Bài giải phần giải mạch P15
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Chapter 15, Solution 1.
(a)
2
ee
)atcosh(
at-at
+
=
[]
=
+
+
−
=
as
1
as
1
2
1
)atcosh(L
22
as
s
−
(b)
2
ee
)atsinh(
at-at
−
=
[]
=
+
−
−
=
as
1
as
1
2
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=
[ ] [ ]
)tsin()cos()tcos()sin()s(F ωθ+ωθ= LL
=)s(F
22
s
)cos()sin(s
ω+
θω−θ
Chapter 15, Solution 3.
(a)
[]
=
)t(u)t3cos(e
-2t
L
9)2s(
2s
2
++
+
(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
−+
+
(d) Since
[]
22
as
a
)atsinh(
−
=L
[
=)t(u)tsinh(e
-4t
L
]
1)4s(
1
2
−+
(e)
[]
4)1s(
2
)t2sin(e
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
22
+⋅
++
=
[]
=
)t2sin(et
-t
L
22
)4)1s((
)1s(4
++
+
Chapter 15, Solution 4.
(a)
16s
se6
e
4s
s
6)s(G
2
s
s
22
+
=
+
=
−
−
(b)
3s
e
5
s
2
)s(F
s2
2
+
+=
−
Chapter 15, Solution 5.
(a)
[]
4s
)30sin(2)30cos(s
)30t2cos(
2
+
°
−°
=°+
L
[]
+
−°
=°+
4s
1)30cos(s
ds
d
)30t2cos(t
22
2
2
L
()
+
−=
1-
2
4s1s
2
3
ds
d
ds
d
() ()
+
−−+=
2-
2
1-
2
4s1s
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
1s
2
3
2
4s
2s-
2
3
+
−
+
+
−
+
−
−
+
=
() ()
3
2
2
2
2
4s
1s
2
3
)s8(
4s
s32s3s3-
+
−
+
+
−+−
=
() ()
3
2
23
3
2
2
4s
s8s34
4s
)42)(ss3(-3
+
−
+
+
++
=
[]
=°+ )30t2cos(t
2
L
()
3
2
32
4s
s3s6s3128
+
+−−
(b)
[]
=
+
⋅=
5
t-4
)2s(
!4
30et30
L
5
)2s(
720
+
(c)
=−⋅−=
δ−
)01s(4
s
2
)t(
dt
d
4)t(ut2
2
L
s4
s
2
2
−
(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
L
s
5
(f)
[]
=
+
=
31s
6
)t(ue6
3t-
L
1s3
18
+
(g)
Let f . Then, )t()t(
δ=
1)s(F
=
.
"
−
′
−−=
=
δ
−−
)0(fs)0(fs)s(Fs)t(f
dt
d
)t(
dt
d
2n1nn
n
n
n
n
LL
"
−⋅−⋅−⋅=
=
δ
−−
0s0s1s)t(f
dt
d
)t(
dt
d
2n1nn
n
n
n
n
LL
=
δ
)t(
dt
d
n
n
L
n
s
Chapter 15, Solution 6.
(a)
[ ]
=−δ
)1t(2
L
-s
e2
(b)
[ ]
=−
)2t(u10
L
2s-
e
s
10
(c)
[ ]
=+
)t(u)4t(
L
s
4
s
1
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
, we use the linearity and shift properties to
obtain
[]
=−−
)1t(u))1t(4cos(10
L
16s
es10
2
s-
+
(b)
Since
[]
3
2
s
2
t
=
L
,
[]
s
1
)t(u
=
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(
2
-3s
3
+
+
Chapter 15, Solution 8.
