Chapter 15, Problem 1. Find the Laplace transform of: (a) cosh at (b ...
Chapter 15, Problem 1. Find the Laplace transform of: (a) cosh at (b ...
Chapter 15, Problem 1. Find the Laplace transform of: (a) cosh at (b ...
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> <strong>1.</strong><br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
(a) <strong>cosh</strong> <strong>at</strong> (b) sinh <strong>at</strong><br />
1 x −x<br />
1 x −x<br />
[Hint: <strong>cosh</strong> x = ( e + e ) , sinh x = ( e − e )<br />
2<br />
<strong>Chapter</strong> <strong>15</strong>, Solution <strong>1.</strong><br />
(a)<br />
<strong>at</strong> -<strong>at</strong><br />
e + e<br />
<strong>cosh</strong>( <strong>at</strong>)<br />
=<br />
2<br />
<strong>cosh</strong>( <strong>at</strong>)<br />
1 ⎡ 1 1 ⎤<br />
= ⎢⎣<br />
+<br />
2 s − a s + a ⎥⎦<br />
(b)<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 2.<br />
[ ] =<br />
L 2 2 s a<br />
e<br />
sinh( <strong>at</strong>)<br />
=<br />
-<strong>at</strong><br />
e<br />
2<br />
sinh( <strong>at</strong>)<br />
1 ⎡ 1 1 ⎤<br />
= ⎢⎣<br />
−<br />
2 s − a s + a ⎥⎦<br />
<strong>at</strong> −<br />
[ ] =<br />
2<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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.]<br />
s<br />
−<br />
L 2 2 s − a<br />
Determine <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
(a) cos( ω t + θ ) (b) sin( ω t + θ )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 2.<br />
(a) f ( t)<br />
= cos( ωt)<br />
cos( θ)<br />
− sin( ωt)<br />
sin( θ)<br />
F( s)<br />
= cos( θ)<br />
L cos( ωt)<br />
− sin( θ)<br />
L sin( ωt)<br />
scos(<br />
θ)<br />
− ωsin(<br />
θ)<br />
F( s)<br />
= 2 2<br />
s + ω<br />
[ ] [ ]<br />
(b) f ( t)<br />
= sin( ωt)<br />
cos( θ)<br />
+ cos( ωt)<br />
sin( θ)<br />
F( s)<br />
= sin( θ)<br />
L cos( ωt)<br />
+ cos( θ)<br />
L sin( ωt)<br />
ssin(<br />
θ)<br />
− ωcos(<br />
θ)<br />
F( s)<br />
=<br />
2 2<br />
s + ω<br />
[ ] [ ]<br />
a
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 3.<br />
Obtain <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each <strong>of</strong> <strong>the</strong> following functions:<br />
(a) e tu()<br />
t<br />
t −2<br />
cos3<br />
(b) e tu(<br />
t)<br />
t −2<br />
sin 4<br />
(c) e tu()<br />
t<br />
t −3<br />
<strong>cosh</strong> 2<br />
(d) e tu(<br />
t)<br />
t −4<br />
sinh<br />
(e) te tu()<br />
t<br />
t −<br />
sin 2<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 3.<br />
(a) [ e cos( 3t)<br />
u(<br />
t)<br />
]=<br />
-2t L<br />
(b) [ e sin( 4t)<br />
u(<br />
t)<br />
]=<br />
-2t L<br />
L<br />
<strong>cosh</strong>( <strong>at</strong>)<br />
( s<br />
( s<br />
=<br />
s<br />
s + 2<br />
+ 2)<br />
+<br />
4<br />
2)<br />
(c) Since [ ] 2<br />
[ e <strong>cosh</strong>( 2t)<br />
u(<br />
t)<br />
]=<br />
-3t L<br />
s<br />
2 − a<br />
( s<br />
(d) Since L [ sinh( <strong>at</strong>)<br />
] = 2<br />
[ e sinh( t)<br />
u(<br />
t)<br />
]=<br />
-4t L<br />
-t<br />
(e) L [ sin( 2t)<br />
]<br />
s<br />
a<br />
2 − a<br />
( s<br />
2<br />
=<br />
( s + 1)<br />
e 2<br />
+<br />
2 +<br />
2 +<br />
s + 3<br />
+ 3)<br />
1<br />
4)<br />
+ 4<br />
2 −<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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9<br />
2 −<br />
16<br />
If f ( t)<br />
← ⎯→ F(<br />
s)<br />
t f ( t)<br />
← ⎯→<br />
- d<br />
F(<br />
s)<br />
ds<br />
Thus,<br />
- d<br />
-t -1<br />
[ ] [ 2<br />
L t e sin( 2t)<br />
= 2 ( ( s + 1)<br />
+ 4)<br />
]<br />
ds<br />
2<br />
=<br />
⋅ 2(<br />
s + 1)<br />
2 2<br />
(( s + 1)<br />
+ 4)<br />
-t [ t e sin( 2t)<br />
]=<br />
L 2 2<br />
(( s + 1)<br />
+ 4)<br />
1<br />
4<br />
4(<br />
s +<br />
1)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 4.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong> following:<br />
g t = 6cos 4t<br />
−1<br />
(a) () ( )<br />
−3(<br />
t−2<br />
)<br />
(b) f () t = 2tu()<br />
t + 5e<br />
u(<br />
t − 2)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 4.<br />
(a)<br />
(b)<br />
G(<br />
s)<br />
F(<br />
s)<br />
= 6<br />
s<br />
2<br />
s<br />
+ 4<br />
2 e<br />
= + 5<br />
s s<br />
2<br />
e<br />
−2s<br />
2 +<br />
3<br />
−s<br />
−s<br />
6se<br />
=<br />
2<br />
s + 16<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 5.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each <strong>of</strong> <strong>the</strong> following functions:<br />
2<br />
4 −2t<br />
(a) t cos(<br />
2t<br />
+ 30°<br />
) u(<br />
t)<br />
(b) 3t<br />
e u(<br />
t)<br />
d<br />
−(<br />
t−1) (c) 2tu() t − 4 δ () t (d) 2e<br />
u(<br />
t)<br />
dt<br />
t 3<br />
(e) 5u ( t 2)<br />
(f) 6e u(<br />
t)<br />
−<br />
n<br />
d<br />
(g) δ () t n<br />
dt<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 5.<br />
(a) L [ 2t<br />
+ 30°<br />
) ]<br />
L<br />
cos(<br />
s cos( 30°<br />
) − 2sin(<br />
30°<br />
)<br />
=<br />
2 s + 4<br />
2 t cos( 2t<br />
+ 30°<br />
)<br />
2 d ⎡s<br />
cos( 30°<br />
) −1⎤<br />
= 2 ds ⎢⎣ 2 s + 4<br />
d d ⎡⎛<br />
= ⎢⎜<br />
ds ds ⎜<br />
⎢⎣<br />
⎝<br />
3 ⎞<br />
s −1⎟<br />
2 ⎟<br />
⎠<br />
d ⎡ 3<br />
⎛<br />
2 -1<br />
= ⎢ s + 4 − 2s⎜<br />
ds<br />
⎜<br />
⎢⎣<br />
2<br />
⎝<br />
[ ] ⎥ ⎦<br />
=<br />
-<br />
=<br />
3<br />
2<br />
( - 2s)<br />
⎛<br />
2⎜<br />
⎜<br />
−<br />
⎝<br />
2 -1<br />
( s + 4)<br />
⎥<br />
⎥<br />
3<br />
2 -2<br />
( ) s −1⎟<br />
( s + 4)<br />
⎥<br />
⎥<br />
3 ⎞ ⎛<br />
s −1⎟<br />
2s⎜<br />
2 ⎟ ⎜<br />
⎠<br />
−<br />
⎝<br />
( ) ( ) ( ) ( ) 3<br />
2 2 2 2 2 2<br />
2<br />
s + 4 s + 4 s + 4 s + 4<br />
(-3<br />
=<br />
3s<br />
−<br />
3s<br />
3 s<br />
+ 2 −<br />
3 s<br />
( 8s<br />
+<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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3<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
( ) ( ) 3<br />
2 2<br />
2<br />
s + 4<br />
s + 4<br />
+ 2)(s<br />
2<br />
+<br />
4)<br />
4<br />
+<br />
2<br />
⎛<br />
) ⎜<br />
⎝<br />
3<br />
3s<br />
− 8s<br />
( ) ( ) 3<br />
2 3<br />
2<br />
s + 4 s + 4<br />
[ t cos( 2t<br />
+ 30°<br />
) ]=<br />
2<br />
8 −<br />
12<br />
L<br />
3 s<br />
( ) 3 2<br />
s<br />
−<br />
6s<br />
+ 4<br />
2<br />
+<br />
2<br />
⎤<br />
⎦<br />
( 8s<br />
+<br />
2<br />
3 ⎞<br />
s −1⎟<br />
2 ⎟<br />
⎠<br />
2<br />
3s<br />
3<br />
⎞<br />
⎟<br />
⎠<br />
⎛<br />
) ⎜<br />
⎝<br />
⎤<br />
⎦<br />
3 ⎞<br />
s −1⎟<br />
2 ⎟<br />
⎠
4<br />
-2t<br />
4!<br />
(b) L [ ] = ⋅ =<br />
3t<br />
e<br />
3<br />
( s + 2)<br />
5<br />
( s<br />
72<br />
5<br />
+ 2)<br />
⎡ d ⎤ 2<br />
2<br />
(c) L⎢⎣ 2t u(<br />
t)<br />
− 4 δ(<br />
t)<br />
⎥⎦<br />
= 2 − 4(<br />
s ⋅1−<br />
0)<br />
= 2 4s<br />
dt s<br />
s<br />
−<br />
(d) 2e<br />
-1)<br />
u(<br />
t)<br />
-t 2e<br />
u(<br />
t)<br />
2e<br />
-(t-1)<br />
L [ 2e u(<br />
t)<br />
]=<br />
s + 1<br />
-(t =<br />
(e) Using <strong>the</strong> scaling property,<br />
L 5u(<br />
t 2)<br />
1<br />
= 5⋅<br />
⋅<br />
1 2 s<br />
1 1<br />
= 5⋅<br />
2 ⋅<br />
( 1 2)<br />
2s<br />
[ ] =<br />
6<br />
s + 1 3<br />
-t<br />
3<br />
(f) L [ 6e u(<br />
t)<br />
] = =<br />
18<br />
3s<br />
+ 1<br />
(g) Let f ( t)<br />
= δ(<br />
t)<br />
. Then, F ( s)<br />
= <strong>1.</strong><br />
n<br />
n<br />
⎡ d ⎤ ⎡ d<br />
L ⎢ n δ(<br />
t)<br />
⎥ = L<br />
⎣ ⎦<br />
⎢ n<br />
dt ⎣ dt<br />
⎤<br />
n<br />
n−<br />
1<br />
n−2<br />
f ( t)<br />
⎥ = s F(<br />
s)<br />
− s f ( 0)<br />
− s f ′ ( 0)<br />
− L<br />
⎦<br />
n<br />
n<br />
⎡ d ⎤ ⎡ d<br />
L ⎢ n δ(<br />
t)<br />
⎥ = L<br />
⎣ ⎦<br />
⎢ n<br />
dt ⎣ dt<br />
n ⎡ d ⎤<br />
n<br />
L⎢<br />
n δ ( t)<br />
⎥ = s<br />
⎣ dt ⎦<br />
⎤<br />
n n−1<br />
n−2<br />
f ( t)<br />
⎥ = s ⋅1−<br />
s ⋅ 0 − s ⋅ 0 − L<br />
⎦<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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5<br />
s
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 6.<br />
<strong>Find</strong> F(s) given th<strong>at</strong><br />
⎧2t,<br />
0 < t < 1<br />
⎪<br />
f () t = ⎨t,<br />
1 < t < 2<br />
⎪<br />
⎩0,<br />
o<strong>the</strong>rwise<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 6.<br />
∞<br />
1 2<br />
−st −st −st<br />
∫ ∫ ∫<br />
F() s = f() t e dt = 2te dt+ 2e<br />
dt<br />
0 0 1<br />
e 1 e 2 2<br />
s 0 −s<br />
1 s<br />
−st −st<br />
2 2 ( −st − 1) + 2 =<br />
−s−2s (1 −e −se<br />
)<br />
2<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 7.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> following signals:<br />
(a) f () t = ( 2 t + 4)<br />
u()<br />
t<br />
−2t<br />
(b) g()<br />
t = ( 4 + 3e<br />
) u()<br />
t<br />
(c) h( t)<br />
= ( 6 sin(<br />
3t)<br />
+ 8cos(<br />
3t)<br />
) u(<br />
t)<br />
−2t<br />
x t = e <strong>cosh</strong> 4t<br />
u t<br />
( ) ()<br />
(d) () ( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 7.<br />
2 4<br />
(a) Fs () = + 2<br />
s s<br />
(b)<br />
(c )<br />
4 3<br />
Gs () = +<br />
s s+<br />
2<br />
H(<br />
s)<br />
= 6<br />
s<br />
2<br />
3<br />
+ 8<br />
+ 9 s<br />
2<br />
s 8s<br />
+ 18<br />
=<br />
2<br />
+ 9 s + 9<br />
(d) From <strong>Problem</strong> <strong>15</strong>.1,<br />
s<br />
L{<strong>cosh</strong> <strong>at</strong>}<br />
= 2 2<br />
s − a<br />
s+ 2 s+<br />
2<br />
X() s = =<br />
2 2 2<br />
( s+ 2) − 4 s + 4s−12 ( a)<br />
2 4<br />
+<br />
2<br />
s s<br />
4 3 8s<br />
+ 18<br />
, ( b)<br />
+ , ( c)<br />
, ( d)<br />
s s + 2 2<br />
s + 9 s<br />
s + 2<br />
+ 4s<br />
−12<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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2
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 8.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> F(s), given th<strong>at</strong> f(t) is:<br />
(a) 2tu ( t − 4)<br />
(b) 5cos() t δ ( t − 2)<br />
(c) e u(<br />
t t)<br />
t −<br />
−<br />
sin 2t<br />
u t −τ<br />
(d) ( ) ( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 8.<br />
(a) 2t=2(t-4) + 8<br />
f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4)<br />
2 −4s8 −4s ⎛ 2 8⎞<br />
−4s<br />
Fs () = e + e = e<br />
2 ⎜ + 2 ⎟<br />
s s ⎝s s⎠<br />
(b)<br />
(c)<br />
∞ ∞<br />
−st −st −st −2s<br />
∫ ∫ δ<br />
5cos(2)e<br />
t = 2<br />
0 0<br />
–2s<br />
Fs ( ) = fte ( ) dt= 5cos t ( t− 2) e dt= 5coste = 5cos 2e<br />
−t −( t−τ)<br />
−τ<br />
e = e e<br />
−τ −( t−τ)<br />
f() t = e e u( t−<br />
τ )<br />
− τ s+<br />
−τ −τs1<br />
e<br />
Fs () = e e =<br />
s+ 1 s+<br />
1<br />
( 1)<br />
(d) sin 2t = sin[2( t− τ ) + 2 τ] = sin 2( t− τ)cos 2τ + cos 2( t−<br />
τ)sin2τ f( t) = cos2τsin2( t−τ) u( t− τ) + sin2τ cos2( t−τ) u( t−<br />
τ)<br />
−τs 2<br />
−τs<br />
s<br />
Fs () = cos2τe+ sin2τe<br />
2 2<br />
s + 4 s + 4<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
written permission <strong>of</strong> <strong>the</strong> publisher, or used beyond <strong>the</strong> limited distribution to teachers and educ<strong>at</strong>ors<br />
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you are using it without permission.
