Chapter 6, Solution 1 - WordPress.com · 2015. 8. 8. · Chapter 15, Solution 7. (a) Since [] s2 42...

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−


(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


(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++

+


(a) 16s

se6e4s

s6)s(G 2

ss

22 +=

+=

−−

(b) 3s

e5s2)s(F

s2

2 ++=

−


(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


(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

+


(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 ++


(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−⋅+


(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

−


(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+++


(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

++++++


)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

+−+−−

−−

−

−−−−−−−−−

+

+=

++

+=

++

+=

−+−−=−=


(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


<<−<<

=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−


)4t(u5)3t(u3)1t(u3)t(u5)t(f −−−+−−=

=)s(F [ ]s4-s3-s- e5e3e35s1

−+−


><<<<

<

=

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 −−−


(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 +−−


Since [ ] 1)t( =δL and , 2T = =)s(F s2-e11−


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


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


(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 −−

+


(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

−+−−


(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


(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

−−− +−=→++

+−+

+=

=−==→−==+−−==+

++++++=+++++=+

++

++

+=

+++

+=


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


(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 −+= −−−−


(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 ++++++=+


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 ++


(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 +++++++=+−


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 ++++++=+


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 −


(a) 1s

C1sBAs

)1s)(1s(6

1s)1s(6

)s(F 224 ++

++

=++

=−−

=

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


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 +++++++++=


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 −−−


(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- −

−+


(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 +++++=


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- −−+−+


(a) 3s

D2s

CsB

sA

)3s)(2s(s1

)s(X 22 ++

+++=

++=

61B = , 41C = , 91-D =

)s2s(D)s3s(C)6s5s(B)s6s5s(A1 2323223 +++++++++=


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 +++++=


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 ++++++++++=


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 ++−


(a) Let 4sCBs

sA

)4s(s12

)s(P 22 ++

+=+

=

3412s)s(PA 0s === =

sCsB)4s(A12 22 +++=


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 +++++++=+


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


(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 ++++=++


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 −+


(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 +++++++++=+


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 −++


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


(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


(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


(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


(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


(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


)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

−+


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 −−−−= −−−


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 +++++++++=+


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- −++


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 −− −=


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 −



[ ] [ ]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- ++−


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 +−


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- −−−+


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



[ ]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 +++++++=++


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- ++


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− − − − = − − − −


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 −+


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 +−+

Chapter 6, Solution 1 - WordPress.com · 2015. 8. 8. · Chapter 15, Solution 7. (a) Since [] s2 42...

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Transcript of Chapter 6, Solution 1 - WordPress.com · 2015. 8. 8. · Chapter 15, Solution 7. (a) Since [] s2 42...