Fourier and wavelet transforms

Let \( f:\mathbb{R}\rightarrow\mathbb{C} \)

Maclaurin representation:

\[ f(x) = a_0 + a_1 x + a_2 x^2 +\cdots \]

Fourier representation:

\[ f(x) = a_0 + a_1 \cos{(w_1 x)} + \cdots + b_1 \sin{(w_1 x)} + \cdots \]

equivalently,

\[ f(x) = a_0 + a_1 e^{i w_1 x} + a_2 e^{i w_2 x} + \cdots \]

The Fourier representation looks different depending on the domain of the function \( f:A\rightarrow\mathbb{C} \), with \( A \) being:

\( \mathbb{R} \): Real numbers

\( \mathbb{T}_p \): Circle of length \( p \)

\( \mathbb{Z} \): Integer numbers

\( \mathbb{P}_N \): Polygon of \( N \) sides

For functions defined on the real line \( \mathbb{R} \).

\( f:\mathbb{R}\rightarrow\mathbb{C} \)

\[ \begin{aligned} f(x) &= \int_{-\infty}^\infty F(s) e^{2\pi i s x} ds \\ F(s) &= \int_{-\infty}^\infty f(x) e^{-2\pi i s x} dx \\ \end{aligned} \]

For functions defined on the integers on an interval \( \mathbb{T}_p=[0,p) \).

\( g:\mathbb{T}_p\rightarrow\mathbb{C} \) with \( g(p) = g(0) \)

\[ \begin{aligned} g(x) &= \sum_{k=-\infty}^\infty G(k) e^{2\pi ikx/p} \\ G(k) &= \frac{1}{p}\int_{0}^p g(x) e^{-2\pi ikx/p} dx \\ \end{aligned} \]

For functions defined on the integers \( \mathbb{Z} \).

\( \phi:\mathbb{Z}\rightarrow\mathbb{C} \)

\[ \begin{aligned} \phi(n) &= \int_{0}^p \Phi(s) e^{2\pi isn/p} ds \\ \Phi(s) &= \frac{1}{p}\sum_{n=-\infty}^\infty \phi(n) e^{-2\pi isn/p} \\ \end{aligned} \]

For functions defined on a polygon with \( N \) vertices \( \mathbb{P}_N = \{0,1,2,\ldots, N-1\} \).

\( \gamma:\mathbb{P}_N\rightarrow\mathbb{C} \) with \( \gamma(N) = \gamma(0) \)

\[ \begin{aligned} \gamma(n) &= \sum_{k=0}^{N-1} \Gamma(k) e^{2\pi ikn/N} \\ \Gamma(k) &= \frac{1}{N}\sum_{n=0}^{N-1} \gamma(n) e^{-2\pi ikn/N} \\ \end{aligned} \]

\[ \begin{aligned} \int_{-\infty}^\infty f(x)\overline{g(x)}dx &= \int_{-\infty}^\infty F(s)\overline{G(x)}ds,\ \ \mathbb{R} \\ \int_{0}^p f(x)\overline{g(x)}dx &= p \sum_{k=-\infty}^\infty F(k)\overline{G(k)},\ \ \mathbb{T}_p \\ \sum_{n=-\infty}^\infty f(n)\overline{g(n)} &= p \int_{0}^p F(s)\overline{G(x)}ds,\ \ \mathbb{Z} \\ \sum_{n=0}^{N-1} f(n)\overline{g(n)} &= N \sum_{k=0}^{N-1} F(k)\overline{G(k)},\ \ \mathbb{P}_N \end{aligned} \]

A continuous function \( f \) can be made discrete \( \phi \) by \( h \)-sampling

\[ \phi(n) = f(nh), n\in\mathbb{Z},\ h>0 \]

From a function \( f \), under some conditions, we can construct a periodic function \( g \) by \( p \)-summation

\[ g(x) = \sum_{m=-\infty}^\infty f(x-mp) \]

\( p \)-summation

\[ g(x) = \sum_{m=-\infty}^\infty f(x-mp) \implies G(k) = \frac{1}{p}F\left(\frac{k}{p}\right) \]

\( p/N \)-sampling

\[ \phi(n) = f\left(\frac{np}{N}\right) \implies \Phi(s) = \sum_{m=-\infty}^\infty F\left(s - \frac{mN}{p}\right) \]

Assume that \( f(t) \) is \( \sigma \)-bandlimited

\[ |\omega|>\sigma \implies F(\omega)=0 \]

Sample it with an interval \( \Delta t \)

\[ g(n) \equiv f(n\Delta t) \]