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 9.<br />
Determine <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong>se functions:<br />
(a) f () t = ( t − 4) u(<br />
t − 2)<br />
(b) () 2 ( 1)<br />
4 − t<br />
g t = e u t −<br />
(c) h() t = 5cos( 2t<br />
−1)<br />
u()<br />
t<br />
p t = 6 u t − 2 − u t − 4<br />
(d) ( ) [ ( ) ( ) ]<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 9.<br />
(a) f ( t)<br />
= ( t − 4)<br />
u(<br />
t − 2)<br />
= ( t − 2)<br />
u(<br />
t − 2)<br />
− 2u(<br />
t − 2)<br />
-2s -2s<br />
e 2e<br />
F( s)<br />
= 2 − 2 s s<br />
-4t -4 -4(t-1)<br />
(b) g(<br />
t)<br />
= 2e<br />
u(<br />
t −1)<br />
= 2e<br />
e u(<br />
t −1)<br />
G(<br />
s)<br />
=<br />
e<br />
2e<br />
( s<br />
4<br />
-s<br />
+<br />
4)<br />
(c) h( t)<br />
= 5cos(<br />
2t<br />
−1)<br />
u(<br />
t)<br />
cos( A − B)<br />
= cos( A)<br />
cos( B)<br />
+ sin( A)<br />
sin( B)<br />
cos( 2t<br />
− 1)<br />
= cos( 2t)<br />
cos( 1)<br />
+ sin( 2t)<br />
sin( 1)<br />
h ( t)<br />
= 5cos(<br />
1)<br />
cos( 2t)<br />
u(<br />
t)<br />
+<br />
5sin(<br />
1)<br />
sin( 2t)<br />
u(<br />
t)<br />
s<br />
2<br />
H( s)<br />
= 5cos(<br />
1)<br />
⋅ 2 + 5sin(<br />
1)<br />
⋅ 2<br />
s + 4 s + 4<br />
2.<br />
702s<br />
8.<br />
4<strong>15</strong><br />
H(<br />
s)<br />
= 2 + 2 s + 4 s + 4<br />
(d) p( t)<br />
= 6u(<br />
t − 2)<br />
− 6u(<br />
t − 4)<br />
P(<br />
s)<br />
=<br />
6<br />
e<br />
s<br />
-2s<br />
6<br />
−<br />
e<br />
s<br />
-4s<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 10.<br />
In two different ways, find <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong><br />
d −t<br />
g()<br />
t = ( te cost<br />
)<br />
dt<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 10.<br />
(a) By taking <strong>the</strong> deriv<strong>at</strong>ive in <strong>the</strong> time domain,<br />
-t -t<br />
-t<br />
g(<br />
t)<br />
= ( -t e + e ) cos( t)<br />
− t e sin( t)<br />
-t -t<br />
-t<br />
g(<br />
t)<br />
= e cos( t)<br />
− t e cos( t)<br />
− t e sin( t)<br />
s + 1 d ⎡ s + 1 ⎤ d ⎡ 1 ⎤<br />
G( s)<br />
= + ⎢ ⎥ +<br />
2<br />
2 ⎢ 2 ⎥<br />
(s + 1) + 1 ds ⎣(<br />
s + 1)<br />
+ 1⎦<br />
ds ⎣(<br />
s + 1)<br />
+ 1⎦<br />
2<br />
2<br />
s + 1 s + 2s 2s + 2 s (s + 2)<br />
G( s)<br />
= 2 − 2<br />
2 − 2<br />
2 = 2<br />
2<br />
s + 2s<br />
+ 2 (s + 2s<br />
+ 2)<br />
(s + 2s<br />
+ 2)<br />
(s + 2s<br />
+ 2)<br />
(b) By applying <strong>the</strong> time differenti<strong>at</strong>ion property,<br />
G( s)<br />
= sF(<br />
s)<br />
− f ( 0)<br />
-t<br />
where f ( t)<br />
= t e cos( t)<br />
, f ( 0)<br />
= 0<br />
⎡ ⎤<br />
2<br />
2<br />
- d s + 1 (s)(s + 2s) s (s +<br />
2)<br />
G( s)<br />
= ( s)<br />
⋅ ⎢ 2 ⎥ = 2<br />
2 = 2<br />
2<br />
ds ⎣(<br />
s + 1)<br />
+ 1⎦<br />
(s + 2s<br />
+ 2)<br />
(s + 2s<br />
+ 2)<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 1<strong>1.</strong><br />
<strong>Find</strong> F(s) if:<br />
(a) f () t<br />
t<br />
6e <strong>cosh</strong> 2t<br />
−<br />
8 <strong>cosh</strong> 2<br />
3<br />
f t<br />
− t<br />
= e tu t −<br />
=<br />
−2t<br />
(b) f ( t)<br />
= 3te<br />
sinh 4t<br />
(c) () ( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 1<strong>1.</strong><br />
s<br />
(a) Since L [ <strong>cosh</strong>( <strong>at</strong>)<br />
] = 2 2 s − a<br />
6(<br />
s + 1)<br />
6(<br />
s + 1)<br />
F( s)<br />
= 2 = 2 ( s + 1)<br />
− 4 s + 2s<br />
− 3<br />
sinh( <strong>at</strong>)<br />
a<br />
= 2 s − a<br />
sinh( 4t)<br />
( 3)(<br />
4)<br />
=<br />
2 =<br />
( s + 2)<br />
−16<br />
s<br />
(b) Since L [ ] 2<br />
L<br />
-2t<br />
[ e<br />
]<br />
3 2<br />
12<br />
+ 4s<br />
−12<br />
F( s)<br />
= L<br />
- d<br />
ds<br />
−<br />
24(<br />
s + 2)<br />
2 -2<br />
( s)<br />
= ( 12)(<br />
2s<br />
+ 4)(<br />
s + 4s<br />
−12)<br />
=<br />
( s + 4s<br />
− 12<br />
- 2t<br />
2<br />
-1<br />
[ t ⋅ 3e<br />
sinh( 4t)<br />
] = [ 12(<br />
s + 4s<br />
12)<br />
]<br />
F 2 2 )<br />
1 t -t<br />
(c) <strong>cosh</strong>( t)<br />
= ⋅ ( e + e )<br />
2<br />
1 -3t t -t<br />
f ( t)<br />
= 8e<br />
⋅ ⋅ ( e + e ) u(<br />
t − 2)<br />
2<br />
-2t<br />
-4t<br />
= 4e<br />
u(<br />
t − 2)<br />
+ 4e<br />
u(<br />
t − 2)<br />
= 4e<br />
-4<br />
e<br />
-2(t<br />
-2)<br />
u(<br />
t<br />
− 2)<br />
+ 4e<br />
-8<br />
− 2)<br />
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e<br />
-4(t-2)<br />
-4 -2(t-2)<br />
-4 -2s -2<br />
[ 4e<br />
e u(<br />
t − 2)<br />
] = 4e<br />
e L [ e u(<br />
t)<br />
]<br />
L ⋅<br />
-(2s+<br />
4)<br />
4e<br />
-4<br />
-2(t-2)<br />
L [ 4e<br />
e u(<br />
t − 2)<br />
] =<br />
s + 2<br />
-8<br />
-4(t-2)<br />
Similarly, L [ 4e<br />
e u(<br />
t − 2)<br />
]<br />
Therefore,<br />
+<br />
4e<br />
F(<br />
s)<br />
=<br />
s + 2<br />
-(2s+<br />
4e<br />
+<br />
s + 4<br />
=<br />
u<br />
( t<br />
-(2s+<br />
4e<br />
=<br />
s + 4<br />
-(2s 4)<br />
8) -(2s+<br />
6) 2 -2<br />
2 -2<br />
e [ ( 4e<br />
+ 4e<br />
) s + ( 16e<br />
+ 8e<br />
) ]<br />
8)<br />
s<br />
2<br />
+<br />
6s<br />
+ 8
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 12.<br />
−2t<br />
If g()<br />
t = e cos 4t<br />
find G(s).<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 12.<br />
s+ 2 s+<br />
2<br />
Gs () = =<br />
2 2 2<br />
( s+ 2) + 4 s + 4s+ 20<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 13.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> following functions:<br />
(a) t u () t<br />
t cos (b) e t t u(<br />
t)<br />
t −<br />
sin<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 13.<br />
(a) tf ( t)<br />
← ⎯→<br />
d<br />
− F(<br />
s)<br />
ds<br />
If f(t) = cost, <strong>the</strong>n<br />
F(<br />
s)<br />
=<br />
s<br />
2<br />
s<br />
+ 1<br />
sin βt<br />
(c) u()<br />
t<br />
t<br />
and<br />
F(<br />
s)<br />
= −<br />
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-<br />
L ( t cos t)<br />
=<br />
(b) Let f(t) = e -t sin t.<br />
1 1<br />
F ( s)<br />
= = 2<br />
2<br />
( s + 1)<br />
+ 1 s + 2s<br />
+ 2<br />
2<br />
dF ( s + 2s<br />
+ 2)(<br />
0)<br />
− ( 1)(<br />
2s<br />
+ 2)<br />
=<br />
2<br />
2<br />
ds ( s + 2s<br />
+ 2)<br />
− dF 2(<br />
s + 1)<br />
( e t sin t)<br />
= − = 2<br />
ds ( s + 2s<br />
+ 2)<br />
t<br />
L<br />
f ( t)<br />
(c ) ←⎯→<br />
∫ F(<br />
s)<br />
ds<br />
t<br />
∞<br />
d<br />
ds<br />
( s<br />
s<br />
2<br />
2<br />
2<br />
−1<br />
+ 1)<br />
Let<br />
s<br />
f ( t)<br />
= sin βt,<br />
<strong>the</strong>n F(<br />
s)<br />
=<br />
β<br />
2 2<br />
∞<br />
s<br />
+ β<br />
2<br />
( s<br />
2<br />
+ 1)(<br />
1)<br />
− s(<br />
2s)<br />
2 2<br />
( s + 1)<br />
⎡sin<br />
βt ⎤ β 1 −1<br />
s ∞ π −1<br />
s −1<br />
β<br />
L<br />
⎢<br />
ds β tan<br />
tan tan<br />
2 2<br />
t ⎥<br />
= ∫ =<br />
= − =<br />
s + β β β s<br />
⎣ ⎦<br />
2 β s<br />
s
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 14.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> signal in Fig. <strong>15</strong>.26.<br />
Figure <strong>15</strong>.26<br />
For Prob. <strong>15</strong>.14.<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 14.<br />
Taking <strong>the</strong> deriv<strong>at</strong>ive <strong>of</strong> f(t) twice, we obtain <strong>the</strong> figures below.<br />
f’(t)<br />
5<br />
0 t<br />
2 4 6<br />
-2.5<br />
f’’(t)<br />
5δ (t) 2.5δ(t-6)<br />
0 2 6<br />
-7.5δ(t-2)<br />
f” = 5δ(t) – 7.5δ(t–2) + 2.5δ(t–6)<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term,<br />
s 2 F(s) = 5 – 7.5e –2s + 2.5e –6s or<br />
F(<br />
s)<br />
Please note th<strong>at</strong> we can obtain <strong>the</strong> same answer by representing <strong>the</strong> function as,<br />
f(t) = 5tu(t) – 7.5u(t–2) + 2.5u(t–6).<br />
=<br />
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5<br />
s<br />
−<br />
7.<br />
5<br />
e<br />
−2s<br />
s<br />
2<br />
+<br />
2.<br />
5<br />
e<br />
−6s<br />
s<br />
2
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> <strong>15</strong>.<br />
Determine <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> function in Fig. <strong>15</strong>.27.<br />
Figure <strong>15</strong>.27<br />
For Prob. <strong>15</strong>.<strong>15</strong>.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution <strong>15</strong>.<br />
This is a periodic function with T=3.<br />
F1() s<br />
Fs () 3s<br />
1 e − =<br />
−<br />
To get F1(s), we consider f(t) over one period.<br />
f1(t) f1’(t) f1’’(t)<br />
5 5<br />
5δ(t)<br />
0 1 t 0 1 t 0 1 t<br />
f1” = 5δ(t) –5δ(t–1) – 5δ’(t–1)<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term,<br />
Hence,<br />
–5δ(t-1)<br />
s 2 F1(s) = 5 –5e –s – 5se –s or F1(s) = 5(1 – e –s – se –s )/s 2<br />
1−<br />
e − se<br />
F(s) = 5<br />
2 −3s<br />
s ( 1−<br />
e )<br />
Altern<strong>at</strong>ively, we can obtain <strong>the</strong> same answer by noting th<strong>at</strong> f1(t) = 5tu(t) – 5tu(t–1) –<br />
5u(t–1).<br />
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−s<br />
−s<br />
–5δ(t-1)<br />
–5δ’(t-1)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 16.<br />
Obtain <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> f(t) in Fig. <strong>15</strong>.28.<br />
Figure <strong>15</strong>.28<br />
For Prob. <strong>15</strong>.16.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 16.<br />
f ( t)<br />
= 5u(<br />
t)<br />
− 3u(<br />
t −1)<br />
+ 3u(<br />
t − 3)<br />
− 5u(<br />
t −<br />
1<br />
-s<br />
-3s<br />
-4s<br />
F ( s)<br />
= [ 5 −<br />
3e<br />
+ 3e<br />
− 5e<br />
]<br />
s<br />
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4)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 17.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> f(t) shown in Fig. <strong>15</strong>.29.<br />
Figure <strong>15</strong>.29<br />
For Prob. <strong>15</strong>.17.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 17.<br />
Taking <strong>the</strong> deriv<strong>at</strong>ive <strong>of</strong> f(t) gives f’(t) as shown below.<br />
f’(t)<br />
2δ(t)<br />
-δ(t-1) – δ(t-2)<br />
f’(t) = 2δ(t) – δ(t–1) – δ(t–2)<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term,<br />
sF(s) = 2 – e –s – e –2s which leads to<br />
F(s) = [2 – e –s – e –2s ]/s<br />
We can also obtain <strong>the</strong> same answer noting th<strong>at</strong> f(t) = 2u(t) – u(t–1) – u(t–2).<br />
t<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 18.<br />
Obtain <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong> functions in Fig. <strong>15</strong>.30.<br />
Figure <strong>15</strong>.30<br />
For Prob. <strong>15</strong>.18.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 18.<br />
(a) g( t)<br />
= u(<br />
t)<br />
− u(<br />
t −1)<br />
+ 2[<br />
u(<br />
t −1)<br />
− u(<br />
t − 2)<br />
] + 3[<br />
u(<br />
t − 2)<br />
− u(<br />
t − 3)<br />
]<br />
= u( t)<br />
+ u(<br />
t −1)<br />
+ u(<br />
t − 2)<br />
− 3u(<br />
t − 3)<br />
1 s<br />
-s<br />
-2s<br />
-3<br />
G ( s)<br />
= ( 1 + e + e − 3e<br />
