convolution shape (full/same/valid)

19 vues (au cours des 30 derniers jours)
Rex Cheang
Rex Cheang le 25 Août 2021
Commenté : Rex Cheang le 26 Août 2021
clear all
close all
%Sampling time
Ts=0.001;
t=0:Ts:2*pi;
%Timeaxis
TimeAxes_x = [0 0.1 -2*pi 2*pi];
TimeAxes_y = [0 60 0 10^7];
%Functions
f1 = 2*cos(100*pi*t)+cos(60*pi*t);
f3 = 2*cos(100*pi*(t-2))+cos(60*pi*(t-2));
f4 = conv(f1,f3,'full');
%Sampling frequency
fs=1/Ts;
fre=(0:length(f4)-1)*fs/length(f4);
%Fourier transform
F4 = fft(f4);
F1 = fft(f1);
F3 = fft(f3);
convf = conv(F1,F3,'full');
%plotting F4
graph_subplot(1,211,fre,abs(F4),TimeAxes_y,'f(Hz)','F4(w)','Frequency domain for conv(f1,f3)');
%plotting F4_2
graph_subplot(1,212,fre,abs(convf),TimeAxes_y,'f(Hz)','F4(w)','F1(w)*F4(w)');
These are the codes for the two plots, I just wonder why I got nothing in the graph when I do 'same' for the convolution but I got the attached graph when I do 'full'.
  2 commentaires
Bjorn Gustavsson
Bjorn Gustavsson le 26 Août 2021
Plot your functions, ponder about what happens and why. It is your homework problem to solve.
Rex Cheang
Rex Cheang le 26 Août 2021
I just want to know why Im not having anything when I do for example, conv(f1,f3, 'same') but I am getting something for 'full'

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