GATE 2021 Control Systems Practice Questions and Solutions

1 mark. A single-loop feedback system has two blocks in cascade in its forward path, \(G_1(s)=\dfrac{1}{s+1}\) followed by \(G_2(s)=\dfrac{2}{s+3}\). The output of \(G_2\) is fed back through \(H(s)=s\) to a summing junction where it is subtracted from the reference. The closed-loop transfer function \(C(s)/R(s)\) is

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

Solution

For a single negative-feedback loop the standard reduction rule of Chapter 4 gives \(T=G/(1+GH)\) with \(G=G_1G_2\). First form the forward-path product:

The loop gain is \(G(s)H(s)=\dfrac{2s}{(s+1)(s+3)}\). Substituting into the reduction rule and clearing the common denominator \((s+1)(s+3)\):

Expanding the denominator, \((s+1)(s+3)=s^{2}+4s+3\), so \((s+1)(s+3)+2s=s^{2}+6s+3\).

2 marks. A signal flow graph has four nodes \(x_1\) (input), \(x_2\), \(x_3\) and \(x_4\) (output). Its branches are: \(x_1\to x_2\) with gain 6, \(x_2\to x_3\) with gain 2, \(x_3\to x_4\) with gain 3, a feedback branch \(x_3\to x_2\) with gain \(-1\), a feedback branch \(x_4\to x_2\) with gain \(-2\), and a self-loop at node \(x_4\) with gain \(-1\). The overall gain \(x_4/x_1\) is _____.

Solution

Apply Mason's gain formula, as developed in Chapter 5. There is one forward path from \(x_1\) to \(x_4\):

Three individual loops exist. The loop \(x_2\to x_3\to x_2\) has gain \(L_1=(2)(-1)=-2\); the loop \(x_2\to x_3\to x_4\to x_2\) has gain \(L_2=(2)(3)(-2)=-12\); the self-loop at \(x_4\) has gain \(L_3=-1\).

Loop \(L_1\) touches only nodes \(x_2\) and \(x_3\), while \(L_3\) touches only node \(x_4\), so this pair is non-touching and contributes the product \(L_1L_3=(-2)(-1)=2\). Loop \(L_2\) passes through \(x_2\), \(x_3\) and \(x_4\), so it touches both of the others and forms no non-touching pair.

Every loop touches the forward path \(P_1\) (which runs through all four nodes), so the cofactor is \(\Delta_1=1\).

2 marks. A unity negative feedback system has the open-loop transfer function \(G(s)=\dfrac{K}{s(s+6)}\). The gain \(K\) is chosen so that the closed-loop damping ratio is exactly \(0.6\). For a unit-step reference, the peak time \(t_p\) (in seconds, rounded to three decimal places) is _____.

Solution

The closed-loop transfer function is \(T(s)=\dfrac{K}{s^{2}+6s+K}\). Comparing with the standard second-order form \(\omega_n^{2}/(s^{2}+2\zeta\omega_n s+\omega_n^{2})\) of Chapter 8 gives \(\omega_n=\sqrt{K}\) and \(2\zeta\omega_n=6\).

So \(\omega_n=5\) rad/s and \(\zeta=0.6\). The damped natural frequency is

The peak of the step response occurs at the first instant the derivative vanishes, giving

As a consistency check, the same \(\zeta\) predicts an overshoot of \(100\,e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}=100\,e^{-\pi(0.6)/0.8}=100\,e^{-2.356}=9.48\%\), and a 2% settling time of \(4/(\zeta\omega_n)=4/3=1.333\) s.

2 marks. A unity negative feedback system has \(G(s)=\dfrac{10(s+3)}{s(s+2)(s+5)}\). The reference input is \(r(t)=(2+4t)\,u(t)\). The steady-state error is

1. \(0\)
2. \(1.33\)
3. \(4\)
4. infinite

Solution

The open-loop transfer function has exactly one pole at the origin, so the system is type 1. Using the static error constants of Chapter 9,

The input is a superposition of a step of size 2 and a ramp of slope 4. By linearity the errors add:

The integrator in the forward path drives the step component of the error to zero, and the ramp component settles at \(4/K_v\).

2 marks. The characteristic equation of a closed-loop system is \(s^{4}+6s^{3}+11s^{2}+6s+K=0\), where \(K\) is a real gain. The value of \(K\) at which the system becomes marginally stable is _____.

Solution

Build the Routh array of Chapter 11:

The \(s^{2}\) row is computed from the two rows above it:

and the \(s^{1}\) entry follows from the \(s^{3}\) and \(s^{2}\) rows:

Stability requires every entry of the first column to be positive, which gives \(K>0\) from the \(s^{0}\) row and \(60-6K>0\), i.e. \(K<10\), from the \(s^{1}\) row. The stable range is therefore \(0<K<10\), and the boundary is reached when \(c_1=0\):

At \(K=10\) the auxiliary polynomial formed from the \(s^{2}\) row is \(10s^{2}+10=0\), so \(s=\pm j1\): the system sustains an oscillation at 1 rad/s. Substituting \(K=10\) confirms the factorisation \(s^{4}+6s^{3}+11s^{2}+6s+10=(s^{2}+1)(s^{2}+6s+10)\), whose remaining roots are \(-3\pm j1\).

