Control Engineering

NPTEL Online Course

Indian Institute of Technology Madras
Assignment 6
(Common data for Q1- Q3)
Consider the following transfer function

(s − 2)
KG(s)H(s) =
(s + 1)2

1. For K = 1 The Nyquist plot for the above transfer function is

1.5

1.0

0.5

-2.0 -1.5 -1.0 -0.5 0.5

-0.5

-1.0

-1.5

2. From the selected Nyquist plot in above question is closed loop system stable?
(a) Yes
(b) No [correct]
3. Determine the range of K for which the system is stable using the Nyquist plot for the following
transfer functions:

−1
(a) 2 <K<2
1
(b) 0 < K < 2
1
(c) −2 < K < 2 [correct]
−1
(d) −2 < K < 2

Solution:

1
 K(s−2)
a. KG(s)H(s) = (s+1)2

Detailed steps for plotting the Nyquist plot is : 1. Put s = jω and find KG(jω)H(jω)
in terms of M ∠θ.

K(s−2)
G(s)H(s) = (s+1)2

K(jω−2) K(jω−2)
i.e., G(jω)H(jω) = (jω+1)2 = 1−ω 2 +2jω
√
K (ω 2 +4)
Here, M = √
(1−ω 2 )2 +(2ω)2 )

and ∠θ = 180o − tan−1 ( ω2 ) − 2tan−1 ( ω1 )

2. Draw the polar plot by varying ω from zero to infinity. If pole or zero present at ‘s0 = 0,
then varying ω from 0+ to infinity for drawing polar plot.
√
K (ω 2 +4)
M |at ω=0 =√ = 2K
(1−ω 2 )2 +(2ω)2 )
√
K (ω 2 +4)
M |at ω=∞ =√ =0
(1−ω 2 )2 +(2ω)2 )

and ∠θ|at ω=0 = 180o − tan−1 ( ω2 ) − 2tan−1 ( ω1 ) = 180o

∠θ|at ω=∞ = 180o − tan−1 ( ω2 ) − 2tan−1 ( ω1 ) = −90o

3. For K=1, find out intersection on the X and Y axis in the plot by multiplying numerator
and denominator of G(jω)H(jω) by complex conjugate.
(jω−2)
G(jω)H(jω) = 1−ω 2 +2jω

(jω−2)(1−ω 2 −2jω)
= (1−ω 2 +2jω)(1−ω 2 −2jω)
(jω−2)(1−ω 2 −2jω)
= (1−ω 2 )2 +(2ω)2
jω−jω 3 +2ω 2 −2+2ω 2 +4jω
= (1−ω 2 )2 +(2ω)2
j(−ω 3 +5ω)−2+4ω 2
= (1−ω 2 )2 +(2ω)2
Here, real part will be equal to 0 at ω = √12 . Imaginary part will be equal to 0 at ω = 0
√
and ω = 5.
The magnitude at ω = 0 is; M = 2K √.
The magnitude at ω = √12 is; M = 2K .
√
The magnitude at ω = 5 is; M = K 2 .

4. For K=1, The magnitude at ω =√0 is; M = 2 .

The magnitude at ω = √12 is; M = 2 .
√
The magnitude at ω = 5 is; M = 21 .
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 direction of the plot can be obtained by checking
angles for some frequencies in between ω = 0 and ∞.

5. 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

2
 Im
1
p ! = p
2
2K

K
2K 2
pRe
! = 0 -1 ! = 1 ! = 5

Figure 1: Nyquist plot

plot starts.
Number of poles at origin=0. Hence, there is No infinite radius half circle.
6. If there are P poles and Z zeros are enclosed by the ‘s0 plane closed path, then the
corresponding G(s)H(s) plane must encircle -1, Z − P times. We can write the number of
encirclements N as, Z = P + N . N will be taken as positive for CW encirclement.
P=0 (Right hand side 0 pole)
N=1 (1 clockwise encirclement of -1 point in L.H.P. on real axis)
Hence, Z=1.
Hence, system is unstable for K=1.
Now, to find range of stability, we can apply R.H. criterion and find out the range of K
for stability. But, directly using Nyquist plot, we can see that for Z=0, we need to get
N=0, we need to find K to make system stable. Hence, The magnitude at ω = 0 is;
M = 2K and should be 2K < 1 and always √ from right hand side (positive x axis)
cut take > −1 i.e., The magnitude at ω = 5 is; M = K 2 and should be K
2 > −1. Hence,
K > −2 and K < 21 or, −2 < K < 12 is the range of stability.

4. The open loop transfer function and closed loop transfer function of a system are stable. Then,
which of the following statement is true?
(a) Contour of GH encircles the origin
(b) Contour of GH does not encircle the origin
(c) Contour of GH does not encircle (−1 + j0) [Correct]
(d) Contour of GH does encircles (−1 + j0)

3
 Solution

If the open loop system is stable, then there are no open loop poles (P ) in RHP. If the closed
loop system is stable, there are no zeros of 1 + GH (Z) in the right half plane. Therefore
N = Z − P = 0 and contour of of GH does not encircle (−1 + j0)
(Common data for Q5-Q7)
Consider a unity feedback control system whose open-loop transfer function is
as + 1
G(s) =
s2
5. What is the gain crossover frequency in terms of a? [Hint: |G(jω)| = 1 at ω = ωgc ]
q √
2 4
(a) ωgc = a + 2a +4 [Correct]
√
4
(b) ωgc = a2 +4
(c) ωgc = a2
√
a2 + a4 +8
(d) ωgc = 4

