Class Wiki - User contributions [en]

Adaptive Filter Application by harrde

2007-12-13T18:21:57Z

Harrde: /* Solution */

[[User:Harrde|Back to my page]]

==Problem Statement==
Come up with your own application and do a simulation in MATLAB.

==Solution==
Suppose that you want to record bird calls so that you can listen to them later and learn to identify birds only by their call. The problem is that you live by a huge freeway and all the noise from the cars and trucks makes it hard for you to get a good recording.
The MATLAB code below can take a recorded bird call and remove the noise. This lets you get good bird call recordings.

<pre>
% Bird call recording simulation where we filter out noise
clear all

% This is the bird call that we want to record.
load chirp;
s = y';
Ls = length(y);

% Creating noise
T = 1/Fs;
t = 0:T:3;
noise = sin(2*pi*100*t)+cos(2*pi*200*t);
n = noise;

bn = 3*[0 0 0 0 0 0 0 0 0 0 0 0 0 0 .5 1 .5]; %This amplifies noise
an = [1];
nf = filter(bn,an,n);

y = s(1:length(s)) + n(1:length(s)); %Here is the noise plus bird call

wavplay(y,Fs,'sync') %Here's what the bird call plus noise sounds like

N = 20; % Length of adaptive filter

% LMS algorithm for adaptive noise cancellation
h = zeros(N,1);
mu = 1/(10*N*var(n));

for k=N:Ls
xk = n(k:-1:(k-N+1));
nhat(k) = h'*xk';
e(k) = - y(k) + nhat(k);
h = h - mu*e(k)*xk';
end

% Here is the cleaned signal with beautiful bird call;
wavplay(e,Fs)

</pre>

Run this code on a computer with speakers or headphones. First you will hear the original recorded bird call with all the noise. Then you will hear the bird call after the noise has been filtered out by an adaptive filter.

User:Harrde

2007-12-13T06:58:14Z

Harrde: /* Homework */

*[[Signals and systems|Back to Signals and Systems]]
=Denver Harris' Page=

==Interests==
I have always found too many things in life to occupy my time. I want to know what it's like to run out of things to do and therefore be bored.

It seems there is never enough time to specialize in anything. "Jack of all trades, master of none" seems to be a description of me.

==Info==
*Senior
*Electrical Engineering
*Married
*harrde@gmail.com

==Homework==
H4 - [[FT|Fourier Transforms]]

H8 - [[Samping by harrde|Samping]]

H11 - [[Alias|Aliasing]]

H12 - [[DFT by harrde|DFT]]

H13 - [[DFT Exploration by harrde|DFT Exploration]]

H14 - [[Wiki Errors by harrde|Wiki Errors]]

H15 - [[Oversample and Predistort by harrde|Oversample and Predistort]]

H16 - [[Adaptive Filter Application by harrde|Adaptive Filter Application]]

H17 - [[Sigma Delta A/D Converter by harrde|Sigma Delta A/D Converter]]

Oversample and Predistort by harrde

2007-12-07T23:07:49Z

Harrde: /* Explanation */

[[User:Harrde|Back to my page]]

==Problem Statement==
Make a MATLAB script to do four times oversampling and filter so as to eliminate as much as possible the effect of the D/A converter that follows the interpolating filter.

==Solution==
MATLAB script and results:
<pre>
f = 10; % Sampling freq.
T = 1/f; % Sampling period
N = 10; % Number of sampling points
t=0:T:(N-1)*T;
x=sin(2*pi*t)+cos(2*pi*2*t); % Signal that is sampled
t_original = 0:.01:1;
x_original = sin(2*pi*t_original)+cos(2*pi*2*t_original);

OSample = 4; % How many times to oversampling

figure(1);
clf
plot(t_original,x_original,'r'); %Original signal sampled
hold on
stem(t,x)
xlabel('Time(s)');
ylabel('x(t)');
title('Orignal Data and Sampled Points')
hold
legend('Original Data','Sampled Points')

figure(2);
N1 = N*OSample;
X = fft(x);
X1 = (N1+N)/N*[X(1:N/2),zeros(1,N1),X(N/2+1:N)]; %This is the interpolation transfer function
x1 = ifft(X1);
t1=0:N/(N+N1)*T:(N-1/(N1+N))*T;
plot(t1,x1,'bo') %Sampled data with interpolation points
title('Sampled Data After Interpolation');
ylabel('x(t)');
xlabel('Time(s)');