(a)
[]
t3-t2-
e10e4)t2(u6)t3(2
−++δ
L
3s
10
2s
4
2s
1
2
1
6
3
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
−+−−=−
[]
=
+
+
+
=−
1s
ee
)1s(
ee
)1t(uet
-s-1
2
-s-1
t-
L
1s
e
)1s(
e
1)-(s
2
1)-(s
+
+
+
++
(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
[ ] [ ]
)t(u))t(4sin()t(u)t4sin(
π−π−−=
LL
=
+
−
+
=
π
16s
e4
16s
4
2
s-
2
)e1(
16s
4
s-
2
π
−⋅
+
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
−=−=
=
)s(G
)4s(e
e2
4
-s
+
(c)
)t(u)1t2cos(5)t(h
−=
)Bsin()Asin()Bcos()Acos()BAcos(
+=−
)1sin()t2sin()1cos()t2cos()1t2cos(
+=−
)t(u)t2sin()1sin(5)t(u)t2cos()1cos(5)t(h
+=
4s
2
)1sin(5
4s
s
)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
s
6
e
s
6
−
Chapter 15, Solution 10.
(a) By taking the derivative in the time domain,
)tsin(et)tcos()ee-t()t(g
-t-t-t
−+=
)tsin(et)tcos(et)tcos(e)t(g
-t-t-t
−−=
++
+
++
+
+
++
+
=
1)1s(
1
ds
d
1)1s(
1s
ds
d
11)(s
1s
)s(G
222
=
++
+
−
++
+
−
++
+
=
2222
2
2
)2s2(s
22s
)2s2(s
2ss
2s2s
1s
)s(G
22
2
)2s2(s
2)(ss
++
+
(b)
By applying the time differentiation property,
)0(f)s(sF)s(G −=
where
, f
)tcos(et)t(f
-t
= 0)0( =
=
++
+
=
++
+
⋅=
22
2
2
)2s2(s
2s)(s)(s
1)1s(
1s
ds
d-
)s()s(G
22
2
)2s2(s
2)(ss
++
+
Chapter 15, Solution 11.
(a)
Since
[]
22
as
s
)atcosh(
−
=L
=
−+
+
=
4)1s(
)1s(6
)s(F
2
3s2s
)1s(6
2
−+
+
(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
=−++=
2-2
)12s4s)(4s2)(12()s(F
22
)12s4s(
)2s(24
−+
+
(c)
)ee(
2
1
)tcosh(
t-t
+⋅=
)2t(u)ee(
2
1
e8)t(f
t-t3t-
−+⋅⋅=
=
)2t(ue4)2t(ue4
-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-
+
=−
+
L
Similarly,
[]
4s
e4
)2t(uee4
8)-(2s
2)-4(t-8-
+
=−
+
L
Therefore,
=
+
+
+
=
++
4s
e4
2s
e4
)s(F
8)-(2s4)-(2s
[ ]
8s6s
)e8e16(s)e4e4(e
2
-22-226)-(2s
++
+++
+
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
+−
+−−
−
−
−
−−−−−−−−−
+
+
=
+
+
+
=
+
+
+
=
−+−−=−=
Chapter 15, Solution 13.
(a)
)()( sF
ds
d
t −→←tf
If f(t) = cost, then
22
2
2
)1(
)12()1)(1(
)( and
1
)(
+
+−+
=
+
=
s
sss
sF
ds
d
s
s
sF
22
2
)1(
1
)cos(
+
−+
=
s
ss
ttL
(b) Let f(t) = e
-t
sin t.
22
1
1)1(
1
)(
22
++
=
++
=
sss
sF
22
2
)22(
)22)(1()0)(22(
++
+−++
=
ss
sss
ds
dF
22
)22(
)1(2
)sin(
++
+
=−=
−
ss
s
ds
dF
tte
t
L
(c )
∫
∞
→←
s
dssF
t
tf
)(
)(
Let
22
)( then ,sin)(
β
β
β
+
==
s
sFttf
s
ss
ds
s
t
t
s
s
β
β
π
ββ
β
β
ββ
111
22
tantan
2
tan
1sin
−−
∞
−
∞
=−==
+
=
∫
L
Chapter 15, Solution 14.
