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Chapter 15, Problem 1. Find the Laplace transform of: (a) cosh at

(b) sinh at

(

)

(

)

1 x 1 e + e − x , sinh x = e x − e − x .] 2 2

[Hint: cosh x =

Chapter 15, Solution 1.

(a)

(b)

e at + e - at cosh(at ) = 2 1⎡ 1 1 ⎤ s L [ cosh(at ) ] = ⎢ + = 2 ⎣ s − a s + a ⎥⎦ s 2 − a 2 e at − e - at sinh(at ) = 2 a 1⎡ 1 1 ⎤ L [ sinh(at ) ] = ⎢ − = 2 ⎥ 2 ⎣ s − a s + a ⎦ s − a2

Chapter 15, Problem 2.

Determine the Laplace transform of: (a) cos( ωt + θ )

(b) sin( ωt + θ )

Chapter 15, Solution 2.

(a)

f ( t ) = cos(ωt ) cos(θ) − sin(ωt ) sin(θ) F(s) = cos(θ) L [ cos(ωt ) ] − sin(θ) L [ sin(ωt ) ] s cos(θ) − ω sin(θ) F(s) = s 2 + ω2

(b)

f ( t ) = sin(ωt ) cos(θ) + cos(ωt ) sin(θ) F(s) = sin(θ) L [ cos(ωt ) ] + cos(θ) L [ sin(ωt ) ] s sin(θ) − ω cos(θ) F(s) = s 2 + ω2

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 3.

Obtain the Laplace transform of each of the following functions: (a) e −2t cos 3tu (t )

(c) e −3t cosh 2tu (t ) (e) te − t sin 2tu (t )

(b) e −2t sin 4tu (t ) (d) e −4t sinh tu (t )

Chapter 15, Solution 3.

(a)

L [ e -2t cos(3t ) u ( t ) ] =

s+2 (s + 2 ) 2 + 9

(b)

L [ e -2t sin(4 t ) u ( t ) ] =

4 (s + 2) 2 + 16

(c)

Since L [ cosh(at ) ] =

(d)

Since L [ sinh(at ) ] =

(e)

L [ e - t sin( 2t ) ] =

s s − a2 s+3 L [ e -3t cosh(2 t ) u ( t ) ] = (s + 3 ) 2 − 4 2

a s − a2 1 L [ e -4t sinh( t ) u ( t ) ] = (s + 4) 2 − 1 2

2 (s + 1) 2 + 4

f (t) ← ⎯→ F(s) -d F(s) t f (t) ← ⎯→ ds -d -1 2 ( (s + 1) 2 + 4) Thus, L [ t e - t sin(2 t ) ] = ds 2 = ⋅ 2 (s + 1) ((s + 1) 2 + 4) 2 4 (s + 1) L [ t e -t sin( 2t ) ] = ((s + 1) 2 + 4) 2 If

[

]

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 4. Find the Laplace transforms of the following: (a) g (t ) = 6 cos(4t − 1) (b) f (t ) = 2tu (t ) + 5e −3(t − 2 )u (t − 2 )

Chapter 15, Solution 4. s

e −s =

(a)

G (s) = 6

(b)

e −2s F(s) = +5 s+3 s2

s2 + 42

6se −s s 2 + 16

2

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Chapter 15, Problem 5. Find the Laplace transform of each of the following functions: (b) 3t 4 e −2t u (t ) (a) t 2 cos(2t + 30°)u (t ) d (c) 2tu (t ) − 4 δ (t ) (d) 2e − (t −1)u (t ) dt (f) 6e − t 3 u (t ) (e) 5u (t 2 ) dn (g) n δ (t ) dt

Chapter 15, Solution 5. (a)

s cos(30°) − 2 sin(30°) s2 + 4 d 2 ⎡ s cos(30°) − 1 ⎤ L [ t 2 cos(2t + 30°) ] = 2 ⎢ ds ⎣ s 2 + 4 ⎥⎦ L [ cos(2t + 30°) ] =

=

⎞ -1 ⎤ d d ⎡⎛ 3 s − 1⎟⎟ (s 2 + 4) ⎥ ⎢⎜⎜ ds ds ⎢⎣⎝ 2 ⎥⎦ ⎠

=

⎛ 3 ⎞ -1 -2 ⎤ d ⎡ 3 2 s − 1⎟⎟ (s 2 + 4) ⎥ ⎢ (s + 4) − 2s ⎜⎜ ds ⎣⎢ 2 ⎝ 2 ⎠ ⎦⎥

⎛ 3 ⎞ ⎛ 3⎞ ⎛ 3 ⎞ 2 3 (- 2s ) 2 ⎜⎜ 2 s − 1⎟⎟ 2s ⎜⎜ 2 ⎟⎟ (8s ) ⎜⎜ 2 s − 1⎟⎟ ⎠− ⎝ ⎠+ ⎝ ⎠ = 2 − ⎝ 3 2 2 2 2 2 2 2 s +4 s +4 s +4 s +4