%Now we will work on predistortion filter
f = linspace(-1/T,1/T,length(X1));
P = sin(pi*f*T)./(pi*f); %Transfer function of A/D distortion
figure(3)
plot(f,P) % A/D distortion transfer function
title('A/D Distortion Transfer Function');
ylabel('P(f)');
xlabel('f(Hz)');

Out_unfiltered = X1.*P;
out_unfiltered = ifft(Out_unfiltered);

Out = X1.*(1./P).*P;
out = ifft(Out);

figure(4)
clf
plot(f./10,real(out_unfiltered)) %This is the output with out filter/predistortion
title('Unfiltered Output');
ylabel('x(t)');
xlabel('time(s)');

figure(5)
clf
plot(f./10,real(out)) %This is the output with the filter
title('Filtered/Predistorted Output');
ylabel('x(t)');
xlabel('time(s)');
</pre>

[[Image:DH15_1.jpg]]
[[Image:DH15_2.jpg]]
[[Image:DH15_3.jpg]]
[[Image:DH15_4.jpg]]
[[Image:DH15_5.jpg]]

===Explanation===
The five graphs show a signal at important times as it passes through this system; which is similar to a signal being stored on a CD and then played back. The first graph shows the original signal and the data gathered from sampling. The next graph shows the sampled data plus the interpolation points found through 4 times oversampling. The third graph is a depiction of the transfer function of the D/A converter. This transfer function distorts the output signal as seen in the fourth graph. This is why we pass the signal through a digital filter to predistort it so that at the output of our system it will not be distorted. The filter's transfer function is simply the reciprocal of the D/A converter's transfer function.

File:DH15 5.jpg

2007-12-07T22:50:09Z

Harrde:

File:DH15 4.jpg

2007-12-07T22:49:59Z

Harrde:

File:DH15 3.jpg

2007-12-07T22:49:48Z

Harrde:

File:DH15 2.jpg

2007-12-07T22:49:33Z

Harrde:

File:DH15 1.jpg

2007-12-07T22:49:23Z

Harrde:

Oversample and Predistort by harrde

2007-12-07T22:46:48Z

Harrde:

Sigma Delta A/D Converter by harrde

2007-12-07T07:42:09Z

Harrde: New page: Back to my page ==Problem Statement== In MATLAB simulate a Sigma Delta A/D Converter. ==Solution== MATLAB script and results:

[[User:Harrde|Back to my page]]

==Problem Statement==
In MATLAB simulate a Sigma Delta A/D Converter.

==Solution==
MATLAB script and results:

Adaptive Filter Application by harrde

2007-12-07T07:41:02Z

Harrde: New page: Back to my page ==Problem Statement== Come up with your own application and do a simulation in MATLAB. ==Solution==

[[User:Harrde|Back to my page]]

==Problem Statement==
Come up with your own application and do a simulation in MATLAB.

==Solution==

Oversample and Predistort by harrde

2007-12-07T07:40:26Z

Harrde: New page: Back to my page ==Problem Statement== Make a MATLAB script to do four times oversampling and filter so as to eliminate as much as possible the effect of the D/A converter ...

User:Harrde

2007-12-07T07:33:23Z

Harrde: /* Homework */

*[[Signals and systems|Back to Signals and Systems]]
=Denver Harris' Page=

==Interests==
I have always found too many things in life to occupy my time. I want to know what it's like to run out of things to do and therefore be bored.

It seems there is never enough time to specialize in anything. "Jack of all trades, master of none" seems to be a description of me.

==Info==
*Senior
*Electrical Engineering
*Married
*harrde@gmail.com

==Homework==
H5 - [[FT|Fourier Transforms]]

H8 - [[Samping by harrde|Samping]]

H11 - [[Alias|Aliasing]]

H12 - [[DFT by harrde|DFT]]

H13 - [[DFT Exploration by harrde|DFT Exploration]]

H14 - [[Wiki Errors by harrde|Wiki Errors]]

H15 - [[Oversample and Predistort by harrde|Oversample and Predistort]]

H16 - [[Adaptive Filter Application by harrde|Adaptive Filter Application]]

H17 - [[Sigma Delta A/D Converter by harrde|Sigma Delta A/D Converter]]

File:DH12 3.jpg

2007-12-07T07:21:05Z

Harrde:

File:DH12 2.jpg

2007-12-07T07:20:56Z

Harrde:

File:DH12 1.jpg

2007-12-07T07:20:44Z

Harrde:

DFT by harrde

2007-12-07T07:19:20Z

Harrde:

[[User:Harrde|Back to my page]]

===Problem Statement===
Use MATLAB, OCTAVE, or SCILAB to show how the DFT is related to the actual Fourier Transform.

