Houjun Liu


For \(f,g : \mathbb{R} \to \mathbb{C}\), we have:

\begin{equation} (f * g)(x) = \int_{\mathbb{R}} f(x-y) g(y) \dd{y} = \int_{\mathbb{R}} f(y) g(x-y) \dd{y} \end{equation}

properties of convolution

  • \((g * f) (x) = (f * g) (x)\)
  • \(\mathcal{F}(f * g) = \mathcal{F}(f)\mathcal{F}(g)\)
  • \(\mathcal{F}^{-1}(\hat{f} \hat{g}) = f * g\)
  • \((f * g)’ = f * g’ = f’ * g\)
  • \(\lambda ( f * g ) = (\lambda f) * g = f * (\lambda g)\)

=> “in a convolution, if ANY ONE of the two functions are Differentiable, both are Differentiable.”; think about smoothing a jagged function using a Gaussian.


rolling average

\begin{align} U_{L}(x) = \begin{cases} L, |x| \leq \frac{1}{2L} \\ 0, |x| > \frac{1}{2L} \end{cases} \end{align}

The width of the area for which the expression is positive is \(2L\), and the height is \(L\), so the area (integral) is \(1\).

So now let’s consider:

\begin{equation} (f * U_{L})(x) \end{equation}

which is:

\begin{equation} \int_{\mathbb{R}} f(x-y) U_{L}(y) \dd{y} \end{equation}


\begin{equation} L \int_{-\frac{1}{2}L}^{\frac{1}{2}L} f(x-y) \dd{y} \end{equation}

You will note that we are sweeping something of window width \(\frac{1}{L}\) over the function, which averages the function \(f\) over the window \(L\).

So convolving with this function essentially smoothes function over a window \(\frac{1}{L}\); as \(L\) decreases, we are averaging over a greater interval; vise versa.

signal compression

Write your signal in terms of its Fourier transform:

\begin{equation} f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{it\lambda} \hat{f}(\lambda) \dd{\lambda} \end{equation}

We can write:

\begin{equation} \hat{f}(\lambda) \cdot 1_{J}(\lambda) \end{equation}

whose inverse Fourier transform would be:

\begin{equation} f(x) * \mathcal{F}\qty(1_{J}(\lambda)) \end{equation}


What if we want the Fourier Transform of \(\hat{f}(\lambda)\hat{g}(\lambda)\) in terms of one expression?


\begin{equation} \hat{f}(\lambda) \hat{g}(\lambda) = \qty(\int_{\mathbb{R}} f(x) e^{-i\lambda x} \dd{x}) \qty(\int_{\mathbb{R}} g(y) e^{-i\lambda y} \dd{y}) \end{equation}

Notice that because neither integral have dependence on the other, we can actually:

\begin{equation} \hat{f}(\lambda) \hat{g}(\lambda) = \int_{\mathbb{R}} \int_{\mathbb{R}} f(x) g(y) e^{-i\lambda (x+y)} \dd{x}\dd{y} \end{equation}

writing this as a change of variable:

\begin{equation} \begin{cases} u = x+y \\ x = u-y \\ \dd{x} = \dd{u} \end{cases} \end{equation}

we can write:

\begin{equation} \hat{f}(\lambda) \hat{g}(\lambda) = \int_{\mathbb{R}} \qty(\int_{\mathbb{R}} f(u-y) g(y) e^{-i\lambda (u)} \dd{u})\dd{y} \end{equation}

Considering they the integrands are isolated and decaying, we can swap them, pulling out also \(e^{-i\lambda(u)}\) because it has no \(y\) dependence:

\begin{equation} \hat{f}(\lambda) \hat{g}(\lambda) = \int_{\mathbb{R}} \qty(\int_{\mathbb{R}} f(u-y) g(y) \dd{y})e^{-i\lambda (u)} \dd{u} \end{equation}

Notice! The inner part is a function, and the outer part is a Fourier transform! This is similar to a convolution (probability)!


\begin{equation} \hat{f}(\lambda) \hat{g}(\lambda) = \mathcal{F}(f * g) = \mathcal{F}(f) \mathcal{F}(g) \end{equation}

Operating on the inverse, we can obtain a similar result:

\begin{equation} \mathcal{F}^{-1}(\hat{f} \hat{g}) = f * g \end{equation}