How to use the convolution function rather than filter for signal

Hi, I would like to ask how I can apply the convolution for the following code, rather than applying the filter function (note: for Fourier Transform)

Could please anyone help me.

B=[0.5 0.5]; % coefficient of the x(n), x(n-1)

A= [1 -1 0.5];% coefficient of the y(n), y(n-1), y(n-2)

fs=10000;

n=0:1000;

f1=100;

f2=3000;

x=2*sin(2*pi*f1/fs*n)+4*sin(2*pi*f2/fs*n); % input signal

y=filter(B,A,x);% filter

nfft_of_fft = length(y);% number of samples

yf1 = fft(y,nfft_of_fft);%fourier transform function

yf1_mag = abs(yf1);% to make the y-axis more than 0 (absolute value)

yf1_mag_norm = yf1_mag/max(yf1_mag);%normlized the magnitude

yf1_xaxis=fs/2*(0:nfft_of_fft/2-1)/(nfft_of_fft/2);%frequency

plot(yf1_xaxis,yf1_mag_norm(1:nfft_of_fft/2),'r');% plot DEFT

title('Fourier Transform y(n)')

xlabel('Frequency (Hz)')

ylabel('Magnitude')

Answer 1

[Chaitanya Mallela]

filter function gives the filter response for both IIR and FIR systems whereas conv function performs linear convolution of the input signals. For FIR filters the functions might perform similar operation but for IIR filters the outputs from both the functions are different.

The output length of filter function is equal to input vector length whereas conv function output length equal to length(filter_vector) + length(input_vector) - 1.

Instead replacement to filter function with conv function in the code can be considered by taking symbolic coeffients

syms z
F = (0.5 + 0.5*z^(-1))/(1-z^(-1)+0.5*z^(-2));   % Symbolic Function
h_sym = iztrans(F);                             % Inverse z transform
h = double(subs(h_sym,n));                      % filter coefficients
y_conv = conv(h,x);                             % convolution 
y_out  = y_conv(1:length(x));

[Abdulaziz Shaqhan]

Thank you very much