Class Wiki - User contributions [en]

Hildebrand, Kurt

2012-12-13T05:20:49Z

Kurt: /* PSK 31 Demodulation */

PSK31 Demodulation (Kurt & Michael)

2012-12-13T05:16:06Z

Kurt: /* Results/Problems */

== Project Description ==

The goal of this project is to design and code a Matlab script that will encode and decode a PSK31 signal including signals with noise. The receiver should be able to read in signals (as a wav file) from other sources as well.

PSK31 is a audible text encoding that can be sent over the air by amateur radio operators. A computer's sound card can be used to send and receive the signal since the signal is audible. For more information regarding PSK31 see the Wikipedia article.[http://en.wikipedia.org/wiki/PSK31]

== Our Approach ==
=== Code Overview ===

==== Transmitter ====

Our code creates a PSK31 signal given an input carrier frequency and message. For testing our receiver code, the transmitter is setup to generate a random carrier frequency and phase. It also adds random noise to the signal before writing it to a wav file.

==== Receiver ====
Our receiver is split into several steps:
# We read in the signal from a wav file and generate utility variables and matrices.
# We run the signal through an FFT and take an average over the range of the FFT spike to obtain a carrier frequency guess.
# A PID controller was implemented to help offset the fact that a carrier frequency derived from the FFT produces an inaccurate carrier frequency. An offset in the carrier frequency causes the constellation diagram to rotate in a circle. The PID tries to compensate for this by adding to or subtracting from the phase of the signal while trying to drive the Q-component of the signal to zero. The feedback loop is described by the code below.
% PID controller constants
Kp = 25;
Ki = 50;
Kd = 1;
output = zeros(1, length(v)); % Init
setpoint = 0; % Sets the feedback point to control to.
previous_error = 0; % In this case, you want to set yt to 0.
integral = 0;
xt_filt = [];
yt_filt = [];
for k=51:length(v)
xt(k) = v(k)*cos(2*pi*u_fc*t(k)+output(k));
yt(k) = v(k)*sin(2*pi*u_fc*t(k)+output(k));

xt_filt(k:-1:k-50) = filter(b, a, xt(k:-1:k-50));
yt_filt(k:-1:k-50) = filter(b, a, yt(k:-1:k-50));

if(sign(xt(k)) == sign(yt(k)))
error = setpoint - yt_filt(k);
else
error = setpoint + yt_filt(k);
end

% PID Feedback Controller compensates for inaccurate fc guess.
% error = setpoint - yt_filt(k);
integral = integral + error*10;
derivative = (error - previous_error)/10;
output(k+1) = Kp*error + Ki*integral + Kd*derivative;
previous_error = error;
end
# We multiply the signal with a cosine wave at our guess frequency. This splits the signal into two parts: a high frequency component (with frequency equal to the sum of the ''actual'' carrier frequency and our ''guess'' frequency) and a low frequency component (due to the difference of the two frequencies).
# We filter the multiplied signal through a butterworth filter with a 75Hz cutoff frequency to remove the high frequency component.
# We mark the locations where the filtered signal changes sign

=== Code ===

== Results/Problems ==

=== PID Controller Problems ===
The first iteration did not take into account the I component of the signal and was constantly trying to drive the signal to the positive I coordinates. A smiple if-else statement solved this problem. However there are some bugs. There might be a singularity at (0,0) which causes the controller to perform in a strange fashion. Also, the code has very bad noise rejection. It can produce baseband but not very well once noise is added. Finally, it doesn't decode signals that we generate ourselves. When we produce our signal in MATLAB, the code works fine. However, the signal from wikipedia.com and Dr. Frohne's signal don't work. One problem with the wikipedia.com signal is that it's native format is .ogg which is lossy. This could've presented problems when we convert the signal to a wave file as high frequencies get chopped in lossy formats.