<<−
<<
=
2t1t510
1t0t5
)t(f
We may write f as )t(
[ ] [ ]
)2t(u)1t(u)t510()1t(u)t(ut5)t(f −−−−+−−=
)2t(u)2t(5)1t(u)1t(10)t(ut5
−−+−−−=
s2-
2
s-
22
e
s
5
e
s
10
s
5
)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
e
s
2
s
1
10)s(F
s2-
s-
2s-
)e1(
s
10
−
Chapter 15, Solution 16.
)4t(u5)3t(u3)1t(u3)t(u5)t(f −−−+−−=
=)s(F
[]
s4-s3-s-
e5e3e35
s
1
−+−
Chapter 15, Solution 17.
>
<<
<<
<
=
3t0
3t11
1t0t
0t0
)t(f
2
[][ ]
)3t(u)1t(u1)1t(u)t(ut)t(f
2
−−−+−−=
)3t(u)1t(u)1t(u)1t2-()1t(u)1t()t(ut
22
−−−+−++−−−=
)3t(u)1t(u)1t(2)1t(u)1t()t(ut
22
−−−−−−−−=
=)s(F
s
e
e
s
2
)e1(
s
2
s-3
s-
2
s-
3
−−−
Chapter 15, Solution 18.
(a)
[ ] [ ]
)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
−−+−+−−−
)4t(u)4t(2)3t(u)3t(2)1t(u)1t(2)t(ut2 −
−+−−−−−−=
=+−−=
s4-
2
s3-
2
s-
2
e
s
2
e
s
2
)e1(
s
2
)s(H
)eee1(
s
2
s4-s3-s-
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=
Note that sin( )tsin(-)tsin())1t(
π=π−π=−π .
So,
1)-u(t1))-t(sin(u(t)t)sin()t(g
1
π+π=
)e1(
s
)s(G
s-
22
1
+
π+
π
=
=
−
=
2s-
1
e1
)s(G
)s(G
)e1)(s(
)e1(
2s-22
-s
−π+
+π
Chapter 15, Solution 21.
π= 2T
Let
[]
)1t(u)t(u
2
t
1)t(f
1
−−
π
−=
)1t(u
2
1
1)1t(u)1t(
2
1
)t(u
2
t
)t(u)t(f
1
−
π
−−−−
π
+
π
−=
[ ]
[]
2
-s-s
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
++
π
+
π
−=
=
−
=
Ts-
1
e1
)s(F
)s(F
[ ]
[ ]
)e1(s2
e1-se)12-(2
s2-2
-s-s
π
−π
+++π+π
Chapter 15, Solution 22.
(a)
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,
e1
)s(G
)s(G
sT-
1
=
−
=
=)s(G
)e1(s
)ese1(2
s-2
-s-s
−
+−
(b)
Let h , where is the periodic triangular wave. )t(uh
0
+=
0
h
Let
h
be within its first period, i.e.
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
2
s
e2
e
s
4
s
2
)s(H −=−−=
)e1(
)e1(
s
2
)s(H
2s-
2-s
2
0
−
−
=
=)s(H
)e1(
)e1(
s
2
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
1
−−−−−−=
)2t(u)1t(u2)t(u)t(f
1
−+−−=
2s-2s-s-
1
)e1(
s
1
)ee21(
s
1
)s(F −=+−=
2T,
)e1(
)s(F
)s(F
sT-
1
=
−
=
=)s(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
s
4
)e1(
s
2
)s(H −−−=
2T,
)e1(
)s(H
)s(H
Ts-
1
=
−
=
=)s(H
)e1(s
)ss(es4)e1(2
2s-3
2-2s-2s
−
+−−
Chapter 15, Solution 24.
(a)
5s6s
ss10
lim)s(sFlim)0(f
2
4
ss
++
+
==
∞→∞→
==
++
+
∞→
0
10
s
5
s
6
s
1
s
1
10
lim
432
3
s
=
∞
=
++
+
==∞
→→
5s6s
ss10
lim)s(sFlim)(f
2
4
0s0s
0
(b)
=
+−
+
==
∞→∞→
6s4s
ss
lim)s(sFlim)0(f
2
2
ss
1
The complex poles are not in the left-half plane.
