GATE 2026 Control Systems Practice Questions and Solutions

A single-input single-output system consists of a constant-gain block \(K_1=5\) in cascade with a block \(G_2(s)=\dfrac{1}{s+3}\). A unity-gain feedback path is taken from the output of \(G_2\) back to its own input summing junction, and that inner path is connected with a positive sign. The output of the cascade is then compared with the reference through a unity-gain negative feedback path that closes the outer loop. The overall transfer function \(\dfrac{C(s)}{R(s)}\) is

1. \(\dfrac{5}{s+2}\)
2. \(\dfrac{5}{s+7}\)
3. \(\dfrac{5}{s+8}\)
4. \(\dfrac{5}{s+9}\)

Solution

Reduce the inner loop first. For a forward block \(G\) with a feedback path \(H\) taken with a positive sign, the equivalent block is \(G/(1-GH)\); the sign in the denominator is opposite to the sign at the summing junction. The reduction rules are collected in Chapter 4.

Positive feedback has moved the pole from \(-3\) towards the origin, to \(-2\). The gain block is in cascade, so the total forward transfer function seen by the outer loop is

Now close the outer unity negative feedback loop:

Check the DC gain: the forward path has a DC gain of \(5/2=2.5\), and unity negative feedback reduces it to \(2.5/(1+2.5)=0.714\), which equals \(5/7\) as required.

A signal flow graph has four nodes: \(x_1\) (input), \(x_2\), \(x_3\) and \(x_4\) (output). The branches are \(x_1\to x_2\) with gain \(2\), \(x_2\to x_3\) with gain \(3\), \(x_3\to x_4\) with gain \(4\), and a second forward branch \(x_2\to x_4\) with gain \(5\). In addition there is a feedback branch \(x_3\to x_2\) with gain \(-1\) and a self-loop at node \(x_4\) with gain \(-2\). The overall gain \(x_4/x_1\) is _____ (round off to 2 decimal places).

Solution

Use Mason's gain formula \(T=\dfrac{1}{\Delta}\sum_k P_k\Delta_k\), developed in Chapter 5.

There are two forward paths from \(x_1\) to \(x_4\):

There are two individual loops. The first is formed by \(x_2\to x_3\) and \(x_3\to x_2\); the second is the self-loop at \(x_4\):

\(L_1\) occupies nodes \(x_2\) and \(x_3\) only, while \(L_2\) occupies node \(x_4\) only, so the two loops share no node and form one non-touching pair:

Path \(P_1\) passes through \(x_2\), \(x_3\) and \(x_4\), so it touches both loops and \(\Delta_1=1\). Path \(P_2\) passes through \(x_2\) (shared with \(L_1\)) and \(x_4\) (shared with \(L_2\)), so it also touches both loops and \(\Delta_2=1\).

The closed-loop transfer function of a second-order system is \(\dfrac{C(s)}{R(s)}=\dfrac{25}{s^{2}+6s+25}\). The percentage peak overshoot of its unit-step response is closest to

1. 4.60 %
2. 9.48 %
3. 16.30 %
4. 25.38 %

Solution

Compare with the standard second-order form \(\dfrac{\omega_n^{2}}{s^{2}+2\zeta\omega_n s+\omega_n^{2}}\) (see Chapter 8).

The system is underdamped, so the peak overshoot is

A plant \(G(s)=\dfrac{5}{(s+1)(s+3)}\) is placed in a unity negative feedback loop with a cascaded PI controller \(G_c(s)=K_p+\dfrac{K_i}{s}\). The steady-state error of the closed-loop system to a unit-ramp input \(r(t)=t\,u(t)\) is required to be \(0.1\). The value of \(K_i\) is _____.

Solution

The integral action of the PI controller adds one pole at the origin, so the open-loop transfer function becomes type 1 (see Chapter 19):

For a type-1 system the ramp error is governed by the velocity error constant (Chapter 9):

Note that \(K_p\) does not appear in \(K_v\); it does however affect stability. With \(K_i=6\) the characteristic equation is \(s^{3}+4s^{2}+(3+5K_p)s+30=0\), and the Routh condition \(4(3+5K_p)>30\) requires \(K_p>0.9\) for the design to be usable.

The characteristic equation of a closed-loop system is \(s^{4}+6s^{3}+11s^{2}+6s+K=0\). The range of the real parameter \(K\) for which all roots lie in the left half of the \(s\)-plane is

1. \(0 < K < 10\)
2. \(0 < K < 6\)
3. \(0 < K < 11\)
4. \(K > 10\)

Solution

Build the Routh array (Chapter 11). The first two rows are read off from the coefficients:

The \(s^{2}\) row follows from the usual determinant rule:

The \(s^{1}\) row uses the \(s^{3}\) and \(s^{2}\) rows:

The last row is simply \(s^{0}: K\). The completed array is

All entries of the first column must be positive for stability:

At \(K=10\) the \(s^{1}\) row vanishes and the auxiliary polynomial \(10s^{2}+10=0\) places a pair of roots at \(s=\pm j1\), confirming that \(K=10\) is the marginal-stability limit.