(

)

(

)

(

)

(

)

⎛ 3 ⎞ (8s 2 ) ⎜⎜ s − 1⎟⎟ - 3s − 3s + 2 − 3s ⎝ 2 ⎠ = + 2 3 2 2 s +4 s +4

(

=

)

(-3 3 s + 2)(s 2 + 4)

(s

2

+4

L [ t 2 cos(2t + 30°) ] =

)

3

(

+

)

4 3 s3 − 8 s 2

(s

2

+4

)

3

8 − 12 3 s − 6s 2 + 3s 3

( s 2 + 4) 3

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

[

]

4!

72

(b)

L 3 t 4 e - 2t = 3 ⋅

(c)

⎡ ⎤ 2 2 d L ⎢ 2t u ( t ) − 4 δ( t ) ⎥ = 2 − 4(s ⋅ 1 − 0) = 2 − 4s ⎣ ⎦ s s dt

(s + 2) 5

=

(s + 2) 5

(d)

2 e -(t-1) u ( t ) = 2 e -t u ( t ) 2e L [ 2 e -(t-1) u ( t ) ] = s+1

(e)

Using the scaling property, 1 1 1 5 L [ 5 u ( t 2) ] = 5 ⋅ ⋅ = 5⋅ 2⋅ = 1 2 s (1 2) 2s s

(f) (g)

L [ 6 e -t 3 u ( t ) ] =

6 18 = s + 1 3 3s + 1

Let f ( t ) = δ( t ) . Then, F(s) = 1 . ⎡ dn ⎤ ⎡ dn ⎤ L ⎢ n δ( t ) ⎥ = L ⎢ n f ( t ) ⎥ = s n F(s) − s n −1 f (0) − s n − 2 f ′(0) − L ⎣ dt ⎦ ⎣ dt ⎦

⎡ dn ⎤ ⎡ dn ⎤ L ⎢ n δ( t ) ⎥ = L ⎢ n f ( t ) ⎥ = s n ⋅ 1 − s n −1 ⋅ 0 − s n − 2 ⋅ 0 − L ⎣ dt ⎦ ⎣ dt ⎦ n ⎡ d ⎤ L ⎢ n δ( t ) ⎥ = s n ⎣ dt ⎦

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 6.

Find F(s) given that ⎧2t , 0 < t < 1 ⎪ f (t ) = ⎨t , 1 < t < 2 ⎪0, otherwise ⎩ Chapter 15, Solution 6. ∞

1

2

0

1

F ( s ) = ∫ f (t )e − st dt = ∫ 2te− st dt + ∫ 2e− st dt 0

− st

2

− st

1 e e 2 2 ( 1) 2 st − − + = 2 (1 − e − s − se −2 s ) 2 0 s −s 1 s

Chapter 15, Problem 7.

Find the Laplace transform of the following signals: (a) f (t ) = (2t + 4 )u (t ) (b) g (t ) = (4 + 3e −2t )u (t ) (c) h(t ) = (6 sin (3t ) + 8 cos(3t ))u (t ) (d) x(t ) = (e −2t cosh (4t ))u (t )

Chapter 15, Solution 7. 2 4 (a) F ( s ) = 2 + s s

(b) G ( s ) =

4 3 + s s+2

(c ) H(s) = 6

3 2

s +9

+8

s 2

s +9

=

8s + 18 s2 + 9

(d) From Problem 15.1, s L{cosh at} = 2 s − a2 s+2 s+2 = 2 X (s) = 2 2 s + 4s − 12 ( s + 2) − 4 (a )

4 4 3 8s + 18 s+2 , (c ) , (d ) + , (b) + s s+2 s2 s s2 + 9 s 2 + 4s − 12 2

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 8.

Find the Laplace transform F(s), given that f(t) is: (a) (b) (c) (d)

2tu (t − 4 ) 5 cos(t )δ (t − 2 )

e − t u (t − t ) sin (2t )u (t − τ )

Chapter 15, Solution 8.

