name: summary class: center,middle, .toc[[✧](../index.html)] .title[Summary] <!-- class: left, .toc[[✧](../index.html)] #In-Class Assignments 1. In Matlab, let `$n=(0:N-1)'$` for some choice `$N$`. Form the array `$x=2*($`<tt>iseven</tt>`$(n)-1/2)$` and understand what this is giving you. Then take `$X=$`<tt>fft</tt>`$(x)$` for both `$N$` even and odd. 2. As a result of this excercise, at what position does the highest resolvable frequency appear in the Fourier transform? Know this in one-based counting, the position in the Matlab array, and in zero-based counting, which we are using for algebra. 4. Run <tt>psi=sleptap(length(x),M);</tt> followed by <tt>[f,S]=mspec(dt,x,psi)</tt> for `$M=4$.` Compute the variance by integrating the spectrum, not forgetting to multiply by the frequency interval. Compare to the variance computed directly from `$x$.` Do these match? 1. Run <tt>psi=sleptap(length(x),M);</tt> for `$M=3$.` Plot `$\psi$` on the left axis of a plot and the log of its <tt>fft</tt> on the right-hand side. --> --- class: left, .toc[[✧](../index.html)] **The sampling model** `\[z_n=z(n \Delta)\]` **Euler's formula** `\[ e^{i\omega t} = \cos(\omega t) + i \sin(\omega t) \]` **The discrete Fourier transform** `\[z_n=\frac{1}{N}\sum_{m=0}^{N-1} Z_m e^{i2\pi nm/N},\quad\quad\quad Z_m \equiv \sum_{n=0}^{N-1}z_n e^{-i2\pi nm /N} \]` --- class: left, .toc[[✧](../index.html)] **Continuous Fourier transform for deterministic signals** `\begin{equation} g(t) =\frac{1}{2\pi} \int_{-\infty}^{\infty} G(\omega)\, e^{i\omega t} d\omega,\quad\quad G(\omega)\equiv \int_{-\infty}^{\infty} g(t)\, e^{-i\omega t} d t \end{equation}` **The Cramér spectral representation of a stochastic process** `\begin{equation} z(t) =\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} dZ(\omega) \end{equation}` **Spectrum and autocovariance, a Fourier transform pair** `\begin{equation} R (\tau)\equiv\mathrm{E}\{z(t+\tau)\,z^*(t)\},\quad\quad S(\omega)\delta(\omega-\nu)\, d\omega d\nu\equiv \frac{1}{2\pi}\mathrm{E}\left\{dZ(\omega) dZ^*(\nu)\right\} \end{equation}` `\[S(\omega) = \int_{-\infty}^\infty e^{-i \omega \tau} R(\tau) d\tau , \quad\quad R (\tau) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{i \omega \tau} S(\omega) \, d\omega \]` --- class: left, .toc[[✧](../index.html)] **The origin of broadbias** We consider a truncated version of our continuously sampled time series, `$z_T(t)\equiv z(t) \Pi_T(t)$`. Its expected autovariance is `\[R_T(\tau) \equiv \mathrm{E}\left\{\frac{1}{T}\int_{-T/2}^{T/2} z_T(t+\tau) z_T^*(t) dt \right\}\]` which corresponds to an expected periodogram-like spectral estimate that is a filtered version of the true spectrum `\[R_T(\tau) = R(\tau) \Lambda_T(\tau) \Longleftrightarrow S_T(\omega) \equiv \int_{-\infty}^\infty S(\omega-\nu) F_T(\nu) d \nu\]` where `$F_T(\omega)\equiv \int_{-\infty}^{\infty}\Lambda_T(t)e^{i\omega\tau} \, d \tau =\frac{1}{T}\frac{\sin^2(\omega T/2)}{(\omega/2)^2}$` is the Fejér kernel. `\begin{equation} \Pi_T(t)\equiv \left\{\begin{array} {cc} 1, & t\le T/2 \\ 0, & t> T/2 \end{array}\right.,\quad\quad\quad \Lambda_T(t)\equiv \left\{\begin{array} {cc} 1-\frac{|\tau|}{T}, & t\le T \\ 0, & t> T \end{array}\right. \end{equation}` --- class: left, .toc[[✧](../index.html)] **Multitaper spectral estimation** Let `$\psi_n^{\{k\}}$` be `$K=2P-1$` length-`$N$` orthogonal functions that are optimally concentrated in a frequency band `$2P$` Rayleigh frequencies wide centered at zero. `$K$` spectral estimates, known as the “eigenspectra”, are defined as `\[\widehat S_m^{\{k\}}\equiv \left|\sum_{n=0}^{N-1} \psi_n^{\{k\}} z_n\, e^{-i 2\pi m n/N}\right|^2,\quad\quad n=0, 1, 2, \ldots,N-1.\]` are averaged to give the *multitaper spectral estimate* `\[\widehat S_m^{\psi}\equiv \frac{1}{K}\sum_{k=1}^K\widehat S_m^{\{k\}}.\]` This reduces bias by tapering the data with functions that minimize broadband leakage from *outside* the `$2P$` band, while reducing variance through approximating an average over independent Fourier coefficients *within* the `$2P$` band. --- class: left, .toc[[✧](../index.html)] **The continuous wavelet transform** The Morse wavelets are defined in terms of their Fourier transform `\[\psi_{\beta,\gamma}(t) \Longleftrightarrow \Psi_{\beta,\gamma}(\omega) = \left\{ \begin{array}{ccc} a_{\beta,\gamma} \,\omega^\beta e^{-\omega^\gamma}, && \omega>0 \\ 0, && \omega\le 0\end{array}\right.\]` Since `$\Psi_{\beta,\gamma}(\omega)$` is real, `$\psi_{\beta,\gamma}(t) = \psi_{\beta,\gamma}^*(-t)$`. The wavelet transform is `\[w(t,s) =\int_{-\infty}^\infty \frac{1}{s} \psi\left(\frac{\tau-t}{s}\right) z(\tau) d\tau=\frac{1}{2\pi} \int_{-\infty}^\infty \Psi(s\omega) Z(\omega) e^{i\omega t} d\omega\]` and scale is mapped to frequency by `$\omega=s/\omega_{\beta,\gamma}$` with `$\omega_{\beta,\gamma}\equiv(\beta/\gamma)^{1/\gamma}$`. Note, this is a *joint function* of the signal and the wavelet!! `\begin{align} &z(t) = \delta(t-t_o) &\!\!\!\!\!\!\! \Longrightarrow \quad \quad & w(t,s) = \frac{1}{s} \psi^*\left(\frac{t-t_o}{s}\right) \\ &\left.\begin{array}{l} z(t) = e^{i\omega_o t}\\ Z(\omega) = 2\pi \delta(\omega-\omega_o) \end{array} \right\}&\!\!\!\!\!\!\!\Longrightarrow \quad \quad & w(t,s) = \Psi(s\omega_o) e^{i\omega_o t} \end{align}`