Question

Consider the state-space description of an LTI system with matrices: \[ A = \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix}, B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, C = [3 \;\; -2], D = 1 \] For the input \(\sin(\omega t)\), \(\omega > 0\), the value of \(\omega\) for which the steady-state output of the system will be zero is ................. (Round off to the nearest integer).

Hint:

For sinusoidal inputs, steady-state output zeros occur when \(G(j\omega)=0\). This is equivalent to finding system transmission zeros.

Answer 1

Step 1: Transfer function. \\ The transfer function is: \[ G(s) = C (sI - A)^{-1} B + D \] \[ sI - A = \begin{bmatrix} s & -1 \\ 1 & s+2 \end{bmatrix}, \det(sI - A) = s(s+2)+1 = s^2+2s+1 = (s+1)^2 \] Inverse: \[ (sI - A)^{-1} = \frac{1}{(s+1)^2} \begin{bmatrix} s+2 & 1 \\ -1 & s \end{bmatrix} \]

Step 2: Compute transfer function. \\ \[ (sI - A)^{-1} B = \frac{1}{(s+1)^2} \begin{bmatrix} 1 \\ s \end{bmatrix} \] Multiply by \(C\): \[ C (sI - A)^{-1} B = \frac{3(1) + (-2)(s)}{(s+1)^2} = \frac{3 - 2s}{(s+1)^2} \] Add \(D=1\): \[ G(s) = 1 + \frac{3 - 2s}{(s+1)^2} \]

Step 3: Steady-state sinusoidal response. \\ For sinusoidal input, steady-state output magnitude is \(|G(j\omega)|\). We want this to be zero, so numerator must vanish: \[ G(j\omega) = 0 \] \[ 1 + \frac{3 - 2j\omega}{(j\omega+1)^2} = 0 \] \[ \frac{(j\omega+1)^2 + (3 - 2j\omega)}{(j\omega+1)^2} = 0 \] \[ (j\omega+1)^2 + 3 - 2j\omega = 0 \]

Step 4: Simplify. \\ \[ (j\omega+1)^2 = (1 + j\omega)^2 = 1 + 2j\omega - \omega^2 \] Equation: \[ (1 - \omega^2 + 2j\omega) + (3 - 2j\omega) = 0 \] \[ (4 - \omega^2) + 0j = 0 \] \[ \omega^2 = 4 \Rightarrow \omega = 2 \]

Final Answer: \\ \[ \boxed{2} \]