Control system miscellaneous Easy Questions and Answers | Page - 36

Control system miscellaneous

The feedback system shown below oscillates at 2 rad/s when

1. K = 2 and a = 0.75
1. K = 3 and a = 0.75
1. K = 4 and a = 0.5
1. K = 2 and a = 0.5

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Characterisitic equation is
1 + G(s)H(s) = 0
∴ 1 + k(s + 1) = 0
s³ + as² + 2s + 1

⇒ s³ + s²a + s(k + 2) + (k + 1) = 0

as² + (k + 1) = 0 ; s = jω ; s² = – ω²; ; ω = 2
– aω² + (k + 1) = 0 ; aω² = k + 1 ; 4a = k + 1 ;
From options, k = 2, a = 0.75

Correct Option: A

Characterisitic equation is
1 + G(s)H(s) = 0
∴ 1 + k(s + 1) = 0
s³ + as² + 2s + 1

⇒ s³ + s²a + s(k + 2) + (k + 1) = 0

as² + (k + 1) = 0 ; s = jω ; s² = – ω²; ; ω = 2
– aω² + (k + 1) = 0 ; aω² = k + 1 ; 4a = k + 1 ;
From options, k = 2, a = 0.75

The frequency response of a linear system G(jω) is provided in the tabular form below
| G(jω)| 1.3 1.2 1.0 0.8 0.5 0.3
∠ G(jω) – 130° – 140° – 150° – 160° – 180° – 200°

The gain margin and phase margin of the system are

1. 6 dB and 30°
1. 6 dB and – 30°
1. – 6 dB and 30°
1. – 6 dB and – 30°

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At ∠ G(jω) = – 180, magnitude M = 0.5
∴ G.M = 20 log 1 = 6 dB
0.5

At | G(jω) | = 1, phase angle ∠ G(jω) = – 150
∴ P.M = 180 + (– 150) = 30°

Correct Option: A

At ∠ G(jω) = – 180, magnitude M = 0.5
∴ G.M = 20 log 1 = 6 dB
0.5

At | G(jω) | = 1, phase angle ∠ G(jω) = – 150
∴ P.M = 180 + (– 150) = 30°

The steady state error of a unity feedback linear system for a unit step input is 0.1. The steady state error of the same system, for a pulse input r(t) having a magnitude of 10 and a duration of one second, as shown in the figure is

1. 0
1. 0.1
1. 1
1. 10

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For step input, e_ss = 0.1 = 1
1 + k

Now input is pulse r(t) = 10 [u(t) – u(t – 1)]
R(s) = 10 1 - e^-s
s

Correct Option: A

For step input, e_ss = 0.1 = 1
1 + k

Now input is pulse r(t) = 10 [u(t) – u(t – 1)]
R(s) = 10 1 - e^-s
s

A point z has been plotted in the complex plane, as shown in figure below.

The plot of the complex number y = 1 is
z

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| Z| < 1, so | Y| > 1
Z is having +ve real part and positive imaginary part (from characteristics)
So,Y should have +ve real part and negative imaginary part.

Correct Option: D

| Z| < 1, so | Y| > 1
Z is having +ve real part and positive imaginary part (from characteristics)
So,Y should have +ve real part and negative imaginary part.

An open loop system represented by the transfer function G(s) = (s - 1) is
(s + 2)(s + 3)

1. stable and of the minimum phase type
1. stable and of the non-minimum phase type
1. unstable and of the minimum phase type
1. unstable and of the non-minimum phase type

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Since, open loop system stability is depends only on pole locations, hence system is stable. There is one zero on right half of s-plane so system is non-minimum phase.

Correct Option: B

Since, open loop system stability is depends only on pole locations, hence system is stable. There is one zero on right half of s-plane so system is non-minimum phase.