Band-Pass-Filter-using-Rectangular-Window

A band-pass filter allows to pass signals with a frequencies between that of two determined cutoff points, while attenuating signals with a frequency lower or higher than these cutoff frequecies. This band-pass filter has been designed using the "Rectangular Window" method, which uses a set of N coefficients all with value of 1. Though being easy to implement, the Rectangular Window method is not really used for digital filter design, due to its low attenuation of the stopbands and producing filters characterized by extreme values (e.g. sudden jump from zero at the initial sample).

Output samples of a FIR filter are produced based on:

$$y(n) = \sum_{j=0}^{N} b_k \times x[n - j]$$

While its frequency response is given by:

$$H(w) = \sum_{j=0}^{N} b_k \times e^{-jwn}$$

The parameters $b_k$ are camculated using the DFT (Discrete Fourier Transform) method.

In order to test the time response of the filter you need to create the mex function in MATLAB using the command mex -R2018a FIR_BP_Rectangular.c. When calling the function you need to do it passing the parameters shown below:

FIR_BP_Rectangular(input_signal, passband_frequency1, stopband_frequency1, passband_frequency2, stopband_frequency2, sampling_frequency, stopband_ripple);

Overall, the following MATLAB code helps plotting the time response:

fs = 20000;
dt = 1/fs;
F = 5000; 
t = 0:dt:0.0025;
x = sin(2*pi*F*t);
plot(x, 'red')
[filt_data, freq_response] = FIR_BP_Rectangular(x, 3000, 2000, 6000, 9000, 20000, 0.01);
hold on
plot(filt_data, 'blue')

Here, a signal with a frequency of 5KHz is passed through the filter, depending on the cutoff frequencies (which is half the sum of every passband and the stopband frequencies), the signal might be allowed or attenuated. In this case, the first cutoff frequency is ~2.5KHz, while the second cutoff frequency is ~7.5KHz, so the signal will be allowed.

On the other hand, the frequency response (with a length of 512 points) is also quite easy to plot:

f = 0:10000/512:10000;
plot(f, 20*log10(abs(freq_response)))

Note: As the order of FIR filters increase, the delay of the output increases based on filter_order/(2 * sampling_frequency). In this implementation the delay is not compensated.