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Introduction

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Signals & Systems

Example

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

1. What is orthonormal? [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.

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

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

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

$\mathbf{c}=0$

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

$\mathbf{a}=\sqrt{\frac{3}{2}}$

5. Now that we know what the first two orthonormal 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. The Fourier transform allows to change a function from the time domain to frequency domain or the inverse fourier transform from frequency domain to time domain.

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

$x(t)={{ℱ}}^{-1}[X(f)]$

How a CD Player Works

The first step on how a CD player works is that it takes data from the cd that is mathmatically represented by $$\begin{gathered} \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }} \\ x(nt)\delta (t-nT) \end{gathered}$$ The data then goes through the Digital to Analog Converter and it is convolved with $p(t)$ ( See figure below)

File:BarnsaDA.jpgThe result is $$\begin{gathered} \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }} \\ x(nt) \\ p(t-nT) \end{gathered}$$

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

2x Oversampling(Interperolating FIR filter)

The benefit of using oversampling is that this allows for more samples to be taken.

We have $$\begin{gathered} \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }} \\ x(nT)\delta (t-nT) \end{gathered}$$ in the time domain and we convolve it with $\sum _{m=-M}^{M}h(\frac{mT}{2})\delta (t-\frac{mT}{2})$ File:2x01.jpg which equals $$\begin{gathered} \sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }} \\ x(nT)h(\frac{mT}{2})\delta (t-nT-\frac{mT}{2}) \end{gathered}$$ File:2x02.jpg If we let $(\frac{l}{2})=(n+(\frac{m}{2}))$ therefore, $n=(\frac{l-m}{2})$ which makes $\hat{y}(t)=\sum _{l=-\mathrm{\infty }}^{\mathrm{\infty }}\left(\sum _{m=-M}^{M}x\left(\frac{l-m}{2}T\right)h\left(\frac{mT}{2}\right)\right)\delta (t-\frac{lT}{2})$