PSK31 Demodulation (Kurt & Michael)

2012-12-13T05:10:50Z

Kurt: /* Receiver */

Hildebrand, Kurt

2012-12-12T22:45:06Z

Kurt: /* PSK 31 Demodulation */

== Contact Information ==

'''Phone:''' (909)528-3558

'''Prefered E-mail:''' kurthildebrand@gmail.com

'''School E-mail:''' kurt.hildebrand@wallawalla.edu

== PSK 31 Demodulation ==

Fall, 2012
*Partner [[Michael von Pohle]]

'''''Abstract'''''
::The goal of this project was to learn more about Phase Shift Keying and to Signals and Systems in general. Since guessing the carrier frequency is slightly inaccurate a PID feedback and control system was used to correct the phase of the received signals. Despite demodulating my own signals created in MATLAB, my script does not work on the signal provided to us by Dr. Frohne and the sample signal on Wikipedia. One problem lies in the scripts inability to remove noise.

'''''Introduction'''''
::PSK31 was developed by Peter Martinez (G3PLX) in 1998. PSK31 varies in amplitude and phase, but not frequency. This is extremely useful in amateur radio. When there is constant phase,
:it is considered to be a 'one', when the phase shifts, it is a 'zero'.

'''''Matlab Script'''''

'''''Matlab Plots'''''

'''''Conclusion'''''
:The end.

Hildebrand, Kurt

2012-12-12T22:44:55Z

Kurt: /* Latex Information */

Signals and Systems

2012-12-12T22:43:05Z

Kurt: /* 2012-2013 Contributors */

== Topics ==
[[Fourier series - by Ray Betz|Overview of Signals and Systems]]

===Individual Subjects===
*[[Linear Time Invariant System|Linear Time Invariant Systems]]
**[[The Game|"The Game"]]
*[[Orthogonal functions|Orthogonal Functions]]
*[[Energy in a signal|Finding the Energy in a Signal]]
**[[Rayleigh's Theorem]]
*[[Fourier series|Fourier Series]]
*[[Fourier transform|Fourier Transforms]]
**[[Discrete Fourier transform]]
*[[Sampling]]
*[[FIR Filter Example]]
*[[Relationship between e, sin and cos]]

== Some Useful Links to Suppliment or Substitute for a Textbook ==
===Books on Signal Processing===
*[https://ccrma.stanford.edu/~jos/sasp/sasp.html Spectral Audio Signal Processing, by Julius O. Smith III]
*[http://www.dspguide.com/ The Scientist and Engineer's Guide to Digital Signal Processing by Steven W. Smith, Ph.D.] The professor likes this one.
*[http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-003-signals-and-systems-spring-2010/readings/ Discrete Time Signals & Systems]

===Fourier Series===
*[http://www.intmath.com/Fourier-series/Fourier-intro.php Interactive Mathematics (like a textbook with some examples)]
*[http://mathworld.wolfram.com/FourierSeries.html Mathworld]
*[http://en.wikipedia.org/wiki/Fourier_series Wikipedia]
*[http://web.mit.edu/2.14/www/Handouts/FreqDom.pdf MIT handout on Fourier Series, Fourier Transform, and Laplace Transform]
*[http://www.maths.mq.edu.au/~bon/Fourier%20Theory.pdf Fourier Theory B..M..N.. Clarke]

===Dirac Delta Function and Convolution===
*[http://web.mit.edu/2.14/www/Handouts/Convolution.pdf MIT handout on Dirac Delta Function and Convolution]

===Multi-rate Filtering===

[http://www.google.com/url?url=http://www.mds.com/tech/filter/multirate_article.pdf&rct=j&sa=U&ei=pD_UTJqtKY6ksQPbzPWMCw&ved=0CBUQFjAA&q=Purcell+Multirate+Filters&usg=AFQjCNFsHM7ROpUdrQ6py9ZH_RhQ_BeigA Multirate Filters Introduction]

