|Mar11-06, 03:02 PM||#1|
how can I calculate the fourier transform for unit step function:
v(t)=1 where 0=<t<+infinity
I applied the general definition relation for FT:
v(w)=integral(v(t)*e^-jwt) ; - infinity<t<+infinity
but i had v(w)=infinity due to the term infinity-displaced e^(+jwt) so that's wrong of course!
I think we could divide this function into two functions for example:
v'(t)=1/2 ; t of all values
v''(t)=1/2 ; 0=<t<+infinity
so we notice v(t)=v'(t)+v''(t)
I don't know what to do , could anyone help!
|Mar11-06, 03:54 PM||#2|
Your v(t) does not have a Fourier transform. When functions like this are encountered in practice, Laplace transform is used instead (You need to describe the application you are working on that needs this transform).
|Mar11-06, 05:18 PM||#3|
The Fourier transform of the Heaviside step function doesn't exist as a function. It does, however, exist as a (somewhat nasty) distribution that involves the the Dirac delta "function".
|Mar12-06, 03:40 AM||#4|
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