GATE 2025 Control Systems Practice Questions and Solutions

A single-input single-output system is built from three blocks. The forward block has transfer function \(G(s)=\dfrac{20}{s+5}\). A minor loop is closed around this block alone through a constant-gain path \(H_1=0.25\) in negative feedback. The output of that minor loop is then fed to a unity-gain negative feedback path that closes the outer loop around the whole arrangement. The overall closed-loop transfer function \(\dfrac{C(s)}{R(s)}\) is

1. \(\dfrac{20}{s+10}\)
2. \(\dfrac{20}{s+25}\)
3. \(\dfrac{20}{s+30}\)
4. \(\dfrac{20}{s+45}\)

Solution

Reduce the inner loop first. For a forward gain \(G\) with negative feedback \(H_1\), the equivalent block is \(G/(1+GH_1)\). See Chapter 4 for the reduction rules.

The inner loop has therefore only moved the pole from \(-5\) to \(-10\); the numerator is unchanged. Now close the outer unity-feedback loop around \(G_{in}(s)\).

Check the DC gain: \(20/30=0.667\), which is consistent with an open-loop DC gain of \(20/10=2\) reduced by unity feedback to \(2/(1+2)=0.667\).

A signal flow graph has five nodes \(x_1\) (input), \(x_2\), \(x_3\), \(x_4\) and \(x_5\) (output). The forward branches are \(x_1\to x_2\) with gain \(1\), \(x_2\to x_3\) with gain \(4\), \(x_3\to x_4\) with gain \(2\), and \(x_4\to x_5\) with gain \(5\). There are two feedback branches: \(x_3\to x_2\) with gain \(-0.5\), and \(x_5\to x_4\) with gain \(-0.2\). The overall gain \(x_5/x_1\) is _____ (round off to 2 decimal places).

Solution

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

There is exactly one forward path, \(x_1\to x_2\to x_3\to x_4\to x_5\):

There are two individual loops. The first uses the branches \(x_2\to x_3\) and \(x_3\to x_2\); the second uses \(x_4\to x_5\) and \(x_5\to x_4\):

\(L_1\) touches only nodes \(x_2, x_3\) and \(L_2\) touches only nodes \(x_4, x_5\), so the two loops share no node and form a non-touching pair. The graph determinant is

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

A unity negative feedback system has forward transfer function \(G(s)=\dfrac{25}{s(s+6)}\). For a unit step input, the percentage peak overshoot of the output is _____ % (round off to 2 decimal places).

Solution

Form the closed-loop transfer function and match it to the standard second-order form \(\dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\) (Chapter 8).

Comparing coefficients, \(\omega_n^2=25\) so \(\omega_n=5\) rad/s, and \(2\zeta\omega_n=6\) so \(\zeta=6/10=0.6\). The system is underdamped, so the peak overshoot formula applies:

For reference the peak occurs at \(t_p=\dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}=\dfrac{\pi}{4}=0.785\) s.

A unity negative feedback system has open-loop transfer function \(G(s)=\dfrac{20(s+2)}{s(s+5)(s+10)}\). The reference input is \(r(t)=4+6t\) for \(t\ge 0\). The steady-state error is _____ (round off to 2 decimal places).

Solution

The loop gain has one pole at the origin, so the system is type 1. Before using the final value theorem, confirm the closed loop is stable; otherwise no steady state exists. The characteristic equation is

All coefficients are positive and \(15\times 70=1050 \gt 40\), so by the Routh criterion the closed loop is stable. Now use superposition on the two input components (Chapter 9).

For the step component \(4u(t)\), the position error constant of a type-1 system is

For the ramp component \(6t\), the velocity error constant is

Adding the two contributions gives \(e_{ss}=0+7.5=7.5\).

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 \(K\) that puts the system on the verge of instability, and the frequency of the resulting sustained oscillation, are

1. \(K=6\), \(\omega=1\) rad/s
2. \(K=10\), \(\omega=1\) rad/s
3. \(K=10\), \(\omega=\sqrt{11}\) rad/s
4. \(K=11\), \(\omega=\sqrt{6}\) rad/s

Solution

Build the Routh array (Chapter 11). The first two rows come straight from the coefficients:

The last row requires \(K \gt 0\). The \(s^1\) row requires \(60-6K \gt 0\), that is \(K \lt 10\). So the closed loop is stable for \(0 \lt K \lt 10\), and marginal stability occurs at \(K=10\), where \(c_1\) vanishes and the \(s^1\) row becomes a row of zeros.

At \(K=10\) the frequency of oscillation comes from the auxiliary equation formed from the row above the vanished row, namely the \(s^2\) row with entries \(b_1=10\) and \(K=10\):

So a pair of roots sits on the imaginary axis at \(\pm j1\) and the sustained oscillation has \(\omega=1\) rad/s. Substituting \(K=10\) into the quartic confirms it factors as \((s^2+1)(s^2+6s+10)\), whose remaining roots \(-3\pm j1\) are in the left half plane.

The root locus of a unity feedback system with open-loop transfer function \(G(s)H(s)=\dfrac{K}{s(s+2)(s+4)}\) is drawn for \(K\) varying from \(0\) to \(\infty\). The breakaway point of the locus lies on the real axis at \(s=-\sigma_b\). The value of \(\sigma_b\) is _____ (round off to 3 decimal places).

