Revision as of 07:34, 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

Section 7.1 Sampling \ Representing a CT signal by Sampling

What is sampling? The process of measuring a CT signal is to take samples ( at the same interval for ECE301 ) so you get a vector of values that fit the equation $ x(nT) $ where n is an int and the distance between the samples is T. This makes a DT signal $ x_d[n] = x(nT) $

Problem: given $ x_d[n] $ can you reconstruct $ x(t) $? In general no. However, we can approximate $ x(t) $ by interpolation values of $ x(t) $ between the samples.

Ex 1: Interpolation by using step function

$ x(t) = \sum_{k =-\infty}^{\infty} x(kt)\left( u(t-kT) - u(t-(k+1)T) \right) $

Ex 2: Interpolation by using piecewise linear function

Take take the current point and the next one and find a line that will connect the two

Conclusion: There are infanately many ways to reconstruct the signal from samples and get an approximation. All of these could have been the initial $ x(t) $. So it is impossible to reconstruct $ x(t) $ from samples.

Sampling Theorem

1. Let $ \omega_m $ be a non-neg number
2. Let $ x(t) $ be a signal such that $ \mathcal{X}(\omega) = 0 $ when $ |\omega| > \omega_m $ (signal is band limited to $ -\omega_m \leq \omega \leq \omega_m $ )

If $ T < \frac{1}{2}\left( \frac{2\pi}{\omega_m} \right) $ then $ x(t) $ can be uniquely recovered from its samples

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