Chapter 15, Solution 1.
(a) 2
ee)atcosh(
at-at +=
[ ] =
++
−=
as1
as1
21
)atcosh(L 22 ass−
(b) 2
ee)atsinh(
at-at −=
[ ] =
+−
−=
as1
as1
21
)atsinh(L 22 asa−
Chapter 15, Solution 2.
(a) )sin()tsin()cos()tcos()t(f θω−θω= [ ] [ ])tsin()sin()tcos()cos()s(F ωθ−ωθ= LL
=)s(F 22s)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 22s)cos()sin(s
ω+θω−θ
Chapter 15, Solution 3.
(a) [ ] =)t(u)t3cos(e-2tL9)2s(
2s2 ++
+
(b) [ ] =)t(u)t4sin(e-2tL16)2s(
42 ++
(c) Since [ ] 22 ass
)atcosh(−
=L
[ ] =)t(u)t2cosh(e-3tL4)3s(
3s2 −+
+
(d) Since [ ] 22 asa
)atsinh(−
=L
[ =)t(u)tsinh(e-4tL ]1)4s(
12 −+
(e) [ ]4)1s(
2)t2sin(e 2
t-
++=L
If )s(F)t(f →←
)s(Fdsd-
)t(ft →←
Thus, [ ] ( )[ ]1-2t- 4)1s(2dsd-
)t2sin(et ++=L
)1s(2)4)1s((
222 +⋅
++=
[ ] =)t2sin(et -tL 22 )4)1s(()1s(4++
+
Chapter 15, Solution 4.
(a) 16s
se6e4s
s6)s(G 2
ss
22 +=
+=
−−
(b) 3s
e5s2)s(F
s2
2 ++=
−
Chapter 15, Solution 5.
(a) [ ]4s
)30sin(2)30cos(s)30t2cos( 2 +
°−°=°+L
[ ]
+−°
=°+4s
1)30cos(sdsd
)30t2cos(t 22
22L
( )
+
−=
1-2 4s1s23
dsd
dsd
( ) ( )
+
−−+=
2-21-2 4s1s23s24s
23
dsd
( )
( ) ( ) ( ) ( )32
2
222222 4s
1s23)s8(
4s
23s2
4s
1s232
4s
2s-23
+
−
++
−+
−
−+
=
( ) ( )32
2
22 4s
1s23)s8(
4ss32s3s3-
+
−
++
−+−=
( ) ( )32
23
32
2
4ss8s34
4s)42)(ss3(-3
+
−+
+
++=
[ ] =°+ )30t2cos(t 2L ( ) 32
32
4ss3s6s3128
+
+−−
(b) [ ] =+
⋅= 5t-4
)2s(!4
30et30L 5)2s(720+
(c) =−⋅−=
δ− )01s(4s2
)t(dtd
4)t(ut2 2L s4s22 −
(d) )t(ue2)t(ue2 -t1)--(t =
[ ] =)t(ue2 1)--(tL1s
e2+
(e) Using the scaling property,
[ ] =⋅⋅=⋅⋅=s2
125
)21(s1
211
5)2t(u5Ls5
(f) [ ] =+
=31s
6)t(ue6 3t-L
1s318+
(g) Let f . Then, )t()t( δ= 1)s(F = .
−′−−=
=
δ −− )0(fs)0(fs)s(Fs)t(fdtd
)t(dtd 2n1nn
n
n
n
n
LL
−⋅−⋅−⋅=
=
δ −− 0s0s1s)t(fdtd
)t(dtd 2n1nn
n
n
n
n
LL
=
δ )t(dtd
n
n
L ns
Chapter 15, Solution 6.
(a) [ ] =−δ )1t(2L -se2
(b) [ ] =− )2t(u10L 2s-es
10
(c) [ ] =+ )t(u)4t(Ls4
s12 +
(d) [ ] [ ] =−=− )4t(uee2)4t(ue2 4)--(t-4-t LL)1s(e
e24
-4s
+
Chapter 15, Solution 7.
(a) Since [ ] 22 4ss
)t4cos(+
=L , we use the linearity and shift properties to
obtain [ ] =−− )1t(u))1t(4cos(10L16ses10
2
s-
+
(b) Since [ ] 32
s2
t =L , [ ]s1
)t(u =L ,
[ ] 3t2-2
)2s(2
et+
=L , and [ ]s
e)3t(u
s3-
=−L
[ ] =−+ )3t(u)t(uet t2-2Ls
e)2s(
2 -3s
3 ++
Chapter 15, Solution 8.
(a) [ ]t3-t2- e10e4)t2(u6)t3(2 −++δL
3s10
2s4
2s1
21
631
2+
−+
+⋅⋅+⋅=
=3s
102s
4s6
32
+−
+++
(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-1t-L
1se
)1s(e 1)-(s
2
1)-(s
++
+
++
(c) [ ] =−− )1t(u))1t(2cos(L4s
es2
-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
16se4
16s4
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
se2
se
−
(b) )1t(uee2)1t(ue2)t(g 1)--4(t-4-4t −=−=
=)s(G)4s(e
e24
-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 +=
4s2
)1sin(54s
s)1cos(5)s(H 22 +
⋅++
⋅=
=)s(H4s
415.84ss702.2
22 ++
+
(d) )4t(u6)2t(u6)t(p −−−=
=)s(P 4s-2s- es6
es6
−
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(
1dsd
1)1s(1s
dsd
11)(s1s)s(G 222
=++
+−
+++
−++
+= 2222
2
2 )2s2(s22s
)2s2(s2ss
2s2s1s
)s(G 22
2
)2s2(s2)(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(s2s)(s)(s
1)1s(1s
dsd-
)s()s(G 22
2
)2s2(s2)(ss+++
Chapter 15, Solution 11.
(a) Since [ ] 22 ass
)atcosh(−
=L
=−+
+=
4)1s()1s(6
)s(F 2 3s2s)1s(6
2 −++
(b) Since [ ] 22 asa
)atsinh(−
=L
[ ]12s4s
1216)2s(
)4)(3()t4sinh(e3 22
2t-
−+=
−+=L
[ ] [ ]1-22t- )12s4s(12dsd-
)t4sinh(e3t)s(F −+=⋅= L
=−++= 2-2 )12s4s)(4s2)(12()s(F 22 )12s4s()2s(24
−++
(c) )ee(21
)tcosh( t-t +⋅=
)2t(u)ee(21
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)-(2s2)-2(t-4-
+=−
+
L
Similarly, [ ]4s
e4)2t(uee4
8)-(2s2)-4(t-8-
+=−
+
L
Therefore,
=+
++
=++
4se4
2se4
)s(F8)-(2s4)-(2s [ ]
8s6s)e8e16(s)e4e4(e
2
-22-226)-(2s
++++++
Chapter 15, Solution 12.