===Solution===

MATLAB code and results:
<pre>
f = 5; % Sampling freq.
T = 1/f; % Sampling period
t = 0:.01:5;
N2 = 100; % Number of sampling points
N3 = 30;
t2 = 0:T:N2*T;
t3 = 0:T:N3*T;

x = sin(2*pi*t); % Signal that is sampled
x2 = sin(2*pi*t2);

X2 = fft(x2); % DFT of signal

figure(1) %Original signal with sampling
plot(t(1:300), x(1:300), 'r')
hold on
stem(t2(1:15), x2(1:15));
hold off;
xlabel('Time (s)')
ylabel('x(t)')
title('Input Signal With Sampling')
legend('Original signal', 'Sampled signal');

figure(2) %DFT
plot(t2/(N2*T*T),abs(X2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT')

figure(3) % Shifted DFT
XS2=fftshift(X2);
f2=-1/(2*T):1/(N2*T):1/(2*T);
plot(f2,abs(XS2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT')
</pre>
[[Image:DH12_1.jpg]]
[[Image:DH12_2.jpg]]
[[Image:DH12_3.jpg]]

===Explaination===
From the graphs above we see that the DFT is similar to the real Fourier Transform but is only an approximation. The actual Fourier Transform of an input signal with a single frequency F would simply have impulses of infinite height at the positive and negative frequency F. But as seen in this example the DFT only gives impulse like "spikes" that have a height of about 48. Furthermore, without shifting the DFT the "spike" on the left is at the right frequency but the one on the left is not. But as shown, a simple shift takes care of this latter problem.

DFT Exploration by harrde

2007-12-07T07:10:39Z

Harrde: /* Explanation */

[[User:Harrde|Back to my page]]

==Problem Statement==
Sample <math> sin( 2 \pi nt) </math> at 3Hz, take the DFT, and explain the results.

==Solution==
Here is the MATLAB code and resulting figures:
<pre>
f = 3; % Sampling freq.
T = 1/f; % Sampling period
t = 0:.01:5;
N2 = 500; % Number of sampling points
N3 = 30;
t2 = 0:T:N2*T;
t3 = 0:T:N3*T;

x = sin(2*pi*t); % Signal that is sampled
x2 = sin(2*pi*t2);
x3 = sin(2*pi*t3);

X2 = fft(x2); % DFT of long signal
X3 = fft(x3); % DFT of short signal

figure(1) %Original signal
plot(t(1:500),x(1:500))
xlabel('Time (s)')
ylabel('x(t)')
title('Original Input Signal')

figure(2)
plot(t2(1:15),x2(1:15)) % Sampled signal
xlabel('Time (s)')
ylabel('x(t)')
title('Sampled Input Signal')

figure(3) %DFT of long signal
plot(t2/(N2*T*T),abs(X2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 500 Samples')

figure(4) % DFT of short signal
plot(t3/(N3*T*T),abs(X3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 30 Samples')

figure(5) % Shifted DFT of long signal
XS2=fftshift(X2);
f2=-1/(2*T):1/(N2*T):1/(2*T);
plot(f2,abs(XS2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 500 Samples')

figure(6) % Shifted DFT of short signal
XS3=fftshift(X3);
f3=-1/(2*T):1/(N3*T):1/(2*T);
plot(f3,abs(XS3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 30 Samples')
</pre>

[[Image:DH13_1.jpg]]
[[Image:DH13_2.jpg]]
[[Image:DH13_3.jpg]]
[[Image:DH13_4.jpg]]
[[Image:DH13_5.jpg]]
[[Image:DH13_6.jpg]]

===Explanation===
The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points.
From the example above we see that the more sampling points we take the closer the DFT resembles the actual transform. e.g. with 30 sample points the DFT has "spikes" with height of about 12 but with 500 sample points they have a height of 250, which is obviously closer the expected Dirac delta funtions with infinite height.

