( )
=ω=ω
ω−
FeY
2j
()
1e
je
j
2j
−
ω
ω−
ω−
(c)
If h(t) = f '(t)
() ()
( )
=−
ω
ω=ωω=ω
ω−
1e
j
jFjH
j
ω−
−
j
e1
(d)
()
ω+
ω=ω
=
5
3
F
5
3
x10
2
3
F
2
3
x4)(G,t
3
5
f
3
ft
+
10t
2
4g
( ) ( )
1e
5
3
j6
1e
2
3
j
6
5/3j2/3j
−
ω
+−
ω
⋅=
ω−ω−
=
( ) ( )
1e
10j
1e
4j
5/3j2/3j
−
ω
+−
ω
ω−ω−
Chapter 18, Solution 25.
(a)
()
()
ω=
+
+=
+
=
js,
2s
B
s2ss
s
A10
F
5
2
10
B,5
2
10
A
−=
−
===
()
2j
5
j
5
F
+ω
−
ω
=ω
f(t) =
() ()
tue5t
2
t2
−
−
sgn
5
(b)
()
()( )
2j
B
1j
A
2j1j
4j
F
+ω
+
+ω
=
+ω+ω
−ω
=ω
()
()( )
ω=
+
+
+
=
++
−
=
js,
2s
B
1s
A
2s1s
4s
sF
A = 5, B = 6
()
ω+
+
ω+
−
=ω
j2
6
j1
5
F
f(t) =
( )
( )
tue6e5
t2t
−−
+−
Chapter 18, Solution 26.
(a)
)t(ue)t(
)2t( −−
=f
(b)
)t(ute)t(
t4−
=h
(c) If
ω
ω
=ω→−−+=
sin
2)(X)1t(u)1t(u)t(x
By using duality property,
