GATE 2022 Control Systems Practice Questions and Solutions

1 mark. A linear time-invariant system with input \(u(t)\) and output \(y(t)\) obeys

with zero initial conditions. The steady-state value of \(y(t)\) for a unit-step input is

1. \(2/3\)
2. \(1\)
3. \(4/3\)
4. \(8/5\)

Solution

Taking the Laplace transform of both sides with zero initial conditions, as in Chapter 3, each derivative becomes a power of \(s\):

Both poles lie at \(s=-2\) and \(s=-3\), strictly in the left half plane, so the final value theorem may be applied. With \(U(s)=1/s\),

The steady-state step response is simply the DC gain of the transfer function.

1 mark. The characteristic polynomial of a closed-loop system is \(s^{3}+2s^{2}+3s+10\). The number of its roots lying in the right half of the \(s\)-plane is

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

Solution

All coefficients are present and positive, so the necessary condition for stability is met, but that is not sufficient. Construct the Routh array of Chapter 11:

The first column reads \(1,\;2,\;-2,\;10\). Moving down it, the sign changes from \(+\) to \(-\) and then from \(-\) to \(+\), giving two sign changes. By the Routh–Hurwitz criterion the number of roots in the right half plane equals the number of sign changes in the first column, so there are two.

Direct factorisation confirms this: the polynomial has one real root near \(s=-2.445\) and a complex pair at \(s=0.223\pm j2.010\), both of which lie in the right half plane.

2 marks. A signal flow graph has nodes \(x_1\) (input), \(x_2\), \(x_3\) and \(x_4\) (output), with branches: \(x_1\to x_2\) of gain 1, \(x_2\to x_3\) of gain 4, \(x_3\to x_4\) of gain 5, a direct branch \(x_2\to x_4\) of gain 3, a feedback branch \(x_3\to x_2\) of gain \(-1\), and a feedback branch \(x_4\to x_3\) of gain \(-1\). The transfer function \(x_4/x_1\) is _____.

Solution

Identify the forward paths first. Following Chapter 5, a forward path visits no node more than once:

There are two loops. The loop \(x_2\to x_3\to x_2\) has gain \(L_1=(4)(-1)=-4\), and the loop \(x_3\to x_4\to x_3\) has gain \(L_2=(5)(-1)=-5\). They share node \(x_3\), so they touch and there are no non-touching combinations:

Path \(P_1\) passes through \(x_2\), \(x_3\) and \(x_4\), so both loops touch it and \(\Delta_1=1\). Path \(P_2\) passes through \(x_2\) and \(x_4\); loop \(L_1\) touches it at \(x_2\) and loop \(L_2\) touches it at \(x_4\), so \(\Delta_2=1\) as well.

2 marks. The closed-loop transfer function of a position servo is \(T(s)=\dfrac{16}{s^{2}+4s+16}\). For a unit-step input, the peak percentage overshoot (rounded to one decimal place) is _____ %.

Solution

Match the denominator to the canonical second-order form \(s^{2}+2\zeta\omega_n s+\omega_n^{2}\) of Chapter 8:

Since \(0<\zeta<1\) the response is underdamped and overshoots. The overshoot depends on \(\zeta\) alone:

With \(\sqrt{0.75}=0.8660\), the exponent is \(-1.5708/0.8660=-1.8138\), and \(e^{-1.8138}=0.16303\):

The peak itself occurs at \(t_p=\pi/\left(\omega_n\sqrt{1-\zeta^{2}}\right)=\pi/(4\times 0.866)=0.907\) s, where the output reaches \(1.163\).

2 marks. A plant \(G(s)=\dfrac{1}{(s+1)(s+4)}\) is placed in a unity negative feedback loop with the PI controller \(C(s)=K_p+\dfrac{K_i}{s}\). The steady-state error for a unit-ramp reference is required to be \(0.05\). The necessary value of \(K_i\) is

1. 20
2. 40
3. 80
4. 100

Solution

The integral term of the PI controller supplies a pole at the origin, so the loop transfer function

is type 1, and a ramp input therefore produces a finite error governed by the velocity error constant of Chapter 9:

The proportional gain drops out of the limit, so only \(K_i\) sets the ramp error. Imposing \(e_{ss}=1/K_v=0.05\):

The gain \(K_p\) is then free to be chosen for transient behaviour and stability, as discussed in Chapter 19.

2 marks. For the unity feedback system with \(G(s)=\dfrac{K}{s(s+2)}\), the value of \(K\) that gives a phase margin of exactly \(45^\circ\) (rounded to two decimal places) is _____.

