How should I fix my convolutional integral

4 views (last 30 days)
Hello all,
Actually, I am trying to convolve two functions (f and y)using the following MATLAB code:
close all;
clear all;
clc;
%set time vector
t0=0;
tf=100;
N=10000;
dt=(tf-t0)/N;
t=t0:dt:tf;
T0=60;
alpha=1;
beta=0.01;
%construct the firxt part of the convolution
for i=1:length(t)
if(t(i)<=T0)
y(i)=0;
else
y(i)=alpha+beta*(t(i)-T0);
end
end
%plot y(t)
subplot(1,2,1);
plot(t,y,'LineWidth',2);
title('y(t)');
%Second function of the convolution
f2= 0.08787*exp(-((t-63.08)/1.593).^2);
%Convolution of these 2 functions
z=conv(y,f2,'full');
con = z*dt;
subplot(1,2,2);
plot(t,f2,'LineWidth',2);
title('f2');
t = (1:length(con))*dt ;
hold on;
plot(t,con,'LineWidth',2);
I tried to use the guidlines in the following link to run a full convolutional integral :
Here are my results:
This result seems not to be reasonable because both my functions start rising at x=60 but my convolution starts rising at x=120. Can someone helps me with this problem.
  1 Comment
Matt J
Matt J on 9 Mar 2020
Edited: Matt J on 9 Mar 2020
This result seems not to be reasonable because both my functions start rising at x=60
There's nothing unreasonable about that. That is theoretically what should happen if both functions start at t=60. For the convolution result to start at t=60, you need one of the signals to start at t=0.

Sign in to comment.

Accepted Answer

Matt J
Matt J on 9 Mar 2020
Edited: Matt J on 9 Mar 2020
This might be what you want,
subplot(1,2,1);
ty=t;
plot(ty,y,'LineWidth',2);
title('y(t)');
%Second function of the convolution
f2= 0.08787*exp(-((t-63.08)/1.593).^2);
%Convolution of these 2 functions
z=conv(y,f2(find(f2,1):end),'same');
con = z*dt;
subplot(1,2,2);
plot(ty,f2,'LineWidth',2);
title('f2');
hold on;
tc=linspace(0,1,numel(con));
tc=tc/tc(2)*dt;
plot(tc,con,'LineWidth',2);
hold off
  3 Comments
Matt J
Matt J on 10 Mar 2020
Edited: Matt J on 10 Mar 2020
As I said in my comment above,
I think you've had the correct result from the very beginning.
Steve Chou
Steve Chou on 15 Mar 2021
This chart is more clear to read.

Sign in to comment.

More Answers (0)

Community Treasure Hunt

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

Start Hunting!