GATE 2020 EE Q.38: solution and answer

GATE 2020 · EEQ.38 · 2 marks · MCQ

The causal realization of a system transfer function $H (s)$ having poles at $(2, - 1)$ , $(- 2, 1)$ and zeroes at $(2, 1)$ , $(- 2, - 1)$ will be

(A)
stable, real, allpass
(B)
unstable, complex, allpass
(C)
unstable, real, highpass
(D)
stable, complex, lowpass

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Answer and solution

ANSWER

(B) unstable, complex, allpass

Official answer: BSource: Official answer key, GATE 2020 EE (IIT Delhi)

Read each coordinate pair as a point $s = σ + j ω$ in the s-plane. The poles are $p_{1} = 2 - j$ and $p_{2} = - 2 + j$ . The zeros are $z_{1} = 2 + j$ and $z_{2} = - 2 - j$ .
Build the transfer function (unity gain). Each pair of factors is a difference of squares:
$H (s) = \frac{( s - 2 - j ) ( s + 2 + j )}{( s - 2 + j ) ( s + 2 - j )} = \frac{s ^{2} - ( 2 + j ) ^{2}}{s ^{2} - ( 2 - j ) ^{2}} = \frac{s ^{2} - 3 - 4 j}{s ^{2} - 3 + 4 j}$
Real or complex: a system with a real impulse response has a real-coefficient $H (s)$ , so its complex poles come in conjugate pairs. The conjugate of the pole $2 - j$ is $2 + j$ , which is a zero, not a pole. The poles are not a conjugate pair, the coefficients $- 3 \mp 4 j$ are complex, and so $h (t)$ is complex-valued. The system is complex.
All-pass test: each zero must be the mirror image of a pole about the $j ω$ axis, $z = - p^{*}$ . Here $- (2 - j)^{*} = - (2 + j) = - 2 - j = z_{2}$ and $- (- 2 + j)^{*} = - (- 2 - j) = 2 + j = z_{1}$ . Both zeros pass.
Confirm on the $j ω$ axis. Put $s = j ω$ , so $s^{2} = - ω^{2}$ :
$∣ H (j ω) ∣^{2} = \frac{- ( ω ^{2} + 3 ) - 4 j ^{2}}{∣ - ( ω ^{2} + 3 ) + 4 j ∣ ^{2}} = \frac{( ω ^{2} + 3 ) ^{2} + 16}{( ω ^{2} + 3 ) ^{2} + 16} = 1 for all ω$
The magnitude is flat, so the system is all-pass. Any other constant gain $K$ gives $∣ H (j ω) ∣ = ∣ K ∣$ , which is still flat.
Stability from the ROC: the ROC of a causal system lies to the right of its rightmost pole. The pole $2 - j$ is in the right half-plane, so the ROC is $Re (s) > 2$ . That region does not contain the $j ω$ axis, so the causal realization is unstable. The residue of $H (s)$ at $s = 2 - j$ is $\frac{( 2 - j ) ^{2} - 3 - 4 j}{( 2 - j ) + 2 - j} = \frac{- 8 j}{4 - 2 j} \neq = 0$ , so $h (t)$ really does contain a term in $e^{(2 - j) t} u (t)$ , whose magnitude $e^{2 t}$ grows without bound.
The system is unstable, complex and all-pass, which is option (B). Every other option fails at least one test: (A) and (D) say stable, (A) and (C) say real, (C) says highpass and (D) says lowpass.

KEY IDEA

Read each property straight from the pole-zero pattern. Zeros that mirror the poles about the $j ω$ axis ( $z = - p^{*}$ ) give an all-pass. Complex poles without their conjugates give complex coefficients, so the system is complex. A causal system with a right-half-plane pole has ROC $Re (s) > 2$ , which misses the $j ω$ axis, so it is unstable.

WATCH OUT FOR

Taking the symmetric-looking pattern as proof the system is real. The poles $2 - j$ and $- 2 + j$ are symmetric about the origin ( $p$ and $- p$ ), not about the real axis. Symmetry about the origin only makes $H$ a function of $s^{2}$ ; it does not make the coefficients real. A second trap is assuming an all-pass is automatically stable. All-pass describes only $∣ H (j ω) ∣$ . Causality combined with the right-half-plane pole at $2 - j$ makes the system unstable, which rules out (A).

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GATE 2020 EE Q.38: solution and answer

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