Revision as of 08:17, 18 March 2011

Lecture 21

Multiplication Property

$ \mathcal{F}(x_1(t) x_2(t)) = \frac{1}{2\pi}\mathcal{F}(x_1(t))*\mathcal{F}(x_2(t)) $

Causal LTI system defined by cst coeff diff equations

$ \sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t) = \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) $

What is the frequency response of this system? Recall:

$ \begin{align} \mathcal{H}(\omega) &= \mathcal{F}(h(t)) \\ \mathcal{Y}(\omega) &= \mathcal{H}(\omega)\mathcal{X}(\omega) \end{align} $

Steps to solve:

$ \begin{align} \mathcal{F}\left(\sum_{k=0}^{N}a_k \frac{d^k}{dt^k}y(t)\right) &= \mathcal{F}\left( \sum_{k=0}^{M}b_k \frac{d^k}{dt^k}x(t) \right) \\ \sum_{k=0}^{N}a_k \mathcal{F}\left(\frac{d^k}{dt^k}y(t)\right) &= \sum_{k=0}^{M}b_k \mathcal{F}\left(\frac{d^k}{dt^k}x(t)\right) \\ & \text{recall: }\mathcal{F}\left(\frac{d^n}{dt^n}y(t)\right) = (j\omega)^n Y(\omega) \\ \sum_{k=0}^{N}a_k \left( j \omega \right)^k \mathcal{Y}(\omega) &= \sum_{k=0}^{M}b_k \left( j \omega \right)^k \mathcal{X}(\omega) \\ \mathcal{Y}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \mathcal{X}(\omega) \\ \mathcal{H}(\omega) &= \frac{\sum_{k=0}^{M}b_k(j\omega)^{k}}{\sum_{k^\prime=0}^{N}a_k(j\omega)^{k^\prime}} \\ h(t) &= \mathcal{F}^{-1}\left(\mathcal{H}(\omega)\right) \end{align} $

Def of DT F.T.

Here are the practice problems that do this: Problem 1, Problem 2, Problem 3

$ \begin{align} x[n] &\xrightarrow{\mathcal{F}} \mathcal{X}(\omega) \text{ in book as }\mathcal{X}(e^{j\omega})& \\ x[n] &\xleftarrow{\mathcal{F}^{-1}} \mathcal{X}(\omega) &\\ \mathcal{X}(\omega) & = \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} &\\ x[n] &= \frac{1}{2\pi}\int_{0}^{2\pi} \mathcal{X}(\omega)e^{j \omega n} d\omega & \end{align} $

Example

Compute the FT of $ x[n] = 2^{-n}u[n] $

$ \begin{align} \mathcal{X}(\omega) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty}2^{-n}u[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty}2^{-n}e^{-j\omega n} \\ &= \sum_{n=0}^{\infty}\left(\frac{1}{2e^{-j\omega}}\right)^n \\ &= \frac{1}{1-\frac{1}{2e^{j\omega}}} \end{align} $

Properties of DT FT

Periodicity

$ \begin{align} \mathcal{X}(\omega + 2\pi) &= \mathcal{X}\text{, for all } \omega \\ \text{because} & \\ \mathcal{X}(\omega + 2 \pi) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j(\omega + 2\pi)n} \\ &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} e^{-j2\pi n} \\ &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \mathcal{X}(\omega) \end{align} $

Linearity

$ \text{for all }a,b,c \in \mathcal{C}\text{ and all } x_1[n],x_2[n] $

$ \mathcal{F}(ax_1[n] + bx_2[n]) = a\mathcal{F}(n_1[n]) + b\mathcal{F}(n_2[n]) $ provided both FT's exist.

The FT of DT periodic signals

$ x[n] \text{ periodic} \Rightarrow x[n] = \sum_{n=0}^{N-1}a_k e^{\omega_0 n k } $

$ \mathcal{F}(x[n]) = \sum_{n=0}^{N-1}a_k \mathcal{F}\left(e^{\omega_0 n k }\right) $, by linearity

so all we need is the FT of $ e^{j k \omega_0 n} $

we want $ \mathcal{X}(\omega) $ such that $ \frac{1}{2\pi}\int_{0}^{2\pi}\mathcal{X}(\omega)e^{j\omega t} d\omega = e^{j\omega_0 n k} $

try $ \mathcal{X}(\omega) = 2\pi \delta(\omega - k\omega_0) $ and it works. So the real answer is

$ \mathcal{X}(\omega) = \sum_{m = -\infty}^{\infty} 2\pi \delta(\omega - \omega_0 k + 2\pi m) $

Lecture 22

Time Shifting and Freq Shifting Property

$ \begin{align} \mathcal{F}(x[n-n_0]) &= e^{-j\omega n_0}\mathcal{F}(x[n]) \\ \mathcal{F}\left( e^{j \omega_0 n}x[n] \right) &= \mathcal{X}(\omega - \omega_0) \end{align} $

Conjugation and Conjugation Symmetry

$ \mathcal{F}\left( x^*[n] \right) = \mathcal{X}^*(-\omega) $

Important Corrilary

if signal is real then $ \mathcal{X}(\omega) = \mathcal{X}^*(-\omega) $ because $ x[n] $ is real.

