UNIT-5
Frequency Domain Analysis
Q1) Plot polar plot for T(S) = 1/S + 1
Sol: 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.
Q2) Plot polar plot for 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 fig
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
Real part
Re(M) = 1/(2-w2)+j3w x (2-w2)-3jw/(2-w2)-3jw
Re(M) = (2-w2)/(2-w2+9w2) - 3jw/(2-w2) +9w2
Equating Real part = 0
(2-w2)/(2-w2)+ 9w2 = 0
W = +-√2
For w=√2 real part on the polar plot becomes zero.
So, the polar plot intersects the imaginary axis at w=√2 at φ = -900
Q3) Plot polar plot for T(S) = 1/(S+1)(S+2)(S+3)
(1). Substitute S =jw
(2). M = 1/√1+w2 √4+w2 √9+w2
(3). Φ = -tan-1w – tan-1 w/2 – tan-1w/3
W M φ
0 0.16 0
1 0.1 -900
2 0.04 -142.10
∞ 0 -270
(4). The intersection of Polar plot with Real axis
M = 1/(S+1)(S+2)(S+3)
= 1/S3+6S2+11S+6
=1/(jw)2+6(jw)2+11jw+6
= 1/(6-6w2)+j(11w-w3)
Re(M) = 0 [Intersection with imaginary axis ]
M = 6-6w2/(6-6w2)2+(11w-w2)2 - j(11w-w3)/(6-6w2)2+(11w-w2)2
Re(M) = 0
6-6w2 = 0
w=1
Im(M) = 0[Intersection with Real axis]
11w = w3
The plot is shown in fig 3(c)
Q4) For T(S) = 1/S(S+1) plot polar plot?
Sol:
(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.
Q5) For T(s) = 1/S2(S+1) plot polar plot?
Sol:
(1). M = 1/w2√1+jw
(2). Φ = -1800 – tan-1W/T
The plot is shown in fig.5(b)
Q6) Sketch the bode plot for the transfer function
G(S) = 
Sol: 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/decade till second corner frequency i.e 
2 after 
2 the slope will be -40 dB/decade (-20+(-20)) as there are poles
= 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 1
Q7) For the given transfer function determines
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 2
Q8. 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)
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 3
Q 9) For the given transfer function plot the bode plot (magnitude plot)
G(S) = 
Given the transfer function
G(S) = 
Converting above transfer function to standard form
G(S) = 
= 
T1 = 
, 
11= 5 (zero)
T2 = 1 , 
2 = 1 (pole)
4. The 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 4
Q 10. For the given plot determine the transfer function
From figure 5 we can conclude that
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
5. is a zero at 
= 8 (-40+(+20)=20)
6. Hence transfer function is
T(S) = 
Q11) For T(S) = 1/S plot polar plot?
Sol:
(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.
Q12. 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 the 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)
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 6
From the plot we can see that -1 is not encircled so, N = 0
But N = Z, Z = 0
So, the system is stable.
Q.13. 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)
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 . From plot N=0, Z=0, system stable.
Q.14. For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?
Soln: As the poles exist at the origin. So, the first time we do not include poles in the Nyquist plot. Then check the stability for the second case we include the poles at the origin in the Nyquist path. Then again check the stability.
PART – 1: Not including poles at the origin in the 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 a mirror image of P1 about a 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 the right half of the S plane so, P = 0
N = Z – 0
But from the plot shown in fig 8(a). it is clear that a number of encirclements in the Anticlockwise direction. So,
N = 2
N = Z – P
2 = Z – 0
Z = 2
Hence, the system unstable.
PART 2 Including poles at the origin in the 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
The plot is shown in fig. from the plot it is clear that there is no encirclement of -1 in the Nyquist path. (N = 0). But the two poles at origin lies to the right half of the S-plane in the Nyquist path.(P = 2)[see path P2]
N = Z – P
0 = Z – 2
Z = 2
Hence, the system is unstable.
Path P2 will be formed by rotating through -π to -2π to -3π
