Control Systems Test 7

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Control Systems Test 7

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Weekly Quiz Competition

Question 1
1 / -0

Consider a plant transfer function \(G\left( s \right) = \frac{1}{s}\). The state feedback gain controller K that places the pole of the regulatory control systems at s = -2 is given by

A
K = 0

B
K = 1

C
K = 2

D
K = 0.5

Solution

Characteristic equation is
CE = 1 + G(S) = 0
\(\Rightarrow 1 + \frac{K}{s} = 0\)
⇒ S + K = 0 ⇒ S = -K
Given that, S = -2
⇒ K = 2
Question 2
1 / -0

A proportional band of a PI controller is 40% and the reset time is 0.25s. The transfer function of the controller is given by

A
\(0.4\left( {1 + \frac{{4}}{s}} \right)\)

B
\(2.5\left( {1 + \frac{4}{s}} \right)\)

C
\(0.4\left( {1 + \frac{{0.25}}{s}} \right)\)

D
\(2.5\left( {1 + \frac{{0.25}}{s}} \right)\)

Solution

The standard transfer function of a PI controller is
\(T\left( s \right) = {K_p} + \frac{{{K_I}}}{s} = {K_p}\;\left( {1 + \frac{1}{{{T_I}s}}} \right)\)
Where \({T_I} = \frac{{{K_p}}}{{{k_I}}}\) is known as integral or reset time.
Proportional band \(\left( {PB} \right) = \frac{{100}}{{{K_p}}}\)
The transfer function of PI controller is
\(T\left( s \right) = 2.5\;\left( {1 + \frac{1}{{0.25s}}} \right)\)
\(= 2.5\;\left( {1 + \frac{4}{s}} \right)\)
Question 3
1 / -0

If we have a control system which is actually unstable or has too low gain or phase-margins, then for a more satisfactory performance of the system, we should resort to which of the following?
i. Reduce the gain of the amplifier
ii. Incorporate the phase advance circuit in the system
iii. Use integral error compensation in the system
Select the correct answer using the codes given below:

A
i, ii and iii

B
i and ii

C
i and iii

D
ii and iii

Solution

i) Reducing the gain of the amplifier in the control system may improve stability
ii) Phase advance circuit improves gain margin and phase margin by adding a zero to the transfer function
iii) Integral error compensation minimizes steady-state error but doesn't improve stability
Integral compensator adds a pole at origin which reduces the stability.
Hence only first 2 options can be used to increase the stability
Question 4
1 / -0

The transfer function of a position servo system is given as \(G\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}}.\) A first order compensator is designed in a unity feedback configuration so that the poles of the compensated system are placed at -1 ± j1 and -4. The transfer function of the compensated system is

A
\(\frac{{s + 3}}{{2\left( {s + 5} \right)}}\)

B
\(\frac{{2s + 3}}{{s + 5}}\)

C
\(\frac{{5\left( {s + 1.6} \right)}}{{s + 5}}\)

D
\(\frac{{3\left( {2s + 3} \right)}}{{s + 4}}\)

Solution

For the given poles the characteristics equation is
(s + 1 + j) (s + 1 - j) (s + 4) = (s² + 2s + 2) (s + 4) = 0
⇒ (s³ + 6s² + 10s + 8) = 0 ______ (1)
After adding compensators to the given system
\(CE = 1 + \frac{1}{{s\left( {s + 1} \right)}}\left[ {T.F.\;of\;compensator} \right]\) ______ (2)
(1) = (2)
For option (3)
\(CE = 1 + \frac{1}{{s\left( {s + 1} \right)}}\frac{{5\left( {s + 1.6} \right)}}{{\left( {s + 5} \right)}} = 0\)
\(CE = {s^3} + 6{s^2} + 10s + 8 = 0\)
Thus option (3) is correct
Question 5
1 / -0

The given uncompensated system \(\frac{{900}}{{s\left( {s + 1} \right)\left( {s + 9} \right)}}\) is to be such that its gain crossover frequency becomes same as its uncompensated phase crossover frequency and provides a 45° phase margin. To achieve this, one may use

A
a lag compensator that provides an attenuation of 20 dB and a phase lag of 45° at the frequency of 3√3 rad/s

B
a lead compensator that provides an amplification of 20 dB and a phase lead of 45° at the frequency of 3 rad/s

C
a lag-lead compensator that provides an amplification of 20 dB and a phase lag of 45° at the frequency of √3 rad/s

D
a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45° at the frequency of 3 rad/s

Solution

A phase crossover frequency, ∠G(jω) = -180°
-tan^-1(ω) - tan^-1(ω/9) - 90° = - 180°
⇒ ω_pc = 3 rad/sec
\(|G{\rm{(}}j{\omega _{pc}}H\left( {j{\omega _{pc}}} \right){\rm{|}} = \frac{{900}}{{s\left( {s + 1} \right)\left( {s + 9} \right)}} = 10\)
Gain margin. GM = 20 log 10 = -20 dB
System is unstable.
ω_pc should become ω_gc
⇒ ω_gc = 3 rad/sec
The magnitude should be 0 dB.
To make the magnitude 0 dB at ω_gc = 3 rad/sec a lag compensator which gives an attenuation of 20 dB is used.
To obtain 45° phase margin at ω_pc = 3 rad/sec.
A lead compensator with a phase lead of 45° used
Therefore, a lag-lead compensator that provides an attenuation of 20 dB and phase lead of 45° at the frequency of 3 rad/s.