DFT Exploration by harrde

2007-12-07T06:54:17Z

Harrde:

[[User:Harrde|Back to my page]]

==Problem Statement==
Sample <math> sin( 2 \pi nt) </math> at 3Hz, take the DFT, and explain the results.

==Solution==
Here is the MATLAB code and resulting figures:
<pre>
f = 3; % Sampling freq.
T = 1/f; % Sampling period
t = 0:.01:5;
N2 = 500; % Number of sampling points
N3 = 30;
t2 = 0:T:N2*T;
t3 = 0:T:N3*T;

x = sin(2*pi*t); % Signal that is sampled
x2 = sin(2*pi*t2);
x3 = sin(2*pi*t3);

X2 = fft(x2); % DFT of long signal
X3 = fft(x3); % DFT of short signal

figure(1) %Original signal
plot(t(1:500),x(1:500))
xlabel('Time (s)')
ylabel('x(t)')
title('Original Input Signal')

figure(2)
plot(t2(1:15),x2(1:15)) % Sampled signal
xlabel('Time (s)')
ylabel('x(t)')
title('Sampled Input Signal')

figure(3) %DFT of long signal
plot(t2/(N2*T*T),abs(X2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 500 Samples')

figure(4) % DFT of short signal
plot(t3/(N3*T*T),abs(X3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('DFT of 30 Samples')

figure(5) % Shifted DFT of long signal
XS2=fftshift(X2);
f2=-1/(2*T):1/(N2*T):1/(2*T);
plot(f2,abs(XS2))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 500 Samples')

figure(6) % Shifted DFT of short signal
XS3=fftshift(X3);
f3=-1/(2*T):1/(N3*T):1/(2*T);
plot(f3,abs(XS3))
xlabel('Frequency (s)')
ylabel('X(F)')
title('Shifted DFT of 30 Samples')
</pre>

[[Image:DH13_1.jpg]]
[[Image:DH13_2.jpg]]
[[Image:DH13_3.jpg]]
[[Image:DH13_4.jpg]]
[[Image:DH13_5.jpg]]
[[Image:DH13_6.jpg]]

===Explanation===
The DFT only gives us an approximation of the actual Fourier Transform, as was found in the last homework. With few sampling points the DFT hardly resembles the expected trasfrom. This is because the edge effects have a greater affect with fewer sample points. But as we increase the number of sample points the DFT more closely resembles the expected transform.

File:DH13 6.jpg

2007-12-07T06:40:41Z

Harrde:

File:DH13 5.jpg

2007-12-07T06:40:30Z

Harrde:

File:DH13 4.jpg

2007-12-07T06:40:18Z

Harrde:

File:DH13 3.jpg

2007-12-07T06:40:01Z

Harrde:

File:DH13 2.jpg

2007-12-07T06:39:48Z

Harrde:

File:DH13 1.jpg

2007-12-07T06:39:29Z

Harrde:

DFT Exploration by harrde

2007-12-07T06:38:45Z

Harrde: /* Solution */

DFT Exploration by harrde

2007-12-07T06:35:23Z

Harrde:

DFT Exploration by harrde

2007-12-06T00:24:50Z

Harrde:

[[User:Harrde|Back to my page]]

==Problem Statement==
Sample <math> sin( 2 \pi nt) </math> at 3Hz, take the DFT, and explain the results.

==Solution==

DFT by harrde

2007-12-03T01:12:35Z

Harrde:

[[User:Harrde|Back to my page]]

===Problem Statement===
Use MATLAB, OCTAVE, or SCILAB to show how the DFT is related to the actual Fourier Transform.

===Solution===

DFT by harrde

2007-12-03T01:12:20Z

Harrde:

[[User:Harrde|Back to my page]]

===Problem Statement==
Use MATLAB, OCTAVE, or SCILAB to show how the DFT is related to the actual Fourier Transform.

