# Unit-step FT

how can I calculate the fourier transform for unit step function:

v(t)=1 where 0=<t<+infinity

v(t)=0 otherwise

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
v''(t)=-1/2; -infinity<t<0

so we notice v(t)=v'(t)+v''(t)

I don't know what to do , could anyone help!

thanks!

## Answers and Replies

mathman
Science Advisor
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).

George Jones
Staff Emeritus
Science Advisor
Gold Member
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".

Regards,
George

George Jones said:
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".

Regards,
George

thanks.Anyway could you clarify more, how did you deduct that Heaviside step function exists as Dirac delta function and how can we describe that distribution practically on diagram