[http://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE521_files/IV-05%20Polyphase%20FIlters_2007.pdf Slides from a Presentation on Polyphase Decimation and Interpolation by Mark Fowler]

===FIR Filters===

[http://www.dspguru.com/dsp/faqs/fir This is a very easy-to-understand summary of FIR basics, properties, design, and implementation]

[http://www.dspguru.com/dsp/faqs/multirate/decimation Another easy-to-understand article about decimation]

[http://www.dspguru.com/dsp/faqs/multirate/interpolation Another easy-to-understand article about interpolation]

===Adaptive FIR Filters===
[http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.42.6386&rep=rep1&type=pdf Introduction to Adaptive Filters, Simon Haykin]

[http://saba.kntu.ac.ir/eecd/taghirad/E%20books/TOC/Adaptive%20Filters.pdf Simon Haykin's book chapter]

[http://www.latticesemi.com/documents/doc22982x20.pdf?jsessionid=f0308ccf0f735471e49a6054323c5c177969 Adaptive LMS in an FPGA]

[[Adaptive Filters In the Frequency Domain]]

====Constant Modulus Algorythm====
[http://ens.ewi.tudelft.nl/Education/courses/et4147/sheets/cma_leus.pdf Using the CMA on antenna arrays]

===Course Pages===
[[2005-2006 Assignments]]

[[2006-2007 Assignments]]

[[2008-2009 Assignments]]

[[2009-2010 Assignments]]

[http://people.wallawalla.edu/~Rob.Frohne/ClassNotes/engr455index.htm Class notes for Signals & Systems]

==Articles==

===Octave Tutorials===
[[Installing Octave on a Mac]] (Chris Lau)

[[Octave and Scilab on a Mac]] (Ben Henry)

[[ASN2 - Octave Tutorial]] (Jodi S. Hodge)

[[A u(t) function example]]

[[FIR Filter Example Code for Octave]]

[[Leakage Example Octave Script]]

[[Interpolation using the DFT Example Script]]

[[Tuner Upper Removal Demonstration]]

[[Airplane Noise Removal Demonstration]]

===Final Project (2011)===
[[Matlab/Octave deMorse.m]]

[[morse.m This is the one from mathworks]]

===[[Table of Fourier Transform Properties]]===

==Homework Assignments==
Please put your name next to the assignment, linking it to your submission
* HW #1 - Make a personal page on this wiki ([[Christopher Garrison Lau I|Chris Lau]])([[Jodi S. Hodge]])([[user:chris.wills|Chris Wills]])[[Shepherd,Victor|(Victor Shepherd)]]
* HW #2 - Write a tutorial about installing and/or using Octave ([[Installing Octave on a Mac|Chris Lau]])([[Jodi S. Hodge]])([[Octave|Victor Shepherd]])
* HW #3 - Show graphically that <math> \int_{-\infty}^{\infty} e^{j2\pi f(t-u)}\, df = \delta (t-u)</math> ([[HW 3|Chris Lau]])([[Jodi S. Hodge]])([[user:chris.wills/HW3|Chris Wills]])([[Hw3|Victor Shepherd)]]

* HW #4 - Given a linear time-invariant system where <math>\ u(t) </math> produces an output <math>\ w(t) </math>, find the output due to any function <math>\ x(t) </math> ([[HW 4|Chris Lau]])
* HW #5: ([[HW 5|Chris Lau]])
** Part 1 - Find <math> \mathcal{F}[e^{- \sigma t} x(t)u(t)] </math> and relate it to the Laplace Transform. Derive the Inverse Laplace Transform of this from the inverse Fourier Transform.
** Part 2 - [[Image:20101006KeyDSCN3161.jpg|thumb|300px|center]]