)2s(2
)2s(s2
2
2s
)1t(22)1t(222)1t(2
e)2s(3s
2s11
2se
2see
)2s(ee)s(f
)1t(uee)1t(uee)1t()1t(uete)t(f
+−+−−
−−
−
−−−−−−−−−
+
+=
++
+=
++
+=
−+−−=−=
Chapter 15, Solution 13.
(a) )()( sFdsdt −→←tf
If f(t) = cost, then 22
2
2 )1()12()1)(1()( and
1)(
++−+
=+
=s
ssssFdsd
sssF
22
2
)1(1)cos(
+−+
=sssttL
(b) Let f(t) = e-t sin t.
221
1)1(1)( 22 ++
=++
=sss
sF
22
2
)22()22)(1()0)(22(
+++−++
=ss
sssdsdF
22 )22()1(2)sin(++
+=−=−
sss
dsdFtte tL
(c ) ∫∞
→←s
dssFttf )()(
Let 22)( then ,sin)(β
ββ+
==s
sFttf
sssds
stt
ss
ββ
πββ
ββ
ββ 11122 tantan
2tan1sin −−∞−
∞
=−==+
=
∫L
Chapter 15, Solution 14.
<<−<<
=2t1t5101t0t5
)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
s5
es10
s5
)s(F +−=
=)s(F )ee21(s5 s2-s-2 +−
Chapter 15, Solution 15. [ ])2t(u)1t(u)1t(u)t(u10)t(f −+−−−−=
=
+−=s
ee
s2
s1
10)s(Fs2-
s- 2s- )e1(s
10−
Chapter 15, Solution 16.
)4t(u5)3t(u3)1t(u3)t(u5)t(f −−−+−−=
=)s(F [ ]s4-s3-s- e5e3e35s1
−+−
Chapter 15, Solution 17.
><<<<
<
=
3t03t111t0t
0t0
)t(f2
[ ] [ ])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(Fs
ee
s2
)e1(s2 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(s1 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
s2
es2
)e1(s2
)s(H )eee1(s2 s4-s3-s-2 +−−
Chapter 15, Solution 19.
Since [ ] 1)t( =δL and , 2T = =)s(F s2-e11−
Chapter 15, Solution 20.
Let 1t0),tsin()t(g1 <<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(g1 π+π=
)e1(s
)s(G s-221 +
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(u2t
1)t(f1 −−
π−=
)1t(u21
1)1t(u)1t(21
)t(u2t
)t(u)t(f1 −
π−−−−
π+
π−=
[ ] [ ]
2
-s-ss-
2
s-
21 s2e1-se)12-(2
s1
e21
1-s2
es2
1s1
)s(Fπ
+++π+π=⋅
π++
π+
π−=
=−
= Ts-1
e1)s(F
)s(F[ ] [ ]
)e1(s2e1-se)12-(2
s2-2
-s-s
π−π+++π+π
Chapter 15, Solution 22.
(a) Let 1t0,t2)t(g1 <<= [ ])1t(u)t(ut2 −−= )1t(u2)1t(u)1t(2)t(ut2 −+−−−=
s-
2
-s
21 es2
se2
s2
)s(G +−=
1T,e1
)s(G)s(G sT-
1 =−
=
=)s(G)e1(s
)ese1(2s-2
-s-s
−+−
(b) Let h , where is the periodic triangular wave. )t(uh 0 += 0h
Let h be within its first period, i.e. 1 0h
<<−<<
=2t1t241t0t2
)t(h1
)2t(u)2t(2)1t(ut2)1t(u4)1t(ut2)t(ut2)t(h1 −−−−−−+−−=
)2t(u)2t(2)1t(u)1t(4)t(ut2)t(h1 −−−−−−=
2s-
22
-2ss-
221 )e1(s2
se2
es4
s2
)s(H −=−−=
)e1()e1(
s2
)s(H 2s-
2-s
20 −−
=
=)s(H)e1(
)e1(s2
s1
2s-
2-s
2 −−
+
Chapter 15, Solution 23.
(a) Let
<<<<
=2t11-1t01
)t(f1
[ ] [ ])2t(u)1t(u)1t(u)t(u)t(f1 −−−−−−=
)2t(u)1t(u2)t(u)t(f1 −+−−=
2s-2s-s-1 )e1(
s1
)ee21(s1
)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-31 e
s4
es4
)e1(s2
)s(H −−−=
2T,)e1(
)s(H)s(H Ts-
1 =−
=
=)s(H)e1(s
)ss(es4)e1(22s-3
2-2s-2s
−+−−
Chapter 15, Solution 24.
(a) 5s6s
ss10lim)s(sFlim)0(f 2
4
ss +++
==∞→∞→
==++
+
∞→ 010
s5
s6
s1
s110
lim432
3
s= ∞
=+++
==∞→→ 5s6s
ss10lim)s(sFlim)(f 2
4
0s0s0
(b) =+−
+==
∞→∞→ 6s4sss
lim)s(sFlim)0(f 2
2
ss1
The complex poles are not in the left-half plane.
)(f ∞ existnotdoes
(c) )5s2s)(2s)(1s(
s7s2lim)s(sFlim)0(f 2
3
ss +++++
==∞→∞→
==
++
+
+
+
∞→ 10
s5
s21
s21
s11
s7
s2
lim
2
3
s= 0
==++++
+==∞
→→ 100
)5s2s)(2s)(1s(s7s2
lim)s(sFlim)(f 2
3
0s0s0
Chapter 15, Solution 25.
(a) )4s)(2s()3s)(1s)(8(
lim)s(sFlim)0(fss ++
++==
∞→∞→
=
+
+
+
+
∞→
s41
s21
s31
s11)8(
lims
= 8
===∞→→ )4)(2(
)3)(1)(8(lim)s(sFlim)(f
0s0s3
(b) 1s
)1s(s6lim)s(sFlim)0(f 4ss −
−==
∞→∞→
==−
−=
∞→ 10
s1
1
s1
s1
6lim)0(f
4
42
s0
All poles are not in the left-half plane.
)(f ∞ existnotdoes Chapter 15, Solution 26.
(a) =++
+==
∞→∞→ 6s4ss3s
lim)s(sFlim)0(f 23
3
ss1
Two poles are not in the left-half plane.
)(f ∞ existnotdoes
(b) )4s2s)(2s(
ss2slim)s(sFlim)0(f 2
23
ss ++−+−
==∞→∞→
=
++
−
+−
∞→
2
2
s
s4
s21
s21
s1
s21
lim= 1
One pole is not in the left-half plane.