===Solution===

Samping by harrde

2007-11-13T04:20:50Z

Harrde: /* How to Sample */

[[User:Harrde|Back to my page]]

==Sampling==
Sampling is the process by which an anolog signal is converted into a digital signal and fed into digital circuit. Sampling is done by measuring the analog signal at a regular time interval. Each measurement of the anolog signal is called a sample. We can store and easily manupulate these samples in a computer and then interpolate between the sample points to fully reconstruct the original analog signal when we want to send the signal out of the computer. But we can only reconstruct orignal signal if we have sampled at a rate greater than twice the highest frequency component of the anolog signal. This restriction is given by the Nyquist Theorem which states:

===Nyquist Theorem===
A band-limited input signal can be recovered without distortion if it is sampled at a rate of at least twice the highest frequency component of interest within the signal.

==How to Sample==
For a given input signal x(t)
[[Image:InputSignal.jpg]]
:we want to sample this at time intervals T. This can be done by multiplying x(t) with a series of impulses
[[Image:ImpulseString.jpg]]
:given by <math>\sum_{n=-\infty}^{\infty} \delta(t - nT) \ </math>.

The product gives us
[[Image:ResultSignal.jpg]]
which is represented by: <math>\sum_{n=-\infty}^{\infty} x(nT) \delta(t - nT) \ </math>

Now we have a digital representation of the signal in a computer as long as the signal is finite in length. The "height" or magnitude of each sample is given by the coeficients x(nT) of each delta funtion.

File:ResultSignal.jpg

2007-11-13T04:15:03Z

Harrde:

Samping by harrde

2007-11-13T04:14:48Z

Harrde: /* How to Sample */

Samping by harrde

2007-11-13T03:36:54Z

Harrde: /* How to Sample */

File:ImpulseString.jpg

2007-11-13T03:36:21Z

Harrde:

File:InputSignal.jpg

2007-11-13T03:35:59Z

Harrde:

Samping by harrde

2007-11-13T03:17:23Z

Harrde:

[[User:Harrde|Back to my page]]

==Sampling==
Sampling is the process by which an anolog signal is converted into a digital signal and fed into digital circuit. Sampling is done by measuring the analog signal at a regular time interval. Each measurement of the anolog signal is called a sample. We can store and easily manupulate these samples in a computer and then interpolate between the sample points to fully reconstruct the original analog signal when we want to send the signal out of the computer. But we can only reconstruct orignal signal if we have sampled at a rate greater than twice the highest frequency component of the anolog signal. This restriction is given by the Nyquist Theorem which states:

===Nyquist Theorem===
A band-limited input signal can be recovered without distortion if it is sampled at a rate of at least twice the highest frequency component of interest within the signal.

==How to Sample==
For a given input signal x(t)
[[Image:InputSignal.jpg]]
we want to sample this at time intervals T. This can be done by convolving x(t) with a series of impulses
[[Image:ImpulseString.jpg]]
given by <math>\int_{-\infty}^\infty \delta(x) \, dx = 1.</math>.

DFT Exploration by harrde

2007-11-13T01:43:53Z

Harrde: New page: Back to my page

[[User:Harrde|Back to my page]]

DFT by harrde

2007-11-13T01:43:42Z

Harrde: New page: Back to my page

[[User:Harrde|Back to my page]]

Samping by harrde

2007-11-13T01:43:31Z

Harrde: New page: Back to my page

[[User:Harrde|Back to my page]]

Wiki Errors by harrde

2007-11-13T01:43:20Z

Harrde:

[[User:Harrde|Back to my page]]

===Problem Statement===
Read Wiki and find three things to fix

===Things I fixed===
Changed variable u to t in GabrielaV's Fourier Transform section to make it more clear.

Changed variable u to t in my Fourier Transform section to make it correct.

Fixed picture for Andrew Lopez.

Wiki Errors by harrde

2007-11-13T01:38:41Z

Harrde:

===Problem Statement===
Read Wiki and find three things to fix

===Things I fixed===
Changed variable u to t in GabrielaV's Fourier Transform section to make it more clear.

Changed variable u to t in my Fourier Transform section to make it correct.

Fixed picture for Andrew Lopez.