)(f ∞ existnotdoes
(c)
)5s2s)(2s)(1s(
s7s2
lim)s(sFlim)0(f
2
3
ss
++++
+
==
∞→∞→
==
++
+
+
+
∞→
1
0
s
5
s
2
1
s
2
1
s
1
1
s
7
s
2
lim
2
3
s
=
0
==
++++
+
==∞
→→
10
0
)5s2s)(2s)(1s(
s7s2
lim)s(sFlim)(f
2
3
0s0s
0
Chapter 15, Solution 25.
(a)
)4s)(2s(
)3s)(1s)(8(
lim)s(sFlim)0(f
ss
++
++
==
∞→∞→
=
+
+
+
+
∞→
s
4
1
s
2
1
s
3
1
s
1
1)8(
lim
s
=
8
===∞
→→
)4)(2(
)3)(1)(8(
lim)s(sFlim)(f
0s0s
3
(b)
1s
)1s(s6
lim)s(sFlim)0(f
4
ss
−
−
==
∞→∞→
==
−
−
=
∞→
1
0
s
1
1
s
1
s
1
6
lim)0(f
4
42
s
0
All poles are not in the left-half plane.
)(f ∞ existnotdoes
Chapter 15, Solution 26.
(a)
=
++
+
==
∞→∞→
6s4s
s3s
lim)s(sFlim)0(f
23
3
ss
1
Two poles are not in the left-half plane.
)(f ∞ existnotdoes
(b)
)4s2s)(2s(
ss2s
lim)s(sFlim)0(f
2
23
ss
++−
+−
==
∞→∞→
=
++
−
+−
∞→
2
2
s
s
4
s
2
1
s
2
1
s
1
s
2
1
lim
=
1
One pole is not in the left-half plane.
)(f ∞ existnotdoes
Chapter 15, Solution 27.
(a)
=)t(f
-t
e2)t(u +
(b)
4s
11
3
4s
11)4s(3
)s(G
+
−=
+
−+
=
=)t(g
-4t
e11)t(3
−δ
(c)
3s
B
1s
A
)3s)(1s(
4
)s(H
+
+
+
=
++
=
2A =
,
-2B =
3s
2
1s
2
)s(H
+
−
+
=
=)t(h
-3t-t
e2e2
−
(d)
4s
C
)2s(
B
2s
A
)4s()2s(
12
)s(J
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
s:
6-2ABBA2C4BA60 ==→+=++=
0
s:
121224-24C4B4A812 =++=++=
4s
3
)2s(
6
2s
3-
)s(J
2
+
+
+
+
+
=
=)t(j
-2t-2t-4t
et6e3e3
+−
Chapter 15, Solution 28.
(a)
)t(ut)e4e2()t(f
5s
4
3s
2
5s
2
)4(2
3s
2
)2(2
)s(F
t5t3 −−
+−=
+
+
+
−
=
+
−
−
+
+
−
=
(b)
()
)t(ut2sine5.1t2cose2e2)t(h
5s2s
1s2
1s
2
)s(H
1C;2B;2A3CB,3CB;BA;11CA5
CA5s)CBA2(s)BA()1s)(CBs()5s2s(A11s3
5s2s
CBs
1s
A
)5s2s)(1s(
11s3
)s(H
ttt
2
22
22
−−−
+−=→
++
+−
+
+
=
=−==→−==+−−==+
++++++=+++++=+
++
+
+
+
=
+++
+
=
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
)s(V
++
−
++
+
−=
−=−=→+=++++
++
+
+=
v(t) = 0t,t3sine
3
2
t3cose2)t(u
t2t2
≥−−
−−
2
Chapter 15, Solution 30.