* HW #6 - Pick a property of the Fourier Transform & present it on the Wiki. Make a table with all your properties. Interpret your property. ([[HW 6|Ben Henry]])([[Table of Fourier Transform Properties|Chris Lau]])([[Table of Fourier Transform Properties|Victor Shepherd]])
* HW #7 - Finish the practice tests
* HW #8 - Make a page about interpolating FIR filters. Note how many multiply/add operations.([[Jodi S. Hodge]])([[Interpolating FIR filters|Chris Lau]])([[Hw8|Victor Shepherd]])
* HW #9 - Add to #8 writeup how to do a decimating filter and figure out how many multiply & adds are needed for a n/2 decimating low pass filter.([[Jodi S. Hodge]])([[Decimating FIR filters|Chris Lau]])([[Hw9|Victor Shepherd)]]
* HW #10 - Use Octave (or Mathlab or Silab) to plot the frequency response of low pass filters with cut off frequencies of 1/32T, 1/8T, and 1/4T and compare how many coeffficients are needed with an eye to answer the question "Is it less calculation to decimate and then filter, or better to put the filter in the pre-decimation filter?" ([[Jodi S. Hodge]])([[Hw10|Victor Shepherd)]]
* HW #11 - Is our method the same as Mark Fowler's? See
[http://www.ws.binghamton.edu/fowler/fowler%20personal%20page/EE521_files/IV-05%20Polyphase%20FIlters_2007.pdf Wiki]. Same # multiply and adds? See Notes 11/3/10. ([[Jodi S. Hodge]])([[Hw11|Victor Shepherd)]]
* HW #12 - Experiment with a variety of signals having a 3Khz bandwidth to determine the resolution you can get when doing a cross correlation <math> \ r(m) =

\displaystyle\sum\limits_{n=0}^{N-1} x(n) x(n+m) </math>. You can generate the signals randomly and filter them to obtain the band-limited signals. ([[Jodi S. Hodge]])
* HW #13 - Derive the following realtions:
**a) <math>DFT(x(k-l))\!</math>
**b) <math> DFT(e^{j2 \pi lk/N}x(k)\!</math>
**c) <math>\sum\limits_{k=0}^{N-1} x(k)y(k)^{*}=c\sum\limits_{k=0}^{N-1} X(n)Y(n)^{*}</math> ([[Hw13|Victor Shepherd)]]

* HW #14 - Come up with a use for an adaptiveFIR filter and make an Octave script to demonstrate it. ([[Jodi S. Hodge]])([[Hw14|Victor Shepherd)]]
* HW #15 - Do Practice Exam II ([[Hw15|Victor Shepherd)]]
* [[CW-Robot Octave Simulation]]

==People Involved with this Wiki==

===2012-2013 Contributors===
[[Brian Haddad]]

[[Kurt Hildebrand]]

[[Denver Lodge]]

===2011-2012 Contributors===
[[Matthew Blaire]]

[[Cody Lorenz]]

===2010-2011 Contributors===
[[Ben Henry|Ben Henry]]

[[Christopher Garrison Lau I]]

[[user:chris.wills|Chris Wills]]

[[Jodi S. Hodge]]

[[Luke Chilson]]

[[Shepherd,Victor|Victor Shepherd]]

===2009-2010 Contributors===
[[Nick Christman]]

[[Joshua Sarris]]

[[Kevin Starkey]]

[[Max Woesner]]

[[Jodi Hodge]]

[[Corneliu Turturica]]

===2008-2009 Contributors===
[[User:eric.clay|Eric Clay]]

[[User:tsung-lin.yang|Chuck Yang]]

[[User:elton.zebron|Elton Zebron]]

[[User:Luke.chilson|Luke Chilson]]

[[User:Brandon.price|Brandon Price]]

[[User:Fonggr|Greg Fong]]

===2007-2008 contributors===

[[User:baldwin.britton|Baldwin Britton]]

[[User:Harrde|Denver Harris]]

[[User:Pridma|Mark Priddy]]

[[User:ChrisRas|Chris Rasmussen]]

[[User:RothMi|Michael Roth]]

[[User:Rothsa|Sally Roth]]

