Plot convolution of two wave signals

i want to plot the convolution of x=cos(wt) with frequency=10^6 and c=0.5(1+square(wt)) with freequency=10^5

i tried with the code below but convolution signal graph wasn't appear

>> fs=10^6;

>> T=1/fs;

>> tt=0:T/100:30*T;

>> m=cos(2*pi*fs*tt);

>> plot(tt,m)

>> fc=10^5;

>> c=0.5*(1+square(2*pi*fc*tt));

>> y=conv(m,c);

>> plot(tt,y)

Error using plot

Vectors must be the same length.

>> t1=0:0.01:10;

>> plot(t1,y)

Error using plot

Vectors must be the same length.

Answer 1

[DGM]

The length of the result of convolving two vectors is the sum of the vector lengths. Try this:

fs=10^6;
T=1/fs;
tt=0:T/100:30*T;
m=cos(2*pi*fs*tt);

fc=10^5;
c=0.5*(1+square(2*pi*fc*tt));
y=conv(m,c,'same'); % conv and conv2 return the full convolution by default

subplot(3,1,1)
plot(tt,m)
subplot(3,1,2)
plot(tt,c)
subplot(3,1,3)
plot(tt,y)

Answer 2

[Paul]

If we assume that m(t) = c(t) = 0 for t < 0, we can show analytically that the convolution integral m(t)*c(t) is periodic with period 1/fc, and one period is

p(t) = (sin(2*pi*fs*t)/(2*pi*fs) , t < (1/(2*fc)

p(t) = 0, 1/(2*fc) < t < 1/fc

When approximating the convlution integral with the convolution sum, don't forget to scale the sum by dt. Here is the code, extending the time vector furher out

fs=10^6;
T=1/fs;
tt=0:T/100:60*T;
m=cos(2*pi*fs*tt);
fc=10^5;
c=0.5*(1+square(2*pi*fc*tt));
yy=conv(m,c)*tt(2); % multiply by dt!

% analytic solution
pfunc = @(t) (sin(2*pi*fs*mod(t,1/fc))/(2*pi*fs).*(mod(t,1/fc) < 1/2/fc));

plot(tt,yy(1:numel(tt)),tt,pfunc(tt))

As you can see, the approximating convolution sum isn't quite accurate and becomes less so as t increases. I'm not exactly sure why that is, some sort of round-off accumulation perhaps?