(a)
222222
1
3)2s(
3
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
)1s(
A
)5s2s()1s(
4s
)s(
2222
2
2
++
+
+
+
+
+
=
+++
+
=H
22222
)1s(D)1s(Cs)5s2s(B)5s2s)(1s(A4s ++++++++++=+
or
)1s2s(D)ss2s(C)5s2s(B)5s7s3s(A4s
2232232
++++++++++++=+
Equating coefficients:
ACCA0:s
3
−=→+=
DBADC2BA31:s
2
++=+++=
2/1C,2/1A2A4D2B2A6D2CB2A70:s =−=→+=++=+++=
4/1D,4/5B1B4A4DB5A54:constant ==→++=++=
++
−+
+
+
+
+
−
=
++
+
+
+
+
+
−
=
)2)1s(
1)1s(2
)1s(
5
)1s(
2
4
1
)5s2s(
1s2
)1s(
5
)1s(
2
4
1
)s(H
22222
2
Hence,
()
)t(ut2sine5.0t2cose2te5e2
4
1
)t(h
tttt
2
−−−−
−++−=
(c )
+
+
+
=
+
+
+
=
++
+
=
−−
−
)3s(
1
)1s(
1
e
2
1
)3s(
B
)1s(
A
e
)3s)(1s(
e)2s(
)s(H
ss
s
3
()
)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
+
+
+
+
+
=
+++
=
-5
2
10-
)1s()s(FA
1-s
==+=
=
20
1-
20-
)2s()s(FB
-2s
==+=
=
-15
2
30-
)3s()s(FC
3-s
==+=
=
3s
15
2s
20
1s
5-
)s(F
+
−
+
+
+
=
=)t(f
-3t-2t-t
e15e20e5-
−+
(b)
323
2
)2s(
D
)2s(
C
2s
B
1s
A
)2s)(1s(
1s4s2
)s(F
+
+
+
+
+
+
+
=
++
++
=
-1)1s()s(FA
1-s
=+=
=
-1)2s()s(FD
-2s
3
=+=
=
)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
+
−
+
+
+
+
+
=
2t-
2
2t-2t-t-
e
2
t
et3ee-f(t) −++=
=)t(f
2t-
2
t-
e
2
t
t31e-
−++
(c)
5s2s
CBs
2s
A
)5s2s)(2s(
1s
)s(F
22
++
+
+
+
=
+++
+
=
5
1-
)2s()s(FA
-2s
=+=
=
)2s(C)s2s(B)5s2s(A1s
22
++++++=+
Equating coefficients :
2
s:
5
1
-ABBA0 ==→+=
1
s
: 1CC0CB2A21 =→+=++=
0
s:
121-C2A51 =+=+=
222222
2)1s(
54
2)1s(
)1s(51
2s
51-
2)1s(
1s51
2s
51-
)s(F
++
+
++
+
+
+
=
++
+⋅
+
+
=
=)t(f
)t2sin(e4.0)t2cos(e2.0e0.2-
-t-t-2t
++
Chapter 15, Solution 32.
(a)
4s
C
2s
B
s
A
)4s)(2s(s
)3s)(1s(8
)s(F
+
+
+
+=
++
++
=
3
)4)(2(
(8)(3)
s)s(FA
0s
===
=
2
)-4(
(8)(-1)
)2s()s(FB
-2s
==+=
=
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)
22
2
)2s(
C
2s
B
1s
A
)2s)(1s(
4s2s
)s(F
+
+
+
+
+
=
++
+−
=
)1s(C)2s3s(B)4s4s(A4s2s
222
+++++++=+−
Equating coefficients :
2
s:
A1BBA1 −=→+=
1
s:
CA3CB3A42- ++=++=
0
s:
-6B2B-CB2A44 =→−=++=
7B1A =−=
-12A--5C ==
2
)2s(
12
2s
6
1s
7
)s(F
+
−
+
−
+
=
=
)t(f
-2t-t
e)t21(6e7
+−
(c)
5s4s
CBs
3s
A
)5s4s)(3s(
1s
)s(F
22
2
++
+
+
+
=
+++
+
=
)3s(C)s3s(B)5s4s(A1s
222
++++++=+
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
++
+
−
+
=
++
+
−
+
=
=)t(f
)tcos(e4e5
-2t-3t
−
Chapter 15, Solution 33.
