Feedback control of dynamic systems 8th edition franklin solutions manual

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6000

Solutions Manual: Chapter 6 8th Edition

Feedback Control of Dynamic Systems . .

Gene F. Franklin .

J. David Powell .

Abbas Emami-Naeini . . . .

Assisted by: H. K. Aghajan H. Al-Rahmani P. Coulot P. Dankoski S. Everett R. Fuller T. Iwata V. Jones F. Safai L. Kobayashi H-T. Lee E. Thuriyasena M. Matsuoka

Copyright (c) 2018 Pearson Education All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of the publisher.

Chapter 6

The Frequency-response Design Method Problems and Solutions for Section 6.1 1. (a) Show that

0 in Eq. (6.2) is given by 0 =

G(s)

s

U0 ! = j! s= j!

and 0 =

G(s)

U0 G( j!)

1 2j

U0 !

1 = U0 G(j!) : s + j! s=+j!

2j

(b) By assuming the output can be written as y(t) =

0e

j!t

j!t ; 0e

+

derive Eqs. (6.4) - (6.6).

Solution: (a) Eq. (6.2):

Y (s) =

1

s

p1

+

2

s

p2

+

+

n

s

pn

+

o

s + j! o

+

o

s

j! o

Multiplying this by (s + j!) :

Y (s)(s+j!) =

1

s + a1

(s+j!)+:::+ 6001

n

s + an

(s+j!)+ o +

o

s

j!

(s+j!)

6002

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

o

)

o = Y (s)(s + j!)

=

o js= j! =

= =

1

s + a1

(s + j!)

Y (s)(s + j!)

(s + j!)

n

s + an :::

(s + j!) o

s

j!

j!) o

=

(s + j!)

s= j!

1

j!) : n

o

(s j!) + s + j!

Uo ! = j!)js=j! = G(s) 2 (s j!)js=j! o js=j! = Y (s)(s s + !2 1 Uo ! js=j! = Uo G(j!) = G(s) s + j!

2j

y(t)

=

y(t)

= =

jG(j!)j

=

\G(j!)

=

jG( j!)j

= =

\G( j!)

s + a1

(s

j!) + ::: +

s + an

(s

j!) +

=

oe

j!t

+

j!t oe

1 1 j!t e + Uo G(j!) ej!t 2j 2j G(j!)ej!t G( j!)e j!t Uo 2j n o1 2 2 2 =A Re [G(j!)] + Im [G(j!)]

Uo G( j!)

tan 1

Im [G(j!)] = Re [G(j!)]

n o1 2 2 2 = jG(j!)j Re [G( j!)] + Im [G( j!)] n o 21 2 2 Re [G(j!)] + Im [G(j!)] =A tan 1

Im [G( j!)]

Im [G(j!)] = tan 1 = Re [G( j!)]

Re [G(j!)]

) G(j!) = Aej ; G( j!) = Ae j Thus,

y(t)

j!

(s + j!)

(b)

y(t)

o

s

Uo ! (s + j!)js= j!

Y (s)(s + j!)js= j! = G(s) 2 s + !2 Uo !

1 G(s) js= j! = Uo G( j!) s j!

2j

Similarly, multiplying Eq. (6.2) by (s

Y (s)(s

1

s + a1

:::

Ae j e j!t ej(!t+ ) e j(!t+ ) = Uo A 2j 2j = Uo A sin(!t + ) = Uo

Aej ej!t

o

6003 where A = jG(j!)j ;

= tan 1

Im [G(j!)] = \G(j!)

Re [G(j!)]

2. (a) Calculate the magnitude and phase of G(s) =

1 s + 10

by hand for ! = 1, 2, 5, 10, 20, 50, and 100 rad/sec. (b) sketch the asymptotes for G(s) according to the Bode plot rules, and compare these with your computed results from part (a).

Solution: (a) G(s)

1 10 j!

1 ; G(j!) = = s + 10 10 + j!

100 + ! 2 1 ! p ; \G(j!) = tan 1 10 100 + ! 2

=

jG(j!)j = !

1 2 5 10 20 50 100

jG(j!)j 0:0995 0:0981 0:0894 0:0707 0:0447 0:0196 0:00995

\G(j!)

