Control System Notes

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

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CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

UNIT-4: Frequency, Domain analysis, Bode plots, Effect of adding, poles and Zeros, Polar plot,
Nyquist stability analysis, Relative stability: Gain and phase margins

4.1Introduction:-In control systems engineering, the frequency domain refers to the analysis of
mathematical functions or signals with respect to frequency, rather than time. To analyze the properties
and design the system based on frequency response, various kinds of plots are used. These are as follows

1. Bode Plot

2. Polar Plot

3. Nyquist Plot

4.2What is Bode Plot

A Bode plot is a graph commonly used in control system engineering to determine the stability of a
control system. A Bode plot maps the frequency response of the system through two graphs – the Bode
magnitude plot (expressing the magnitude in decibels) and the Bode phase plot (expressing the phase
shift in degrees).The slope of a straight line on a Bode magnitude plot is measured in units
of dB/Decade, because the units on the vertical axis are dB, and the units on the horizontal axis are
decades.
Bode plot consists of two plot-

(i)Magnitude plot expressed in dB against logω
(ii)Phase angle plot against logω

4.3Gain Margin
Gain Margin is defined as the margin in gain allowable by which gain can be increased till system
reaches on the verge of instability. It is usually expressed as a magnitude in dB.

We can usually read the gain margin directly from the Bode plot (as shown in the diagram above).

This is done by calculating the vertical distance between the magnitude curve (on the Bode magnitude
plot) and the x-axis at the frequency where the Bode phase plot = 180°. This point is known as the phase
crossover frequency.

4.4Phase Margin

The amount of additional phase lag which can be introduced in the system till system reaches on the
verge of instability is called phase margin. It is usually expressed as a phase in degrees.

We can usually read the phase margin directly from the Bode plot (as shown in the diagram
above). This is done by calculating the vertical distance between the phase curve (on the Bode phase plot)

and the x-axis at the frequency where the Bode magnitude plot = 0 dB. This point is known as the gain
crossover frequency.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig.4.1Typical Bode plot
4.5 Important Points about Bode Plot Stability

1. Gain Margin: Greater will the gain margin greater will be the stability of the system. It refers to
the amount of gain, which can be increased or decreased without making the system unstable. It is
usually expressed in dB.

2. Phase Margin: Greater will the phase margin greater will be the stability of the system. It refers
to the phase which can be increased or decreased without making the system unstable. It is
usually expressed in phase.

3. Gain Crossover Frequency: It refers to the frequency at which magnitude curve cuts the zero dB
axis in the bode plot.

4. Phase Crossover Frequency: It refers to the frequency at which phase curve cuts the negative
times the 180o axis in this plot.

5. Corner Frequency: The frequency at which the two asymptotes cuts or meet each other is known
as break frequency or corner frequency.

6. Resonant Frequency: The value of frequency at which the modulus of G (jω) has a peak value is
known as the resonant frequency.

7. Factors: Every loop transfer function {i.e. G(s) × H(s)} product of various factors like constant
term K, Integral factors (jω), first-order factors ( 1 + jωT)(± n) where n is an integer, second order
or quadratic factors.

8. Slope: There is a slope corresponding to each factor and slope for each factor is expressed in the
dB per decade.

9. Angle: There is an angle corresponding to each factor and angle for each factor is expressed in the
degrees.

10. Constant term K: This factor has a slope of zero dB per decade. There is no corner frequency
corresponding to this constant term. The phase angle associated with this constant term is also
zero.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

11. Integral factor 1/(jω)n: This factor has a slope of -20 × n (where n is any integer)dB per decade.
There is no corner frequency corresponding to this integral factor. The phase angle associated
with this integral factor is -90 × n here n is also an integer.

12. First order factor 1/ (1+jωT): This factor has a slope of -20 dB per decade. The corner frequency
corresponding to this factor is 1/T radian per second. The phase angle associated with this first
factor is -tan– 1(ωT).

13. First order factor (1+jωT): This factor has a slope of 20 dB per decade. The corner frequency
corresponding to this factor is 1/T radian per second. The phase angle associated with this first
factor is tan– 1(ωT) .

