Unit-4
Frequency Domain Analysis
The calculation of system parameters for higher order systems is difficult in time domain analysis. This can be overcome by frequency domain analysis. As the frequency response of a linear time invariant system is independent of amplitude and phase of the input test signal. The effect of noise and parameter variation in frequency domain can be easily computed.
In frequency response can be determine by calculating the phase and amplitude oof the given system transfer function. If a sinusoidal signal XG(j
)sin
t then its amplitude will be
C(t) = XG(j
)sin
t
= X|G(j
)|
sin
t
Magnitude = |G(j
)|
Phase = 
In polar form = |G(j
)|
The transfer function of second order system is shown as
C(S)/R(S) = W2n / S2 + 2ξWnS + W2n - - (1)
ξ = Ramping factor
Wn = Undamped natural frequency for frequency response let S = jw
C(jw) / R(jw) = W2n / (jw)2 + 2 ξWn(jw) + W2n
Let U = W/Wn above equation becomes
T(jw) = W2n / 1 – U2 + j2 ξU
So,
| T(jw) | = M = 1/√(1 – u2)2 + (2ξU)2 - - (2)
T(jw) = φ = -tan-1[ 2ξu/(1-u2)] - - (3)
For sinusoidal input the output response for the system is given by
C(t) = 1/√(1-u2)2 + (2ξu)2Sin[wt - tan-1 2ξu/1-u2] - - (4)
The frequency where M has the peak value is known as Resonant frequency Wn. This frequency is given as (from eqn (2)).
DM/du|u=ur = Wr = Wn√(1-2ξ2) - - (5)
From equation(2) the maximum value of magnitude is known as Resonant peak.
Mr = 1/2ξ√1-ξ2 - - (6)
The phase angle at resonant frequency is given as
Φr = - tan-1 [√1-2ξ2/ ξ] - - (7)
As we already know for step response of second order system the value of damped frequency and peak overshoot are given as
Wd = Wn√1-ξ2 - - (8)
Mp = e- πξ2|√1-ξ2 - - (9)
Fig 1 Frequency Domain Specification
The comparison of Mr and Mp is shown in figure(1). The two performance indices are correlated as both are functions of the damping factor ξ only. When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.
Similarly the correlation of Wr and Wd is shown in fig(1) for the given input step response [ from eqn(5) & eqn(8) ]
Wr/Wd = √(1- 2ξ2)/(1-ξ2)
Mp = Peak overshoot of step response
Mr = Resonant Peak of frequency response
Wr = Resonant frequency of Frequency response
Wd = Damping frequency of oscillation of step response.
From fig(1) it is clear that for ξ> 1/2, value of Mr does not exists.
Key takeaway
i) Mr and Mp are correlated as both are functions of the damping factor ξ only
Ii) When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.
In polar plot any point gives the magnitude phase of the transfer function in bode we split magnitude and 
plot.
Advantages
- By looking at bode plot we can write the transfer function of system
Q. G(S) = 
1. Substitute S = j
G(j
) = 
M = 
= tan-1
= -900
Magnitude varies with ‘w’ but phase is constant.
MdB = +20 log10
Decade frequency :-
W present = 10 
past
Then 
present is called decade frequency of 
past
2 = 10 
1
2 is decade frequency of 
1
MdB
0.01 40
0.1 20
1 0 (shows pole at origin)
0 -20
10 -40
100 -60
Slope = (20db/decade)
Fig 1 MAGNITUDE PLOT
Fig 2 PHASE PLOT
Que.G(S) = 
Sol: G(j
) = 
M = 
; 
= -1800 (-20tan-1
)
MdB = +20 log 
-2
MdB = -40 log10
MdB
0.01 80
0.1 40
1 0 (pole at origin)
10 -40
100 -80
Slope = 40dbdecade
Que. G(S) = S
Sol: M= W
= 900
MdB = 20 log10
MdB
0.01 -40
0.1 -20
1 0
10 20
100 90
1000 60
Fig 3 Bode Plot G(S) = S
Que. G(S) = S2
Sol: M= 
2
MdB = 20 log10
2 == 40 log10
= 1800
W MdB
0.01 -80
0.1 -40
1 0
10 40
100 80
Fig 4 Magnitude Plot G(S) = S2
Que. G(S) = 
Sol: G(j
) = 
M = 
MdB = 20 log10 K-20 log10
= tan-1(
) –tan-1(
)
= 0-900 = -900
K=1 K=10
MDb MdB
=-20 log10
=20 -20 log10
0.01 40 60
0.1 20 40
1 0 20
10 -20 0
100 -40 -20
Fig 5 Bode Plot G(S) = 
Fig 6 All bode plots in one plot
Fig 7 Variation in K shifts magnitude plot by +20db
As we vary K 
then plot shift by 20 log10K
i.e adding a d.c. To a.c. Quantity
Approximation of Bode Plot:
If poles and zeros are not located at origin
G(S) = 
TF = 
M = 
MdB = -20 log10 (
= -tan-1
Approximation: 
T >> 1. So, we can neglect 1.
