Using MATLAB to get the fourier transform of dirac delta function

[luckycool]

Homework Statement

Dear all,

I have a problem when I using MATLAB to get the Fourier transform of dirac delta function. below is my code.

Homework Equations

clear all;
clc;
close all;

% t=0:0.002:2;
t=0:0.002:4;
dt=t(2)-t(1);
u=zeros(size(t));
pos0=find(t>=1,1);
u(pos0)=1/dt;

f=linspace(-.5/dt,.5/dt,length(t));
U=fftshift(fft(u))/length(t);

figure;subplot(2,1,1);hold on;plot(t,u);
subplot(2,1,2);hold on;plot(f,abs(U));

The Attempt at a Solution

I have a doubt that why when I change the end of time to 4, the absolute value of U(f) would be the half when I set the time as 2?

Thank you very much.

 

[bob1182006]

That part's tricky and it took me a looong time to figure it out.
What they're pretty much doing in the examples where you see them divide by L is that they're actually multiplying by the sampling rate dt. It's just that in those examples where they divide by L the time ends at 1s so 1s/length(t) = dt.

 

[luckycool]

Thanks so much.

However, if I using this way for cosine function, then it doesn't work again.
u2=1*cos(20*t);
U2=fftshift(fft(u2))*dt;

This time, when I change the time setting, the value of U2 will change.

It's so strange.

 

[bob1182006]

You mean when you change the end time the value given changes from 0.5 to something else?

Sorry I had that problem too and I don't understand why...

I haven't been able to find an answer to this either, but I know they discuss this topic in Digital Signal Processing classes/books.

 

[luckycool]

Hi, thanks again :)

If I change from t(end)=2 to other value then U2(f) will change its amplitude if I use dt instead of length(t).
But it's ok, I will check it with books. Do you know in which book or which section they discuss this topic?

 

[bob1182006]

Nope sorry, I haven't taken that class yet.
But I think it should be discussed in the same section as the sampling theorem or maybe the Fourier integrals.