)(f ∞ existnotdoes Chapter 15, Solution 27.
(a) =)t(f -te2)t(u +
(b) 4s
113
4s11)4s(3
)s(G+
−=+
−+=
=)t(g -4te11)t(3 −δ
(c) 3s
B1s
A)3s)(1s(
4)s(H
++
+=
++=
2A = , -2B =
3s2
1s2
)s(H+
−+
=
=)t(h -3t-t e2e2 −
(d) 4s
C)2s(
B2s
A)4s()2s(
12)s(J 22 +
++
++
=++
=
62
12B == , 3
)-2(12
C 2 ==
2)2s(C)4s(B)4s()2s(A12 ++++++=
Equating coefficients :
2s : -3-CACA0 ==→+=1s : 6-2ABBA2C4BA60 ==→+=++=0s : 121224-24C4B4A812 =++=++=
4s3
)2s(6
2s3-
)s(J 2 ++
++
+=
=)t(j -2t-2t-4t et6e3e3 +−
Chapter 15, Solution 28. (a)
)t(ut)e4e2()t(f
5s4
3s2
5s2
)4(2
3s2
)2(2
)s(F
t5t3 −− +−=
++
+−
=+−−
++
−
=
(b)
( ) )t(ut2sine5.1t2cose2e2)t(h5s2s
1s21s
2)s(H
1C;2B;2A3CB,3CB;BA;11CA5
CA5s)CBA2(s)BA()1s)(CBs()5s2s(A11s3
5s2sCBs
1sA
)5s2s)(1s(11s3)s(H
ttt2
22
22
−−− +−=→++
+−+
+=
=−==→−==+−−==+
++++++=+++++=+
++
++
+=
+++
+=
Chapter 15, Solution 29.
2222
2222
3)2s(3
32
3)2s()2s(2
s2)s(V
6Band2A26s2BsAs26s8s2;3)2s(
BAss2)s(V
++−
++
+−=
−=−=→+=++++++
++=
v(t) = 0t,t3sine32t3cose2)t(u t2t2 ≥−− −−2
Chapter 15, Solution 30.
(a) 22222213)2s(
332
3)2s()2s(2
3)2s(2)2s(2)s(H
+++
++
+=
++
++=
t3sine32t3cose2)t(h t2t2
1−− +=
(b) )5s2s(
DCs)1s(
B)1s(
A)5s2s()1s(
4s)s( 2222
22
++
++
++
+=
+++
+=H
22222 )1s(D)1s(Cs)5s2s(B)5s2s)(1s(A4s ++++++++++=+
or
)1s2s(D)ss2s(C)5s2s(B)5s7s3s(A4s 2232232 ++++++++++++=+ Equating coefficients:
ACCA0:s3 −=→+=
DBADC2BA31:s2 ++=+++=
2/1C,2/1A2A4D2B2A6D2CB2A70:s =−=→+=++=+++=
4/1D,4/5B1B4A4DB5A54:constant ==→++=++=
++
−++
++
+−
=
++
++
++
+−
=)2)1s(
1)1s(2)1s(
5)1s(
241
)5s2s(1s2
)1s(5
)1s(2
41)s(H 222222
Hence,
( ) )t(ut2sine5.0t2cose2te5e241)t(h tttt
2−−−− −++−=
(c )
++
+=
++
+=
++
+= −−
−
)3s(1
)1s(1e
21
)3s(B
)1s(Ae
)3s)(1s(e)2s()s(H ss
s3
( ) )1t(uee21)t(h )1t(3)1t(
3 −+= −−−−
Chapter 15, Solution 31.
(a) 3s
C2s
B1s
A)3s)(2s)(1s(
s10)s(F
++
++
+=
+++=
-5210-
)1s()s(FA 1-s ==+= =
201-20-
)2s()s(FB -2s ==+= =
-15230-
)3s()s(FC 3-s ==+= =
3s15
2s20
1s5-
)s(F+
−+
++
=
=)t(f -3t-2t-t e15e20e5- −+
(b) 323
2
)2s(D
)2s(C
2sB
1sA
)2s)(1s(1s4s2
)s(F+
++
++
++
=++++
=
-1)1s()s(FA 1-s =+= =
-1)2s()s(FD -2s3 =+= =
)4s4s)(1s(B)4s4s)(2s(A1s4s2 222 +++++++=++
)1s(D)2s)(1s(C +++++ Equating coefficients :
3s : 1 -ABBA0 ==→+=2s : 3 A2CCACB5A62 =−=→+=++=1s : DC3A4DC3B8A124 ++=+++=
-1A-2DDA64 =−=→++=0s : 116-4D2C4ADC2B4A81 =−+=++=+++=
32 )2s(1
)2s(3
2s1
1s1-
)s(F+
−+
++
++
=
2t-
22t-2t-t- e
2t
et3ee-f(t) −++=
=)t(f 2t-2
t- e2t
t31e-
−++
(c) 5s2s
CBs2s
A)5s2s)(2s(
1s)s(F 22 ++
++
+=
++++
=
51-
)2s()s(FA -2s =+= =
)2s(C)s2s(B)5s2s(A1s 22 ++++++=+
Equating coefficients :
2s : 51
-ABBA0 ==→+=
1s : 1CC0CB2A21 =→+=++=0s : 121-C2A51 =+=+=
222222 2)1s(54
2)1s()1s(51
2s51-
2)1s(1s51
2s51-
)s(F++
++++
++
=+++⋅
++
=
=)t(f )t2sin(e4.0)t2cos(e2.0e0.2- -t-t-2t ++
Chapter 15, Solution 32.
(a) 4s
C2s
BsA
)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 ==+= =
4s3
2s2
s3
)s(F+
++
+=
=)t(f -4t-2t e3e2)t(u3 ++
(b) 22
2
)2s(C
2sB
1sA
)2s)(1s(4s2s
)s(F+
++
++
=+++−
=
)1s(C)2s3s(B)4s4s(A4s2s 222 +++++++=+−
Equating coefficients :
2s : A1BBA1 −=→+=1s : CA3CB3A42- ++=++= 0s : -6B2B-CB2A44 =→−=++=
7B1A =−= -12A--5C ==
2)2s(12
2s6
1s7
)s(F+
−+
−+
=
=)t(f -2t-t e)t21(6e7 +−
(c) 5s4s
CBs3s
A)5s4s)(3s(
1s)s(F 22
2
+++
++
=+++
+=
)3s(C)s3s(B)5s4s(A1s 222 ++++++=+
Equating coefficients :
2s : A1BBA1 −=→+=1s : -3CACA3CB3A40 =+→++=++=0s : 5A2A-9C3A51 =→+=+=
-4A1B =−= -83A-C =−=
1)2s()2s(4
3s5
1)2s(8s4
3s5
)s(F 22 +++
−+
=++
+−
+=
=)t(f )tcos(e4e5 -2t-3t −
Chapter 15, Solution 33.
