Revision as of 08:11, 7 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) $

--- 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