---
title: "LTI properties; is y[n] = e^(x[n]) linear"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:lti-properties-exponential-system"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/lti-properties-exponential-system
site: BCT Engineering
---

# LTI properties; is y[n] = e^(x[n]) linear

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2072 Magh, 4+5 marks
- Video: [Properties Of Systems | Example 1](https://www.youtube.com/watch?v=NYtQ542JRw0) — 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-properties-exponential-system
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

What are the properties of LTI systems? Determine whether the given system is linear or not? y[n] = e^(x[n])

## Short answer

LTI systems are linear and time-invariant; convolution is commutative, associative, and distributive, and causality, stability, invertibility, and eigenfunction behavior follow from h[n]. The system y[n]=e^(x[n]) is nonlinear because e^(a1x1+a2x2) does not equal a1 e^(x1)+a2 e^(x2), and zero input gives y=1 instead of 0.

## Answer

#### Properties of LTI systems
An LTI system is linear and time-invariant; its output is $y[n]=x[n]*h[n]$. Its main properties:

1. **Commutative:** $x*h=h*x$.
2. **Associative:** $(x*h_1)*h_2=x*(h_1*h_2)$; a cascade equals one system $h_1*h_2$, in any order.
3. **Distributive:** $x*(h_1+h_2)=x*h_1+x*h_2$; parallel systems add their impulse responses.
4. **Memoryless** only if $h[n]=K\delta[n]$.
5. **Causal** if and only if $h[n]=0$ for $n<0$.
6. **BIBO stable** if and only if $\sum_n|h[n]|<\infty$.
7. **Invertible** if some $h_i[n]$ satisfies $h*h_i=\delta[n]$.
8. **Step response** $s[n]=\sum_{k\le n}h[k]$, and $h[n]=s[n]-s[n-1]$.
9. **Eigenfunction:** $e^{j\omega n}\rightarrow H(e^{j\omega})e^{j\omega n}$.

#### Is $y[n]=e^{x[n]}$ linear?
A system is linear if $a_1x_1+a_2x_2\rightarrow a_1y_1+a_2y_2$.

Let $y_1[n]=e^{x_1[n]}$ and $y_2[n]=e^{x_2[n]}$. For $x_3=a_1x_1+a_2x_2$:

$$
y_3[n]=e^{a_1x_1[n]+a_2x_2[n]}=\left(e^{x_1[n]}\right)^{a_1}\left(e^{x_2[n]}\right)^{a_2}
$$

while

$$
a_1y_1[n]+a_2y_2[n]=a_1e^{x_1[n]}+a_2e^{x_2[n]}
$$

These are not equal in general (one is a product, the other a sum).

Also, zero input gives $y[n]=e^0=1\ne0$, which a linear system cannot do.

Numerical check: $x=1\rightarrow y=e\approx2.718$; $x=2\rightarrow y=e^2\approx7.389\ne2(2.718)=5.437$.

**Answer: $y[n]=e^{x[n]}$ is non-linear.** (It is, however, time-invariant, memoryless, causal, and BIBO stable, since $|x|\le B$ gives $|y|\le e^{B}$.)

## Check yourself

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

1. What does the associative property say about a cascade of LTI systems?
   A. It equals one system with h = h1*h2, in any order
   B. It equals one system with h = h1+h2
   C. It equals one system with h = h1/h2
   D. Order matters, only h1 then h2 works
   Answer: A. A cascade convolves impulse responses; commutativity makes the order irrelevant. (Key point: Associative property)

2. Which pair of conditions for LTI systems is correct?
   A. Causal iff h[n]=0 for n>0; stable iff sum h[n]=0
   B. Causal iff h[n]=0 for n<0; BIBO stable iff sum|h[n]| is finite
   C. Causal iff h[n]=delta[n]; stable iff h[n]>0
   D. Causal iff sum|h|<infinity; stable iff h[n]=0 for n<0
   Answer: B. These are the standard necessary and sufficient conditions. (Key point: Causality and stability conditions)

3. Why does y[n]=e^(x[n]) fail the linearity test?
   A. Because it is time-varying
   B. Because it is not causal
   C. e^(a1x1+a2x2) is a product of powers, but a1y1+a2y2 is a weighted sum
   D. Because it is unstable for bounded inputs
   Answer: C. Exponent of a sum becomes a product, which differs from the weighted sum of outputs. (Key point: Why e^x[n] is non-linear)

4. What does x[n]=0 give for y[n]=e^(x[n])?
   A. y=0, so it is linear
   B. y=e, so it is unstable
   C. y=1, which proves it is linear
   D. y=1, which is not 0, so the system is non-linear
   Answer: D. A linear system must map zero input to zero output. (Key point: Zero input check)

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