5:71 11:3 26:6 45:0 63:4 78:7 84:3

(b) To plot the asymptotes, you …rst note that n = 0; as de…ned in Section 6.1.1. That signi…es that the leftmost portion of the asymptotes will have zero slope. That portion of the asymptotes will be located at the DC gain of the transfer function, which, in this case it can be seen by inspection to be 0.1.

So the asymptote starts with a straght horizontal line at 0.1 and that continues until the breakpoint at ! = 10, at which point the asymptote has a slope of n = 1 that continues until forever, at least until the edge of the paper. The values computed above by "hand" (at least we hope you didn’t cheat) are plotted on the graph below and you see they match quite well except very near the breakpoint, a you should have expected. The Bode plot is :

6004

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

10

Bode plot for Problem 6.2

0

-1

-2

-3

10

-1

10

0

1

10 ω (rad/sec)

10

2

10

3

20 Phase (deg)

0 -20 -40 -60 -80 -100 -1 10

10

0

1

10 ω (rad/sec)

10

2

3. Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

2000 s(s + 200) 100 (b) L(s) = s(0:1s + 1)(0:5s + 1) 1 (c) L(s) = s(s + 1)(0:02s + 1) 1 (d) L(s) = (s + 1)2 (s + 10)2 10(s + 4) (e) L(s) = s(s + 1)(s + 200) 1000(s + 0:1) (f) L(s) = s(s + 1)(s + 8)2 (a) L(s) =

10

3

6005 (g) L(s) =

(s + 5)(s + 10) s(s + 1)(s + 100)

4s(s + 10) (s + 100)(s + 500) s (i) L(s) = (s + 1)(s + 10)(s + 200)

(h) L(s) =

Solution: 10 s +1 s 200 The asymptote will have magnitude = 10 at ! = 1 and have a -1 slope at the low frequencies. So the low frequency asymptote will pass through magnitude = 1, at ! = 10: At ! = 200; the slope changes to -2, and that slope continues forever. The phase asymptotes are even simpler and shown in the plot below along with the Matlab generated exact curve.

(a) L(s) =

Magnitude

Bode plot for Prob. 6.3 (a)

10

10

0

-5

10

1

10

2

10

3

10

4

ω (rad/sec) -80 Phase (deg)

-100 -120 -140 -160 -180 -200 1 10

10

2

10 ω (rad/sec)

3

10

4

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) L(s) =

100 s

s 10 + 1

s 2 +1

Breakpoints are at 2 and 10. The slope starts at n = -1, the changes to -2 at ! = 2; then changes to -3 at ! = 10: The magnitude on the -1 slope goes through 100 at ! = 1:Thus, will be 10,000 at ! = 0:01

Magnitude

Bode plot for Prob. 6.3 (b)

10

0

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

-50 -100 Phase (deg)

6006

-150 -200 -250 -300 -2 10

10

-1

0

10 ω (rad/sec)

10

1

1 (c) L(s) = s(s+1)(0:02s+1)

Slope starts at n = -1, becomes -2 at ! = 1; then -3 at n = 50.

Magnitude = 1 at ! = 1:

10

2

6007

Magnitude

Bode plot for Prob. 6.3 (c)

10

10

0

-5

10

-2

10

-1

10

0

10

1

10

2

10

3

ω (rad/sec) -50 Phase (deg)

-100 -150 -200 -250 -300 -2 10

10

-1

10

0

10

1

ω (rad/sec)

(d) L(s) =

1 100 2 s (s + 1)2 10 +1

Initially, slope = 0, then becomes n = -2 at ! = 1; then n = -4 at ! = 10: (e) L(s) =

0:2 4s + 1 s s(s + 1)(s 200 + 1) 50 32

(f) L(s) =

(10s + 1) 2 s 8 +1

=

(1:56) (10s + 1) 2

s(s + 1) s(s + 1) 8s + 1 Breakpoints are at ! = 0:1; 1; 8 So, to position the low frequency asymptote, you need to extrapolate the line from ! = 1 backward.

