convolution shape (full/same/valid)
18 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
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 Kommentare
Bjorn Gustavsson
am 26 Aug. 2021
Plot your functions, ponder about what happens and why. It is your homework problem to solve.
Antworten (0)
Siehe auch
Kategorien
Mehr zu Fourier Analysis and Filtering finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!