Control Systems (K-Wiki_Frequency Response Analysis).pdf

Control Systems (Frequency Response Analysis)
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Frequency Response Analysis
Objective
Upon completion of this chapter you will be able to:
 Determine the output of a system for a sinusoidal input.
 Determine Frequency Response Specifications like Resonance Peak, Resonance
Frequency and Bandwidth.
 Draw the Polar Plot of a closed loop system.
 Draw the Nyquist Plot of a system to judge stability using Nyquist Stability Criteria.
 Draw Bode Plot for Magnitude and Phase of a Transfer Function.
 Determine relative stability of a system in terms of Gain Margin and Phase Margin.
Introduction
When any system is subjected to sinusoidal input, the output is also sinusoidal having
different magnitude & phase angle but same frequency ω rad/sec.
Frequency response analysis implies varying ω from 0 to  & observing corresponding
variations in the magnitude & phase angle of the response.
For system with very large time constants, the frequency test is cumbersome to perform as
the time required for the outputs to reach steady state for each frequency of test signal is
excessively long.
Therefore, frequency response analysis is not recommended for system with large time
constant.
Advantages of Frequency Response Analysis
1) Analysis is easy even for higher order systems as there exists a number of plots which are
not limited by size of systems.
2) Time domain approaches like Routh criterion establish stability of system easily and
relative stability is very difficult to determine but relative stability is much easier to determine
using frequency domain analysis.
3) Systems noise sensitivity can be analyzed with frequency response.

Sinusoidal Response of a System
Let  
 
 
C s
F s Transfer Function
R s
 
Put s = jω
F(jω) = Sinusoidal Transfer Function
     
F j F j F j
    
  2 2
F j Re Im
  
  Im
1
F j tan
Re
 

    
 
Where Re = Real part of F(jω) and Im = Imaginary part of F(jω)
Solved Examples
Problem: Find the system response c(t) if it is excited by r(t) = sin2t.
Solution:    
c t Msin 2t
  
1 1
T( j )
1 s 1 j
s j
  
  
 
1
M
2
1

 
1
tan ( )

   
  1 1
c t sin 2t tan
2
1
 

  
 
 
 
Here ω = 2
   
c t 0.44sin 2t 63.43
 

Second order systems
 
 
2
C s n
2 2
R s s 2 s
n n


   
Where  = damping factor
n
 = natural frequency
 
 
 
   
2
C j n
T j
2
R j 2
j 2 j
n n


  

       
1
2
1 u j2 u

  
Where u
n



 
 
1
T j M
2 2
2
1 u 2 u
  
 
  
 
 
  2 u
1
T j tan
2
1 u
 
 
 
 
 
 
 
 
 


   

Frequency Domain Specifications
1. Resonance Frequency( r
 )
It is the frequency at which magnitude of sinusoidal Transfer Function has maximum value
2
r n
1 2 rad / sec
    
If r
 is to be real & positive
2
2 1
1
2
 
 
Which means Resonance is only exhibited if
1
2
  .

2. Resonant Peak  
r
M
It is maximum value of magnitude M occurring resonant frequency r
 .
r 2
1
M
2 1

  
The phase angle r
 at r
 is
2
1
r
1 2
tan
 
 
 
  

 
 
r
r
1
,M 1
2
1
,M 1
2
  
  
For
1
,
2
  the maximum value of magnitude occurs at zero frequency and it is equal to the
value of DC Gain.
3. Bandwidth (BW)
It is the range of frequencies over which the magnitude has a value of
1
2
, it indicates the
speed of response of the system.
Wider BandwidthFaster response
r
1
BW
t
 where r
t =rise time
4. Cut-off Frequency
It is the frequency at which the magnitude has a value of
1
2
. It indicates the ability of
system to distinguish signal from noise.
2 4 2
BW or w w 1 2 4 4 2 rad/ sec
0 n
       

Solved Examples
Problem: An under damped second order system having a transfer function of the form.
 