MdB = -20 log10
MdB = -20 log10
T . ; 
= -tan-1(
T)
Approximation: 
T << 1. So, we can neglecting 
T.
MdB= 0dB, 
= 00
At a point both meet so equal i.e a time will come hence both approx become equal
-20 log10
T= 0
T= 1
corner frequency
At this frequency both the cases are equal
MdB = -20 log10
Now for 
MdB = -20 log10
= -20 log10
= -10 log102
MdB = 10
Fig 8 Approximation in bode plot
When we increase the value of 
in app 2 and decrease the 
of app 1 so a RT comes when both cases are equal and hence for that value of 
where both app are equal gives max. Error we found above and is equal to 3dB
At corner frequency we have max error of -3dB
Que. G(S) = 
Sol: TF = 
M = 
MdB = -20 log10 (
at T=2
MdB
1 -20 log10
10 -20 log10
100 -20 log10
MdB 
=
= 
0.1 -20 log10
= 1.73 
10-3
0.1 -20 log10
= -0.1703
0.5 -20 log10
= -3dB
1 -20 log10
= -6.98
10 -20 log10
= -26.03
100 -20 log10
= -46.02
Fig 9 Magnitude Plot with approximation
Without approximation
For second order system
TF = 
TF = 
= 
= 
= 
M= 
MdB= 
Case 1 
<<
<< 1
MdB= 20 log10
= 0 dB
Case 2 
>>
>> 1
MdB = -20 log10
= -20 log10
= -20 log10
< 1 
is very large so neglecting other two terms
MdB = -20 log10
= -40 log10
Case 3 . When case 1 is equal to case 2
-40 log10
= 0
= 1
The natural frequency is our corner frequency
Fig 10 Magnitude Plot
Max error at 
i.e at corner frequency
MdB = -20 log10
For 
MdB = -20 log10 
error for
Completely the error depends upon the value of 
(error at corner frequency)
The maximum error will be
MdB = -20 log10 
M = -20 log10 
= 0
is resonant frequency and at this frequency we are getting the maximum error so the magnitude will be
M = -
+ 
= 
Mr =
MdB = -20 log10
MdB = -20 log10 
= tan-1
Mr = 
Type of system | Initial slope | Intersection |
0 | 0 dB/decade | Parallel to 0 axis |
1 | -20 dB/decade | =K1 |
2 | -40 dB/decade | =K1/2 |
3 | -60 dB/decade | =K1/3 |
. | . | 1 |
. | . | 1 |
. | . | 1 |
N | -20N dB/decade | =K1/N |
Gain Cross Over Frequency
The frequency at which the bode plot culls the 0db axis is called as Gain Cross Over Frequency.
Fig 11 Gain cross over frequency
Phase Cross Over Frequency
The Frequency at which the phase plot culls the -1800 axis.
Fig 12 Phase Margin and Gain Margin
GM=MdB= -20 log [ G (jw)]
.: 
.: 
Key takeaway
i) More the difference b/WPC and WGC core are the stability of system
Ii) If GM is below 0dB axis than take +ve and stable. If GM above 0dB axis, that is take -ve
GM= ODB - 20 log M
Iii) The IM should also lie above -1800 for making the system (i.e. pm=+ve
Iv) For a stable system GM and PM should be -ve
v) GM and PM both should be +ve more the value of GM and PM more the system is stable.
Vi) If Wpc and Wgc are in same line Wpc= Wgc than system is marginally stable as we get GM=0dB.