The magnitude at ! = 1 would be 1.56, and to position the asymptote at ! = 0:1; we need to multiply by 10 due to the -1 slope Therefore, it is located at a magnitude of 15.6 at ! = 0:1:and will be 156 at ! = 0:01:

(g) L(s) =

1 2

s 5 +1

s 10 + 1

s(s + 1)(s + 100)

10

2

10

3

6008

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot for Prob 6.3(d)

Magnitude

10

10

0

-5

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

0 -100 -200 -300 -400 -1 10

10

0

10 ω (rad/sec)

1

10

2

6009

Bode plot for Prob 6.3(e) 0

Magnitude

10

10

-5

10

-1

10

0

1

10 ω (rad/sec)

10

2

10

3

Phase (deg)

-50

-100

-150

-200 -1 10

10

0

1

10 ω (rad/sec)

10

2

10

3

6010

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob 6.3(f)

10

0

10

-2

10

-1

0

10 ω(rad/sec)

10

1

10

2

Phase (deg)

0 -100 -200 -300 10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

6011

Magnitude

Bode plot for Prob 6.3(g)

10

10

0

-2

10

-1

10

0

1

10 ω(rad/sec)

10

2

10

3

50 Phase (deg)

0 -50 -100 -150 -200 -1 10

10

0

1

10 ω (rad/sec)

10

2

10

3

6012

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode plot for Prob 6.3(h)

Magnitude

10

10

0

-2

10

0

10

1

2

10 ω (rad/sec)

10

3

10

4

200 Phase (deg)

150 100 50 0 -50 0 10

(h) L(s) =

10

4 5000 s s 100 + 1

1

2

10 ω (rad/sec)

10

3

s 10 + 1 s 500 + 1

This one is unusual in that the slope is generally rising with increasing frequency. Finite output at extremely high frequencies doesn’t really happen in nature, but hey, this is a theoretical homework problem to check whether you can follow the rules and do what the math tells you to do. (i) L(s) =

(s + 1)

1 2000 s s s 10 + 1 ( 200 + 1)

4. Real poles and zeros. Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. (a) L(s) =

1 s(s + 1)(s + 5)(s + 10)

10

4

6013

Magnitude

10

10

10

Bode plot for Prob 6.3(i)

-2

-4

-6

10

-1

10

0

1

10 ω (rad/sec)

10

2

10

3

Phase (deg)

100

0

-100

-200 -1 10

10

0

1

10 ω (rad/sec)

10

2

10

3

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) L(s) =

(s + 2) s(s + 1)(s + 5)(s + 10)

(c) L(s) =

(s + 2)(s + 6) s(s + 1)(s + 5)(s + 10)

(d) L(s) =

(s + 2)(s + 8) s(s + 1)(s + 5)(s + 10)

Solution: 1 50

(a) L(s) =

s(s + 1)

s 5 +1

s 10 + 1

Bode plot for Prob. 6.4 (a)

Magnitude

10

10

0

-10

10

-2

10

-1

10

0

10

1

10

2

10

3

ω (rad/sec) 0 Phase (deg)

6014

-100 -200 -300 -400 10

(b) L(s) =

-2

10

1 25

s(s + 1)

-1

s 2 +1 s s 5 +1 10 + 1

0

1

10 10 Frequency (rad/sec)

10

2

10

3

6015 Bode plot for Prob. 6.4 (b)

Magnitude

10

10

0

-10

10

-2

10

-1

10

0

10

1

10

2

10

3

ω (rad/sec)

Phase (deg)

0 -100 -200 -300 10

-2

10

-1

10

0

10 ω (rad/sec)

(c) L(s) =

s s 2 +1 6 +1 s s(s + 1) 5s + 1 10 +1 6 25

1

10

2

10

3

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Bode plot for Prob. 6.4 (c)

Magnitude

10

10

10

0

-5

-10

10

-2

10

-1

10

0

10

1

10

2

10

3

ω (rad/sec) 0 -50 Phase (deg)

6016

-100 -150 -200 -250 10

-2

10

-1

10

0

10 ω (rad/sec)

(d) L(s) =

s s 2 +1 4 +1 s(s + 1) 5s + 1 8s + 1 1 5

1

10

2

10

3

6017 Bode Plot for Prob. 6.4(d)

Magnitude

10

10

10

0

-5

-10

10

-2

10

-1

0

1

0

1

10 10 Frequency (rad/sec)

10

2

10

3

Phase (deg)

