Fourier Transform #

It’s good to start out with the definition of a Fourier transform:

[]

Fourier Transform

\[\mathcal{F}(f(t)) \equiv F(\lambda) \equiv \int_{-\infty}^{\infty}e^{i \lambda t} f(t) \dd t\]

Note: \( t \) does not necessarily denote time, as it generally does in the Laplace transform. It’s just a dummy variable in the integration and is often a spatial coordinate.

To recover \( f(t) \) we also have the inverse Fourier transform

\[f(t) = \mathcal{F} ^{-1} (F(\lambda)) = \frac{1}{2 \pi} \int_{-\infty}^{\infty}e^{- i \lambda t} F(\lambda) \dd \lambda\]

\[\]

Fourier’s Theorem

This is what gives us the ability to write down the inverse transform. We won’t derive the theorem here since that is usually done in an earlier course. The general approach starts with the Fourier series expansion and takes the limit to converge to the Riemann sum of the integral. The theorem states:

\[f(t) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \dd \lambda \int_{-\infty}^{\infty}\dd \tau e^{i \lambda ( \tau - t)} f(\tau)\]

This is valid for any piecewise smooth \( f(t) \) which is integrable.

If we think about it, it’s amazing that we can recover \( f(t) \) after all of that. The key is that \( e^{i \lambda t}, e^{-i \lambda t} \) form a complete, orthogonal basis for all piecewise smooth integrable functions.

For consistency, we need to make sure we consistently place the \( \pm \) exponent on the transform and the factor of \( 1/2 \pi \), since they are arbitrary choices.

“Piecewise smooth” means that over any interval, we can subdivide into a finite number of pieces where the function is smooth. We can have a finite number of jumps, but the function can never jump to infinity.

Example:

As a real integral, this isn’t the easiest thing to do. But we can use Jordan’s lemma to make it simple for us. \( 1 / z^2 + 4 \) gotes to zero as \( |z| \rightarrow \infty \), so

Let’s take our inverse transform for a spin to show that it works!

This is pretty cool, but in real life the powerful use of the transform is in solving partial differential equations.

Fourier Transform of Derivatives #

We’ve required that \( f(t) \) is integrable, so the boundary term must vanish. \( f(t) |^{\pm \infty} = 0 \). This is actually a pretty major restriction on Fourier transforms. Many familiar functions are not totally integrable, but we often want to use their Fourier transform. The generalized Fourier transform solves this by restricting the transform in the direction that is integrable, then setting the function to zero in the other direction. This is very related to the Laplace transform, which is always one-sided.

Let’s do some examples

\[f(t) = e^{- t} \quad \rightarrow \quad F(\lambda) = \int_{-\infty}^{\infty}e^{i \lambda t - t} \dd t = \left.\frac{1}{i \lambda - 1} e^{i (\lambda - 1)t} \right|_{-\infty} ^{\infty}\] This limit does not exist as \( t \rightarrow -\infty \). Similarly, the transform of \( f(t) = e^{t} \) blows up as \( t \rightarrow \infty \). To overcome this we, can consider a one-sided version of the function

These functions do have a well-defined Fourier transform

\[f(t) = \begin{cases} e^{-i \lambda_0 t} & t \geq 0 \\ 0 & t < 0 \end{cases}\] \[F(\lambda) = \int _0 ^\infty e^{i (\lambda - \lambda _0)t} \dd t = \frac{e^{i (\lambda - \lambda_0)t}}{i (\lambda - \lambda_0)}\]

This limit does not exist for purely real \( \lambda \), since it oscillates forever in the \( + \infty \) direction. But if there is a small positive imaginary part of \( \lambda \) then the limit is \[F(\lambda) = \frac{i}{\lambda - \lambda_0}\]

Now we want an inverse Fourier transform for this:

Now we’ve made ourselves an improper integral. But at least we know how we got here! (it was by letting \( \text{Im}(\lambda) > 0 \) ). We should not try to convert this into a principal value integral, because we will get a different answer that way.