---
title: "Transfer function of DT low pass filter"
subject: "Signals and Systems"
chapter: 6
key: "semester_4/signals_and_systems:dt-lowpass-transfer-function"
source: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-lowpass-transfer-function
site: BCT Engineering
---

# Transfer function of DT low pass filter

Signals and Systems · Chapter 6: Discrete-Time Systems

- Asked in: 2075 Bhadra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/signals_and_systems/dt-lowpass-transfer-function
- Whole chapter: https://www.bctengineer.com/md/semester_4/signals_and_systems/chapter-6

## Question

Derive the transfer function for discrete time low pass filter.

## Short answer

A first-order recursive low-pass filter is y[n]-a y[n-1]=x[n], 0<a<1. Its transfer function is H(z)=1/(1-a z^-1)=z/(z-a), with magnitude 1/sqrt(1+a^2-2a cos w). It is low-pass because gain is maximum at w=0 and minimum at w=pi; impulse response is h[n]=a^n u[n].

## Answer

A simple discrete-time low pass filter is the **first-order recursive filter**

$$
y[n]-a\,y[n-1]=x[n], \qquad 0<a<1
$$

(the output is the input plus a fraction $a$ of the previous output, so fast changes are smoothed).

#### Derivation of the transfer function
Take the z-transform (zero initial conditions), $y[n-1]\leftrightarrow z^{-1}Y(z)$:

$$
\begin{aligned}
Y(z)-az^{-1}Y(z)&=X(z)\\
H(z)=\frac{Y(z)}{X(z)}&=\frac{1}{1-az^{-1}}=\frac{z}{z-a}, \qquad |z|>|a|
\end{aligned}
$$

Frequency response (put $z=e^{j\omega}$):

$$
H(e^{j\omega})=\frac{1}{1-ae^{-j\omega}}, \qquad |H(e^{j\omega})|=\frac{1}{\sqrt{1+a^2-2a\cos\omega}}
$$

Impulse response: $h[n]=a^n u[n]$.

#### Why it is low pass
- At $\omega=0$: $|H|=\frac{1}{1-a}$ (maximum).
- At $\omega=\pi$: $|H|=\frac{1}{1+a}$ (minimum).

For $0<a<1$ the gain decreases from low to high frequency. Example $a=0.5$: gain 2 at $\omega=0$ and 0.667 at $\omega=\pi$. (For $-1<a<0$ the same equation is a high pass filter.) The pole at $z=a$ lies on the positive real axis, closer to $z=1$ (i.e. $\omega=0$) than to $z=-1$, so it boosts low frequencies.

```text
 |H|
 1/(1-a) *
          *
            *  *
                  *  *  *  1/(1+a)
 ---------------------------> w
  0                     pi
```

#### Ideal low pass filter (for reference)
The ideal LPF has $H(e^{j\omega})=1$ for $|\omega|\le\omega_c$ and $0$ for $\omega_c<|\omega|\le\pi$, which gives $h[n]=\frac{\sin\omega_c n}{\pi n}$. It is non-causal and not realizable, so first-order (or higher-order) recursive filters like the one above are used in practice.

## Check yourself

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

1. Which difference equation was used as the simple discrete-time low pass filter?
   A. y[n]-a y[n-1]=x[n] with 0<a<1
   B. y[n]=x[n]-x[n-1]
   C. y[n]+a y[n-1]=x[n] with 0<a<1
   D. y[n]=a x[n]+x[n+1]
   Answer: A. It is a first-order recursive filter with a positive coefficient a below 1. (Key point: Filter structure)

2. What is H(z)?
   A. 1/(1+a z^(-1))
   B. 1/(1-a z^(-1)) = z/(z-a)
   C. (1-a z^(-1))
   D. z/(z+a)
   Answer: B. Taking the z-transform gives Y(1-a z^(-1))=X. (Key point: Transfer function)

3. What are the gains at ω=0 and ω=π?
   A. 1/(1+a) at ω=0 and 1/(1-a) at ω=π
   B. 1 at both
   C. 1/(1-a) at ω=0 (maximum) and 1/(1+a) at ω=π (minimum)
   D. 0 at ω=0 and 1/(1-a) at ω=π
   Answer: C. Evaluating 1/sqrt(1+a^2-2a cosω) gives maximum at ω=0 and minimum at ω=π. (Key point: Low pass behaviour)

4. Why does a pole at z=a close to 1 make the filter low pass?
   A. It lies near ω=π and boosts high frequencies
   B. It lies at the origin and removes all frequencies
   C. It lies outside the unit circle and makes the filter unstable
   D. It lies near ω=0 on the unit circle and boosts low frequencies
   Answer: D. The gain is large near the pole, which for a close to 1 is the low frequency region. (Key point: Effect of the pole)

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