GATE 2020 EC Q.14: solution and answer

GATE 2020 · ECQ.14 · 1 mark · MCQ

Which one of the following pole-zero plots corresponds to the transfer function of an LTI system characterized by the input-output difference equation given below?

y [n] = k = 0 \sum 3 (- 1)^{k} x [n - k]

GATE 2020 EC question 14: Each option is a z-plane pole-zero plot: horizontal axis Re, vertical axis Im, with the unit circle drawn and the points… — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
Figure — zeros (o) at $z = 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "3rd order pole".
(B)
Figure — zeros (o) at $z = - 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "3rd order pole".
(C)
Figure — zeros (o) at $z = - 1$ and at two conjugate points on the unit circle in the right half-plane, at about $\pm 6 0^{\circ}$ (i.e. $e^{\pm j π /3}$ ); a cross at the origin labelled "3rd order pole".
(D)
Figure — zeros (o) at $z = 1$ , $z = - 1$ , $z = j$ and $z = - j$ on the unit circle; a cross at the origin labelled "4th order pole".

Answer and solution

ANSWER

(A): $H (z) = \frac{( z - 1 ) ( z ^{2} + 1 )}{z ^{3}}$ , so the zeros are at $z = 1$ and $z = \pm j$ and there is a 3rd-order pole at the origin.

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

Expand the sum: $y [n] = x [n] - x [n - 1] + x [n - 2] - x [n - 3]$ . This is a 4-tap FIR filter with impulse response $h [n] = {1, - 1, 1, - 1}$ .
Take the z-transform using the time-shift property $x [n - k] \leftrightarrow z^{- k} X (z)$ : $H (z) = \frac{Y ( z )}{X ( z )} = 1 - z^{- 1} + z^{- 2} - z^{- 3}$
Factor by grouping: $H (z) = (1 - z^{- 1}) + z^{- 2} (1 - z^{- 1}) = (1 - z^{- 1}) (1 + z^{- 2})$
Multiply numerator and denominator by $z^{3}$ so the poles and zeros can be read off in positive powers of $z$ : $H (z) = \frac{z ^{3} - z ^{2} + z - 1}{z ^{3}} = \frac{( z - 1 ) ( z ^{2} + 1 )}{z ^{3}}$
Zeros: $z - 1 = 0 \Rightarrow z = 1$ , and $z^{2} + 1 = 0 \Rightarrow z = \pm j$ . All three lie on the unit circle.
Poles: the denominator $z^{3}$ gives a 3rd-order pole at $z = 0$ . The numerator is not zero at $z = 0$ (it equals $- 1$ there), so nothing cancels. In general, a causal FIR filter with taps $h [0], \dots, h [N - 1]$ and $h [N - 1] \neq = 0$ has $N - 1$ poles at the origin; here $N = 4$ .
Cross-check with the geometric series: $H (z) = k = 0 \sum 3 (- z^{- 1})^{k} = \frac{1 - z ^{- 4}}{1 + z ^{- 1}}$ The numerator vanishes at the fourth roots of unity $1, j, - 1, - j$ , but the factor at $z = - 1$ is cancelled by the denominator $1 + z^{- 1}$ . That leaves zeros at $1$ and $\pm j$ , which rules out B (zero at $- 1$ instead of $1$ ), C (zeros at $- 1$ and $e^{\pm j π /3}$ ) and D (four zeros, 4th-order pole).
Only option A shows zeros at $1$ , $j$ , $- j$ with a 3rd-order pole at the origin.

KEY IDEA

For an FIR difference equation, $H (z)$ is just the tap polynomial $\sum h [k] z^{- k}$ . Multiply through by $z^{N - 1}$ to get $N - 1$ poles at the origin and the finite zeros as roots of $z^{N - 1} H (z)$ . With alternating signs the taps factor neatly: $1 - z^{- 1} + z^{- 2} - z^{- 3} = (1 - z^{- 1}) (1 + z^{- 2})$ .

WATCH OUT FOR

Each wrong option matches a specific slip. Dropping the $(- 1)^{k}$ gives $1 + z^{- 1} + z^{- 2} + z^{- 3} = \frac{( z + 1 ) ( z ^{2} + 1 )}{z ^{3}}$ , which is option B. Using the geometric-series form $\frac{1 - z ^{- 4}}{1 + z ^{- 1}}$ but forgetting that the denominator cancels the zero at $z = - 1$ gives four zeros and a 4th-order pole, which is option D. The sum has four terms ( $k = 0$ to $3$ ), so the highest delay is $z^{- 3}$ and there are exactly 3 poles at the origin, not 4.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 12 z-transform questions, with notes

GATE 2020 EC Q.14: solution and answer

Answer and solution

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