GATE 2007 Control Systems Practice Questions and Solutions

[1 mark] The closed-loop transfer function of a unity-feedback position control system is

The peak overshoot of the unit-step response is closest to

1. 16.3 %
2. 25.4 %
3. 30.5 %
4. 40.0 %

Solution

Compare with the standard second-order form \(\omega_n^{2}/(s^{2}+2\zeta\omega_n s+\omega_n^{2})\) discussed in Chapter 8.

The peak overshoot of an underdamped second-order system depends on \(\zeta\) alone:

So the overshoot is \(25.38\%\), which rounds to 25.4 %.

[2 marks] A unity-negative-feedback system has the open-loop transfer function

The input is the ramp \(r(t)=5t\,u(t)\). The value of \(K\) that makes the steady-state error equal to 0.5 is _____.

Solution

The open loop has one pole at the origin, so the system is Type 1 and the relevant static error constant is the velocity constant \(K_v\) (see Chapter 9).

For a ramp of slope \(A=5\), the steady-state error is \(e_{ss}=A/K_v\):

This value is only meaningful if the closed loop is stable. The characteristic equation is \(s^{3}+14s^{2}+40s+K=0\), and the Routh condition for a cubic \(s^3+a_2s^2+a_1s+a_0\) is \(a_2a_1 \gt a_0\), i.e. \(K \lt 14\times 40=560\). Since \(400 \lt 560\), the closed loop is stable and the answer is admissible.

[2 marks] The characteristic equation of a feedback system is

The range of \(K\) for which all roots lie in the left half of the \(s\)-plane is

1. \(0 \lt K \lt 6\)
2. \(0 \lt K \lt 10\)
3. \(0 \lt K \lt 16\)
4. \(K \gt 10\)

Solution

Build the Routh array as in Chapter 11. The first two rows come from the coefficients:

The \(s^{2}\) row:

The \(s^{1}\) row:

The \(s^{0}\) row is \(K\). For no sign change in the first column we need

A spot check confirms it: at \(K=9\) the roots are \(-2.973\pm j0.960\) and \(-0.027\pm j0.960\), all in the left half-plane, while at \(K=11\) two roots move to \(+0.025\pm j1.037\).

[2 marks] A signal flow graph has five nodes \(R, x_2, x_3, x_4, C\) connected in a chain. The transmittances along the chain are \(R\to x_2\) of gain 1, \(x_2\to x_3\) of gain \(G_1\), \(x_3\to x_4\) of gain \(G_2\) and \(x_4\to C\) of gain \(G_3\). Three feedback branches are present: \(x_3\to x_2\) of gain \(-H_1\), \(x_4\to x_3\) of gain \(-H_2\), and \(C\to x_2\) of gain \(-1\). For \(G_1=2\), \(G_2=5\), \(G_3=1\), \(H_1=0.5\) and \(H_2=0.4\), the overall gain \(C/R\) is _____ (round to three decimal places).

Solution

Apply Mason's gain formula from Chapter 5. There is a single forward path from \(R\) to \(C\):

The individual loop gains are

Every pair of loops shares at least one node: \(L_1\) and \(L_2\) share \(x_3\), and \(L_3\) passes through \(x_2, x_3, x_4\), so it touches both. There are therefore no non-touching loop products and

The forward path touches every loop, so \(\Delta_1=1\) and

Substituting the numbers, \(G_1G_2G_3=2\times 5\times 1=10\), \(G_1H_1=2\times 0.5=1\) and \(G_2H_2=5\times 0.4=2\):

[2 marks] The root locus of a unity-feedback system with

crosses the imaginary axis at a gain \(K\) and frequency \(\omega\) given by

1. \(K=8,\ \omega=1.41\) rad/s
2. \(K=16,\ \omega=2.00\) rad/s
3. \(K=24,\ \omega=2.45\) rad/s
4. \(K=48,\ \omega=2.83\) rad/s

Solution

The characteristic equation is

The locus meets the imaginary axis where the Routh array develops a zero row (Chapter 12):

The \(s^{1}\) entry vanishes at \(K=48\). The crossing frequency comes from the auxiliary equation formed by the row above:

Substituting \(K=48\) into the cubic gives roots \(-6\) and \(\pm j2.828\), confirming the crossing.

[2 marks] For the open-loop transfer function

the phase margin of the closed-loop system is _____ degrees (round to one decimal place).

Solution

The phase margin is measured at the gain crossover frequency \(\omega_{gc}\), where the magnitude is unity (Chapter 17).

Put \(u=\omega^{2}\), so \(u^{2}+u-100=0\):

The phase at that frequency is

Carrying more digits gives \(\tan^{-1}(3.0842)=72.036^{\circ}\) and PM \(=17.96^{\circ}\).

[2 marks] A unity-feedback system has

The gain margin and the phase crossover frequency are

1. 10 dB at 1.73 rad/s
2. 14 dB at 2.45 rad/s
3. 20 dB at 3.32 rad/s
4. 26 dB at 4.58 rad/s

Solution

Expand the denominator and substitute \(s=j\omega\):

The phase crossover is where the loop transfer function is real and negative, i.e. where the imaginary part of the denominator vanishes (Chapter 16):

At that frequency the denominator is purely real, \(6-6(11)=-60\), so

The Nyquist plot therefore crosses the negative real axis at \(-0.1\), well to the right of \(-1\), so the closed loop is stable with 20 dB of gain margin.

[2 marks] A linear time-invariant system is described by

The value of the first state variable \(x_1\) at \(t=\ln 2\) seconds is _____.

Solution

The zero-input response is \(x(t)=\Phi(t)x(0)\) with \(\Phi(t)=\mathcal{L}^{-1}\{(sI-A)^{-1}\}\) (Chapter 24).

With \(x(0)=[1\ \ 0]^{T}\), only the first column matters:

At \(t=\ln 2\), \(e^{-t}=\tfrac12\) and \(e^{-2t}=\tfrac14\):

[2 marks] A lead compensator of the form

must supply its maximum phase lead of \(30^{\circ}\) at \(\omega=4\) rad/s. The required value of \(T\) is _____ seconds (round to three decimal places).

Solution

For this compensator the maximum phase lead and the frequency at which it occurs are (Chapter 21)

First find \(\alpha\) from the required \(30^{\circ}\):

Then place \(\omega_m\) at 4 rad/s:

The compensator is therefore \(G_c(s)=(1+0.433s)/(1+0.144s)\), with corner frequencies at \(1/(\alpha T)=2.31\) rad/s and \(1/T=6.93\) rad/s, whose geometric mean is \(\sqrt{2.31\times 6.93}=4\) rad/s as required.