===2006-2007 contributors===

[[User:Smitry|Ryan J Smith]]

[[User:Nathan|Nathan Ferch]]

[[User:Andrew|Andrew Lopez]]

[[User:Sherna|Nathan Sherman]]

[[User:Adkich|Chris Adkins]]

===2005-2006 contributors===

[[User:GabrielaV|Gabriela Valdivia]]

[[User:SDiver|Raymond Betz]]

[[User:chrijen|Jenni Christensen]]

[[User:wonoje|Jeffrey Wonoprabowo]]

[[User:wilspa|Paul Wilson]]

[[User:Frohro|Instructor: Rob Frohne]]

===2004-2005 contributors===

[[User:Barnsa|Sam Barnes]]

[[User:Santsh|Shawn Santana]]

[[User:Goeari|Aric Goe]]

[[User:Caswto|Todd Caswell]]

[[User:Andeda|David Anderson]]

[[User:Guenan|Anthony Guenterberg]]

Kurt's Assignment

2010-11-02T07:11:01Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\
&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\
&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side}
\end{align}</math>

<math>\begin{align}
test function &=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

TODO:
*finish the triangle wave derivation
*start sawtooth wave derivation
*put pictures in
*implement triangle wave Fourier Series in OCTAVE
*implement sawtooth wave Fourier Series in OCTAVE

Kurt's Assignment

2010-11-02T07:08:41Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\
&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\
&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side}
\end{align}</math>

<math>\begin{align}
test function &=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

TODO:
*finish the triangle wave derivation
*start sawtooth wave derivation
*implement triangle wave Fourier Series in OCTAVE
*implement sawtooth wave Fourier Series in OCTAVE

Kurt's Assignment

2010-11-02T07:06:21Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\
&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\
&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side}
\end{align}</math>

<math>\begin{align}
test function &=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

TODO:
*finish the triangle wave derivation
*start sawtooth wave derivation
*implement triangle wave Fourier Series in OCTAVE
*implement sawtooth wave Fourier Series in OCTAVE

Kurt's Assignment

2010-11-02T06:53:24Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\
&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\
&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side}
\end{align}</math>

<math>\begin{align}
test function &=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

TODO:
*finish the triangle wave derivation
*start sawtooth wave derivation
*implement triangle wave Fourier Series in OCTAVE
*implement sawtooth wave Fourier Series in OCTAVE

Kurt's Assignment

2010-11-02T06:50:43Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{4H}{T}\left[\frac{-\frac{2}{T}t+1}{n\omega_0} \sin(n\omega_0 t) - \frac{2}{Tn^2\omega_0^2} \cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)\right]\right] + \frac{4H}{T}\left[\left[\frac{-\frac{2}{T}\left(\frac{3}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\frac{3}{4}T\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{3}{4}T\right)\right] - \left[\frac{-\frac{2}{T}\left(-\frac{1}{4}T\right)+1}{n\frac{2\pi}{T}} \sin\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right) - \frac{2}{Tn^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\left(-\frac{1}{4}T\right)\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)\right] - \left[\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\frac{T^2}{n^2 4 \pi^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right] + \frac{4H}{T} \left[\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(\frac{3}{2}n\pi\right)\right] - \left[\frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n \pi\right)-\frac{T}{2\pi^2 n^2}\cos\left(-\frac{1}{2}n\pi\right)\right]\right]\\
&= \frac{8H}{T^2}\left[\frac{T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)-\frac{T^2}{8\pi n}\sin\left(-\frac{1}{2}n\pi\right)+\underbrace{\frac{T^2}{n^2 4\pi^2}\cos\left(\frac{1}{2}n\pi\right)-\frac{T^2}{n^2 4\pi^2}\cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right] + \frac{4H}{T}\left[-\frac{T}{4\pi n}\sin\left(\frac{3}{2}n\pi\right) - \frac{3T}{4\pi n}\sin\left(-\frac{1}{2}n\pi\right) - \underbrace{\frac{T}{2\pi^2 n^2} \cos\left(\frac{3}{2}n\pi\right) + \frac{T}{2\pi^2 n^2} \cos\left(-\frac{1}{2}n\pi\right)}_\text{0}\right]\\
&= \frac{8H}{T^2}\left[\frac{2T^2}{8\pi n}\sin\left(\frac{1}{2}n\pi\right)\right] + \text{integrate other side}\\
&= \frac{2H}{\pi n}\sin\left(\frac{1}{2}n\pi\right) + \text{integrate other side}
\end{align}</math>

