The different Fourier Transforms

[Identity]

In my waves course, the Fourier transform we learn is:

$$X(\omega)=\int_{-\infty}^\infty x(t)e^{-i\omega t}\,dt$$

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^\infty X(\omega) e^{i\omega t}\,d\omega$$

In my PDEs course, we learn

$$X(\omega)=\int_{-\infty}^\infty x(t)e^{-i2\pi\omega t}\,dt$$

$$x(t)=\int_{-\infty}^\infty X(\omega) e^{i2\pi\omega t}\,d\omega$$

What is the difference between them? Given $$x(t)$$ they obviously give different answers for $$X(\omega)$$ so what does this mean?

Thx

 

[K^2]

In my waves course, the Fourier transform we learn is:

$$X(\omega)=\int_{-\infty}^\infty x(t)e^{-i\omega t}\,dt$$

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^\infty X(\omega) e^{i\omega t}\,d\omega$$

In my PDEs course, we learn

$$X(\omega)=\int_{-\infty}^\infty x(t)e^{-i2\pi\omega t}\,dt$$

$$x(t)=\int_{-\infty}^\infty X(\omega) e^{i2\pi\omega t}\,d\omega$$

What is the difference between them? Given $$x(t)$$ they obviously give different answers for $$X(\omega)$$ so what does this mean?

Thx

In the "waves" your ω is angular frequency. (T=2π/ω) In PDEs, you are defining it as ordinary frequency. (T=1/ω). I'm not really sure why, in the later case, as the angular frequency definitions are more common.

The reason for factor out front in the first case is to make sure the numbers work out if you do a reverse transform. Suppose you skip the factor.

$$x(t) = \int_{-\infty}^{\infty} d\omega e^{i\omega t}X(\omega) = \int_{-\infty}^{\infty} d\omega e^{i\omega t} \int_{-\infty}^{\infty} dt' e^{-i\omega t'} x(t') = \int_{-\infty}^{\infty} dt' \int_{-\infty}^{\infty} d\omega e^{i\omega t} e^{-i\omega t'} x(t') = \int_{-\infty}^{\infty} dt' 2\pi \delta(t-t') x(t') = 2\pi x(t)$$

And that's bad. So one of these has to have a 1/2π in it. Or, alternatively, both of these could be defined to have 1/sqrt(2π). This is purely a matter of choice.

If you have 2π in the exponent, the 2π factor in front of the delta function in the last step goes away, and you are left with x(t)=x(t), so you don't need any factors in definition to fix it.

 

[mathman]

When I was first exposed to this subject, we used the form without the 2π in the exponent, but for the sake of symmetry we used 1/√(2π) as the coefficient for both integrals.

 

[Identity]

Ah I see, thanks :)