About this set. These are original practice questions written
in GATE style for the 2023 Control Systems syllabus, with fully worked
solutions. They are not reproductions of the official 2023 question paper.
1 mark. A control system is built from two cascaded blocks. The first block has transfer function \(G_1(s)=\dfrac{5}{s+1}\). The second block has transfer function \(G_2(s)=\dfrac{1}{s+3}\) and is enclosed in a negative feedback loop of constant gain \(H_1=2\), so that the signal fed back from the output of \(G_2\) is subtracted from the input of \(G_2\). The cascade of \(G_1\) followed by this inner loop forms the forward path of an outer unity negative feedback loop. The overall closed-loop transfer function \(C(s)/R(s)\) is
- \(\dfrac{5}{s^{2}+6s+10}\)
- \(\dfrac{5}{s^{2}+6s+5}\)
- \(\dfrac{5}{s^{2}+4s+10}\)
- \(\dfrac{5}{s^{2}+6s+15}\)
Solution
Reduce the inner loop first. For a negative feedback loop the closed-loop gain is the forward gain divided by one plus the loop gain (Chapter 4):
Equation
\[G_{in}(s)=\frac{G_2}{1+G_2H_1}=\frac{\frac{1}{s+3}}{1+\frac{2}{s+3}}=\frac{1}{s+3+2}=\frac{1}{s+5}\]
The forward path of the outer loop is therefore
Equation
\[G(s)=G_1(s)\,G_{in}(s)=\frac{5}{s+1}\cdot\frac{1}{s+5}=\frac{5}{(s+1)(s+5)}\]
Closing the outer unity feedback loop:
Equation
\[\frac{C(s)}{R(s)}=\frac{G}{1+G}=\frac{5}{(s+1)(s+5)+5}=\frac{5}{s^{2}+6s+5+5}=\frac{5}{s^{2}+6s+10}\]
A
Final Answer
Correct answer: (A) \(\dfrac{5}{s^{2}+6s+10}\).
2 marks. A signal flow graph has five nodes \(R,\;x_1,\;x_2,\;x_3,\;C\) and the following branches, each written as (from node \(\to\) to node : transmittance):
- \(R\to x_1:\;1\)
- \(x_1\to x_2:\;G_1=2\)
- \(x_2\to x_3:\;G_2=5\)
- \(x_3\to C:\;G_3=1\)
- \(x_1\to x_3:\;G_4=3\)
- \(x_3\to x_2:\;-H_1=-1\)
- \(C\to x_1:\;-H_2=-1\)
Using Mason's gain formula, the overall transmittance \(C/R\) is _____ (round to two decimal places).
Solution
List the forward paths from \(R\) to \(C\) (Chapter 5):
Equation
\[P_1 = G_1G_2G_3 = 2\times 5\times 1 = 10,\qquad P_2 = G_4G_3 = 3\times 1 = 3\]
Now the individual loops. Branch 6 with branch 3 forms \(x_2\to x_3\to x_2\); branches 2, 3, 4, 7 form \(x_1\to x_2\to x_3\to C\to x_1\); branches 5, 4, 7 form \(x_1\to x_3\to C\to x_1\):
Equation
\[L_1=-G_2H_1=-5,\qquad L_2=-G_1G_2G_3H_2=-10,\qquad L_3=-G_4G_3H_2=-3\]
Every pair of loops shares node \(x_3\), so there are no non-touching loops and the determinant is
Equation
\[\Delta = 1-(L_1+L_2+L_3) = 1-(-5-10-3) = 19\]
Path \(P_1\) passes through every node, so \(\Delta_1=1\). Path \(P_2\) passes through \(x_1,x_3,C\); loop \(L_1\) contains \(x_3\), so it touches \(P_2\) as well, and \(\Delta_2=1\).
Equation
\[\frac{C}{R}=\frac{P_1\Delta_1+P_2\Delta_2}{\Delta}=\frac{10+3}{19}=\frac{13}{19}=0.6842\]
✓
Final Answer
Correct answer: 0.68.
2 marks. A unity negative feedback system has the open-loop transfer function \(G(s)=\dfrac{K}{s(s+8)}\). The gain \(K\) is set so that the unit-step response of the closed-loop system has a peak overshoot of 16.3 %. The peak time \(t_p\) of that response is _____ seconds (round to two decimal places).