2
k
n
T s
2 2
s 2 s
n n


   
. Has a frequency response plot shown below.
System gain, k is ____________
Solution: Magnitude of Sinusoidal Transfer Function is
 
 
2
k
n
T j
2 2
2 2 2
n n

 
 
     
 
 
At ω = 0;  
T j 1
 
2
k
n
1 k
2
n

 

=> k=1
Problem: The damping factor  is approximately for the previous problem.
Solution: The resonant peak is given by:
1
M
r 2
2 1

  
=> 1
2.5
2
2 1

  
 
2 2
0.04 1
    =>
4 2 0.04 0
    
0.2
 

Frequency domain approximation of dead time
 
 
 
Y s sT
F s e
X s

 
  j T
F j e cos t jsin T
 
     
     
2 2
F j cos T sin T 1
     
   
sin T
1 1
F j tan tan tan T
cos T
 
 
 
     
 

 
 
T radians
 
 
j T
e 1 T radians
 
  
Polar Plot
It is a plot of absolute values of magnitude & phase angle in degrees of open loop transfer
function G(jω)H(jω) Vs ω drawn on polar co–ordinates.
1. Suppose we consider a Type-0 and Order-1 system for drawing Polar Plot.
    1
G s H s
1 sT


    1
G j H j
1 j T
  
 
    1
G j H j
2 2
1 T
  
 
 
1
GH tan T

   
The value of magnitude and phase for different values of frequency are shown in table below
and Polar Plot is shown in figure below:

GH GH
0
0 1 0
1 1 0
45
T 2
0
0 90
 

 
Since the phase is always negative for all values of frequency and it varies between 0 and -
900
so Polar Plot lies in 4th
Quadrant.
The point corresponding to 0
  is called as Tail of Polar Plot and point corresponding to
is called as Head of the Polar Plot.
2. If we add a pole at origin then the system becomes Type-1 and Order-2.
   
 
1
G s H s
s 1 sT


   
 
1
G j H j
j 1 j T
  
  
    1
G s H s
2 2
1 T

  
0 1
GH 90 tan T

    
Conclusion: When a pole is added at origin, the head and tail of polar plot shift by
0
90 in
clockwise direction.

3. If a pole is added in a Transfer Function away from origin, then it becomes Type-0 Order-2
System.
   
  
1
G s H s
1 sT 1 sT
1 2

 
    1
G s H s
2 2 2 2
1 T 1 T
1 2

  
   
  
  
     
1 1
G j H(j ) tan T tan T
1 2
 
       
If    
1 1 0
tan T tan T 90
1 2
 
     
T T
1 0
1 2
tan 90
2
1 T T
1 2
 
  
 
 
 
 
 
 
1
T T
1 2
 
At this value of Frequency, the Polar Plot intersects the negative imaginary axis.
Conclusion: When a pole on real axis is added, head of polar plot shift by
0
90 in clockwise
direction, and tail remain the same.
Also, from the above examples it is clear that tail of Polar Plot is determined by the Type of
the system and Head of Polar Plot is determined by Order of the system.
So, for Order-2 system the head is always oriented at -1800
irrespective of the Type of
system.

So, based on Type and Order Polar Plots would lie in following quadrants:
General shapes of polar plots
Type/order Polar plot
   1
Type 0 / order 1, G s
1 sT


  
  
1
1 T s 1 T s
1 2

 
  
   
1
1 T s 1 T s 1 T s
1 2 3

  

   1 1 1 0
Type1 / order 1, G s 90
s j
    
 
 Type 0 / order 4
 
    
1
G s
1 T s 1 T s 1 T s 1 T s
1 2 3 4

   
  
 
1
s 1 T s
1


  
  
1
Type1 / order 3, G s
s 1 T s 1 T s
1 2

 

  
   
1
Type1 / order 4, G s
s 1 T s 1 T s 1 T s
1 2 3

  
  
 
1 1 1 0
Type 2 / order 2, G s 180
2 2 2
s j
    


  
 
1
2
s 1 Ts


  
  
1
2
s 1 sT 1 sT
1 2

 
   1 1 0
Type 3 / order 3, G s 270
3 3
s
   


  
 
1
3
s 1 Ts


Limitations of polar plot
1) Drawing poles plot involves numbers of calculations if it has multiple intersection with
co-ordinate axis.
2) Not feasible to find magnitude &  of transfer function at any desired frequency using
polar plot.
3) Stability of a system can be analyzed if the loop transfer function is of minimum phase
and there is no information about degree of unstability which means we cannot identify the
number of poles lying in the Right Half Plane.
Concept of enclosement and encirclement
 A point is said to be enclosed by a contour if it lies to the right side of direction of
contour.
 A point is said to be encircled if it lies inside a contour which is a closed path.
 In above figure, point ‘y’ is said to be enclosed whereas point x is said to be encircled in
anti – clockwise direction.

Nyquist Criteria
 Using Nyquist criteria, stability of non – minimum phase transfer function can also be
determined.
 Nyquist criteria uses GH to find roots of characteristics equation 1+GH = 0.
 Finding stability using Nyquist stability criteria involves drawing Nyquist plots which is
extension of polar plot.
Nyquist Path
 Nyquist path is a closed path which encloses the entire right half plane but does not pass
through any poles of G(s)H(s). Therefore, we draw an infinitesimal semi-circle around poles at
origin or on imaginary axis.
 Therefore, we can express Nyquist Path as a combination of four paths and we have to
map this path in G(s)H(s) plane.
 Since, G(s)H(s) is a single valued function which means for any value of ‘s’ there is single
value of G(s)H(s) and since we have a closed path in s-plane so we will get a closed path in
G(s)H(s) plane.
The mapping of Nyquist Path in G(s)H(s) plane is called as Nyquist Plot. To draw Nyquist Plot
we map each of the four paths from the Nyquist Path to G(s)H(s) plane.
 To map S
1
Put s j
  which leads to Polar Plot in G(s)H(s) plane.
 To map S
2
Put
R
j
s limRe


 ,
2 2
 
 
   
 
 To map S
3
Put s j
   which leads to mirror image of Polar Plot about real axis in G(s)H(s) plane.

 To map S
4
Put j
s lim Re 2 2
R 0
  
 
    
 

This mapping is illustrated using the example below:
 
  
10
G s
s 2 s 4

 
This is Type-0 Order-2 System and hence it has no pole at origin so there is no need to draw
a semi-circle around origin. Hence there is no S4 present and we need to only map S1, S2 and
S3. The Nyquist Path for this system is shown below:
For S s j
1
  , we will draw polar plot for Type-0 Order-2 System.
For  
j
S s Re R
2

   
 
   R
10 10
G s lim 0 (origin)
s 2 s 4 j j
Re 2 Re 4

  
  
   
 
  
  
For S s j
3
    , we will draw mirror image of polar plot
about real axis.
So, the Nyquist Plot looks like as shown:

Solved Examples
Problem: Plot Nyquist Plot for the system whose open loop Transfer Function is given by
 
 
10
G s
s s 2


Solution: In this case, the system is Type-1 Order-2 System and hence it has a pole at origin
and thus we need to draw a small semi-circle around origin.
The Nyquist Path for this system looks like:
For S s j
1
For  
j
S s Re R
2

   
 
  R
10 10
G s lim 0 (origin)
s s 2 j j
Re Re 2

  
 
  

 
 
For S s j
3
    , we will draw mirror image of polar plot about real axis.
For  
j
S s Re R 0
4

  
 
  1
R 0 R 0
R
10 10 10 5 j
G s lim lim lim e
j R
s s 2 j j Re x2
Re Re 2
 

 
   

 
  

 
 

Since for S
2 2
4
 
 
   
 
 
, so
2 2
 
 
   
 
 
which means mapping of S4 is a
semicircle of infinite radius whose orientation is clockwise which is apparent by limits of “–θ”.
So, the Nyquist Plot looks like as shown below:
Problem: Plot Nyquist Plot for the system whose open loop Transfer Function is given by
 
  
10
G s
s s 2 s 4

 
Solution: In this case, the system is Type-1 Order-3 System and hence it has a pole at origin
and thus we need to draw a small semi-circle around origin.
The Nyquist Path for this system looks like:
For S s j
1

For  
j
S s Re R
2

   
 
   R
10 10
G s lim 0 (origin)
s s 2 s 4 j j j
Re Re 2 Re 4

  
  
    
 
  
  
For S s j
3
    , we will draw mirror image of polar plot about real axis.
For  
j
S s Re R 0
4

  
 
   1
R 0 R 0
R
10 10 10 1.25 j
G s lim lim lim e
j R
s s 2 s 4 j j j Re x8
Re Re 2 Re 4
 

 
   

  
    
 
  
  
Since for S
2 2
4
 
 
   
 
 
, so
2 2
 
 
   
 
 
which means mapping of S4 is a
semicircle of infinite radius whose orientation is clockwise which is apparent by limits of “–θ”.
So, the Nyquist Plot looks like as shown below:
Note: Nyquist Plot = Polar plot + Mirror image of polar plot w.r.t. real axis + No. of infinite
large semicircle equal to type of GH (oriented clockwise).

