GATE 2023 EE Q.46: solution and answer

GATE 2023 · EEQ.46 · 2 marks · NAT

Consider the state-space description of an LTI system with matrices

A = [0 - 1 1 - 2], B = [01], C = [3 - 2], D = 1

For the input, $sin (ω t)$ , $ω > 0$ , the value of $ω$ for which the steady-state output of the system will be zero, is ________ (Round off to the nearest integer).

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Answer and solution

ANSWER

$ω = 2 rad/s$

Official answer: 2Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Transfer function from state space: $G (s) = C (s I - A)^{- 1} B + D$ .
$s I - A = [s 1 - 1 s + 2]$ , so $det (s I - A) = s (s + 2) + 1 = s^{2} + 2 s + 1 = (s + 1)^{2}$ .
Adjoint times B: $(s I - A)^{- 1} B = \frac{1}{( s + 1 ) ^{2}} [s + 2 - 1 1 s] [01] = \frac{1}{( s + 1 ) ^{2}} [1 s]$
Multiply by C: $C (s I - A)^{- 1} B = \frac{3 ( 1 ) + ( - 2 ) ( s )}{( s + 1 ) ^{2}} = \frac{3 - 2 s}{( s + 1 ) ^{2}}$ .
Add the feedthrough $D = 1$ : $G (s) = \frac{3 - 2 s}{( s + 1 ) ^{2}} + 1 = \frac{s ^{2} + 2 s + 1 + 3 - 2 s}{( s + 1 ) ^{2}} = \frac{s ^{2} + 4}{( s + 1 ) ^{2}}$
Stability check: both poles are at $s = - 1$ (left half-plane), so the natural response dies out and a sinusoidal steady state exists: $y_{ss} (t) = ∣ G (j ω) ∣ sin (ω t + ∠ G (j ω))$ .
Frequency response: $G (j ω) = \frac{4 - ω ^{2}}{( 1 + j ω ) ^{2}}$ . The steady-state output is zero only when $∣ G (j ω) ∣ = 0$ , i.e. $4 - ω^{2} = 0$ . With $ω > 0$ this gives $ω = 2 rad/s$ . Physically, the zeros $s = \pm j 2$ lie on the $j ω$ -axis and block a sinusoid of exactly that frequency.

KEY IDEA

For a stable system, the steady-state response to $sin (ω t)$ is $∣ G (j ω) ∣ sin (ω t + ∠ G (j ω))$ . It vanishes only at a frequency where $G (s)$ has a zero on the $j ω$ -axis. So find $G (s) = C (s I - A)^{- 1} B + D$ and look for purely imaginary zeros. Here $G (s) = \frac{s ^{2} + 4}{( s + 1 ) ^{2}}$ , whose zeros are at $s = \pm j 2$ .

WATCH OUT FOR

Dropping the feedthrough term $D = 1$ . Without it, $G (s) = \frac{3 - 2 s}{( s + 1 ) ^{2}}$ , whose only zero is the real $s = 1.5$ , and $∣ G (j ω) ∣$ is never zero for $ω > 0$ . Other slips: getting a sign wrong in the adjoint of $s I - A$ (it is $[s + 2 - 1 1 s]$ ), and not checking stability before using the frequency-response argument.

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GATE 2023 EE Q.46: solution and answer

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