<math>\begin{align}
test function &=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

Kurt's Assignment

2010-11-02T01:59:17Z

Kurt:

=Common Synthesizer Waveforms=
Many synthesizers employ a variety of waveforms to produce varied sounds. The most common waveform is the sine wave. However, in additive synthesis, multiple waveforms can be added together to create a different waveform with different characteristics. The basis for this form of synthesis is the Fourier series:

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t)\, dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t)\, dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t)\, dt\\
\end{align}
</math>

The four basic waveforms are Sine Waves, Square Waves, Triangle Waves, and Sawtooth Waves.

===Square Wave===
By inspection of the waveform, the DC component of the wave will be 0. Also, since the waveform is odd, an will be 0. Here is the proof:

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H\, dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H\, dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin\left(n\omega_0t\right)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{2}T\right)-0\right] + \frac{2}{T}\left[-0+\frac{H}{n\frac{2\pi}{T}}\sin\left(n\frac{2pi}{T}\frac{1}{2}T\right)\right]\\
&=\frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\sin(n\pi)}_\text{0}\right]\\
&=0
\end{align}</math>

This just leaves the sine component of the waveform found below.

<math>\begin{align}
b_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\sin(n\omega_0t)\, dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\sin(n\omega_0t)\, dt\\
&=\frac{2}{T}\left[\frac{-H}{n\omega_0}\cos(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{H}{n\omega_0}\cos(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)+\frac{H}{n\frac{2\pi}{T}}\right] + \frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}T)-\frac{H}{n\frac{2\pi}{T}}\cos(n\frac{2\pi}{T}\frac{1}{2}T)\right]\\
&=\frac{2}{T}\left[-\frac{TH}{2\pi n}\cos(n\pi)+\frac{TH}{2\pi n}\right] + \frac{2}{T}\left[\frac{TH}{2\pi n}\underbrace{\cos(2 \pi n)}_\text{1}-\frac{TH}{2\pi n}\cos(n\pi)\right]\\
&=-\frac{H}{\pi n}\cos(n\pi)+\frac{H}{\pi n} + \frac{H}{\pi n}-\frac{H}{\pi n}\cos(n\pi)\\
&=\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\\
\end{align}</math>

Finally, resulting in the Fourier series for a Square Wave.

<math>\text{Square Wave Fourier Series: }x(t) = x(t+T) = \sum_{n=1}^\infty \left(\frac{2H}{\pi n}-\frac{2H}{\pi n}\cos(n\pi)\right) \sin(n\omega_0t) </math>

===Triangle Wave===
Like the Square wave, the DC component of the Triangle Wave is 0 by inspection. Also, since the triangle wave is odd, it is made up only by sine components.