Then, \( f(t) \) can be uniquely reconstructed only when

\[ 2\sigma \Delta t <1 \]

\[ \sigma < \frac{1}{2\Delta t} \equiv \omega_{Nyq} \]

\[ \begin{aligned} f(t) &= a_1\sin(2\pi \omega_1 t) \\ F(\omega) &= \int_{-\infty}^{\infty}a_1\sin(2\pi \omega_1 t)e^{-2\pi i \omega t} dt\\ &= a_1\int_{-\infty}^{\infty} \left(\frac{e^{2\pi i\omega_1 t} - e^{-2\pi i\omega_1 t}}{2i}\right) e^{-2\pi i \omega t}dt\\ &= \frac{a_1}{2i}\left[\delta(\omega-\omega_1) - \delta(\omega+\omega_1)\right] \end{aligned} \]

\[ P(\omega) \equiv \left|F(\omega)\right|^2= \left(\frac{a_1}{2}\right)^2\left[\delta(\omega-\omega_1) - \delta(\omega+\omega_1)\right]^2 \]

Make Fourier Transform of short segments in the timeseries using a window function centered at \( \tau \), \( w(t-\tau) \)

\[ \hat{F}(\omega, \tau) = \int_{-\infty}^\infty f(t) w(t-\tau) e^{-2\pi i \omega t} dt \]

Problem: Choose an appropriate window width

Frequency vs. time resolution

Idea:

• Narrow windows for large frequencies (good time resolution)
• Wide windows for small frequencies (good frequency resolution)

Choose a localized wave (mother wavelet) \( \psi(t) \) and define

\[ \psi_{a,b}(t)=\frac{1}{\sqrt{a}}\psi\left(\frac{t-b}{a}\right) \]

Given \( f\in L^2(\mathbb{R}) \), we define its continuous wavelet transform with respect to the wavelet \( \psi \) as

\[ \mathcal{W}_{\psi}[f] (a,b) = \int_{-\infty}^\infty f(t)\overline{\psi_{a,b}(t)}dt \]

For \( \mathcal{W}_\psi[f] \) to be invertible we require

\[ 0 < C_{\psi}\equiv\int_{-\infty}^\infty \frac{\left|\hat{\psi}(\omega)\right|^2}{\left|\omega\right|} d\omega <\infty \]

\[ \int_{-\infty}^\infty\int_{-\infty}^\infty \mathcal{W}_\psi f (a,b)\overline{\mathcal{W}_\psi g (a,b)} \frac{da db}{a^2} = C_{\psi}\langle f,g\rangle \]

where

\[ C_\psi \equiv \int_{-\infty}^\infty \frac{|\hat{\psi}(\omega)|^2}{|\omega|}d\omega \]

\[ f(t) = \frac{1}{C_\psi}\int_{-\infty}^\infty\int_{-\infty}^\infty \mathcal{W}_\psi[f](a,b) \psi_{a,b}(t) \frac{da\ db}{a^2} \]

only when

\[ 0 < C_{\psi}\equiv\int_{-\infty}^\infty \frac{\left|\hat{\psi}(\omega)\right|^2}{\left|\omega\right|} d\omega <\infty \]

Change the continuous version

\[ \psi_{a,b}(t) = \frac{1}{\sqrt{a}} \psi\left(\frac{t-b}{a}\right),\ \ a,b\in\mathbb{R},\ a\neq 0 \]

to a discrete version

\[ \psi_{m,n}(t)=2^{-m/2} \psi \left( 2^{-m} t - n \right),\ \ \ n,m \in \mathbb{Z} \]

When can we recover \( f(t) \) from \( \mathcal{W}_\psi[f](m,n) \) ?

Find \( \psi_{m,n} \) such that they form a complete and orthonormal basis in \( L^2(\mathbb{R}) \):

\[ f(t) = \sum_{m,n=-\infty}^\infty \langle f,\psi_{m,n}\rangle\ \psi_{m,n}(t) \]

\[ \int_{-\infty}^\infty e^{2\pi i (x-x')}dx' = \delta(x) \]

\[ \int_0^p e^{2\pi i x (k-l)/p}dx = \begin{cases} p &\mbox{if}\ \ k=l \\ 0 &\mbox{otherwise} \end{cases} \]

\[ \sum_{n=0}^{N-1} e^{2\pi i n (k-l)/N} = \begin{cases} N &\mbox{if}\ \ k = mN\ ,m\in\mathbb{Z} \\ 0 &\mbox{otherwise} \end{cases} \]