Class Wiki - User contributions [en]

User:Rothsa

2007-12-07T06:39:01Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
[[Image:fish.jpg |thumb|What I actually look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Religion
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*Messenger: sallyq_2@hotmail.com
*[http://www.myspace.com/rogueulette Myspace]
*[http://www.facebook.com/profile.php?id=726710413 Facebook]

........That should be enough so, if you can't get a hold of me, I'm probably dead.
==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

[[HW17_SigmaDelta|Homework #17- First Order Sigma-Delta A/D Converter]]

HW16 Application

2007-12-07T06:35:09Z

Rothsa:

==Problem Statement==

Come up with your own application. Do a simulation in Matlab, Octave, or what have you.

==Solution==

HW16 Application

2007-12-07T06:33:08Z

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

==Problem Statement==

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

==Solution==

The Nyquist theorem states: "Exact reconstruction of a continuous-time baseband signal from its samples is possible if the signal is bandlimited and the sampling frequency is greater than twice the signal bandwidth."

sin(2*\pi*t)\!

HW15 D/A Converter Compensation

2007-12-05T22:02:56Z

Rothsa:

HW15 D/A Converter Compensation

2007-12-05T22:02:06Z

Rothsa: New page: Make a MATLAB script to do four times oversampling and a filter so as to eliminate as much as possible the effect of the D/A converter that follows the interpolation filter.

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

HW13 DFT/Sampling Assignment

2007-12-05T21:19:12Z

Rothsa:

===Problem Statement===
1.Sample <math>sin(2*\pi*t)\!</math> at <math> 3 Hz;\!</math>
 2.Take the DFT
 3.Explain the results.
===Solution===
The Nyquist theorem states: "Exact reconstruction of a continuous-time baseband signal from its samples is possible if the signal is bandlimited and the sampling frequency is greater than twice the signal bandwidth."

<math>sin(2*\pi*t)\!</math>

is a sine wave with a frequency of 1 Hz. This signal is bandlimited, because it consists of a single frequency sine wave, and the requested sampling frequency, <math> 3 Hz;\!</math>
is greater than twice the signal bandwidth, as required in the Nyquist Theroem.

HW13 DFT/Sampling Assignment

2007-12-05T21:18:10Z

Rothsa:

===Problem Statement===
1.Sample <math>sin(2*\pi*t)\!</math> at <math> 3 Hz;\!</math>
 2.Take the DFT
 3.Explain the results.
===Solution===
The Nyquist theorem states: "Exact reconstruction of a continuous-time baseband signal from its samples is possible if the signal is bandlimited and the sampling frequency is greater than twice the signal bandwidth."

<math>sin(2*\pi*t)\!</math>

is a sine wave with a frequency of 1 Hz. This signal is bandlimited, because it consists of a single frequency sine wave, and the sampling frequency is greater than twice the signal bandwidth, as required in the Nyquist Theroem.

HW13 DFT/Sampling Assignment

2007-12-04T20:23:45Z

Rothsa:

===Problem Statement===
1.Sample <math>sin(2*\pi*t)\!</math> at <math> 3 Hz;\!</math>
 2.Take the DFT
 3.Explain the results.
===Solution===

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:14:33Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.
Heres the what the DFT of a 1Hz Sine Wave looks like:
::[[Image:hw12.jpg]] 
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:
::[[Image:hw12b.jpg]]
NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs that pass visual inspection, not numerical inspection.
::DFT;
::clear;
::T=.25;
::t=(0:T:10);
::N=length(t);
::f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
::x=(sin(2*pi*t));
::y=abs(fft(x))
::figure(1);
::plot(f,y);
::title(' DFT')
::xlabel('frequency');
::ylabel('magnitude)');

::CFT;
::clear;
::T=.25;
::t=(0:T:1000);
::N=length(t);
::f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
::x=(sin(2*pi*t));
::y=abs(fft(x))
::figure(1);
::plot(f,y);
::title(' CFT')
::xlabel('frequency');
::ylabel('magnitude)');

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:13:53Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.
Heres the what the DFT of a 1Hz Sine Wave looks like:
::[[Image:hw12.jpg]] 
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:
::[[Image:hw12b.jpg]]
NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs that pass visual inspection, not numerical inspection.
DFT;
::clear;
::T=.25;
::t=(0:T:10);
::N=length(t);
::f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
::x=(sin(2*pi*t));
::y=abs(fft(x))
::figure(1);
::plot(f,y);
::title(' DFT')
::xlabel('frequency');
::ylabel('magnitude)');