<math>\begin{align}
a_0 &= \frac{1}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t \right)\,dt + \frac{1}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right) \,dt\\
&= \frac{1}{T}\left[\frac{1}{2}\frac{4H}{T}t^2\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \frac{1}{T}\left[-\frac{1}{2}\frac{4H}{T}t^2+2Ht\right]_{\frac{1}{4}T}^{\frac{3}{4}T}\\
&= \frac{1}{T}\left[\frac{2H}{T}\left(\frac{1}{4}T\right)^2-\frac{2H}{T}\left(-\frac{1}{4}T\right)^2\right] + \frac{1}{T}\left[-\frac{2H}{T}\left(\frac{3}{4}T\right)^2+2H\left(\frac{3}{4}T\right)-\left(-\frac{2H}{T}\left(\frac{1}{4}T\right)^2+2H\left(\frac{1}{4}T\right)\right)\right]\\
&= \frac{1}{T}\left[\underbrace{\frac{2H}{T}\frac{1}{16}T^2-\frac{2H}{T}\frac{1}{16}T^2}_\text{0}\right] + \frac{1}{T}\left[-\frac{2H}{T}\frac{9}{16}T^2+\frac{3}{2}HT+\frac{2H}{T}\frac{1}{16}T^2-\frac{1}{2}HT\right]\\
&= 0 + \frac{1}{T}\left[-\frac{18H}{16T}T^2+\frac{3}{2}HT+\frac{2H}{16T}T^2-\frac{1}{2}HT\right]\\
&= \frac{1}{T}\left[\underbrace{-\frac{18}{16}HT+\frac{24}{16}HT+\frac{2}{16}HT-\frac{8}{16}HT}_\text{0}\right]\\
&= 0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}\left(\frac{4H}{T}t\right)\cos\left(n\omega_0t\right) \,dt + \frac{2}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{4H}{T}t+2H\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\int_{-\frac{1}{4}T}^{\frac{1}{4}T}t\,\cos\left(n\omega_0t\right) \,dt + \frac{4H}{T}\int_{\frac{1}{4}T}^{\frac{3}{4}T}\left(-\frac{2}{T}t+1\right)\cos\left(n\omega_0t\right)\,dt\\
&= \frac{8H}{T^2}\left[\frac{t}{n\omega_0}\sin(n\omega_0 t)+\frac{1}{n^2\omega_0 ^2}\cos(n\omega_0 t)\right]_{-\frac{1}{4}T}^{\frac{1}{4}T} + \text{integrate other side}\\
&= \frac{8H}{T^2}\left[\left[\frac{\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}\frac{1}{4}T\right)\right] - \left[\frac{-\frac{1}{4}T}{n\frac{2\pi}{T}}\sin\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)+\frac{1}{n^2\frac{4\pi^2}{T^2}}\cos\left(n\frac{2\pi}{T}-\frac{1}{4}T\right)\right]\right]
&=
\end{align}</math>

===Sawtooth Wave===

==OCTAVE Scripts to Plot Fourier Series==
====Square Wave====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=100 %Number of iterations to undergo
sum1=0; %Initialize the sum to 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms');
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

Kurt's Assignment

2010-11-02T01:29:18Z

Kurt:

Kurt's Assignment

2010-11-02T00:54:12Z

Kurt:

Kurt's Assignment

2010-11-01T23:54:27Z

Kurt:

Kurt's Assignment

2010-11-01T23:53:41Z

Kurt:

Kurt's Assignment

2010-11-01T23:48:18Z

Kurt:

Kurt's Octave Assignment

2010-11-01T23:04:41Z

Kurt:

=Useful Octave Templates=

=====Fourier Series=====
clf; %Clear Figure
t=0:.01:10; %Limits of the graph
T=2*pi %Definition of the period
M=1000 %Number of iterations to undergo
sum1=0; %Initialize the sum to 0 
%----------FOURIER SERIES----------%
for m=1:1:M, %For m=1, increment by 1 until you get to M
if(m!=0)
sum1 = sum1 + ((2/(pi*m))-(2/(pi*m))*cos(m*pi))*sin(m*2*pi/T*t);
end
end 
%---------------PLOT---------------%
plot(t,real(sum1),'b-')
title('Fourier Series Representation of a Wave')
xlabel('time (seconds)')
ylabel('Function')
grid on; 
legend(num2str(M) ' terms ')
print("squarewave.png","-dpng") % Prints the plot to a png file called squarewave.png

Kurt's Octave Assignment

2010-11-01T22:50:43Z

Kurt:

