GATE 2020 Control Systems Practice Questions and Solutions

In a control system the error signal \(E(s) = R(s) - C(s)\) drives a block \(G_1(s) = \dfrac{5}{s+3}\). The output of \(G_1\) drives a second block \(G_2(s) = \dfrac{2}{s}\), and the output of \(G_2\) is the system output \(C(s)\). An inner negative feedback path of constant gain \(H_1 = 4\) is taken from the output of \(G_2\) back to the summing junction placed just before \(G_2\). The overall transfer function \(C(s)/R(s)\) is

1. \(\dfrac{10}{s^2 + 11s + 34}\)
2. \(\dfrac{10}{s^2 + 11s + 24}\)
3. \(\dfrac{10}{s^2 + 3s + 10}\)
4. \(\dfrac{10}{s^2 + 8s + 34}\)

Solution

Reduce the inner loop first (see Chapter 4). For a negative feedback loop of forward gain \(G_2\) and feedback gain \(H_1\):

The inner block is now in cascade with \(G_1\):

Closing the outer unity negative feedback loop:

A unity negative feedback system has the open-loop transfer function

The steady-state error for the input \(r(t) = 4\,u(t)\) is _____.

Solution

There is no pole at the origin, so the system is Type 0 and a step input leaves a finite non-zero error governed by the position error constant (see Chapter 9):

For a step of amplitude \(A\), the steady-state error is \(e_{ss} = A/(1+K_p)\). Here \(A = 4\):

A unity negative feedback system has the open-loop transfer function

The range of \(K\) for which the closed-loop system is stable is

1. \(0 < K < 23\)
2. \(0 < K < 192\)
3. \(0 < K < 207\)
4. \(K > 192\)

Solution

Form the characteristic equation \(1 + G(s) = 0\). Expanding the denominator,

Construct the Routh array (see Chapter 11):

The \(s^0\) row requires \(15 + K > 0\), i.e. \(K > -15\). The \(s^1\) row requires

Taking \(K\) positive as a gain, the stable range is \(0 < K < 192\). At \(K = 192\) the \(s^1\) row vanishes and the auxiliary equation \(9s^2 + 207 = 0\) gives sustained oscillation at \(\omega = \sqrt{23} = 4.796\) rad/s.

For the unity negative feedback system with

the centroid of the root-locus asymptotes and the angles of the asymptotes are

1. \(-3.5\) and \(\pm 90^\circ\)
2. \(-1.5\) and \(\pm 60^\circ, 180^\circ\)
3. \(-1.5\) and \(\pm 90^\circ\)
4. \(-2.33\) and \(\pm 60^\circ, 180^\circ\)

Solution

The open-loop poles are \(0, -1, -6\) so \(n = 3\), and there is one finite zero at \(-4\) so \(m = 1\). The number of asymptotes is \(n - m = 2\) (see Chapter 12).

The asymptote angles are

So the two branches that go to infinity leave along the vertical asymptotes at \(\pm 90^\circ\) through \(\sigma_A = -1.5\).

A unity negative feedback system has the open-loop transfer function

The value of \(K\) that produces a phase margin of \(60^\circ\) is _____ (rounded off to two decimal places).

Solution

The phase of the loop is independent of \(K\), so the phase-margin specification fixes the gain crossover frequency first (see Chapter 17):

Now enforce unity magnitude at \(\omega_{gc}\):

With \(\omega_{gc}^2 = 16/3\), the term under the root is \(16/3 + 16 = 64/3\), so \(\sqrt{\cdot} = 8/\sqrt{3}\):

A plant \(G_p(s) = \dfrac{5}{(s+1)(s+5)}\) is placed in a unity negative feedback loop with the PI controller

The largest value of \(K_i\) for which the closed-loop system remains stable is _____.

Solution

The PI controller adds a pole at the origin, raising the system to Type 1 (see Chapter 19). The characteristic equation is \(1 + G_c(s)G_p(s) = 0\):

Expanding \(s(s+1)(s+5) = s^3 + 6s^2 + 5s\) and substituting \(K_p = 2\):

Routh array:

Both conditions must hold: \(5K_i > 0\) gives \(K_i > 0\), and

The stable range is \(0 < K_i < 18\), so the limiting value is \(K_i = 18\). At that value the auxiliary equation \(6s^2 + 90 = 0\) gives sustained oscillation at \(\omega = \sqrt{15} = 3.873\) rad/s.

The loop transfer function of a unity negative feedback system is

Using the Nyquist criterion, the closed-loop system is

1. unstable, with two encirclements of \(-1+j0\)
2. stable, with a gain margin of \(13.06\) dB
3. stable, with a gain margin of \(6.02\) dB
4. marginally stable

Solution

The loop transfer function has no poles in the right half plane, so \(P = 0\) and the Nyquist criterion \(N = Z - P\) requires zero encirclements of \(-1+j0\) for stability (see Chapter 16). Find where the plot crosses the negative real axis. Expanding,

With \(s = j\omega\):

At that frequency \(D = 8 - 7(14) = -90\), so

The crossing occurs at \(-0.222\), which is to the right of \(-1\). The critical point is therefore not enclosed, \(N = 0\), and with \(P = 0\) the number of closed-loop right-half-plane poles is \(Z = 0\): the system is stable. The gain margin is

A system is described by

The value of \(\alpha\) for which the system is not completely observable is

1. \(\alpha = -1\)
2. \(\alpha = 0\)
3. \(\alpha = 1\)
4. \(\alpha = 3\)

Solution

For a second-order system the observability matrix is \(Q_o = \begin{bmatrix} C \\ CA \end{bmatrix}\), and complete observability requires \(\det Q_o \neq 0\) (see Chapter 25). Compute \(CA\):

The determinant is zero only at \(\alpha = 1\), where \(Q_o\) has rank one and the mode associated with the eigenvalue \(-3\) becomes invisible at the output.

A unity negative feedback system has the open-loop transfer function

For a unit-step input, the percentage peak overshoot is _____ % (rounded off to two decimal places), and the \(2\%\) settling time is _____ s.

Solution

Close the loop:

Comparing with \(s^2 + 2\zeta\omega_n s + \omega_n^2\) (see Chapter 8):

Percentage peak overshoot:

The \(2\%\) settling time uses the envelope \(e^{-\zeta\omega_n t}\):

The unforced state equation of a system is \(\dot{x} = Ax\) with

The value of \(x_1(t)\) at \(t = 1\) s is _____ (rounded off to three decimal places).

Solution

The matrix has a repeated eigenvalue at \(-1\). Split it as \(A = -I + N\) where

Since \(-I\) and \(N\) commute and \(N\) is nilpotent, the series for the state transition matrix terminates after two terms (see Chapter 24):

The zero-input response is \(x(t) = \phi(t)x(0)\), so

At \(t = 1\) s: