Chapter 15, Solution 1. (a)
2
ee
)atcosh(
at-at
+
=

[]
=






+
+
−
=
as
1
as
1
2
1
)atcosh(L
22

1
)atsinh(L
22
as
a
−
Chapter 15, Solution 2. (a)
)sin()tsin()cos()tcos()t(f θω−θω=

[ ] [ ]
)tsin()sin()tcos()cos()s(F ωθ−ωθ= LL=)s(F
22
s
)sin()cos(s
ω+
θω−θ(b)
)sin()tcos()cos()tsin()t(f θω+θω=
[ ] [ ]

(b)

[]
=
)t(u)t4sin(e
-2t
L
16)2s(
4
2
++(c)

Since
[]
22
as
s
)atcosh(
−
=L

[]
=)t(u)t2cosh(e
-3t
L
4)3s(
3s

2
t-
++
=
L

If
)s(F)t(f →←)s(F
ds
d-
)t(ft →←

Thus,
[]()
[ ]
1-
2t-
4)1s(2
ds
d-
)t2sin(et ++=
L
)1s(2
)4)1s((

2
s
s
22
+
=
+
=
−
−(b)
3s
e
5
s
2
)s(F
s2
2
+
+=
−Chapter 15, Solution 5. (a)

2
L
()








+








−=
1-
2
4s1s
2
3
ds

2
3
s24s
2
3
ds
d()
()()() ()
3
2
2
2
2
2
2
2
2
4s
1s
2
3
)s8(
4s
2
3
s2
4s









−
−
+
=() ()
3
2
2
2
2
4s
1s
2
3
)s8(
4s
s32s3s3-
+




[]
=°+ )30t2cos(t
2
L
()
3
2
32
4s
s3s6s3128
+
+−−(b)

[]
=
+
⋅=
5
t-4
)2s(
!4
30et30
L
5
)2s(
720
(d)

)t(ue2)t(ue2
-t1)--(t
=
[]
=
)t(ue2
1)--(t
L
1s
e2
+(e)

Using the scaling property,
[]
=⋅⋅=⋅⋅=
s2
1
25
)21(s
1
21
1
5)2t(u5

−
′
−−=






=






δ
−−
)0(fs)0(fs)s(Fs)t(f
dt
d
)t(
dt
d
2n1nn
n
n
n
n
LL

LL=






δ
)t(
dt
d
n
n
L
n
s
Chapter 15, Solution 6. (a)

[ ]
=−δ
)1t(2


(d)

[][ ]
=−=−
)4t(uee2)4t(ue2
4)--(t-4-t
LL
)1s(e
e2
4
-4s
+
Chapter 15, Solution 7. (a)

Since
[]
22
4s
s
)t4cos(
+
=
L

L
,

[]
3
t2-2
)2s(
2
et
+
=
L
, and
[]
s
e
)3t(u
s3-
=−
L[]
=−+
)3t(u)t(uet
t2-2
L
s
e
)2s(

1
2
+
−
+
+⋅⋅+⋅==
3s
10
2s
4
s
6
3
2
+
−
+
++(b)

)1t(ue)1t(ue)1t()1t(uet
-t-t-t
−+−−=−
)1t(uee)1t(uee)1t()1t(uet
-11)--(t-11)--(t-t
(c)

[ ]
=−−
)1t(u))1t(2cos(L
4s
es
2
-s
+ (d)

Since )t4sin()t4cos()4sin()4cos()t4sin())t(4sin(
=π−π=π−

)t(u))t(4sin()t(u)t4sin(
π−π−=π−[ ][ ]
)t(u)t(u)t4sin(
π−−
L
Chapter 15, Solution 9. (a)

)2t(u2)2t(u)2t()2t(u)4t()t(f
−−−−=−−==
)s(F
2
-2s
2
-2s
s
e2
s
e
−(b)

)1t(uee2)1t(ue2)t(g
1)--4(t-4-4t
−=−=


)1cos(5)s(H
22
+
⋅+
+
⋅==
)s(H
4s
415.8
4s
s702.2
22
+
+
+(d)

