GATE 2024 Control Systems Practice Questions and Solutions

1 mark. A single-input single-output system has the state model

The transfer function \(Y(s)/U(s)\) is

1. \(\dfrac{1}{s^{2}+5s+6}\)
2. \(\dfrac{s}{s^{2}+5s+6}\)
3. \(\dfrac{1}{s^{2}+6s+5}\)
4. \(\dfrac{s+1}{s^{2}+5s+6}\)

Solution

For \(D=0\) the transfer function is \(C(sI-A)^{-1}B\) (Chapter 23).

The matrix is in phase-variable (controllable canonical) form, so this result could also be written down directly from the last row of \(A\).

2 marks. A signal flow graph has nodes \(R,\;x_1,\;x_2,\;x_3,\;C\) and the following branches, written as (from node \(\to\) to node : transmittance):

1. \(R\to x_1:\;1\)
2. \(x_1\to x_2:\;G_1=2\)
3. \(x_2\to x_3:\;G_2=5\)
4. \(x_3\to C:\;1\)
5. \(x_3\to x_3:\;G_3=0.5\) (a self-loop at node \(x_3\))
6. \(x_2\to x_1:\;-H_1=-1\)
7. \(x_3\to x_1:\;-H_2=-1\)

The overall transmittance \(C/R\) is _____ (round to two decimal places).

Solution

There is a single forward path, \(R\to x_1\to x_2\to x_3\to C\) (Chapter 5):

The individual loops are \(x_1\to x_2\to x_1\), \(x_1\to x_2\to x_3\to x_1\), and the self-loop at \(x_3\):

Loop \(L_1\) touches only nodes \(x_1,x_2\); the self-loop \(L_3\) touches only \(x_3\). They share no node, so \(L_1\) and \(L_3\) are the one non-touching pair. \(L_2\) touches both. Hence

The forward path passes through every node, including \(x_3\), so it touches all three loops and \(\Delta_1=1\).

2 marks. The closed-loop transfer function of a second-order system is

The percentage peak overshoot of the unit-step response is _____ % (round to two decimal places).

Solution

Compare with the standard second-order form (Chapter 8):

Since \(\zeta \lt 1\) the response is underdamped and overshoot exists.

As a percentage, \(M_p=9.48\,\%\). For reference the same data give a peak time \(t_p=\pi/(\omega_n\sqrt{1-\zeta^{2}})=\pi/4=0.785\) s and a 2 % settling time \(t_s=4/(\zeta\omega_n)=4/3=1.333\) s.

1 mark. A unity negative feedback system has the open-loop transfer function \(G(s)=\dfrac{100}{(s+2)(s+5)}\). The steady-state error for a unit-step input is

1. \(0\)
2. \(0.0909\)
3. \(0.1\)
4. \(1\)

Solution

There is no pole at the origin, so the system is Type 0 and a step input leaves a finite error (Chapter 9). The position error constant is

Option (C) is the trap: \(1/K_p=0.1\) omits the unity term in the denominator.

2 marks. The characteristic equation of a closed-loop system is

The number of roots of this equation that lie on the imaginary axis is _____.

Solution

Start the Routh array (Chapter 11):

because \(\dfrac{2\times 3-1\times 6}{2}=0\) and \(\dfrac{2\times 2-1\times 4}{2}=0\). A complete row of zeros signals roots symmetrically placed about the origin. Form the auxiliary polynomial from the \(s^{4}\) row and differentiate it:

Replace the zero row by these coefficients and continue:

The first column \(\{1,\,2,\,8,\,3,\,4/3,\,4\}\) has no sign change, so there are no roots in the right half-plane. The symmetric roots are those of the auxiliary polynomial:

Dividing the original polynomial by \((s^{2}+1)(s^{2}+2)=s^{4}+3s^{2}+2\) leaves the factor \((s+2)\), so the full factorisation is \((s+2)(s^{2}+1)(s^{2}+2)\). Four roots lie on the imaginary axis, one lies at \(-2\), and none lies in the right half-plane; the system is marginally stable.

