Duality

If we take the Fourier integral, and compare it to the inverse Fourier integral:

 

you can see that the two are very similar. In fact, if we replace t by -w and w by t in the second, then we get the integral

And from this you should be able to see that if we can do a transform one way, we can do it the other way dead easily! Just replace t by -w and w by t, and multiply the function in terms of w by 2p, and you're there.

The Impossible (A Non-Dirichlet Function)

Earlier I mentioned the Dirichlet conditions for Fourier series, and I noted that these guarantee a Fourier transform is possible. Well, it would be nice to extend that to functions which go to any constant value, instead of to zero.

So, what we basically want is a transform of the function

x(t) = 1

which can be accomplished by finding the integral

Eek. This doesn't quite work. However, we know that the Fourier transform of a Dirac impulse function d(t) is 1, so using duality we can conclude that the Fourier transform of x(t) = k is

X(w) = 2pkd(-w) = 2pkd(w)

because d(w) is an even function (d(w) = d(-w)).

A function which approaches a constant can now be expressed as the sum of a constant and a function that approaches zero. The constant we now know how to deal with, and the rest satisfies the Dirichlet conditions.

Transforming the Kitchen Sinc

Sometimes you may just have a function that you'd rather not integrate. (It's okay to be lazy!) So, let's try transforming our old friend from the last section, the sinc function. You may recall that the Fourier transform of P(t) is:

sinc (w / 2p)

So, therefore, the Fourier transform of sinc (t / 2p) is

2p P(-w) = 2p P(w)

We can apply scaling (see next section) here, as well, to get the Fourier transform of x(t) = sinc kt:

X(w) = (2p / 2pk) P(w / 2pk) = (1 / k) P(w / 2pk)

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