GATE 2020 EC Q.53: solution and answer

GATE 2020 · ECQ.53 · 2 marks · NAT

The transfer function of a stable discrete-time LTI system is $H (z) = \frac{K ( z - α )}{z + 0.5}$ , where $K$ and $α$ are real numbers. The value of $α$ (rounded off to one decimal place) with $∣ α ∣ > 1$ , for which the magnitude response of the system is constant over all frequencies, is ________.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$α = - 2.0$ . The zero at $z = - 2$ is the reciprocal of the pole at $z = - 0.5$ , which makes $H (z)$ all-pass with $∣ H (e^{j ω}) ∣ = 2∣ K ∣$ at every frequency.

Official answer: -2Source: Official answer key, GATE 2020 EC (IIT Delhi)

Magnitude response means $∣ H (e^{j ω}) ∣$ , i.e. $H (z)$ evaluated on the unit circle $z = e^{j ω}$ . If it is constant for every $ω$ , $H (z)$ is an all-pass system.
All-pass rule for a first-order real section: the zero must sit at the reciprocal of the pole, $z_{0} = 1/ p$ . For real $p$ , $∣ e^{j ω} - \frac{1}{p} ∣ = \frac{1}{∣ p ∣} ∣ p e^{j ω} - 1∣ = \frac{1}{∣ p ∣} ∣ p - e^{- j ω} ∣ = \frac{1}{∣ p ∣} ∣ e^{j ω} - p ∣,$ where the last step uses the fact that a complex number and its conjugate have the same magnitude. So $∣ e^{j ω} - z_{0} ∣ / ∣ e^{j ω} - p ∣ = 1/∣ p ∣$ at every point on the unit circle.
Here the pole is $p = - 0.5$ . It lies inside the unit circle, so the system is stable, as the question says. The zero is at $z = α$ , so the rule gives $α = \frac{1}{- 0.5} = - 2$ , and $∣ α ∣ = 2 > 1$ as required.
(The only other way to get a constant magnitude is $α = - 0.5$ : the zero cancels the pole and $H (z) = K$ . The condition $∣ α ∣ > 1$ rules that out.)
Check on the unit circle: $∣ H (e^{j ω}) ∣^{2} = K^{2} \frac{∣ e ^{j ω} + 2 ∣ ^{2}}{∣ e ^{j ω} + 0.5 ∣ ^{2}} = K^{2} \frac{( 2 + cos ω ) ^{2} + sin ^{2} ω}{( 0.5 + cos ω ) ^{2} + sin ^{2} ω} = K^{2} \frac{5 + 4 cos ω}{1.25 + cos ω} = 4 K^{2}$
So $∣ H (e^{j ω}) ∣ = 2∣ K ∣$ at every frequency, which confirms $α = - 2.0$ .

KEY IDEA

A first-order section has constant magnitude response (all-pass) when its zero is the reciprocal of its pole, $z_{0} = 1/ p$ . On the unit circle, $∣ e^{j ω} - 1/ p ∣ = ∣ e^{j ω} - p ∣/∣ p ∣$ for real $p$ , so the ratio of the two distances does not depend on $ω$ . Pole at $- 0.5$ gives zero at $- 2$ .

WATCH OUT FOR

Answering $α = - 0.5$ (pole-zero cancellation). That does give a constant magnitude, but it breaks the condition $∣ α ∣ > 1$ . Another common slip is to take the reciprocal as $+ 2$ , putting the zero at the mirror image through the origin instead of at $1/ p$ . That gives $∣ H ∣^{2} \propto (5 - 4 cos ω) / (1.25 + cos ω)$ , which changes with frequency.

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.53: solution and answer

Answer and solution

More z-transform questions