(a) 2t=2(t-4) + 8 f(t) = 2tu(t-4) = 2(t-4)u(t-4) + 8u(t-4) 2 8 ⎛ 2 8⎞ F ( s ) = 2 e −4 s + e−4 s = ⎜ 2 + ⎟ e−4 s s s s⎠ ⎝s ∞

∞

0

0

(b) F ( s ) = ∫ f (t )e − st dt = ∫ 5cos tδ (t − 2)e− st dt =5cos te− st (c)

t=2

5cos(2)e = 5cos 2e−2 s –2s

e − t = e− ( t −τ ) e−τ f (t ) = e −τ e− (t −τ )u (t − τ ) −τ

F (s) = e e

−τ s

1 e −τ ( s +1) = s +1 s +1

(d) sin 2t = sin[2(t − τ ) + 2τ ] = sin 2(t − τ ) cos 2τ + cos 2(t − τ )sin 2τ f (t ) = cos 2τ sin 2(t − τ )u (t − τ ) + sin 2τ cos 2(t − τ )u (t − τ ) 2 s F ( s ) = cos 2τ e −τ s 2 + sin 2τ e−τ s 2 s +4 s +4

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 9.

Determine the Laplace transforms of these functions: (a) f (t ) = (t − 4 )u (t − 2 )

(b) g (t ) = 2e −4t u (t − 1) (c) h(t ) = 5 cos(2t − 1)u (t ) (d) p (t ) = 6[u (t − 2 ) − u (t − 4 )]

Chapter 15, Solution 9.

(a)

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

(b)

g( t ) = 2 e -4t u ( t − 1) = 2 e -4 e -4(t -1) u ( t − 1) 2 e -s G (s) = 4 e (s + 4)

(c)

h ( t ) = 5 cos(2 t − 1) u ( t ) cos(A − B) = cos(A) cos(B) + sin(A) sin(B) cos(2t − 1) = cos(2t ) cos(1) + sin(2t ) sin(1) h ( t ) = 5 cos(1) cos(2 t ) u ( t ) + 5 sin(1) sin(2t ) u ( t ) s 2 + 5 sin(1) ⋅ 2 s +4 s +4 2.702 s 8.415 H(s) = 2 + s + 4 s2 + 4 H(s) = 5 cos(1) ⋅

(d)

2

p( t ) = 6u ( t − 2) − 6u ( t − 4) P(s) =

6 - 2s 6 -4s e − e s s

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 10.

In two different ways, find the Laplace transform of d −t g (t ) = te cos t dt

(

)

Chapter 15, Solution 10.

(a)

(b)

By taking the derivative in the time domain, g( t ) = (-t e -t + e -t ) cos( t ) − t e -t sin( t ) g( t ) = e -t cos( t ) − t e -t cos( t ) − t e -t sin( t ) G (s) =

⎤ s +1 d ⎡ s +1 ⎤ d ⎡ 1 + ⎢ ⎥+ ⎢ ⎥ 2 2 2 (s + 1) + 1 ds ⎣ (s + 1) + 1⎦ ds ⎣ (s + 1) + 1⎦

G (s) =

s +1 s 2 + 2s 2s + 2 s 2 (s + 2) − − = s 2 + 2s + 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

By applying the time differentiation property, G (s) = sF(s) − f (0) where f ( t ) = t e -t cos( t ) , f (0) = 0 - d ⎡ s +1 ⎤ (s)(s 2 + 2s) s 2 (s + 2) = G (s) = (s) ⋅ ⎢ = ds ⎣ (s + 1) 2 + 1 ⎥⎦ (s 2 + 2s + 2) 2 (s 2 + 2s + 2) 2

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 11.

Find F(s) if: (b) f (t ) = 3te −2t sinh 4t (a) f (t ) = 6e − t cosh 2t (c) f (t ) = 8e −3t cosh tu (t − 2) Chapter 15, Solution 11.

s s − a2 6 (s + 1) 6 (s + 1) F(s) = = 2 2 (s + 1) − 4 s + 2s − 3

(a)

Since L [ cosh(at ) ] =

(b)

Since L [ sinh(at ) ] =

2

a s − a2 (3)(4) 12 L [ 3 e -2t sinh(4t ) ] = = 2 2 (s + 2) − 16 s + 4s − 12 2

-d [ 12 (s 2 + 4s − 12) -1 ] ds 24 (s + 2) F(s) = (12)(2s + 4)(s 2 + 4s − 12) -2 = 2 (s + 4s − 12) 2 F(s) = L [ t ⋅ 3 e -2t sinh(4t ) ] =

(c)

1 ⋅ (e t + e - t ) 2 1 f ( t ) = 8 e -3t ⋅ ⋅ (e t + e - t ) u ( t − 2) 2 -2t = 4 e u ( t − 2) + 4 e-4t u ( t − 2) = 4 e-4 e-2(t - 2) u ( t − 2) + 4 e-8 e-4(t - 2) u ( t − 2) cosh( t ) =