)<br />
s<br />
(b) h( t)<br />
= 2t<br />
[ u(<br />
t)<br />
− u(<br />
t −1)<br />
] + 2[<br />
u(<br />
t −1)<br />
− u(<br />
t − 3)<br />
]<br />
+ ( 8 − 2t)<br />
[ u(<br />
t − 3)<br />
− u(<br />
t − 4)<br />
]<br />
= 2t u(<br />
t)<br />
− 2(<br />
t −1)<br />
u(<br />
t −1)<br />
− 2u(<br />
t −1)<br />
+ 2u(<br />
t −1)<br />
− 2u(<br />
t − 3)<br />
− 2( t − 3)<br />
u(<br />
t − 3)<br />
+ 2u(<br />
t − 3)<br />
+ 2(<br />
t − 4)<br />
u(<br />
t − 4)<br />
= 2t u(<br />
t)<br />
− 2(<br />
t −1)<br />
u(<br />
t −1)<br />
− 2(<br />
t − 3)<br />
u(<br />
t − 3)<br />
+ 2(<br />
t − 4)<br />
u(<br />
t − 4)<br />
2 2 2 2 -s<br />
-3s<br />
-4s<br />
-s<br />
-3s<br />
-4s<br />
H ( s)<br />
= 2 ( 1−<br />
e ) − 2 e + 2 e = 2 ( 1 −<br />
e − e + e )<br />
s s s s<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 19.<br />
Calcul<strong>at</strong>e <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> train <strong>of</strong> unit impulses in Fig. <strong>15</strong>.3<strong>1.</strong><br />
Figure <strong>15</strong>.31<br />
For Prob. <strong>15</strong>.19.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 19.<br />
1<br />
Since L [ δ(<br />
t)<br />
] = 1 and T = 2 , ( s)<br />
=<br />
1 − e<br />
F -2s<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 20.<br />
The periodic function shown in Fig. <strong>15</strong>.32 is defined over its period as<br />
() ⎨<br />
⎩ ⎧sinπ<br />
t,<br />
0 < t < 1<br />
g t<br />
0,<br />
1 < t < 2<br />
<strong>Find</strong> G(s)<br />
Figure <strong>15</strong>.32<br />
For Prob. <strong>15</strong>.20.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 20.<br />
Let g1 ( t)<br />
= sin( πt),<br />
0 < t < 1<br />
= sin( πt)<br />
[ u(<br />
t)<br />
− u(<br />
t −1)<br />
]<br />
= sin( πt)<br />
u(<br />
t)<br />
− sin( πt)<br />
u(<br />
t −1)<br />
Note th<strong>at</strong> sin( π ( t −1))<br />
= sin( πt<br />
− π)<br />
= -sin(<br />
πt)<br />
.<br />
So, ( t)<br />
= sin( πt)<br />
u(t) + sin( π(<br />
t -1))<br />
u(t -1)<br />
g 1<br />
G<br />
1<br />
( s)<br />
G(<br />
s)<br />
=<br />
s<br />
2<br />
π<br />
+ π<br />
G1<br />
( s)<br />
- 1−<br />
e<br />
= 2s<br />
2<br />
- ( 1+<br />
e<br />
=<br />
( s<br />
2<br />
s<br />
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)<br />
-s π<br />
( 1 + e )<br />
2 + π )( 1 − e<br />
-2s<br />
)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 2<strong>1.</strong><br />
Obtain <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> periodic waveform in Fig. <strong>15</strong>.33.<br />
Figure <strong>15</strong>.33<br />
For Prob. <strong>15</strong>.2<strong>1.</strong><br />
<strong>Chapter</strong> <strong>15</strong>, Solution 2<strong>1.</strong><br />
T = 2π<br />
Let ( t)<br />
= 1−<br />
[ u(<br />
t)<br />
− u(<br />
t − 2π)<br />
]<br />
f 1<br />
f1( t)<br />
⎛ t ⎞<br />
⎜ ⎟<br />
⎝ 2π<br />
⎠<br />
t 1<br />
= u(<br />
t)<br />
− u(<br />
t)<br />
+ ( t − 2π)<br />
u(<br />
t − 2π)<br />
2π<br />
2π<br />
-2πs<br />
[ -1<br />
+ e ]<br />
-2πs<br />
1 1 e 2π<br />
s +<br />
F1<br />
( s)<br />
= − + =<br />
s 2 2<br />
2<br />
2πs<br />
2πs<br />
2πs<br />
F(<br />
s)<br />
F1<br />
( s)<br />
- 1−<br />
e<br />
= Ts<br />
=<br />
2πs<br />
−1<br />
+ e<br />
2πs<br />
2<br />
( 1−<br />
e<br />
−2πs<br />
-2πs<br />
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)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 22.<br />
<strong>Find</strong> <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong> functions in Fig. <strong>15</strong>.34.<br />
Figure <strong>15</strong>.34<br />
For Prob. <strong>15</strong>.22.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 22.<br />
(a) Let g1 ( t)<br />
= 2t,<br />
0 < t < 1<br />
= 2t [ u(<br />
t)<br />
− u(<br />
t −1)<br />
]<br />
= 2t u(<br />
t)<br />
− 2(<br />
t −1)<br />
u(<br />
t −1)<br />
+ 2u(<br />
t −1)<br />
-s<br />
2 2e<br />
2 -s<br />
G 1(<br />
s)<br />
= 2 − 2 + e<br />
s s s<br />
G1(<br />
s)<br />
G( s)<br />
= -sT<br />
,<br />
1−<br />
e<br />
T = 1<br />
-s -s<br />
2(<br />
1 − e + se<br />
)<br />
G(<br />
s)<br />
= 2 -s<br />
s ( 1 − e )<br />
(b) Let h = h 0 + u(<br />
t)<br />
, where h 0 is <strong>the</strong> periodic triangular wave.<br />
Let 1 h be h 0 within its first period, i.e.<br />
⎧ 2t<br />
h ⎨<br />
1 ( t)<br />
=<br />
⎩4<br />
− 2t<br />
0 < t < 1<br />
1 < t < 2<br />
h1 ( t)<br />
= 2t<br />
u(<br />
t)<br />
− 2t<br />
u(<br />
t −1)<br />
+ 4u<br />
( t −1)<br />
− 2t<br />
u(<br />
t −1)<br />
− 2(<br />
t − 2)<br />
u(<br />
t<br />
h1 ( t)<br />
= 2t<br />
u(<br />
t)<br />
− 4(<br />
t −1)<br />
u(<br />
t −1)<br />
− 2(<br />
t − 2)<br />
u(<br />
t − 2)<br />
-2s<br />
2 4 2e<br />
2<br />
-s<br />
-s<br />
2<br />
H1 ( s)<br />
= 2 − 2 e − 2 = 2 ( 1−<br />
e )<br />
s s s s<br />
-s 2 2 ( 1−<br />
e )<br />
H 0 ( s)<br />
= 2 -2s<br />
s ( 1−<br />
e )<br />
H(<br />
s)<br />
1 2 ( 1 − e<br />
= +<br />
2 s s ( 1 − e<br />
-s<br />
2 )<br />
)<br />
-2s<br />
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−<br />
2)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 23.<br />
Determine <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong> periodic functions in Fig. <strong>15</strong>.35.<br />
Figure <strong>15</strong>.35<br />
For Prob. <strong>15</strong>.23.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 23.<br />
(a) Let<br />
f1 ( t)<br />
f1( t)<br />
=<br />
f1( t)<br />
= u(<br />
t)<br />
⎧ 1<br />
= ⎨<br />
⎩-1<br />
0 < t < 1<br />
1 < t < 2<br />
[ u(<br />
t)<br />
− u(<br />
t −1)<br />
] − [ u(<br />
t −1)<br />
− u(<br />
t − 2)<br />
]<br />
1<br />
s<br />
−<br />
2 u(<br />
t<br />
−1)<br />
+<br />
u(<br />
t<br />
1<br />
s<br />
−<br />
2)<br />
-s<br />
-2s<br />
F1 ( s)<br />
= ( 1−<br />
2e<br />
+ e ) = ( 1−<br />
F1<br />
( s)<br />
F( s)<br />
= -sT<br />
,<br />
( 1−<br />
e )<br />
-s 2 ( 1 − e )<br />
F(<br />
s)<br />
=<br />
-2s<br />
s(<br />
1 − e )<br />
T = 2<br />
2<br />
2<br />
2<br />
(b) Let h ( t)<br />
= t [ u(<br />
t)<br />
− u(<br />
t − 2)<br />
] = t u(<br />
t)<br />
− t u(<br />
t − 2)<br />
1<br />
2<br />
2<br />
h1<br />
( t)<br />
= t u(<br />
t)<br />
− ( t − 2)<br />
u(<br />
t − 2)<br />
−<br />
2 4 4<br />
-2s<br />
-2s<br />
H1( s)<br />
= 3 ( 1−<br />
e ) − 2 e − e<br />
s<br />
s s<br />
H1(<br />
s)<br />
H( s)<br />
= -Ts<br />
, T = 2<br />
( 1−<br />
e )<br />
-2s<br />
-2s 2<br />
2(<br />
1 −<br />
e ) − 4se<br />
( s + s )<br />
H(<br />
s)<br />
=<br />
3 -2s<br />
s ( 1 − e )<br />
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e<br />
-s<br />
)<br />
2<br />
4(<br />
t<br />
-2s<br />
−<br />
2)<br />
u(<br />
t<br />
−<br />
2)<br />
−<br />
4u(<br />
t<br />
−<br />
2)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 24.<br />
Given th<strong>at</strong><br />
s<br />
2<br />
+<br />
10 2<br />
s + 6<br />
F () s =<br />
s(<br />
s + 1)<br />
( s + 3)<br />
Evalu<strong>at</strong>e f(0) and f ( ∞)<br />
if <strong>the</strong>y exist.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 24.<br />
f(0) = lim sF( s)<br />
= lim<br />
2<br />
s + 10s+ 6<br />
2 3<br />
1/ s+ 10/ s + 6/ s 0<br />
= lim = = 0<br />
s→∞ s→∞ 2<br />
( s+ 1) ( s+ 2) s→∞<br />
(1+ 1/ s)(1+ 2/ s)<br />
1<br />
2<br />
s + 10s+ 6 6<br />
f( ∞ ) = lim sF( s)<br />
= lim = = 3<br />
s→0 s→0<br />
2<br />
( s+ 1) ( s+<br />
2) (1)(2)<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 25.<br />
Let<br />
5<br />
()<br />
( s + 1)<br />
F s =<br />
( s + 2)(<br />
s + 3)<br />
(a) Use <strong>the</strong> initial and final value <strong>the</strong>orems to find f(0) and ( ∞)<br />
f .<br />
(b) Verify your answer in part (a) by finding f(t), using partial fractions.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 25.<br />
5( ss+ 1) 5(1+ 1/ s)<br />
(a) f(0) = lim sF( s)<br />
= lim = lim = 5<br />
s→∞ s→∞ ( s+ 2)( s+ 3) s→∞<br />
(1+ 2/ s)(1+ 3/ s)<br />
5( ss+<br />
1)<br />
f( ∞ ) = lim sF( s)<br />
= lim = 0<br />
s→0 s→0<br />
( s+ 2)( s+<br />
3)<br />
(b)<br />
5( s+ 1) A B<br />
Fs () = = +<br />
( s+ 2)( s+ 3) s+ 2 s+<br />
3<br />
5( −1) A= =− 5,<br />
1<br />
5( −2)<br />
B=<br />
= 10<br />
−1<br />
−5<br />
10<br />
Fs () = +<br />
s+ 2 s+<br />
3<br />
⎯⎯→ ft () =− 5e + 10e<br />
f(0) = -5 + 10 = 5<br />
f( ∞ )= -0 + 0 = 0.<br />
−2t −3t<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 26.<br />
Determine <strong>the</strong> initial and final values <strong>of</strong> f(t), if <strong>the</strong>y exist, given th<strong>at</strong>:<br />
2<br />
s + 3<br />
(a) F () s = 3 2<br />
s + 4s<br />
+ 6<br />
2<br />
s − 2s<br />
+ 1<br />
(b) F () s =<br />
2<br />
s − 2 s + 2s<br />
+ 4<br />
( )( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 26.<br />
3<br />
s + 3s<br />
(a) f ( 0)<br />
= limsF(<br />
s)<br />
= lim 3 2 = 1<br />
s→∞<br />
s→∞<br />
s + 4s<br />
+ 6<br />
(b)<br />
Two poles are not in <strong>the</strong> left-half plane.<br />
f ( ∞) does not exist<br />
3 2 s − 2s<br />
+ s<br />
( 0)<br />
= limsF(<br />
s)<br />
= lim<br />
→∞<br />
s→∞<br />
( s − 2)(<br />
s + 2s<br />
2 1<br />
1−<br />
+ 2<br />
= lim s s = 1<br />
s→∞<br />
⎛ 2 ⎞⎛<br />
2 4 ⎞<br />
⎜1−<br />
⎟⎜1<br />
+ + 2 ⎟<br />
⎝ s ⎠⎝<br />
s s ⎠<br />
f 2<br />
s +<br />
One pole is not in <strong>the</strong> left-half plane.<br />
f ( ∞) does<br />
not exist<br />
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4)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 27.<br />
Determine <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each <strong>of</strong> <strong>the</strong> following functions:<br />
1 2<br />
(a) F () s = +<br />
s s + 1<br />
3s<br />
+ 1<br />
(b) G () s =<br />
s + 4<br />
4<br />
(c) H () s =<br />
( s + 1)(<br />
s + 3)<br />
12<br />
(d) J<br />
() s = 2<br />
s + 2 s + 4<br />
( ) ( )<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 27.<br />
(a) f ( t)<br />
= u(<br />
t)<br />
+ 2e<br />
u(<br />
t)<br />
(b)<br />
(c)<br />
(d)<br />
G(<br />
s)<br />
=<br />
3(<br />
s<br />
-t<br />
+ 4)<br />
−11<br />
11<br />
= 3 −<br />
s + 4 s + 4<br />
-4t<br />
g ( t)<br />
= 3δ<br />
( t)<br />
−11e<br />
u(<br />
t)<br />
4 A B<br />
H(<br />
s)<br />
=<br />
= +<br />
( s + 1)(<br />
s + 3)<br />
s + 1 s + 3<br />
A = 2,<br />
B = -2<br />
2 2<br />
H(<br />
s)<br />
= −<br />
s + 1 s + 3<br />
h(<br />
t)<br />
-t -3t<br />
= 2e − 2e<br />
u(t)<br />
12 A B<br />
( s)<br />
=<br />
= +<br />
( s + 2)<br />
( s + 4)<br />
s + 2 ( s + 2)<br />
12<br />
12<br />
B = = 6 , C 2 3<br />
2<br />
( -2)<br />
= =<br />
C<br />
+<br />
s 4<br />
J 2 2 +<br />
12 = A ( s + 2)<br />
( s + 4)<br />
+ B(<br />
s + 4)<br />
+ C(<br />
s +<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 0 = A + C ⎯⎯→<br />
A = -C = -3<br />
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2 2)<br />
1 s: 0 = 6A<br />
+ B + 4C<br />
= 2A<br />
+ B ⎯⎯→<br />
B = -2A = 6<br />
0 s : 12 = 8A<br />
+ 4B<br />
+ 4C<br />
= -24 + 24 + 12 = 12<br />
- 3 6 3<br />
J( s)<br />
= + 2 +<br />
s + 2 ( s + 2)<br />
s + 4<br />