Solution
The closed-loop transfer function is
Equation
\[\frac{C(s)}{R(s)}=\frac{K}{s^{2}+8s+K}\equiv\frac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}\]
so \(\omega_n=\sqrt{K}\) and \(2\zeta\omega_n=8\). Invert the overshoot formula (Chapter 8) to get \(\zeta\):
Equation
\[M_p=e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}=0.163\;\Rightarrow\;\ln 0.163=-1.814=-\frac{\pi\zeta}{\sqrt{1-\zeta^{2}}}\]
Equation
\[\zeta=\frac{1.814}{\sqrt{\pi^{2}+1.814^{2}}}=\frac{1.814}{\sqrt{9.870+3.291}}=\frac{1.814}{3.628}=0.5\]
With \(\zeta=0.5\), the relation \(2\zeta\omega_n=8\) gives \(\omega_n=8\) rad/s, and hence \(K=\omega_n^{2}=64\). The damped natural frequency is
Equation
\[\omega_d=\omega_n\sqrt{1-\zeta^{2}}=8\sqrt{1-0.25}=8\times 0.8660=6.928~\text{rad/s}\]
Equation
\[t_p=\frac{\pi}{\omega_d}=\frac{3.1416}{6.928}=0.4534~\text{s}\]
✓
Final Answer
Correct answer: 0.45 s (with \(K=64\), \(\zeta=0.5\), \(\omega_n=8\) rad/s).
1 mark. A unity negative feedback system has the open-loop transfer function
Equation
\[G(s)=\frac{20(s+2)}{s(s+3)(s+5)}\]
The steady-state error for the input \(r(t)=\big(4+6t\big)u(t)\) is
- \(0\)
- \(1.5\)
- \(2.25\)
- infinite
Solution
\(G(s)\) has one pole at the origin, so the system is Type 1 (Chapter 9). Superposition applies to the two input components.
For the step component of amplitude 4, the position error constant is \(K_p=\lim_{s\to 0}G(s)=\infty\) because of the free integrator, so that component contributes zero error.
For the ramp component \(6t\), the velocity error constant is
Equation
\[K_v=\lim_{s\to 0}sG(s)=\lim_{s\to 0}\frac{20(s+2)}{(s+3)(s+5)}=\frac{20\times 2}{3\times 5}=\frac{40}{15}=\frac{8}{3}\]
Equation
\[e_{ss}=0+\frac{6}{K_v}=\frac{6}{8/3}=\frac{18}{8}=2.25\]
C
Final Answer
Correct answer: (C) 2.25.
2 marks. The characteristic equation of a closed-loop system is
Equation
\[s^{4}+6s^{3}+11s^{2}+6s+K=0\]
The value of \(K\) for which the system sustains oscillations of constant amplitude is _____.
Solution
Build the Routh array (Chapter 11). The first two rows carry the alternate coefficients:
Equation
\[\begin{array}{c|ccc}
s^{4} & 1 & 11 & K\\
s^{3} & 6 & 6 & 0\\
s^{2} & b_1 & b_2 & \\
s^{1} & c_1 & & \\
s^{0} & K & &
\end{array}\]
Equation
\[b_1=\frac{6\times 11-1\times 6}{6}=\frac{60}{6}=10,\qquad b_2=\frac{6\times K-1\times 0}{6}=K\]
Equation
\[c_1=\frac{10\times 6-6\times K}{10}=\frac{60-6K}{10}\]
All first-column entries must be positive for stability, which requires \(K \gt 0\) and \(60-6K \gt 0\), i.e. \(0 \lt K \lt 10\). Sustained oscillation occurs at the boundary, where the \(s^{1}\) row vanishes:
Equation
\[60-6K=0\;\Rightarrow\;K=10\]
The oscillation frequency follows from the auxiliary polynomial formed by the row above the vanishing row, using \(b_1=10\) and \(b_2=K=10\):
Equation
\[A(s)=10s^{2}+10=0\;\Rightarrow\;s=\pm j1\;\Rightarrow\;\omega=1~\text{rad/s}\]
Substituting \(K=10\) into the characteristic polynomial confirms it factors as \((s^{2}+1)(s^{2}+6s+10)\), whose remaining roots are \(-3\pm j1\).
