GATE 2024 EC Q.48: solution and answer

GATE 2024 · ECQ.48 · 2 marks · MCQ

A continuous time signal $x (t) = 2 cos (8 π t + π /3)$ is sampled at a rate of 15 Hz. The sampled signal $x_{s} (t)$ when passed through an LTI system with impulse response

h (t) = (\frac{sin 2 π t}{π t}) cos (38 π t - π /2)

produces an output $x_{o} (t)$ . The expression for $x_{o} (t)$ is ______.

(A)
$15 sin (38 π t + π /3)$
(B)
$15 sin (38 π t - π /3)$
(C)
$15 cos (38 π t - π /6)$
(D)
$15 cos (38 π t + π /6)$

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Answer and solution

ANSWER

$x_{o} (t) = 15 cos (38 π t - π /6)$ . That is option (C), and option (A), $15 sin (38 π t + π /3)$ , is the same function. With two correct options printed, the official key awarded marks to all candidates (MTA).

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Input: $x (t) = 2 cos (2 π \cdot 4 t + π /3)$ , so $f_{0} = 4 Hz$ and $X (f) = e^{j π /3} δ (f - 4) + e^{- j π /3} δ (f + 4)$
Ideal (impulse-train) sampling at $f_{s} = 15 Hz$ gives $X_{s} (f) = f_{s} \sum_{k} X (f - k f_{s})$ . The spectral lines sit at $\pm 4 + 15 k$ , i.e. at $4, 11, 19, 26, \dots Hz$ and their negatives. Each line has magnitude $15$ . Lines at $4 + 15 k$ carry phase $+ π /3$ , and lines at $- 4 + 15 k$ carry phase $- π /3$ .
The filter, by the modulation property: $\frac{sin 2 π t}{π t} \leftrightarrow G (f)$ , a rectangle of height $1$ for $∣ f ∣ < 1 Hz$ . Writing $cos (2 π \cdot 19 t - π /2) = \frac{1}{2} e^{- j π /2} e^{j 2 π 19 t} + \frac{1}{2} e^{j π /2} e^{- j 2 π 19 t}$ gives $H (f) = \frac{1}{2} e^{- j π /2} G (f - 19) + \frac{1}{2} e^{j π /2} G (f + 19)$ This is an ideal band-pass filter passing $18$ to $20 Hz$ . Its gain is $\frac{1}{2}$ and its phase is $- π /2$ on the positive-frequency side.
Only the line at $19 = 15 + 4 Hz$ (with its mirror at $- 19 Hz$ ) falls in the passband. The lines at $11$ and $26 Hz$ are rejected. At $+ 19 Hz$ the sampled spectrum has coefficient $15 e^{j π /3}$ and at $- 19 Hz$ it has $15 e^{- j π /3}$ . In the time domain this surviving image is $f_{s} \cdot 2 cos (38 π t + π /3) = 30 cos (38 π t + π /3)$ .
Apply the filter's gain and phase at $19 Hz$ . The coefficient at $+ 19 Hz$ becomes $15 e^{j π /3} \cdot \frac{1}{2} e^{- j π /2} = 7.5 e^{- j π /6}$ , and the one at $- 19 Hz$ is its conjugate. So $x_{o} (t) = \frac{1}{2} \cdot 30 cos (38 π t + \frac{π}{3} - \frac{π}{2}) = 15 cos (38 π t - \frac{π}{6})$ which is option (C).
Check the other options with $sin θ = cos (θ - π /2)$ . Option (A): $15 sin (38 π t + π /3) = 15 cos (38 π t + π /3 - π /2) = 15 cos (38 π t - π /6)$ , so (A) is the same function as (C). Option (B): $15 sin (38 π t - π /3) = 15 cos (38 π t - 5 π /6)$ , which has the wrong phase. Option (D): $15 cos (38 π t + π /6)$ also has the wrong phase. Two options describe the correct output, which explains why the official key awarded marks to all candidates (MTA).

KEY IDEA

Ideal impulse-train sampling copies the spectrum to every multiple of $f_{s}$ and scales each copy by $f_{s}$ . The impulse response is a sinc low-pass multiplied by a 19 Hz cosine, so it is a narrow band-pass filter centred at 19 Hz. That filter picks out only the image at $f_{s} + f_{0} = 15 + 4 = 19$ Hz. The factor $f_{s} = 15$ in every option confirms the ideal impulse-sampling model, and the $- π /2$ inside the modulating cosine shows up as a $- π /2$ phase shift on the passed tone.

WATCH OUT FOR

Students often drop the $\frac{1}{2}$ from modulation and get amplitude 30, or forget the $f_{s}$ scaling of the images and get amplitude 1. Some take the image at 19 Hz to be $- 4 + 15 k$ and so give it phase $- π /3$ instead of $+ π /3$ (19 is $4 + 15$ ). Others ignore the $- π /2$ phase of the band-pass filter. Finally, many do not notice that $sin (θ + π /3) = cos (θ - π /6)$ , which makes (A) and (C) the same function.

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GATE 2024 EC Q.48: solution and answer

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