Solution

The open-loop poles are at \(0, -2, -4\) and there are no finite zeros. Real-axis segments belong to the locus where the number of poles and zeros to the right is odd, which gives the segments \([-2,\,0]\) and \((-\infty,\,-4]\). A breakaway point can therefore only occur between \(0\) and \(-2\) (Chapter 12).

From the characteristic equation \(1+G(s)H(s)=0\),

Breakaway points satisfy \(dK/ds=0\):

The root \(-3.1547\) lies in \((-4,-2)\), which is not part of the locus, so it is discarded. The valid breakaway point is \(s=-0.8453\), and the gain there is

A positive \(K\) confirms the point is a genuine breakaway on the \(K \gt 0\) locus, so \(\sigma_b=0.845\).

A unity negative feedback system has open-loop transfer function \(G(s)H(s)=\dfrac{10}{s(s+1)(s+5)}\). 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 open-loop phase equals \(-180^\circ\) (Chapter 17).

Two angles sum to \(90^\circ\) when the product of their tangents is 1, so

An equivalent route is to expand the denominator and set its imaginary part to zero: \(j\omega(j\omega+1)(j\omega+5)=-6\omega^2+j(5\omega-\omega^3)\), which is purely real when \(\omega^2=5\). The magnitude at that frequency is

The gain margin is positive, so the gain may be raised by a factor of 3 before the closed loop becomes unstable.

A unity negative feedback system has open-loop transfer function \(G(s)H(s)=\dfrac{K(s+2)}{s(s-1)}\) with \(K \gt 0\). Using the Nyquist stability criterion, the closed-loop system is stable for

1. all \(K \gt 0\)
2. \(K \gt 1\)
3. \(0 \lt K \lt 1\)
4. no value of \(K\)

Solution

The Nyquist criterion states \(N=P-Z\), where \(P\) is the number of open-loop poles in the right half plane, \(Z\) the number of closed-loop poles there, and \(N\) the number of anticlockwise encirclements of the \(-1+j0\) point (Chapter 16). Here the open-loop poles are at \(s=0\) and \(s=1\), so \(P=1\) and stability (\(Z=0\)) demands exactly one anticlockwise encirclement of \(-1\).

The encirclement count is fixed by where the closed-loop poles sit, so evaluate the characteristic equation directly:

For a second-order polynomial, all roots lie in the left half plane if and only if every coefficient is positive. The constant term gives \(2K \gt 0\), satisfied for all \(K \gt 0\). The middle coefficient gives

Consistency check with the Nyquist count: at \(K=2\) the roots are \(s^2+s+4=0\), i.e. \(s=-0.5\pm j1.94\), so \(Z=0\) and \(N=P-Z=1\), one anticlockwise encirclement, as required. At \(K=0.5\) the polynomial is \(s^2-0.5s+1\) with roots \(0.25\pm j0.968\); now \(Z=2\) and \(N=1-2=-1\), a clockwise encirclement, and the loop is unstable.

A compensator has transfer function \(G_c(s)=\dfrac{s+2}{s+8}\). The maximum phase that this compensator can contribute is _____ degrees (round off to 2 decimal places).

Solution

The zero at \(-2\) is closer to the origin than the pole at \(-8\), so the network is a lead compensator and its phase contribution is positive (Chapter 21). Write it in the standard lead form \(G_c(s)=\dfrac{1}{\alpha}\cdot\dfrac{1+\alpha Ts}{1+Ts}\) with \(\alpha \lt 1\), where the zero is at \(-1/(\alpha T)\) and the pole at \(-1/T\):

The phase of \(G_c(j\omega)\) is \(\phi(\omega)=\tan^{-1}(\omega/2)-\tan^{-1}(\omega/8)\). Setting \(d\phi/d\omega=0\) gives the geometric mean of the two corner frequencies:

Direct substitution confirms it: \(\phi(4)=\tan^{-1}(2)-\tan^{-1}(0.5)=63.43^\circ-26.57^\circ=36.87^\circ\).

A linear time-invariant system is described by \(\dot{x}=Ax\) with \(A=\begin{bmatrix} 0 & 1 \\ -2 & -3\end{bmatrix}\) and initial state \(x(0)=\begin{bmatrix} 1 \\ 0\end{bmatrix}\). The value of the first state variable at \(t=1\) s, that is \(x_1(1)\), is _____ (round off to 3 decimal places).

Solution

The zero-input response is \(x(t)=\phi(t)x(0)\), where \(\phi(t)=e^{At}\) is the state transition matrix, obtained from \(\phi(t)=\mathcal{L}^{-1}\{(sI-A)^{-1}\}\) (Chapter 24).

Only the first column is needed because \(x(0)=[1\;\;0]^T\) selects it. Expanding the \((1,1)\) entry in partial fractions,

because at \(s=-1\) the residue is \((-1+3)/(-1+2)=2\) and at \(s=-2\) it is \((-2+3)/(-2+1)=-1\). Taking the inverse Laplace transform,

A quick sanity check: \(x_1(0)=2-1=1\), matching the given initial condition, and \(\dot{x}_1(0)=-2+2=0=x_2(0)\), as the state equation requires. Evaluating at \(t=1\):