14. Second order or quadratic factor : [{1/(1+(2ζ/ω)} × (jω) + {(1/ω2)} × (jω)2)]: This factor has a
slope of -40 dB per decade. The corner frequency corresponding to this factor is ωn radian per
second. The phase angle associated with this first factor is

Keeping all the above points in mind, we are able to draw a Bode plot for any kind of control system.

Fig4.2. Magnitude &Phase Plot of Standard Functions /Factors

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig.4.3 Magnitude &Phase Plot of Standard Functions /Factors

Table No.4.1Table for Initial slope of Bode Plot

Type Of System Initial Slope Intersection with 0db

axis at

0 0db/decade Parallel to 0db axis

1 -20db/decade ω=K

2 -40db/decade ω=K1/2

3 -60db/decade ω=K1/3

4.6Now let us discuss the procedure of drawing a Bode plot:

1. Substitute the s = jω in the open loop transfer function G(s) × H(s).
2. Find the corresponding corner frequencies and tabulate them.
3. Now we are required one semi-log graph chooses a frequency range such that the plot should start

with the frequency which is lower than the lowest corner frequency. Mark angular frequencies on
the x-axis, mark slopes on the left hand side of the y-axis by marking a zero slope in the middle
and on the right hand side mark phase angle by taking -180o in the middle.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

4. Calculate the gain factor and the type or order of the system.
5. Now calculate slope corresponding to each factor.
4.7 For drawing the Bode magnitude plot:

 Mark the corner frequency on the semi-log graph paper.
 Tabulate these factors moving from top to bottom in the given sequence.

1. Constant term K.
2. Integral factor 1/jωn
3. First order factor 1/(1+jωT).
4. First order factor (1+jωT).

5. Second order or quadratic factor:

 Now sketch the line with the help of the corresponding slope of the given factor. Change the slope
at every corner frequency by adding the slope of the next factor. You will get the magnitude plot.

 Calculate the gain margin.
4.8 For drawing the Bode phase plot:

1. Calculate the phase function adding all the phases of factors.
2. Substitute various values to the above function in order to find out the phase at different points

and plot a curve. You will get a phase curve.
3. Calculate the phase margin.

4.9 Bode Stability Criterion

Stability conditions are given below:
1. For Stable System: Both the margins should be positive(G.M. &P.M. +ve) or phase crossover
frequency should be greater than the gain crossover frequency(ωpc>ωgc).
2. For Marginal Stable System: Both the margins should be zero(G.M.=P.M.=0) or phase crossover
frequency should be equal to the gain crossover frequency(ωpc=ωgc). .
3. For Unstable System: Both the margins should be negative or phase crossover frequency be less
than the gain crossover frequency(ωpc<ωgc).

4.10 Advantages of Bode Plot
1. It is based on the asymptotic approximation, which provides a simple method to plot the
logarithmic magnitude curve.
2. The multiplication of various magnitude appears in the transfer function can be treated as an
addition, while division can be treated as subtraction as we are using a logarithmic scale.
3. With the help of this plot only we can directly comment on the stability of the system without
doing any calculations.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

4. Bode plots provides relative stability in terms of gain margin and phase margin.

5. It also covers from low frequency to high frequency range.

Note: The locations of every pole and every zero are called break points. At a zero breakpoint,

the slope of the line increases by 20dB/Decade. At a pole, the slope of the line decreases by

20dB/Decade.

Example 1.Draw the Bode Diagram for the transfer function:

G(s)H(s)= 80

s(s+2)(s+20)

Step 1: Rewrite the transfer function in proper form. Bring the given G(s)H(s) transfer function into

standard time constant form (convert all the term in the numerator or/and denominator into (1+sT) ).

G(s)H(s) = 80
s (2)(1+s/2)(20)(1+s/20)

=2
s (1+s/2)(1+s/20)

=2
s (1+0.5s)(1+0.05s)

Step 2 : Replace all s by jω to get the frequency domain transfer function:

G(jω)H(jω) = 2
jω (1+0.5 jω)(1+0.05 jω)

Step 3: Separate the transfer function into its constituent parts.

