GATE 2024 EE Q.38: solution and answer

GATE 2024 · EEQ.38 · 2 marks · MCQ

Consider the discrete-time systems $T_{1}$ and $T_{2}$ defined as follows:

{T_{1} x} [n] = x [0] + x [1] + \dots + x [n]

{T_{2} x} [n] = x [0] + \frac{1}{2} x [1] + \dots + \frac{1}{2 ^{n}} x [n]

Which one of the following statements is true?

(A)
$T_{1}$ and $T_{2}$ are BIBO stable.
(B)
$T_{1}$ and $T_{2}$ are not BIBO stable.
(C)
$T_{1}$ is BIBO stable but $T_{2}$ is not BIBO stable.
(D)
$T_{1}$ is not BIBO stable but $T_{2}$ is BIBO stable.

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Answer and solution

ANSWER

(D) $T_{1}$ is not BIBO stable but $T_{2}$ is BIBO stable. With the constant input $x [n] = 1$ , $T_{1}$ gives $n + 1$ , which grows without limit. For any input with $∣ x [n] ∣ \leq B$ , $T_{2}$ is bounded by $B \sum_{k} (1/2)^{k} = 2 B$ .

Official answer: DSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

BIBO stability means that every bounded input, $∣ x [n] ∣ \leq B$ for all $n$ , gives a bounded output. One bounded input with an unbounded output is enough to prove instability. Proving stability needs a bound that holds for every bounded input.
$T_{1}$ is a running sum (accumulator) starting at $k = 0$ : $y_{1} [n] = \sum_{k = 0}^{n} x [k]$ . Test it with the bounded input $x [n] = 1$ ( $B = 1$ ).
Then $y_{1} [n] = n + 1 terms 1 + 1 + \dots + 1 = n + 1 \to \infty$ as $n \to \infty$ . The output is unbounded, so $T_{1}$ is not BIBO stable.
$T_{2}$ is $y_{2} [n] = \sum_{k = 0}^{n} (\frac{1}{2})^{k} x [k]$ . The weight $2^{- k}$ depends on the absolute index $k$ , not on $n - k$ , so $T_{2}$ is time-varying and the LTI test $\sum ∣ h [n] ∣ < \infty$ does not apply. Bound the output directly instead.
For any input with $∣ x [k] ∣ \leq B$ , the triangle inequality and the geometric series give $∣ y_{2} [n] ∣ \leq k = 0 \sum n (\frac{1}{2})^{k} ∣ x [k] ∣ \leq B k = 0 \sum \infty (\frac{1}{2})^{k} = \frac{B}{1 - \frac{1}{2}} = 2 B .$
The bound $2 B$ does not depend on $n$ , so every bounded input gives an output bounded by $2 B$ . $T_{2}$ is BIBO stable.
So $T_{1}$ is not BIBO stable and $T_{2}$ is BIBO stable. The answer is option (D).

KEY IDEA

To show a system is BIBO unstable, you only need one bounded input that gives an unbounded output. A constant fed into a running sum gives $n + 1$ . To show it is stable, bound $∣ y [n] ∣$ for every bounded input. For $T_{2}$ , the weights $(1/2)^{k}$ add up to a finite total of 2, so $∣ y_{2} [n] ∣ \leq 2 B$ for every $n$ . This works directly even though $T_{2}$ is time-varying, so no impulse-response test is needed.

WATCH OUT FOR

Calling $T_{2}$ unstable because the number of terms grows with $n$ . What decides stability is whether the weights have a finite total, and $\sum (1/2)^{k} = 2$ . Another mistake is to treat $T_{2}$ as LTI and use $\sum ∣ h [n] ∣$ . $T_{2}$ is time-varying, because its weights depend on $k$ and not on $n - k$ . That test does not apply, even though here it happens to give the right answer. A third mistake is calling $T_{1}$ stable because every term $x [k]$ is bounded. A sum of more and more bounded terms can still grow without limit.

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GATE 2024 EE Q.38: solution and answer

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