Why the step function has a Fourier transform?

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safwatonline

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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.
 

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
 

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
 

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.
 

Fourier

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

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