j(<br />
t)<br />
-4t -2t -2t<br />
= 3 e − 3e<br />
+ 6t<br />
e u(t)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 28.<br />
<strong>Find</strong> <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> following functions:<br />
20<br />
(a) ()<br />
( s + 2)<br />
F s = 2<br />
s(<br />
s + 6s<br />
+ 25)<br />
6 36 20<br />
(b) ()<br />
1 2 3<br />
2<br />
s + s +<br />
P s =<br />
s + s + s +<br />
( )( )( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 28.<br />
20( s+ 2) A Bs+ C<br />
(a) Fs () = = +<br />
2 2<br />
ss ( + 6s+ 25) s s + 6s+ 25<br />
2 2<br />
20( s+ 2) = A( s + 6s+ 25 s) + Bs + Cs<br />
Equ<strong>at</strong>ing components,<br />
s 2 : 0 = A + B or B= - A<br />
s: 20 = 6A + C<br />
constant: 40 – 25 A or A = 8/5, B = -8/5, C= 20 – 6A= 52/5<br />
8 52 8 24 52<br />
− s+ − ( s+<br />
3) + +<br />
8 5 5 8<br />
Fs () = + = + 5 5 5<br />
2 2 2 2<br />
5 s ( s+ 3) + 4 5 s ( s+<br />
3) + 4<br />
8 8 −3t 19 −3t<br />
f () t = u() t − e cos4t+ e sin4t<br />
5 5 5<br />
2<br />
6s + 36s+ 20 A B C<br />
(b) Ps () = = + +<br />
( s+ 1)( s+ 2)( s+ 3) s+ 1 s+ 2 s+<br />
3<br />
6− 36+ 20<br />
A = =−5<br />
( − 1+ 2)( − 1+ 3)<br />
24 − 72 + 20<br />
B = = 28<br />
( −1)(1)<br />
54 − 108 + 20<br />
C = =−17<br />
( −2)( −1)<br />
−5<br />
28 17<br />
Ps () = + −<br />
s+ 1 s+ 2 s+<br />
3<br />
pt e e e ut<br />
−t −2t −3t<br />
( ) = ( − 5 + 28 −<br />
17 ) ( )<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
written permission <strong>of</strong> <strong>the</strong> publisher, or used beyond <strong>the</strong> limited distribution to teachers and educ<strong>at</strong>ors<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 29.<br />
<strong>Find</strong> <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
2s<br />
+ 26<br />
V () s = 2<br />
s s + 4s<br />
+ 13<br />
( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 29.<br />
V(<br />
s)<br />
V(<br />
s)<br />
=<br />
=<br />
2<br />
s<br />
2<br />
s<br />
As + B 2<br />
2<br />
+<br />
; 2s<br />
+ 8s<br />
+ 26 + As + Bs = 2s<br />
+ 26 → A = −2and<br />
B = −6<br />
2 2<br />
( s + 2)<br />
+ 3<br />
2(<br />
s + 2)<br />
2 3<br />
−<br />
−<br />
2 2 3 2 2<br />
( s + 2)<br />
+ 3 ( s + 2)<br />
+ 3<br />
−2t<br />
2 −2t<br />
v(t) = ( 2 − 2e<br />
cos3t<br />
− e sin 3t)<br />
u(<br />
t),<br />
t ≥ 0<br />
3<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 30.<br />
<strong>Find</strong> <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
2<br />
6s + 8s<br />
+ 3<br />
(a) F 1()<br />
s = 2<br />
s(<br />
s + 2s<br />
+ 5)<br />
2<br />
s + 5s<br />
+ 6<br />
(b) F 2 () s = 2 ( s + 1)<br />
( s + 4)<br />
10<br />
(c) F 3 () s =<br />
2<br />
s + 1 s + 4s<br />
+ 8<br />
( )( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 30.<br />
(a)<br />
2<br />
6 8 3<br />
1() = = +<br />
2 2<br />
F s<br />
s + s+ A Bs+ C<br />
ss ( + 2s+ 5) s s + 2s+ 5<br />
6 + 8 + 3 = ( + 2 + 5) + +<br />
We equ<strong>at</strong>e coefficients.<br />
s 2 : 6 = A + B<br />
s: 8= 2A + C<br />
constant: 3=5A or A=3/5<br />
B=6-A = 27/5, C=8-2A = 34/5<br />
2 2 2<br />
s s A s s Bs Cs<br />
3/5 27 s/5+ 34/5 3/5 27( s+<br />
1)/5+ 7/5<br />
F1() s = + = +<br />
2 2 2<br />
s s + 2s+ 5 s ( s+<br />
1) + 2<br />
⎡3 27 −t 7 −t<br />
⎤<br />
f1() t =<br />
⎢<br />
+ e cos2t+ e sin2 t u() t<br />
⎣5 5 10 ⎥<br />
⎦<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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(b)<br />
+ 5 + 6<br />
( s+ 1) ( s+ 4) s+ 1 ( s+ 1) s+<br />
4<br />
2<br />
s s A B C<br />
2() = = + +<br />
2 2<br />
F s<br />
s + 5s+ 6 = A( s+ 1)( s+ 4) + B( s+ 4) + C( s+<br />
1)<br />
Equ<strong>at</strong>ing coefficients,<br />
2 2<br />
s 2 : 1=A+C<br />
s: 5=5A+B+2C<br />
constant: 6=4A+4B+C<br />
Solving <strong>the</strong>se gives<br />
A=7/9, B= 2/3, C=2/9<br />
7/9 2/3 2/9<br />
F2() s = + + 2<br />
s+ 1 ( s+ 1) s+<br />
4<br />
⎡7 2 2 ⎤<br />
f t =<br />
⎢<br />
e + te + e u t<br />
⎣9 3 9 ⎥<br />
⎦<br />
−t −t −4t<br />
2()<br />
()<br />
10 A Bs+ C<br />
(c ) F 3() s = = +<br />
2 2<br />
( s+ 1)( s + 4s+ 8) s+ 1 s + 4s+ 8<br />
2 2<br />
10 = As ( + 4s+ 8) + Bs ( + s) + Cs ( + 1)<br />
s 2 : 0 = A + B or B = -A<br />
s: 0=4A+ B + C<br />
constant: 10=8A+C<br />
Solving <strong>the</strong>se yields<br />
A=2, B= -2, C= -6<br />
2 −2s− 6 2 2( s+<br />
1) 4<br />
F 3() s = + = − −<br />
2 2 2 2 2<br />
s+ 1 s + 4s+ 8 s+ 1 ( s+ 1) + 2 ( s+<br />
1) + 2<br />
f3(t) = (2e –t – 2e –t cos(2t) – 2e –t sin(2t))u(t).<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 3<strong>1.</strong><br />
<strong>Find</strong> f(t) for each F(s):<br />
10s<br />
s + 1 s + 2 s + 3<br />
(a) ( )( )( )<br />
2s<br />
2<br />
+ 4s<br />
+ 1<br />
(b) ( )( ) 3<br />
s + 1 s + 2<br />
s + 1<br />
s + 2<br />
2<br />
s + 2s<br />
+ 5<br />
(c) ( )( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 3<strong>1.</strong><br />
(a)<br />
F(<br />
s)<br />
=<br />
( s<br />
+<br />
1)(<br />
s<br />
10s<br />
+<br />
2)(<br />
s<br />
-10<br />
A = F(<br />
s)<br />
( s + 1)<br />
s = -1<br />
= = -5<br />
2<br />
- 20<br />
B = F(<br />
s)<br />
( s + 2)<br />
s = -2 = = 20<br />
-1<br />
- 30<br />
C = F(<br />
s)<br />
( s + 3)<br />
s = -3<br />
= = -<strong>15</strong><br />
2<br />
F(<br />
s)<br />
- 5 20 <strong>15</strong><br />
= + −<br />
s + 1 s + 2 s + 3<br />
-t<br />
-2t<br />
A B C<br />
= + +<br />
+ 3)<br />
s + 1 s + 2 s + 3<br />
-3t<br />
f ( t)<br />
= (-5e<br />
+<br />
20e<br />
−<strong>15</strong>e<br />
) u(<br />
t)<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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2s<br />
+ 4s<br />
+ 1<br />
( s + 1)(<br />
s + 2)<br />
A<br />
s + 1<br />
B<br />
s + 2<br />
C<br />
( s + 2)<br />
D<br />
( s + 2)<br />
2<br />
(b) F(<br />
s)<br />
=<br />
3 = + + 2 + 3<br />
(c)<br />
A = F(<br />
s)<br />
( s + 1)<br />
s = -1<br />
= -1<br />
D = F(<br />
s)<br />
( s + 2)<br />
s=<br />
-2<br />
3 =<br />
2s<br />
2<br />
+ 4s<br />
+ 1 = A(<br />
s + 2)(<br />
s + 4s<br />
+ 4)<br />
+ B(<br />
s + 1)(<br />
s<br />
+ C ( s + 1)(<br />
s + 2)<br />
+ D(<br />
s + 1)<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s: 0 = A + B ⎯⎯→<br />
B = -A = 1<br />
-1<br />
+ 4s<br />
2 2 +<br />
2 s : 2 = 6A<br />
+ 5B<br />
+ C = A + C ⎯⎯→<br />
C = 2 − A = 3<br />
1 s: 4 = 12A<br />
+ 8B<br />
+ 3C<br />
+ D = 4A<br />
+ 3C<br />
+ D<br />
4 = 6 + A + D ⎯⎯→<br />
D = -2 − A = -1<br />
0 s : 1 = 8A<br />
+ 4B<br />
+ 2C<br />
+ D = 4A + 2C + D = -4 + 6 −1<br />
= 1<br />
F(<br />
s)<br />
-1<br />
1<br />
= + +<br />
s + 1 s + 2<br />
( s<br />
3<br />
2)<br />
+<br />
−<br />
2 3 ( s 2)<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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1<br />
+<br />
2<br />
t<br />
-t -2t<br />
-2t<br />
-2t<br />
f(t) = -e<br />
+ e + 3t<br />
e − e<br />
2<br />
⎛ 2<br />
t ⎞<br />
-t<br />
-2t<br />
f ( t)<br />
= (-e<br />
+ ⎜1+<br />
3t<br />
− ⎟ e ) u(<br />
t)<br />
⎜ 2 ⎟<br />
⎝ ⎠<br />
s + 1 A Bs + C<br />
F( s)<br />
=<br />
2 = + 2<br />
( s + 2)(<br />
s + 2s<br />
+ 5)<br />
s + 2 s + 2s<br />
+ 5<br />
-1<br />
A = F(<br />
s)<br />
( s + 2)<br />
s = -2 =<br />
5<br />
2 2<br />
s + 1 = A ( s + 2s<br />
+ 5)<br />
+ B(<br />
s + 2s)<br />
+ C(<br />
s + 2)<br />
Equ<strong>at</strong>ing coefficients :<br />
1<br />
2 s : 0 = A + B ⎯⎯→<br />
B = -A =<br />
5<br />
1 s : 1 = 2A<br />
+ 2B<br />
+ C = 0 + C ⎯⎯→<br />
C = 1<br />
0 s : 1 = 5A<br />
+ 2C<br />
= -1+<br />
2 = 1<br />
F(<br />
s)<br />
-1<br />
5 1 5⋅<br />
s + 1<br />
= +<br />
s + 2 ( s + 1)<br />
+<br />
-2t<br />
-1<br />
5<br />
= +<br />
+<br />
1 5(<br />
s<br />
+ 1)<br />
4)<br />
4 5<br />
2 2<br />
2 2<br />
2 2<br />
2 s 2 ( s + 1)<br />
+ 2 ( s + 1)<br />
+ 2<br />
-t<br />
f ( t)<br />
= (-0.2e<br />
+<br />
0.<br />
2e<br />
cos( 2t)<br />
+ 0.<br />
4e<br />
sin( 2t))<br />
u(<br />
t)<br />
-t<br />
+
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 32.<br />
Determine <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each <strong>of</strong> <strong>the</strong> following functions:<br />
2<br />
2<br />
8(<br />
s + 1)(<br />
s + 3)<br />
s − 2s<br />
+ 4<br />
s + 1<br />
(a)<br />
(b) 2<br />
2<br />
s s + 2 s + 4 ( s + 1)(<br />
s + 2)<br />
s + 3 s + 4s<br />
+ 5<br />
( )( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 32.<br />
(a)<br />
F(<br />
s)<br />
8(<br />
s + 1)(<br />
s + 3)<br />
=<br />
=<br />
s ( s + 2)(<br />
s + 4)<br />
(c) ( )( )<br />
A B C<br />
+ +<br />
s s + 2 s + 4<br />
F(<br />
s)<br />
s s<br />
(8)(3)<br />
= = 3<br />
( 2)(<br />
4)<br />
F(<br />
s)<br />
( s + 2)<br />
s<br />
(8)(-1)<br />
= = 2<br />
( -4)<br />
F(<br />
s)<br />
( s + 4)<br />
s<br />
(8)(-1)(-3)<br />
= = 3<br />
( -4)<br />
(-2)<br />
A = = 0<br />
B = = -2<br />
C = = -4<br />
F(<br />
s)<br />
f ( t)<br />
3 2 3<br />
= + +<br />
s s + 2 s + 4<br />
= 3 u(<br />
t)<br />
+ 2e<br />
+ 3<br />
s − 2s<br />
+ 4<br />
( s + 1)(<br />
s + 2)<br />
-2t -4t e<br />
A<br />
s + 1<br />
B<br />
s + 2<br />
C<br />
( s + 2)<br />
2<br />
(b) F(<br />
s)<br />
=<br />
2 = + + 2<br />
s<br />
− 2s<br />
+ 4 = A ( s<br />
+ 4s<br />
+ 4)<br />
+ B(<br />
s<br />
+ 3s<br />
+ 2)<br />
+ C(<br />
s<br />
2 2<br />
2<br />
+<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 1 = A + B ⎯⎯→<br />
B = 1−<br />
A<br />
1 s : - 2 = 4A<br />
+ 3B<br />
+ C = 3 + A + C<br />
0 s : 4 = 4A<br />
+ 2B<br />
+ C = -B<br />
− 2 ⎯⎯→<br />
B = -6<br />
A = 1−<br />
B = 7 C = -5 - A = -12<br />
F(<br />
s)<br />
f ( t)<br />
=<br />
7 6<br />
− −<br />
s + 1 s + 2<br />
( s<br />
= 7 e −<br />
6(<br />
1 + 2t)<br />
12<br />
2 + 2)<br />
-t -2t e<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
written permission <strong>of</strong> <strong>the</strong> publisher, or used beyond <strong>the</strong> limited distribution to teachers and educ<strong>at</strong>ors<br />
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1)
(c)<br />
2 s + 1 A Bs + C<br />
( s)<br />
=<br />
2 = +<br />
( s + 3)(<br />
s + 4s<br />
+ 5)<br />
s + 3 s + 4s<br />
+ 5<br />
F 2<br />
s<br />
+ 1 = A ( s<br />
+ 4s<br />
+ 5)<br />
+ B(<br />
s<br />
+ 3s)<br />
+ C(<br />
s<br />
2 2<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 1 = A + B ⎯⎯→<br />
B = 1−<br />
A<br />
1 s: 0 = 4A<br />
+ 3B<br />
+ C = 3 + A + C ⎯⎯→<br />
A + C = -3<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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3)<br />
0 s : 1 = 5A<br />
+ 3C<br />
= -9 + 2A ⎯⎯→<br />
A = 5<br />
B = 1−<br />
A = -4 C = -A<br />
− 3 = -8<br />
5 4s<br />
+ 8 5 4(<br />
s + 2)<br />
( s)<br />
= − 2 = −<br />
s + 3 ( s + 2)<br />
+ 1 s + 3 ( s + 2)<br />
+ 1<br />
F 2<br />
-3t -2t<br />
f ( t)<br />
= 5e<br />
−<br />
4e<br />
cos( t)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 33.<br />
Calcul<strong>at</strong>e <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
6( s −1)<br />
(a) 4<br />
s −1<br />
−<br />
se<br />
(b) 2<br />
s + 1<br />
s π<br />
(c)<br />
( ) 3<br />
8<br />
