bode plot

UNIT - III

UNIT - IIIFrequency Domain Analysis

Frequency Domain Plots are

1. Bode plot

2. Polar plot

Input is SINUSOIDAL SIGNAL

BODE PLOT db Magnitude Vs Frequencyand

Phase angle Vs Frequency

APPLICATION OF BODE PLOT

to analyze the stability of control systems in frequency domain.

to design closed loop control systems in frequency domain.

Bode plot is drawn using GH function.

Consider the general form of the transfer function

)2.(..))...(...()2.(..))...(...()()( 22

21

2221

nnnN

nnmM

SSPSPSSSSZSZSKSSHSG

++++++++=

21 1

1

21

1

2( ) 1 1( )( )( ) 21 1

k k

n n

jj jK j zNHD jj j

p

+ + + = = + + +

K

K

Bode plot of G(S) = K (Constant factor)Substitute S = j

G(j) = K

magnitude KjGM == )( db magnitude KMA log20log20 ==

A

Phase angle 00tan 1 === Kreal

img

Single Integral factor G(S) = K/S

Substitute S = j

G(j) = K/j

magnitude KjGM == )( db magnitude

KMA log20log20 ==

Phase angle o90tan0/tantan 111 ==== K

realimg

Substitute =0.1K dbK

KA 201.0

log20 ==

dbKKA 0log20 ==Substitute =K

dbKKA 20

10log20 ==Substitute =10K

-20 db/decade

G(S) = K/S

Single first order factor G(S) = 1/(1+TS)

Substitute S = j

G(j) = 1/(1+jT)

magnitude221

1)(T

jGM +==

db magnitude 2222

1log201

1log20log20 TT

MA +=+==

at low frequencies 1

TTA log20log20 22 ==

Phase angle TTrealimg 111 tan

1tantan ===

Substitute =1/T dbA 01log20 ==

dbA 2010log20 ==Substitute =10/T

Substitute =0 o00tan 1 == Substitute =1/T o451tan 1 == Substitute = o90tan 1 ==

=1/T is called as corner frequency. It is the frequency at which asymptotes meet.

=1/T

G(S) = 1/(1+TS)

PROCEDURE FOR DRAWING BODE PLOT

Step 1: Convert the given transfer function to standard form (Time constant form) and Substitute K=1 and S=j.

PROCEDURE FOR MAGNITUDE PLOT

Step 2: Calculate corner frequencies (c), slope, and change in slope

where corner frequency is the reciprocal of coefficient of S term in standard form of transfer function

Change in SlopeChange in Slopeinin

db/decadedb/decade

SlopeSlopeinin

db/decadedb/decade

Corner Corner frequencyfrequencyRadRad/sec/sec

FactorsFactors

Write the factors in the increasing order of the corner frequencies

First factor is K (or) K/(j)n (or) K(j)n

Step 3: Choose two arbitrary frequencies (l and h), one decade less than the lowest corner frequency and one decade greater than the highest corner frequency.

Write the frequencies in the increasing order

Step 4: Calculate db magnitude at first two frequencies using first factor

Step 5: Calculate db magnitude at 3rd and higher frequencies using the following formula

x

yyxxy tofromslopeatGainatGain

log

+=

Step 6: Tabulate frequency and db magnitude values. Mark the values in semilog graph sheet and join the points by straight lines.

db magnitudedb magnitudeFrequencyFrequencyIn In radrad/sec/sec

Write the phase angle function of the given transfer function and calculate phase angle values at different frequencies. Tabulate frequency and phase angle values. Mark the values in semilog graph sheet and join the points by a smooth curve.

PROCEDURE FOR PHASE ANGLE PLOT

Phase anglePhase angle

FrequencyFrequencyIn In radrad/sec/sec

Frequency Domain Specifications1. Resonant Peak (Mr): Maximum value of the magnitude of closed loop

transfer function.2. Resonant frequency (r): The frequency at which resonant peak

occurs.3. Bandwidth (b): Range of frequencies for which gain of the system is

more than -3db.4. Gain Margin (Kg): This the gain value by which system gain can be

increased, beyond which system is unstable.

)(1

pcg jG

K =

)(1log20

pcg jG

K =in db

Where pc is Phase cross over frequency, the frequency at which phase angle is 180.

5. Phase Margin (): This the phase angle value by which system phase angle can be increased, beyond which system is unstable.

gc += o180Where gc is Gain cross over frequency, the frequency at which gain is 0 db.