Solution

Phase margin is defined at the gain-crossover frequency \(\omega_{gc}\), where \(|G(j\omega)|=1\). Work backwards: first find the frequency at which the phase alone gives a \(45^\circ\) margin. The phase is

By the definition in Chapter 17, \(PM = 180^\circ + \angle G(j\omega_{gc})\). Setting \(PM=45^\circ\) requires \(\angle G(j\omega_{gc})=-135^\circ\):

Now force the magnitude to be unity at that frequency. Note that \(K\) shifts the magnitude curve without altering the phase, so this step fixes \(K\) uniquely:

2 marks. For the unity feedback system with open-loop transfer function \(G(s)H(s)=\dfrac{K(s+1)}{s^{2}(s+9)}\), \(K>0\), the root locus has a break-in point on the real axis at

1. \(s=-1\)
2. \(s=-3\)
3. \(s=-4\)
4. \(s=-9\)

Solution

Solve the characteristic equation for the gain, then locate the stationary point, the standard breakaway/break-in procedure of Chapter 12. From \(1+G(s)H(s)=0\),

Differentiate using the quotient rule and set the numerator to zero:

The root \(s=0\) is the double open-loop pole itself, from which the two branches leave immediately; it is not a break-in point. The candidate \(s=-3\) lies on the real-axis segment between the zero at \(s=-1\) and the pole at \(s=-9\), which belongs to the locus because an odd number of real poles and zeros lies to its right (the zero at \(-1\) plus the double pole at the origin counted twice, giving three). The gain there is

Since \(K=27>0\), the point is genuinely on the locus, and the two complex branches return to the real axis there. As a cross-check, with \(n-m=2\) the asymptotes make angles of \(\pm 90^\circ\) about the centroid \(\sigma_a=\dfrac{(0+0-9)-(-1)}{2}=-4\).

2 marks. A plant is described by \(\dot{x}=Ax+Bu\) with

State feedback \(u=-Kx\) with \(K=\begin{bmatrix}k_1 & k_2\end{bmatrix}\) is applied to place the closed-loop poles at \(s=-4\) and \(s=-5\). The value of \(k_1\) is _____.

Solution

Check controllability first, since pole placement is possible only if the pair \((A,B)\) is controllable. With \(AB=\begin{bmatrix}1\\ -3\end{bmatrix}\),

The system is controllable, so any pole locations are achievable. Form the closed-loop matrix, as in Chapter 26:

Because the pair is already in controllable canonical form, the characteristic polynomial is read from the bottom row:

The desired polynomial is \((s+4)(s+5)=s^{2}+9s+20\). Matching coefficients term by term:

2 marks. A compensator has the transfer function \(G_c(s)=\dfrac{1+0.2s}{1+0.05s}\). Its maximum phase lead, and the frequency at which that maximum occurs, are

1. \(30^\circ\) at 5 rad/s
2. \(36.87^\circ\) at 10 rad/s
3. \(45^\circ\) at 10 rad/s
4. \(53.13^\circ\) at 20 rad/s

Solution

Write the compensator in the standard lead form \(G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}\) with \(\alpha<1\). Here \(T=0.2\) s and \(\alpha T=0.05\), so

Since \(\alpha<1\) the zero at \(s=-1/T=-5\) precedes the pole at \(s=-1/(\alpha T)=-20\), and the network is indeed phase-leading. The maximum lead of Chapter 21 is

It occurs at the geometric mean of the corner frequencies, which on a logarithmic axis is the midpoint between the zero at 5 rad/s and the pole at 20 rad/s:

2 marks. A block diagram consists of an inner loop and an outer loop. In the inner loop the forward block is \(G_1(s)=\dfrac{5}{s+2}\) and its output is fed back through \(H_1=0.2\) to a summing junction where it is subtracted. The output of this inner loop drives an integrator \(G_2(s)=1/s\), whose output is the system output and is also fed back with unity gain and subtracted from the reference. The damping ratio of the overall closed-loop system (rounded to three decimal places) is _____.

Solution

Reduce the inner loop first using \(G_1/(1+G_1H_1)\), from Chapter 4:

Cascade this with the integrator to obtain the open-loop transfer function of the outer loop:

Close the outer unity-feedback loop:

Compare with \(\omega_n^{2}/(s^{2}+2\zeta\omega_n s+\omega_n^{2})\):

Since \(\zeta<1\) the response is underdamped, overshooting by \(100\,e^{-\pi(0.671)/\sqrt{1-0.671^{2}}}=5.8\%\).