ch-8 bode plot slides

7/31/2019 Ch-8 Bode Plot Slides

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Chapter 8


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Input is sinusoidal

Steady state output will be sinusoidal with

same frequency for linear systems

amplitude and phase will vary with

frequency


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Example :

1s

1

(s)G ++++====

r (t) = sin

t


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In steady state

)-t(sin

1

1(t)C2

+

=

where = tan 1

tsin1tcos-e)(1

t-2

++

C (t) =


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Examine

G (s) with s = j

1j

1)(jG

++++====

tan

1

1 -1

2

++++

====


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Hence when input isSin

t

)t(sin)j(G

statesteadyinputout

++++====)j(Gwhere =


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Hence we plotmagnitude and phase

angle of as a

function of frequency

G ( j )

- frequency response

- sinusoidal transfer function.

put s = j in the transfer function


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Bode plot

magnitude as a

function of frequency phase angle as a

function of frequency


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Polar plot

For a particularfrequency, mark ( r, )on the polar graph,

Where r is magnitude & is phase angle.


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Example

1j

1

)(jG ++++====

1

1

)(M 2++++==== ( ) = - tan

1


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= 0 M = 1 = 0

= M = 0 = - 90

= 1 M = 21

= - 45


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Polar plot of 1s

1

(s)G ++++====

. .

==== 0====

6K-

Polar plot of


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Bode plot

)(je)(jG)j(G ====

)(j434.0)(jGlog +=


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graphsemilogonlogversus)(and)(jGlog20plot

scalelinearmagnitude

scalelog

.dbordecibelin)j(Glog20


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Ex:

Tj1

1)j(G

++++====

tan-

1

1

1-

22

++++====


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(((( )))) 21-22 1log20)j(Glog20

++++====

)1(log10-22

++++====


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for frequencies

,1


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For high frequencies

,1 >>>>>>>>

)T(gol01-

)j(Ggol0222

====


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Tgol02-====Tgol02-gol02- =

Value is zero db

,T1At ====


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. .

.

0

db

-20

0 db line

T1====

T10====

log


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The db magnitude plot for ,1 >>>>>>>>

is a straight line with -20 per unit change in log

This slop is - 20 db/decade or -6 db / octave

The straight line approximation is

called asymptotic plot Two asymptotesmeet at the corner frequency 1 / T.


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2log10-error

,T

1at

====

====

= - 3 db

1log10)411(log10-

T21aterror

++=

=

= - 1 db

To get actual plot,

T10for


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1log20(2)log10-error

,T

1at

+=

=

= - 3 db

2log204)1(log10-error

,T2at

++=

=

= - 1 db


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Ex:Bode plot for

)1s(s

)2s(5(s)G ++++

++++====


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1)(jj

2)(j5)j(G

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

)1j(1j

)2j(110

++++++++

====


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10 20 log 10

= 20 db

constant for allfrequencies


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log20-

1

log20j

1

====


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It is a straight line

with - 20 db/decade

slope ; passes

through 0 db line

for= 1


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rad/sec2frequencycorner)2j1(

====++++

decade/db20,2

db0,2

++++>>>>>>>>


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rad/sec1frequencycorner

)j(1

1

====

++++

decade/db20-,1

db0,1

>>>>>>>>


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db

20

log


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db

.log

1

- 20 db/decade


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db

. .

log

2

+ 20 db/decade

0 db


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db

.log

1

- 20 db/decade

0 db


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20

log

db

.

.

12

- 20 db/decade

- 20 db/decade

- 40 db/decade


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0

log

32 db.

.

1

+12

5

6

. . .

.

0.5

-6

Slopes in db/octave

db


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Corner frequencies :0.5 , 1, 5

+ 12 db / octave : s2

)0.5/s(1

1

:octave/db6 ++++++++


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)s1(

1:db0

++++

)5/s1(

1

:db6- ++++


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..

0.5 1

26 3212

6

38

log

db


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20 log K = 38K = 79.8

)s0.21()s1()2s1(

s8.97

(s)G

2

++++++++++++

====


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Gain margin & phasemargin :

..

>

. - a

- 1


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Gain margin =

a

1

for stable system a


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Phase margin()

=G(j)H(j) + 180

phase margin is + ve for

stable system

= 1

G i C O F


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log180-

0

db

PM

GH

log

GM

Gain Cross Over Freq.

Phase Cross Over Freq.


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Put s = j to get

the sinusoidal transfer function

)s2s()s(R

)s(C

2nn

2

2

n

++

=

Frequency Response of Second Order

Under damped Systems

2


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u

n

=

Putnormalised driving

Signal frequency

2

nn

2

2

n

)(j2)(j

)R(j)C(j ++=

u2ju1

12

+

=

2/122

])u2()u1[(

1M

+

=


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]u1

u2[tan2

1

=

u = 0 M = 1 = 0

u = 1 M = 1/2 = -/2u = M = 0 = -


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( )

uuu

u

for

u

u

log40log201

01

4)u-(1log20

)j(Glog20

2

212222

=>>


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Correlation between time &

frequency response:

Steady state output

for r(t) = Sin t

c(t) = M Sin(t+)

0dM

will give resonant Frequency where M is max


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2r

2nr

2r

121M

21

21u

====

====

====

Whereur : Normalised resonant freq

r : resonant freqM : resonant peak value

0

du

= will give resonant Frequency where M is max.


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M

Mr

1.0

0.707

0 r c

Bandwidth

Typical magnification curve of a feedback control system.

where


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r resonant frequencyc cut off frequency

Bandwidth

where

The range of frequencies over which M 1/2

2


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0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0.2 0.6 1 1.4 1.8

Normalized

bandwidth

ub

Bandwidth versus damping factor

0.707

Time domain


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2nd

d

p

1

p

1

t

eM

2

====

====

====

Frequency domain

2

nr

2

r

21

12

1M

====

====

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