The following factors are present:

1. K=2
2. Pole at origin = 1/ jω =1/ jω
3. First order pole = 1/(1+0.5 jω)
4. First order pole = 1/(1+0.05 jω)

Step 4: Find the corner Frequencies

ω1= 1/0.5=2
ω2= 1/0.05=20

Table No.4.2 Slope & Phase Calculation for Bode plot

Sr no. Factor Frequency Slope Phase Remark

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

1 K/ jω= 2/ jω ω =K* -20dB/dec Φ = -90 Slope of -20 dB/dec passing
ω =2 through 0 dB at ω = 2 rad/sec

2 1/( 1 +0.5 jω) ω =2 -20dB/dec Draw resultant Slope of

-20-20 =(-40) from lower
Φ = -tan-1(0.5 ω) corner frequency ω =2

Draw resultant Slope of

-20-20-20 =(-60) from lower
3 1/( 1 +0.05 jω) ω =20 -20dB/dec Φ = -tan-1(0.05 ω) corner frequency ω =20

 For K/ jω frequency ω =K; For K/ jω2 frequency ω =K1/2

Step 5 : Phase plot table:
Resultant Phase Φ =-90 -tan-1(0.5 ω) -tan-1(0.05 ω)

Take any No of value of ω in between from lower corner frequency to upper corner frequency.

Table No.4.3 Resultant Phase Calculation for Bode plot

ω Φ =-90 -tan-1(0.5 ω) -tan-1(0.05 ω)
0.2 -96.270
2 -140.70
8 -187.760
10 -195.290
20 -219.280
40 -240.580

Step 6 : The bode plot is shown in below figure :

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig.4.4 Bode Plot of Example 1

Example 2.Draw the Bode Plot for the transfer function:

G(s) = 36(1+0.2s)

s2(1+0.05s) (1+0.01s)

From the Bode plot determine

(a)Phase crossover frequency (b)Gain crossover frequency

(c)Phase Margin (d)Gain Margin

Comment on the stability of the system.

Solution
Step 1: Replace all s by jω to get the frequency domain transfer function:

G(jω) = 36(1+0.2 jω)
( jω) 2(1+0.05 jω) (1+0.01 jω)

Step 2: Separate the transfer function into its constituent parts.The following factors are present:

1. K=36

2. Double Pole at origin = 1/ jω2
3. First order Zero=(1+0.2 jω)

4. First order pole = 1/(1+0.05 jω)
5. First order pole = 1/(1+0.01 jω)

Step 3: Find the corner Frequencies

ω1 = 1/0.2 = 5 Lower Corner Frequency

ω2 = 1/0.05 = 20 Intermediate Frequency

ω3 =1/0.01 = CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

100 Upper Corner Frequency

Table No4.4 Slope & Phase Calculation for Bode plot

Sr no. Factor Frequency Slope Phase Remark

ω =(36)1/2 Slope of -40 dB/dec passing
through 0 dB at ω = 6 rad/sec
1 36/ jω2 ω =6 -40dB/dec Φ = -180

2. (1+0.2 jω) ω1 = 5 +20dB/dec Draw resultant Slope of

-40+20 =(-20) from lower
Φ = tan-1(0.2 ω) corner frequency ω =5

Draw resultant Slope of

-20-20 =(-40) from lower
2 1/( 1 +0.05 jω) ω =20 -20dB/dec Φ = -tan-1(0.05 ω) corner frequency ω =20

Draw resultant Slope of
-20-20-20 =(-60) from lower
3 1/( 1 +0.01 jω) ω =100 -20dB/dec Φ = -tan-1(0.01 ω) corner frequency ω =20

 For K/ jω frequency ω =K; For K/ (jω)2 frequency ω =K1/2

Step 5 : Phase plot table:

Resultant Phase Φ = [ -180+ tan-1(0.2 ω)-tan-1(0.05 ω) -tan-1(0.01ω)]

Take any No of value of ω in between from lower corner frequency to upper corner frequency.

Table No.4.5 Resultant Phase Calculation for Bode plot

ω Φ = [ -180+ tan-1(0.2 ω)-tan-1(0.05 ω) -tan-1(0.01ω)]

1 -172.10
5 -151.90
10 -148.80
30 -172.50
50 -190.40
100 -216.60

Step 6 : The bode plot is shown in below figure :

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig.4.5 Bode Plot of Example 2

Practice Questions

1. Sketch Bode plot for the following transfer function and determine gain K for the gain cross over
frequency wgc to be 5rad/sec

G(s)H(s) = Ks2

(1+0.2s)(1+0.02s)

2. For a unity feedback system system

G(s)= 800(s+2)

s2(s+10)(s+40)

Find Gain Margin & Phase Margin by bode Plot. also comment on stability.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

3. For a unity feedback system G(s)= K

s(s+2)(s+10)

Determine marginally value of ‘K’ for which system will be marginally stable.