File:AndrewLopez.jpg

2007-11-13T01:37:50Z

Harrde:

User:Andrew

2007-11-13T01:37:36Z

Harrde: /* The Comatose Stare */

==The Comatose Stare==
[[Image:AndrewLopez.jpg]]

'''Andrew - Origin: Greek, French | Meaning: Manly, Valiant, Courageous'''

For more information about me click here --- > [http://www.mask.wwc.edu/profile/show/826 Mask]

Add Me to your [http://www.myspace.com/andrewlop Myspace]

FT

2007-11-13T01:33:48Z

Harrde: /* Fourier Transform Derivation */

[[User:Harrde|Back to my page]]

==Introduction==
'''Why''' do we want to find the fourier transform function? This question should be answered first so that we can have motivation for learning '''how''' to perform this transformation. Once we see why we want the learn about the fourier transform it will be useful to work through the derivation of the transform. This will give us a greater understanding of the transform and greatly increase the chances that we will remember how to use it in the future.

==Why==
The reasoning behind performing the fourier transform is fairly simple compared to actually performing the transform. We want this transform so that for any given real signal x(t) we can easily see the frequency and amplititude of each sine and cosine component. These components when added together using the principle of superposition sum to exactly x(t). This transform is very useful in many applications from radio signal processing to image processing, and this is '''why''' we want to learn about it.

==How==
===Fourier Series===
The fourier transform is a generalization of the fourier series. The fourier series is given by
<math> \sum_{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math>, where <math> \alpha_n = 1/T \int_{-T/2} ^{T/2} x(t) e^{-j2 \pi nt/T} dt </math>, and can be used to represent periodic functions.

===Fourier Transform Derivation===
Now from the Fourier Series we want to derive the Fourier Transform, which is very similar except that it can be used for nonperiodic funtions. This can be done by simply letting the period <math> T </math> go to infinity.

So for <math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math>

We'll let:

<math> T </math> -> <math> \infty </math>

<math> n/T </math> -> <math> f </math>

<math> 1/T </math> -> <math> df </math>

<math> \sum _{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math> -> <math> \int_{-\infty}^\infty \alpha_n e^{j2 \pi ft}df </math>

<math> \alpha_n </math> -> <math> X(f) = \int_{-\infty} ^{\infty} x(t) e^{-j2 \pi ft} dt </math>

Now we have:

<math> x(t) = \int_{-\infty}^\infty (\int_{-\infty} ^{\infty} x(u) e^{-j2 \pi ft} du) e^{j2 \pi ft}df </math> -> <math> x(t) = \mathcal{F}^{-1}[X(f)] </math>

Where <math> \int_{-\infty} ^{\infty} x(u) e^{-j2 \pi ft} du = X(f) = \mathcal{F}[x(t)] </math>

and

<math> \int_{-\infty}^\infty X(f) e^{j2 \pi ft}df = x(t) = \mathcal{F}^{-1}[X(f)] </math>

===Some Fourier Transform Identities===

=====Differentiation=====
<math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math>

=====Time Shift=====
<math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math>

<math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math>

<math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math>

FT

2007-11-13T01:33:17Z

Harrde: /* Fourier Transform Derivation */

[[User:Harrde|Back to my page]]

==Introduction==
'''Why''' do we want to find the fourier transform function? This question should be answered first so that we can have motivation for learning '''how''' to perform this transformation. Once we see why we want the learn about the fourier transform it will be useful to work through the derivation of the transform. This will give us a greater understanding of the transform and greatly increase the chances that we will remember how to use it in the future.

==Why==
The reasoning behind performing the fourier transform is fairly simple compared to actually performing the transform. We want this transform so that for any given real signal x(t) we can easily see the frequency and amplititude of each sine and cosine component. These components when added together using the principle of superposition sum to exactly x(t). This transform is very useful in many applications from radio signal processing to image processing, and this is '''why''' we want to learn about it.

==How==
===Fourier Series===
The fourier transform is a generalization of the fourier series. The fourier series is given by
<math> \sum_{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math>, where <math> \alpha_n = 1/T \int_{-T/2} ^{T/2} x(t) e^{-j2 \pi nt/T} dt </math>, and can be used to represent periodic functions.

===Fourier Transform Derivation===
Now from the Fourier Series we want to derive the Fourier Transform, which is very similar except that it can be used for nonperiodic funtions. This can be done by simply letting the period <math> T </math> go to infinity.