Nyquist Stability Criteria
 In addition, to providing absolute stability like Routh Criteria, Nyquist Stability Criteria also
gives information on Relative Stability as well as degree of instability of unstable systems.
 It also gives an indication as to how system stability can be improved if needed. But unlike
Root-Locus Method, this method does not give exact location of roots of Characteristic
Equation.
 If there is any pole of the system lying inside the Nyquist Path, then there is one clockwise
encirclement of origin and if there is any zero enclosed by Nyquist Path, then there is one
counter-clockwise encirclement of origin.
 We wish to determine the roots of characteristic equation so the function that we can use
is “1+GH”. The poles of this system is same as the poles of GH function and zeroes of this
function gives the closed loop poles.
 Since, we are plotting Nyquist Plot on GH Plane so instead of looking at encirclements of
origin we can rather use (-1+j0) point because we are interested in finding the roots of
Characteristic Equation which is 1 GH 0
  , which implies GH 1
  , so to determine stability
by Nyquist Plot in GH plane, the critical point is -1+j0 point.
Nyquist Stability Criteria is given as:
N = P – Z
N = Number of encirclements of (-1 + j0) point
N is positive for anti – clockwise encirclements & negative for clockwise encirclements.
P = Number of open loop poles of G(s)H(s) in right half s – plane.
Z = Number of zeroes of 1 + G(s)H(s) = 0 or closed loop poles in RHP.
 An Open Loop System is stable if there are no open-loop poles lying in the Right Half
Plane which means P = 0.
 A closed loop system is stable if Z = 0. i.e. N = P and unstable if N P
 & degrees of
instability is given by Z P N
  .

If we consider the number of encirclements of origin in GH Plane then number of
encirclements of origin will depend on poles and zeroes of GH in Nyquist Path which is Right
Half Plane.
N = P – Z
N = Number of encirclements of origin
N is positive for anti – clockwise encirclements & negative for clockwise encirclements.
P = Number of open loop poles of G(s)H(s) in right half s – plane.
Z = Number of open loop zeroes in RHP.
Conclusion:
 There is no encirclement of (-1 + j0) point that is N=0 and P =Z. This implies that the
system is stable if there are no poles of G(s) H(s) in the right half s – plane otherwise system
is unstable.
 There is a counter – clockwise encirclement or encirclement of (-1 + j0) point which means
N>0. In this case system is stable if the number of counter clockwise encirclement is same as
number of poles of G(s)H(s) in RHP else system is unstable.
 There is clockwise encirclement or encirclement of the (-1 + j0) point which means N<0.
In this case system is unstable.
Note: The criteria for Stability of a system using Polar Plot is enclosement of -1+j0 point
which means if the point is enclosed by the Polar Plot then the system would be Unstable
and otherwise Stable but it does not tell about the degree of instability that is the number of
closed loop poles in the right half plane.
The effect of the value of gain on the stability of the system can be understood by
considering the Nyquist Plot of the system whose Transfer Function is given below:
   
  
k
G s H s
s T s 1 T s 1
1 2

 

 There are no Open Loop Poles in the Right Half Plane which means P = 0.
 For small k, There is no encirclement of (-1 + j0) point which means N=0 and since P=0
thus Z = 0 so there is no closed loop pole in the Right Half Plane.
 But for large K there are two encirclement in clockwise direction which means N = -2 and
since P = 0 so Z = 2 & thus two closed loop poles in RHP so system is unstable.
 For good accuracy we tend to increase k but it may lead to instability.
Effect of adding zeroes on shape of Nyquist & Polar Plot
Suppose, initially the Open Loop Transfer Function is given as :  
  
1
G s
2
s 1 s 1 2s

 
To draw Polar Plot of the system:
   
1 1
G j 180 tan tan 2
 
       
If     0
0, G j , G j 180
        
    0
, G j 0, G j 360
        
To map S
4
 
R 0
1 1 j2
G s e
2 j2
2
s limR e

 
   

, to
2 2
 
 
 
 
So there will be two infinite semi circles at s
4 will have phase from to
 .

So, the Nyquist Plot of the system looks like as shown below:
Suppose, now we add a zero to the Open Loop Transfer Function, then Transfer Function
becomes:
 
  
1 4s
G s
2
s 1 s 1 2s


 
     
0 1 1 1
G j 180 tan tan 2 tan 4
  
         
    0
0, G j G j 180
        
    0
, G j 0 G j 270
        
To find the intersection of Nyquist Plot with negative real axis, we can equate the phase of
Transfer Function to -1800
.
0 1 1 1 0
180 tan tan 2 tan 4 180
  
        
1 1 1
tan 4 tan tan 2
  
    
2
4
2
1 2
  
 
 
=>
1
2
4 8 3 rad s
8
     
2
4
1
8
G( j ) 10.6
2 2 2
1 1 2
1 1
8 8 8
 
  
 
  
     
 
     
     