s s + 1<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 33.<br />
(a)<br />
6(<br />
s −1)<br />
6 As + B C<br />
F( s)<br />
= 4 = 2 = 2 +<br />
s −1<br />
( s + 1)(<br />
s + 1)<br />
s + 1 s + 1<br />
(b)<br />
2 2<br />
6 = A ( s + s)<br />
+ B(<br />
s + 1)<br />
+ C(<br />
s + 1)<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 0 = A + C ⎯⎯→<br />
A = -C<br />
1 s : 0 = A + B ⎯⎯→<br />
B = -A = C<br />
0 s : 6 = B + C = 2B ⎯⎯→<br />
B = 3<br />
A = -3 , B = 3 , C = 3<br />
3 - 3s + 3 3 - 3s 3<br />
F( s)<br />
= + 2 = + 2 + 2<br />
s + 1 s + 1 s + 1 s + 1 s + 1<br />
-t<br />
f ( t)<br />
= ( 3e<br />
+ 3sin(<br />
t)<br />
− 3cos(<br />
t))<br />
u(<br />
t)<br />
-πs<br />
s e<br />
F( s)<br />
= 2 s + 1<br />
f ( t)<br />
= cos( t − π)<br />
u(<br />
t − π)<br />
F(<br />
s)<br />
=<br />
8<br />
A<br />
= +<br />
(c) 3 2<br />
3<br />
s ( s + 1)<br />
s s + 1 ( s + 1)<br />
( s + 1)<br />
A = 8 , D = -8<br />
8 = A ( s 2<br />
3<br />
+ 3s<br />
+ 3s<br />
+ 1)<br />
+ B(<br />
s + 2s<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s : 0 = A + B ⎯⎯→<br />
B = -A<br />
B<br />
+<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
written permission <strong>of</strong> <strong>the</strong> publisher, or used beyond <strong>the</strong> limited distribution to teachers and educ<strong>at</strong>ors<br />
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C<br />
+<br />
D<br />
+ s)<br />
+ C(<br />
s<br />
+ s)<br />
3 2<br />
2 +<br />
2 s : 0 = 3A<br />
+ 2B<br />
+ C = A + C ⎯⎯→<br />
C = -A = B<br />
1 s: 0 = 3A<br />
+ B + C + D = A + D ⎯⎯→<br />
D = -A<br />
0 s : A = 8,<br />
B = −8,<br />
C = −8,<br />
D = −8<br />
F(<br />
s)<br />
8 8<br />
= − −<br />
s s + 1<br />
( s<br />
8<br />
+ 1)<br />
−<br />
8<br />
+<br />
2 3 ( s 1)<br />
-t -t 2 -t<br />
f ( t)<br />
= 8[<br />
1 −<br />
e − t e − 0.<br />
5t<br />
e ] u(<br />
t)<br />
Ds
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 34.<br />
<strong>Find</strong> <strong>the</strong> time functions th<strong>at</strong> have <strong>the</strong> following <strong>Laplace</strong> <strong>transform</strong>s:<br />
2<br />
s + 1<br />
e<br />
(a) F () s = 10 +<br />
(b) G()<br />
s =<br />
2<br />
s + 4<br />
s<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 34.<br />
(a)<br />
(b)<br />
−s<br />
2<br />
s + 4 − 3<br />
( s)<br />
= 10 + 2 = 11−<br />
s + 4 s<br />
F 2<br />
f ( t)<br />
= 11δ ( t)<br />
− <strong>1.</strong><br />
5sin(<br />
2t)<br />
G(<br />
s)<br />
Let<br />
=<br />
e<br />
( s<br />
( s<br />
-s<br />
+<br />
+<br />
+ 4e<br />
2)(<br />
s<br />
1<br />
2)(<br />
s<br />
-2s<br />
+<br />
+<br />
4)<br />
4)<br />
A = 1 2 B = 1 2<br />
G(<br />
s)<br />
(c) Let<br />
2<br />
−2s<br />
+ 4e<br />
+ 6s<br />
+ 8<br />
( )<br />
( )( )<br />
−2s<br />
s + 1 e<br />
(c) H () s =<br />
s s + 3 s + 4<br />
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3<br />
+ 4<br />
A B<br />
= +<br />
s + 2 s + 4<br />
-s e ⎛ 1 1 ⎞<br />
= ⎜ + ⎟ + 2e<br />
2 ⎝s<br />
+ 2 s + 4⎠<br />
-2s<br />
⎛ 1 1 ⎞<br />
⎜ + ⎟<br />
⎝s<br />
+ 2 s + 4⎠<br />
-2(t-1) -4(t-1)<br />
-2(t-2)<br />
-4(t-2)<br />
g ( t)<br />
= 0.<br />
5[<br />
e − e ] u(<br />
t − 1)<br />
+ 2[<br />
e − e ] u(<br />
t − 2)<br />
s ( s<br />
+<br />
s + 1<br />
3)(<br />
s<br />
+<br />
4)<br />
=<br />
A B C<br />
+ +<br />
s s + 3 s + 4<br />
A = 1 12,<br />
B = 2 3,<br />
C = - 3 4<br />
⎛ 1 1 2 3 3 4 ⎞<br />
H( s)<br />
= ⎜ ⋅ + − ⎟e ⎝12<br />
s s + 3 s + 4⎠<br />
-2s<br />
⎛ 1 2 3 ⎞<br />
-3(t-2)<br />
-4(t-2)<br />
h ( t)<br />
= ⎜ +<br />
e − e ⎟ u(<br />
t − 2)<br />
⎝12<br />
3 4 ⎠
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 35.<br />
Obtain f(t) for <strong>the</strong> following <strong>transform</strong>s:<br />
(a) F()<br />
s =<br />
−6s<br />
( s + 3)<br />
e<br />
( s + 1)(<br />
s + 2)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 35.<br />
(a) Let<br />
(b) Let<br />
G(<br />
s)<br />
=<br />
−2s<br />
4 − e<br />
(b) F()<br />
s = 2<br />
s + 5s<br />
+ 4<br />
( s<br />
+<br />
s + 3<br />
1)(<br />
s<br />
A = 2,<br />
B = -1<br />
+<br />
2)<br />
A B<br />
= +<br />
s + 1 s + 2<br />
2 1<br />
G( s)<br />
= − ⎯⎯→<br />
g(<br />
t)<br />
= 2e<br />
−<br />
s + 1 s + 2<br />
-t -2t<br />
e<br />
-6s F( s)<br />
= e G(<br />
s)<br />
⎯⎯→<br />
f ( t)<br />
= g(<br />
t − 6)<br />
u(<br />
t − 6)<br />
-(t-6) -2(t-6)<br />
f ( t)<br />
= [ 2e<br />
− e ] u(<br />
t − 6)<br />
G(<br />
s)<br />
=<br />
( s<br />
+<br />
1<br />
1)(<br />
s<br />
A = 1 3 , B = -1<br />
3<br />
G(<br />
s)<br />
1<br />
= −<br />
3(<br />
s + 1)<br />
1<br />
g( t)<br />
= −<br />
3<br />
F(<br />
s)<br />
=<br />
-t -4t<br />
[ e e ]<br />
4G(<br />
s)<br />
− e<br />
1<br />
3(<br />
s +<br />
-2t<br />
G(<br />
s)<br />
+<br />
4)<br />
4)<br />
f ( t)<br />
= 4g(<br />
t)<br />
u(<br />
t)<br />
− g(<br />
t − 2)<br />
u(<br />
t −<br />
A B<br />
= +<br />
s + 1 s + 4<br />
(c) F()<br />
s =<br />
4 1<br />
-t<br />
-4t<br />
-(t-2)<br />
-4(t-2)<br />
f ( t)<br />
= [ e −<br />
e ] u(<br />
t)<br />
− [ e − e ] u(<br />
t − 2)<br />
3<br />
3<br />
2 ( s + 3)(<br />
s + 4)<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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2)<br />
se<br />
−s
(c) Let<br />
A = - 3<br />
s =<br />
A ( s<br />
s<br />
( s)<br />
=<br />
( s + 3)(<br />
s<br />
A Bs + C<br />
= +<br />
+ 4)<br />
s + 3 s 4<br />
G 2 2 +<br />
13<br />
+ 4)<br />
+ B(<br />
s<br />
+ 3s)<br />
+ C(<br />
s<br />
2 2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 0 = A + B ⎯⎯→<br />
B = -A<br />
1 s: 1 = 3B<br />
+ C<br />
0<br />
s : 0 = 4A<br />
+ 3C<br />
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3)<br />
A = - 3 13 , B = 3 13 , C = 4 13<br />
- 3 3s<br />
+ 4<br />
13G( s)<br />
= + 2 s + 3 s + 4<br />
-3t 13g( t)<br />
= -3e<br />
+ 3cos(<br />
2t)<br />
+<br />
F(<br />
s)<br />
= e<br />
-s<br />
G(<br />
s)<br />
f ( t)<br />
= g(<br />
t −1)<br />
u(<br />
t −1)<br />
1 1)<br />
2sin(<br />
2t)<br />
-3(t-<br />
f ( t)<br />
= [ - 3e<br />
+<br />
3cos(<br />
2(<br />
t − 1))<br />
+ 2sin(<br />
2(<br />
t − 1))<br />
] u(<br />
t − 1)<br />
13
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 36.<br />
Obtain <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong>s <strong>of</strong> <strong>the</strong> following functions:<br />
1<br />
(a) X () s = 2<br />
s s + 2<br />
(b) ()<br />
( ) 2<br />
1<br />
Y s =<br />
s s + 1<br />
(c) Z () s =<br />
s s + 1 s<br />
( )( s + 3)<br />
1<br />
2 ( )( + 6s<br />
+ 10)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 36.<br />
(a)<br />
1 A B C D<br />
= + + +<br />
( s + 2)(<br />
s + 3)<br />
s s s + 2 s 3<br />
X( s)<br />
= 2 2<br />
s<br />
+<br />
B = 1 6,<br />
C = 1 4 , D = -1<br />
9<br />
1 =<br />
A ( s<br />
+ 5s<br />
+ 6s)<br />
+ B(<br />
s<br />
+ 5s<br />
+ 6)<br />
+ C(<br />
s<br />
2 + 3s<br />
) + D(<br />
s<br />
3 2<br />
2<br />
3<br />
3 +<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s : 0 = A + C + D<br />
2 s : 0 = 5A<br />
+ B + 3C<br />
+ 2D<br />
= 3A<br />
+ B + C<br />
1 s : 0 = 6A<br />
+ 5B<br />
0 s : 1 = 6B<br />
⎯⎯→<br />
B = 1 6<br />
A = - 5 6 B =<br />
- 5<br />
36<br />
- 5 36 1 6 1 4 1 9<br />
X( s)<br />
= + 2 + −<br />
s s s + 2 s + 3<br />
x(<br />
t)<br />
- 5 1<br />
= u(<br />
t)<br />
+<br />
t +<br />
36 6<br />
1<br />
e<br />
4<br />
1<br />
−<br />
9<br />
-2t -3t<br />
e<br />
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2s<br />
2<br />
)
Y(<br />
s)<br />
=<br />
1<br />
A<br />
= +<br />
(b) 2 2<br />
s ( s + 1)<br />
s s + 1 ( s + 1)<br />
(c)<br />
A = 1,<br />
C = -1<br />
1 =<br />
A ( s<br />
+ 2s<br />
+ 1)<br />
+ B(<br />
s<br />
B<br />
+<br />
+ s)<br />
2 2 +<br />
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C<br />
Cs<br />
Equ<strong>at</strong>ing coefficients :<br />
2 s : 0 = A + B ⎯⎯→<br />
B = -A<br />
1 s : 0 = 2A<br />
+ B + C = A + C ⎯⎯→<br />
C = -A<br />
0 s : 1 = A,<br />
B = -1, C = -1<br />
Y(<br />
s)<br />
y(<br />
t)<br />
1 1 1<br />
= − −<br />
s s + 1 ( s + 1<br />
= u( t)<br />
− e − t<br />
-t -t e<br />
2 )<br />
A B Cs + D<br />
Z( s)<br />
= + + 2 s s + 1 s + 6s<br />
+ 10<br />
A = 1 10,<br />
B = -1<br />
5<br />
1 =<br />
A ( s<br />
+ 7s<br />
+ 16s<br />
+ 10)<br />
+ B(<br />
s<br />
+ 6s<br />
+ 10s)<br />
+ C(<br />
s<br />
2 + s ) + D(<br />
s<br />
3 2<br />
3 2<br />
3<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s: 0 = A + B + C<br />
2 s : 0 = 7A<br />
+ 6B<br />
+ C + D = 6A<br />
+ 5B<br />
+ D<br />
1 s: 0 = 16A<br />
+ 10B<br />
+ D = 10A<br />
+ 5B<br />
⎯⎯→<br />
B = -2A<br />
0 s : 1 = 10A<br />
⎯⎯→<br />
A = 1 10<br />
A = 1 10,<br />
B = -2A = -1<br />
5,<br />
C = A = 1 10 ,<br />
1 2 s + 4<br />
10 Z(<br />
s)<br />
= − + 2 s s + 1 s + 6s<br />
+ 10<br />
1 2 s + 3 1<br />
Z(<br />
s)<br />
= − + +<br />
s s + 1 ( s + 3)<br />
+ 1 ( s + 3)<br />
10 2 2 +<br />
-t -3t<br />
-3t<br />
z ( t)<br />
= 0.<br />
1[<br />
1 −<br />
2e<br />
+ e cos( t)<br />
+ e sin( t)<br />
] u(<br />
t)<br />
1<br />
s)<br />
D = 4A<br />
=<br />
4<br />
10
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 37.<br />
<strong>Find</strong> <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong>:<br />
(a) H () s =<br />
(b) G () s =<br />
s<br />
s<br />
+<br />
4<br />
2<br />
( s + )<br />
s<br />
2<br />
+ 4s<br />
+ 5<br />
2 ( s + 3)(<br />
s + 2s<br />
+ 2)<br />
−4s<br />
e<br />
(c) F()<br />
s =<br />
s + 2<br />
(d) D () s = 2<br />
s + 1<br />
s<br />
10 2<br />
( )( s + 4)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 37.<br />
(a)<br />
s+ 4 A B<br />
H() s = = +<br />
ss ( + 2) s s+<br />
2<br />
s+4 =A(s+2) + Bs<br />
Equ<strong>at</strong>ing coefficients,<br />
s: 1 = A + B<br />
constant: 4= 2 A A =2, B=1-A = -1<br />
2 1<br />
H() s = −<br />
s s+<br />
2<br />
−2t −2t<br />
ht () = 2 ut () − e ut () = (2 −<br />
e ) ut ()<br />
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Gs<br />
A Bs+ C<br />
= +<br />
s+ 3 s + 2s+ 2<br />
(b) ()<br />
2<br />
2 2<br />
s s Bs C s A s s<br />
+ 4 + 5 = ( + )( + 3) + ( + 2 + 2)<br />
Equ<strong>at</strong>ing coefficients,<br />
s 2 : 1= B + A (1)<br />
s: 4 = 3B + C + 2A (2)<br />
Constant: 5 =3C + 2A (3)<br />
Solving (1) to (3) gives<br />
2 3 7<br />
A= , B= , C =<br />
5 5 5<br />
0.4 0.6s+ <strong>1.</strong>4 0.4 0.6( s+<br />
1) + 0.8<br />
Gs () = + = +<br />
2 2<br />
s+ 3 s + 2s+ 2 s+ 3 ( s+<br />
1) + 1<br />
−3t −t −t<br />
gt ( ) = 0.4e + 0.6e cost+ 0.8e sin t<br />
(c)<br />
f t e u t<br />
−2( t−4)<br />
() = ( − 4)<br />
10s<br />
As+ B Cs+ D<br />
Ds () = = +<br />
( s + 1)( s + 4) s + 1 s + 4<br />
(d) 2 2 2 2<br />
s s As B s Cs D<br />
Equ<strong>at</strong>ing coefficients,<br />
s 3 : 0 = A + C<br />
s 2 : 0 = B + D<br />
s: 10 = 4A + C<br />
constant: 0 = 4B+D<br />
Solving <strong>the</strong>se leads to<br />
A = -10/3, B = 0, C = -10/3, D = 0<br />
10 s/ 3 10 s/<br />
3<br />
Ds () == − 2 2<br />
s + 1 s + 4<br />
10 10<br />
dt () = cost− cos2t<br />
3 3<br />
2 2<br />
