Why is the Fourier transform of 1 the 2pi*dirac(w) function

[thomas49th]

Homework Statement

Hi, I was wondering why the Fourier transform of 1 is

$$2\pi\delta(w)$$

I would of though that one would be of infinite frequencies (like a square wave).

Further more if g(t) = 1, for all t, g(t) = 1. Why does the Fourier transform have the argument of g(t) = 1 have an argument (w). g(t) has no frequency!

Thanks
Thomas

 

[Dick]

Wouldn't g(t)=1 have only zero frequency? It never oscillates at all. And isn't delta(w) the function in the frequency domain that only has zero frequency? It's only 'nonzero' at w=0. Make perfect sense to me.

 

[thomas49th]

good intuition. May I ask where the 2pi comes from again :)

 

[Dick]

good intuition. May I ask where the 2pi comes from again :)

It's a normalization of the Fourier transform. There's more than one convention for doing that. What's yours? I thought this question was more about intuition. The 2*pi has nothing to do with that. Take the inverse Fourier transform of 2*pi*delta(w). Do you get 1?

 

[myth_kill]

http://mathworld.wolfram.com/FourierTransform1.html