---
title: "Time invariance of y(t) = y(t−4) + x(t−4)"
subject: "Communication Systems"
chapter: 1
key: "semester_6/communication_systems:time-invariance-y-t-4"
source: https://www.bctengineer.com/q/semester_6/communication_systems/time-invariance-y-t-4
site: BCT Engineering
---

# Time invariance of y(t) = y(t−4) + x(t−4)

Communication Systems · Chapter 1: Introduction

- Asked in: 2080 Chaitra, 3 marks
- Video: [Time-Invariant and Time-Variant Systems](https://www.youtube.com/watch?v=LezLNMznZm4) — Neso Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/time-invariance-y-t-4
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-1

## Question

Determine whether the signal y(t) = y(t−4) + x(t−4) is time invariant or not.

## Short answer

The recursive equation has constant coefficients and no explicit time dependence. A delayed input x(t−t0) yields the same output as shifting the original response by t0, given rest initial conditions, so the system is time-invariant.

## Answer

A system is **time-invariant** if a time shift of the input causes the same time shift of the output: if $x(t) \to y(t)$, then $x(t-t_0) \to y(t-t_0)$.

Given: $y(t) = y(t-4) + x(t-4)$ (a recursive system; assume it starts at rest).

**Step 1:** apply a delayed input $x_1(t) = x(t-t_0)$. Its output $y_1(t)$ satisfies

$$
y_1(t) = y_1(t-4) + x_1(t-4) = y_1(t-4) + x(t-t_0-4)
$$

**Step 2:** shift the original equation by $t_0$ (replace $t$ by $t-t_0$):

$$
y(t-t_0) = y(t-t_0-4) + x(t-t_0-4)
$$

**Step 3:** compare. The delayed output $y(t-t_0)$ satisfies exactly the same equation, driven by the same input $x(t-t_0-4)$, as $y_1(t)$. With the same (rest) initial condition, the solutions are equal:

$$
y_1(t) = y(t-t_0)
$$

The reason is that the coefficients (both equal to 1) and the delay (4 s) are constants; time $t$ does not appear explicitly in the equation.

**Answer:** the system $y(t) = y(t-4) + x(t-4)$ is **time-invariant**.

## Check yourself

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

1. A system is time-invariant if x(t) -> y(t) implies:
   A. x(t - t0) -> y(t) + t0
   B. x(t - t0) -> y(t - t0)
   C. x(t - t0) -> y(t0) only
   D. x(t - t0) -> y(-t + t0)
   Answer: B. Delaying the input must just delay the output by the same amount. (Key point: Time-invariance definition)

2. How is time invariance shown for y(t) = y(t-4) + x(t-4)?
   A. By checking that the output is bounded
   B. By computing the Fourier transform of y
   C. By showing x is periodic
   D. By showing the delayed output y(t - t0) satisfies the same equation with the delayed input as the response to x(t - t0)
   Answer: D. Both y1(t) and y(t - t0) satisfy the same equation driven by x(t - t0 - 4) with the same rest condition, so they are equal. (Key point: Method for recursive systems)

3. Why is this system time-invariant?
   A. It is non-causal
   B. The coefficients (both 1) and delay (4 s) are constants; t does not appear explicitly
   C. Because it is non-linear
   D. Because the output is zero at t = 0
   Answer: B. A time-varying multiplier such as t would break time invariance; constant coefficients and delays do not. (Key point: Reason for time invariance)

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