Transfer function of DT low pass filter

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.

 |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.