(a) 1s
C1sBAs
)1s)(1s(6
1s)1s(6
)s(F 224 ++
++
=++
=−−
=
)1s(C)1s(B)ss(A6 22 +++++=
Equating coefficients :
2s : -CACA0 =→+=1s : C -ABBA0 ==→+=0s : 3B2BCB6 =→=+=
-3A = , 3B = , 3C =
1s3
1s3s-
1s3
1s33s-
1s3
)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
1sB
sA
)1s(s8
)s(F+
++
++
+=+
=
8A = , -8D =
sD)ss(C)ss2s(B)1s3s3s(A8 22323 +++++++++=
Equating coefficients :
3s : -ABBA0 =→+=2s : B-ACCACB2A30 ==→+=++=1s : -ADDADCBA30 =→+=+++=0s : 8D,8C,8B,8A −=−=−==
32 )1s(8
)1s(8
1s8
s8
)s(F+
−+
−+
−=
=)t(f [ ] )t(uet5.0ete18 -t2-t-t −−−
Chapter 15, Solution 34.
(a) 4s
311
4s34s
10)s(F 22
2
+−=
+−+
+=
=)t(f )t2sin(5.1)t(11 −δ
(b) )4s)(2s(
e4e)s(G
-2s-s
+++
=
Let 4s
B2s
A)4s)(2s(
1+
++
=++
21A = 21B =
++
++
++
+=
4s1
2s1
e24s
12s
12
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
C3s
BsA
)4s)(3s(s1s
++
++=
+++
121A = , 32B = , 43-C =
2s-e
4s43
3s32
s1
121
)s(H
+−
++⋅=
=)t(h )2t(ue43
e32
121 2)-4(t-2)-3(t- −
−+
Chapter 15, Solution 35.
(a) Let 2s
B1s
A)2s)(1s(
3s)s(G
++
+=
+++
=
2A = , -1B =
2t-t- ee2)t(g
2s1
1s2
)s(G −=→+
−+
=
)6t(u)6t(g)t(f)s(Ge)s(F -6s −−=→=
=)t(f [ ] )6t(uee2 6)--2(t6)--(t −−
(b) Let 4s
B1s
A)4s)(1s(
1)s(G
++
+=
++=
31A = , 31-B =
)4s(31
)1s(31
)s(G+
−+
=
[ ]4t-t- ee31
)t(g −=
)s(Ge)s(G4)s(F -2t−=
)2t(u)2t(g)t(u)t(g4)t(f −−−=
=)t(f [ ] [ ] )2t(uee31
)t(uee34 2)-4(t-2)-(t-4t-t- −−−−
(c) Let 4sCBs
3sA
)4s)(3s(s
)s(G 22 ++
++
=++
=
133-A =
)3s(C)s3s(B)4s(As 22 +++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : CB31 +=0s : C3A40 +=
133-A = , 133B = , 134C =
4s4s3
3s3-
)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(3e3-131 1)-3(t- −−+−+
Chapter 15, Solution 36.
(a) 3s
D2s
CsB
sA
)3s)(2s(s1
)s(X 22 ++
+++=
++=
61B = , 41C = , 91-D =
)s2s(D)s3s(C)6s5s(B)s6s5s(A1 2323223 +++++++++=
Equating coefficients :
3s : DCA0 ++= 2s : CBA3D2C3BA50 ++=+++= 1s : B5A60 +=0s : 61BB61 =→=
365-B65-A ==
3s91
2s41
s61
s365-
)s(X 2 +−
+++=
=)t(x 3t-2t- e91
e41
t61
)t(u36
5-−++
(b) 22 )1s(C
1sB
sA
)1s(s1
)s(Y+
++
+=+
=
1A = , 1-C =
sC)ss(B)1s2s(A1 22 +++++=
Equating coefficients :
2s : -ABBA0 =→+=1s : -ACCACBA20 =→+=++=0s : -1C-1,B,A1 ===
2)1s(1
1s1
s1
)s(Y+
−+
−=
=)t(y -t-t ete)t(u −−
(c) 10s6s
DCs1s
BsA
)s(Z 2 +++
++
+=
101A = , 51-B =
)ss(D)ss(C)s10s6s(B)10s16s7s(A1 2232323 ++++++++++=
Equating coefficients :
3s : CBA0 ++= 2s : DB5A6DCB6A70 ++=+++= 1s : A 2-BB5A10DB10A160 =→+=++=0s : 101AA101 =→=
101A = , 51--2AB == , 101AC == , 104
A4D ==
10s6s4s
1s2
s1
)s(Z10 2 +++
++
−=
1)3s(1
1)3s(3s
1s2
s1
)s(Z10 22 +++
+++
++
−=
=)t(z [ ] )t(u)tsin(e)tcos(ee211.0 -3t-3t-t ++−
Chapter 15, Solution 37.
(a) Let 4sCBs
sA
)4s(s12
)s(P 22 ++
+=+
=
3412s)s(PA 0s === =
sCsB)4s(A12 22 +++=
Equating coefficients :
0s : 3AA412 =→=1s : C0 =2s : -3-ABBA0 ==→+=
4ss3
s3
)s(P 2 +−=
)t2cos(3)t(u3)t(p −=
)s(Pe)s(F -2s=
=)t(f [ ] )2t(u))2t(2cos(13 −−−
(b) Let 9sDCs
1sBAs
)9s)(1s(1s2
)s(G 2222 ++
+++
=++
+=
)1s(D)ss(C)9s(B)s9s(A1s2 2323 +++++++=+
Equating coefficients :
3s : -ACCA0 =→+=
2s : -BDDB0 =→+=1s : 82-C,82AA8CA92 ==→=+= 0s : 81-D,81BB8DB91 ==→=+=
++
−
++
=9s1s2
81
1s1s2
81
)s(G 22
9s1
81
9ss
41
1s1
81
1ss
41
)s(G 2222 +⋅−
+⋅−
+⋅+
+⋅=
=)t(g )t3sin(241
)t3cos(41
)tsin(81
)tcos(41
−−+
(c) Let 13s4s
117s369
13s4ss9
)s(H 22
2
+++
−=++
=
2222 3)2s(3
153)2s(
2s369)s(H
++⋅−
+++
⋅−=
=)t(h )t3sin(e15)t3cos(e36)t(9 t2-t2- −−δ
Chapter 15, Solution 38.