)4t(u6)2t(u6)t(p
−−−==
)s(P
4s-2s-
e




++
+
+
++
+
=
1)1s(
1
ds
d
1)1s(
1s
ds
d
11)(s
1s
)s(G
222=
++
+
−
++
+
−


=
++
+
=






++
+
⋅=
22
2
2
)2s2(s
2s)(s)(s
1)1s(
1s
ds
d-
)s()s(G
22
2
)2s2(s
2)(ss
++
+
(b) Since
[]
22
as
a
)atsinh(
−
=L

[]
12s4s
12
16)2s(
)4)(3(
)t4sinh(e3
22
2t-
−+
=
−+
=L

[][]
1-22t-
)12s4s(12
ds
d-
)t4sinh(e3t)s(F −+=⋅= L

-4t-2t
−+−

=
)2t(uee4)2t(uee4
2)--4(t-82)--2(t-4
−+−

[][ ]
)t(ueee4)2t(uee4
-2-2s-42)--2(t-4
LL ⋅=−

[]
2s
e4
)2t(uee4
4)-(2s
2)-2(t-4-
+
=−
+
LSimilarly,
[]
4s
e4
)2t(uee4


Chapter 15, Solution 12.

)2s(
2
)2s(s
2
2
2
s
)1t(22)1t(222)1t(2
e
)2s(
3s
2s
1
1
2s
e
2s
e
e
)2s(
e
e)s(f
)1t(uee)1t(uee)1t()1t(uete)t(f
+−
+−−
−
−


If f(t) = cost, then
22
2
2
)1(
)12()1)(1(
)( and
1
)(
+
+−+
=
+
=
s
sss
sF
ds
d
s
s
sF22
2
)1(
1
)cos(

++
+−++
=
ss
sss
ds
dF22
)22(
)1(2
)sin(
++
+
=−=
−
ss
s
ds
dF
tte
t
L

(c )
∫
∞
→←
s

ββ
β
β
ββ
111
22
tantan
2
tan
1sin
−−
∞
−
∞
=−==
+
=






∫
L
Chapter 15, Solution 14.


)s(F +−=

=)s(F
)ee21(
s
5
s2-s-
2
+−
Chapter 15, Solution 15.
[]
)2t(u)1t(u)1t(u)t(u10)t(f −+−−−−==






+−=
s
e


Chapter 15, Solution 17. 






>
<<
<<
<
=
3t0
3t11
1t0t
0t0
)t(f
2[][ ]
)3t(u)1t(u1)1t(u)t(ut)t(f
2
−−−+−−=

[ ] [ ]
)3t(u)2t(u3)2t(u)1t(u2)1t(u)t(u)t(g −−−+−−−+−−=)3t(u3)2t(u)1t(u)t(u −−−+−+==)s(G
)e3ee1(
s
1
s3-s2-s-
−++(b)

[ ] [ ]
)3t(u)1t(u2)1t(u)t(ut2)t(h −−−+−−=

[ ]
)4t(u)3t(u)t28( −−−−+
)3t(u2)1t(u2)1t(u2)1t(u)1t(2)t(ut2 −−−+−−−−−=

)4t(u)4t(2)3t(u2)3t(u)3t(2
−−+−+−−−

Chapter 15, Solution 19. Since
[ ]
1)t( =δ
L
and ,
2T = =)s(F
s2-
e1
1
−
Chapter 15, Solution 20. Let
1t0),tsin()t(g
1
<<π=[]
)1t(u)t(u)tsin( −−π=

)1t(u)tsin()t(u)tsin(
−π−π=

2s-22
-s
−π+
+π
Chapter 15, Solution 21.
π= 2T

Let
[]
)1t(u)t(u
2
t
1)t(f
1
−−






π
−=


s-
2
s-
2
1
s2
e1-se)12-(2
s
1
e
2
1
1-
s2
e
s2
1
s
1
)s(F
π
+++π+π
=⋅






π


Let
1t0,t2)t(g
1
<<=[ ]
)1t(u)t(ut2 −−=)1t(u2)1t(u)1t(2)t(ut2 −+−−−=s-
2
-s
2
1
e
s
2
s
e2
s
2
)s(G +−=1T,

