GATE 2024 IN Q.22: solution and answer

GATE 2024 · INQ.22 · 1 mark · NAT

Consider a filter defined by the difference equation

y [n] - 0.5 y [n - 2] = a x [n - 4]

where $x [n]$ and $y [n]$ represent the input and output, respectively. If the magnitude response of the filter at $ω = \frac{π}{2}$ is $H (\frac{π}{2}) = 0.5$ , the value of $a$ is ________ (rounded off to two decimal places).

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$a = 0.75$ .

Official answer: 0.70 to 0.80Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Take the DTFT of both sides, using the time-shift property $x [n - k] \leftrightarrow e^{- j ω k} X (e^{j ω})$ : $Y (e^{j ω}) (1 - 0.5 e^{- j 2 ω}) = a e^{- j 4 ω} X (e^{j ω})$
Frequency response: $H (e^{j ω}) = \frac{Y ( e ^{j ω} )}{X ( e ^{j ω} )} = \frac{a e ^{- j 4 ω}}{1 - 0.5 e ^{- j 2 ω}}$ (The equivalent $H (z) = \frac{a z ^{- 4}}{1 - 0.5 z ^{- 2}}$ has poles at $z = \pm \frac{1}{2}$ , inside the unit circle, so the causal filter is stable and its frequency response exists.)
At $ω = \frac{π}{2}$ : $e^{- j 2 ω} = e^{- j π} = - 1$ and $e^{- j 4 ω} = e^{- j 2 π} = 1$ .
So $H (\frac{π}{2}) = \frac{a \cdot 1}{1 - 0.5 ( - 1 )} = \frac{a}{1.5}$
Magnitude condition: $H (\frac{π}{2}) = \frac{∣ a ∣}{1.5} = 0.5 \Rightarrow ∣ a ∣ = 0.75$
The magnitude fixes only $∣ a ∣$ ; the value asked for is the positive gain, $a = 0.75$ . (A gain of $- 0.75$ would give the same magnitude, since it only adds a phase of $π$ .)

KEY IDEA

Delays in the difference equation become factors $e^{- j ω k}$ in the frequency response. At $ω = π /2$ , $e^{- j 2 ω} = - 1$ and $e^{- j 4 ω} = 1$ , so the input delay $x [n - 4]$ has magnitude 1 and drops out, and the denominator becomes $1 + 0.5 = 1.5$ .

WATCH OUT FOR

Taking $e^{- j 2 ω}$ as $+ 1$ at $ω = π /2$ (it is $e^{- j π} = - 1$ ). That gives a denominator of $0.5$ and the wrong answer $a = 0.25$ . Another mistake is thinking the delay $x [n - 4]$ changes the magnitude. It only adds phase, because $∣ e^{- j 4 ω} ∣ = 1$ at every frequency.

From the GATE 2024 IN paper. Every question of this paper · Download the full paper · All 2 FIR & IIR filters questions, with notes