4.11 Effect of adding, poles and Zeros

Addition of poles to the transfer function has the effect of pulling the root locus to the right, making the
system less stable. Addition of zeros to the transfer function has the effect of pulling the root locus to the
left, making the system more stable.This can be proved by Following Examples.

4.11.1 Addition of poles
Consider, G(s)H(s) =K/s(s+2) Corresponding root locus is shown in Fig 4.6

Fig 4.6 root locus of K/s(s+2)

Now If pole at s=-4 is added to G(s)H(s) root locus becomes as shown in Fig 4.7,
G(s)H(s) =K/s(s+2)(s+4)

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 4.7 root locus of K/s(s+2)(s+4)

So for any value of K before addition of poles in left half, system is totally stable but after addition of
pole in left half two branches of root locus after some value of K moves in right half of s plane so system
up to this value of K is stable. After this value of K system becomes unstable so stability of system gets
restricted.
If Now one more pole at s=-6 is added to the system, G(s)H(s) =K/s(s+2)(s+4)(s+6)
Breakaway point in section s=0and s=-2 gets shifted towards right as compared to previous case.so
system stability further gets restricted. This is shown in the Fig.4.8

Fig 4.8 root locus of K/s(s+2)(s+4)(s+6)

From this we can conclude that due to addition of poles, root locus shifts towards R.H.S. of s plane so
system stability decreases.
4.11.2 Effect of Addition of Open Loop Poles can be summarised as
1.Root locus shifts towards imaginary axis.
2.System stability relatively decreases
3.System becomes more oscillatory in nature.
4. Range of operating values of ‘K’for stability of the system decreases.
4.11.3Addition of Zeros
Let us introduce a zero at s=-4 G(s)H(s) =K/s(s+2), Root locus for G(s)H(s) =K/s(s+2) is shown in Fig
4.6. Root locus for G(s)H(s) =K(s+4)/s(s+2) is shown in Fig 4.9.it can be seen that root locus shift
towards left i.e. towards zero which is added. So as roots move towards left half of s plane relative
stability increases as compared to previous case.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 4.9 root locus of K(s+4)/s(s+2)

Let us introduce one more zero at s=-6
G(s)H(s) =K(s+4)(s+6)/s(s+2

In this case breakaway point’s shifts towards left half of s plane as shown. so relative stability of
system gets further increased.

As we go on adding zeros to left resultant root loci bend towards left half of s plane. As roots
move towards left half of s plane, relative stability of system improves. Due to addition of zeros towards
left half of s plane system stability increases. Also it increases the range of operating values of K for
system stability.

Fig 4.10 root locus of K(s+4)(s+6)/s(s+2)

4.11.4 Effect of Addition of Open Loop Poles can be summarized as
1. Root locus shifts to left away from imaginary axis.
2. Relatively stability of the system increases.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

3. System becomes less oscillatory in nature.

4. Range of operating values of ‘K’for stability of the system increases.

4.12 Polar plot

Polar plot is a plot which can be drawn between magnitude and phase. Here, the magnitudes are
represented by normal values only.

The polar form of G(jω)H(jω) is
G(jω)H(jω)=|G(jω)H(jω)|∠G(jω)H(jω)

The Polar plot is a plot, which can be drawn between the magnitude and the phase angle of G(jω)H(jω)
by varying ω from zero to ∞

4.13 Rules for Drawing Polar Plots
Follow these rules for plotting the polar plots.

 Substitute, s=jω in the open loop transfer function.
 Write the expressions for magnitude and the phase of G(jω)H(jω).
 Find the starting magnitude and the phase of G(jω)H(jω) by substituting ω=0. So, the polar plot

starts with this magnitude and the phase angle.
 Find the ending magnitude and the phase of G(jω)H(jω) by substituting ω=∞. So, the polar plot

ends with this magnitude and the phase angle.

 Check whether the polar plot intersects the real axis, by making the imaginary term
of G(jω)H(jω) equal to zero and find the value(s) of ω.

 Check whether the polar plot intersects the imaginary axis, by making real term
of G(jω)H(jω) equal to zero and find the value(s) of ω.

