User:GabrielaV

Introduction

Do you want to know how to contact me or find out some interesting things about me? [[1]]

Signals & Systems

Example

Find the first two orthogonormal polynomials on the interval [-1,1]

1. What is orthogonormal? [2]

2. What is orthogonal? [3]

3. What is a polynomial? [4]

        $\mathbf{a}$
        $\mathbf{b}t+c$

4. Now we can find the values for the unknown variables.

$<a|a>=\underset{-1}{\overset{1}{\int }}aadt=1$

$\mathbf{a}=\sqrt{\frac{1}{2}}$

$<bt+c|a>=\underset{-1}{\overset{1}{\int }}a(bt+c)dt=0$

$\mathbf{c}=0$

$<bt+c|bt+c>=\underset{-1}{\overset{1}{\int }}(bt+c{)}^{2}dt=1$

$\mathbf{b}=\sqrt{(}\frac{3}{2})$

5. Now that we know what the first two orthogonormal polynomials!

Fourier Transform

As previously discussed, Fourier series is an expansion of a periodic function therefore we can not use it to transform a non-periodic funciton from time to the frequency domain. Fortunately the Fourier transform allows for the transformation to be done on a non-periodic function.

In order to understand the relationship between a non-periodic function and it's counterpart we must go back to Fourier series. Remember the complex exponential signal? [5]

$x(t)=x(t+T)=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\alpha }_k{e}^{\frac{j2\pi kt}{T}}$

where

${\mathit{\alpha }}_k=1/T\underset{-\frac{T}{2}}{\overset{\frac{T}{2}}{\int }}x(u)e^{\frac{-j2\pi ku}{T}}du$

If we let

$\mathbf{T}\mathbf{\to }\mathrm{\infty }$

The summation becomes integration, the harmoinic frequency becomes a continuous frequency, and the incremental spacing becomes a differential separation.

$\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\to \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}$

$\frac{k}{T}\to f$

$1/T\to df$

The result is

$\underset{T\to \mathrm{\infty }}{\lim }=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}[\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x(u)e^{-j2\pi fu}du]e^{j2\pi ft}df$

The term in the brackets is the Fourier transfrom of x(t)

${ℱ}[x(t)]=X(f)$

Inverse Fourier transform

$x(t)={{ℱ}}^{-1}[X(f)]$