(a)
1s
C
1s
BAs
)1s)(1s(
6
1s
)1s(6
)s(F
224
+
+
+
+
=
++
=
−
−
=
)1s(C)1s(B)ss(A6
22
+++++=
Equating coefficients :
2
s: -CACA0 =→+=
1
s: C -ABBA0 ==→+=
0
s
: 3B2BCB6 =→=+=
-3A =
,
3B =
,
3C =
1s
3
1s
3s-
1s
3
1s
33s-
1s
3
)s(F
222
+
+
+
+
+
=
+
+
+
+
=
=)t(f
)tcos(3)tsin(3e3
-t
−+
(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
)s(F
+
+
+
+
+
+=
+
=
8A =
,
-8D =
sD)ss(C)ss2s(B)1s3s3s(A8
22323
+++++++++=
Equating coefficients :
3
s: -ABBA0 =→+=
2
s
: B-ACCACB2A30 ==→+=++=
1
s: -ADDADCBA30 =→+=+++=
0
s:
8D,8C,8B,8A −=−=−==
32
)1s(
8
)1s(
8
1s
8
s
8
)s(F
+
−
+
−
+
−=
=)t(f
[]
)t(uet5.0ete18
-t2-t-t
−−−
Chapter 15, Solution 34.
(a)
4s
3
11
4s
34s
10)s(F
22
2
+
−=
+
−+
+=
=)t(f
)t2sin(5.1)t(11
−δ
(b)
)4s)(2s(
e4e
)s(G
-2s-s
++
+
=
Let
4s
B
2s
A
)4s)(2s(
1
+
+
+
=
++
21A =
21B =
+
+
+
+
+
+
+
=
4s
1
2s
1
e2
4s
1
2s
1
2
e
)s(G
2s-
-s
=)t(g
[][ ]
)2t(uee2)1t(uee5.0
2)--4(t2)--2(t-1)-4(t-1)-2(t
−−+−−
(c)
Let
4s
C
3s
B
s
A
)4s)(3s(s
1s
+
+
+
+=
++
+
121A =
,
32B =
,
43-C =
2s-
e
4s
43
3s
32
s
1
12
1
)s(H
+
−
+
+⋅=
=)t(h
)2t(ue
4
3
e
3
2
12
1
2)-4(t-2)-3(t-
−
−+
Chapter 15, Solution 35.
(a)
Let
2s
B
1s
A
)2s)(1s(
3s
)s(G
+
+
+
=
++
+
=
2A = , -1B
=
2t-t-
ee2)t(g
2s
1
1s
2
)s(G −=→
+
−
+
=
)6t(u)6t(g)t(f)s(Ge)s(F
-6s
−−=→=
=)t(f
[]
)6t(uee2
6)--2(t6)--(t
−−
(b)
Let
4s
B
1s
A
)4s)(1s(
1
)s(G
+
+
+
=
++
=
31A
=
,
31-B
=
)4s(3
1
)1s(3
1
)s(G
+
−
+
=
[]
4t-t-
ee
3
1
)t(g −=
)s(Ge)s(G4)s(F
-2t
−=
)2t(u)2t(g)t(u)t(g4)t(f −−−=
=)t(f
[][]
)2t(uee
3
1
)t(uee
3
4
2)-4(t-2)-(t-4t-t-
−−−−
(c)
Let
4s
CBs
3s
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
:
C3A40
+=
133-A =
,
133B =
,
134C =
4s
4s3
3s
3-
)s(G13
2
+
+
+
+
=
)t2sin(2)t2cos(3e-3)t(g13
-3t
++=
)s(Ge)s(F
-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
+
+
+
++=
++
=
61B =
,
41C =
,
91-D =
)s2s(D)s3s(C)6s5s(B)s6s5s(A1
2323223
+++++++++=
Equating coefficients :
3
s:
DCA0 ++=
2
s
:
CBA3D2C3BA50 ++=+++=
1
s:
B5A60 +=
0
s
: 61BB61 =→=