CFT;
::clear;
::T=.25;
::t=(0:T:1000);
::N=length(t);
::f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
::x=(sin(2*pi*t));
::y=abs(fft(x))
::figure(1);
::plot(f,y);
::title(' CFT')
::xlabel('frequency');
::ylabel('magnitude)');

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:12:18Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.
Heres the what the DFT of a 1Hz Sine Wave looks like:
::[[Image:hw12.jpg]] 
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:
::[[Image:hw12b.jpg]]
NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs that pass visual inspection, not numerical inspection.
::clear;
::T=.25;
::t=(0:T:1000);
::N=length(t);
::f = -1/(2*T):1/(N*T):1/(2*T)-1/(N*T);
::x=(sin(2*pi*t));
::y=abs(fft(x))
::figure(1);
::plot(f,y);
::title(' DFT')
::xlabel('frequncy');
::ylabel('dft of x)');

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:10:11Z

Rothsa:

File:Hw12b.jpg

2007-12-04T20:09:32Z

Rothsa:

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:09:21Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.
Heres the what the DFT of a 1Hz Sine Wave looks like:
[[Image:hw12.jpg]]
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:
[[Image:hw12b.jpg]]
NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs I wanted,
not be used for real DFT computations.

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:06:52Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.

Heres the what the DFT of a 1Hz Sine Wave looks like:
[[Image:hw12.jpg]]
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:

NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs I wanted,
not be used for real DFT computations.

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T20:06:11Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.

Heres the what the DFT of a 1Hz Sine Wave looks like:
<math>hw12.jpg</math>
Heres the what the CFT of a 1Hz Sine Wave looks like, in comparison:

NOTE: The following Matlab script is not computatively correct -
it was constructed to only get the graphs I wanted,
not be used for real DFT computations.

File:Hw12.jpg

2007-12-04T20:03:11Z

Rothsa:

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T19:36:49Z

Rothsa:

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-04T19:36:16Z

Rothsa:

===Problem Statement===
Use Matlab or Octave to show how the DFT is related to the actual Fourier Transform.

===Solution===
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

So what does this look like graphically? Well, basically, the Discrete is a bit less accurate on the frequencies it shows since by definition it is the sampled version
of the continuous. So, if a simple sine wave is a particular frequency, it's continuous fourier transform (which, by definition, is a graph of the frequencies present in a in that wave) will spike at that particular frequency. The discrete frequency will also spike, but taper off a bit more gradually.

Heres the graphs:

HW11 Aliasing Example

2007-12-03T22:47:34Z

Rothsa:

Aliasing - This is when you're sampling a function, but you don't sample often enough to reconstruct the waveform as it was before.

How often is often enough? Well, the Nyquist-Shannon sampling thereom says twice as much as the highest frequency. To be exact, the theorem states: "Exact reconstruction of a continuous-time baseband signal from its samples is possible if the signal is bandlimited and the sampling frequency is greater than twice the signal bandwidth."

So what happens when you don't sample as the Nyquist-Shannon thereom suggests? I hate Matlab, but in the self-sacrificing spirit I have towards you, my dear reader, I will use everything in my power including Matlab to clarify things.

In the following picture, I have sampled the waveform at a rate higher than it's frequency, but less than twice the frequency. As you can see, the reconstructed signal is
nowhere near the original signal.

[[Image:sally1.jpg]]

Octave Script:

frequency = 1;

sample = 1.4;

T = 1/sample %period

t = 0:0.01:3.14; %graph accuracy to .01

signal = cos(2*pi*t); %original signal

ts = 0:T:3.14;

sampledpoints = cos(2*pi*frequency*ts);

reconstructed = cos(2*pi*(frequency-sample)*t);

plot(t, signal, 'g', ts, sampledpoints, 'p',t, reconstructed, 'b')

legend('Original Signal', 'Sampled Points', 'Reconstructed Signal')

xlabel('Time')

ylabel('Amplitude')

title('Aliasing Example')

So it has the same shape, but as you can see the frequency is alot lower than the original. Basically what not sampling enough does is that it recognizes multiple frequencies as the same frequency, so you lose crucial details in the reproduction of a sound.