(a) 26s10s
26s626s10s26s10s
s4s)s(F 2
2
2
2
++−−++
=++
+=
26s10s26s6
1)s(F 2 +++
−=
2222 1)5s(4
1)5s()5s(6
1)s(F++
+++
+−=
=)t(f )t5sin(e4)t5cos(e6)t( -t-t +−δ
(b) 29s4s
CBssA
)29s4s(s29s7s5
)s(F 22
2
+++
+=++++
=
sCsB)29s4s(A29s7s5 222 ++++=++
Equating coefficients :
0s : 1AA2929 =→=1s : 3A47CCA47 =−=→+=2s : 4 A5BBA5 =−=→+=
1A = , 4B = , 3C =
22222 5)2s(5
5)2s()2s(4
s1
29s4s3s4
s1
)s(F++
−++
++=
+++
+=
=)t(f )t5sin(e)t5cos(e4)t(u -2t-2t −+
Chapter 15, Solution 39.
(a) 20s4s
DCs17s2s
BAs)20s4s)(17s2s(
1s4s2)s(F 2222
23
+++
+++
+=
++++++
=
)20s4s(B)s20s4s(A1s4s 22323 +++++=++
)17s2s(D)s17s2s(C 223 ++++++Equating coefficients :
3s : CA2 +=2s : DC2BA44 +++= 1s : D2C17B4A200 +++= 0s : D17B201 +=
Solving these equations (Matlab works well with 4 unknowns),
-1.6A = , -17.8B = , 6.3C = , 21D =
20s4s21s6.3
17s2s17.81.6s-
)s(F 22 +++
+++
−=
22222222 4)2s()4((3.45)
4)2s()2(3.6)(s
4)1s()4((-4.05)
4)1s(1)(-1.6)(s
)s(F++
++++
+++
++++
=
=)t(f )t4sin(e45.3)t4cos(e6.3)t4sin(e05.4)t4cos(e6.1- -2t-2t-t-t ++−
(b) 3s6s
DCs9sBAs
)3s6s)(9s(4s
)s(F 2222
2
+++
+++
=+++
+=
)9s(D)s9s(C)3s6s(B)s3s6s(A4s 232232 +++++++++=+
Equating coefficients :
3s : -ACCA0 =→+=2s : DBA61 ++= 1s : A -CBC6B6C9B6A30 ==→+=++=0s : D9B34 +=
Solving these equations,
121A = , 121B = , 121-C = , 125D =
3s6s5s-
9s1s
)s(F12 22 +++
+++
=
5.449--0.551,2
12-366-03s6s2 =
±→=++
Let 449.5s
F551.0s
E3s6s
5s-)s(G 2 +
++
=++
+=
133.1449.5s5s-
E -0.551s =+
+= =
2.133- 551.0s5s-
F -5.449s =++
= =
449.5s133.2
551.0s133.1
)s(G+
−+
=
449.5s133.2
551.0s133.1
3s3
31
3ss
)s(F12 2222 +−
++
+⋅+
+=
=)t(f -5.449t-0.551t e1778.0e0944.0)t3sin(02778.0)t3cos(08333.0 −++
Chapter 15, Solution 40.
Let 5s2s
CBs2s
A)5s2s)(2s(
13s7s4)s( 22
2
++
++
+=
+++
++=H
)2s(C)s2s(B)5s2s(A13s7s4 222 ++++++=++
Equating coefficients gives:
BA4:s2 +=
1CCB2A27:s −=→++=
1B 3,Aor 15A5C2A513:constant ===→+=
222 2)1s(2)1s(
2s3
5s2s1s
2s3)s(H
++
−++
+=
++
−+
+=
Hence,
)t2sinsinAt2coscosA(ee3t2sinet2cosee3)t(h tt2ttt2 α−α+=−+= −−−−−
where o45,2A1sinA,1cosA =α=→=α=α Thus,
[ ] )t(ue3)45t2cos(e2)t(h t2ot −− ++= Chapter 15, Solution 41. Let y(t) = f(t)*h(t). For 0 < t < 1, f )t( λ− )(λh 0 t 1 2 λ
2t0
t
0
2 t22d4)1()t(y =λ=λλ= ∫
For 1 <t<3, )t(f λ− )(h λ 0 1 t 2
4t2t8)28(2d)48)(1(d4)1()t(y 2t1
2t0
2t
1
1
0−−=λ−λ+λ=λλ−+λλ= ∫∫
For 3 < t < 4 h )(λ )t(f λ− 0 1 t-2 2 3 t 4 λ
222t
2
2t
2 t2t163228d)48()t(y +−=λ−λ=λλλ−= −−∫
Thus,
<<+<<−
<<
=
otherwise 0,4 t 3 ,2t16t-32
3 t 1 ,42t-8t1 t 0 ,t2
)t(y 2
2
2
Chapter 15, Solution 42.
(a) For , 1t0 << )t(f1 λ− and f )(2 λ overlap from 0 to t, as shown in Fig. (a).
2t
2d))(1()t(f)t(f)t(y
2t0
2t
021 =λ
=λλ=∗= ∫
1
λt-1 t 1 0
1
λt-1 t 1
f2(λ)f1(t - λ)
0
(a) (b) For 1 , 2t << )t(f1 λ− and f )(2 λ overlap as shown in Fig. (b).
2t
t2
d))(1()t(y2
11t
21
1t−=
λ=λλ= −−
∫
For , there is no overlap. 2t > Therefore,
=)t(y
<<−<<
otherwise,02t1,2tt1t0,2t
2
2
(b) For , the two functions overlap as shown in Fig. (c). 1t0 <<
td)1)(1()t(f)t(f)t(yt
021 =λ=∗= ∫
1
λt-1 t 1 0
1
λt-1 t 1
f2(λ)f1(t - λ)
0
(c) (d)
For 1 , the functions overlap as shown in Fig. (d). 2t <<
t2d)1)(1()t(y 11t
1
1t−=λ=λ= −−
∫ For , there is no overlap. 2t > Therefore,
=)t(y
<<−<<
otherwise,02t1,t21t0,t
(c) For , there is no overlap. For -1-t < 0t1 << , f )t(1 λ− and overlap
as shown in Fig. (e). )(f2 λ
t1-
2t
1-21 2d)1)(1()t(f)t(f)t(y
λ+λ
=λ+λ=∗= ∫
22 )1t(
21
)1t2t(21
)t(y +=++=
-1
1
λ t 1 0-1
1
λt 1
f1(λ)f2(t - λ)
0
(e) (f) For , the functions overlap as shown in Fig. (f). 1t0 <<
∫∫ λλ−+λ+λ=t
0
0
1-d)1)(1(d)1)(1()t(y
t0
201-
2
22)t(y
λ
−λ+
λ+λ
=
)tt21(21
)t(y 2−+=
For , the two functions overlap. 1t >
∫∫ λλ−+λ+λ=1
0
0
1-d)1)(1(d)1)(1()t(y
121
121
221
)t(y 10
2
=−+=
λ
−λ+=
Therefore,
=)t(y
><<++<<++
<
1t1,1t0),1t2t-(5.00t1-),1t2t(5.0
-1t,0
2
2
Chapter 15, Solution 43.