Vii) When gain cross over frequency is smaller then phase curves over frequency the system is stable and vice versa.
The effect can be understood by plotting few bode plots as shown below
Q.1 sketch the bode plot for transfer function
G(S) = 
- Replace S = j
G(j
= 
This is type 0 system . So initial slope is 0 dB decade. The starting point is given as
20 log10 K = 20 log10 1000
= 60 dB
Corner frequency 
1 = 
= 10 rad/sec
2 = 
= 1000 rad/sec
Slope after 
1 will be -20 dB/decadetill second corner frequency i.e 
2 after 
2 the slope will be -40 dB/decade (-20+(-20)) as there are poles
2. For phase plot
= tan-1 0.1
- tan-1 0.001 
For phase plot
100 -900
200 -9.450
300 -104.80
400 -110.360
500 -115.420
600 -120.00
700 -124.170
800 -127.940
900 -131.350
1000 -134.420
The plot is shown in figure 14
Fig 14 Magnitude Plot for G(S) = 
Q.2 For the given transfer function determine
G(S) = 
Gain cross over frequency phase cross over frequency phase mergence and gain margin
Initial slope = 1
N = 1 , (K)1/N = 2
K = 2
Corner frequency
1 = 
= 2 (slope -20 dB/decade
2 = 
= 20 (slope -40 dB/decade
2. phase
= tan-1
- tan-1 0.5 
- tan-1 0.05 
= 900- tan-1 0.5 
- tan-1 0.05 
1 -119.430
5 -172.230
10 -195.250
15 -209.270
20 -219.30
25 -226.760
30 -232.490
35 -236.980
40 -240.570
45 -243.490
50 -245.910
Finding 
gc (gain cross over frequency
M = 
4 = 
2 (
(
6 (6.25
104) + 0.252
4 +
2 = 4
Let 
2 = x
X3 (6.25
104) + 0.252
2 + x = 4
X1 = 2.46
X2 = -399.9
X3 = -6.50
For x1 = 2.46
gc = 3.99 rad/sec(from plot )
For phase margin
PM = 1800 - 
= 900 – tan-1 (0.5×
gc) – tan-1 (0.05 × 
gc)
= -164.50
PM = 1800 - 164.50
= 15.50
For phase cross over frequency (
pc)
= 900 – tan-1 (0.5 
) – tan-1 (0.05 
)
-1800 = -900 – tan-1 (0.5 
pc) – tan-1 (0.05 
pc)
-900 – tan-1 (0.5 
pc) – tan-1 (0.05 
pc)
Taking than on both sides
Tan 900 = tan-1
Let tan-1 0.5 
pc = A, tan-1 0.05 
pc = B
= 00
= 0
1 =0.5 
pc 0.05
pc
pc = 6.32 rad/sec
The plot is shown in figure 15.
Fig 15 Magnitude Plot G(S) = 
Q3. For the given transfer function
G(S) = 
Plot the rode plot find PM and GM
T1 = 0.5 
1 = 
= 2 rad/sec
Zero so, slope (20 dB/decade)
T2 = 0.2 
2 = 
= 5 rad/sec
Pole , so slope (-20 dB/decade)
T3 = 0.1 = T4 = 0.1
3 = 
4 = 10 (2 pole ) (-40 db/decade)
- Initial slope 0 dB/decade till
1 = 2 rad/sec - From
1 to
2 (i.e. 2 rad /sec to 5 rad/sec) slope will be 20 dB/decade - From
2 to 
3 the slope will be 0 dB/decade (20 + (-20)) - From
3 ,
4 the slope will be -40 dB/decade (0-20-20)
Phase plot
= tan-1 0.5
- tan-1 0.2
- tan-1 0.1
- tan-1 0.1
500 -177.30
1000 -178.60
1500 -179.10
2000 -179.40
2500 -179.50
3000 -179.530
3500 -179.60
GM = 00
PM = 61.460
The plot is shown in figure 16
Fig 16 Magnitude and phase Plot for G(S) = 
Q 4. For the given transfer function plot the bode plot (magnitude plot) G(S) = 
Sol: Given transfer function
G(S) = 
Converting above transfer function to standard from
G(S) = 
= 
- As type 1 system , so initial slope will be -20 dB/decade
- Final slope will be -60 dB/decade as order of system decides the final slope
- Corner frequency
T1 = 
, 
11= 5 (zero)
T2 = 1 , 
2 = 1 (pole)
4. Initial slope will cut zero dB axis at
(K)1/N = 10
i.e 
= 10
5. finding 
n and 
T(S) = 
T(S)= 
Comparing with standard second order system equation
S2+2
ns +
n2
n = 11 rad/sec
n = 5
11 = 5
= 
= 0.27
5. Maximum error
M = -20 log 2
= +6.5 dB
6. As K = 10, so whole plot will shift by 20 log 10 10 = 20 dB
The plot is shown in figure 17
Fig 17 Magnitude Plot for G(S) = 
Q5. For the given plot determine the transfer function
Fig 18 Magnitude Plot
Sol: From figure 18 we can conclude that
- Initial slope = -20 dB/decade so type -1
- Initial slope all 0 dB axis at
= 10 so
K1/N N = 1
(K)1/N = 10.
