Why the step function has a Fourier transform?

[safwatonline]

kind of silly question, but anyway, i was wondering why the step function has a Fourier transform given that it is not bounded (i.e. absolutely integrable) which means that it doesn't satisfy Dirichlet's conditions.

 

[laktronics]

Re: Fourier

Hi,
I think physically, anything having a rate of change, should have frequency components and wherever there is a frequency component, there is Fourier. The step function has a steep rate of change at t = zero, and, had it continued backwards without this change, Fourier would have spared it.
Regards,
Laktronics

 

[raghuram_msc]

Re: Fourier

Hi Friend,

I agree with laktronics, in a sense that a step function indicates a sudden "step" increase. Now, suppose this step were sent through an Inductor/Capacitor the o/p fuction will not change instantaneously... but, rather "sluggisly". Fourier Transform components (ie., Fourier Expansion) actually indicate those transiton frequencies.

Hope I was helpful....!!

Sai

 

[maxwellequ]

Re: Fourier

I don't remember the math details (and, quite honestly, I don't want to have to remember them again ), but I think that is where you enter the realm of generalized functions (Dirac Delta is the most famous one), where the integral definition is "extended" to accommodate those difficulties.

 

[safwatonline]

Fourier

well, actually i was wrong, as the U(t) does obey the Dirichlet condition, as the integration is undetermined not undefined !