---
title: "Linear/TI/stable/memoryless: y(t), y[k]"
subject: "Data Communication"
chapter: 2
key: "semester_5/data_communication:check-1-minus-e-4t-and-sin"
source: https://www.bctengineer.com/q/semester_5/data_communication/check-1-minus-e-4t-and-sin
site: BCT Engineering
---

# Linear/TI/stable/memoryless: y(t), y[k]

Data Communication · Chapter 2: Signals and Systems

- Asked in: 2072 Kartik, 3+3 marks
- Video: [Determine whether the system is memoryless,  causal and stable](https://www.youtube.com/watch?v=AF4Swi-H_jM) — Engg-Course-Made-Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/check-1-minus-e-4t-and-sin
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-2

## Question

Determine if the following systems are linear, time-invariant, stable and memoryless. a) y(t) = [1 - e^(-4t)][U(t)] where U(t) is the continuous-time unit step function. b) y[k] = sin(x[k-4])

## Short answer

a) y(t) = (1 - e^{-4t})u(t) depends only on t, not input, so it is non-linear, time-variant, stable, and memoryless. b) y[k] = sin(x[k-4]) is non-linear (sin is not additive), time-invariant, stable (|y|≤1), and has memory (uses past sample).

## Answer

#### a) y(t) = [1 - e^(-4t)] U(t)
The expression has no input x(t); it is the given output function. I take the system as one whose output is this fixed function of time, y(t) = (1 - e^(-4t))U(t) (independent of the input), and test it as such.

- **Linear?** No. With zero input, the output is not zero (y ≠ 0 for t > 0), so superposition (zero in, zero out) fails.
- **Time-invariant?** No. The output depends explicitly on t: shifting the input by t₀ does not shift the output, since the output does not follow the input at all.
- **Stable?** Yes. 0 ≤ y(t) < 1 for all t, so the output is always bounded.
- **Memoryless?** Yes. The output at t depends only on t, not on past or future values of any input.

(If the signal is meant to be the step response of a system, then h(t) = 4e^(-4t)U(t): that system is linear, time-invariant, stable, causal, and has memory.)

#### b) y[k] = sin(x[k - 4])
- **Linear?** No. sin(a·x₁ + b·x₂) ≠ a·sin x₁ + b·sin x₂.
- **Time-invariant?** Yes. A shift of n in the input gives sin(x[k - n - 4]) = y[k - n].
- **Stable?** Yes. |y[k]| ≤ 1 for any input.
- **Memoryless?** No. The output depends on the input four samples earlier, so the system has memory.
- It is also causal (it uses a past value).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is system a) treated as non-linear?
   A. It contains an exponential
   B. It has the step function
   C. It is unbounded
   D. Its output is nonzero even with zero input
   Answer: D. Linear systems give zero output for zero input. (Key point: a) Linearity)

2. Is system a) stable?
   A. Yes, 0 <= y(t) < 1 always
   B. No, y grows
   C. No, it is not linear
   D. Only for t < 0
   Answer: A. The output is always bounded. (Key point: a) Stability)

3. Is y[k] = sin(x[k-4]) linear?
   A. Yes, it is a delay
   B. No, sin of a sum is not the sum of sines
   C. Yes, since |y| <= 1
   D. Only for small x
   Answer: B. sin is a nonlinear function. (Key point: b) Linearity)

4. Does y[k] = sin(x[k-4]) have memory?
   A. No, it is memoryless
   B. No, it is nonlinear
   C. Yes, it depends on the input four samples earlier
   D. Only for k > 4
   Answer: C. A dependence on past input means memory. (Key point: b) Memory)

## Related in Data Communication

- [Unit step response for h(t) = exp(-at)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aunit-step-response-exp) — Data Communication, chapter 2
- [Stable/unstable systems; test stability](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Astability-define-test-72c) — Data Communication, chapter 2
- [Define unit impulse and unit step](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aunit-impulse-step-define) — Data Communication, chapter 2
- [Properties of continuous-time Fourier series](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afourier-series-properties) — Data Communication, chapter 3
- [Check linearity/TI: log(x(t)), t·x(t)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acheck-log-xt-and-t-xt) — Data Communication, chapter 2
- [Unit impulse function and its significance](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aunit-impulse-significance) — Data Communication, chapter 2

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