So, by adding a zero in the Open Loop Transfer Function, the head of the polar plot shifts by
900
in the anticlockwise direction and also we get an intersection of the Nyquist Plot with the
negative real axis. So, number of intersections with the negative real axis is same as number
of zeroes in the Open Loop Transfer Function.
Systems with Dead Time or Transportation Lag
To determine stability for systems with dead time polar/Nyquist plots is not a good idea as
its cuts –ve real axis infinite times. Suppose we consider the Transfer Function mentioned
below:
 
 
sT
e
G s
s s 1



 
 
j T
e
G j
j 1 j
 
 
  
  0
0
T 1
G j T 90 tan
1
90 tan
 
       
 

  
   
 
To find intersections of Polar Plot with negative real axis:
At   0
G j 180
pc
      

0 0
1
T 90 tan 180
pc pc

     
0 1
90 tan
pc pc

    
0
tan 90 T cot T
pc pc pc
   
     
 
 
   
Solutions are infinite & not manually possible to draw Nyquist Plot and judge stability of the
system.
Solved Examples
Problem: The Nyquist plot of a type 1 system which has one right half of s-plane pole, for
0
  to 
   is given below. Discuss stability of Closed Loop System.

Solution: The Nyquist plot is given below
From the plot N= 1,
P = 1(Type 1 system)
Z=P – N= 1 – 1 = 0
Z = 0
system is stable
Problem: The Nyquist plot of a certain G(s)H(s) which has no right hand pole is given
below. Discuss Stability of Closed Loop System.
Solution: N = P – Z
Z = P – N = 0 – (- 1) = 1
So, closed loop system is unstable with one pole in the right half plane.
Relative Stability
 The curve below shows polar plots for different value of k. For a large value of k system is
unstable as (-1 + j0) point is enclosed by polar plot.
 The second curve just crosses (-1 + j0) points & so it is just verge of instability & so it
exhibits sustained oscillations. System in this condition is termed as Marginally Stable.
 Thus, closeness of polar plot to (-1 + j0) points corresponds to relative stability so it can
be used as a measure for relative stability.

 Relative Stability in Frequency domain can be analysed using Gain Margin and Phase
Margin.
Phase Margin
The phase margin is that amount of additional phase lag at the gain crossover frequency
required to bring the system on the verge of instability.
Gain Crossover frequency
 It is the frequency at which  
g
G j , the magnitude of open loop transfer function is
unity.
 The phase margin  is
0
180 plus the phase angle  of the open loop transfer function at
gain crossover frequency. 180
  

From above figure we can observe that positive phase margin represents a stable system and
a negative phase margin represents an unstable system.
Higher Phase Margin indicates that higher phase shift is required to make the system
unstable and thus the system is relatively more stable.

Gain Margin
The gain margin is the reciprocal of the magnitude  
G j at the phase crossover frequency.
Phase crossover frequency
The frequency at which the phase angle of the open loop transfer function equals
0
180

gives gain margin k
g
.
 
p
1
k
g G j


In terms of decibels,
 
k dB 20logk 20log G j
g g p
 
   
 
 
 The gain margin expressed in decibels is positive, if G j
p
 

 
 
is less than 1 and negative if
G j
p
 

 
 
is greater than 1.
 If the Gain of Transfer Function is greater than 1 at the point of intersection with Negative
Real axis then (-1+j0) point is enclosed by the Polar Plot and hence the system is unstable.
 If the Gain of Transfer Function is less than 1 at the point of intersection with Negative
Real axis then (-1+j0) point is not enclosed by the Polar Plot and hence the system is stable.
 Thus, positive gain margin indicates a stable system and negative gain margin indicates
an unstable system.
 Higher Gain Margin indicates that gain must be increased by a large factor to bring the
system on the verge of instability and thus system is relatively more stable.
 Gain Margin and Phase Margin can also be computed by Bode Plot for the system where
Gain is plotted in dB and Phase is plotted in degrees.
 In Bode Plot, we can identify the Gain Crossover frequency as the point where Gain is 0dB
and Phase Crossover frequency as the point where phase is -1800
.

 For a stable minimum phase system, the gain margin is indicates of how much gain can be
increased before the system becomes unstable.
 For an unstable system, it is indicative of how much the gain must be decreased to make the
system stable.
 The gain margin of second order system is infinite as the Polar Plot does not cross real axis.
So, theoretically first and second order system cannot be unstable. This is also obtained from
Routh Criteria where positiveness of coefficients of Characteristic Equation guarantees
stability of First and Second Order System.
 For non – minimum phase system, the stability condition will not be satisfied unless the G(ω)
plot encircle ( -1 + j0) point. Hence, a stable non – minimum phase system will have negative
gain and phase margins. So, Gain Margin and Phase Margin cannot be used to judge the
stability of Non-Minimum Phase systems.
 Conditionally stable systems will have two or more phase cross – over frequencies as shown
below and the points of intersection with negative real axis depend on the value of Gain, K.
By changing the points of intersection we can enclose -1+j0 point and also make it
unenclosed leading to change in stability of the system based on the value of K.