$ \begin{align} x^*[n] &= x[n] \\ \mathcal{X}^*[-\omega] &= \mathcal{X}(\omega) \end{align} $

This mean that $ x[n] $ real

=> Re $ \mathcal{X}(\omega) $ is an odd function

=> Im $ \mathcal{X}(\omega) $ is an odd function

Panseval's relation

$ \sum_{n=-\infty}^{\infty}| x[n] |^2 = \frac{1}{2\pi} \int_{0}^{2\pi} | x[\omega] |^2 d\omega $

Convolution Property

$ \begin{align} \mathcal{F}(x[n]*y[n]) &= \mathcal{F}(x[n])\mathcal{F}(y[n]) \\ &= \mathcal{X}(\omega) \mathcal{X}(\omega) \end{align} $

so for any LTI system $ x \rightarrow h[n] \rightarrow y[n] = x[n]*h[n] $

Lecture 23

Multiplication Property

$ x[n]y[n] \xrightarrow{\mathcal{F}} \frac{1}{2\pi} \mathcal{X}(\omega)*\mathcal{Y}(\omega) $

Differentiation in frequency property

$ nx[n] \xrightarrow{\mathcal{F}} j\frac{d}{d\omega}\mathcal{X}(\omega) $

Example

Assume $ |\alpha | < 1 $

1. Compute the FT of $ x_1[n] = \alpha^n u[n] $
2. Use your andwer to compute the FT of $ x_2[n] = (n+1)\alpha^n u[n] $

Answer: 1)

$ \begin{align} \mathcal{X}_1(\omega) &= \sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n} \\ &= \sum_{n=-\infty}^{\infty} \alpha^nu[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty} \alpha^nu[n]e^{-j\omega n} \\ &= \sum_{n=0}^{\infty} \left(\alpha u[n]e^{-j\omega } \right) ^n \\ &= \frac{1}{1-\alpha e^{-j\omega}} \end{align} $

2)

$ \begin{align} \mathcal{X}_2(\omega) &= \mathcal{F}((x+1)\alpha^n u[n])\\ &= \mathcal{F}(n\alpha^n u[n]) + \mathcal{F}(\alpha^n u[n]) \\ &= j\frac{d}{d\omega}\mathcal{X}_1(\omega) + \mathcal{X}_1(\omega) \\ &= j\frac{d}{d\omega}\left( \frac{1}{1-\alpha e^{-j\omega}} \right) + \frac{1}{1-\alpha e^{-j\omega}} \\ &= \frac{1}{\left( 1 - \alpha e^{-j\omega} \right)^2} \end{align} $

LTI systems defined by linear, constant coef diff eq's

$ \sum_{k=0}^{N}a_k y[n-k] = \sum_{k=0}^{M}b_k y[n-k] $

Example: $ y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n] = 2x[n] $ can look at this eq in freq domain.

$ \begin{align} \mathcal{F}(y[n] - \frac{3}{4} y[n-1] + \frac{1}{8}y[n]) &= 2\mathcal{F}(x[n]) \\ \mathcal{Y} - \frac{3}{4}\mathcal{F}(y[n-1]) + \frac{1}{8}\mathcal{F}(y[n-8]) &= 2\mathcal{X}(\omega) \\ \mathcal{Y}(\omega) - \frac{3}{4}e^{-j\omega}\mathcal{Y}(\omega) + \frac{1}{8}e^{-2j\omega}\mathcal{Y}(\omega) &= 2\mathcal{X}(\omega) \end{align} $

$ \begin{align} \mathcal{Y}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}}\mathcal{X}(\omega) \\ \mathcal{H}(\omega) &= \frac{2}{1 - \frac{3}{4}e^{-j\omega} + \frac{1}{8}e^{-2j\omega}} \\ &= \frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}} \\ h[n] &= 4\left(\frac{1}{2}\right)^n u[n] - 2 \left( \frac{1}{4} \right)^n u[n] \end{align} $

Now find the systems responce to $ x[n] = \left( \frac{1}{4} \right)^n u[n] $ $ \begin{align} \mathcal{Y}(\omega) &= \mathcal{X}(\omega)\mathcal{Y}(\omega) \\ &= \left(\frac{4}{1-\frac{1}{2}e^{-j\omega}} + \frac{-2}{1-\frac{1}{4}e^{-j\omega}}\right)\frac{1}{1-\frac{1}{4}e^{-j\omega}} \\ &= \frac{4}{\left(1-\frac{1}{2}e^{-j\omega}\right) \left(1-\frac{1}{4}e^{-j\omega}\right)} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\ &= \frac{-4}{1-\frac{1}{2}e^{-j\omega}} + \frac{8}{1-\frac{1}{4}e^{-j\omega}} - \frac{2}{\left(1-\frac{1}{4}e^{-j\omega}\right)^2} \\ y[n] &= -4\left(\frac{1}{4}\right)^n u[n]+8\left(\frac{1}{2}\right)^n u[n] -2(n+1)\left(\frac{1}{4}\right)^n u[n] \end{align} $

Lecture 24

Lecture 25

Here are my lecture notes from ECE301 you can download both files from my dropbox account by Prof. Boutin

There are bound to be a few errors in the document, if you find them please let me know and I'll fix them ASAP.

Lecture.pdf contains all lectures after lecture 5.

Lecture.pdf

Lecture.tex

Lecture5.pdf

Lectures 1 - 4