---
title: "Is y(t) = x(t−2) + x(2−t) LTI"
subject: "Signals and Systems"
chapter: 5
key: "semester_4/signals_and_systems:lti-check-time-reversal-system"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-check-time-reversal-system
site: BCT Engineering
---

# Is y(t) = x(t−2) + x(2−t) LTI

Signals and Systems · Chapter 5: Continuous-Time Systems

- Asked in: 2077 Chaitra, 5 marks
- Video: [Q3.b. Whether the system is Time Invariant & Invertible? | Part 2](https://www.youtube.com/watch?v=issScMDkj1c) — EnggClasses
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-check-time-reversal-system
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-5

## Question

Consider a continuous time system with input x(t) and output y(t) related by y(t) = x(t−2) + x(2−t). Is the system Linear Time invariant? Explain.

## Short answer

The system is linear because superposition holds. It is time-variant because x(2-t) changes differently under input delay; delaying input gives x(2-t-t0) but delayed output has x(2-t+t0). Thus it is not LTI.

## Answer

Given $y(t) = x(t-2) + x(2-t)$. Check linearity and time invariance separately.

#### Linearity
Let $x_1(t) \to y_1(t) = x_1(t-2) + x_1(2-t)$ and $x_2(t) \to y_2(t) = x_2(t-2) + x_2(2-t)$.

For $x_3(t) = a x_1(t) + b x_2(t)$:
$$
\begin{aligned}
y_3(t) &= x_3(t-2) + x_3(2-t) \\
&= a x_1(t-2) + b x_2(t-2) + a x_1(2-t) + b x_2(2-t) \\
&= a\big[x_1(t-2) + x_1(2-t)\big] + b\big[x_2(t-2) + x_2(2-t)\big] \\
&= a y_1(t) + b y_2(t)
\end{aligned}
$$
Superposition holds, so the system is **linear**.

#### Time invariance
Apply a delayed input $x_1(t) = x(t - t_0)$. The output is obtained by replacing $x(\cdot)$ with $x(\cdot - t_0)$:
$$
y_1(t) = x_1(t-2) + x_1(2-t) = x(t - 2 - t_0) + x(2 - t - t_0)
$$
Now delay the original output by $t_0$ (replace $t$ by $t - t_0$):
$$
y(t - t_0) = x(t - t_0 - 2) + x(2 - t + t_0)
$$
The first terms match, but the second terms differ: $x(2 - t - t_0) \ne x(2 - t + t_0)$. So $y_1(t) \ne y(t - t_0)$ and the system is **time-variant**.

**Counter-example:** let $x(t) = \delta(t)$. Then $y(t) = \delta(t-2) + \delta(2-t) = 2\delta(t-2)$. For $x(t) = \delta(t-1)$: $y(t) = \delta(t-3) + \delta(1-t) = \delta(t-3) + \delta(t-1)$, which is not $y(t-1) = 2\delta(t-3)$.

The cause is the time-reversal term $x(2 - t)$: a delay of the input moves this part of the output *earlier*, not later.

**Answer: the system is linear but time-variant, so it is not an LTI system.** (It is also non-causal: $y(0) = x(-2) + x(2)$ needs a future input.)

## Check yourself

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

1. Is y(t)=x(t-2)+x(2-t) linear?
   A. No, due to the time reversal
   B. No, due to the delay
   C. Yes, superposition holds
   D. No, since it has two terms
   Answer: C. Both terms are linear operations on x, so a·y1+b·y2 results. (Key point: Linearity)

2. Which term breaks time invariance?
   A. x(t-2), because of the delay
   B. Neither term
   C. Both terms equally
   D. x(2-t), because time reversal moves the output the opposite way
   Answer: D. Delayed input gives x(2-t-t0), but y(t-t0) has x(2-t+t0). (Key point: Time invariance)

3. For x=δ(t), y=2δ(t-2). For x=δ(t-1), what is y?
   A. δ(t-3)+δ(t-1)
   B. 2δ(t-3)
   C. 2δ(t-1)
   D. δ(t-2)
   Answer: A. x(t-2)=δ(t-3) and x(2-t)=δ(1-t)=δ(t-1); not equal to y(t-1)=2δ(t-3). (Key point: Counter-example)

4. The system is:
   A. LTI
   B. Linear but time-variant, so not LTI
   C. Time-invariant but non-linear
   D. Neither linear nor time-invariant
   Answer: B. Linear but fails time invariance. (Key point: Conclusion)

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