10 = ( + 4)( + ) + ( + 1)( + )<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 38.<br />
<strong>Find</strong> f(t) given th<strong>at</strong>:<br />
2<br />
s + 4s<br />
(a) F () s = 2<br />
s + 10s<br />
+ 26<br />
2<br />
5s + 7s<br />
+ 29<br />
(b) F () s = 2<br />
s s + 4s<br />
+ 29<br />
( )<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 38.<br />
(a)<br />
(b)<br />
2<br />
2<br />
s + 4s<br />
s + 10s<br />
+ 26 − 6s<br />
− 26<br />
F( s)<br />
= 2 = 2<br />
s + 10s<br />
+ 26 s + 10s<br />
+ 26<br />
6s<br />
+ 26<br />
F( s)<br />
= 1−<br />
2 s + 10s<br />
+ 26<br />
6(<br />
s + 5)<br />
4<br />
F(<br />
s)<br />
= 1−<br />
2 2 + 2 2<br />
( s + 5)<br />
+ 1 ( s + 5)<br />
+ 1<br />
-t -t<br />
f ( t)<br />
= δ ( t)<br />
− 6e<br />
cos( 5t)<br />
+ 4e<br />
sin( 5t)<br />
5s<br />
( s)<br />
=<br />
s ( s<br />
2<br />
+ 7s<br />
+ 29 A<br />
= +<br />
+ 4s<br />
+ 29)<br />
s s<br />
F 2<br />
2<br />
5s<br />
+ 7s<br />
+ 29 = A ( s<br />
Bs + C<br />
+ 4s<br />
+ 29<br />
+ 4s<br />
+ 29)<br />
+ Bs<br />
2 2<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
0 s : 29 = 29A<br />
⎯⎯→<br />
A = 1<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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Cs<br />
1 s: 7 = 4A<br />
+ C ⎯⎯→<br />
C = 7 − 4A<br />
= 3<br />
2 s : 5 = A + B ⎯⎯→<br />
B = 5 − A = 4<br />
A = 1,<br />
B = 4 , C = 3<br />
F(<br />
s)<br />
1<br />
= +<br />
s s<br />
4s<br />
+<br />
+ 3<br />
+<br />
1<br />
= +<br />
4(<br />
s<br />
+<br />
2)<br />
+<br />
2 2 2<br />
2 2<br />
4s<br />
29 s ( s 2)<br />
5 ( s 2)<br />
5<br />
-2t -2t<br />
f ( t)<br />
= u(<br />
t)<br />
+<br />
4e<br />
cos( 5t)<br />
− e sin( 5t)<br />
+<br />
−<br />
+<br />
5<br />
+
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 39.<br />
*Determine f(t) if:<br />
(a) F () s =<br />
(b) F () s =<br />
2s<br />
3<br />
+ 4s<br />
+ 1<br />
2<br />
2<br />
( s + 2s<br />
+ 17)(<br />
s + 4s<br />
+ 20)<br />
s<br />
2<br />
+ 4<br />
2 2 ( s + 9)(<br />
s + 6s<br />
+ 3)<br />
* An asterisk indic<strong>at</strong>es a challenging problem.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 39.<br />
(a)<br />
( s)<br />
=<br />
( s<br />
2<br />
3 2s<br />
+ 4s<br />
+ 2s<br />
+ 17)(<br />
s<br />
2<br />
+ 1<br />
=<br />
+ 4s<br />
+ 20)<br />
s<br />
As + B<br />
+<br />
+ 2s<br />
+ 17 s<br />
F 2<br />
2<br />
2<br />
2<br />
2<br />
3 2<br />
2<br />
s + 4s<br />
+ 1 = A(<br />
s + 4s<br />
+ 20s)<br />
+ B(<br />
s + 4s<br />
3 2<br />
2<br />
+ C(<br />
s + 2s<br />
+ 17s)<br />
+ D(<br />
s + 2s<br />
+ 17)<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s: 2 = A + C<br />
3 +<br />
2 s : 4 = 4A<br />
+ B + 2C<br />
+ D<br />
1<br />
s: 0 = 20A<br />
+ 4B<br />
+ 17C<br />
+ 2D<br />
0 s : 1 = 20B<br />
+ 17D<br />
Cs + D<br />
+ 4s<br />
+ 20<br />
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20)<br />
Solving <strong>the</strong>se equ<strong>at</strong>ions (M<strong>at</strong>lab works well with 4 unknowns),<br />
A = -<strong>1.</strong>6 , B = -17.8 , C = 3.<br />
6 , D = 21<br />
-<strong>1.</strong>6s<br />
−17.8<br />
3.<br />
6s<br />
+ 21<br />
F( s)<br />
= 2 + 2<br />
s + 2s<br />
+ 17 s + 4s<br />
+ 20<br />
(-<strong>1.</strong>6)(s + 1) (-4.05) ( 4)<br />
F(<br />
s)<br />
=<br />
2 + 2 2<br />
( s + 1)<br />
+ 4 ( s + 1)<br />
+ 4<br />
+<br />
(3.6)(s + 2)<br />
+ +<br />
+<br />
(3.45) ( 4)<br />
+ +<br />
2 2 2<br />
2 2<br />
( s 2)<br />
4 ( s 2)<br />
4<br />
-t -t<br />
-2t<br />
-2t<br />
f ( t)<br />
= - <strong>1.</strong><br />
6e<br />
cos( 4t)<br />
−<br />
4.<br />
05e<br />
sin( 4t)<br />
+ 3.<br />
6e<br />
cos( 4t)<br />
+ 3.<br />
45e<br />
sin( 4t)
(b)<br />
( s)<br />
=<br />
( s<br />
2 s + 4 As + B Cs + D<br />
2 = 2 +<br />
+ 9)(<br />
s + 6s<br />
+ 3)<br />
s + 9 s + 6s<br />
+ 3<br />
F 2<br />
2<br />
s<br />
+ 4 = A ( s<br />
+ 6s<br />
+ 3s)<br />
+ B(<br />
s<br />
+ 6s<br />
+ 3)<br />
+ C(<br />
s<br />
+ 9s)<br />
+ D(<br />
s<br />
2 3 2<br />
2<br />
3<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s: 0 = A + C ⎯⎯→<br />
C = -A<br />
2 s : 1 = 6A<br />
+ B + D<br />
1<br />
s: 0 = 3A<br />
+ 6B<br />
+ 9C<br />
= 6B<br />
+ 6C<br />
⎯⎯→<br />
B = -C = A<br />
0 s : 4 = 3B<br />
+ 9D<br />
Solving <strong>the</strong>se equ<strong>at</strong>ions,<br />
A = 1 12,<br />
B = 1 12 , C = -1<br />
12 , D = 5 12<br />
s + 1 - s + 5<br />
F(<br />
s)<br />
= 2 +<br />
s + 9 s + 6s<br />
+ 3<br />
12 2<br />
- 6 36 -12<br />
s 6s<br />
3 0<br />
2<br />
2 ±<br />
+ + = ⎯⎯→<br />
=<br />
Let<br />
- s + 5 E<br />
G( s)<br />
= 2 = +<br />
s + 6s<br />
+ 3 s + 0.<br />
551 s +<br />
- s + 5<br />
s + 5.<br />
449<br />
- s + 5<br />
s + 0.<br />
551<br />
E = s = -0.551<br />
F = s = -5.449<br />
G(<br />
s)<br />
=<br />
<strong>1.</strong><br />
133<br />
= - 2.133<br />
<strong>1.</strong><br />
133 2.<br />
133<br />
= −<br />
s + 0.<br />
551 s + 5.<br />
449<br />
-0.551, - 5.449<br />
F<br />
5.<br />
449<br />
s 1 3 <strong>1.</strong><br />
133 2.<br />
133<br />
F(<br />
s)<br />
= 2 2 + ⋅ 2 + −<br />
s + 3 3 s + 3 s + 0.<br />
551 s + 5.<br />
449<br />
12 2<br />
f ( t)<br />
= 0. 08333cos(<br />
3t)<br />
+<br />
0.<br />
02778sin(<br />
3t)<br />
+ 0.<br />
0944e<br />
− 0.<br />
1778<br />
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9)<br />
-0.551t -5.449t e
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 40.<br />
Show th<strong>at</strong><br />
L<br />
−1<br />
⎡<br />
⎢<br />
⎣<br />
2 ( s + 2)(<br />
s + 2s<br />
+ 5)<br />
−t<br />
2t<br />
[ 2e<br />
cos(<br />
2t<br />
+ 45°<br />
) + 3e<br />
] u()<br />
t<br />
2<br />
4s + 7s<br />
+ 13 ⎤<br />
−<br />
⎥ =<br />
⎦<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 40.<br />
Let<br />
H(<br />
s)<br />
4s<br />
=<br />
2<br />
⎡<br />
⎢<br />
⎢⎣<br />
( s<br />
2<br />
4s<br />
+ 7s<br />
+ 13 ⎤ A Bs + C<br />
⎥ = +<br />
2<br />
2)(<br />
s 2s<br />
5)<br />
s 2 2<br />
+ + + ⎥⎦<br />
+ s + 2s<br />
+ 5<br />
+ 7s<br />
+ 13 =<br />
A(<br />
s<br />
Equ<strong>at</strong>ing coefficients gives:<br />
s : 4 A B<br />
2<br />
= +<br />
2<br />
+<br />
2s<br />
+ 5)<br />
+<br />
B(<br />
s<br />
2<br />
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+<br />
2s)<br />
s : 7 = 2A<br />
+ 2B<br />
+ C ⎯⎯→<br />
C = −1<br />
+<br />
C(<br />
s<br />
constant : 13 = 5A<br />
+ 2C<br />
⎯⎯→<br />
5A<br />
= <strong>15</strong> or A = 3, B = 1<br />
Hence,<br />
h(<br />
t)<br />
=<br />
where<br />
Thus,<br />
H(<br />
s)<br />
3e<br />
−2t<br />
3<br />
= +<br />
s + 2 s<br />
+ e<br />
−t<br />
2<br />
cos 2t<br />
s −1<br />
3<br />
= +<br />
+ 2s<br />
+ 5 s + 2<br />
− e<br />
−t<br />
sin 2t<br />
=<br />
3e<br />
( s<br />
−2t<br />
( s<br />
+ 1)<br />
− 2<br />
+ 1)<br />
+ e<br />
2<br />
−t<br />
+ 2<br />
2<br />
+<br />
( A cos<br />
2)<br />
α<br />
cos 2t<br />
A cosα<br />
= 1,<br />
Asin<br />
α = 1 ⎯⎯→<br />
A = 2,<br />
α =<br />
h(<br />
t)<br />
=<br />
−t<br />
o −2t<br />
[ 2e<br />
cos( 2t<br />
+ 45 ) + 3e<br />
] u(<br />
t)<br />
−<br />
A sin<br />
o<br />
45<br />
αsin<br />
2t)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 4<strong>1.</strong><br />
* Let x(t) and y(t) be as shown in Fig. <strong>15</strong>.36. <strong>Find</strong> z ( t)<br />
x(<br />
t)<br />
* y(<br />
t)<br />
Figure <strong>15</strong>.36<br />
For Prob. <strong>15</strong>.4<strong>1.</strong><br />
* An asterisk indic<strong>at</strong>es a challenging problem.<br />
= .<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 4<strong>1.</strong><br />
We fold x(t) and slide on y(t). For t
For 2 < t < 6, <strong>the</strong> two functions overlap, as shown below.<br />
2<br />
∫ ∫<br />
0 0<br />
t<br />
t<br />
4<br />
-4<br />
y( λ )<br />
zt ( ) = (2)(4) dλ+ (2)( − 4) dλ = 16 −8t<br />
For 6
For 8< t
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 42.<br />
Suppose th<strong>at</strong> f () t = u()<br />
t − u(<br />
t − 2)<br />
. Determine ( t)<br />
f ( t)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 42.<br />
For 0
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 43.<br />
<strong>Find</strong> y () t x()<br />
t * h()<br />
t<br />
Figure <strong>15</strong>.37<br />
For Prob. <strong>15</strong>.43.<br />
= for each paired x(t) and h(t) in Fig. <strong>15</strong>.37.<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 43.<br />
(a) For 0 < t < 1,<br />
x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (a).<br />
2<br />
t λ<br />
y(<br />
t)<br />
= x(<br />
t)<br />
∗ h(<br />
t)<br />
= ∫ ( 1)(<br />
λ)<br />
dλ<br />
=<br />
0<br />
2<br />
2<br />
t t<br />
0 =<br />
2<br />
x(t - λ)<br />
For 1 < t < 2 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (b).<br />
2<br />
1<br />
t λ -1<br />
1 t 2<br />
y(<br />
t)<br />
= ∫ ( 1)(<br />
λ)<br />
dλ<br />
+ ∫ ( 1)(<br />
1)<br />
dλ<br />
= t 1 + λ 1 = t + 2t<br />
t−1<br />
1<br />
−<br />
2 2<br />
For t > 2 , <strong>the</strong>re is a complete overlap so th<strong>at</strong><br />
y(<br />
t)<br />
Therefore,<br />
y(<br />
t)<br />
1<br />
t-1 0 t 1 λ<br />
∫<br />
t<br />
t<br />
= ( 1)(<br />
1)<br />
dλ<br />
= λ t−<br />
t−1<br />
⎧<br />
⎪<br />
- ( t<br />
= ⎨<br />
⎪<br />
⎪<br />
⎩<br />
2<br />
(a)<br />
t<br />
2<br />
2)<br />
+<br />
1,<br />
0,<br />
2,<br />
2t<br />
= t − ( t −1)<br />
= 1<br />
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1<br />
− 1,<br />
h(λ)<br />
0 < t < 1<br />
1 < t < 2<br />
t > 2<br />
1<br />
o<strong>the</strong>rwise<br />
0<br />
t-1 1 t λ<br />
(b)<br />
−1
(b) For t > 0 , <strong>the</strong> two functions overlap as shown in Fig. (c).<br />
t<br />
-λ -λ<br />
y(<br />
t)<br />
= x(<br />
t)<br />
∗ h(<br />
t)<br />
= ∫ ( 1)<br />
2e<br />
dλ<br />
= -2e<br />
Therefore,<br />
-t y( t)<br />
= 2( 1 − e ), t > 0<br />
0<br />
(c) For - 1 < t < 0 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (d).<br />
2<br />
t+<br />
1 λ<br />
y ( t)<br />
= x(<br />
t)<br />
∗ h(<br />
t)<br />
= ∫ ( 1)(<br />
λ)<br />
dλ<br />
=<br />
0<br />
2<br />
1 t+<br />
1<br />
2<br />
0 = ( t + 1)<br />
2<br />
For 0 < t < 1,<br />
x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (e).<br />
y ( t)<br />
y(<br />
t)<br />
x(t - λ)<br />
=<br />
∫<br />
1<br />
0<br />
2 λ<br />
2<br />
x(t-λ)<br />
2<br />
1<br />
( 1)(<br />
λ)<br />
dλ<br />
+<br />
∫<br />
t+<br />
1<br />
1<br />
2 ⎛ λ ⎞<br />
+ ⎜2λ<br />
− ⎟<br />
⎝ 2 ⎠<br />
( 1)(<br />
2<br />
1<br />
t+<br />
1<br />
= 0<br />
1<br />
1<br />
− λ)<br />
dλ<br />
-1<br />
= t<br />
2<br />
+ t +<br />
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(c)<br />
2<br />
t<br />
0<br />
h(λ) = 2e -λ<br />
0 t λ<br />
1<br />
t-1 -1 t 0 t+1 1 2<br />
λ<br />
1<br />
2<br />
-1 t-1 0 t 1 t+1 2 λ<br />
(e)<br />
(d)<br />
h(λ)
For 1 < t < 2 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (f).<br />
1<br />
2<br />
y ( t)<br />
= ∫ ( 1)(<br />
λ)<br />
dλ<br />
+ ∫ ( 1)(<br />
2 − λ)<br />
dλ<br />
t−1<br />
1<br />
2<br />
2<br />
λ ⎛ λ ⎞ -1<br />
1<br />
1<br />
2 2<br />
y(<br />