 If (-1+j0) point lies in region 1 or region 3 then it is enclosed by the Polar Plot and the
system is unstable but if it lies in region 2 then it is not enclosed by the Polar Plot and
system is unstable.
Solved Examples
Problem: The open loop transfer function of a unit feedback system is G(s) H(s) = 3
1
(1 s)

.
The Gain margin of the system is _____.
Solution: 3
1
G(j )H(j )
(j 1)
  
 
1
G(j )H(j ) 3tan
     
For phase crossover frequency  
pc
 ,
0
G(j )H(j ) 180
    
0 1
pc
180 3tan
    => 0
pc
tan60 3
   rad/s
 
pc
3
2
1
Gain G(j )H(j )
8
1
3 1

    
 
 
 
 

1
GM 8
Gain
 

Problem: The polar plot of a unity feedback system is shown below. Calculate the phase
margin and gain margin respectively.
Solution: Gain margin=
1 1
4
0.25
G(S)
 
Phase margin= 0 0 0
180 30 150
 
Problem: Common Data Question for 1 and 2
Nyquist plot of a stable transfer function G(s) is shown in the figure. We are interested in the
stability of the closed loop system in the feedback configuration shown.

1. Which of the following statements is true?
(a) G(s) is an all-pass filter
(b) G(s) has a zero in the right-half plane
(c) G(s) is the impedance of a passive network
(d) G(s) is marginally stable
Solution: There is one clockwise encirclement of origin so N = -1
Since N = P – Z, where P is number of Poles of Open Loop System in RHP and Z is number of
zeroes in RHP.
Since Open Loop System is stable, so there is no pole in RHP, P=0.
Since N=-1, thus Z =1 so there is a zero of Open Loop System in RHP.
2. The gain and phase margins of G(s) for closed loop stability are
Solution: When Nyquist Plot intersects negative real axis Gain = 0.5
Gain Margin =
1 1
2
Gain 0.5
 
In decibels,
GM = 20 log 2 = 6 dB
When magnitude or Gain is 1 then, phase is -900
PM = 180 + (-90) = 900
Categorization of Transfer Function
According to pole and zero locations, the transfer functions can be categorized as
 Minimum phase Transfer Function
 Non – minimum phase Transfer Function
 All pass Transfer Function

Difference between minimum & non – minimum phase Transfer Function
MPTF NMPTF
1) All poles and zeroes should lie in left half
plane.
At least one pole or zero must lie in right half
plane.
2) There exists a unique relation between
magnitude and phase of transfer function.
1
, 0
1 j
 
 
No unique relationship exists. That is more
than one possible phase for same magnitude.
1
, real
1 j
 
 
 
 
1
tan 0
1
M ,
1
2 2 tan 0
1
 
    

   

    
   
3) The total phase variations are minimum in
phase plot.
 
  
s
1 2
TF
s s
1 1
3 4


 
1 1 1
tan tan tan
2 3 4
  
  
   
0
0, 0,
, 90
   
     
The total phase variations are more in non –
minimum phase plot.
 
  
s
1
2
TF
s s
1 1
3 4


 
0 1 1 1
180 tan tan tan
2 3 3
  
  
    
0
0
0, 180 ,
, 90
   
     
4) Stability can be analyzed using Routh criteria,
Root locus, polar plots, bode & Nyquist plots
Stability can be analyzed using Nyquist or
Routh criteria.

All pass Transfer Function
In case of all pass transfer functions, all TF poles and zeroes are symmetric about imaginary
axis.
As for an example, if sinusoidal Transfer Function is   1 j T
G j
1 j T
 
 
 
, the Pole-Zero Plot is
It has a magnitude of unity at all frequencies & a phase angle of
1
2tan T
 

 
 
 
Which varies
from
0 0
0 to 180 as ω varies from 0 to 
Non – minimum phase transfer function
 
 
  
1 j T
G j
1 1 j T 1 j T
1 2
 
 
   
Whose pole – zero pattern as shown
The transfer function can be rewritten as
 
 
  
 
 
1 j T 1 j T
G j
1 1 j T
1 j T 1 j T
1 2
 
 
   
 
   
 
 
 
   
