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.

$<a|a>=\int _{-1}^{1}aadt=1$

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

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

${\mathbf {c}}=0$

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

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

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]

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

where

${\mathbf {\alpha }}_{k}={1/T}\int _{-{T \over 2}}^{T \over 2}x(u)e^{-j2\pi ku \over T}du$

If we let

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

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

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

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

${1/T}\to df$

The result is

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

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

${\mathcal {F}}[x(t)]=X(f)$

Inverse Fourier transform

$x(t)={\mathcal {F}}^{-1}[X(f)]$

How a CD Player Works

The first step on how a CD player works is that it reads $\sum _{n=-\infty }^{\infty }\ x(nt)\delta (t-nT)$ from the CD.

The data then goes through the Digital to Analog Converter and it is convolved with $\ p(t)$.

[[6]]

The result is $\sum _{n=-\infty }^{\infty }\ x(nt)\ p(t-nT)$ [[7]]

As you can see in the Frequency domain the final result does not appear to look like the original signal. Therefore we pass $P(f)\cdot frac{1}{T}\sum _{m=-infty}^{\infty }X(f-{\frac {m}{T}}$