GATE 2018 Control Systems Practice Questions and Solutions

[1 mark] The response of a linear time-invariant system to a unit step applied at \(t=0\) is \(y(t)=2\left(1-e^{-4t}\right)\) for \(t\ge 0\). The transfer function of the system is

1. \(\dfrac{2}{s+4}\)
2. \(\dfrac{8}{s+4}\)
3. \(\dfrac{2}{s+2}\)
4. \(\dfrac{8}{s+8}\)

Solution

Take the Laplace transform of the measured output (Chapter 7):

The input was a unit step, so \(R(s)=1/s\) and

The check is consistent: the DC gain is \(8/4=2\), matching the final value of \(y(t)\), and the pole at \(s=-4\) gives the time constant \(0.25\) s seen in the exponential.

[1 mark] A unity-negative-feedback system has the open-loop transfer function

The steady-state error for a unit-step input is _____ (round to four decimal places).

Solution

There is no pole at the origin, so the system is Type 0 and the position error constant applies (Chapter 9):

The closed loop is stable, since its characteristic equation \(s^{2}+6s+55=0\) has both coefficients positive with roots \(-3\pm j6.78\).

[2 marks] In a feedback arrangement, the reference \(R\) enters an outer summing junction. Its output feeds a block \(G_1(s)=10/s\), whose output enters a second summing junction. That second junction drives a block \(G_2(s)=1/(s+2)\), and the output of \(G_2\) is the system output \(C\). Two negative feedback paths exist: the output of \(G_2\) returns through a block \(H_1=4\) to the second summing junction, and \(C\) returns directly (unity gain) to the outer summing junction. The closed-loop transfer function \(C/R\) is

1. \(\dfrac{10}{s^{2}+2s+10}\)
2. \(\dfrac{10}{s^{2}+4s+10}\)
3. \(\dfrac{10}{s^{2}+6s+10}\)
4. \(\dfrac{10}{s^{2}+6s+14}\)

Solution

Reduce the inner loop first (Chapter 4). The inner loop is \(G_2\) with negative feedback \(H_1\):

The forward path is now \(G_1G_{inner}\), closed by unity negative feedback:

Equivalently, in the general form \(G_1G_2/(1+G_2H_1+G_1G_2)\), clearing \(s(s+2)\) gives \(10/\left(s(s+2)+4s+10\right)=10/(s^{2}+6s+10)\), the same result.

[2 marks] A closed-loop system has the transfer function

For a unit-step input, the time at which the output reaches its first peak is _____ seconds (round to three decimal places).

Solution

Match the standard second-order form (Chapter 8):

Since \(\zeta \lt 1\) the response is underdamped and a peak exists. The damped natural frequency is

For reference, the overshoot at that instant is \(\exp\!\left(-\pi(0.8)/0.6\right)=e^{-4.189}=0.0152\), i.e. 1.52 %, and the 2 % settling time is \(4/(\zeta\omega_n)=4/4=1\) s.

[2 marks] The number of roots of the polynomial

that lie in the right half of the \(s\)-plane is

1. 0
2. 1
3. 2
4. 3

Solution

Construct the Routh array (Chapter 11). The first two rows are the alternating coefficients:

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

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

The \(s^{0}\) row is \(5\). The first column reads \(1,\,2,\,1,\,-6,\,5\). The sign changes from \(+1\) to \(-6\) and back from \(-6\) to \(+5\), giving two sign changes, hence two roots in the right half-plane.

Direct root extraction agrees: the roots are \(-1.288\pm j0.858\) and \(+0.288\pm j1.416\).

[2 marks] A plant \(G(s)=1/(s+2)\) is controlled by a PI controller \(G_c(s)=K_p+K_i/s\) in a unity-negative-feedback loop. The gains are chosen so that the closed-loop poles have \(\omega_n=4\) rad/s and \(\zeta=0.5\). The value of \(K_p\) is _____.

Solution

The loop transfer function is (Chapter 19)

The characteristic equation \(1+G_cG=0\) becomes

Match this against \(s^{2}+2\zeta\omega_n s+\omega_n^{2}=0\):

The integral action also raises the system to Type 1, so the steady-state error to a step becomes zero.

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

The closed-loop system is stable for

1. all \(K \gt 0\)
2. \(K \gt 1\)
3. \(0 \lt K \lt 1\)
4. no value of \(K\)

Solution

The open loop has one pole in the right half-plane, at \(s=+1\), so \(P=1\). By the Nyquist criterion (Chapter 16) stability requires \(N=-P=-1\), that is one anticlockwise encirclement of the \(-1\) point. The algebraic form of the same condition is simplest here. The characteristic equation is

For a quadratic, all roots lie in the left half-plane exactly when every coefficient is positive:

Checking two values confirms it: at \(K=2\) the roots are \(-0.5\pm j1.323\), stable; at \(K=0.5\) they are \(+0.25\pm j0.661\), unstable. The gain must therefore be pushed above unity for the Nyquist plot to encircle \(-1\) once in the anticlockwise sense.

[2 marks] A system is described in state-space form by

The DC gain of the system, that is the value of its transfer function at \(s=0\), is _____ (round to three decimal places).

Solution

The transfer function is \(G(s)=C(sI-A)^{-1}B+D\) (Chapter 23).

Setting \(s=0\):

The poles are at \(s=-1\) and \(s=-3\), both stable, so this DC gain is also the final value of the unit-step response.

[2 marks] Consider the state model \(\dot{x}=Ax+Bu\) with

The system fails to be completely state controllable for

1. \(\alpha=0\) only
2. \(\alpha=1\) or \(\alpha=2\)
3. \(\alpha=-1\) or \(\alpha=-2\)
4. every real \(\alpha\)

Solution

For a second-order single-input system the controllability matrix is \(Q_c=[B\ \ AB]\) (Chapter 25).

Controllability is lost when the determinant vanishes:

These are precisely the two eigenvector directions of \(A\), whose eigenvalues are \(-1\) and \(-2\). When \(B\) is aligned with one eigenvector, the other mode receives no input and cannot be steered.

[1 mark] The open-loop transfer function of a system is

The slope of the asymptotic Bode magnitude plot at frequencies well below 10 rad/s is

1. 0 dB/decade
2. \(-20\) dB/decade
3. \(-40\) dB/decade
4. \(-60\) dB/decade

Solution

Write the loop gain in time-constant form (Chapter 14):

For \(\omega\) well below the corner frequency \(1/0.1=10\) rad/s, the factor \((1+j0.1\omega)\) is approximately 1 and

A decade increase in \(\omega\) changes \(40\log_{10}\omega\) by 40, so the magnitude falls by 40 dB per decade. Each pole at the origin contributes \(-20\) dB/decade, and this system has two, so its type number is 2.