0 -50 -100 -150 -200 -250 10

-2

10

-1

10 10 Frequency (rad/sec)

10

2

5. Complex poles and zeros Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions and approximate the transition at the second order break point based on the value of the damping ratio. After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. (a) L(s) =

1 s2 + 5s + 20

(b) L(s) =

1 s(s2 + 3s + 10)

(c) L(s) =

(s2 + 2s + 8) s(s2 + 2s + 10)

(d) L(s) =

(s2 + 1) s(s2 + 4)

(e) L(s) =

(s2 + 4) s(s2 + 1)

10

3

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD Solution:

1 20

(a) L(s) = ps 20

Magnitude

10

10

10

10

2

5 s+1 + 20

Bode plot for Prob. 6.5(a)

-1

-2

-3

-4

10

-1

10

0

10

1

10

2

Frequency (rad/sec) 50 0 Phase (deg)

6018

-50 -100 -150 -200 10

-1

10

0

10 Frequency (rad/sec)

(b) L(s) = s

ps 10

1 10 2

3 + 10 s+1

1

10

2

6019 Bode plot for Prob. 6.5 (b)

Magnitude

10

10

10

0

-5

-10

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

-50 -100 -150 -200 -250 -300 10

-1

10

0

10 ω (rad/sec)

4 5

s p

2 2

(c) L(s) = s

2

ps 10

2

+ 14 s + 1 + 15 s + 1

1

10

2

6020

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.5 (c)

10

0

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

0 -50 -100 -150 -1 10

10

0

10

1

ω (rad/sec) 1

(d) L(s) = s 4 (e) L(s) =

h4 h

(s2 + 1)

i

s 2 +1 2

i

s 2 +1 2

s(s2 + 1)

6. Multiple poles at the origin Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions.

After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales.

1 s2 (s + 2) 1 (b) L(s) = 3 s (s + 8) 1 (c) L(s) = 4 s (s + 10) (a) L(s) =

10

2

6021

Magnitude

Bode Plot for Prob 6.5(d)

10

10

0

-2

10

-1

0

10 Frequency (rad/sec)

10

1

150 Phase (deg)

100 50 0 -50 -100 -150 -1 10

0

10 Frequency (rad/sec)

10

1

6022

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode Plot for Prob 6.5(e)

2

0

-2

10

-1

0

10 Frequency (rad/sec)

10

1

Phase (deg)

0

-100

-200

-300 -1 10

0

10 Frequency (rad/sec)

10

1

6023

Bode Plot for 6.6(a) 0

Magnitude

10

10

-5

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

-150

-200

-250

-300 -1 10

10

0

10 ω (rad/sec)

(d) L(s) =

(s + 3) s2 (s + 10)

(e) L(s) =

(s + 3) s3 (s + 5)

(f) L(s) = (g) L(s) =

(s + 1)2 s3 (s + 10) (s + 1)2 s3 (s + 10)2

Solution: (a) L(s) = (b) L(s) = (c) L(s) =

s2 s3 s4

1 2 s 2 +1 1 8 s 8 +1 1 10 s 10 + 1

1

10

2

6024

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Bode Plot for 6.6(b) 0

Magnitude

10

10

-10

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

-250 -300 -350 -400 10

-1

10

0

10 Frequency (rad/sec)

1

10

2

6025

Bode Plot for 6.6(c)

Magnitude

10

10

10

0

-5

-10

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase + 360 (deg)

50

0

-50

-100 -1 10

10

0

10 ω (rad/sec)

1

10

2

6026

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

10

Bode Plot for 6.6(d)

2

0

-2

-4

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

-50

-100

-150

-200 -1 10

10

0

10

1

ω(rad/sec)

s 3 +1 s 10 + 1 3 s +1 L(s) = 53 3s s 5 +1 2 1 (s + 1) L(s) = 10 s s3 10 +1 2 1 (s + 1) L(s) = 100 2 s +1 s3 10

(d) L(s) = (e) (f) (g)

3 10 s2

7. Mixed real and complex poles Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions. After completing the hand sketches verify your result using Matlab. Turn in your hand sketches and the Matlab results on the same scales. (a) L(s) =

(s + 2) s(s + 10)(s2 + 2s + 2)

10

2

6027

Magnitude

Bode Plot for 6.6(e)