✓
Final Answer
Correct answer: \(K=10\), with oscillation frequency \(\omega=1\) rad/s.
2 marks. The root locus of a unity feedback system with \(G(s)H(s)=\dfrac{K}{s(s+2)(s+4)}\) is drawn for \(K\) varying from 0 to \(\infty\). The breakaway point on the real axis and the value of \(K\) at which the locus crosses the imaginary axis are respectively
- \(-0.845\) and \(48\)
- \(-3.155\) and \(48\)
- \(-0.845\) and \(24\)
- \(-2.0\) and \(48\)
Solution
The open-loop poles are at \(0,\,-2,\,-4\) and there are no finite zeros. Counting poles to the right, the real-axis locus occupies \([-2,\,0]\) and \((-\infty,\,-4]\); the segment between \(-2\) and \(-4\) has two poles to its right and is not on the locus (Chapter 12).
From the characteristic equation \(1+G(s)H(s)=0\),
Equation
\[K=-s(s+2)(s+4)=-\left(s^{3}+6s^{2}+8s\right)\]
Equation
\[\frac{dK}{ds}=-\left(3s^{2}+12s+8\right)=0\;\Rightarrow\;s=\frac{-12\pm\sqrt{144-96}}{6}=\frac{-12\pm 6.928}{6}\]
This gives \(s=-0.845\) and \(s=-3.155\). Only \(-0.845\) lies on a locus segment, so it is the breakaway point; \(-3.155\) is rejected.
For the imaginary-axis crossing, apply Routh to \(s^{3}+6s^{2}+8s+K=0\):
Equation
\[\begin{array}{c|cc}
s^{3} & 1 & 8\\
s^{2} & 6 & K\\
s^{1} & \dfrac{48-K}{6} & \\
s^{0} & K &
\end{array}\]
Equation
\[48-K=0\;\Rightarrow\;K=48,\qquad A(s)=6s^{2}+48=0\;\Rightarrow\;s=\pm j2\sqrt{2}=\pm j2.828\]
A
Final Answer
Correct answer: (A) breakaway at \(s=-0.845\), imaginary-axis crossing at \(K=48\).
2 marks. A unity negative feedback system has the open-loop transfer function \(G(s)=\dfrac{K}{s(s+1)}\) with \(K \gt 0\). The value of \(K\) that gives a phase margin of \(45^{\circ}\) is _____ (round to two decimal places).
Solution
Write the frequency response and separate magnitude and phase (Chapter 17):
Equation
\[G(j\omega)=\frac{K}{j\omega(1+j\omega)},\qquad |G(j\omega)|=\frac{K}{\omega\sqrt{1+\omega^{2}}},\qquad \angle G(j\omega)=-90^{\circ}-\tan^{-1}\omega\]
The phase margin is measured at the gain crossover frequency \(\omega_{gc}\), where the magnitude equals unity:
Equation
\[PM=180^{\circ}+\angle G(j\omega_{gc})=180^{\circ}-90^{\circ}-\tan^{-1}\omega_{gc}=90^{\circ}-\tan^{-1}\omega_{gc}\]
Setting \(PM=45^{\circ}\) gives \(\tan^{-1}\omega_{gc}=45^{\circ}\), so \(\omega_{gc}=1\) rad/s. The gain is then fixed by the unity-magnitude condition at that frequency:
Equation
\[\frac{K}{\omega_{gc}\sqrt{1+\omega_{gc}^{2}}}=1\;\Rightarrow\;K=1\times\sqrt{1+1}=\sqrt{2}=1.414\]
✓
Final Answer
Correct answer: \(K=1.41\) (at \(\omega_{gc}=1\) rad/s).