(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).
2t
2d))(1()t(h)t(x)t(y
2t0
2t
0=
λ=λλ=∗= ∫
1
λt-1 t 1
h(λ)
x(t - λ)
0
1
λ t-1 t 1 0
(a) (b) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).
1t2t21-
2d)1)(1(d))(1()t(y 2t
11
1t
2t
1
1
1t−+=λ+
λ=λ+λλ= −−
∫∫
For , there is a complete overlap so that 2t >
1)1t(td)1)(1()t(y t1t
t
1t=−−=λ=λ= −−
∫
Therefore,
=)t(y
><<−+<<
otherwise,02t,1
2t1,1t2)2t(-1t0,2t
2
2
(b) For , the two functions overlap as shown in Fig. (c). 0t >
t0
-t
0- e-2de2)1()t(h)t(x)t(y λλ =λ=∗= ∫
2
1
h(λ) = 2e-λ
λ0 t
x(t-λ)
(c) Therefore,
=)t(y 0t),e1(2 -t >−
(c) For , 0t1- << )t(x λ− and )(h λ overlap as shown in Fig. (d).
21t0
21t
0)1t(
21
2d))(1()t(h)t(x)t(y +=
λ=λλ=∗= +
+∫
2-1 t+1
1
λ t-1 t 1
h(λ)x(t - λ)
0
(d) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (e).
∫∫ +λλ−+λλ=
1t
1
1
0d)2)(1(d))(1()t(y
21
tt21-
22
2)t(y 21t
1
210
2
++=
λ
−λ+λ
= +
t 2t+1
1
λt-1-1 10
(e) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (f).
∫∫ λλ−+λλ=−
2
1
1
1td)2)(1(d))(1()t(y
21
tt21-
22
2)t(y 22
1
21
1t
2
++=
λ
−λ+λ
= −
1
t 2 λt-1 10 t+1
(f) For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (g).
221t
22
1tt
21
t329
22d)2)(1()t(y +−=
λ
−λ=λλ−= −−∫
1
t2 λ t-110 t+1
(g)
Therefore,
=)t(y
<<+−<<++<<++
otherwise,03t2,29t3)2t(2t0,21t)2t(-0t1-,21t)2t(
2
2
2
Chapter 15, Solution 44.
(a) For , 1t0 << )t(x λ− and )(h λ overlap as shown in Fig. (a).
td)1)(1()t(h)t(x)t(yt
0=λ=∗= ∫
0
h(λ)1x(t - λ)
-1
t 2 λt-1 1
(a) For 1 , 2t << )t(x λ− and )(h λ overlap as shown in Fig. (b).
t23d)1)(1-(d)1)(1()t(y t1
11t
t
1
1
1t−=λ−λ=λ+λ= −−
∫∫ For , 3t2 << )t(x λ− and )(h λ overlap as shown in Fig. (c).
3t-d)-1)(1()t(y 21t
2
1t−=λ=λ= −−
∫
0 λ1 t-1 2 t
-1
1
0 2 λ1 t-1 t
-1
1
(b) (c) Therefore,
=)t(y
<<−<<−<<
otherwise,03t2,3t2t1,t231t0,t
(b) For , there is no overlap. For 2t < 3t2 << , f )t(1 λ− and overlap,
as shown in Fig. (d). )(f2 λ
∫ λλ−=∗=t
221 d)t)(1()t(f)t(f)t(y
2t22t
2t
2t2
2
+−=
λ−λ=
t-1 t
1
1 5432 λ
f2(λ) f1(t - λ)
0
(d)
0 1
1
2 3 4 5tt-1 λ (e)
For , 5t3 << )t(f1 λ− and )(f2 λ overlap as shown in Fig. (e).
21
2td)t)(1()t(y t
1t
2t
1t=
λ
−λ=λλ−= −−∫
For , the functions overlap as shown in Fig. (f). 6t5 <<
12t5t21-
2td)t)(1()t(y 25
1t
25
1t−+=
λ
−λ=λλ−= −−∫
0 1 λ
1
2 3 4 5 t t-1
(f) Therefore,
=)t(y
<<−+<<<<+−
otherwise,06t5,12t5)2t(-5t3,213t2,2t2)2t(
2
2
Chapter 15, Solution 45.
(a) t
t
0)(fd)t()(f)t()t(f =λλ=λλ−δλ=δ∗ ∫
)t(f)t()t(f =δ∗
(b) ∫ λλ−λ=∗t
0d)t(u)(f)t(u)t(f
Since
>λ<λ
=λ−t0t1
)t(u
∫ λλ=∗t
od)(f)t(u)t(f
Alternatively,
{ }s
)s(F)t(u)t(f =∗L
∫ λλ=∗=
−t
o1 d)(f)t(u)t(f
s)s(F
L
Chapter 15, Solution 46.
(a) Let ∫ λλ−=∗=t
0 1221 d)t(x)t(x)t(x)t(x)t(y For , 3t0 << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (a).
)ee(4dee4dee4)t(y t2-t-t
0-t-t
0)-(t-2- −=λ=λ= ∫∫ λλλ
4
3t0 λ
(a) For , the two functions overlap as shown in Fig. (b). 3t >
( ) e1(e4e-e4dee4)t(y 3-t-30
-t-3
0)-(t-2- −==λ= λλλ∫
λ
4
3 t0
(b)
x1(t-λ)
x2(λ)
)
Therefore,
=)t(y
>−<<−3t),e1(e4
3t0),ee(43-t-
t-2t-
(b) For 1 , 2t << )(x1 λ and )t(x 2 λ− overlap as shown in Fig. (c).
1td)1)(1()t(x)t(x)t(y t1
t
121 −=λ=λ=∗= ∫
1 3
1
λt-1 t 2
x1(λ)x2(t - λ)
0
(c) For , the two functions overlap completely. 3t2 <<
1)1t(td)1)(1()t(y t1t
t
1t=−−=λ=λ= −−
∫ For , the two functions overlap as shown in Fig. (d). 4t3 <<
t4d)1)(1()t(y 31t
3
1t−=λ=λ= −−
∫
1 3
1
λ t-1 t20
(d)Therefore,
=)t(y
<<−<<<<−
otherwise,04t3,t43t2,12t1,1t
(c) For , 0t1- << )t(x1 λ− and )(x 2 λ overlap as shown in Fig. (e).