HW13 DFT/Sampling Assignment

2007-12-03T18:18:30Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2*\pi*t)\!</math> at <math> 3 Hz;\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:18:17Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2*\pi*t)\!</math> at <math>sin(2*\pi*t) 3 Hz;\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:17:23Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2*\pi*t) at 3 Hz;\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:16:45Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2*\pi*t)\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:16:34Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2\pi\t)\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:16:25Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2\pit)\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:16:15Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2\pi*t)\!</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:15:30Z

Rothsa:

===Problem Statement===
Sample

::<math>sin(2\pi*t)</math>

at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:14:36Z

Rothsa:

===Problem Statement===
Sample <math>sin(2\pi*t)</math> at 3 Hz; take the DFT; explain the results.
===Solution===

HW13 DFT/Sampling Assignment

2007-12-03T18:14:07Z

Rothsa:

Problem Statement: Sample <math>sin(2*pi*t)</math> at 3 Hz; take the DFT; explain the results.
===Problem Statement===
Sample <math>sin(2\pi*t)</math> at 3 Hz; take the DFT; explain the results.
===Solution===

HW12 DFT/Continuous Fourier Transform Relationship

2007-12-03T18:12:43Z

Rothsa:

HW13 DFT/Sampling Assignment

2007-12-03T18:07:15Z

Rothsa: New page: Problem Statement: Sample <math>sin(2*pi*t)</math> at 3 Hz; take the DFT; explain the results.

Problem Statement: Sample <math>sin(2*pi*t)</math> at 3 Hz; take the DFT; explain the results.

HW11 Aliasing Example

2007-11-30T22:47:21Z

Rothsa:

Aliasing - This is when you're sampling a function, but you don't sample often enough to reconstruct the waveform as it was before.

How often is often enough? Well, the Nyquist-Shannon sampling thereom says twice as much as the highest frequency. To be exact, the theorem states: "Exact reconstruction of a continuous-time baseband signal from its samples is possible if the signal is bandlimited and the sampling frequency is greater than twice the signal bandwidth."

So what happens when you sample as the Nyquist-Shannon thereom suggests, and what happens if you dont? I hate Matlab, but in the self-sacrificing spirit I have towards you, my dear reader, I will use everything in my power including Matlab to clarify things. Hence:

[[Image:sally1.jpg]]

Octave Script:

frequency = 1;

sample = 1.4;

T = 1/sample %period

t = 0:0.01:3.14; %graph accuracy to .01

signal = cos(2*pi*t); %original signal

ts = 0:T:3.14;

sampledpoints = cos(2*pi*frequency*ts);

reconstructed = cos(2*pi*(frequency-sample)*t);

plot(t, signal, 'g', ts, sampledpoints, 'p',t, reconstructed, 'b')

legend('Original Signal', 'Sampled Points', 'Reconstructed Signal')

xlabel('Time')

ylabel('Amplitude')

title('Aliasing Example')

So it has the same shape, but as you can see the frequency is alot lower than the original. Basically what not sampling enough does is that it recognizes multiple frequencies as the same frequency, so you lose crucial details in the reproduction of a sound.

File:Sally1.jpg

2007-11-30T22:45:48Z

Rothsa:

User:Rothsa

2007-11-30T18:53:05Z

Rothsa:

User:Rothsa

2007-11-30T18:42:41Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
[[Image:fish.jpg |thumb|What I actually look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Religion
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:38:05Z

Rothsa:

The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one?
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Marilyn Manson Anthologies.

But, if you're like me, this probably isn't enough description or motivation to truly understand the relationship between DFT's and CFT's. You want pictures, loud and clear. Here they are:

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:37:17Z

Rothsa:

The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one?
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a "sampled"(see HW #8) Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent "aliasing"(see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Michael Jackson Anthologies.

But, if you're like me, this probably isn't enough description or motivation to truly understand the relationship between DFT's and CFT's. You want pictures, loud and clear. Here they are:

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:35:51Z

Rothsa:

The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one?
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a sampled Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent aliasing (see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Michael Jackson Anthologies.

