IIR Filters

IIR Filters

An Infinite Impulse Response (IIR) filter uses feedback — the output depends on both the input and previous outputs. The simplest first-order IIR:

$y[n] = b \cdot x[n] + a \cdot y[n-1]$

with $y[-1] = 0$ (causal initialization).

Stability

The filter is stable when $|a| < 1$. If $a \geq 1$, the output diverges.

Exponential Smoother

A common application is the exponential moving average (EMA):

$y[n] = \alpha \cdot x[n] + (1-\alpha) \cdot y[n-1]$

This is iir_filter(x, a=1-alpha, b=alpha). The parameter $\alpha \in (0,1)$ controls smoothing:

• $\alpha$ close to 1: fast response, little smoothing
• $\alpha$ close to 0: slow response, heavy smoothing

Example

Signal $[1, 0, 0, 0]$ through filter with $a=0.5$, $b=1$:

$y[0] = 1 \cdot 1 + 0.5 \cdot 0 = 1.0$ $y[1] = 1 \cdot 0 + 0.5 \cdot 1 = 0.5$ $y[2] = 1 \cdot 0 + 0.5 \cdot 0.5 = 0.25$ $y[3] = 1 \cdot 0 + 0.5 \cdot 0.25 = 0.125$

The impulse response decays exponentially — hence "infinite" (theoretically never reaches zero).

Your Task

Implement:

• iir_filter(x, a, b) — causal first-order IIR: $y[n] = b \cdot x[n] + a \cdot y[n-1]$
• exponential_smoother(x, alpha) — calls iir_filter(x, 1-alpha, alpha)

def iir_filter(x, a, b):
    y = []
    y_prev = 0.0
    for xi in x:
        yi = b * xi + a * y_prev
        y.append(yi)
        y_prev = yi
    return y

def exponential_smoother(x, alpha):
    return iir_filter(x, 1 - alpha, alpha)