Fourier transform of a Fourier series
The previous post showed how we can take the Fourier transform of functions that don't have a Fourier transform in the classical sense.
The classical definition of the Fourier transform of a function f requires the integral of |f| over the real line to be finite. This implies f(x) must approach zero as x goes to and -. A constant function won't do, and yet we got around that in the previous post. Distribution theory even lets you take the Fourier transform of functions that grow as their arguments go off to infinity, as long as they don't grow too fast, i.e. like a polynomial but not like an exponential.
In this post we want to take the Fourier transform of functions like sine and cosine. If you read that sentence as saying Fourier series, you have the right instinct for classical analysis: you take the Fourier series of periodic functions, not the Fourier transform. But with distribution theory you can take the Fourier transform, unifying Fourier series and Fourier transforms.
For this post I'll be defining the classical Fourier transform using the convention
and generalizing this definition to distributions as in the previous post.
With this convention, the Fourier transform of 1 is , and the Fourier transform of is 2.
One can show that the Fourier transform of a cosine is a sum of delta functions, and the Fourier transform of a sine is a difference of delta functions.
It follows that the Fourier transform of a Fourier series is a sum of delta functions shifted by integers. In fact, if you convert the Fourier series to complex form, the coefficients of the deltas are exactly the Fourier series coefficients.
- Reverse engineering Fourier conventions
- Computing Fourier series coefficients with FFTs
- Fourier uncertainty principle