GATE 2026 EC Q.15: solution and answer

GATE 2026 · ECQ.15 · 1 mark · MCQ

Two analog signals $x_{1} (t)$ and $x_{2} (t)$ ( $t$ in second), are sampled at a rate $F_{s} = 40 Hz$ , where

$x_{1} (t) = cos (20 π t), t \geq 0$ , and $x_{2} (t) = cos (100 π t), t \geq 0$ .

The first ten samples (starting from $t = 0$ ) are considered for the analysis.

Which of the following statements is TRUE?

(A)
All of the first three samples of $x_{1} (t)$ are greater than the corresponding samples of $x_{2} (t)$ .
(B)
All of the last three samples of $x_{1} (t)$ are greater than the corresponding samples of $x_{2} (t)$ .
(C)
All of the samples of $x_{2} (t)$ are greater than the corresponding samples of $x_{1} (t)$ .
(D)
All of the fourth to seventh samples of $x_{1} (t)$ are equal to the corresponding samples of $x_{2} (t)$ .

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Answer and solution

ANSWER

(D) All of the fourth to seventh samples of $x_{1} (t)$ are equal to the corresponding samples of $x_{2} (t)$ . In fact, all ten samples are equal, because the $50 Hz$ signal aliases to $10 Hz$ at $F_{s} = 40 Hz$ .

Official answer: DSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Sampling at $F_{s} = 40 Hz$ gives sampling instants $t = n T_{s} = \frac{n}{40} s$ . The first ten samples, starting from $t = 0$ , are $n = 0, 1, \dots, 9$ .
Samples of $x_{1}$ : $x_{1} [n] = cos (20 π \cdot \frac{n}{40}) = cos (\frac{π n}{2})$ .
Samples of $x_{2}$ : $x_{2} [n] = cos (100 π \cdot \frac{n}{40}) = cos (\frac{5 π n}{2})$ .
Because $n$ is an integer, $2 π n$ is a whole number of cycles, and cosine has period $2 π$ : $cos (\frac{5 π n}{2}) = cos (2 π n + \frac{π n}{2}) = cos (\frac{π n}{2}) .$ So $x_{2} [n] = x_{1} [n]$ for every $n$ .
Aliasing view of the same fact: $x_{2}$ is a $50 Hz$ tone, above $F_{s} /2 = 20 Hz$ , so after sampling it appears at $∣50 - 40∣ = 10 Hz$ , which is exactly the frequency of $x_{1}$ . The two sampled sequences cannot be told apart.
Both sequences, for $n = 0, \dots, 9$ , are $1, 0, - 1, 0, 1, 0, - 1, 0, 1, 0$ .
Every pair of corresponding samples is equal, so no sample of one signal is greater than the other: (A), (B) and (C) are all false. The fourth to seventh samples ( $n = 3$ to $6$ ) are $0, 1, 0, - 1$ for both signals, so they are equal and (D) is TRUE.

KEY IDEA

A sinusoid at $f_{0}$ sampled at $F_{s}$ gives the same samples as one at $f_{0} - k F_{s}$ for any integer $k$ , because the discrete-time frequency $2 π f_{0} / F_{s}$ is only defined modulo $2 π$ . Here $50 Hz - 40 Hz = 10 Hz$ , so $x_{2}$ aliases exactly onto $x_{1}$ and every sample matches.

WATCH OUT FOR

Comparing the continuous-time waveforms instead of the samples. $cos (100 π t)$ oscillates five times faster than $cos (20 π t)$ , so at most instants the two signals differ, and that tempts students towards (A), (B) or (C). At the sampling instants $t = n /40$ , though, the extra $80 Hz$ ( $2 π n$ in phase) adds a whole number of cycles, and the samples are identical. A second slip is counting the fourth sample as $n = 4$ ; since sampling starts at $n = 0$ , the fourth to seventh samples are $n = 3$ to $6$ . Here that makes no difference because every sample is equal.

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GATE 2026 EC Q.15: solution and answer

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