FIR Filters

FIR Filters

A Finite Impulse Response (FIR) filter is defined by its impulse response $h[n]$ of length $M+1$. The output is a causal convolution of the input with $h$:

$y[n] = \sum_{k=0}^{M} h[k] \cdot x[n-k]$

where $x[n-k] = 0$ for $n-k < 0$. The output has the same length as $x$.

Why FIR?

FIR filters are:

• Always stable — bounded input → bounded output
• Linear phase — symmetric $h$ gives constant group delay (no phase distortion)
• Flexible — any frequency response can be approximated

Sinc Lowpass Filter

The ideal lowpass filter with cutoff $f_c$ (normalized, $0 < f_c < 0.5$) has the impulse response:

$h[n] = 2 f_c \cdot \text{sinc}(2 f_c (n - N/2))$

where $\text{sinc}(x) = \sin(\pi x) / (\pi x)$ and $h[N/2] = 2f_c$ when $x=0$.

For $f_c = 0.25$ (quarter-band) with $N=4$: $h = [0, 0.3183, 0.5, 0.3183, 0]$

Your Task

Implement:

• fir_filter(x, h) — causal FIR convolution, output same length as $x$
• sinc_lowpass(fc, N) — returns $N+1$ coefficients for ideal lowpass

import math

def fir_filter(x, h):
    result = []
    for n in range(len(x)):
        val = 0.0
        for k in range(len(h)):
            if n - k >= 0:
                val += h[k] * x[n - k]
        result.append(val)
    return result

def sinc_lowpass(fc, N):
    h = []
    center = N // 2
    for n in range(N + 1):
        x = n - center
        if x == 0:
            h.append(2 * fc)
        else:
            h.append(2 * fc * math.sin(math.pi * 2 * fc * x) / (math.pi * 2 * fc * x))
    return h