 For drawing polar plot more clearly, find the magnitude and phase of G(jω)H(jω) by considering
the other value(s) of ω.

Example 3

Consider the open loop transfer function of a closed loop control system.

G(s)H(s)= 5

s(s+1)(s+2)

Let us draw the polar plot for this control system using the above rules.

Step 1 − Substitute, s=jω in the open loop transfer function.

G(jω)H(jω)= 5

jω(jω+1)(jω+2)

The magnitude of the open loop transfer function is

|G(jω)H(jω)|= 5

ω√ (ω2+1)√(ω2+4)

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

The phase angle of the open loop transfer function is
∠G(jω)H(jω)=−900−tan−1ω−tan−1ω/2
Step 2 − The following table shows the magnitude and the phase angle of the open loop transfer
function at ω=0 rad/sec and ω=∞ rad/sec.

Table No.4.6 Value of Magnitude & Phase

Frequency (rad/sec) Magnitude Phase angle(degrees)

0 ∞ -90

∞ 0 -270

So, the polar plot starts at (∞,−900) and ends at (0,−2700). The first and the second terms with in the
brackets indicate the magnitude and phase angle respectively.

Step 3 Rationalising the function & equate the real & imaginary part is equal to zero.

G(jω)H(jω)= 5[-jω(-jω+1)(-jω+2)]
[jω(jω+1)(jω+2)][-jω(-jω+1)(-jω+2)]

5[-jω(-jω+1)(-jω+2)]

[ω2(ω2+1)(ω2+4)]
5[-3ω2 - jω(2-ω2)]
[ω2(ω2+1)(ω2+4)]

Intersecting point on Real Axis

Imaginary part =0

5[- jω(2-ω2)] =0

[ω2(ω2+1)(ω2+4)]

5[- jω(2-ω2)] =0

ω =√2 (take non zero real value)

By substituting ω=√2 in the magnitude of the open loop transfer function,

|G(jω)H(jω)| ω =√2 = 5
ω√ (ω2+1)√(ω2+4) ω =√2

we will get |G(jω)H(jω)| ω =√2 ==0.83

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Note:
 Check whether the polar plot intersects the real axis, by making the imaginary term of G(jω)H(jω) equal to zero and

find the value(s) of ω.
 Check whether the polar plot intersects the imaginary axis, by making real term of G(jω)H(jω) equal to zero and find

the value(s) of ω.

So, we can draw the polar plot with the above information.

Fig.4.11 polar plot of Example 3

Example 4

Consider the open loop transfer function of a closed loop control system. Draw the polar plot.

G(s)H(s) = K

s2(1+sT1)(1+sT2)

Step 1 : The first step would be convert this transfer function to the frequency domain.This can be done

by converting ‘s’ by ‘jω’

G(jω)H(jω) = K

(jω)2(1+ jωT1)(1+ jωT2)

We don’t need to convert this to time constant form..

Step 2 : We now find the magnitude and phase

|G(jω)H(jω)| = K

ω2√ (1+ ω2 T12) √ (1+ ω2 T22)

 ∠ G(jω)H(jω) = -1800-tan-1 ωT1-tan-1 ωT2

Step 3 − The following table shows the magnitude and the phase angle of the open loop transfer
function at ω=0 rad/sec and ω=∞ rad/sec.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Table No.4.7 Value of Magnitude & Phase

Frequency (rad/sec) Magnitude Phase angle(degrees)

0 ∞ -180

∞ 0 -360

So, the polar plot starts at (∞,−1800) and ends at (0,−3600). The first and the second terms with in the
brackets indicate the magnitude and phase angle respectively

Step 4: Rationalising the function & equate the real & imaginary part is equal to zero.