∫ λ=∗= λt
1-)-(t-
21 de4)1()t(x)t(x)t(y
[ ]1)(t-t
1-t- e14dee4)t(y +λ −=λ= ∫
λ1-1 t-1
1
x1(t - λ)
x2(λ)
(e) For , 1t0 <<
∫∫ λ+λ= λλt
0)-(t-0
1-)-(t- de4)1-(de4)1()t(y
4e4e8ee4ee4)t(y 1)-(t-tt0
-t01-
-t −−=−= +λλ For , the two functions overlap completely. 1t >
∫∫ λ+λ= λλ1
0)-(t-0
1-)-(t- de4)1-(de4)1()t(y
1)-(t1)-(t-t10
-t01-
-t e4e4e8ee4ee4)t(y −+λλ −−=−= Therefore,
=)t(y[ ]
>−−<<−−<<−
−+
+
+
1t,e4e4e81t0,4e4e80t1-,e14
1)(t-1)(t-t-
1)(t-t-
1)-(t
Chapter 15, Solution 47.
)tcos()t(f)t(f 21 ==
[ ] ∫ λλ−λ=t
0211- d)tcos()cos()s(F)s(FL
[ ])BAcos()BAcos(21
)Bcos()Acos( −++=
[ ] ∫ λλ−+=t
0211- d)]2tcos()t[cos(
21)s(F)s(FL
[ ] t0
t021
1-
2-)2tsin(
21
)tcos(21
)s(F)s(Fλ−
⋅+λ⋅=L
[ ] =)s(F)s(F 21
1-L )tsin(5.0)tcos(t5.0 + Chapter 15, Solution 48.
(a) Let 222 2)1s(2
5s2s2
)s(++
=++
=G
)t2sin(e)t(g -t=
)s(G)s(G)s(F =
[ ] λλ−λ== ∫ d)t(g)(g)s(G)s(G)t(f
t
01-L
∫ λλ−λ= λ−λt
0)t(-- d))t(2sin(e)2sin(e)t(f
[ ])BAcos()BAcos(21
)Bsin()Asin( +−−=
[ ]∫ λλ−−= λt
0-t- d))2t(2cos()t2cos(ee
21
)t(f
∫∫ λλ−−λ= λλt
02-
-tt
02-
-t
d)4t2cos(e2e
de)t2cos(2e
)t(f
[ ]∫ λλ+λ−⋅= λλ t
02-
-tt0
-2-t
d)4sin()t2sin()4cos()t2cos(e2e
2-e
)t2cos(2
e)t(f
∫ λλ−+= λt
02-
-tt2-t- d)4cos(e)t2cos(
2e
)1e-()t2cos(e41
)t(f
∫ λλ− λt
02-
-t
d)4sin(e)t2sin(2e
)e1()t2cos(e41
)t(f t2-t- −=
( ) t0
-2-t
)4sin(4)2cos(4-164
e)t2cos(
2e
λ−λ+
−λ
( ) t0
-2-t
)4cos(4)2sin(4-164
e)t2sin(
2e
λ+λ+
−λ
=)t(f )t4cos()t2cos(20e
)t2cos(20e
)t2cos(4
e)t2cos(
2e -3t-t-3t-t
+−−
)t2sin(10e
)t4sin()t2cos(10e -t-3t
++
)t4cos()t2sin(10e
)t4sin()t2sin(20e -t-t
−+
(b) Let 1s
2)s(X
+= ,
4ss
)s(Y+
=
)t(ue2)t(x -t= , )t(u)t2cos()t(y =
)s(Y)s(X)s(F =
[ ] ∫∞ λλ−λ==
01- d)t(x)(y)s(Y)s(X)t(f L
∫ λ⋅λ= λ−t
0)(t- de2)2cos()t(f
( ) t0
t- )2sin(2)2cos(41
ee2)t(f λ+λ
+⋅=
λ
( )[ ]1)t2sin(2)t2cos(ee52
)t(f tt- −+=
=)t(f t-e52
)t2sin(54
)t2cos(52
−+
Chapter 15, Solution 49.
Let x(t) = u(t) – u(t-1) and y(t) = h(t)*x(t).
[ ]
+−
=
−
+==
−−
−−−
)2s(s)e1(4)
se
s1()s(X)s(H)t(y
s1
s11 L
2s4LL
But
+−=
++=
+ 2s1
s1
21
2sB
sA
)2s(s1
++−
+−=
−−
2se
se
2s1
s12)s(Y
ss
)1t(u]e1[4)t(u]e1[2)t(y )1t(2t2 −−−−= −−−
Chapter 15, Solution 50.
Take the Laplace transform of each term.
[ ] [ ]4s
s3)s(V10)0(v)s(Vs2)0(v)0(vs)s(Vs 2
2
+=+−+′−−
4ss3
)s(V102)s(Vs22s)s(Vs 22
+=+−++−
4ss7s
4ss3
s)s(V)10s2s( 2
3
22
++
=+
+=++
10s2sDCs
4sBAs
)10s2s)(4s(s7s
)s(V 2222
3
+++
+++
=+++
+=
)4s(D)s4s(C)10s2s(B)s10s2s(As7s 232233 +++++++++=+
Equating coefficients :
3s : A1CCA1 −=→+=2s : DBA20 ++= 1s : 4B2A6C4B2A107 ++=++=
0s : B-2.5DD4B100 =→+=
Solving these equations yields
269
A = , 2612
B = , 2617
C = , 2630-
D =
++−
+++
=10s2s
30s174s
12s9261
)s(V 22
++
−++
+⋅+
+⋅+
+= 222222 3)1s(
473)1s(
1s17
4s2
64s
s9261
)s(V
=)t(v )t3sin(e7847
)t3cos(e2617
)t2sin(266
)t2cos(269 t-t- −++
Chapter 15, Solution 51.
Taking the Laplace transform of the differential equation yields
[ ] [ ]1s
10)s(V6)]0(v)s(sV5)0('v)0(sv)s(Vs2+
=+−+−−
or ( ))3s)(2s)(1s(
24s16s2)s(V1s
10104s2)s(V6s5s2
2+++
++=→
+=−−−++
Let 3C,0B,5A,3s
C2s
B1s
A)s( −===+
++
++
=V
Hence,
)t(u)e3e5()t(v t3t −− −=
Chapter 15, Solution 52.
