# Fourier transform of unit step

I was reading pdf by caltech and in one of its section, Fourier transform of Unit step signal is calculated but I am confused, how this can be possible if region of convergence for Laplace transform ($$1/s$$) of unit step signal does not contain imaginary axis?

And if above case is possible then if it given that impulse response of a system is Unit step then frequency response of it should also exist and equal to $$H(ω)= πδ(ω) + 1/jω$$ and then can we calculate Fourier transform of output by computing $$H(ω)X(ω)$$ where $$X(ω)$$ is Fourier transform of input?

## 1 Answer

The Fourier transform can be generalized for functions that are not absolutely integrable. We can define a Fourier transform for functions with a constant envelope (e.g., sine, cosine, complex exponential), and even for functions with polynomial growth (but not with exponential growth). In these cases we must be prepared to deal with generalized functions in the expression of the Fourier transform, such as the Dirac delta impulse or its derivatives. This is also true for the Fourier transform of the step function.

The multiplication property of the Fourier transform remains true, albeit with certain restrictions, so we can generally compute the Fourier transform of the convolution of two functions by multiplying their Fourier transforms, provided that the convolution exists.

• Is there any sufficient and necessary condition exist by which we can predict whether fourier transform of function exist or not? Jun 1, 2020 at 9:14
• @user215805: That's a pretty complex topic, but the Fourier transform is defined for tempered distributions, including functions in $L^1$ and $L^2$ (absolutely and square integrable functions), and distributions of compact support. For some math background start on this wikipedia page. Jun 1, 2020 at 18:49