10

10

0

-5

10

-1

10

0

10

1

10

2

ω (rad/sec)

Phase (deg)

-150

-200

-250

-300 -1 10

10

0

10 ω (rad/sec)

1

10

2

6028

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

Bode Plot for 6.6(f)

5

0

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

-50 Phase (deg)

-100 -150 -200 -250 -300 -2 10

10

-1

0

10 Frequency (rad/sec)

10

1

10

2

6029

Magnitude

Bode plot for 6.6(g)

10

10

0

-5

10

-2

10

-1

0

10 ω(rad/sec)

10

1

10

2

-50 Phase (deg)

-100 -150 -200 -250 -300 -2 10

10

-1

0

10 ω (rad/sec)

10

1

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD (b) L(s) = (c) L(s) = (d) L(s) = (e) L(s) =

(s + 2) s2 (s + 10)(s2 + 6s + 25) (s + 2)2 s2 (s + 10)(s2 + 6s + 25) (s + 2)(s2 + 4s + 68) s2 (s + 10)(s2 + 4s + 85) [(s + 1)2 + 1] s2 (s + 2)(s + 3)

Solution: 1 10

(a) L(s) = s

s 10 + 1

s 2 +1 2 ps +s+1 2

Magnitude

Bode plot for Prob. 6.7 (a)

10

10

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

-50 Phase (deg)

6030

-100 -150 -200 -250 -300 10

(b) L(s) = s2

-2

10

-1

1 s 125h 2 + 1 i s 2 6 s + 1 + s + 1 10 5 25

0

10 ω (rad/sec)

10

1

10

2

6031

Magnitude

Bode plot for Prob. 6.7 (b)

10

10

10

0

-5

-10

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

Phase (deg)

-100 -200 -300 -400 10

(c) L(s) = s2

-2

10

-1

2 2 s 125 h 2 + 1 i s s 2 6 + 25 s+1 10 + 1 5

0

10 ω (rad/sec)

10

1

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.7 (c)

10

10

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

0 Phase (deg)

6032

-100 -200 -300 10

-2

10

4 25

s 2 +1

ps 2 17

s2

s 10 + 1

ps 85

(d) L(s) =

2

2

-1

1 + 17 s+1 4 + 85 s+1

0

10 ω (rad/sec)

10

1

10

2

6033

Magnitude

10

10

10

Bode plot for Prob. 6.7 (d)

5

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

Phase (deg)

100 0 -100 -200 10

(e) L(s) =

-2

10

2

1 3

ps 2

s2

s 2 +1

+s+1 s 3 +1

-1

0

10 ω (rad/sec)

10

1

10

2

6034

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

10

Bode plot for Prob. 6.7 (e)

5

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

Phase (deg)

0 -50 -100 -150 -200 10

-2

10

-1

0

10 ω (rad/sec)

10

1

8. Right half plane poles and zeros Sketch the asymptotes of the Bode plot magnitude and phase for each of the following open-loop transfer functions.

Make sure the phase asymptotes properly take the RHP singularity into account by sketching the complex plane to see how the \L(s) changes as s goes from 0 to +j1: After completing the hand sketches verify your result using MATLAB. Turn in your hand sketches and the MATLAB results on the same scales. s+2 1 ; The model for a case of magnetic levitation s + 10 s2 4 with lead compensation. s+2 1 (b) L(s) = ; The magnetic levitation system with ins(s + 10) (s2 1) tegral control and lead compensation. s 1 (c) L(s) = s2 s2 + 2s + 1 (d) L(s) = s(s + 20)2 (s2 2s + 2) (a) L(s) =

10

2

6035 (e) L(s) =

(f) L(s) =

(s + 2) s(s 1)(s + 6)2 1 1)[(s + 2)2 + 3]

(s

Solution:

1 s +1 1 (a) L(s) = 20 s 2 ( 10 + 1) [( 2s )2

1]

Bode plot for Prob. 6.8 (a)

Magnitude

10

0

10

-5

10

-1

10

0

10

1

10

2

Phase (deg)

ω (rad/sec)

-100 -150 -200

10

-1

10

0

10 ω (rad/sec)

1 s +1 1 (b) L(s) = 5 2 s(s + 10) s2 1

1

10

2

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.8 (b)

