GATE 2022 IN Q.14: solution and answer

GATE 2022 · INQ.14 · 1 mark · MCQ

The output of a system $y (t)$ is related to its input $x (t)$ according to the relation $y (t) = x (t) sin (2 π t)$ . This system is ________

(A)
Linear and time-variant
(B)
Non-linear and time-invariant
(C)
Linear and time-invariant
(D)
Non-linear and time-variant

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Answer and solution

ANSWER

(A) Linear and time-variant. Superposition holds because the output is the input multiplied by a fixed function of time. Shifting the input does not shift the multiplier $sin (2 π t)$ , so the system fails the shift test.

Official answer: ASource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Linearity (superposition test). Let $x_{1} (t) \to y_{1} (t) = x_{1} (t) sin (2 π t)$ and $x_{2} (t) \to y_{2} (t) = x_{2} (t) sin (2 π t)$ . Apply $x (t) = a x_{1} (t) + b x_{2} (t)$ for arbitrary constants $a, b$ : $y (t) = [a x_{1} (t) + b x_{2} (t)] sin (2 π t) = a x_{1} (t) sin (2 π t) + b x_{2} (t) sin (2 π t) = a y_{1} (t) + b y_{2} (t)$ Additivity and homogeneity both hold, so the system is linear.
Check: a zero input gives $y (t) = 0 \cdot sin (2 π t) = 0$ , as every linear system must.
Time invariance (shift test). Delay the input by $t_{0}$ and pass it through the system. The multiplier $sin (2 π t)$ is not part of the input, so it is not shifted: $y_{1} (t) = x (t - t_{0}) sin (2 π t)$
Now delay the original output by $t_{0}$ , replacing every $t$ by $t - t_{0}$ : $y (t - t_{0}) = x (t - t_{0}) sin (2 π (t - t_{0}))$
These two must be equal for every input and every shift $t_{0}$ if the system is time-invariant. One counterexample is enough to show it is not. Take $x (t) = 1$ and $t_{0} = 0.25$ . Then $y_{1} (t) = sin (2 π t)$ , while $y (t - 0.25) = sin (2 π t - \frac{π}{2}) = - cos (2 π t)$ At $t = 0$ : $y_{1} (0) = 0$ but $y (0 - 0.25) = - 1$ . So $y_{1} (t) \neq = y (t - t_{0})$ , and the system is time-variant.
The reason is that the gain $sin (2 π t)$ depends explicitly on time, outside the argument of $x$ . This is an amplitude modulator, which is linear but time-variant. Answer: (A).

KEY IDEA

If $y (t) = g (t) x (t)$ with a fixed function $g (t)$ , the system is always linear, because multiplying by any fixed function distributes over sums and scalings. It is time-invariant only when $g (t)$ is a constant. Any explicit $t$ outside the argument of $x$ (here $sin (2 π t)$ ) makes the system time-variant.

WATCH OUT FOR

There are two common mistakes. The first is to shift the $sin (2 π t)$ factor along with the input when checking time invariance. Only $x$ is delayed. The multiplier belongs to the system and stays at $sin (2 π t)$ . The second is to test with an integer shift such as $t_{0} = 1$ . Because $sin (2 π (t - 1)) = sin (2 π t)$ , the outputs agree for that shift, and the student wrongly concludes the system is time-invariant. Time invariance must hold for every shift, so one failing shift like $t_{0} = 0.25$ settles the question. Seeing $sin$ and calling the system non-linear is also wrong: the sine acts on $t$ , not on $x$ .

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GATE 2022 IN Q.14: solution and answer

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