Add third pset

_notes/fourier_transform.md

@@ -8,15 +8,12 @@ notes are based on my prior knowledge and some helpful websites:
 - [Properties of Fourier Transform](http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html)
 - [symmetry.pdf](https://www.cs.unm.edu/~williams/cs530/symmetry.pdf)
 
-
-## Properties of the Fourier Transform
-
 Note: I'm sloppy with the proofs here since all physical functions will have the nice properties
 that make the relevant operations valid, but I don't always call of these properties out when they
 are used.
 
 
-### Basics
+## Basics
 
 For a function $$f : \mathbb{R} \rightarrow \mathbb{C}$$, I use the definitions
 
@@ -55,7 +52,7 @@ $$
 $$
 
 
-### Fourier Flips
+## Fourier Flips
 
 The Fourier transform has a number of interesting properties related to the flip operator
 $$(\mathcal{R} f)(x) = f(-x)$$. By definition
@@ -105,7 +102,7 @@ f(x)
 $$
 
 
-### Conjugate Symmetries
+## Conjugate Symmetries
 
 Recall that the [conjugate](https://en.wikipedia.org/wiki/Complex_conjugate) of a complex number
 $$a + b i$$ is defined as $$\overline{a + bi} = a - bi$$. Conjugation distributes over addition and
@@ -143,7 +140,7 @@ a horizontal arrow by taking the inverse Fourier transform), and vertical arrows
 conjugates.
 
 
-### Even More Odd
+## Even More Odd
 
 Recall that a function $$f$$ is even if $$f(-x) = f(x)$$, and odd if $$f(-x) = -f(x)$$. Note that a
 function is real if and only if it's equal to its own complex conjugate, and a function is purely
@@ -176,3 +173,46 @@ In particular, since conjugation doesn't change magnitude, the magnitude of the
 function is an even function. Because of this it's common to see the negative frequencies ignored
 when we're dealing with a real function and only care about the magnitude of the transform (like
 for spectral power analysis).
+
+
+## Transforms of Gaussians
+
+A Fourier Transform that comes up frequently is that of a Gaussian. It can be calculated by
+completing a square.
+
+$$
+\begin{align*}
+\mathcal{F}(N(0, \sigma^2))(f)
+&= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty e^{\frac{-x^2}{2 \sigma^2}}
+    e^{- 2 \pi i f x} \mathrm{d} x \\
+&= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty e^{-\frac{1}{2 \sigma^2}
+    \left( x^2 + 4 \pi i \sigma^2 f x \right)} \mathrm{d} x \\
+&= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty e^{-\frac{1}{2 \sigma^2}
+    \left( x + 2 \pi i \sigma^2 f \right)^2 + \frac{(2 \pi i \sigma^2 f)^2}{2 \sigma^2}}
+    \mathrm{d} x \\
+&= e^{-2 \pi^2 f^2 \sigma^2} \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty
+    e^{-\frac{\left( x + 2 \pi i \sigma^2 f \right)^2}{2 \sigma^2}} \mathrm{d} x \\
+&= e^{-2 \pi^2 f^2 \sigma^2} \\
+&= e^{-\frac{(2 \pi f)^2}{2 / \sigma^2}}
+\end{align*}
+$$
+
+This is an unnormalized Gaussian with variance $$1/\sigma^2$$. Note that the exponent wants to be
+expressed in radians instead of cycles, so $$f$$ is scaled by $$2 \pi$$.
+
+This function integrates to $$(2 \pi \sigma^2)^{-1/2}$$. One might have hoped it would be
+normalized. One reason this could not be true is Plancherel's Theorem. The "power" of a normalized
+Gaussian is
+
+$$
+\begin{align*}
+\int_{-\infty}^\infty \left( \frac{1}{\sqrt{2 \pi \sigma^2}} e^{\frac{-x^2}{2 \sigma^2}} \right)^2
+\mathrm{d} x
+&= \frac{1}{2 \pi \sigma^2} \int_{-\infty}^\infty e^{\frac{-x^2}{\sigma^2}} \mathrm{d} x \\
+&= \frac{1}{2 \pi \sigma} \int_{-\infty}^\infty e^{-x^2} \mathrm{d} x \\
+&= \frac{1}{2 \sqrt{\pi} \sigma}
+\end{align*}
+$$
+
+This depends on the variance, which is inverted by the Fourier Transform. So since the power is
+invariant, the normalization cannot in general be conserved.

_sass/main.scss

@@ -32,3 +32,16 @@ a:visited {
 .question {
     font-style: italic;
 }
+
+table {
+    margin: 1.6em auto;
+}
+
+table, th, td {
+    border: 1px solid black;
+    border-collapse: collapse;
+}
+
+th, td {
+    padding: 5px 10px;
+}