t)<br />
= ⎜ t 1 2 ⎟<br />
− + λ − 1 = t + t +<br />
2 ⎝ 2 ⎠ 2 2<br />
For 2 < t < 3 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (g).<br />
Therefore,<br />
2<br />
2<br />
⎛ λ ⎞ 9 2<br />
y ( t)<br />
= ∫ ( 1)(<br />
2 − λ)<br />
dλ<br />
= ⎜2λ<br />
− ⎟<br />
t 1<br />
t 1 = − 3t<br />
+<br />
−<br />
−<br />
⎝ 2 ⎠ 2<br />
y(<br />
t)<br />
1<br />
1<br />
0<br />
0<br />
⎧ ( t<br />
⎪<br />
- ( t<br />
= ⎨ 2<br />
⎪(<br />
t<br />
⎪<br />
⎩<br />
2<br />
2<br />
2)<br />
2)<br />
2)<br />
t-1 1 t 2 t+1 λ<br />
+ t + 1<br />
−<br />
+ t + 1<br />
3t<br />
0,<br />
+ 9<br />
< t < 0<br />
0 < t < 2<br />
2 < t < 3<br />
o<strong>the</strong>rwise<br />
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2,<br />
(f)<br />
1 t-1 2 t t+1 λ<br />
2,<br />
2,<br />
(g)<br />
- 1<br />
1<br />
2<br />
t<br />
2
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 44.<br />
Obtain <strong>the</strong> convolution <strong>of</strong> <strong>the</strong> pairs <strong>of</strong> signals in Fig. <strong>15</strong>.38.<br />
Figure <strong>15</strong>.38<br />
For Prob. <strong>15</strong>.44.<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 44.<br />
(a) For 0 < t < 1,<br />
x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (a).<br />
t<br />
y(<br />
t)<br />
= x(<br />
t)<br />
∗ h(<br />
t)<br />
= ∫ ( 1)(<br />
1)<br />
dλ<br />
= t<br />
0<br />
For 1 < t < 2 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (b).<br />
y(<br />
t)<br />
∫<br />
1<br />
1<br />
= ( 1)(<br />
1)<br />
dλ<br />
+ ( -1)(<br />
1)<br />
dλ<br />
= λ t−<br />
t−1<br />
∫<br />
1<br />
t<br />
= 3 −<br />
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1<br />
− λ<br />
For 2 < t < 3 , x( t − λ)<br />
and h( λ ) overlap as shown in Fig. (c).<br />
1<br />
0<br />
-1<br />
y(<br />
t)<br />
Therefore,<br />
x(t - λ) 1<br />
h(λ)<br />
y(<br />
t)<br />
∫<br />
2<br />
2<br />
= ( 1)(<br />
-1)<br />
dλ<br />
= -λ<br />
t−<br />
t−1<br />
t-1 0 t 1 2 λ<br />
-1<br />
⎧ t,<br />
⎪<br />
3 −<br />
= ⎨<br />
⎪ t −<br />
⎪<br />
⎩ 0,<br />
2t,<br />
3,<br />
(a)<br />
t-1 1 t 2 λ<br />
(b)<br />
1<br />
0 < t < 1<br />
1 < t < 2<br />
2 < t < 3<br />
o<strong>the</strong>rwise<br />
= t − 3<br />
1<br />
0<br />
-1<br />
t<br />
1<br />
(c)<br />
2t<br />
1 t-1 2 t λ
(b) For t < 2 , <strong>the</strong>re is no overlap. For 2 < t < 3 , f1( t − λ)<br />
and<br />
as shown in Fig. (d).<br />
( )<br />
y ( t)<br />
= f ( t)<br />
∗ f<br />
t<br />
( t)<br />
= ∫ ( 1)(<br />
t − λ)<br />
dλ<br />
1<br />
1<br />
1<br />
0<br />
0<br />
0<br />
⎧ ( t<br />
⎪<br />
Therefore, y(<br />
t)<br />
= ⎨<br />
⎪-<br />
( t<br />
⎪<br />
⎩<br />
1<br />
2<br />
2<br />
2<br />
2<br />
f1(t - λ) f2(λ)<br />
1<br />
1<br />
1<br />
t-1<br />
2 t 3 4 5<br />
(d)<br />
2 t-1 3 t 4 5<br />
(e)<br />
2<br />
2<br />
⎛ λ ⎞ t t<br />
= ⎜<br />
⎜λt<br />
− 2 = − 2t<br />
+ 2<br />
2 ⎟<br />
⎝ ⎠ 2<br />
For 3 < t < 5 , f1( t − λ)<br />
and ( )<br />
2<br />
t<br />
⎛ λ ⎞ 1 t<br />
y(<br />
t)<br />
= ∫ ( 1)(<br />
t − λ)<br />
dλ<br />
= ⎜λt<br />
− ⎟<br />
t 1<br />
t 1 =<br />
−<br />
−<br />
⎝ 2 ⎠ 2<br />
For 5 < t < 6 , <strong>the</strong> functions overlap as shown in Fig. (f).<br />
2<br />
5<br />
⎛ λ ⎞ 5 -1<br />
2<br />
y( t)<br />
= ∫ ( 1)(<br />
t − λ)<br />
dλ<br />
=<br />
1 = + 5 −12<br />
1<br />
⎜ λt<br />
−<br />
2 ⎟ t − t t<br />
t −<br />
⎝ ⎠ 2<br />
f 2 λ overlap as shown in Fig. (e).<br />
2 3 4 t-1 5<br />
2)<br />
2)<br />
− 12,<br />
5 < t < 6<br />
f 2 λ overlap,<br />
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−<br />
1 2,<br />
+<br />
0,<br />
(f)<br />
2t<br />
5t<br />
+<br />
2,<br />
2 < t < 3<br />
3 < t < 5<br />
o<strong>the</strong>rwise<br />
λ<br />
λ<br />
t<br />
λ
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 45.<br />
2t<br />
Given h()<br />
t 4e u(<br />
t)<br />
−<br />
2t<br />
= and x()<br />
t ( t)<br />
2e u(<br />
t)<br />
−<br />
= δ − , find y ( t)<br />
x(<br />
t)<br />
* h(<br />
t)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 45.<br />
= .<br />
−2t −2t<br />
yt () = ht ()* xt () = ⎡<br />
⎣4 e ut () ⎤<br />
⎦* ⎡<br />
⎣δ() t −2<br />
e ut () ⎤<br />
⎦<br />
t<br />
−2t −2t −2t −2t −2t<br />
o<br />
= 4 e u( t)* δ ( t) − 4 e u( t)*2 e u( t) = 4 e u( t) −8e ∫ e dλ<br />
= e u t − te u t<br />
−2t −2t<br />
4 ( ) 8 ( )<br />
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 46.<br />
Given <strong>the</strong> following functions<br />
−2t<br />
() t = 2 () t , y() t = 4u()<br />
t , z(<br />
t)<br />
e u(<br />
t)<br />
x δ<br />
evalu<strong>at</strong>e <strong>the</strong> following convolution oper<strong>at</strong>ions.<br />
(a) x () t * y()<br />
t<br />
(b) x () t * z()<br />
t<br />
(c) y () t * z()<br />
t<br />
y t * y t + z t<br />
(d) () [ () () ]<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 46.<br />
0<br />
= ,<br />
(a) x()* t yt () = 2 δ ()*4 t ut () = 8 ut ()<br />
−2t −2t<br />
(b) x()* t z() t = 2 δ ()* t e u() t = 2 e u() t<br />
t<br />
−2λ<br />
2 2 4 t<br />
− t − λ e<br />
−2t<br />
(c ) yt ( )* zt ( ) = 4 ut ( )* e ut ( ) = 4∫ e dλ= = 2(1 −e)<br />
−2<br />
0<br />
0<br />
−2t−2λ (d) yt ()*[ yt () + zt ()] = 4 ut ()*[4 ut () + e ut ()] = 4 ∫ [4 u( λ) + e u( λ)] dλ<br />
t<br />
−2λ<br />
−2λet−2t = 4 ∫<br />
[4 + e ] dλ= 4[4 t+ ] = 16t− 2e + 2<br />
−2<br />
0<br />
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0
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 47.<br />
A system has <strong>the</strong> transfer function<br />
H<br />
() s<br />
=<br />
s<br />
( s + 1 )( s + 2)<br />
(a) <strong>Find</strong> <strong>the</strong> impulse response <strong>of</strong> <strong>the</strong> system.<br />
(b) Determine <strong>the</strong> output y(t), given th<strong>at</strong> <strong>the</strong> input is x ( t)<br />
= u(<br />
t)<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 47.<br />
s A B<br />
(a) H() s = = +<br />
( s+ 1)( s+ 2) s+ 1 s+<br />
2<br />
s=A(s+2) + B(s+1)<br />
We equ<strong>at</strong>e <strong>the</strong> coefficients.<br />
s: 1= A+B<br />
constant: 0 =2A +B<br />
Solving <strong>the</strong>se, A = -1, B= 2.<br />
−1<br />
2<br />
H() s = +<br />
s+ 1 s+<br />
2<br />
−t −2t<br />
ht () = ( − e + 2 e ) ut ()<br />
Y() s s 1<br />
(b) H() s = ⎯⎯→ Y() s = H() s X() s =<br />
X () s ( s+ 1)( s+ 2) s<br />
1 C D<br />
Y() s = = +<br />
( s+ 1)( s+ 2) s+ 1 s+<br />
2<br />
C=1 and D=-1 so th<strong>at</strong><br />
1 1<br />
Y() s = −<br />
s+ 1 s+<br />
2<br />
yt = e −<br />
e ut<br />
−t −2t<br />
() ( ) ()<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 48.<br />
<strong>Find</strong> f(t) using convolution given th<strong>at</strong>:<br />
(a) F () s =<br />
(b) F<br />
() s =<br />
4<br />
( ) 2<br />
2<br />
s + 2s<br />
+ 5<br />
2s 2<br />
+<br />
( s + 1)(<br />
s 4)<br />
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<strong>Chapter</strong> <strong>15</strong>, Solution 48.<br />
2<br />
2<br />
(a) Let G(<br />
s)<br />
= 2 = 2 2<br />
s + 2s<br />
+ 5 ( s + 1)<br />
+ 2<br />
-t g(<br />
t)<br />
= e sin( 2t)<br />
F ( s)<br />
=<br />
f ( t)<br />
f ( t)<br />
G(<br />
s)<br />
G(<br />
s)<br />
t<br />
[ G(<br />
s)<br />
G(<br />
s)<br />
] = g(<br />
λ)<br />
g(<br />
t − λ)<br />
λ<br />
-1<br />
= L ∫ d<br />
t<br />
-λ -(<br />
t−λ)<br />
= ∫ e sin( 2λ)<br />
e sin( 2(<br />
t − λ))<br />
dλ<br />
0<br />
0<br />
1<br />
sin( A)<br />
sin( B)<br />
= [ cos( A − B)<br />
− cos( A + B)<br />
]<br />
2<br />
1 t<br />
-t -λ<br />
f ( t)<br />
= e ∫ e [ cos( 2t)<br />
− cos( 2(<br />
t − 2λ))<br />
] dλ<br />
2 0<br />
-t<br />
-t<br />
e<br />
t e t<br />
-2λ<br />
-2λ<br />
f ( t)<br />
= cos( 2t)<br />
∫ e dλ<br />
− ∫ e cos( 2t<br />
− 4λ)<br />
dλ<br />
2<br />
0 2 0<br />
-t<br />
-2λ<br />
-t<br />
e e e t<br />
t<br />
-2λ<br />
f ( t)<br />
= cos( 2t)<br />
⋅ − ∫ [ λ +<br />
λ ]<br />
0 e cos( 2t)<br />
cos( 4 ) sin( 2t)<br />
sin( 4 ) dλ<br />
2 - 2 2 0<br />
-t<br />
1<br />
e<br />
t<br />
-t<br />
-2t<br />
-2λ<br />
f ( t)<br />
= e cos( 2t)<br />
( -e<br />
+ 1)<br />
− cos( 2t)<br />
∫ e cos( 4λ)<br />
dλ<br />
4<br />
2<br />
0<br />
-t e<br />
t<br />
-2λ<br />
− sin( 2t)<br />
∫ e sin( 4λ)<br />
dλ<br />
2<br />
0<br />
1 -t -2t<br />
f ( t)<br />
= e cos( 2t)<br />
( 1−<br />
e )<br />
4<br />
-t<br />
-2λ<br />
e ⎡ e<br />
⎤ t<br />
− cos( 2t)<br />
( - 2cos(4 ) 4sin(<br />
4 ) ) 0<br />
2<br />
⎢<br />
λ − λ<br />
⎣ 4 + 16<br />
⎥<br />
⎦<br />
-t<br />
-2λ<br />
e ⎡ e<br />
− sin( 2t)<br />
2<br />
⎢<br />
⎣ 4 + 16<br />
t<br />
( - 2sin(4λ)<br />
+ 4cos(<br />
4λ)<br />
)<br />
-t<br />
-3t<br />
-t<br />
-3t<br />
e e<br />
e e<br />
f ( t)<br />
= cos( 2t)<br />
− cos( 2t)<br />
− cos( 2t)<br />
+ cos( 2t)<br />
cos( 4t)<br />
2 4 20 20<br />
-3t<br />
e<br />
+<br />
10<br />
cos( 2t)<br />
sin( 4t)<br />
+<br />
-t<br />
e<br />
sin( 2t)<br />
10<br />
-t<br />
-t<br />
e<br />
e<br />
+<br />
sin( 2t)<br />
sin( 4t)<br />
− sin( 2t)<br />
cos( 4t)<br />
20<br />
10<br />
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⎤<br />
⎥<br />
⎦<br />
0
(b) Let<br />
x(<br />
t)<br />
F ( s)<br />
=<br />
f ( t)<br />
= L<br />
f ( t)<br />
X(<br />
s)<br />
2<br />
= ,<br />
s + 1<br />
Y(<br />
s)<br />
s<br />
=<br />
s + 4<br />
-t = 2e<br />
u(<br />
t)<br />
, y ( t)<br />
= cos( 2t)<br />
u(<br />
t)<br />
X(<br />
s)<br />
Y(<br />
s)<br />
[ ] ∫ ∞<br />
-1<br />
X(<br />
s)<br />
Y(<br />
s)<br />
= y(<br />
λ)<br />
x(<br />
t − λ)<br />
0<br />
t<br />
0<br />
-(t−λ)<br />
cos( 2λ)<br />
⋅ 2e<br />
d<br />
( ) t<br />
- t e ⋅<br />
λ e<br />
cos( 2λ)<br />
+ 2sin(<br />
2 ) 0<br />
= ∫ λ<br />
f ( t)<br />
= 2<br />
1+<br />
4<br />
λ<br />
2 -t f ( t)<br />
= e<br />
5<br />
t e cos( 2t)<br />
+ 2sin(<br />
2t)<br />
−1<br />
2 4 2 -t<br />
f ( t)<br />
= cos( 2t)<br />
+<br />
sin( 2t)<br />
− e<br />
5 5 5<br />
[ ( ) ]<br />
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dλ
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 49.<br />
* Use <strong>the</strong> convolution integral to find:<br />
<strong>at</strong><br />
(a) t * e u()<br />
t<br />
cos t * cos t u t<br />
(b) () () ()<br />
* An asterisk indic<strong>at</strong>es a challenging problem.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 49.<br />
(b)<br />
(a) t*e αt u(t) =<br />
t<br />
aλ<br />
t<br />
aλ<br />
t<br />
<strong>at</strong><br />
aλ<br />
e e<br />
t <strong>at</strong> 1 e<br />
e ( t − λ)<br />
dλ<br />
= t − ( aλ<br />
−1)<br />
= ( e −1)<br />
− −<br />
0 a 2<br />
a<br />
2 2<br />
0<br />
a<br />
0<br />
a a<br />
∫<br />
t t<br />
∫ ∫<br />
0 0<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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( <strong>at</strong><br />
−1)<br />
cos t*cos tu( t) = cos λ cos( t− λ) dλ = {costcosλcos λ+ sin tsinλcos λ} dλ<br />
t t<br />
⎡ 1 ⎤ ⎡1 sin2λ t cosλ<br />
t⎤<br />
= ⎢cos t∫ [1+ cos 2 λ] dλ+ sin t cos λd( cos λ) cos t[ λ ] sin t<br />
2 ∫ − ⎥ = ⎢ + − ⎥<br />
⎢2 2 0 2 0⎥<br />
⎣ 0 0<br />
⎦ ⎣ ⎦<br />
= 0.5cos(t)(t+0.5sin(2t)) – 0.5sin(t)(cos(t) – 1).