10

10

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

-150 Phase (deg)

6036

-200 -250 -300 10

(c) L(s) =

-2

s

1 s2

10

-1

0

10 ω (rad/sec)

10

1

10

2

6037

Magnitude

10

10

Bode plot for Prob. 6.8 (c)

5

0

10

-2

10

-1

10

0

10

1

ω (rad/sec)

Phase (deg)

50 0 -50 -100 -150 -2 10

10

-1

10 ω (rad/sec)

(d) L(s) = s

1 2 40 (s + 2s + 1) 2 2 s ps s+1 20 + 1 2

0

10

1

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

10

10

Bode plot for Prob 6.8(d)

0

-5

10

-2

10

-1

0

10 Frequency (rad/sec)

10

1

10

2

300 200

Phase (deg)

6038

100 0 -100 10

-2

1 18

(e) L(s) = s(s

10

s 2 +1 2 1) 6s + 1

-1

0

10 Frequency (rad/sec)

10

1

10

2

6039

Magnitude

Bode plot for Prob. 6.8 (e)

10

10

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

Phase (deg)

-100 -150 -200 -250 -300 10

-2

10

-1

1 7

(f) L(s) = (s

1)

ps 7

2

+ 47 s + 1

0

10 ω (rad/sec)

10

1

10

2

6040

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD

Magnitude

Bode plot for Prob. 6.8 (f)

10

10

0

-5

10

-2

10

-1

0

10 ω (rad/sec)

10

1

10

2

Phase (deg)

-100 -150 -200 -250 -300 10

-2

10

-1

0

10 ω (rad/sec)

10

1

9. A certain system is represented by the asymptotic Bode diagram shown in Fig. 6.85. Find and sketch the response of this system to a unit step input (assuming zero initial conditions).

Fig. 6.85 Magnitude portion of Bode plot for Problem 6.9

10

2

6041 Solution: By inspection, the given asymptotic Bode plot is from Therefore, G(s) =

10(s=10 + 1) s + 10 = s s

The response to a unit step input is : Y (s)

= G(s)U (s) s + 10 1 1 10 = = + 2 s s s s y(t) = $ 1 [Y (s)] = 1(t) + 10t (t 0)

Prob. 6.9: Unit Step Response 25

20

y(t)

15

10

5

0

0

0.5

10. Prove that a magnitude slope of per decade or -6 db per octave.

Solution:

1 Time (sec)

1.5

1 in a Bode plot corresponds to

2

20 db

6042

CHAPTER 6. THE FREQUENCY-RESPONSE DESIGN METHOD The de…nition of db is db = 20 log jGj

(1)

Assume slope = d(logjGj) d(log !) =

(2)

(2) =) log jGj =

1

log ! + c (c is a constant.)

(1) and (3) =) db =

(3)

20 log ! + 20c

Di¤erentiating this, d (db) = d (log !)

20

Thus, a magnitude slope of -1 corresponds to -20 db per decade.

Similarly, d (db) d (db) = d (log2 !) d log !

+

6

log 2

Thus, a magnitude slope of -1 corresponds to -6 db per octave.

11. A normalized second-order system with a damping ratio additional zero is given by G(s) =

s=a + 1 s2 + s + 1

= 0:5 and an

:

Use MATLAB to compare the Mp from the step response of the system for a = 0:01; 0:1, 1, 10, and 100 with the Mr from the frequency response of each case. Is there a correlation between Mr and Mp ?

Solution: 0:01 0:1 1 10 100

Resonant peak, Mr 98:8 9:93 1:46 1:16 1:15

Overshoot, Mp 54:1 4:94 0:30 0:16 0:16

As is reduced, the resonant peak in frequency response increases.

This leads us to expect extra peak overshoot in transient response. This e¤ect is signi…cant in case of = 0:01; 0:1; 1, while the resonant peak in frequency response is hardly changed in case of = 10. Thus, we do not have considerable change in peak overshoot in transient response for 10.

The response peak in frequency response and the peak overshoot in transient response are correlated.

6043

Unit Step Response 60

50 a=0.01 40

y(t)

30

20

10

a=0.1 a=1

a=100

0

-10

a=10

0

2

4

6 Time (sec)

8

10