Question

ON MATLAB:  ii. Using FIR low-pass filter, remove signal S2, considering fc = 20 Hz as a cut-off frequency and consider two sets of filter coefficients: 11 and 301. Plot the time and frequency domain of the filtered signal, and comment.

the process x(n) is sum of two signals S1 and S2; mathematically be described as:

?(?) = ?1 + ?2

where ?1 = ?1cos(2??1??? ) and ?2 = ?2cos(2??2??? ),

A1 = A2 = 1; f1 = 10Hz; f2 = 50; fs =0.5 kHz (sampling frequency).

Answer 1

Script:

fc=20;fs=500;
%For N=301
N=301
wc=2*fc/fs
h=fir1(N-1,wc)

%input
n=0:1:N-1
s1=cos(2*pi*10*n/fs)
s2=cos(2*pi*50*n/fs)
x=s1+s2
f=[-fs/2:fs/N:fs/2-fs/N];
subplot(211)
plot(f,fftshift(abs(fft(x))),'m');grid;
xlim([0,max(f)])
xlabel('f in Hz')
ylabel('|X(f)|')
title('Frequency domain of input')

y=filter(h,1,x)
subplot(212)
plot(f,fftshift(abs(fft(y))),'m');grid;
xlim([0,max(f)])
xlabel('f in Hz')
ylabel('|Y(f)|')
title('Frequency domain of Output')

Conclsuion:

The high frequency component 50 Hz(signal S2) is removed from the signal x(N).

The output spectrum has only one peak at 10 Hz. The 50 Hz peak is not present.