Calculating Fourier Transform of Unit Step Function

In summary, the Fourier transform of the Heaviside step function does not exist as a function, but can be expressed as a distribution involving the Dirac delta function. This can be seen by dividing the function into two separate functions and taking the Fourier transform of each, resulting in a distribution that cannot be graphically represented. The use of Laplace transform may be more suitable for practical applications.
  • #1
electronic engineer
145
3
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!
 
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  • #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).
 
  • #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".

Regards,
George
 
  • #4
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
 

1. What is the unit step function?

The unit step function, also known as the Heaviside step function, is a mathematical function that is defined as 0 for negative input values and 1 for positive input values.

2. Why is the Fourier transform of the unit step function important?

The Fourier transform of the unit step function is important because it helps us understand the frequency components of a signal and how they are affected by the unit step function. It is also used in many applications, such as signal processing, image processing, and control systems.

3. How do you calculate the Fourier transform of the unit step function?

The Fourier transform of the unit step function can be calculated using the formula: F(ω) = 1/ (ω + j0), where ω is the frequency variable and j is the imaginary unit. This formula can also be expressed in terms of the Dirac delta function as F(ω) = πδ(ω) + j1/ω.

4. What are the properties of the Fourier transform of the unit step function?

Some of the properties of the Fourier transform of the unit step function include time shifting, frequency shifting, time scaling, and time differentiation. These properties allow us to easily manipulate and analyze signals in the frequency domain.

5. Can the Fourier transform of the unit step function be used for non-periodic signals?

Yes, the Fourier transform of the unit step function can be used for both periodic and non-periodic signals. However, for non-periodic signals, the Fourier transform is often defined as a continuous function rather than a discrete series of frequencies.

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