Control Systems Test 4

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Control Systems Test 4

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Question 1
1 / -0

A plant has an open-loop transfer function
\({G_P}\left( s \right) = \frac{{20}}{{\left( {s + 0.1} \right)\left( {s + 2} \right)\left( {s + 100} \right)}}\)
The approximate model obtained by retaining only one of the above poles, which is closest to the frequency response of the original transfer function at low frequency is

A
\(\frac{{0.1}}{{s + 0.1}}\)

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

C
\(\frac{{100}}{{s + 100}}\)

D
\(\frac{{20}}{{s + 0.1}}\)

Solution

According to dominant pole concept, we can neglect the poles which are far from origin but the DC gain should be constant.
Approximate transfer function \(= \frac{{20}}{{\left( {s + 0.1} \right)\left( 2 \right)\left( {100} \right)}} = \frac{{0.1}}{{\left( {s + 0.1} \right)}}\)
Question 2
1 / -0

A second order system has the following properties.
The damping ratio, ξ = 0.8 damped frequency, ω_d = 6 rad/sec
The DC gain of the system is 2.
The transfer function of the system is

A
\(\frac{{200}}{{{s^2} + 16s + 100}}\)

B
\(\frac{{100}}{{{s^2} + 16s + 50}}\)

C
\(\frac{{50}}{{{s^2} + 12.8s + 64}}\)

D
\(\frac{{128}}{{{s^2} + 12.8\;s + 64}}\)

Solution

Damping ratio (ξ) = 0.8
Damped frequency (ω_d) = 6 rad/sec
\( \Rightarrow {\omega _n}\;\sqrt {1 - {\xi ^2}} = 6\)
\( \Rightarrow {\omega _n}\;\sqrt {1 - {{0.8}^2}} = 6\)
⇒ ω_n = 10 rad/sec
Standard second order system transfer function
\(T\left( s \right) = \frac{{k\;\omega _n^2}}{{{s^2} + 2\xi {\omega _n}s + \omega _n^2}}\)
\( = \frac{{100\;k}}{{{s^2} + 16s + 100}}\)
DC gain = 2
\( \Rightarrow \frac{{100k}}{{100}} = 2\)
⇒ k = 2
Now, \(T\left( s \right) = \frac{{200}}{{{s^2} + 16s + 100}}\)
Question 3
1 / -0

For a system having transfer function \(G\left( s \right) = \frac{{2 - s}}{{2 + s}}\), a step input is applied at time t = 0. The value of the system at t = 2 sec is ______ (rounded off to three decimal places)

A
0.96-0.97

B
0.964-0.971

C
0.967-0.976

D
0.969-0.972

Solution

\(G\left( s \right) = \frac{{2 - s}}{{2 + s}}\)
\(r\left( t \right) = u\left( t \right)\)
\( \Rightarrow R\left( s \right) = \frac{1}{s}\)
\( \Rightarrow \frac{{C\left( s \right)}}{{R\left( s \right)}} = G\left( s \right) = \frac{{2 - s}}{{2 + s}}\)
\( \Rightarrow C\left( s \right) = \left( {\frac{{2 - s}}{{2 + s}}} \right)\left( {\frac{1}{s}} \right)\)
\( \Rightarrow C\left( s \right) = \frac{1}{s} = \frac{2}{{2 + s}}\)
⇒ C(t) = u(t) – 2e^-2t u(t)
At t = 2 sec,
c(t = 2s) = 1 – 2e^-4 = 0.9633