)( gcgc jG =where

Resonant Peak

Resonant frequency

Bandwidth

)5)(10(250)( ++= SSS

KSG

1. The open loop transfer function of certain unity feedback control system is given by

Draw the bode plot and determine gain margin and phase margin.

Determine the value of K for the desired specifications

a) Gain margin = 20 db

b) Phase margin = 45

gc

pc

)104()5(7)( 2 ++

+=SSS

SSG







CALCULATION FOR MAGNITUDE PLOT

)101

1041(*10

)511(7*5

)(2SSS

SSG

+++

=

)1.04.01)(()2.01(35.0)( 2

+

+=jj

jSG

)104()5(7)( 2 ++

+=SSS

SSG

transfer function in standard form (Time constant form)

Substitute K=1 and S=j


c1 = n= 10 =3.16 rad/sec (for 2nd order factor)

and c2=5 rad/sec

----

--6060

--4040

--2020

--4040

2020

----

3.163.16

55

Change in Change in SlopeSlope

inindb/decadedb/decade

SlopeSlopeinin

db/decadedb/decade


FactorsFactors

j35.0

21.04.011

+ j2.01 j+

Step 3: l = 0.3 and h = 50


l = 0.3, c1 =3.16, c2=5 and h = 50


1

22112 log

c

ccccc tofromslopeatGainatGain

+=

dbAatGain l 0)3.03.0log(20)3.0log(201 ====

dbAatGain c 45.20)16.33.0log(20)3.0log(2021 ====

Step 5: Calculate db magnitude at c2=5 using the following formula

[ ]16.35log6045.2032 +== AatGain c = -32.407 db

00

--20.4520.45

--32.40732.407

--68.40768.407

0.30.3

3.163.16

55

5050



222 log

c

hhcch tofromslopeatGainatGain

+=

Step 6: Calculate db magnitude at h=50 using the following formula

[ ]550log40407.324 +== AatGain h = -68.407 db

CALCULATION FOR PHASE ANGLE PLOT

--(180(180 + ) + )

--(180(180 + )+ )

--(180(180 + )+ )

--(180(180 + )+ )

--93.47 93.47

--190.6 190.6

0.30.3

22

n = 3.16n = 3.16

44

55

2525

5050



211

1.014.0tan2.0tan90)( +==

ojG For For nn

)1.01

4.0tan180(2.0tan90)( 211

++==

oojG For For >>nn

2.0tan 1 21 1.014.0tan






2

KG(S)=S(S+2)

2

KG(S)=S (S+1.5)


Determine the value of K for the new gain crossover frequency to be 2 rad/sec.



CALCULATION FOR MAGNITUDE PLOT

2)211(4

)(SS

KSG+

=

2)5.01(25.0)( jjjG +=

2)2()( += SS

KSG

transfer function in standard form (Time constant form)

Substitute K=1 and S=j


c = 2 rad/sec

----

--6060

--2020

--4040

----

22

Change in Change in SlopeSlope

inindb/decadedb/decade

SlopeSlopeinin

db/decadedb/decade


FactorsFactors

j25.0

2)5.01(1

j+

Step 3: l = 0.2 and h = 20


l = 0.2, c =2, and h = 20


dbAatGain l 463.4)2.025.0log(20)25.0log(201 ====

dbAatGain c 589.41)225.0log(20)25.0log(202 ====

4.4634.463

--41.58941.589

--179.74179.74

0.20.2

22

2020



c

hhcch tofromslopeatGainatGain

log

+=

Step 5: Calculate db magnitude at h=20 using the following formula

[ ]2

20log60589.413 +== AatGain h = -179.74 db

CALCULATION FOR PHASE ANGLE PLOT

--91.42191.421

--258.58 258.58

0.20.2

11

22

55

1010

2020



5.0tan*290)( 1== ojG

5.0tan*2 1

POLAR PLOT

It is a plot between magnitude and phase angle of

G(S)H(S) in polar graph when is varied from 0 to .

It is a plot between real and imaginary part of G(S)H(S) in

rectangular graph when is varied from 0 to .

APPLICATION OF POLAR PLOT


to design closed loop control systems in frequency domain.

