---
title: "Is y[n] = nx[n] − x[n−1] time invariant"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:time-invariance-check"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/time-invariance-check
site: BCT Engineering
---

# Is y[n] = nx[n] − x[n−1] time invariant

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2080 Asoj, 3 marks; 2079 Asoj, 3 marks; 2078 Chaitra, 3 marks (asked 3 times)
- Video: [TIME VARIANT/ INVARIANT SYSTEM- complete steps and sums](https://www.youtube.com/watch?v=BvQIrR2vLZE) — Shrenik Jain
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/time-invariance-check
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Check if the system is time invariant y[n] = nx[n] − x[n−1]

## Short answer

The system is time variant. The time-dependent coefficient n multiplying x[n] makes the delayed input response differ from the delayed output, as shown by y₁[n] ≠ y[n−k].

## Answer

A system is **time invariant** if delaying the input by $k$ only delays the output by $k$: $x[n-k] \rightarrow y[n-k]$.

**Given:** $y[n] = n\,x[n] - x[n-1]$

**Step 1 – Output for the delayed input** $x_1[n] = x[n-k]$ (replace $x$ only, not the $n$ multiplying it):

$$
y_1[n] = n\,x[n-k] - x[n-1-k]
$$

**Step 2 – Delayed output** (replace every $n$ by $n-k$):

$$
y[n-k] = (n-k)\,x[n-k] - x[n-k-1]
$$

**Step 3 – Compare:**

$$
y_1[n] - y[n-k] = n\,x[n-k] - (n-k)\,x[n-k] = k\,x[n-k] \ne 0
$$

Since $y_1[n] \ne y[n-k]$ (for $k \ne 0$), the system is **time variant**.

**Reason:** the coefficient $n$ multiplying $x[n]$ changes with time, so the system's gain is different at different instants. The second term $-x[n-1]$ alone is time invariant, but the first term makes the whole system time variant.

**Numerical check:** take $x[n] = \delta[n]$: $y[n] = 0\cdot\delta[n] - \delta[n-1] = -\delta[n-1]$. For $x[n] = \delta[n-1]$: $y_1[n] = n\,\delta[n-1] - \delta[n-2] = \delta[n-1] - \delta[n-2]$, which is not $y[n-1] = -\delta[n-2]$.

## Check yourself

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

1. What is compared to test time invariance?
   A. Response to a1x1 + a2x2 with the weighted outputs
   B. Response to x[n-k] with the delayed output y[n-k]
   C. h[n] with h[-n]
   D. The magnitude of x with y
   Answer: B. A time-invariant system satisfies x[n-k] -> y[n-k]. (Key point: Test for time invariance)

2. For y[n] = n x[n] - x[n-1], what is the response to x[n-k]?
   A. (n-k) x[n-k] - x[n-k-1]
   B. n x[n-k] - x[n-1-k]
   C. n x[n] - x[n-1-k]
   D. n x[n-k] - x[n-1]
   Answer: B. Only x is shifted, the coefficient n stays unchanged. (Key point: Output for delayed input)

3. What is the result?
   A. Time invariant since -x[n-1] is time invariant
   B. Time invariant since the difference is zero
   C. Non-linear
   D. Time variant, because the difference is k x[n-k] which is not zero
   Answer: D. The explicit coefficient n makes the gain depend on time. (Key point: Conclusion)

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