1 0
h



<<−
<<
=
2t1t24
1t0t2
)t(h
1)2t(u)2t(2)1t(ut2)1t(u4)1t(ut2)t(ut2)t(h
1
−−−−−−+−−=

)2t(u)2t(2)1t(u)1t(4)t(ut2)t(h
1
−−−−−−=2s-
22
-2s
s-
22
1
)e1(

s
1
2s-
2-s
2
−
−
+
Chapter 15, Solution 23. (a)

Let



<<
<<
=
2t11-
1t01
)t(f
1

[ ] [ ]
)2t(u)1t(u)1t(u)t(u)t(f

)e1(s
)e1(
2s-
2-s
−
−(b)

Let
[ ]
)2t(ut)t(ut)2t(u)t(ut)t(h
222
1
−−=−−=
)2t(u4)2t(u)2t(4)2t(u)2t()t(ut)t(h
22
1
−−−−−−−−=

2s-2s-
2
2s-
3
1
e
s
4
e

5s6s
ss10
lim)s(sFlim)0(f
2
4
ss
++
+
==
∞→∞→
==
++
+
∞→
0
10
s
5
s
6
s
1
s
1
10
lim
432

ss
1The complex poles are not in the left-half plane.
)(f ∞ existnotdoes(c)
)5s2s)(2s)(1s(
s7s2
lim)s(sFlim)0(f
2
3
ss
++++
+
==
∞→∞→
==






++

7
s
2
lim
2
3
s
=
0==
++++
+
==∞
→→
10
0
)5s2s)(2s)(1s(
s7s2
lim)s(sFlim)(f
2
3
0s0s
0Chapter 15, Solution 25.




+






+
∞→
s
4
1
s
2
1
s
3
1
s
1
1)8(
lim
s
=
8



−
=
∞→
1
0
s
1
1
s
1
s
1
6
lim)0(f
4
42
s
0All poles are not in the left-half plane.
)(f ∞ existnotdoes
Chapter 15, Solution 26. (a)

=






++






−
+−
∞→
2
2
s
s
4
s
2
1
s
2

11)4s(3
)s(G
+
−=
+
−+
=

=)t(g
-4t
e11)t(3
−δ

(c)

3s
B
1s
A
)3s)(1s(
4
)s(H
+
+
+
=
++
=

2A =

22
+
+
+
+
+
=
++
=

6
2
12
B == ,
3
)-2(
12
C
2
==

2
)2s(C)4s(B)4s()2s(A12 ++++++=

Equating coefficients :
2
s
:
-3-CACA0 ==→+=
1


(a)
)t(ut)e4e2()t(f
5s
4
3s
2
5s
2
)4(2
3s
2
)2(2
)s(F
t5t3 −−
+−=
+
+
+
−
=
+
−
−
+
+
−
=
++++++=+++++=+
++
+
+
+
=
+++
+
=

Chapter 15, Solution 29. 2222
22
22
3)2s(
3
3
2
3)2s(
)2s(2
s
2
)s(V
6Band2A26s2BsAs26s8s2;
3)2s(
BAs
s
2

3
2
3)2s(
)2s(2
3)2s(
2)2s(2
)s(H
++
+
++
+
=
++
++
=

t3sine
3
2
t3cose2)t(h
t2t2
1
−−
+=

(b)
)5s2s(
DCs
)1s(
B

−=→+=
DBADC2BA31:s
2
++=+++=
2/1C,2/1A2A4D2B2A6D2CB2A70:s =−=→+=++=+++=

4/1D,4/5B1B4A4DB5A54:constant ==→++=++=









++
−+
+
+
+
+
−
=








()
)t(ut2sine5.0t2cose2te5e2
4
1
)t(h
tttt
2
−−−−
−++−=

(c )








+
+
+
=








()
)1t(uee
2
1
)t(h
)1t(3)1t(
3
−+=
−−−−
Chapter 15, Solution 31. (a)