L [ 4 e -4 e -2(t -2) u ( t − 2)] = 4 e -4 e -2s ⋅ L [ e -2 u ( t )] 4 e -(2s+ 4) L [ 4 e -4 e -2(t -2) u ( t − 2)] = s+2

Similarly, L [ 4 e

-8

e

- 4(t - 2)

4e u ( t − 2) ] =

-(2s+ 8)

s+4

Therefore, 4 e -(2s+ 4) 4 e -(2s+8) e -(2s+ 6) [ (4 e 2 + 4 e -2 ) s + (16 e 2 + 8 e -2 )] + = F(s) = s+2 s+4 s 2 + 6s + 8 PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 12.

If g (t ) = e −2t cos 4t find G(s). Chapter 15, Solution 12. G(s) =

s+2 s+2 = 2 2 2 ( s + 2) + 4 s + 4s + 20

Chapter 15, Problem 13.

Find the Laplace transform of the following functions: (b) e − t t sin t u (t )

(a) t cos t u (t )

(c)

sin βt u (t ) t

Chapter 15, Solution 13.

← ⎯→

(a) tf (t )

−

d F (s) ds

If f(t) = cost, then F(s)=

s s2 + 1

and -

L ( t cos t ) =

d (s 2 + 1)(1) − s(2s) F(s)= − ds (s 2 + 1) 2 s2 −1 (s 2 + 1) 2

(b) Let f(t) = e-t sin t. 1 1 = 2 F (s) = 2 ( s + 1) + 1 s + 2s + 2 dF ( s 2 + 2s + 2)(0) − (1)(2s + 2) = ds ( s 2 + 2s + 2) 2 dF 2( s + 1) L (e −t t sin t ) = − = 2 ds ( s + 2s + 2) 2

(c )

f (t ) t

∞

← ⎯→

∫ F (s)ds s

Let f (t ) = sin βt , then F ( s ) = ∞

β s +β2 2

β 1 s ⎡ sin βt ⎤ L⎢ =∫ 2 ds = β tan −1 2 ⎥ β β ⎣ t ⎦ s s +β

∞ s

=

π 2

− tan −1

s

β

= tan −1

β s

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 14.

Find the Laplace transform of the signal in Fig. 15.26.

Figure 15.26

For Prob. 15.14.

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Chapter 15, Solution 14.

Taking the derivative of f(t) twice, we obtain the figures below. f’(t) 5 0

t 2

4

6

-2.5

f’’(t) 5 δ (t)

0

2.5δ(t-6)

2

6

-7.5δ(t-2) f” = 5δ(t) – 7.5δ(t–2) + 2.5δ(t–6) Taking the Laplace transform of each term, s2F(s) = 5 – 7.5e–2s + 2.5e–6s or F(s) =

5 e −2s e −6s − 7.5 + 2.5 s s2 s2

Please note that we can obtain the same answer by representing the function as, f(t) = 5tu(t) – 7.5u(t–2) + 2.5u(t–6).

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 15.

Determine the Laplace transform of the function in Fig. 15.27.

Figure 15.27

For Prob. 15.15. Chapter 15, Solution 15.

This is a periodic function with T=3. F ( s) F ( s ) = 1 −3 s 1− e To get F1(s), we consider f(t) over one period. f1(t)

f1’(t)

5

f1’’(t)

5 5δ(t)

0

1

t

0

1

t

–5δ(t-1)

0

1

t

–5δ(t-1) –5δ’(t-1)

f1” = 5δ(t) –5δ(t–1) – 5δ’(t–1) Taking the Laplace transform of each term, s2F1(s) = 5 –5e–s – 5se–s or F1(s) = 5(1 – e–s – se–s)/s2 Hence, F(s) = 5

1 − e −s − se −s s 2 (1 − e − 3s )

Alternatively, we can obtain the same answer by noting that f1(t) = 5tu(t) – 5tu(t–1) – 5u(t–1). PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 16.

Obtain the Laplace transform of f(t) in Fig. 15.28.

Figure 15.28

For Prob. 15.16.

Chapter 15, Solution 16.

f ( t ) = 5 u ( t ) − 3 u ( t − 1) + 3 u ( t − 3) − 5 u ( t − 4)

F(s) =

1 [ 5 − 3 e -s + 3 e - 3 s − 5 e - 4 s ] s

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 17.

Find the Laplace transform of f(t) shown in Fig. 15.29.