G(jω)H(jω)= K[ (1-jωT1) 1-jωT2) ]

(-ω2)[ (1+jωT1) (1+jωT2) ] [ (1-jωT1) (1-jωT2) ]

G(jω)H(jω)= K[ (1-jωT1-jωT2- ω2 T1T2) ]
(-ω2)[ (1+ω2T1) (1+ω2T2) ]
+ j K[(ω(T1+T2)]
-K[ (1 - ω2 T1T2) ]

(ω2)[ (1+ω2T1) (1+ω2T2) ] (ω2)[ (1+ω2T1) (1+ω2T2) ]

Intersecting point on imaginary Axis

Real part =0

-K[ (1 - ω2 T1T2) ] =0

(ω2)[ (1+ω2T1) (1+ω2T2) ]

(1 - ω2 T1T2) =0

ω =1/ √T1T2

By substituting ω=1/ √T1T2 in the magnitude of the open loop transfer function,

|G(jω)H(jω)| ω=1/ √T1T2 = K
ω2√ (1+ ω2 T12) √ (1+ ω2 T22)
ω=1/ √T1T2

|G(jω)H(jω)| ω=1/ √T1T2 = (KT1T2)2
T1T2

Note:
 Check whether the polar plot intersects the real axis, by making the imaginary term of G(jω)H(jω) equal to zero and

find the value(s) of ω.
 Check whether the polar plot intersects the imaginary axis, by making real term of G(jω)H(jω) equal to zero and find

the value(s) of ω.

So, we can draw the polar plot with the above information.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig.4.12 polar plot for Example 2

4.14 Stability on Polar plots:

We can check stability using polar plot in very simple manner. Imagine yourself walking on the polar
plot, along the direction of the arrow i.e from = 0 to ∞ . The entire area to the right of you up to the real

axis represented by an infinite radius is said to be enclosed by the polar plot. The portion inside the

shaded area is said to be enclosed by the polar plot. A system is stable if the (-1,0) point is not enclosed

by the polar plot.

Fig.4.13 Typical Polar plot

4.15 Rules for Drawing Nyquist Plots
Follow these rules for plotting the Nyquist plots.

 Locate the poles and zeros of open loop transfer function G(s)H(s) in ‘s’ plane.
 Draw the polar plot by varying ω from zero to infinity. If pole or zero present at s = 0, then

varying ω from 0+ to infinity for drawing polar plot.
 Draw the mirror image of above polar plot for values of ω ranging from −∞ to zero (0− if any

pole or zero present at s=0).
 The number of infinite radius half circles will be equal to the number of poles or zeros at origin.

The infinite radius half circle will start at the point where the mirror image of the polar plot ends.
And this infinite radius half circle will end at the point where the polar plot starts.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

After drawing the Nyquist plot, we can find the stability of the closed loop control system using the

Nyquist stability criterion. If the critical point (-1+j0) lies outside the encirclement, then the closed loop

control system is absolutely stable.

Stability Analysis using Nyquist Plots

From the Nyquist plots, we can identify whether the control system is stable, marginally stable or

unstable based on the values of these parameters.

 Gain cross over frequency and phase cross over frequency

 Gain margin and phase margin

4.18 Phase Cross over Frequency

The frequency at which the Nyquist plot intersects the negative real axis (phase angle is 1800) is known

as the phase cross over frequency. It is denoted by ωpc or ωΦ

At phase cross over frequency ∠ G(j ωΦ)H(j ωΦ) =-1800

4.19 Gain Cross over Frequency

The frequency at which the Nyquist plot is having the magnitude of one is known as the gain cross over

frequency. It is denoted by ωgc/ ωg. |G(jωg)H(jωg)|=1
At Gain cross over frequency

The stability of the control system based on the relation between phase cross over frequency and gain

cross over frequency is listed below.
 If the phase cross over frequency ωpc is greater than the gain cross over frequency ωgc, then the

control system is stable.

 If the phase cross over frequency ωpc is equal to the gain cross over frequency ωgc, then the

control system is marginally stable.
 If phase cross over frequency ωpc is less than gain cross over frequency ωgc, then the control

system is unstable.

4.20 Gain Margin

The gain margin GM is equal to the reciprocal of the magnitude of the Nyquist plot at the phase cross

over frequency. GM=1/Mpc

Where, Mpc is the magnitude in normal scale at the phase cross over frequency

In decibel the gain margin is given by GM= 20 log10 1 db

|G(jωΦ)H(jωΦ)|

For stable system |G(jωΦ)H(jωΦ)|<1,the gain margin in decibel is positive.

For marginally stable system |G(jωΦ)H(jωΦ)|=1,the gain margin in decibel is zero.

