---
title: "H(e^(jω)) and h[n] for y[n] = 0.5y[n−1] + x[n]"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:first-order-difference-equation"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-difference-equation
site: BCT Engineering
---

# H(e^(jω)) and h[n] for y[n] = 0.5y[n−1] + x[n]

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2081 Asoj, 8 marks; 2077 Chaitra, 4+2 marks; 2075 Baisakh, 5 marks; 2072 Asoj, 6 marks (asked 4 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/first-order-difference-equation
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Find the frequency response and impulse response for a system characterized by linear constant coefficient difference equation y[n] = 0.5y[n−1] + x[n].

## Short answer

Taking the DTFT gives H(e^jw)=1/(1-0.5e^(-jw)), a lowpass response. The impulse response is h[n]=(0.5)^n u[n], which is causal, stable, and IIR; DC gain is 2.

## Answer

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

#### Frequency response

Take the DTFT of both sides, using the shift property $x[n-1] \leftrightarrow e^{-j\omega}X(e^{j\omega})$:

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

Writing $e^{-j\omega} = \cos\omega - j\sin\omega$, the denominator is $(1 - 0.5\cos\omega) + j\,0.5\sin\omega$.

**Magnitude response:**

$$
|H(e^{j\omega})| = \frac{1}{\sqrt{(1-0.5\cos\omega)^2 + (0.5\sin\omega)^2}} = \frac{1}{\sqrt{1.25 - \cos\omega}}
$$

**Phase response:**

$$
\angle H(e^{j\omega}) = -\tan^{-1}\left(\frac{0.5\sin\omega}{1 - 0.5\cos\omega}\right)
$$

| $\omega$ | 0 | $\pi/4$ | $\pi/2$ | $3\pi/4$ | $\pi$ |
|---|---|---|---|---|---|
| $\lvert H\rvert$ | 2.000 | 1.357 | 0.894 | 0.715 | 0.667 |
| Phase | 0° | −28.7° | −26.6° | −14.6° | 0° |

The gain is largest at $\omega = 0$ and smallest at $\omega = \pi$, so the system is a **low pass filter**. $H(e^{j\omega})$ is periodic with period $2\pi$.

#### Impulse response

Using the DTFT pair $a^n u[n] \leftrightarrow \frac{1}{1 - a e^{-j\omega}}$ for $|a| < 1$, with $a = 0.5$:

$$
h[n] = (0.5)^n\,u[n]
$$

**Check by recursion** ($x[n] = \delta[n]$, $h[-1] = 0$):

| $n$ | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| $h[n] = 0.5h[n-1] + \delta[n]$ | 1 | 0.5 | 0.25 | 0.125 | 0.0625 |

This matches $(0.5)^n$. The system is causal ($h[n] = 0$ for $n < 0$), stable ($\sum|h[n]| = \frac{1}{1-0.5} = 2 < \infty$), and IIR (infinite-length $h[n]$). Note $\sum h[n] = 2 = H(e^{j0})$, as expected.

## Check yourself

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

1. Using x[n-1] <-> e^(-jw) X, what is H for y[n] = 0.5 y[n-1] + x[n]?
   A. 1/(1 + 0.5 e^(-jw))
   B. 0.5/(1 - e^(-jw))
   C. 1/(1 - 0.5 e^(jw))
   D. 1/(1 - 0.5 e^(-jw))
   Answer: D. Y(1 - 0.5e^(-jw)) = X. (Key point: DTFT of difference equation)

2. What is |H| and what does it show?
   A. 1/sqrt(1.25 - cos w); maximum 2 at w = 0 and minimum 2/3 at w = pi (low pass)
   B. 1/sqrt(1.25 + cos w); maximum at w = pi (high pass)
   C. 1/(1.25 - cos w); constant
   D. sqrt(1.25 - cos w); maximum at w = pi
   Answer: A. At w = 0, |H| = 1/0.5 = 2; at w = pi, |H| = 1/1.5 = 0.667. (Key point: Magnitude response)

3. What is h[n]?
   A. 0.5 delta[n]
   B. (0.5)^n u[n]
   C. (-0.5)^n u[n]
   D. 2^n u[n]
   Answer: B. a^n u[n] <-> 1/(1 - a e^(-jw)) with a = 0.5; recursion gives 1, 0.5, 0.25, ... (Key point: Impulse response)

4. Which properties does this system have?
   A. Non-causal and unstable
   B. Causal, stable and FIR
   C. Causal, stable and IIR (sum of h = 2 = H at w = 0)
   D. Anti-causal and stable
   Answer: C. h[n] = 0 for n < 0; sum |h| = 1/(1 - 0.5) = 2 and h has infinite length. (Key point: System properties)

## Close variants set in other exams

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