Figure 15.29

For Prob. 15.17. Chapter 15, Solution 17.

Taking the derivative of f(t) gives f’(t) as shown below. f’(t) 2δ(t)

t -δ(t-1) – δ(t-2) f’(t) = 2δ(t) – δ(t–1) – δ(t–2) Taking the Laplace transform of each term, sF(s) = 2 – e–s – e–2s which leads to F(s) = [2 – e–s – e–2s]/s We can also obtain the same answer noting that f(t) = 2u(t) – u(t–1) – u(t–2).

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 18.

Obtain the Laplace transforms of the functions in Fig. 15.30.

Figure 15.30

For Prob. 15.18.

Chapter 15, Solution 18.

(a)

g ( t ) = u ( t ) − u ( t − 1) + 2 [ u ( t − 1) − u ( t − 2)] + 3 [ u ( t − 2) − u ( t − 3)] = u ( t ) + u ( t − 1) + u ( t − 2) − 3 u ( t − 3) 1 G (s) = (1 + e -s + e - 2s − 3 e - 3s ) s

(b)

h ( t ) = 2 t [ u ( t ) − u ( t − 1)] + 2 [ u ( t − 1) − u ( t − 3)] + (8 − 2 t ) [ u ( t − 3) − u ( t − 4)]

= 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 u ( t − 1) + 2 u ( t − 1) − 2 u ( t − 3) − 2 ( t − 3) u ( t − 3) + 2 u ( t − 3) + 2 ( t − 4) u ( t − 4) = 2t u ( t ) − 2 ( t − 1) u ( t − 1) − 2 ( t − 3) u ( t − 3) + 2 ( t − 4) u ( t − 4) H(s) =

2 2 - 3s 2 - 4 s 2 -s + 2 e = 2 (1 − e -s − e - 3s + e -4s ) 2 (1 − e ) − 2 e s s s s

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 19.

Calculate the Laplace transform of the train of unit impulses in Fig. 15.31.

Figure 15.31

For Prob. 15.19. Chapter 15, Solution 19.

Since L[ δ( t )] = 1 and T = 2 , F(s) =

1 1 − e - 2s

Chapter 15, Problem 20.

The periodic function shown in Fig. 15.32 is defined over its period as ⎧sin π t , 0 < t < 1 g (t )⎨ 1< t < 2 ⎩0, Find G(s)

Figure 15.32

For Prob. 15.20. Chapter 15, Solution 20. Let g 1 ( t ) = sin(πt ), 0 < t < 1 = sin( πt ) [ u ( t ) − u ( t − 1)] = sin(πt ) u ( t ) − sin(πt ) u ( t − 1)

Note that sin(π( t − 1)) = sin(πt − π) = - sin(πt ) . g1 ( t ) = sin( πt) u(t) + sin( π( t - 1)) u(t - 1) So, G 1 (s) =

π (1 + e -s ) s + π2 2

G 1 (s) π (1 + e -s ) G (s) = = 1 − e -2s (s 2 + π 2 )(1 − e - 2s ) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 21.

Obtain the Laplace transform of the periodic waveform in Fig. 15.33.

Figure 15.33

For Prob. 15.21.

Chapter 15, Solution 21.

T = 2π Let

t ⎞ ⎛ f1 ( t ) = ⎜1 − ⎟ [ u ( t ) − u ( t − 2π)] ⎝ 2π ⎠ t 1 f1 ( t ) = u ( t ) − u(t) + ( t − 2 π) u ( t − 2 π) 2π 2π

[

1 1 e - 2πs 2π s + - 1 + e -2πs + = F1 (s) = − s 2πs 2 2πs 2 2πs 2 F(s) =

]

F1 (s) 2πs − 1 + e −2πs = 1 − e -Ts 2πs 2 (1 − e - 2πs )

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 15, Problem 22.

Find the Laplace transforms of the functions in Fig. 15.34.

Figure 15.34

For Prob. 15.22. Chapter 15, Solution 22. (a) Let g1 ( t ) = 2t, 0 < t < 1 = 2 t [ u ( t ) − u ( t − 1)] = 2t u ( t ) − 2 ( t − 1) u ( t − 1) + 2 u ( t − 1) 2 2 e -s 2 G 1 (s) = 2 − 2 + e -s s s s G 1 (s) G (s) = , T =1 1 − e -sT 2 (1 − e -s + s e -s ) G (s) = s 2 (1 − e -s )

(b)

Let h = h 0 + u ( t ) , where h 0 is the periodic triangular wave. Let h 1 be h 0 within its first period, i.e. ⎧ 2t 0 < t