S-plane

j

0 to

PROCEDURE FOR CONSTRUCTING POLAR PLOT

Convert the transfer function to TIME CONSTANT FORM

Substitute S = j and K = 1

Write magnitude and phase angle of the given transfer function

Choose arbitrary frequencies near corner frequencies

Tabulate frequency, magnitude and phase angle values

Mark the values in polar graph sheet

To use rectangular graph sheet convert magnitude and phase angle to real and imaginary values

1.The open loop transfer function of a unity feedback system is given by

1( )( 1)(2 1)

G SS S S

= + +Draw the polar plot and determine gain margin and phase margin.




)21)(1(1)(

SSSSG ++=

The given transfer function is in time constant form

Substitute S = j

)12)(1(1)( ++= jjjjG

Corner frequencies are c1 = 0.5 and c2 = 1

22 4111

++=MMagnitude

Phase angle 2tantan90 11 = o

--270270--180180--9090Phase angle Phase angle

in degin deg

000.660.66MagnitudeMagnitude

0.7070.70700FrequencyFrequencyRadRad/sec/sec

)()( jHjG

)()( jHjG

0000--ImagImag PartPart

00--0.660.6600Real PartReal Part

0.7070.70700FrequencyFrequencyRadRad/sec/sec

)]()(Re[ jHjG

)]()(Im[ jHjG

0.66G)G(j Bpc ==pc

0

gc

Unit circle

0.0063G)G(j Bpc ==pc

0

gc=91

Unit circle

)(1log20

pcg jG

K =

Gain margin

66.01log20= db52.3=

Phase margino4.11=

Gain and Phase margin values are positive, hence the given closed system is stable.

S-plane

j

0 to

GH plane

Real part of GH

I

m

a

g

p

a

r

t

o

f

G

H

GH plot

NYQUIST PLOTAPPLICATION OF Nyqiust plot


Stability of the closed loop system is determined from the location of poles of G(S)H(S) and the G(S)H(S) contour.

NYQUIST STABILITY CRITERIONfor a stable closed loop system, GH contour should

encircle -1+j0 as many times as number of right half open loop (GH) poles in anticlockwise direction.

for no right half open loop (GH) poles, GH contour should not encircle -1+j0 in anticlockwise direction.

If GH contour encircles in clockwise direction, the closed loop system is unstable and the numder of encirclements in clockwise direction is equal to number of closed loop poles on right half of s-plane.

Section C1:s=j

Polar plot 0 to +

Section C2:s=Rej

where R is from +/2 to- /2

Section C3:s=-j

Inverse Polar plot 0 to -

Section C4:s=Rej

where R 0 is from -/2 to +/2

To draw GH contour, in s-plane a contour is chosen, which has four sections.

C4X

If this s-plane contour is mapped (substituting) to GH plane, GH contour is obtained.

1.The open loop transfer function of a unity feedback system is given by

Draw the Nyquist plot and determine the stability of the given closed loop system.

)21)(1()(

SSSKSG ++=

Step 1: mapping of section C1

Soln: Convert the given transfer function to time constant form.

Number of right half open loop poles = 0.

Therefore if the given closed loop system is to be stable, GH contour should not encircle -1+j0.

C4XSubstitute S = j 0 to +

0

Step 2: mapping of section C2

s=Rejwhere R is from +/2 to -/2)21)(1(

)(SSS

KSG ++=

)1)(1()( jjj eee

KSG ++=

30..

)( jjjj eeeeKSG ==

In s-plane is from +/2 to- /2

In GH plane is from -3/2 to +3/2

Step 3: mapping of section C3 s=-j

Inverse Polar plot 0 to -

-

0

Step 4: mapping of section C4 s=Rejwhere R 0 is from -/2 to-+/2

)21)(1()(

SSSKSG ++=

)01)(01(0)( jjj eee

KSG ++=

jj ee

KSG ==0

)(

In s-plane is from -/2 to +/2

In GH plane is from +/2 to -/2

R

-0.66K = -1

K = 1.5

-0.66K

GH plane

GH contour

Real[G(s)H(s)]

I

m

a

g

[

G

(

s

)

H

(

s

)

]

when K1.5, GH contour encircles -1+j0 two times along clockwise direction. Therefore, closed loop system is unstable.

Hence two closed loop poles on right half of s-plane.

)2()( += SS

KSG

2. The open loop transfer function of a unity feedback system is given by

Draw the Nyquist plot and determine the stability of the given closed loop system.

Polar plot is not intersecting -180 line. Therefore, the closed loop system is stable for all values of K from 0 to .

bode plot

Documents

db magnitude values

tabulate frequency

phase angle function

phase angle values

lowest corner frequency

highest corner frequency

slopewhere corner frequency

db magnitude kma log20log20