Unit - 4
Time and Frequency Domain Analysis
Q1) The open loop transfer function of a system with unity feedback gain G (S) = 20 / S2 + 5S + 4. Determine the ξ, Mp, tr, tp.
A1) Finding closed loop transfer function,
C (S) / R (S) = G (S) / 1 + G (S) + H (S)
As it is unity feedback so, H(S) = 1
C(S)/R(S) = G(S)/1 + G(S)
= 20/S2 + 5S + 4/1 + 20/S2 + 5S + 4
C(S)/R(S) = 20/S2 + 5S + 24
Standard equation for second order system,
S2 + 2ξWnS + Wn2 = 0
We have,
S2 + 5S + 24 = 0
Wn2 = 24
Wn = 4.89 rad/sec
2ξWn = 5
(a). ξ = 5/2 x 4.89 = 0.511
(b). Mp% = e-∏ξ / √1 –ξ2 x 100
= e-∏ x 0.511 / √1 – (0.511)2 x 100
Mp% = 15.4%
(c). tr = ∏ - φ / Wd
φ = tan-1√1 – ξ2 / ξ
φ= tan-1√1 – (0.511)2 / (0.511)
φ = 1.03 rad.
tr = ∏ - 1.03/Wd
Wd = Wn√1 – ξ2
= 4.89 √1 – (0.511)2
Wd = 4.20 rad/sec
tr = ∏ - 1.03/4.20
tr = 502.34 msec
(d). tp = ∏/4.20 = 747.9 msec
Q2) A second order system has Wn = 5 rad/sec and is ξ = 0.7 subjected to unit step input. Find (i) closed loop transfer function. (ii) Peak time (iii) Rise time (iv) Settling time (v) Peak overshoot.
A2) (i) The closed loop transfer function is
C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2
= (5)2 / S2 + 2 x 0.7 x S + (5)2
C(S)/R(S) = 25 / S2 + 7s + 25
(ii). tp = ∏ / Wd
Wd = Wn√1 - ξ2
= 5√1 – (0.7)2
= 3.571 sec
(iii). tr = ∏ - φ/Wd
φ= tan-1√1 – ξ2 / ξ = 0.795 rad
tr = ∏ - 0.795 / 3.571
tr = 0.657 sec
(iv). For 2% settling time
ts = 4 / ξWn = 4 / 0.7 x 5
ts = 1.143 sec
(v). Mp = e-∏ξ / √1 –ξ2 x 100
Mp = 4.59%
Q3) The open loop transfer function of a unity feedback control system is given by
G(S) = K/S (1 + ST)
Calculate the value by which k should be multiplied so that damping ratio is increased from 0.2 to 0.4?
A3) C(S)/R(S) = G(S) / 1 + G(S)H(S) H(S) = 1
C(S)/R(S) = K/S (1 + ST) / 1 + K/S (1 + ST)
C(S)/R(S) = K/S (1 + ST) + K
C(S)/R(S) = K/T / S2 + S/T + K/T
For second order system,
S2 + 2ξWnS + Wn2
2ξWn = 1/T
ξ = 1/2WnT
Wn2 = K/T
Wn =√K/T
ξ = 1 / 2√K/T
ξ = 1 / 2 √KT
Forξ1 = 0.2, for ξ2 = 0.4
ξ1 = 1 / 2 √K1T
ξ2 = 1 / 2 √K2T
ξ1/ ξ2 = √K2/K1
K2/K1 = (0.2/0.4)2
K2/K1 = 1 / 4
K1 = 4K2
Q4) Consider the transfer function C(S)/R(S) = Wn2 / S2 + 2ξWnS + Wn2
Find ξ, Wn so that the system responds to a step input with 5% overshoot and settling time of 4 sec?
A4)
(i) Mp = 5% = 0.05
Mp = e-∏ξ / √1 –ξ2
0.05 = e-∏ξ / √1 –ξ2
Cn 0.05 = - ∏ξ / √1 –ξ2
-2.99 = - ∏ξ / √1 –ξ2
8.97(1 – ξ2) = ξ2∏2
0.91 – 0.91 ξ2 = ξ2
0.91 = 1.91 ξ2
ξ2 = 0.69
(ii). ts = 4/ ξWn
4 = 4/ ξWn
Wn = 1/ ξ = 1/ 0.69
Wn = 1.45 rad/sec
Q5) For the CLTF G(s) = 
. Find Kp, Kv and Ka?
A5) Kp = 
G(s)
= 
= 1
Kv = 
SG(s)
= 
S 
= 0
Ka = 
s2 G(s)
= 
s2 
= 0
Q6) Sketch the bode plot for transfer function G(S) = 
A6)
1. 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
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 below
Fig: Magnitude Plot for G(S) = 
Q7) For the given transfer function determine G(S) = 
Gain cross over frequency phase cross over frequency phase mergence and gain margin
A7)
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
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 below
Fig: Magnitude Plot G(S) = 
Q8) For the given transfer function G(S) = 
.Plot the rode plot find PM and GM.
A8)
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)
1. Initial slope 0 dB/decade till 
1 = 2 rad/sec
2. From 
1 to
2 (i.e., 2 rad /sec to 5 rad/sec) slope will be 20 dB/decade
3. From 
2 to 
3 the slope will be 0 dB/decade (20 + (-20))
4. 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 below
Fig: Magnitude and phase Plot for G(S) = 
Q8) For the given transfer function plot the bode plot (magnitude plot) G(S) = 
A9)
Given transfer function
G(S) = 
Converting above transfer function to standard from
G(S) = 
= 
1. As type 1 system, so initial slope will be -20 dB/decade
2. Final slope will be -60 dB/decade as order of system decides the final slope
3. 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
6. Maximum error
M = -20 log 2
= +6.5 dB
As K = 10, so whole plot will shift by 20 log 10 10 = 20 dB
The plot is shown in figure below
Fig: Magnitude Plot for G(S) = 
Q9) For the given plot determine the transfer function
Fig: Magnitude plot for Q10
A10)
From figure, we can conclude that
1. Initial slope = -20 dB/decade so type -1
2. 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
4. 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
5. At 
= 8 the slope changes from -40 dB/decade to -20 dB/decade hence is a zero at 
= 8 (-40+(+20) = 20)
6. Hence transfer function is T(S) = 
Q10) For a unity feedback system G(s) = 
an input t3u(t) is applied. Find the steady state error?
A11) ess= 
r(t) = t3u(t)
R(s) = 6/s4
H(s) =1
ess= 
= 
= 
= 5/3
Q11) For the OLTF with unity feedback is G(s)= 
. Determine the damping ratio, maximum overshoot, rise time?
A12) The CLTF will be T(s) = 
T(s) = 
For second order system,
S2 + 2ξWnS + Wn2
wn = 
= 5.1
2ξWn = 5
i) Damping Ratio 
= 0.49
Ii) Maximum overshoot Mp = e-∏ξ / √1 –ξ2 x 100
= e-∏x0.49 / √1 –(0.49)2
= 17.1%
Iii) Rise Time tr = ∏ - φ/Wd
Wd = Wn√1 - ξ2
= 5.1 √1 – (0.49)2
= 4.45 sec
φ= tan-1√1 – ξ2 / ξ = 1.059 rad
tr = ∏ - φ/Wd
= ∏ - 1.059/4.45
=468.53msec
Q12) Define and derive the time domain specifications for a second order system?
A13)
1) Rise Time (tp): The time taken by the output to reach the already status value for the first time is known as Rise time.
C(t) = 1-e-
wnt/
1-
2 sin (wdt+ø)
Sin (wd +ø) = 0
Wdt +ø = n
tr =n
-ø/wd
For first time so, n=1.
tr = 
-ø/wd
T=1/
2) Peak Time (tp)
The peak value attained by the output is called peak time. The time required by the output to reach this value is lp.
d(cct) /dt = 0 (maxima)
d(t)/dt = peak value
tp = n
/wd for n=1
tp = 
wd
3) Peak Overshoot Value:
Maximum deviation of output from steady state value is called peak overshoot value (Mp).
(ltp) = 1 = Mp
(
Sin (Wat + φ)
(
Sin (Wd∏/Wd + φ)
Mp = e-∏ξ / √1 –ξ2
Condition 3 ξ = 1
C (S) = R (S) Wn2 / S2 + 2ξWnS + Wn2
C (S) = Wn2 / S (S2 + 2WnS + Wn2) [ R(S) = 1/S]
C (S) = Wn2 / S (S2 + Wn2)
C (t) = 1 – e-Wnt + tWne-Wnt
The response is critically damped.
4) Settling Time (ts):
ts = 3 / ξWn (5%)
ts = 4 / ξWn (2%)
Q13) Determine the type and order of the system G(s)= K/S(S+1)
G(S)= K(S+1)/S2(S+2)
A14)
G(s)= K/S(S+1)
It is order 2 and type 1 system
G(S)= K(S+1)/S2(S+2)
It is order 3 and type 2 system
Q14) Derive position error coefficient?
A15) R(s)= Unit step
R(t) = u(t)
R(s) = 1/s
Ess = 
s [ R(s)/1+G(s)]
= 
s[ys/1+G(s)]
ess= 
1+1/1+G(s)
=1/1+lt G(s) s- 0
Kp = 
G(s)
(Position error coefficient)