2 marks. The open-loop transfer function of a unity feedback system is

The angle of departure of the root locus from the open-loop pole at \(s=-1+j1\) is

1. \(-90^{\circ}\)
2. \(-45^{\circ}\)
3. \(+45^{\circ}\)
4. \(+90^{\circ}\)

Solution

Factor the quadratic: \(s^{2}+2s+2=(s+1)^{2}+1\), giving poles at \(-1\pm j1\). The complete pole set is \(\{0,\,-2,\,-1+j1,\,-1-j1\}\) and there are no finite zeros.

The angle of departure follows from the angle criterion (Chapter 12):

Take the vectors drawn from each remaining pole to \(p=-1+j1\):

The branch therefore leaves the pole \(-1+j1\) heading straight down; by conjugate symmetry the branch at \(-1-j1\) departs at \(+90^{\circ}\).

2 marks. A unity negative feedback system has the open-loop transfer function \(G(s)=\dfrac{100}{s(s+10)}\). The phase margin of the system is _____ degrees (round to one decimal place).

Solution

First locate the gain crossover frequency, where \(|G(j\omega)|=1\) (Chapter 14):

Put \(u=\omega^{2}\), so \(u^{2}+100u-10^{4}=0\):

The phase at that frequency is

The phase never reaches \(-180^{\circ}\) at any finite frequency, so the gain margin is infinite and the closed-loop system is stable.

2 marks. A unity negative feedback system has \(G(s)=\dfrac{4}{s^{2}(s+2)}\). Its Nyquist plot, traced for the standard clockwise Nyquist contour, encircles the point \(-1+j0\) twice in the clockwise direction. The number of closed-loop poles in the right half of the \(s\)-plane is

1. \(0\)
2. \(1\)
3. \(2\)
4. \(3\)

Solution

The Nyquist criterion states \(Z=N+P\), where \(N\) counts clockwise encirclements of \(-1+j0\), \(P\) is the number of open-loop poles strictly inside the right half-plane, and \(Z\) is the number of closed-loop poles there (Chapter 16).

The open-loop poles are at \(s=0\) (double) and \(s=-2\). Poles on the imaginary axis are indented around by the contour and are not counted in \(P\), so \(P=0\). With \(N=2\),

Confirm this independently with the Routh array. The characteristic equation is

The first column runs \(1,\,2,\,-2,\,4\): two sign changes, hence two right-half-plane roots, matching the Nyquist count. The missing \(s^{1}\) coefficient already warns that the system cannot be stable for any gain.

2 marks. A plant \(G(s)=\dfrac{1}{s+2}\) is placed in a unity negative feedback loop with a PI controller \(G_c(s)=K_p+\dfrac{K_i}{s}\). The gains that place both closed-loop poles at \(s=-3\pm j3\) are

1. \(K_p=4,\;K_i=18\)
2. \(K_p=6,\;K_i=18\)
3. \(K_p=4,\;K_i=9\)
4. \(K_p=2,\;K_i=18\)

Solution

The loop transfer function is (Chapter 19)

so the closed-loop characteristic equation is

The desired pole pair gives the target polynomial

Matching coefficients term by term:

The integrator raises the system to Type 1, so the steady-state error to a step input becomes zero; the resulting damping ratio is \(\zeta=6/(2\sqrt{18})=0.707\).

2 marks. A system is described by

The system is

1. controllable and observable
2. controllable but not observable
3. observable but not controllable
4. neither controllable nor observable

Solution

Form the controllability matrix \(Q_c=[\,B\;\;AB\,]\) (Chapter 25):

Now the observability matrix \(Q_o=\begin{bmatrix}C\\ CA\end{bmatrix}\):

The loss of observability shows up as a cancellation in the transfer function. Using the inverse computed the usual way,

The mode at \(s=-2\) is cancelled by the zero of \(C\), so it never appears at the output even though the input can still excite it.