Take the Laplace transform of each term. [ ] [ ] 01)s(I2)0(i)s(Is3)0(i)0(is)s(Is2 =++−+′−−
0133s)s(I)2s3s( 2 =+−−−++
2sB
1sA
)2s)(1s(5s
)s(I+
++
=++
+=
4A = , 3-B =
2s3
1s4
)s(I+
−+
=
=)t(i )t(u)e3e4( -2t-t −
Chapter 15, Solution 53.
Take the Laplace transform of each term.
[ ] [ ]4s
s)s(V6)0(y)s(Ys5)0(y)0(ys)s(Ys 2
2
+=+−+′−−
4ss
54s)s(Y)6s5s( 22
+=−−−++
4s)4s)(9s(s
4ss
9s)s(Y)6s5s( 2
2
22
++++
=+
++=++
4sDCs
3sB
2sA
)4s)(3s)(2s(36s5s9s
)s(Y 22
23
++
++
++
=+++
+++=
427
)s(Y)2s(A -2s =+= = , 1375-
)s(Y)3s(B -3s =+= =
When , 0s =
265
D4D
3B
2A
)4)(3)(2(36
=→++=
When , 1s =
521C
5D
5C
4B
3A
)5)(12(546
=→+++=+
Thus,4s
265s5213s
13752s427
)s(Y 2 ++⋅
++
−+
=
=)t(y )t2sin(525
)t2cos(521
e1375
e4
27 3t-2t- ++−
Chapter 15, Solution 54.
Taking the Laplace transform of the differential equation gives
[ ] [ ]3s
5)s(V2)0(v)s(Vs3)0(v)0(vs)s(Vs2
+=+−+′−−
3ss2
13s
5)s(V)2s3s( 2
+−
=−+
=++
)3s)(2s)(1s(s2
)2s3s)(3s(s2
)s(V 2 +++−
=+++
−=
3sC
2sB
1sA
)s(V+
++
++
=
23A = , -4B = , 25C =
3s25
2s4
1s23
)s(V+
++
−+
=
=)t(v )t(u)e5.2e4e5.1( -3t-2t-t +−
Chapter 15, Solution 55.
Take the Laplace transform of each term. [ ] [ ])0(y)0(ys)s(Ys6)0(y)0(ys)0(ys)s(Ys 223 ′−−+′′−′−−
[ ] 22 2)1s(1s)0(y)s(Ys8++
+=−+
Setting the initial conditions to zero gives
5s2s1s
)s(Y)s8s6s( 223
+++
=++
5s2sEDs
4sC
2sB
sA
)5s2s)(4s)(2s(s)1s(
)s(Y 22 +++
++
++
+=++++
+=
401
A = , 201
B = , 104
3-C = ,
653-
D = , 657-
E =
22 2)1s(7s3
651
4s1
1043
2s1
201
s1
401
)s(Y++
+⋅−
+⋅−
+⋅+⋅=
2222 2)1s(4
651
2)1s()1s(3
651
4s1
1043
2s1
201
s1
401
)s(Y++
⋅−++
+⋅−
+⋅−
+⋅+⋅=
=)t(y )t2sin(e652
)t2cos(e653
e104
3e
201
)t(u401 t-t-4t-2t- −−−+
Chapter 15, Solution 56.
Taking the Laplace transform of each term we get:
[ ] 0)s(Vs
12)0(v)s(Vs4 =+−
8)s(Vs
12s4 =
+
3ss2
12s4s8
)s(V 22 +=
+=
=)t(v ( )t3cos2
Chapter 15, Solution 57.
Take the Laplace transform of each term.
[ ]4s
s)s(Y
s9
)0(y)s(Ys 2 +=+−
4s4ss
4ss
1)s(Ys
9s2
2
2
2
+++
=+
+=
+
9sDCs
4sBAs
)9s)(4s(s4ss
)s(Y 2222
23
++
+++
=++
++=
)4s(D)s4s(C)9s(B)s9s(As4ss 232323 +++++++=++
Equating coefficients :
0s : D4B90 +=1s : C4A94 +=2s : DB1 +=3s : CA1 +=
Solving these equations gives
0A = , 54-B = , 1C = , 59D =
9s59
9ss
4s54-
9s59s
4s54-
)s(Y 22222 ++
++
+=
++
++
=
=)t(y )t3sin(6.0)t3cos()t2sin(0.4- ++
Chapter 15, Solution 58.
We take the Laplace transform of each term and obtain
10s6sse)s(Ve)s(V
s10)]0(v)s(sV[)s(V6 2
s2s2
++=→=+−+
−−
1)3s(e3e)3s()s(V
2
s2s2
++
−+=
−−
Hence, 3( 2) 3( 2)( ) cos( 2) 3 sin( 2) ( 2)t tv t e t e t u t− − − − = − − − −
Chapter 15, Solution 59.
Take the Laplace transform of each term of the integrodifferential equation.
[ ]2s
6)s(Y
s3
)s(Y4)0(y)s(Ys+
=++−
−+
=++ 12s
6s)s(Y)3s4s( 2
)3s)(2s)(1s(s)s4(
)3s4s)(2s()s4(s
)s(Y 2 +++−
=+++
−=
3sC
2sB
1sA
)s(Y+
++
++
=
5.2A = , 6B = , -10.5C =
3s5.10
2s6
1s5.2
)s(Y+
−+
++
=
=)t(y -3t-2t-t e5.10e6e5.2 −+
Chapter 15, Solution 60.
Take the Laplace transform of each term of the integrodifferential equation.
[ ]16s
4s4
)s(Xs3
)s(X5)0(x)s(Xs2 2 +=+++−
16s64s36s4s2
16ss4
4s2)s(X)3s5s2( 2
23
22
+−+−
=+
+−=++
)16s)(5.1s)(1s(32s18s2s
)16s)(3s5s2(64s36s4s2
)s(X 2
23
22
23
+++−+−
=+++−+−
=
16sDCs
5.1sB
1sA
)s(X 2 ++
++
++
=
-6.235)s(X)1s(A -1s =+= =
329.7)s(X)5.1s(B -1.5s =+= =
When , 0s =
2579.0D16D
5.1B
A(1.5)(16)
32-=→++=
)16s16ss(B)24s16s5.1s(A32s18s2s 232323 +++++++=−+−
)5.1s5.2s(D)s5.1s5.2s(C 223 ++++++
Equating coefficients of the s terms, 3
0935.0-CCBA1 =→++=
16s0.25790.0935s-
5.1s329.7
1s6.235-
)s(X 2 ++
++
++
=
=)t(x )t4sin(0645.0)t4cos(0935.0e329.7e6.235- -1.5t-t +−+
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