This article discusses binomial filters and some of its properties. These filters are common in the signal and image processing domain. The best-known instance of binomial filters is the moving average filter \([\frac{1}{2},\frac{1}{2}]\). The article shows:
To keep things simple, only unnormalized filters (e.g. \([1,1]\) instead of \([\frac{1}{2},\frac{1}{2}]\)) are discussed.
The filter coefficients of an order \(i\) binomial filter are the coefficients found in the \(i\)-th row of Pascal's triangle. The first few unnormalized filters \(b^i\) are listed in the following (superscript number denotes filter order):
Computing the filter coefficients (or equally, the rows of Pascal's triangle) can be interpreted in terms of convolution: by convolving the filter \(b^1=[1, 1]\) \(i\)-times with itself, we get an order \(i\) filter. Let's assume we already have the coefficients for an order \(i-1\) filter \(b^{i-1}\) and we want to compute an order \(i\) filter \(b^i\):
\[b^i_n=(b^1 * b^{i-1})_n = \sum_k{b^1_k \cdot b^{i-1}_{n-k}} = b^{i-1}_n + b^{i-1}_{n-1} \]Here, \(*\) denotes convolution and the subscript number is the sequence index. The result \(b^{i-1}_n + b^{i-1}_{n-1}\) essentially computes Pascal's triangle when we start with \(b^0=[1]\) in the first row: take the two elements \(n\) and \(n-1\) from the row above (and zero if the element does not exist), and add them together. As an example, if we want to compute the order 2 filter (see Fig. 1), we get the sequence: \([0+1, 1+1, 1+0]=[1,2,1]\). Some observations:
Here we compute the frequency response of the unnormalized order 1 filter where the superscript number is the filter order and the subscript number is the frequency index:
\[B_k^1= \sum_n^{N-1}{b_n^1 \cdot e^{-j \cdot 2 \cdot \pi \cdot \frac{n \cdot k}{N}}} = e^0 + e^{-j \cdot 2 \cdot \pi \cdot \frac{k}{N}} = 1 + cos(2 \cdot \pi \cdot \frac{k}{N}) - j \cdot sin(2 \cdot \pi \cdot \frac{k}{N}) \]We are interested in how certain frequencies \(k\) get amplified/attenuated, so we compute the absolute value of \(B_k^1\). To keep things simple, we compute the squared absolute value \(|B_k|^2\) and set \(\alpha=2 \cdot \pi \cdot \frac{k}{N}\).
\[|B_k^1|^2 =(1+cos(\alpha))^2 + sin(\alpha)^2 = 1 + 2 \cdot cos(\alpha) + cos(\alpha)^2 +sin(\alpha)^2 = 2 + 2 \cdot cos(\alpha)\]Some observations:
There are two ways to compute the 2D filters. Either take the outer product of two 1D filters, or start with the 2D filter \([[1,1], [1,1]]\) and convolve it with itself as in the 1D case. The first few unnormalized filters \(b^i\) are listed in the following (see Fig. 3 for two samples):
The frequency response is shown in Fig. 4. It gets closer to a 2D Gaussian distribution as the order \(i\) gets larger. Again, as in the 1D case, the frequency response contains no ripple.
Here is the code to compute the filter coefficients. In the main function, the order 2 filter is applied to two signals. The first signal gets erased completely, while the second one gets damped by a factor of 2.
from scipy.signal import convolve import numpy as np def compute_kernel(order): bi = np.asarray([1]) for _ in range(order): bi = convolve(bi, np.asarray([1, 1])) return bi / bi.sum() def main(): bi = compute_kernel(order=2) xs = [1, -1, 1, -1, 1, -1, 1, -1], [1, 0, -1, 0, 1, 0, -1, 0] for x in xs: y = convolve(x, bi, mode='valid') print(f'{x} * {bi} = {y}') if __name__ == '__main__': main()
It was shown how a binomial filters of arbitrary order \(i\) is computed by convolving the moving average filter \([1,1]\) \(i\)-times with itself. Further, the frequency response for the moving average filter was derived. Higher order frequency responses of order \(i\) are computed by taking the response of the moving average filter to the power of \(i\).
Harald Scheidl, 2021