---
title: "Check properties of y(t) = 2x(t+1)"
subject: "Data Communication"
chapter: 2
key: "semester_5/data_communication:check-2x-t-plus-1"
source: https://www.bctengineer.com/q/semester_5/data_communication/check-2x-t-plus-1
site: BCT Engineering
---

# Check properties of y(t) = 2x(t+1)

Data Communication · Chapter 2: Signals and Systems

- Asked in: 2073 Shrawan, 6 marks
- Video: [Check System for Linearity, Causality, Time-Invariance and Stability | Lecture-57](https://www.youtube.com/watch?v=VS84_wPfxlE) — Engineers choice tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/check-2x-t-plus-1
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-2

## Question

Check linearity, causality, stability and time invariance characteristics of system y(t) = 2x(t+1).

## Short answer

y(t)=2x(t+1) is linear, time-invariant, and stable (bounded input yields bounded output). It is non-causal because the output depends on the future input x(t+1).

## Answer

y(t) = 2x(t + 1)

#### Linearity
Response to a·x₁ + b·x₂ is 2[a·x₁(t+1) + b·x₂(t+1)] = a·y₁ + b·y₂. **Linear.**

#### Causality
At time t the output uses x(t + 1), the input one second in the future. **Non-causal.**

#### Stability
If |x(t)| ≤ M, then |y(t)| = 2|x(t + 1)| ≤ 2M. A bounded input gives a bounded output. **Stable.** (h(t) = 2δ(t + 1), and ∫|h| dt = 2.)

#### Time invariance
Delay the input: x₁(t) = x(t - t₀) gives y₁(t) = 2x₁(t + 1) = 2x(t + 1 - t₀).
Delay the output: y(t - t₀) = 2x(t - t₀ + 1).
These are equal. **Time-invariant.**

**Answer: linear, time-invariant, non-causal and stable (an LTI system that advances the signal by 1 s and doubles it).**

## Check yourself

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

1. Why is y(t) = 2x(t+1) non-causal?
   A. It amplifies the input by 2
   B. It is time-invariant
   C. It is unstable
   D. The output at t uses the input at t+1, a future value
   Answer: D. Causal systems cannot use future input. (Key point: Causality)

2. Is y(t) = 2x(t+1) stable?
   A. Yes, |x| <= M gives |y| <= 2M
   B. No, since it is non-causal
   C. No, since it advances the signal
   D. Only if x is periodic
   Answer: A. A bounded input gives a bounded output regardless of causality. (Key point: Stability)

3. What does the time-invariance test give?
   A. They differ, so it is time-variant
   B. Both delayed-input and delayed-output responses equal 2x(t+1-t0), so it is time-invariant
   C. It cannot be tested
   D. Time-variant because of the factor 2
   Answer: B. A constant gain and a fixed shift commute with delay. (Key point: Time invariance)

4. What is the impulse response of this system?
   A. 2 delta(t-1)
   B. delta(t) + 2
   C. 2 delta(t+1)
   D. 2u(t+1)
   Answer: C. Putting x = delta gives 2 delta(t+1), a non-causal advance. (Key point: Impulse response)

## Related in Data Communication

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