But, if you're like me, this probably isn't enough description or motivation to truly understand the relationship between DFT's and CFT's. You want pictures, loud and clear. Here they are:

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:35:26Z

Rothsa:

The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one?
For those of you who recognize the squiggly codes of mathematical formulas, observe:

Here is the continuous Fourier Transform 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Here is the Discrete Fourier Transform 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

In words, you may describe the Discrete Fourier Transform (DFT) as a sampled Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent aliasing (see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

The Discrete Transform allows us to chop that bit ol' continuous ones into a bunch of ones and zeros that will fit on your computer along with your Vanilla Ice and Mariah Carey and Michael Jackson Anthologies.

Fourier Transform looks like this: 
<math>X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt
</math> 
Which uses an integral, while the DFT which looks like this: 
<math>X(f)=\sum _{n=0}^{N-1} x(n) e^{\frac{-j2\pi nm}{N}}
</math> 

But, if you're like me, this probably isn't enough description or motivation to truly understand the relationship between DFT's and CFT's. You want pictures, loud and clear. Here they are:

User:Rothsa

2007-11-30T18:29:07Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
[[Image:fish.jpg |thumb|What I actually look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

File:Fish.jpg

2007-11-30T18:28:35Z

Rothsa:

User:Rothsa

2007-11-30T18:26:46Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
[[Image:244021375 l.jpg |thumb|What I actually look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

User:Rothsa

2007-11-30T18:26:26Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

User:Rothsa

2007-11-30T18:26:06Z

Rothsa:

=Sally Roth=
[[Image:244021375 l.jpg |thumb|What I want you to think I look like]]

===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

User:Rothsa

2007-11-30T18:23:51Z

Rothsa:

=Sally Roth=

[[Image:244021375 l.jpg | What I want you to think I look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

User:Rothsa

2007-11-30T18:23:35Z

Rothsa:

=Sally Roth=

[[Image:244021375 l.jpg "What I want you to think I look like"]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

User:Rothsa

2007-11-30T18:23:22Z

Rothsa:

=Sally Roth=

[[Image:244021375 l.jpg|What I want you to think I look like]]
===About Me===

Major: Electrical Engineering 
Graduation Date: March 2009 
Interests:
* Logic and Structural Problems
* Economics, Politics, and Culture
* Running, Rock Climbing, Snowboarding
* Music, Movies, Local and Foreign Culture

===Contact Info===

*Phone: (503)-703-3482
*Email: sally.roth@wallawalla.edu
*[http://www.myspace.com/rogueulette Myspace]

==Homework==

Primary Sources: Wikipedia, ClassNotes, ClassWiki 
Programs utilized: Octave, Gnuplot, qtOctave, Matlab 

[[HW#4 Fourier Transform Applications: The Fast Fourier Transform|Homework #4 - FFT Applications]]

[[Homework: Sampling: A class review|Homework #8 - Sampling]]

[[HW11 Aliasing Example|Homework #11 - Aliasing]]

[[HW12 DFT/Continuous Fourier Transform Relationship|Homework #12 - DFT/FFT Relationship]]

[[HW13 DFT/Sampling Assignment|Homework #13 - DFT/Sampling]]

[[HW14_What_I_Fixed|Homework #14 - What I fixed]]

[[HW15 D/A Converter Compensation|Homework #15 - DAC Compensation]]

[[HW16_Application|Homework #16 - Application]]

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:11:05Z

Rothsa:

The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one?

In words, you may describe the Discrete Fourier Transform (DFT) as a sampled Continuous Fourier Transform. The Discrete represents a finite amount of points
on the continuous transform - enough so that you can reconstruct the signal to the correct amount of accuracy that you need, and prevent aliasing (see HW #11)

So why would you make a Discrete Fourier Transform? Why not just keep all the information you had in the first place with the continuous? Well, you could, you'd just
need an infinite amount of memory on your computer to store the infinite amount of points on the continuous waveform. And that would leave no room for your precious Mp3's and movie files. You wouldn't want that, would you?

But, if you're like me, this probably isn't enough description or motivation to truly understand the relationship between DFT's and CFT's. You want pictures, loud and clear. Here they are:

HW12 DFT/Continuous Fourier Transform Relationship

2007-11-30T18:08:00Z

Rothsa: New page: The Question to answer here is: What is the relationship between the continuous Fourier Transform, and the discrete one? In words, you may describe the Discrete Fourier Transform (DFT) as...