3s
C
2s
B
1s
A
)3s)(2s)(1s(
s10
)s(F
+
+
+

=3s
15
2s
20
1s
5-
)s(F
+
−
+
+
+
=

=)t(f
-3t-2t-t
e15e20e5-
−+

(b)

323
2
)2s(
D
)2s(
C

)4s4s)(1s(B)4s4s)(2s(A1s4s2
222
+++++++=++
)1s(D)2s)(1s(C +++++

Equating coefficients :
3
s: 1 -ABBA0 ==→+=
2
s: 3 A2CCACB5A62 =−=→+=++=
1
s:
DC3A4DC3B8A124 ++=+++=

-1A-2DDA64 =−=→++=
0
s:
116-4D2C4ADC2B4A81 =−+=++=+++=32
)2s(
1
)2s(
3
2s
1
1s
1-
)s(F


−++(c)

5s2s
CBs
2s
A
)5s2s)(2s(
1s
)s(F
22
++
+
+
+
=
+++
+
=

5
1-
)2s()s(FA
-2s
=+=
=


2s
51-
)s(F
++
+
++
+
+
+
=
++
+⋅
+
+
=

=)t(f
)t2sin(e4.0)t2cos(e2.0e0.2-
-t-t-2t
++

Chapter 15, Solution 32. (a)
4s
C

=3
(-2))-4(
)(8)(-1)(-3
)4s()s(FC
-4s
==+=
=4s
3
2s
2
s
3
)s(F
+
+
+
+=

=)t(f
-4t-2t
e3e2)t(u3
++

(b)

A1BBA1 −=→+=
1
s:
CA3CB3A42- ++=++=

0
s:
-6B2B-CB2A44 =→−=++=

7B1A =−=

-12A--5C ==2
)2s(
12
2s
6
1s
7
)s(F
+
−
+
−
+
=

=
Equating coefficients :
2
s
: A1BBA1
−=→+=
1
s: -3CACA3CB3A40
=+→++=++=
0
s: 5A2A-9C3A51
=→+=+=

-4A1B =−=

-83A-C =−=1)2s(
)2s(4
3s
5
1)2s(
8s4
3s
5
)s(F
22
++

224
+
+
+
+
=
++
=
−
−
=

)1s(C)1s(B)ss(A6
22
+++++=

Equating coefficients :
2
s: -CACA0 =→+=
1
s: C -ABBA0 ==→+=
0
s
: 3B2BCB6 =→=+=

-3A =
,
3B =
,
3C =


(b)

1s
es
)s(F
2
s-
+
=
π=)t(f
)t(u)tcos(
π−π−

(c)

323
)1s(
D
)1s(
C
1s
B
s
A
)1s(s
8
32
)1s(
8
)1s(
8
1s
8
s
8
)s(F
+
−
+
−
+
−=

=)t(f
[]
)t(uet5.0ete18
-t2-t-t
−−−
Chapter 15, Solution 34.

Let
4s
B
2s
A
)4s)(2s(
1
+
+
+
=
++21A =

21B =





+
+
+
+

−−+−−

(c)

Let
4s
C
3s
B
s
A
)4s)(3s(s
1s
+
+
+
+=
++
+121A =
,
32B =
,
43-C =2s-
e

−






−+
Chapter 15, Solution 35.

(a)

Let
2s
B
1s
A
)2s)(1s(
3s
)s(G
+
+
+
=
++
+
=

1s
A
)4s)(1s(
1
)s(G
+
+
+
=
++
=

31A
=
,
31-B
=)4s(3
1
)1s(3
1
)s(G
+
−
+
=

[]

A
)4s)(3s(
s
)s(G
22
+
+
+
+
=
++
=

133-A
=)3s(C)s3s(B)4s(As
22
+++++=

Equating coefficients :
2
s: -ABBA0 =→+=
1
s:
CB31 +=
0
s
:

)1t(u)1t(g)t(f
−−=

=)t(f
[]
)1t(u))1t(2sin(2))1t(2cos(3e3-
13
1
1)-3(t-
−−+−+Chapter 15, Solution 36.

(a)

3s
D
2s
C
s
B
s
A
)3s)(2s(s
1
)s(X
22
+
+

s
: 61BB61 =→=