---
title: "Time-invariance of y[n]=y[n−4]+x[n−4]"
subject: "Digital Signal Analysis and Processing"
chapter: 1
key: "semester_7/digital_signal_analysis_and_processing:time-invariance-check"
source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/time-invariance-check
site: BCT Engineering
---

# Time-invariance of y[n]=y[n−4]+x[n−4]

Digital Signal Analysis and Processing · Chapter 1: Discrete Time Signals and Systems

- Asked in: 2074 Chaitra, 3 marks
- Video: [Check for Linearity, Causality, Time Invariance, and Dynamicity of a System (DSP Lecture-54)](https://www.youtube.com/watch?v=Ddf5Z6XBP00) — Engineers choice tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/time-invariance-check
- Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-1

## Question

Determine whether the system y[n] = y[n−4] + x[n−4] is time-invariant or not.

## Short answer

The system y[n]=y[n-4]+x[n-4] is time-invariant. Shifting the input by n0 gives the same shifted difference equation as the delayed output, and with initial rest conditions the solutions match; equivalently, unrolling gives y[n]=sum_{k=0}^∞ x[n-4-4k], so a shift in x shifts y equally.

## Answer

A system is time-invariant if a shift of the input by n₀ shifts the output by the same n₀.

**System:** y[n] = y[n − 4] + x[n − 4]. It is a recursive (difference-equation) system, taken with initial rest conditions (output is zero before the input starts).

**Step 1.** Apply input x₁[n] = x[n − n₀]. Output y₁[n] satisfies the same equation:

y₁[n] = y₁[n − 4] + x₁[n − 4] = y₁[n − 4] + x[n − n₀ − 4]

**Step 2.** Delay the original output by n₀. From y[n] = y[n − 4] + x[n − 4], replacing n by n − n₀:

y[n − n₀] = y[n − n₀ − 4] + x[n − n₀ − 4]

**Step 3.** Compare. Let w[n] = y[n − n₀]. Then w[n] = w[n − 4] + x[n − n₀ − 4], which is exactly the equation satisfied by y₁[n]. With the same initial rest condition both have the same solution, so

y₁[n] = y[n − n₀].

Equivalently, unrolling the recursion: y[n] = Σ(k=0 to ∞) x[n − 4 − 4k], and a shift of x shifts every term, so y shifts equally.

**Answer:** The system is **time-invariant** (all coefficients are constants, and with initial rest it is also linear).

## Check yourself

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

1. To test time invariance we compare
   A. response to x[n] with y[n]
   B. y[n] with x[n]
   C. response to x[n-n0] with y[n-n0]
   D. response to 2x[n] with 2y[n]
   Answer: C. Shift the input, find the output, and compare with the shifted original output. (Key point: Test for time invariance)

2. y[n] = y[n-4] + x[n-4] with initial rest is
   A. time-variant
   B. time-invariant
   C. non-causal
   D. memoryless
   Answer: B. Constant coefficients mean shifts commute through. (Key point: Recursive system)

3. Unrolling gives y[n] =
   A. sum over k>=0 of x[n-4-4k]
   B. sum of x[k] only
   C. x[n-4] only
   D. x[n]^2
   Answer: A. A shift of x shifts every term, so y shifts equally. (Key point: Unrolled form)

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