2 marks. The root locus of a unity feedback system with open-loop transfer function \(G(s)H(s)=\dfrac{K}{s(s+2)(s+4)}\), \(K>0\), crosses the imaginary axis at a gain of

1. 8
2. 24
3. 48
4. 64

Solution

The characteristic equation \(1+G(s)H(s)=0\) expands to

The locus meets the imaginary axis at the gain that makes the system marginally stable, so apply the Routh test:

The \(s^{1}\) entry vanishes when \(48-K=0\), that is \(K=48\). The auxiliary equation from the \(s^{2}\) row is then

So the locus crosses the imaginary axis at \(\omega=2.828\) rad/s with \(K=48\); the third root at that gain is \(s=-6\). For reference, the technique of Chapter 12 also gives the centroid \(\sigma_a=(0-2-4)/3=-2\) with asymptotes at \(\pm 60^\circ\) and \(180^\circ\), and a breakaway point from \(3s^{2}+12s+8=0\) at \(s=-0.845\) (the root \(s=-3.155\) does not lie on the locus).

2 marks. A unity feedback system has open-loop transfer function \(G(s)=\dfrac{K}{s(1+0.1s)(1+0.2s)}\). The value of \(K\) that yields a gain margin of exactly 20 dB is _____.

Solution

The gain margin is measured at the phase-crossover frequency \(\omega_{pc}\), where the open-loop phase is \(-180^\circ\). The phase is

Setting this to \(-180^\circ\) requires \(\tan^{-1}(0.1\omega)+\tan^{-1}(0.2\omega)=90^\circ\). Two angles sum to \(90^\circ\) when their tangents are reciprocal, so \((0.1\omega)(0.2\omega)=1\):

Evaluate the magnitude at this frequency:

Since \(\sqrt{1.5}\sqrt{3}=\sqrt{4.5}=2.121\) and \(7.071\times 2.121=15\),

A gain margin of 20 dB means \(GM=10^{20/20}=10\) as a ratio, and \(GM=1/|G(j\omega_{pc})|\) by the definition in Chapter 17:

2 marks. A second-order system is described by \(\dot{x}=Ax+Bu\) with

The system fails to be state controllable for

1. \(\alpha=-1\) or \(\alpha=-6\)
2. \(\alpha=-2\) or \(\alpha=-3\)
3. \(\alpha=2\) or \(\alpha=3\)
4. \(\alpha=1\) or \(\alpha=6\)

Solution

For a second-order system the Kalman controllability matrix of Chapter 25 is \(Q_c=[\,B \;\; AB\,]\), and controllability fails exactly when \(\det Q_c=0\). Compute \(AB\):

The determinant vanishes at \(\alpha=-2\) and \(\alpha=-3\). These are precisely the values at which \(B\) aligns with an eigenvector of \(A\): the eigenvalues of \(A\) are the roots of \(\lambda^{2}+5\lambda+6=0\), namely \(\lambda=-2\) and \(\lambda=-3\), with eigenvectors \([1\;\;-2]^{T}\) and \([1\;\;-3]^{T}\). An input direction lying along a single eigenvector cannot excite the other mode, so that mode is uncontrollable.

2 marks. A system has the state matrix \(A=\begin{bmatrix}0 & 1\\ 0 & -2\end{bmatrix}\). Let \(\phi(t)=e^{At}\) be its state transition matrix. The value of the element \(\phi_{12}(t)\) at \(t=0.5\) s (rounded to three decimal places) is _____.

Solution

Use the Laplace method of Chapter 24, \(\phi(t)=\mathcal{L}^{-1}\{(sI-A)^{-1}\}\). First form \(sI-A\):

Inverting a \(2\times 2\) matrix by the adjugate rule,

The off-diagonal term expands in partial fractions as \(\dfrac{1}{s(s+2)}=\dfrac{1/2}{s}-\dfrac{1/2}{s+2}\), so its inverse transform is \(\tfrac12\left(1-e^{-2t}\right)\). Hence

This satisfies the required check \(\phi(0)=I\). Evaluating the \((1,2)\) entry at \(t=0.5\):

1 mark. The Nyquist plot of \(G(s)=\dfrac{1}{(s+1)(s+2)(s+3)}\) intersects the negative real axis at the point

1. \(-1/6\)
2. \(-1/11\)
3. \(-1/50\)
4. \(-1/60\)

Solution

The plot meets the real axis where the imaginary part of \(G(j\omega)\) is zero, which is the same as the frequency where the phase is \(-180^\circ\). Expand the denominator with \(s=j\omega\):

The imaginary part vanishes when \(\omega(11-\omega^{2})=0\), so the non-zero crossing is at \(\omega=\sqrt{11}=3.317\) rad/s. At that frequency the denominator is purely real:

Therefore

Consistently with Chapter 16, a gain \(K\) placed in cascade would push this point to \(-K/60\), so the closed loop is stable for \(K<60\).