The open-loop transfer function of a unity negative feedback system is \(G(s)H(s)=\dfrac{K}{s(s+2)(s+6)}\) with \(K>0\). The breakaway point of the root locus on the real axis is at \(s=\) _____ (round off to 2 decimal places).

Solution

The open-loop poles are at \(0,-2,-6\) and there are no finite zeros. A point on the real axis belongs to the locus when the number of real poles and zeros to its right is odd (Chapter 12). Counting: the segment \(-2<\sigma<0\) has one pole to its right and is on the locus; the segment \(-6<\sigma<-2\) has two and is not; the segment \(\sigma<-6\) has three and is on the locus. Two branches leave the origin and \(-2\) towards each other, so a breakaway point must exist between \(-2\) and \(0\).

Breakaway points satisfy \(dK/ds=0\), which for a pole-only open loop reduces to

Multiply through by \(\sigma(\sigma+2)(\sigma+6)\):

Only \(\sigma_1\) lies on a locus segment; \(\sigma_2=-4.43\) falls in \(-6<\sigma<-2\), which was shown above not to be part of the locus, so it is rejected.

A unity negative feedback system has open-loop transfer function \(G(s)=\dfrac{40}{s(s+2)(s+4)}\). The gain margin of the system is _____ dB (round off to 2 decimal places).

Solution

The gain margin is measured at the phase-crossover frequency \(\omega_{pc}\), where the phase reaches \(-180^\circ\) (Chapter 17). The phase of \(G(j\omega)\) is

Setting this to \(-180^\circ\) gives \(\tan^{-1}(\omega/2)+\tan^{-1}(\omega/4)=90^\circ\). Two angles sum to \(90^\circ\) only when their tangents are reciprocals, so

Evaluate the magnitude there. With \(\omega^{2}=8\),

So \(|G(j\omega_{pc})|=0.8333\) and the gain margin as a ratio is its reciprocal:

The gain margin is positive but small, so the loop gain could only be raised by 20 % before the system becomes unstable. This is consistent with the Routh limit: the characteristic equation \(s^{3}+6s^{2}+8s+K=0\) is marginally stable at \(K=48\), and the given gain is \(K=40\).

The open-loop transfer function \(G(s)H(s)\) of a feedback system has two poles in the right half of the \(s\)-plane and no right-half-plane zeros. When the complete Nyquist contour is mapped, the resulting Nyquist plot of \(G(s)H(s)\) encircles the point \(-1+j0\) twice in the anticlockwise direction. The number of closed-loop poles in the right half of the \(s\)-plane is

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

Solution

The Nyquist criterion applies the principle of the argument to \(F(s)=1+G(s)H(s)\), whose zeros are the closed-loop poles and whose poles are the open-loop poles (Chapter 16). Counting anticlockwise encirclements of \(-1+j0\) as positive,

where \(P\) is the number of open-loop poles and \(Z\) the number of closed-loop poles inside the right half plane. Here \(P=2\) and \(N=+2\), so

No closed-loop pole lies in the right half plane, so this open-loop-unstable plant has been stabilised by the feedback loop.

A lead compensator of the form \(G_c(s)=\dfrac{1+\alpha T s}{1+T s}\), with \(\alpha>1\), is to provide a maximum phase lead of \(30^\circ\) at a frequency of \(6\) rad/s. The compensator pole lies at \(s=-p\). The value of \(p\) is _____ rad/s (round off to 2 decimal places).

Solution

For this compensator the maximum phase lead and the frequency at which it occurs are (Chapter 21)

First find \(\alpha\) from the required lead of \(30^\circ\):

Now impose \(\omega_m = 6\) rad/s to find \(T\):

The compensator has a zero at \(s=-1/(\alpha T)\) and a pole at \(s=-1/T\):

As a check, \(\omega_m\) must be the geometric mean of the zero and pole corner frequencies: \(\sqrt{3.4641\times 10.3923}=\sqrt{36}=6\) rad/s, as required.

A system is described in state-variable form by \(\dot{x}=Ax+Bu\) with \(A=\begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix}\) and \(B=\begin{bmatrix} 0 \\ 1 \end{bmatrix}\). Full state feedback \(u=-Kx\) with \(K=\begin{bmatrix} k_1 & k_2 \end{bmatrix}\) is applied to place the closed-loop poles at \(s=-4\pm j4\). The value of \(k_1\) is _____.

Solution

First confirm that the poles can be placed arbitrarily by checking controllability (Chapter 25):

\(\det Q_c = (0)(-3)-(1)(1) = -1 \neq 0\), so the pair is controllable. Form the closed-loop matrix (Chapter 26):

The desired characteristic polynomial for poles at \(-4\pm j4\) is

Matching coefficients term by term:

So \(K=\begin{bmatrix} 30 & 5 \end{bmatrix}\).