convolution shape (full/same/valid)

18 次查看（过去 30 天）
Rex Cheang
Rex Cheang 2021-8-25
评论： Rex Cheang 2021-8-26
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 个评论
Bjorn Gustavsson
Bjorn Gustavsson 2021-8-26
Plot your functions, ponder about what happens and why. It is your homework problem to solve.
Rex Cheang
Rex Cheang 2021-8-26
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'

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Fourier Analysis and Filtering 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by