   
 
   
G j G j
2
  

It is a product of two transfer functions, one  
G j
2
 having no poles and zeroes in right
half plane which is Minimum Phase Transfer Function and other an all pass transfer function.
The Phase variation vs frequency for all three Transfer Functions is shown below:
Bode plot or Logarithmic Plot
A sinusoidal transfer function may represented by two separate plots, one giving magnitude
versus frequency and other the phase angle (in degrees) versus frequency.
 Bode plot consists of two graphs:
1. Logarithmic of the magnitude of sinusoidal transfer function.
2. Phase angle.
Both are plotted against frequency on logarithmic scale.
 The logarithmic representation is useful becomes it allows to plot both low frequency and
high frequency components on same graph.
Requirements:
1) The GH should be in time constant form
  
k 1 sT 1 sT
1 2
GH
1 sT 1 sT
p1 p2
 

  
 
  
  
2) The transfer function must be a minimum phase transfer function.

Magnitude Plot
 The Bode magnitude plot is, M in dB vs frequency which is drawn in semi log graph
sheet.
 The unit for slope in bode magnitude plot is dB/decade.
1 decade means 10 times change in frequency per unit change in magnitude.
Other unit is dB/octave where octave means 2 times change in frequency.
If
10000 3
2 1000 10 3 decades
10
1

   

3
2 8 2 3 octaves
1

  

Solved Examples
Problem: Find approximately how many octaves are present per decade?
Solution:
3
10 2

3 octave per decade.
Therefore, slope of 20 dB/decade is equivalent to a slope of 6 dB/decade.
 Drawing Bode Magnitude Plot
  
T
ds
K 1 sT 1 sT .......................e
1 2
GH
2
2 s s
n n
s 1 sT 1 sT .......................... 1
p1 p2 2 2
n n

 

 

 
  
   
    
    
 
 
Magnitude in dB
2 2 2 2 n 2 2
20logk 20log 1 T 20log 1 T ......................... 20log 20log 1 T ......................
1 1 p1
            
Since, after logarithm all terms come in addition, so we can draw Bode Plot for each term
and then add up those curves to obtain the Bode Plot for entire Transfer Function.

Terms Contained
1) A constant factor, k
2) A pole or zero at origin   p
s

3) Pole or zeroes away from origin   q
1 sT


4) Complex poles or complex zeroes
r
2
2 s s
n
1
2 2
n n

 

 
 
 
 
 
 
Constant Factor, k
 A number a greater than unity has positive value in decibels, while a number smaller than
unity has negative value.
 The log magnitude plot is a straight line of magnitude 20log k dB and phase angle is zero.
Integral & Derivative Factors   n
j


 If the transfer function contains the factor  
n
1 n
or j
j
 

 

 
, the log magnitude becomes,
 
1
20log n 20log j
n
j
   

20nlog dB
  
 n
20log j n 20log j
   

 The slopes of log magnitude curves for the factors  
n
1 n
or j
j
 

 

 
are then
20n dB dec
 and 20n dB dec .
 The phase angle of
n
1
j
 
 

 
is equal to 0
90 n
 & for  n
j it is equal to 0
90 n .
Magnitude Plot
Phase Plot
Poles & Zeroes away from origin  q
1 sT 

q
2 2
M 1 T

 
  
 
 
,
1
T
= corner frequency

Corner Frequency is the frequency at which Real and Imaginary parts of a term is equal.
Log magnitude plot 2 2
20log M 20q log 1 T
 
    
 
   
0 , T 1
20q log T , T 1
  

 
   


Thus Magnitude is 0dB at low frequencies and it is a straight line of slope 20q
 dB/decade.
Phase Plot
0
0 , T 1
q 90 , T 1
  
  
  

So, for approximation below corner frequencies we neglect the imaginary terms or term
containing ‘s’ and above corner frequencies we neglect the real part of term or the constant
value
Note: These curves are approximations & are different from actual plots. These are called as
asymptotic approximation.
Exact & Asymptotic approximation for   1
1 sT 


Summary
   
 
0
0
0
0
0
Term Slope Contributed Phase contributed
K 0dB / decade 0
s 20dB / decor 6dB / oct 90
1
20dB / decor 6dB / oct 90
s
p
s 20p dB / decor 6p dB / oct 90 P
1
20p dB / decor 6p dB / oct 90 P
p
s
q 1
1 sT 20q dB / decor 6q dB / oct 90q
T
q
1 sT 20q dB / decor
  

  
  

   
 
