Calculate the Fourier transform of the Heaviside function H(x)?

Answer 1

This is a sort of tricky question, and I think there are a couple of ways to do it. Before I start I'll warn you that depending on your definition of the Fourier Transform, factors of \pi may appear in various places where I haven't written them. The first thing that you can use is the fact that the weak derivative of the Heaviside function is the delta function, and it's easy to calculate the Fourier Transform of the delta function- its simply 1. Thus, if \psi is a test function and F indicates Fourier Transform:

= <1,\psi> = \int \psi dx

However, we can calculate this another way using a property of the Fourier Transform:

= = \int \psi dx.

Thus we have discoverred that i \xi F(H) = 1, and so we are left with a division problem. Unfortunately division is difficult with distributions. Away from zero we can just divide by i \xi, but near 0 this will not work. In fact you can use a partition of unity argument to say that F(H) must equal -i p.v.1/(\xi) + d where p.v. means principle value and d is a distribution supported at the the origin. By the theory of distributions, d must be a linear combination of derivatives of the delta function. To figure out d, we can use the following fact: if \psi is an even function then = 0 and thus = = . However, the Fourier Transform of an even function is even, so

= = 1/2 \int F(\psi) = 1/(\pi) \psi(0)

by the Fourier inversion formula. This of course only shows that d = the delta function + some odd derivatives of the delta function. However, you can show there are no odd derivatives of delta by considering = \int \psi dx = <-i pv 1/\xi, i \xi \psi> for an even function \psi by the above.

So, the Fourier Transform of H is:

-i pv 1/\xi + 1/(\pi) \delta.

Once again, there are different definitions of the Fourier Transform which put the factors of \pi in different places, and so the constants may be differ depending on your definition. I used:

F(\psi)(\xi) = \int e^{-ix \xi} \psi(x) dx.

Note: edited, I originally wrote 1/(4 \pi), but it should just be 1/\pi.

Answer 2

Fourier Transform Of Heaviside Function

Answer 3

on the Fourier grow to be is suitable to each and each periodic and aperiodic indicators. The Fourier sequence is barely ideal to periodic indicators, for that reason why the Fourier grow to be is a extra bodily effectual gadget as quickly as you're able to remodel a time-sign to a frequency-sign. a thorough derivation of the Fourier sequence isn't now no longer consumer-friendly, inspite of if is composed of determining matters in Linear Algebra, mutually with Linear indepence, and what's talked approximately as a beginning place. undergo in concepts inspite of the shown fact that that integration is comfortably an infinite sum, so the Fourier redesign is likewise derived using the Fourier sequence.

Answer 4

There isn't a Fourier transform of the Heaviside step function H(x) because it isn't square integrable. The integral of H(x) from negative infinity to infinity is infinite. It has to be finite for a Fourier transform to exist.