3. corner frequency
1 = 
= 0.2 rad/sec
2 = 
= 0.125 rad/sec
3. At 
= 5 the slope becomes -40 dB/decade, so there is a pole at 
= 5 as
Slope changes from -20 dB/decade to -40 dB/decade
4. At 
= 8 the slope changes from -40 dB/decade to -20 dB/decade hence is a zero at 
= 8 (-40+(+20) = 20)
5. Hence transfer function is T(S) = 
It gives the frequency response of the system. If the transfer function is given, then from the plot number of poles and zeros can be calculated.
Polar plot of some standard functions: -
# TYPE ‘O’
Ex : 1T(S) = 1/S + 1
(1). For polar plot substitute S=jw.
TF = 1/1 + jw
(2). Magnitude M = 1 + 0j / 1 + jw = 1/√1 + w2
(3). Phase φ = tan-1(0)/ tan-1w = - tan-1w
W M φ
0 1 00
1 0.707 -450
∞ 0 -900
The plot is shown in fig. 19(a)
Fig 19(a) Polar Plot T(S) = 1/S + 1
Ex.2>. T(S) = 1/(S+1)(S+2)
(1). S = jw
TF = 1/(1+jw)(2+jw)
(2). M = 1/(1+jw)(2+jw) = 1/-w2 + 3jw + 2
M = 1/√1 + w2√4 + w2
(3). Φ = - tan-1 w - tan-1(w/2)
W M Φ
0 0.5 00
1 0.316 -71.560
2 0.158 -108.430
∞ 0 -1800
The plot is shown in fig19(b)
Fig19(b) Polar Plot T(S) = 1/(S+1)(S+2)
Intersection of polar plot with imaginary axis will be when real part of Transfer function = 0
M = 1/(jw + 1)(jw + 2)
= 1/-w2 + j3w + 2
TYPE ‘1’
Ex.1 T(S) = 1/S
(1). S = jw
(2). M = 1/W
(3). Φ = -tan-1(W/O) = -900
W M φ
0 ∞ -900
1 1 -900
2 0.5 -900
∞ 0 -900
The plot is shown in fig.20(a)
Fig 20(a) Polar Plot T(S) = 1/S
Ex.2 T(S) = 1/S2
(1). S = jw
(2). M = 1/w2
(3). Φ = -tan-1(W/O)-tan-1(W/O) = -1800
W M Φ
0 ∞ -1800
1 1 -1800
2 0.25 -1800
∞ 0 -1800
The plot is shown in fig.20(b)
Fig 20(b) Polar Plot T(S) = 1/S2
Key takeaway
- For Polar Plot starting point depends upon type of system
- The terminating phase depends on order of the system.
TYPE 1 ORDER 2
Ex.1 T(S) = 1/S(S+1)
(1). M = 1/W√1+w2
(2). Φ = -900 - tan -1(W/T)
W M φ
0 ∞ -900
1 0.707 -1350
2 0.45 -153.40
∞ 0 -1800
The plot is shown in fig.21(a)
Fig 21(a) Polar Plot T(S) = 1/S(S+1)
Ex.2TYPE 2 ORDER 3
T(s) = 1/S2(S+1)
(1). M = 1/w2√1+jw
(2). Φ = -1800 – tan-1W/T
The plot is shown in fig.21(b)
Fig 21(b) Polar Plot T(s) = 1/S2(S+1)
It gives the frequency response as well as comments on the stability and the relative stability of the system.
