Moving Average Filter

Moving Average Filter

The moving average is the simplest digital filter. It smooths a signal by replacing each sample with the average of itself and the $M-1$ preceding samples:

$y[n] = \frac{1}{M} \sum_{k=0}^{M-1} x[n-k]$

This is a causal filter — it only uses past and current samples.

Zero-Padding at the Start

For $n < M-1$, not enough past samples exist. We treat missing samples as zero:

$y[n] = \frac{1}{M}(x[0] + x[1] + \cdots + x[n] + \underbrace{0 + \cdots + 0}_{M-1-n \text{ zeros}})$

This means the output array has the same length as the input.

Example: $M = 3$

Signal: $[1, 2, 3, 4, 5]$

$y[0] = (0 + 0 + 1)/3 = 0.3333$ $y[1] = (0 + 1 + 2)/3 = 1.0$ $y[2] = (1 + 2 + 3)/3 = 2.0$ $y[3] = (2 + 3 + 4)/3 = 3.0$ $y[4] = (3 + 4 + 5)/3 = 4.0$

Frequency Response

The moving average is a low-pass filter — it attenuates high-frequency components. A longer window ($M$ larger) gives stronger smoothing but slower response.

Your Task

Implement moving_average(x, M) with causal zero-padding. Return a list of the same length as $x$.

def moving_average(x, M):
    result = []
    for n in range(len(x)):
        window = x[max(0, n - M + 1):n + 1]
        padded_sum = sum(window)  # missing samples are 0
        result.append(padded_sum / M)
    return result

print([round(v, 4) for v in moving_average([1, 2, 3, 4, 5], 3)])
# [0.3333, 1.0, 2.0, 3.0, 4.0]