GATE 2019 Control Systems Practice Questions and Solutions

A signal flow graph has four nodes \(x_1, x_2, x_3, x_4\), where \(x_1\) is the input node and \(x_4\) is the output node. The branches are: \(x_1 \to x_2\) with gain \(1\), \(x_2 \to x_3\) with gain \(2\), and \(x_3 \to x_4\) with gain \(3\). In addition there is a self-loop at node \(x_3\) of gain \(-2\), and a feedback branch \(x_4 \to x_2\) of gain \(-1\). The overall transmittance \(x_4/x_1\) is

1. \(6\)
2. \(\dfrac{2}{3}\)
3. \(-\dfrac{6}{5}\)
4. \(\dfrac{6}{7}\)

Solution

Apply Mason's gain formula (see Chapter 5). There is one forward path, \(x_1 \to x_2 \to x_3 \to x_4\):

There are two individual loops. The self-loop at \(x_3\) gives \(L_1 = -2\). The outer loop \(x_2 \to x_3 \to x_4 \to x_2\) gives

Both loops share node \(x_3\), so no pair of loops is non-touching:

Every loop touches the forward path, so the path cofactor is \(\Delta_1 = 1\):

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

The steady-state error of the system for the input \(r(t) = 3t\,u(t)\) is _____ (rounded off to two decimal places).

Solution

The loop transfer function has exactly one pole at the origin, so the system is Type 1 (see Chapter 9). A Type 1 system tracks a ramp with a finite error governed by the velocity error constant:

For a ramp \(r(t) = A t\,u(t)\) the steady-state error is \(e_{ss} = A/K_v\). Here \(A = 3\):

The characteristic equation of a closed-loop system is

The number of roots of this equation that lie in the right half of the \(s\)-plane is

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

Solution

All coefficients are present and positive, so the necessary condition is met but stability is not yet decided. Build the Routh array (see Chapter 11).

The \(s^2\) row:

The \(s^1\) row:

The \(s^0\) row is \(d_1 = 5\). The first column is therefore

There are two sign changes, \(1 \to -6\) and \(-6 \to 5\). By the Routh–Hurwitz criterion the number of sign changes in the first column equals the number of roots with positive real part, so two roots lie in the right half plane.

The closed-loop transfer function of a position control system is

For a unit-step input, the peak overshoot and the peak time of the response are respectively

1. \(9.48\%\) and \(0.785\) s
2. \(16.30\%\) and \(0.785\) s
3. \(9.48\%\) and \(0.628\) s
4. \(4.60\%\) and \(0.628\) s

Solution

Match the denominator with the standard second-order form \(s^2 + 2\zeta\omega_n s + \omega_n^2\) (see Chapter 8):

The damped natural frequency is

Peak overshoot:

which is \(9.48\%\). Peak time:

The open-loop transfer function of a unity negative feedback system is

The breakaway point of the root locus on the real axis is at \(\sigma =\) _____ (rounded off to three decimal places).

Solution

The open-loop poles are at \(0, -2, -4\) and there are no finite zeros. On the real axis the locus exists where the total count of poles and zeros to the right is odd, which gives the segments \([-2,\,0]\) and \((-\infty,\,-4]\). A breakaway point must lie on one of these segments (see Chapter 12).

Write the characteristic equation \(1 + G(s)H(s) = 0\) as \(K = -s(s+2)(s+4)\) and set \(dK/ds = 0\):

The point \(\sigma_2 = -3.155\) lies between \(-4\) and \(-2\), which is not part of the root locus, so it is rejected. The point \(\sigma_1 = -0.845\) lies inside \([-2,\,0]\) and is the breakaway point. The gain there is \(K = -(-0.845)(1.155)(3.155) = 3.079\), which is positive, confirming the point is valid for \(K > 0\).

For the unity negative feedback system with open-loop transfer function

the gain margin, in dB, is _____ (rounded off to two decimal places).

Solution

The gain margin is measured at the phase crossover frequency \(\omega_{pc}\), where the phase equals \(-180^\circ\) (see Chapter 17). The phase is

Setting this to \(-180^\circ\) requires \(\tan^{-1}(\omega/2) + \tan^{-1}(\omega/10) = 90^\circ\), which happens when the two arguments are reciprocals:

Evaluate the magnitude at that frequency:

The gain margin is the reciprocal of this magnitude:

The loop transfer function of a unity negative feedback system is

The value of \(K\) for which the Nyquist plot of \(G(s)H(s)\) passes through the point \(-1 + j0\), and the corresponding frequency, are

1. \(K = 6\), \(\omega = 3.317\) rad/s
2. \(K = 60\), \(\omega = 3.317\) rad/s
3. \(K = 60\), \(\omega = 1.732\) rad/s
4. \(K = 90\), \(\omega = 3.317\) rad/s

Solution

The Nyquist plot crosses the negative real axis where the imaginary part of \(G(j\omega)H(j\omega)\) vanishes (see Chapter 16). Expand the denominator:

Substituting \(s = j\omega\):

The ratio \(K/D(j\omega)\) is real when \(D(j\omega)\) is real, so

At that frequency the denominator is purely real:

At \(K = 60\) the plot passes exactly through \(-1+j0\), so the closed-loop system is marginally stable with sustained oscillation at \(3.317\) rad/s.

A single-input system is described by

The system fails to be completely state controllable for

1. \(\beta = 0\) only
2. \(\beta = -1\) or \(\beta = -6\)
3. \(\beta = -2\) or \(\beta = -3\)
4. no real value of \(\beta\)

Solution

For a second-order system the controllability matrix is \(Q_c = [\,B \;\; AB\,]\) and the system is completely controllable when \(\det Q_c \neq 0\) (see Chapter 25). First compute \(AB\):

The determinant vanishes at \(\beta = -2\) and \(\beta = -3\), so the rank of \(Q_c\) drops to one and the system loses complete controllability at exactly those two values. (These are precisely the values of \(\beta\) for which \(B\) becomes an eigenvector of \(A\), whose eigenvalues are \(-2\) and \(-3\).)

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

The value of the first state variable \(x_1(t)\) at \(t = \ln 2\) seconds is _____ (rounded off to two decimal places).

Solution

The state transition matrix is \(\phi(t) = \mathcal{L}^{-1}\{(sI-A)^{-1}\}\) (see Chapter 24). First form \(sI-A\):

With \(x(0) = [1 \;\; 0]^T\), only the first column of \(\phi(t)\) is needed:

Taking the inverse Laplace transform:

At \(t = \ln 2\) we have \(e^{-t} = 1/2\) and \(e^{-2t} = 1/4\):

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

The value of \(K\) that gives a phase margin of \(45^\circ\) is _____ (rounded off to three decimal places).

Solution

Phase margin is defined at the gain crossover frequency \(\omega_{gc}\), where \(|G(j\omega_{gc})| = 1\), as \(\text{PM} = 180^\circ + \angle G(j\omega_{gc})\) (see Chapter 17). The phase here is

Impose the phase-margin requirement first, since it does not involve \(K\):

Now force the magnitude to be unity at that frequency:

As a check, the closed-loop characteristic equation \(s^2 + s + \sqrt{2} = 0\) gives \(\omega_n = 2^{1/4} = 1.189\) rad/s and \(\zeta = 1/(2\omega_n) = 0.420\), a damping ratio consistent with a phase margin near \(45^\circ\).