Nyquist Stability Criteria:
The Nyquist criteria is a semi graphical method that determines stability of CL system investigating the properties of the frequency domain plot (Polar plot), the Nyquist plot of the OLTF G(S) H(S) is represented as L(S)
L(S) = G(S)H(S)
Specially the Nyquist plot of L(S) is a plot drawn by substituting S=jw and varying the value of w as per in polar plot. In polar plot we take one sided frequency response ( 0 - ∞) in Nyquist plot we will vary the frequency in entire range possible from ( -∞ to 0 ) and (0 to ∞ )
Nyquist Criteria also gives :-
(1). In addition to providing the absolute stability like other plots, the Nyquist criteria also gives information on the relative stability of a stable system and the degree of instability of an unstable system.
(2). It also gives indications on how the system stability can be improved.
(3). The Nyquist plot of G(S) H(S) is the polar plot of G(S) H(S) drawn with wider range of frequency ( -∞ to ∞ ) and along the Nyquist path.
(4). The Nyquist plot of G(S) H(S) gives information on frequency domain characteristics such as B.W, gain margin and phase margin.
Construction of Nyquist Plot
Encircled : A point or region in a complex function phase i.e. S-plane is said to be encircled by a closed path if it is found inside the path.
Assumption :-
Fig 22 Encirclement
In this example point A is encircled by the closed path Y. Since, A is inside the closed path point B is not encircled by y. It is outside the path. Furthermore, when the closed path Y, has a direction assign to it, encirclement, if made can be in the clockwise direction or in the anti-clockwise direction.
Point A is encircled by Y by anticlockwise direction. We can say that the region inside the path is encircled in the prescribed direction and the region outside the path is not encircled.
Enclosed :-
A point or region is said to be enclosed by a closed path if it is encircled in the counter clockwise direction, or the point or region lies to the left of the path (always), when the path is traveling in the prescribed direction.
The concept of enclosure is particularly useful, if only a portion of a closed path is shown.
In this example the shaded region are
Fig 23 Enclosure
Considered to be enclosed by the closed path Y. In other words, point A is enclosed by Y in fig a. But is not enclosed by Y in fig b. And for point B it is viceversa.
No of encirclements and enclosures:
For A line is cut once
For B line is cut twice
Fig 24 Encirclement and Enclosure with example
As it’s overlapping but 2 times in Same direction
When a point is encircled by a closed path Y, a no. N can be assigned to the no. Of times it is encircled. The magnitude of N can be determined by drawing an arrow around the closed path Y.
Taking an arbitrary point S, and moving around in clockwise direction and anti-clockwise direction respectively. We are getting a direction.
The path followed by S1 gives us the direction and this path which covers the total number of revolution travelled by this point S1 is N or the net angle is ‘ 2 π N ’.
For B = 2 = N for A = 1 = N
In this e.g. Point A is encircled ones (or 2 π radians) by function Y and point B is encircled twice (or 4 π radians) all in clockwise direction.
In diagram b again A and B are encircled but in counter clockwise direction thus for this diagram A is enclosed one’s and B is enclosed twice.
By definition M is +ve for anticlockwise(direction) encirclement and –ve for clockwise encirclement.
OLTF G(S) = (S + Z1)(S + Z2)/(S + P1)(S + P2) H(S) = 1 - - (1)
Fig 25 Unity feedback control system
CLTF G(S)/1 + G(S)
CE = 1 + G(S)
= 1 + (S + Z1)(S + Z2)/(S + P1)(S + P2)
CE = (S + P1)(S + P2) + (S + Z1)(S + Z2)/ (S + P1)(S + P2) - - (2)
Key takeaway:
# OLTF poles is equal to CE poles.
CE = (S + Z’1)( S + Z’2)/( S + P1)( S + P2) - - (3)
CLTF = G(S) –(1) / 1 + G(S) –(3)
= (S + Z1)( S + Z2) / (S + Z’1)( S + Z’2) - - (4)
Key takeaway:
# Zeros of characteristic equation is poles of CLTF (3 and 4).
For the closed loop system to be stable zeros of CE(i.e. poles of CLTF) should not be located at right half of the S-plane.
Consider a contour, which covers the entire right half of S-plane.
Fig 26 Contour For Right half of S-plane
P1 W(0 - ∞)
P2 RejR ∞
ϴ - π/2 to 0 to + π/2
P3 W(∞ to 0 )
If each and every point along the boundary of contour is mapped in q(S) where q(S) is 1+G(S)H(S)[CE]. The CE is drawn in S-domain. Now, as the CE : q(s) = 1 + G(S)H(S) contour is drawn into S-plane.
This q(S) contour may encircle the origin. Thus, the number of encirclement of q(S) contour with respect to origin is given by
N = Z – P
Where : Z1P zeros and poles of q(S)[CE]
N Total no of encirclement of origin
- Z1P Zeros and poles of CE in the right half of S-plane
** for the CL system to be stable Z=0 always.
Important :-
# Open loop System(stable) :- When OL system is stable P=0 i.e. no of poles on right half
N = Z – P
If P = 0
N = Z
- Now for CL system to be stable Z = 0
- N = 0
- i.e. q(s) contour should not encircle the origin.
# Open loop system(unstable) :-Let P = 1 i.e. one OL pole is located in right half of S-plane i.e. OLTF is unstable.
As N = Z – P
N = Z – 1
For CL system to be stable the only criteria is (Z=0) i.e.
N = -1
Which means q(S) contour should encircle the origin one’s in CW direction.
Key takeaway:
(1). When OL system is unstable then corresponding CL will be stable only when q(S) contour will encircle origin in CW direction.
(2). The no of encirclements should be equal to no of open loop poles located in right half of S-plane.
** the no of encirclements(N) can also be calculated by using G(S) contour (instead of q(S) contour) but the reference is -1+j0 instead of 0+j0 i.e. the no of encirclements should be considered w.r.t -1+j0 and not with the origin.
Explanation Mapping
q(S) = 1 + G(S)
G(S) is always given to us, so we can relate G(S) with q(S).
G(S) = q(S) – 1
But q(S) can be drawn by adding 1 real part to the q(S).
Fig 27 Mapping
G(S) given then q(S) shift to right side.
Q1. For the transfer function below plot the Nyquist plot and also comment on stability?
G(S) = 1/S+1
Sol:- N = Z – P ( No pole of right half of S plane P = 0 )
P = 0, N = Z
NYQUIST PATH :-
P1 = W – (0 to - ∞)
P2 = ϴ( - π/2 to 0 to π/2 )
P3 = W(+∞ to 0)
Fig 28 Nyquist path
Substituting S = jw
G(jw) = 1/jw + 1
M = 1/√1+W2
Φ = -tan-1(W/I)
For P1 :- W(0 to -∞)
W M φ
0 1 0
-1 1/√2 +450
-∞ 0 +900
Path P2 :-
W = Rejϴ R ∞ϴ -π/2 to 0 to π/2
G(jw) = 1/1+jw
= 1/1+j(Rejϴ) (neglecting 1 as R ∞)
M = 1/Rejϴ = 1/R e-jϴ
M = 0 e-jϴ = 0
Path P3 :-
W = -∞ to 0
M = 1/√1+W2 , φ = -tan-1(W/I)
W M φ
∞ 0 -900
1 1/√2 -450
0 1 00
The Nyquist Plot is shown in fig 29
Fig 29 Nyquist Plot G(S) = 1/S+1
From plot we can see that -1 is not encircled so, N = 0
But N = Z, Z = 0
So, system is stable.
Q.2. For the transfer function below plot the Nyquist Plot and comment on stability G(S) = 1/(S + 4)(S + 5)
Soln :- N = Z – P , P = 0, No pole on right half of S-plane
N = Z
NYQUIST PATH
P1 = W(0 to -∞)
P2 = ϴ(-π/2 to 0 to +π/2)
P3 = W(∞ to 0)
Fig 30 Nyquist Path
Path P1 W(0 to -∞)
M = 1/√42 + w2 √52 + w2
Φ = -tan-1(W/4) – tan-1(W/5)
W M Φ
0 1/20 00
-1 0.047 25.350
-∞ 0 +1800
Path P3 will be the mirror image across the real axis.
