Control Systems - GATE EE 2012 Solved Paper

For a lead compensator \(G_c(s) = \frac{s+a}{s+b}\), the frequency at which maximum phase occurs is given by: \[\omega_m = \sqrt{a \cdot b}\]

From the given transfer function: \[G_c(s) = \frac{s+1}{s+2} \Rightarrow a = 1, b = 2\] \[\omega_m = \sqrt{1 \times 2} = \sqrt{2} \text{ rad/s}\]

Therefore, the phase is maximum at \(\omega = \sqrt{2}\) rad/s.

For \(G_c(s) = \frac{s+a}{s+b}\) to be a lead compensator:

• Both poles and zeros must be in LHP for physical realizability

• Zero must be closer to origin than pole: \(|a| < |b|\)

• For positive \(a\) and \(b\): \(a < b\)

Checking the options:

• A: \(a = 1, b = 2\) \(\Rightarrow\) \(a < b\) ✓ (Lead compensator)

• B: \(a = 3, b = 2\) \(\Rightarrow\) \(a > b\) ✗ (Lag compensator)

• C: \(a = -3, b = -1\) \(\Rightarrow\) Poles/zeros in RHP ✗

• D: \(a = 3, b = 1\) \(\Rightarrow\) \(a > b\) ✗ (Lag compensator)

Therefore, only option A satisfies the conditions for a lead compensator.

Given: \(G(s) = \frac{K(s+1)}{s^3 + as^2 + 2s + 1}\), \(H(s) = 1\)

Characteristic equation: \[1 + G(s)H(s) = 0 \Rightarrow 1 + \frac{K(s+1)}{s^3 + as^2 + 2s + 1} = 0\] \[\Rightarrow s^3 + as^2 + (K+2)s + (K+1) = 0\]

Routh array: \[\begin{array}{c|cc} s^3 & 1 & K+2 \\ s^2 & a & K+1 \\ s^1 & \frac{a(K+2)-(K+1)}{a} & 0 \\ s^0 & K+1 & 0 \end{array}\]

For oscillation at \(\omega = 2\) rad/s:

• Auxiliary equation from \(s^2\) row: \(as^2 + (K+1) = 0\)

• Substitute \(s = j\omega = j2\): \(-a(4) + (K+1) = 0\)

• \(K = 4a - 1\)

Also, for marginal stability, \(s^1\) row must be zero: \[\frac{a(K+2)-(K+1)}{a} = 0 \Rightarrow a(K+2) = K+1\]

Solving both equations: \[a(4a - 1 + 2) = (4a - 1) + 1 \Rightarrow a(4a + 1) = 4a\] \[4a^2 + a = 4a \Rightarrow 4a^2 - 3a = 0 \Rightarrow a(4a - 3) = 0\] \[a = 0.75 \quad (\text{since } a > 0)\] \[K = 4(0.75) - 1 = 3 - 1 = 2\]

Therefore, \(K = 2\) and \(a = 0.75\).

Given: \[A = \begin{bmatrix} 0 & a_1 & 0 \\ 0 & 0 & a_2 \\ a_3 & 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}\]

Controllability matrix: \[Q_c = \begin{bmatrix} B & AB & A^2B \end{bmatrix}\]

Calculate \(AB\): \[AB = \begin{bmatrix} 0 & a_1 & 0 \\ 0 & 0 & a_2 \\ a_3 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ a_2 \\ 0 \end{bmatrix}\]

Calculate \(A^2\): \[A^2 = \begin{bmatrix} 0 & 0 & a_1a_2 \\ a_2a_3 & 0 & 0 \\ 0 & a_1a_3 & 0 \end{bmatrix}\]

Calculate \(A^2B\): \[A^2B = \begin{bmatrix} 0 & 0 & a_1a_2 \\ a_2a_3 & 0 & 0 \\ 0 & a_1a_3 & 0 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a_1a_2 \\ 0 \\ 0 \end{bmatrix}\]

Controllability matrix: \[Q_c = \begin{bmatrix} 0 & 0 & a_1a_2 \\ 0 & a_2 & 0 \\ 1 & 0 & 0 \end{bmatrix}\]

Determinant: \[\det(Q_c) = -a_1a_2^2\]

For controllability, \(\det(Q_c) \neq 0\): \[a_1 \neq 0 \quad \text{and} \quad a_2 \neq 0\] (\(a_3\) can be any value)

Therefore, the system is controllable when \(a_1 \neq 0, a_2 \neq 0, a_3 = 0\).

Given: \(G(s) = \dfrac{(s^2 + 9)(s + 2)}{(s + 1)(s + 3)(s + 4)}\)

For sinusoidal input \(\sin(\omega t)\), the steady-state output is zero when \(|G(j\omega)| = 0\).

Substitute \(s = j\omega\): \[G(j\omega) = \frac{((j\omega)^2 + 9)(j\omega + 2)}{(j\omega + 1)(j\omega + 3)(j\omega + 4)} = \frac{(-\omega^2 + 9)(j\omega + 2)}{(j\omega + 1)(j\omega + 3)(j\omega + 4)}\]

Magnitude: \[|G(j\omega)| = \frac{|-\omega^2 + 9| \cdot |j\omega + 2|}{|j\omega + 1| \cdot |j\omega + 3| \cdot |j\omega + 4|}\]

For \(|G(j\omega)| = 0\), the numerator must be zero: \[|-\omega^2 + 9| = 0 \Rightarrow \omega^2 = 9 \Rightarrow \omega = 3 \text{ rad/s}\]

At \(\omega = 3\) rad/s, the term \((s^2 + 9)\) in numerator becomes zero, making the transfer function magnitude zero.

Therefore, the steady-state output is zero at \(\omega = 3\) rad/s.