0
1
6q dB / oct 90q
T
1
40r dB / decor 12r dB / oct 180 r
n
r
2
2 s
1 s
2
n n
   
     
 
 

 
 
 
 
 
Solved Examples
Problem: Draw Bode Plot for the system whose Open Loop Transfer Function is given by
   
 
  
5 s 20
G s H s
s s 100 s 1


 
Solution: Expressing the Transfer Function in time constant form:
   
s
1 1
20
G s H s
s s
s 1 1
100 1
 

 
 

  
 
  
  
Sinusoidal Transfer Function is given by:
   
j
1 1
20
G j H j
j j
j 1 1
100 1
 


 
 
  
  
 
  
  
  
Corner Frequencies of this Transfer Function are 1, 20, 100 rad/s

Due to the pole at origin we cannot start the curve from origin so we rather start at a low
frequency like 0.1 rad/sec.
Making Asymptotic approximations in different frequency ranges:
0.1 < ω < 1
   
 
  
1 1 1
G j H j
j
j 1 1
   


At ω = 0.1, Gain =
1
10
0.1
 = 20 dB
Here we neglected imaginary parts of the terms whose corner frequency was greater than
the frequency range being used.
1 < ω < 20
   
 
   2
1 1 1
G j H j
j j 1
    
  
20 < ω < 100
   
  
j
1
20 1
G j H j
20j
j j 1
 

 
 
   

 
ω > 100
   
 
2
j
1
20 5
G j H j
j
j j
100
 

 
 
    
 
 
   
 
So, the Bode Plot looks like as shown in above figure
Note: The slope of highest frequency asymptote in Bode Plot is given by -20 (P-Z)
dB/decade. Where, P is the total number of Poles and Z is the total number of zeroes of
Open Loop Transfer Function.
For finding the slope of lowest frequency asymptote same expression is used but we
consider poles and zeroes at origin.

Steps for Finding Transfer Function from Bode Plot
Step-1: Find the corner frequencies from the Bode Plot by observing the points where the
slope of Bode Plot changes.
Step-2: Find order of Pole and Zero lying at each corner frequency, if there is a positive
change in slope of 20n dB/decade then there is a zero of order n lying at that point.
Otherwise, if there is a negative change in slope of -20n dB/decade then there is pole of
order n at that point.
Step-3: Obtain Transfer Function in time constant form with a gain K.
Step-4: Determine K using the Magnitude at any point on the Bode Plot.
Solved Examples
Problem: The Bode magnitude plot of system is given below, find the Transfer Function of
the system.
Solution: The Initial Slope of Bode Plot is -40dB/decade which indicates that there are two
poles present at origin.
The corner frequencies are 2, 10 rad/sec.
The change in slope at 2 rad/sec is 20dB/decade so there is a zero of order 1 at 2 rad/sec.
The change in slope at 10 rad/sec is -20dB/decade so there is a pole of order 1 at 10 rad/sec.

The Transfer Function of the system is of the form:
2
s
k(1 )
2
G(s)H(s)
s
s (1 )
10



We see that the straight line in Bode Plot between 0 and 2 rad/sec intersects x-axis at 4
rad/sec
Asymptotic approximation at 0rad / sec and 2rad / sec
    is
 
2 2
k(1) k
G( j )H( j )
j (1)
    


At 4 rad / sec
 
k
M
16

In decibels, M = 20 log
k
16
= 0 dB

k
M 1
16
 
 K =16
So, Transfer Function becomes,
2
s
16(1 )
2
G(s)H(s)
s
s (1 )
10



Problem: For the asymptotic Bode magnitude plot shown below, the system transfer
function is?

Solution: The Initial Slope of Bode Plot is -6dB/octave or -20dB/decade which indicates that
there is a pole of order 1 present at origin.
The corner frequencies are 2, 5 rad/sec.
The change in slope at 2 rad/sec is 6dB/octave or 20dB/decade so there is a zero of order 1
at 2 rad/sec.
The change in slope at 5 rad/sec is –12 dB/octave or -40 dB/decade so there is a pole of
order 1 at 5 rad/sec.
The Transfer Function of the system is of the form:
s
k 1
2
G(s)H(s)
2
s
s 1
5
 

 
 

 

 
 
The gain at 0.1rad / sec
  is 26dB.
Asymptotic approximation at 0rad / sec and 2rad / sec
    is
 
k(1) k
G( j )H( j )
j
j (1)
   


At 0.1rad / sec
 
M 10k

In decibels, M = 20 log 10k = 26 dB
K = 2
s
2 1
2
G(s)H(s)
2
s
s 1
5
 

 
 

 

 
 

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