# User:GabrielaV

## Introduction

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

## Signals & Systems

#### Example

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

1. What is orthonormal? 

2. What is orthogonal? 

3. What is a polynomial? $\bold a$ $\bold bt+c$


4. Now we can find the values for the unknown variables. $= \int_{-1}^1 a a dt =1$ $\bold a = \sqrt{\frac {1}{2}}$ $= \int_{-1}^1 a (bt+c) dt =0$ $\bold c=0$ $= \int_{-1}^1 (bt+c)^2 dt=1$ $\bold a = \sqrt{\frac {3}{2}}$

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? $x(t)=x(t+T)=\sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T}$

where $\bold \alpha_k={1/T}\int_{-{T\over 2}}^{{T\over 2}} x(t) e^{-j2\pi kt\over T}dt$

If we let $\bold {T\to\infty}$

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 takes data from the cd that is mathmatically represented by $\sum_{n=-\infty}^\infty \ x(nt) \delta (t-nT)$ 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 $\sum_{n=-\infty}^\infty \ x(nt) \ p(t-nT)$

As you can see in the Frequency domain the final result does not appear to look like the original signal. Therefore we pass $P(f)\cdot \frac{1} {T}\sum_{m=-\infty}^\infty X(f-\frac{m} {T})$ 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 $\sum_{k= -\infty}^ \infty \ x(nT) \delta (t-nT)$ 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 $\sum_{k= -\infty}^ \infty \ x(nT) h(\frac {mT}{2}) \delta (t-nT- \frac{mT}{2})$ File:2x02.jpg If we let $(\frac{l}{2}) = (n +(\frac{m}{2}))$ therefore, $n = (\frac{l-m}{2})$ which makes $\hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)$

## FIR filter

FIR stands for finite impulse response and it is one of two digital signal filters used. The FIR has no feedback so eventually it will have new data and the old one will be thrown away. $h(mT) = T \int_{-f}^f \hat H (f) e^{j 2 \pi f m t} df$ is the coefficient that are sent out of the D/A converter. H(f) is the frequency response which mathmatically is $H(f)=\sum_{m=-M}^{M}h(mT)e^{-j 2 \pi f m T}$

### FIR Application

Design an FIR low pass filter to pass between $-\frac{1}{4T} < f < \frac{1}{4T}$ and reject the rest. $H_{hat} = 1 \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases}$ $h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}$ $= \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}$

The actual frequency response is $\sum_{m=-M}^M h(mT) e^{- j 2 \pi f m T} = \sum_{m=-M}^M \frac{ sin \left ( \frac{\pi m}{2} \right) }{\pi m} e^{- j 2 \pi f m T}$

 shows the example in Matlab

An Adaptive Fitler is useful when filtering needs to be done and the transfer function of the noise signal is not known. The adaptive filter uses feedback to filter out the noise signal. The objective is to drive the error to zero. File:Afilter.jpg

A pilot and a co-pilot are trying to communicate, but there is a problem because the microphone picks up the engine noise. What to do? Use an adaptive filter to eliminate the unwanted noise. We need two microphones.

• engine noise+voice ( unknown transfer function)
• engine noise ( microphone by the engine)

both signals will pass through a low pass filter and then through a Analog to Digital converter.

r(n)=s(n)+ nf(n) engine noise+voice
x(n) engine noise

Following the adaptive filter block diagram x(n) will go to FIR $h_{n}(k)$ and the coefficient adjust.

-r(n) and y(n) will both go to the summation.

e(n) will then go to the co-pilot's earphone and the coefficient adjust until e(n) is zero.