2 marks. The open-loop transfer function of a unity negative feedback system is \(G(s)H(s)=\dfrac{3}{s(s+1)(s+3)}\). The gain margin of the system is
- \(6.02\) dB
- \(12.04\) dB
- \(20\) dB
- \(4\) dB
Solution
The gain margin is read at the phase crossover frequency \(\omega_{pc}\), where the polar plot cuts the negative real axis. Expand the denominator at \(s=j\omega\) (Chapter 15):
Equation
\[j\omega(j\omega+1)(j\omega+3)=j\omega\left(3-\omega^{2}+j4\omega\right)=-4\omega^{2}+j\omega\left(3-\omega^{2}\right)\]
The imaginary part vanishes when \(\omega(3-\omega^{2})=0\), i.e. at \(\omega_{pc}=\sqrt{3}=1.732\) rad/s. At that frequency the denominator is purely real:
Equation
\[G(j\omega_{pc})H(j\omega_{pc})=\frac{3}{-4\times 3}=-\frac{3}{12}=-0.25\]
Equation
\[GM=\frac{1}{|G(j\omega_{pc})H(j\omega_{pc})|}=\frac{1}{0.25}=4\]
Equation
\[GM_{dB}=20\log_{10}4=12.04~\text{dB}\]
Since the plot crosses the negative real axis to the right of \(-1+j0\), the closed-loop system is stable, and the gain may be raised by a factor of 4 before oscillation begins.
B
Final Answer
Correct answer: (B) 12.04 dB.
2 marks. A linear time-invariant system is described by \(\dot{x}(t)=Ax(t)\) with
Equation
\[A=\begin{bmatrix}0 & 1\\ -2 & -3\end{bmatrix},\qquad x(0)=\begin{bmatrix}1\\ 0\end{bmatrix}\]
The value of the first state variable at \(t=1\) s, that is \(x_1(1)\), is _____ (round to two decimal places).
Solution
The zero-input solution is \(x(t)=\phi(t)x(0)\) with \(\phi(t)=\mathcal{L}^{-1}\left\{(sI-A)^{-1}\right\}\) (Chapter 24).
Equation
\[sI-A=\begin{bmatrix}s & -1\\ 2 & s+3\end{bmatrix},\qquad \det(sI-A)=s(s+3)+2=s^{2}+3s+2=(s+1)(s+2)\]
Equation
\[(sI-A)^{-1}=\frac{1}{(s+1)(s+2)}\begin{bmatrix}s+3 & 1\\ -2 & s\end{bmatrix}\]
Because \(x(0)=[1\;\;0]^{T}\), only the first column of \(\phi(t)\) is needed, and \(x_1(1)\) requires only its top-left entry:
Equation
\[\phi_{11}(s)=\frac{s+3}{(s+1)(s+2)}=\frac{2}{s+1}-\frac{1}{s+2}\]
Equation
\[\phi_{11}(t)=2e^{-t}-e^{-2t}\;\Rightarrow\; x_1(t)=2e^{-t}-e^{-2t}\]
Equation
\[x_1(1)=2e^{-1}-e^{-2}=2(0.36788)-0.13534=0.73576-0.13534=0.6004\]
✓
Final Answer
Correct answer: 0.60.
2 marks. A compensator has the transfer function \(G_c(s)=\dfrac{1+0.4s}{1+0.1s}\). The maximum phase lead it provides and the frequency at which that maximum occurs are respectively
- \(36.87^{\circ}\) at \(5\) rad/s
- \(36.87^{\circ}\) at \(2.5\) rad/s
- \(53.13^{\circ}\) at \(5\) rad/s
- \(30^{\circ}\) at \(5\) rad/s
Solution
Write the compensator in the standard lead form \(G_c(s)=\dfrac{1+a T s}{1+T s}\) with \(a \gt 1\) (Chapter 21). Matching coefficients, \(T=0.1\) and \(aT=0.4\), so \(a=4\). The zero at \(-2.5\) lies closer to the origin than the pole at \(-10\), confirming phase lead.
The phase is \(\phi(\omega)=\tan^{-1}(aT\omega)-\tan^{-1}(T\omega)\). Differentiating and setting the derivative to zero gives the geometric mean of the two corner frequencies:
Equation
\[\omega_m=\frac{1}{T\sqrt{a}}=\frac{1}{0.1\times\sqrt{4}}=\frac{1}{0.2}=5~\text{rad/s}\]
Evaluating the phase there, or using the standard result,
Equation
\[\sin\phi_m=\frac{a-1}{a+1}=\frac{4-1}{4+1}=0.6\;\Rightarrow\;\phi_m=\sin^{-1}(0.6)=36.87^{\circ}\]
Direct check: \(\tan^{-1}(0.4\times 5)-\tan^{-1}(0.1\times 5)=63.43^{\circ}-26.57^{\circ}=36.87^{\circ}\).
A
Final Answer
Correct answer: (A) \(36.87^{\circ}\) at \(5\) rad/s.