Path P2 :ϴ(-π/2 to 0 to +π/2)
S = Rejϴ
G(S) = 1/(Rejϴ + 4)( Rejϴ + 5)
R∞
= 1/ R2e2jϴ = 0.e-j2ϴ = 0
The plot is shown in fig 31. From plot N=0, Z=0, system stable.
Fig 31 Nyquist Plot G(S) = 1/(S + 4)(S + 5)
Q.3. For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?
Soln: As the poles exists at origin. So, first time we do not include poles in Nyquist plot. Then check the stability for second case we include the poles at origin in Nyquist path. Then again check the stability.
PART – 1 : Not including poles at origin in the Nyquist Path.
Fig 32 Nyquist Path
P1 W(∞ Ɛ) where Ɛ 0
P2 S = Ɛejϴ ϴ(+π/2 to 0 to -π/2)
P3 W = -Ɛ to -∞
P4 S = Rejϴ, R ∞, ϴ = -π/2 to 0 to +π/2
For P1
M = 1/w.w√102 + w2 = 1/w2√102 + w2
Φ = -1800 – tan-1(w/10)
W M Φ
∞ 0 -3 π/2
Ɛ ∞ -1800
Path P3 will be mirror image of P1 about Real axis.
G(Ɛ ejϴ) = 1/( Ɛ ejϴ)2(Ɛ ejϴ + 10)
Ɛ 0, ϴ = π/2 to 0 to -π/2
= 1/ Ɛ2 e2jϴ(Ɛ ejϴ + 10)
= ∞. e-j2ϴ [ -2ϴ = -π to 0 to +π ]
Path P2 will be formed by rotating through -π to 0 to +π
Path P4 S = Rejϴ R ∞ ϴ = -π/2 to 0 to +π/2
G(Rejϴ) = 1/ (Rejϴ)2(10 + Rejϴ)
= 0
N = Z – P
No poles on right half of S plane so, P = 0
N = Z – 0
Fig 33 Nyquist Plot for G(S) = k/S2(S + 10)
But from plot shown in fig 32. It is clear that number of encirclements in Anticlockwise direction. So,
N = 2
N = Z – P
2 = Z – 0
Z = 2
Hence, system unstable.
PART 2 Including poles at origin in the Nyquist Path.
Fig 34 Nyquist Path
P1 W(∞ to Ɛ) Ɛ 0
P2 S = Ɛejϴ Ɛ 0 ϴ(+π/2 to +π to +3π/2)
P3 W(-Ɛ to -∞) Ɛ 0
P4 S = Rejϴ, R ∞, ϴ(3π/2 to 2π to +5π/2)
M = 1/W2√102 + W2 , φ = - π – tan-1(W/10)
P1 W(∞ to Ɛ)
W M φ
∞ 0 -3 π/2
Ɛ ∞ -1800
P3( mirror image of P1)
P2 S = Ɛejϴ
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10 + Ɛejϴ)
Ɛ 0
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10)
= ∞. e-j2ϴϴ(π/2 to π to 3π/2)
-2ϴ = (-π to -2π to -3π)
P4 = 0
Fig 35 Nyquist Plot G(S) = k/S2(S + 10)
The plot is shown in fig 35. From the plot it is clear that there is no encirclement of -1 in Nyquist path. (N = 0). But the two poles at origin lies to the right half of S-plane in Nyquist path.(P = 2)[see path P2]
N = Z – P
0 = Z – 2
Z = 2
Hence, system is unstable.
Path P2 will be formed by rotating through -π to -2π to -3π
NOTE : The sign convention for angles is shown
Fig 36 Sign for Angle Directions
Angles are considered –ve for anticlockwise directions and +ve for clockwise directions.
1) The more the value of gain margin more is the stability.
2) The more the value of phase margin more is the stability.
3) The gain cross over frequency, phase cross over frequency is used in determining the stability of system.
4) The frequency at which the two asymptotes cuts or meet each other is known as break frequency or corner frequency.
Reference
1 A. Ananadkumar, “Control system Engineering” PHI publication 2nd edition.
2 R. Anandanatarajan, P. Ramesh Babu , “Control Systems Engineering”, Scitech Publications .
3 John R. Hackworth,Fredrick D. Hackworth “ Programmable Logic Controller” Pearson publication.
4 I.J. Nagrath, M.Gopal “Control Systems Engineering”, 5th Edition, New Age International