6. What is the relation between ωgc and a when the phase margin is 45o ?
(a) ωgc a = 1 [Correct]
(b) ωgc = a
(c) ωgc a2 = 1
(d) ωgc a2 = 1
7. Combining both the relations obtained in previous two questions, what is the value of a rounded
to one decimal point ? [Ans = 0.8] (Answer range between 0.7 and 0.9)

Solution

At the gain crossover frequency, |G(jω)| = 1. Substituting jω in the transfer function, we get:
aωj + 1
G(jω) =
−ω 2
√
1 + a2 ω 2
|G(jω)| = 2
√ ω
1 + a2 ω 2
1=
ω2
ω 4 = 1 + a2 ω 2
Setting ω 2 = x, we get:
x2 = 1 + a2 x
√
a2 ± a4 + 4
x=
2
Since x = ω 2 > 0,
√
a2 + a4 + 4
x=
2
s √
a2 + a4 + 4
ωgc =
2

4
 Given phase margin to be 45o . Therefore,
P M = 180 + ∠G(jωgc )
45 = 180 − 180 + tan−1 (aωgc )
45 = tan−1 (aωgc )
∴ a ωgc = 1
q √
a2 + a4 +4
Substituting ωgc = 2 in a ωgc = 1, we get:
a = 0.51/4 = 0.84
(Common data for Q8-Q10)

Considering the frequency response of the standard 2nd order closed loop system, with transfer
ω2
function T (s) = s2 +2ζωnn +ω2 , the closed loop system with performance characteristics : ζ =
n
0.2, TS = 0.5s
8. The resonant peak frequency (ωP ) in rad/second [Answer range: 36 - 40]
9. Find magnitude of the resonant peak (MP ) [Answer range: 1.8 - 3.2]
10. Find bandwidth (ωBW ) in rad/second [Answer range: 56 - 64]
Solution:

(a) the peak value of magnitude, MP = √1

2ζ 1−ζ 2
p
(b) the frequency at which this peak occurs, ωP = ωn 1 − 2ζ 2
q p
(c) the bandwith, ωBW = ωn (1 − 2ζ 2 ) + (4ζ 4 − 4ζ 2 + 2)
ζ = 0.2 , TS = 0.5s =⇒ ζω4n = 0.5 =⇒ ωn = 0.5×0.2 4
= 40.
Plugging the values of ζ and ωn into the expressions given above we get the required values
as :

MP = 2.55 , ωP = 38.367 , ωBW = 60.38

(Common data for Q11 - Q12)

Given a stable unity feedback system whose open-loop transfer function has the Bode magnitude
plot shown below. (Note: magnitude at 0.01 rad/sec is ≈ 120dB)

5
11. The type of the system is . [Answer range 1.99 to 2.01]

Solution
First determine the type of the system
Using the Slope condition,
∂M (dB) 120 − 50
slope = = ≈ −40dB/dec
∂(log ω) log(0.01) − log(0.55)
Thus system is type 2

√
12. The initial
√ −40dB/dec slope intersects the frequency axis at Ka
=⇒ Ka = 9.5
=⇒ Ka = 90.25
13. Consider a second-order system expressed as
1
G(s)H(s) =
s2 + 2ζs + 1
For different values of ζ, Nyquist plot of such a system is shown below

Im(z)

Re(z)
-1.0 -0.5 0.5 1.0 1.5

-1

-2

6
 Which of the following statement(s) is(are) true?
(a) As ζ increases, the size of the lobes in the plot decreases. [Correct]
(b) As ζ increases, the size of the lobes in the plot increases.
(c) The plot size is affected due to change in the natural frequency only.
(d) The plot size is not affected by value of ζ.

Solution:
As ζ increases, the plot size reduces and vice-versa.
14. If the gain margin of a certain stable feedback system is 20 dB, the Nyquist plot will cross the negative
real axis at the point

(a) −0.05
(b) −0.02
(c) −0.1 [Correct]
(d) −0.01

Solution:
 
1
20 log = 20
|GH(jω)|
1
⇒ = 10
|GH(jω)|
⇒ |GH(jω)| = 0.1

Since the system is stable, it will cross at s = −0.1.

15. The Bode magnitude plot of

104 (1 + jω)
H(jω) = , is
(10 + jω)(100 + jω)2

Correct option (a)

7
Solution:
Given transfer function is
104 (1 + jω)
H(jω) =
(10 + jω)(100 + jω)2
0.1(1 + jω)
=
ω ω 2



1 + j 10 1 + j 100

The system is type 0. So, initial slope of the Bode plot is 0 dB/decade.
Corner frequencies are

ω1 = 1 rad/s
ω2 = 10 rad/s
ω3 = 100 rad/s

As the initial slope of Bode plot is 0 dB/decade and corner frequency ω1 = 1 rad/s, the slope after
ω = 1 rad/s is (0 + 20) = +20 dB/decade.
After corner frequency, ω2 = 10 rad/s, the slope is (20 − 20) = 0 dB/decade.
Similarly, after ω3 = 100 rad/s, the slope is (0 − 20 × 2) = −40 dB/decade.