So for <math> x(t) = \sum_{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math>

We'll let:

<math> T </math> -> <math> \infty </math>

<math> n/T </math> -> <math> f </math>

<math> 1/T </math> -> <math> df </math>

<math> \sum _{n=-\infty}^\infty \alpha_n e^{j2 \pi nt/T} </math> -> <math> \int_{-\infty}^\infty \alpha_n e^{j2 \pi ft}df </math>

<math> \alpha_n </math> -> <math> X(f) = \int_{-\infty} ^{\infty} x(u) e^{-j2 \pi ft} dt </math>

Now we have:

<math> x(t) = \int_{-\infty}^\infty (\int_{-\infty} ^{\infty} x(u) e^{-j2 \pi ft} du) e^{j2 \pi ft}df </math> -> <math> x(t) = \mathcal{F}^{-1}[X(f)] </math>

Where <math> \int_{-\infty} ^{\infty} x(u) e^{-j2 \pi ft} du = X(f) = \mathcal{F}[x(t)] </math>

and

<math> \int_{-\infty}^\infty X(f) e^{j2 \pi ft}df = x(t) = \mathcal{F}^{-1}[X(f)] </math>

===Some Fourier Transform Identities===

=====Differentiation=====
<math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math>

=====Time Shift=====
<math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math>

<math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math>

<math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math>

Wiki Errors by harrde

2007-11-11T21:54:50Z

Harrde: New page: ===Problem Statement=== Read Wiki and find three things to fix ===Things I fixed===

===Problem Statement===
Read Wiki and find three things to fix

===Things I fixed===

User:GabrielaV

2007-11-11T21:49:03Z

Harrde: /* Fourier Transform */

== Welcome to Gabriela's Wiki page ==

==Introduction==

Do you want to know how to contact me or find out some interesting
things about me?
[[http://mask.wwc.edu/2005-2006/person.html?PersonId=20051034]]

==Signals & Systems==

====Example====

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

1. What is orthonormal?
[http://cubex.wwc.edu/~frohro/wiki/index.php/Orthogonal_functions#Normalization]

2. What is orthogonal?
[http://cubex.wwc.edu/~frohro/wiki/index.php/Orthogonal_functions#Orthogonality_for_functions]

3. What is a polynomial?
[http://en.wikipedia.org/w/index.php?title=Polynomial&redirect=no]
<math>\bold a </math>
<math>\bold bt+c </math>

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

::<math> <a|a>= \int_{-1}^1 a a dt =1 </math>
::<math>\bold a = \sqrt{\frac {1}{2}} </math>

::<math> <bt+c|a>= \int_{-1}^1 a (bt+c) dt =0 </math>
::<math>\bold c=0 </math>

::<math> <bt+c|bt+c>= \int_{-1}^1 (bt+c)^2 dt=1 </math>
::<math>\bold a = \sqrt{\frac {3}{2}} </math>

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?
[http://cubex.wwc.edu/~frohro/wiki/index.php/Fourier_series#Periodic_Functions]

<math> x(t)=x(t+T)=\sum_{k= -\infty}^ \infty \alpha_k e^ \frac{j 2 \pi k t}{T}</math>

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

If we let
::<math>\bold {T\to\infty}</math>
The summation becomes integration, the harmoinic frequency becomes a continuous frequency, and the incremental spacing becomes a differential separation.
::<math> \sum_{k=-\infty}^\infty\to\int_{-\infty}^\infty</math>
::<math> {{k\over T}\to f}</math>
::<math> {1/T}\to df</math>

The result is
::<math>\lim_{T\to\infty}=\int_{-\infty}^\infty {\left[\int_{-\infty}^\infty x(u)e^{-j2\pi fu}du\right]} e^{j2\pi ft}df</math>

The term in the brackets is the ''Fourier transfrom'' of ''x(t)''
::<math>
\mathcal{F}[x(t)]=X(f)
</math>

Inverse ''Fourier transform''
::<math>x(t)=\mathcal{F}^{-1}[X(f)]</math>

==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
<math>\sum_{n=-\infty}^\infty \ x(nt) \delta (t-nT)</math>
The data then goes through the Digital to Analog Converter and it is convolved with
<math>\ p(t) </math> ( See figure below)

[[Image:barnsaDA.jpg|Digital to analog conversion]]The result is <math>\sum_{n=-\infty}^\infty \ x(nt) \ p(t-nT) </math>

