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. 




f"(t) = δ(t) - δ(t - 1) - δ'(t - 1)

Taking the Fourier transform gives

-ω
2
F(ω) = 1 - e
-jω
- jωe
-jωF(ω) =
2
j
1e)j1(
ω
−ω+
ωor
F

∫
ω−
=ω
1
0

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
1
ω−ωωω
−++ω−
ω
−=
()
ω+ωω−
ω
2sin2j2cos4j
2
1
2
−= F(
ω
) =
)2cos22(sin
j
2
ωω−ω
ω

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−=
′′

δ(t+1)
–δ(t–1)
1
0
h”(t)
–1

–2δ’(t)
1
t ω−ω=ω−−=ωω
δ
′

+−=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
=






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

(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

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) ]


)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.



∫∫∫
−−
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+−
−

[ ]
)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)(j
2
0
t)(j
)(j
e
e
)(j
1
j2
1
=








ω+π
−
+
ω−π

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−

j
1
jj
ω+−
ω
=−
ω
+−
ω
ω−ω
−= =
ω
ω−
=
j
2/sin4
2
2
2/
2/sin
j







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

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.
)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
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=

ω+
=ω
ω

)(He)(X
2
ω−=ω
)(Xe)(H
2
ω−=ω
−=

()
()
2
2
2j
j2
ej2
π+ω+
ω+−

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−


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−⋅=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
−

4



ωπ4−(b)
ta
e
−

22
a
a2
ω+
22
ta
a2
+

ω−
π
a
e2Chapter 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+ω
−ω++ω
−−ωδ++ωδ
π
=
22
22
2
j
22
2 H(ω) =
()()
[]
4
j
22
2
2
−ω

1
tu

() ()
()
()
()






−ω
−−ωπδ−
+ω
++ωπδ=ω
10j
1
10
10j
1
10
2
j
G ()()
[]

2
j
−ω
−−ωδ−+ωδ
πChapter 18, Solution 18.

Let f
() ()
tuet
t
−
=
()
ω+
=ω
jj
1
F

()
tcostf

()(
[]
1F1F
2
1

[]






+ω+−ω+
−ω+++ω+
=
1j11j1
jj1jj1
2
1
1jjjj1
j1
2
+ω−−ω++ω+
ω+
=

=
2j2
j1
2
+ω−ω
ω+

2
1
F
()
()
()
()
1
0
t2jt2j
e
2j
1
e
2j
1
2
1






π−ω−
+
π+ω−


But
π−π
==π+π=
2j2j
e12sinj2cose ()






π−ω
+
π+ω








−
−=ω
ω−
2

F
(c
n
) = c
n
δ(ω) F

( )
( )
on
tjn
n
ncec
o
ω−ωδ
ω
= F



=




()
∫∫






+⋅
π
==
π
−
ω−
T
00
jnt
tjn
n
0dte1
2
1
dtetf
T
1
c
o

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


j
2
1
)t(f F(ω) =
()
∑
∞
=
≠
−∞=
−ωδ
π
−δω
oddn
0n
n
n
n
j
2
1
Chapter 18, Solution 21.
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


tj
tjtj
o
oo
() ()






−=
∫∫
∞
∞−
∞
∞−
ω+ω−ω−ω−
dtedte)t(f
j2
1
tjtj
oo =
()(

()
=− 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
ω+ω+
ω
=ωω=
j5j2
10j
Fjt'f

(e)
()
∫
f

∞−
t
dtt
( )
()
()( )
ωδπ+
ω
ω
0F
j
F
()()
()
5x2

1e
j
ω
+ωπδ
ω−
j
6 −(b)

() ( )
2tfty −=() ( )
=ω=ω
ω−
FeY
2j
()
1e
je
j
2j
−
ω
ω−
ω−






ω=ω










=
5
3
F
5
3
x10
2
3
F
2
3
x4)(G,t
3
5

( ) ( )
1e
10j
1e
4j
5/3j2/3j
−
ω
+−
ω
ω−ω−
Chapter 18, Solution 25. (a)
()
()
ω=
+
+=
+
=
js,
2s
B
s2ss
s

() ()
tue5t
2
t2
−
−
sgn
5(b)

()
()( )
2j
B
1j
A
2j1j
4j
F
+ω
+
+ω
=
+ω+ω
−ω
=ω
6
j1
5
F

f(t) =
( )
( )
tue6e5
t2t
−−
+−
Chapter 18, Solution 26. (a)
)t(ue)t(
)2t( −−
=f(b)
)t(ute)t(
t4−
=h