GATE 2023 EC Q.26: solution and answer

GATE 2023 · ECQ.26 · 1 mark · MCQ

Consider a system with input $x (t)$ and output $y (t) = x (e^{t})$ . The system is

(A)
Causal and time invariant.
(B)
Non-causal and time varying.
(C)
Causal and time varying.
(D)
Non-causal and time invariant.

See the question in the paper

Answer and solution

ANSWER

The system $y (t) = x (e^{t})$ is non-causal (since $e^{t} > t$ , it always reads a future input value) and time varying (a delayed input gives $x (e^{t} - t_{0})$ , not $x (e^{t - t_{0}})$ ). Answer: (B).

Official answer: BSource: Official answer key, GATE 2023 EC (IIT Kanpur)

Causality test: a system is causal if $y (t)$ depends only on $x (τ)$ for $τ \leq t$ . Here $y (t)$ is the input read at the single instant $τ = e^{t}$ .
Check one instant: at $t = 1$ , $y (1) = x (e) \approx x (2.718)$ , a future value of the input. In fact $e^{t} > t$ for every real $t$ (the function $e^{t} - t$ has its minimum value $1$ at $t = 0$ ), so the output always depends on a future input value. The system is non-causal.
Time-invariance test: delay the input by $t_{0}$ to get $x_{1} (t) = x (t - t_{0})$ . Pass it through the system: $y_{1} (t) = x_{1} (e^{t}) = x (e^{t} - t_{0})$ .
Now delay the original output by $t_{0}$ : $y (t - t_{0}) = x (e^{t - t_{0}}) = x (e^{t} e^{- t_{0}})$ .
Compare the two arguments: $e^{t} - t_{0}$ and $e^{t - t_{0}}$ are different functions of $t$ . For example, at $t = 0$ they are $1 - t_{0}$ and $e^{- t_{0}}$ , and $e^{- t_{0}} > 1 - t_{0}$ for every $t_{0} \neq = 0$ . A concrete input shows the outputs really differ: with $x (t) = t$ and $t_{0} = 1$ , $y_{1} (0) = e^{0} - 1 = 0, y (0 - 1) = e^{- 1} \approx 0.368$ The time axis is warped by $e^{t}$ , not simply shifted, so a delay at the input does not appear as the same delay at the output. The system is time varying.
Non-causal and time varying, which is option (B).

KEY IDEA

When the output is the input evaluated at a warped time $y (t) = x (f (t))$ : it is causal only if $f (t) \leq t$ for every $t$ , and time invariant only if $f (t) = t - c$ (a pure shift). Here $f (t) = e^{t}$ is greater than $t$ everywhere and is not a pure shift, so the system is non-causal and time varying.

WATCH OUT FOR

Delaying the output wrongly by replacing $t$ with $t - t_{0}$ only in the input and writing $y (t - t_{0}) = x (e^{t} - t_{0})$ , which makes the system look time invariant. Another slip is checking causality only at $t = 0$ , where $e^{0} = 1 > 0$ is easy to overlook, or assuming a very negative $t$ makes $e^{t}$ lie in the past. In fact $e^{t} > t$ for every $t$ .

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GATE 2023 EC Q.26: solution and answer

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