As you can see in the Frequency domain the final result does not appear to look like the original signal.
Therefore we pass <math> P(f)\cdot \frac{1} {T}\sum_{m=-\infty}^\infty X(f-\frac{m} {T}) </math> 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 <math> \sum_{k= -\infty}^ \infty \ x(nT) \delta (t-nT) </math> in the time domain and we convolve it with <math> \sum_{m= -M}^ M h(\frac {mT}{2}) \delta (t-\frac {mT}{2}) </math>
[[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]] which equals <math>\sum_{k= -\infty}^ \infty \ x(nT) h(\frac {mT}{2}) \delta (t-nT- \frac{mT}{2})</math>
[[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]
If we let <math> (\frac{l}{2}) = (n +(\frac{m}{2}))</math> therefore, <math> n = (\frac{l-m}{2}) </math> which makes <math> \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)</math>

==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.
<math>h(mT) = T \int_{-f}^f \hat H (f) e^{j 2 \pi f m t} df </math>
is the coefficient that are sent out of the D/A converter.
H(f) is the frequency response which mathmatically is <math> H(f)=\sum_{m=-M}^{M}h(mT)e^{-j 2 \pi f m T} </math>

===FIR Application===
Design an FIR low pass filter to pass between <math> -\frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.
<math> H_{hat} = 1 \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math>
:<math> 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}</math>

:<math> = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math>

The actual frequency response is
<math>\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} </math>

[https://fweb.wwc.edu/index.php/Image:Frequency_Response_Pict.png] shows the example in Matlab

==Adaptive Filter==
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.
[[Image:Afilter.jpg]]
===Adaptive Filter Application===
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 <math> h_{n}(k) </math> 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.

[http://www.wwc.edu/~frohro/ClassNotes/ENGR455/2005/Keystone/thumb/80020051130KeyPB300080.jpg]

User:Harrde

2007-11-11T21:33:48Z

Harrde:

*[[Signals and systems|Back to Signals and Systems]]
=Denver Harris' Page=

==Interests==
I have always found too many things in life to occupy my time. I want to know what it's like to run out of things to do and therefore be bored.

It seems there is never enough time to specialize in anything. "Jack of all trades, master of none" seems to be a description of me.

==Info==
*Senior
*Electrical Engineering
*Married
*harrde@gmail.com

==Homework==
H5 - [[FT|Fourier Transforms]]

H8 - [[Samping by harrde|Samping]]

H11 - [[Alias|Aliasing]]

H12 - [[DFT by harrde|DFT]]

H13 - [[DFT Exploration by harrde|DFT Exploration]]

H14 - [[Wiki Errors by harrde|Wiki Errors]]

Alias

2007-11-11T21:18:01Z

Harrde:

[[User:Harrde|Back to my page]]

==Aliasing==
The Nyquist Theorem states that you must sample at a frequency which is twice that of the highest frequency components of the sampled signal in order to recover the sampled signal. If you sample at rate lower than this, you get aliasing. Below is a figure which shows how aliasing occurs and the Matlab code for it.

[[Image:harrdeH11Fig.jpg]]

<pre>
clear all;

Ts = .2; %Sampling time (s)
ws = 2*pi/Ts; %Sampling frequency (rad/s)

t = [0:0.005:2]; %Time vector

w1 = 7; w2 = 23;
y = cos(w1*t) + cos(w2*t); %Original Signal

t1 = [0:Ts:2];
xs = cos(w1*t1) + cos(w2*t1); %Sampled points

w2s = w2 - ws;
x1 = cos(w1*t) + cos(w2s*t); %Signal from Aliasing

figure(1); %Plot
clf;
plot(t,y,'k',t,x1,'r');
hold on
stem(t1,xs);
hold off;
legend('Original Signal', 'Signal from Aliasing', 'Sampled Points');
xlabel('Time (s)');
ylabel('Amplitude');
title('Aliasing: Signal With Frequencies up to 3.66Hz Sampled at 5Hz');
</pre>

The figure above shows that if the original signal (black) is not sampled fast enough then another signal (blue) can be incorrectly "recovered" which passes through all the sampled points but is clearly not the same as the orignal signal. This incorrect signal from aliasing has correct frequency components for those components which are lower then half the sampling frequency. The components which are higher than half the sampling frequency end up with frequencies equal to the original signals' frequency components minus the sampling frequency.

It should be noted that other signals can be found that pass through each of the sampled points, but their incorrect frequency components will all have frequencies higher than the original signals' highest frequency components. Thus, they can be easily filter out by a low pass filter.