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 50.<br />
Use <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> to solve <strong>the</strong> differential equ<strong>at</strong>ion<br />
2<br />
d v<br />
dt<br />
() t dv()<br />
t<br />
2<br />
+ 2<br />
dt<br />
+ 10v()<br />
t = 3 cos 2t<br />
subject to ( 0) = 1,<br />
dv(<br />
0)<br />
/ dt = −2<br />
v .<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 50.<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term.<br />
3s<br />
2 [ s V(<br />
s)<br />
− s v(<br />
0)<br />
− v′<br />
( 0)<br />
] + 2[<br />
s V(<br />
s)<br />
− v(<br />
0)<br />
] + 10V(<br />
s)<br />
= 2 s + 4<br />
3s<br />
2 s V(<br />
s)<br />
− s + 2 + 2s<br />
V(<br />
s)<br />
− 2 + 10V(<br />
s)<br />
= 2 s + 4<br />
3<br />
3s<br />
s + 7s<br />
2<br />
( s + 2s<br />
+ 10)<br />
V(<br />
s)<br />
= s + 2 = 2 s + 4 s + 4<br />
3 s + 7s<br />
As + B Cs + D<br />
V( s)<br />
= 2 2 = 2 + 2<br />
( s + 4)(<br />
s + 2s<br />
+ 10)<br />
s + 4 s + 2s<br />
+ 10<br />
s<br />
+ 7s<br />
= A ( s<br />
+ 2s<br />
+ 10s)<br />
+ B(<br />
s<br />
+ 2s<br />
+ 10)<br />
+ C(<br />
s<br />
+ 4s)<br />
+ D(<br />
s<br />
3 3 2<br />
2<br />
3<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
3 s : 1 = A + C ⎯⎯→<br />
C = 1−<br />
A<br />
2 s : 0 = 2A<br />
+ B + D<br />
1<br />
s : 7 = 10A<br />
+ 2B<br />
+ 4C<br />
= 6A<br />
+ 2B<br />
+ 4<br />
0 s : 0 = 10B<br />
+ 4D<br />
⎯⎯→<br />
D = -2.5B<br />
Solving <strong>the</strong>se equ<strong>at</strong>ions yields<br />
9<br />
12<br />
A = , B = ,<br />
26 26<br />
17<br />
C = ,<br />
26<br />
1 ⎡9s<br />
+ 12 17s<br />
− 30 ⎤<br />
V( s)<br />
= ⎢⎣ 2 + 2<br />
26 s + 4 s + 2s<br />
+ 10⎥⎦<br />
1 ⎡ 9s<br />
2 s + 1<br />
V ( s)<br />
⎢ + 6 ⋅ 2 + 17 ⋅ 2<br />
26 ⎣s<br />
+ 4 s + 4 ( s + 1)<br />
+ 3<br />
- 30<br />
D =<br />
26<br />
47<br />
− 2 ( s + 1)<br />
+<br />
= 2 2<br />
2 3<br />
9 6 17<br />
47<br />
-t<br />
-t<br />
v ( t)<br />
= cos( 2t)<br />
+<br />
sin( 2t)<br />
+ e cos( 3t)<br />
− e sin( 3t)<br />
26 26 26<br />
78<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
written permission <strong>of</strong> <strong>the</strong> publisher, or used beyond <strong>the</strong> limited distribution to teachers and educ<strong>at</strong>ors<br />
permitted by McGraw-Hill for <strong>the</strong>ir individual course prepar<strong>at</strong>ion. If you are a student using this Manual,<br />
you are using it without permission.<br />
⎤<br />
⎥<br />
⎦<br />
4)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 5<strong>1.</strong><br />
Given th<strong>at</strong> v ( 0 ) = 2 and ( 0 ) / dt = 4<br />
d −t<br />
2<br />
v dv<br />
+ 5 + 6v<br />
= 10e<br />
2<br />
dt dt<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 5<strong>1.</strong><br />
u<br />
dv , solve<br />
() t<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> <strong>the</strong> differential equ<strong>at</strong>ion yields<br />
2<br />
[ s V(<br />
s)<br />
sv(<br />
0)<br />
v'(<br />
0)<br />
] 5[<br />
sV(<br />
s)<br />
v(<br />
0)]<br />
]<br />
2<br />
− − + −<br />
10<br />
6V(<br />
s)<br />
10<br />
=<br />
s + 1<br />
or ( s + 5s<br />
+ 6)<br />
V(<br />
s)<br />
− 2s<br />
− 4 −10<br />
= ⎯⎯→<br />
V(<br />
s)<br />
=<br />
s + 1<br />
( s + 1)(<br />
s + 2)(<br />
s + 3)<br />
A B C<br />
Let V( s)<br />
= + + , A = 5,<br />
B = 0,<br />
C = −3<br />
s + 1 s + 2 s + 3<br />
Hence,<br />
v(<br />
t)<br />
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+<br />
−t −3t<br />
=<br />
( 5e<br />
− 3e<br />
) u(<br />
t)<br />
2s<br />
2<br />
+ 16s<br />
+ 24
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 52.<br />
Use <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> to find i(t) for t > 0 if<br />
2<br />
d i di<br />
+ 3 + 2i<br />
+ δ () t = 0 ,<br />
2<br />
dt dt<br />
i ( 0 ) = 0 , i ' ( 0)<br />
= 3<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 52.<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term.<br />
[ s I(<br />
s)<br />
si(<br />
0)<br />
i ( 0)<br />
] 3[<br />
s I(<br />
s)<br />
i(<br />
0)<br />
] 2I(<br />
s)<br />
1 0<br />
2 − − + − + + =<br />
( s 3s<br />
2)<br />
I(<br />
s)<br />
s 3 3 1 0<br />
2 + + − − − + =<br />
I(<br />
s)<br />
=<br />
( s<br />
+<br />
s + 5<br />
1)(<br />
s<br />
+<br />
′<br />
2)<br />
A = 4,<br />
B = -3<br />
I(<br />
s)<br />
4 3<br />
= −<br />
s + 1 s + 2<br />
A B<br />
= +<br />
s + 1 s + 2<br />
-t -2t<br />
i ( t)<br />
= ( 4e<br />
−<br />
3e<br />
) u(<br />
t)<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 53.<br />
* Use <strong>Laplace</strong> <strong>transform</strong>s to solve for x(t) in<br />
x<br />
λ −t<br />
() t = cos t + e x(<br />
λ)<br />
dλ<br />
∫<br />
t<br />
0<br />
* An asterisk indic<strong>at</strong>es a challenging problem.<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 53.<br />
Transform each term.<br />
We begin by noting th<strong>at</strong> <strong>the</strong> integral term can be rewritten as,<br />
t<br />
0<br />
∫<br />
−(<br />
t −λ)<br />
x ( λ)<br />
e dλ<br />
which is convolution and can be written as e –t *x(t).<br />
Now, <strong>transform</strong>ing each term produces,<br />
X(s) =<br />
X(<br />
s)<br />
s<br />
+<br />
2<br />
s + 1<br />
s + 1<br />
= =<br />
2<br />
s + 1 s<br />
1<br />
X(<br />
s)<br />
s + 1<br />
2<br />
s 1<br />
+<br />
2<br />
+ 1 s + 1<br />
If partial fraction expansion is used we obtain,<br />
⎛ s + 1−<br />
1⎞<br />
s<br />
→ ⎜ ⎟X(<br />
s)<br />
=<br />
⎝ s + 1 ⎠ 2<br />
s + 1<br />
x(t) = cos(t) + sin(t).<br />
x(t) = <strong>1.</strong>4141cos(t–45˚).<br />
This is <strong>the</strong> same answer and can be proven by using trigonometric identities.<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 54.<br />
Using <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong>, solve <strong>the</strong> following differential equ<strong>at</strong>ion for<br />
2<br />
d i di<br />
−2t<br />
+ 4 + 5i<br />
= 2e<br />
2<br />
dt<br />
dt<br />
i ′ .<br />
Subject to ( 0 ) = 0,<br />
i ( 0)<br />
= 2<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 54.<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term gives<br />
2<br />
2<br />
⎡<br />
⎣sIs ( ) −si(0) − i'(0) ⎤<br />
⎦ + 4 [ sIs ( ) − i(0) ] + 5 Is ( ) =<br />
s + 2<br />
2<br />
2<br />
⎡<br />
⎣sIs () −0− 2⎤ ⎦ + 4 [ sIs () − 0] + 5() Is =<br />
s + 2<br />
2 2 2s+ 6<br />
Is ()( s+ 4s+ 5) = + 2=<br />
s+ 2 s+<br />
2<br />
2s+ 6 A Bs+ C<br />
Is () = = +<br />
2 2<br />
( s+ 2)( s + 4s+ 5) s+ 2 s + 4s+ 5<br />
2 2<br />
2s+ 6 = A( s + 4s+ 5) + B( s + 2 s) + C( s+<br />
2)<br />
We equ<strong>at</strong>e <strong>the</strong> coefficients.<br />
s 2 : 0 = A+ B<br />
s: 2= 4A + 2B + C<br />
constant: 6 = 5A + 2C<br />
Solving <strong>the</strong>se gives<br />
A = 2, B= -2, C = -2<br />
2 2s+ 2 2 2( s+<br />
2) 2<br />
Is () = − = − +<br />
s+ s + s+ s+ s+ + s+<br />
+<br />
2 2 2<br />
2 4 5 2 ( 2) 1 ( 2) 1<br />
Taking <strong>the</strong> inverse <strong>Laplace</strong> <strong>transform</strong> leads to:<br />
( )<br />
it e e t e t ut e t tut<br />
−2t −2t −2t −2t<br />
() = 2 − 2 cos+ 2 sin () = 2 (1− cos+ sin ) ()<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 55.<br />
Solve for y(t) in <strong>the</strong> following differential equ<strong>at</strong>ion if <strong>the</strong> initial conditions are zero.<br />
3<br />
2<br />
d y d y dy −t<br />
+ 6 + 8 e cos 2t<br />
3<br />
2<br />
dt dt dt<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 55.<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term.<br />
3<br />
2<br />
2<br />
[ s Y(<br />
s)<br />
− s y(<br />
0)<br />
− s y′<br />
( 0)<br />
− y′<br />
′ ( 0)<br />
] + 6[<br />
s Y(<br />
s)<br />
− s y(<br />
0)<br />
− y′<br />
( 0)<br />
]<br />
+ 8[<br />
s Y(<br />
s)<br />
− y(<br />
0)<br />
] =<br />
s + 1<br />
2 2<br />
( s + 1)<br />
Setting <strong>the</strong> initial conditions to zero gives<br />
s + 1<br />
3 2 ( s + 6s<br />
+ 8s)<br />
Y(<br />
s)<br />
= 2 s + 2s<br />
+ 5<br />
( s + 1)<br />
( s)<br />
=<br />
s ( s + 2)(<br />
s + 4)(<br />
s<br />
+ 2<br />
A B C Ds + E<br />
= + + +<br />
+ 2s<br />
+ 5)<br />
s s + 2 s + 4 s + 2s<br />
5<br />
Y 2 2 +<br />
1<br />
A = ,<br />
40<br />
Y(<br />
s)<br />
Y(<br />
s)<br />
=<br />
=<br />
1<br />
40<br />
1<br />
40<br />
1<br />
⋅ +<br />
s<br />
1<br />
⋅ +<br />
s<br />
1<br />
B = ,<br />
20<br />
1<br />
20<br />
1<br />
20<br />
1<br />
⋅ −<br />
s + 2<br />
1<br />
⋅ −<br />
s + 2<br />
3<br />
104<br />
3<br />
104<br />
- 3<br />
C = ,<br />
104<br />
1<br />
⋅ −<br />
s + 4<br />
1<br />
⋅ −<br />
s + 4<br />
- 3<br />
D = ,<br />
65<br />
3s<br />
+ 7<br />
+ 1)<br />
2 2 + 2<br />
- 7<br />
E =<br />
65<br />
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1<br />
65<br />
1<br />
65<br />
⋅<br />
( s<br />
3(<br />
s + 1)<br />
⋅<br />
( s + 1)<br />
+<br />
−<br />
1<br />
⋅<br />
+<br />
4<br />
+<br />
2 2<br />
2 2<br />
2 65 ( s 1)<br />
2<br />
1 1 3 3<br />
2<br />
-2t<br />
-4t<br />
-t<br />
-t<br />
y ( t)<br />
= u(<br />
t)<br />
+<br />
e − e − e cos( 2t)<br />
− e sin( 2t)<br />
40 20 104 65<br />
65
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 56.<br />
Solve for v () t in <strong>the</strong> integrodifferential equ<strong>at</strong>ion<br />
dv<br />
t<br />
4 + 12∫<br />
v dt = 0<br />
−∞<br />
dt<br />
Given th<strong>at</strong> v ( 0 ) = 2 .<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 56.<br />
Taking <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term we get:<br />
12<br />
4 =<br />
s<br />
[ s V(<br />
s)<br />
− v(<br />
0)<br />
] + V(<br />
s)<br />
0<br />
⎡ 12⎤<br />
⎢⎣<br />
4 s + V(<br />
s)<br />
= 8<br />
s ⎥⎦<br />
8s<br />
2s<br />
( s)<br />
= 2 =<br />
4s<br />
+ 12 s + 3<br />
V 2<br />
v ( t)<br />
= 2<br />
cos(<br />
3t)<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 57.<br />
Solve <strong>the</strong> following integrodifferential equ<strong>at</strong>ion using <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> method:<br />
dy<br />
dt<br />
() t<br />
t<br />
+ ∫ y τ<br />
9 0<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 57.<br />
( τ ) d = cos 2t<br />
, y ( 0 ) = 1<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term.<br />
9 s<br />
s + Y(<br />
s)<br />
= 2 s s + 4<br />
[ Y(<br />
s)<br />
− y(<br />
0)<br />
]<br />
2<br />
2<br />
⎛ s + 9⎞<br />
s s + s + 4<br />
⎜ ⎟Y(<br />
s)<br />
= 1+<br />
2 = 2<br />
⎝ s ⎠ s + 4 s + 4<br />
3 2 s + s + 4s<br />
As + B Cs + D<br />
( s)<br />
= 2 2 = 2 +<br />
( s + 4)(<br />
s + 9)<br />
s + 4 s + 9<br />
Y 2<br />
s<br />
+ s<br />
+ 4s<br />
= A ( s<br />
+ 9s)<br />
+ B(<br />
s<br />
+ 9)<br />
+ C(<br />
s<br />
+ 4s)<br />
+ D(<br />
s<br />
3 2<br />
3<br />
2<br />
3<br />
2 +<br />
Equ<strong>at</strong>ing coefficients :<br />
0 s : 0 = 9B<br />
+ 4D<br />
1 s : 4 = 9A<br />
+ 4C<br />
2 s : 1 = B + D<br />
3 s : 1 = A + C<br />
Solving <strong>the</strong>se equ<strong>at</strong>ions gives<br />
A = 0 , B = - 4 5,<br />
C = 1,<br />
D = 9 5<br />
- 4 5 s + 9 5 - 4 5 s 9 5<br />
( s)<br />
= 2 + 2 = 2 + 2 +<br />
s + 4 s + 9 s + 4 s + 9 s + 9<br />
Y 2<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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4)<br />
y ( t)<br />
= - 0.4sin(<br />
2t)<br />
+<br />
cos( 3t)<br />
+ 0.<br />
6sin(<br />
3t)
<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 58.<br />
Given th<strong>at</strong><br />
dv<br />
t<br />
+ 2v<br />
+ 5 v<br />
0<br />
dt ∫ λ<br />
with ( ) 1<br />
( λ ) d = 4u(<br />
t)<br />
v 0 = − , determine v () t for t > 0 .<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 58.<br />
We take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term.<br />
5 4<br />
[ sV( s) − v(0)] + 2 V( s) + V( s)<br />
=<br />
s s<br />
5 4<br />
[ sV () s + 1] + 2 V () s + V () s =<br />
s s<br />
⎯⎯→<br />
4−s<br />
V () s = 2<br />
s + 2s+ 5<br />
− ( s+ 1) + 5 − ( s+<br />
1) 5 2<br />
V() s = = +<br />
( s+ 1) + 2 ( s+ 1) + 2 2 ( s+<br />
1) + 2<br />
2 2 2 2 2 2<br />
−t −t<br />
vt () = ( − e cos2t+ 2.5esin2) tut ()<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 59.<br />
Solve <strong>the</strong> integrodifferential equ<strong>at</strong>ion<br />
dy<br />
4y<br />
+ 3<br />
t<br />
0<br />
−2t<br />
y dt = 6e<br />
dt<br />
+ ∫ , y ( 0) = −1<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 59.<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term <strong>of</strong> <strong>the</strong> integrodifferential equ<strong>at</strong>ion.<br />
[ s Y(<br />
s)<br />
− y(<br />
0)<br />
]<br />
+<br />
4Y(<br />
s)<br />
3<br />
+<br />
s<br />
Y(<br />
s)<br />
⎛ 6 ⎞<br />
( s + 4s<br />
+ 3)<br />
Y(<br />
s)<br />
= s⎜<br />
−1⎟<br />
⎝s<br />
+ 2 ⎠<br />
2<br />
6<br />
=<br />
s + 2<br />
s ( 4 − s)<br />
( 4 − s)<br />
s<br />
Y( s)<br />
=<br />
2 =<br />
( s + 2)(<br />
s + 4s<br />
+ 3)<br />
( s + 1)(<br />
s + 2)(<br />
s + 3)<br />
Y(<br />
s)<br />
A B C<br />
= + +<br />
s + 1 s + 2 s + 3<br />
A = –2.5, B = 12, C = -10.5<br />
Y(<br />
s)<br />
− 2.<br />
5 12 10.<br />
5<br />
= + −<br />
s + 1 s + 2 s + 3<br />
y ( t)<br />
= –2.5e –t + 12e –2t – 10.5e –3t<br />
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part<br />
<strong>of</strong> this Manual may be displayed, reproduced or distributed in any form or by any means, without <strong>the</strong> prior<br />
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<strong>Chapter</strong> <strong>15</strong>, <strong>Problem</strong> 60.<br />
Solve <strong>the</strong> following integrodifferential equ<strong>at</strong>ion<br />
dx<br />
t<br />
2 + 5x<br />
+ 3 x dt + 4 = sin 4t<br />
dt ∫ , x ( 0 ) = 1<br />
0<br />
<strong>Chapter</strong> <strong>15</strong>, Solution 60.<br />
Take <strong>the</strong> <strong>Laplace</strong> <strong>transform</strong> <strong>of</strong> each term <strong>of</strong> <strong>the</strong> integrodifferential equ<strong>at</strong>ion.<br />
3 4 4<br />
2[ s X(<br />
s)<br />
− x(<br />
0)<br />
] + 5X(<br />
s)<br />
+ X(<br />
s)<br />
+ = 2 s s s + 16<br />
3 2<br />
4s<br />
2s<br />
− 4s<br />
+ 36s<br />
− 64<br />
2<br />
( 2s<br />
+ 5s<br />
+ 3)<br />
X(<br />
s)<br />
= 2s<br />
− 4 + 2 =<br />
2<br />
s + 16 s + 16<br />
2s<br />
( s)<br />
=<br />
( 2s<br />
2<br />
3 2<br />
− 4s<br />
+ 36s<br />
− 64 s − 2s<br />
+ 18s<br />
− 32<br />
2 =<br />
+ 5s<br />
+ 3)(<br />
s + 16)<br />
( s + 1)(<br />
s + <strong>1.</strong><br />
5)(<br />
s + 16)<br />
X<br />
3<br />
2<br />
2<br />
A B Cs + D<br />
X( s)<br />
= + + 2 s + 1 s + <strong>1.</strong><br />
5 s + 16<br />
A = ( s + 1)<br />
X(<br />
s)<br />
s = -1 = -6.235<br />
B = ( s + <strong>1.</strong><br />
5)<br />
X(<br />
s)<br />
s = -<strong>1.</strong>5 =<br />
When s = 0 ,<br />
- 32 B<br />
= A + +<br />
(<strong>1.</strong>5)(16) <strong>1.</strong><br />
5<br />
s<br />
D<br />
16<br />
7.<br />
329<br />
⎯⎯→<br />
D =<br />
0.<br />
2579<br />
2<br />
3 2<br />
3 2<br />
− 2s<br />
+ 18s<br />
− 32 = A(<br />
s + <strong>1.</strong><br />
5s<br />
+ 16s<br />
+ 24)<br />
+ B(<br />
s + s + 16s<br />
3 2<br />
2<br />
+ C(<br />
s + 2.<br />
5s<br />
+ <strong>1.</strong><br />
5s)<br />
+ D(<br />
s + 2.<br />
5s<br />
+ <strong>1.</strong><br />
5)<br />
3 +<br />
Equ<strong>at</strong>ing coefficients <strong>of</strong> <strong>the</strong> 3 s terms,<br />
1 = A + B + C ⎯⎯→<br />
C =<br />
-0.<br />
0935<br />
- 6.235 7.<br />
329 - 0.0935s + 0.2579<br />
X( s)<br />
= + +<br />
2 s + 1 s + <strong>1.</strong><br />
5 s + 16<br />
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16)<br />
-t -<strong>1.</strong>5t<br />
x ( t)<br />
= - 6.235e<br />
+<br />
7.<br />
329e<br />
− 0.<br />
0935cos(<br />
4t)<br />
+ 0.<br />
0645sin(<br />
4t)