For unstable system |G(jωΦ)H(jωΦ)|>11,the gain margin in decibel is negative.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

4.21 Phase Margin

The phase margin PM is equal to the sum of 1800 and the phase angle at the gain cross over frequency.
PM=1800+ϕgc

Where, ϕgc=∠ G(j ωg)H(j ωg) is the phase angle at the gain cross over frequency.

PM=1800+∠ G(j ωg)H(j ωg)

Fig.4.14 gain & Phase margin from Nyquist Plot

Example 5

Consider the open loop transfer function of a closed loop control system.

G(s)H(s)= 1

s(s+3)(s+5)

Let us draw the Nyquist plot for this control system using the above rules.

Step 1 − Substitute, s=jω in the open loop transfer function.

G(jω)H(jω)= 1

jω(jω+3)(jω+5)

The magnitude of the open loop transfer function is

|G(jω)H(jω)|= 1

ω√ (ω2+9)√(ω2+25)

The phase angle of the open loop transfer function is
∠G(jω)H(jω)=−900−tan−1ω/3−tan−1ω/5
Step 2 − The following table shows the magnitude and the phase angle of the open loop transfer

function at ω=0 rad/sec and ω=∞ rad/sec.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Table No. 4.8 Values of Magnitude & Phase

Frequency (rad/sec) Magnitude Phase angle(degrees)

0 ∞ -90

∞ 0 -270

So, the polar plot starts at (∞,−900) and ends at (0,−2700). The first and the second terms with in the
brackets indicate the magnitude and phase angle respectively.

Step 3 Rationalising the function & equate the real & imaginary part is equal to zero.

G(jω)H(jω)= [jω(-jω+3)(-jω+5)]

[jω(jω+3)(jω+5)][jω(-jω+3)(-jω+5)]

[-8ω2 - jω(15-ω2)]

[ω2(ω2+9)(ω2+25)]

Intersecting point on Real Axis

Imaginary part =0

[- jω(15-ω2)] =0

[ω2(ω2+1)(ω2+4)]

[- jω(15-ω2)] =0

ω =√15(take Non zero real value)

By substituting ω=√15 in the magnitude of the open loop transfer function,

|G(jω)H(jω)| ω =√15 = 1
ω√ (ω2+9)√(ω2+25)
ω =√15

we will get |G(jω)H(jω)| ω =√15 ==0.0083

Note:
 Check whether the polar plot intersects the real axis, by making the imaginary term of G(jω)H(jω) equal to zero and

find the value(s) of ω.
 Check whether the polar plot intersects the imaginary axis, by making real term of G(jω)H(jω) equal to zero and find

the value(s) of ω.

So, we can draw the polar plot with the above information.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

 Draw the polar plot by varying ω from 0+

to ∞+.(as shown in blue curve in plot.)

 Draw the mirror image of above polar plot
for values of ω ranging from −∞ to 0−. (as
shown in orange curve in plot.)

 Remember mirror image should be about
real axis or X axis.

 After drawing the mirror image of polar
plot, you have to close the plot to get the
Nyquist plot. You have to remember
following points for closing the plots.
 Close the curve ω =0- to ω =0+
 In clockwise direction,

 With nπ angle (K/Sn) for

type 0= 00
type 1= π=1800(for this problem)
type 2=2 π=3600
 Infinite Radius

Fig 4.15 Nyquist Plot for Example 5

Practice Question

1.Sketch the polar plot K
G(s)H(s) =
s2(1+sT1)(1+sT2)
2. Sketch the polar plot.
K
G(s)H(s) =
s(s+1)(s+2)
3. Sketch the Nyquist plot
G(s)H(s) = (1+4s)
s2(s+1)(1+2s)
4. Sketch the Nyquist plot
G(s)H(s) = K

s(sT-1)

5. Sketch the Nyquist plot

G(s)H(s) = K

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

REFERENCES
1. B.S.Manke, Control system Engineering, Khanna Publishers
2. M. Gopal, ‘Control system engineering’, McGraw Hill
3. K. Ogata, ‘Modern Control Engineering’, Pearson
4. D. Roy, Chaudhary,‘Modern Control Systems’, PHI.
5. B.C. Kuo and FaridGolnaraghi, ‘Automatic Control Systems’, Wiley India.

6. www.tutorialspoint.com

7.Electronics guide4u.com

8.U A Bakshi & S.C.Goyal Feedback Control System,Technical Publication.

CONTROL SYSTEM EE-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering