---
title: "Impulse and frequency response of y[n] = 0.5y[n−1] + x[n] + x[n+1]"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:difference-equation-noncausal"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-noncausal
site: BCT Engineering
---

# Impulse and frequency response of y[n] = 0.5y[n−1] + x[n] + x[n+1]

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2070 Magh, 7 marks
- Video: [Find the impulse response of the system for difference equation y(n)-3y(n-1)-4y(n-2)=x(n)+2x(n-1)](https://www.youtube.com/watch?v=xoWNFCkFnJI) — Dr.Shakunthala Masi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/difference-equation-noncausal
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Given a system y[n] = 0.5y[n−1] + x[n] + x[n+1]. Find out its impulse response and frequency response.

## Short answer

Frequency response H(e^jω)=(1+e^jω)/(1-0.5e^-jω). Impulse response h[n]=(0.5)^n u[n]+(0.5)^(n+1) u[n+1] = δ[n+1] + 1.5(0.5)^n u[n]. System is non-causal because h[-1]≠0.

## Answer

System: $y[n]=0.5y[n-1]+x[n]+x[n+1]$, i.e. $y[n]-0.5y[n-1]=x[n]+x[n+1]$ (initially at rest).

#### Frequency response
Take the DTFT, using $x[n-n_0]\leftrightarrow e^{-j\omega n_0}X(e^{j\omega})$ (so $x[n+1]\leftrightarrow e^{j\omega}X$):

$$
\begin{aligned}
Y(e^{j\omega})\left(1-0.5e^{-j\omega}\right)&=X(e^{j\omega})\left(1+e^{j\omega}\right)\\
H(e^{j\omega})&=\frac{1+e^{j\omega}}{1-0.5e^{-j\omega}}
\end{aligned}
$$

Magnitude and phase: since $1+e^{j\omega}=2\cos(\omega/2)e^{j\omega/2}$,

$$
|H(e^{j\omega})|=\frac{2|\cos(\omega/2)|}{\sqrt{1.25-\cos\omega}},\qquad
\angle H=\frac{\omega}{2}-\tan^{-1}\frac{0.5\sin\omega}{1-0.5\cos\omega}
$$

| $\omega$ | 0 | $\pi/2$ | $\pi$ |
|---|---|---|---|
| $\lvert H\rvert$ | 4 | 1.265 | 0 |
| $\angle H$ (deg) | 0 | 18.43 | — |

It is a low pass response (gain 4 at dc, zero at $\omega=\pi$).

#### Impulse response
Write $H=\frac{1}{1-0.5e^{-j\omega}}+e^{j\omega}\frac{1}{1-0.5e^{-j\omega}}$. Using $(0.5)^n u[n]\leftrightarrow\frac{1}{1-0.5e^{-j\omega}}$ and the time-advance property ($e^{j\omega}$ means $n\to n+1$):

$$
h[n]=(0.5)^n u[n]+(0.5)^{n+1}u[n+1]
$$

Simplified: at $n=-1$, $h[-1]=1$; for $n\ge0$, $h[n]=(0.5)^n+(0.5)^{n+1}=1.5(0.5)^n$. So

$$
h[n]=\delta[n+1]+1.5(0.5)^n u[n]
$$

Values: $h[-1]=1,\ h[0]=1.5,\ h[1]=0.75,\ h[2]=0.375,\ h[3]=0.1875,\dots$ (checked by running the recursion with $x[n]=\delta[n]$).

**Answer: $H(e^{j\omega})=\dfrac{1+e^{j\omega}}{1-0.5e^{-j\omega}}$, $h[n]=(0.5)^n u[n]+(0.5)^{n+1}u[n+1]$.** Since $h[-1]\ne0$ the system is **non-causal**; it is stable because $\sum|h|=1+3=4<\infty$.

## Check yourself

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

1. Which transform applies to x[n+1]?
   A. e^(jω)X(e^(jω))
   B. e^(-jω)X(e^(jω))
   C. X(e^(jω)) / jω
   D. -e^(jω)X(e^(jω))
   Answer: A. A time advance multiplies by e^(jω). (Key point: Advance in DTFT)

2. What is H(e^(jω)) for y[n]=0.5y[n-1]+x[n]+x[n+1]?
   A. (1+e^(-jω))/(1-0.5e^(-jω))
   B. (1+e^(jω))/(1-0.5e^(-jω))
   C. (1+e^(jω))/(1+0.5e^(-jω))
   D. 1/((1-0.5e^(-jω))(1+e^(jω)))
   Answer: B. Y(1-0.5e^(-jω))=X(1+e^(jω)). (Key point: Frequency response)

3. What is h[n] in simplified form?
   A. (0.5)^n u[n]
   B. delta[n]+1.5(0.5)^n u[n]
   C. delta[n+1]+1.5(0.5)^n u[n]
   D. delta[n-1]+1.5(0.5)^n u[n]
   Answer: C. The advanced term (0.5)^(n+1)u[n+1] gives h[-1]=1 and 1.5(0.5)^n for n>=0. (Key point: Impulse response)

4. Why is the system non-causal, and what is |H| at ω=π?
   A. h[0]=0; |H(π)|=4
   B. h[-1]=1 is nonzero for n<0; |H(π)|=4
   C. It is unstable; |H(π)|=0
   D. h[-1]=1 is nonzero for n<0; |H(π)|=0
   Answer: D. The x[n+1] term uses a future input. Also 1+e^(jπ)=0, so gain is 0 at ω=π. (Key point: Causality and gain)

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