=Useful Octave Templates=

A Fourier series example:

Kurt's Assignment

2010-11-01T22:49:25Z

Kurt:

Kurt's Assignment

2010-11-01T22:29:33Z

Kurt:

Kurt's Assignment

2010-11-01T21:44:20Z

Kurt:

=Common Synthesizer Waveforms=

<math>\begin{align}
x(t) &= x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)\\
a_0 &= \frac{1}{T}\int_0^T f(t) dt\\
a_n &= \frac{2}{T}\int_0^T f(t)\cos(n\omega_0t) dt\\
b_n &= \frac{2}{T}\int_0^T f(t)\sin(n\omega_0t) dt\\
\end{align}
</math>

==Square Wave==

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^{\frac{1}{2}T} H dt + \frac{1}{T}\int_{\frac{1}{2}T}^T -H dt\\
&=\frac{1}{T}\left[Ht\right]\bigg|_{t=0}^{\frac{1}{2}T} - \frac{1}{T}\left[Ht\right]\bigg|_{t={\frac{1}{2}T}}^T\\
&=\frac{1}{T}H\frac{1}{2}T-0 - \left[\frac{1}{T}HT - \frac{1}{T}H\frac{1}{2}T\right]\\
&=\frac{H}{2} - \left[H-\frac{1}{2}H\right]\\
&=\frac{H}{2}-\frac{H}{2}\\
&=0
\end{align}</math>

<math>\begin{align}
a_n &= \frac{2}{T}\int_0^{\frac{1}{2}T} H\cos(n\omega_0t) dt + \frac{2}{T}\int_{\frac{1}{2}T}^T -H\cos(n\omega_0t) dt\\
&=\frac{2}{T}\left[\frac{H}{n\omega_0}\sin(n\omega_0t)\right]_0^{\frac{1}{2}T} + \frac{2}{T}\left[\frac{-H}{n\omega_0}\sin(n\omega_0t)\right]_{\frac{1}{2}T}^T\\
&=\frac{2}{T}\left[\frac{H}{n\frac{2\pi}{T}}\sin(n\pi)\right]

&=0
\end{align}</math>

TODO: finish

Kurt's Assignment

2010-11-01T21:32:49Z

Kurt:

Kurt's Assignment

2010-11-01T17:46:00Z

Kurt:

=Common Synthesizer Waveforms=

==Square Wave==

<math>x(t) = x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)</math>

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^T f(t) dt\\
&=\frac{1}{T}\int_0^{\frac{1}{2}T}H dt + \frac{1}{T}\int_{\frac{1}{2}T}^T 0 dt \\
&=\frac{1}{T}[Ht]
\end{align}</math>

TODO: finish

Fall 2010

2010-11-01T17:42:41Z

Kurt:

Put links for your reports here.

Links to the posted reports are found under the publishing author's name.

Please number and sort Authors alphabetically.

# [[Banton, Alex]]
#*[[Alex's Octave Assignment]]
#*[[Alex's Assignment #8]]
# [[Bidwell, Kelvin]]
#*[[Kelvin's Octave Assignment]]
# [[Blaire, Matthew]]
#*[[Matthew's Octave Assignment]]
#*[[Matthew's Asgn #8]]
# [[Boyd, Aaron]]
#*[[ Aaron's Octave Assignment]]
#*[[Aaron Boyd's Assignment 8]]
# [[Bryson, David]]
#*[[David's Octave Assignment]]
# [[Colls, David]]
#*[[Colls Octave Assignment]]
#*[[Fourier Series Assignment]]
# [[Fullerton, Colby]]
#*[[Colby's Octave Assignment]]
# [[Hildebrand, Kurt]]
#*[[Kurt's Octave Assignment]]
#*[[Kurt's Assignment #8]]
# [[Martinez, Jonathan]]
#*[[Martinez's Octave Assignment]]
#*[[Martinez's Fourier Assignment]]
# [[Morgan, David]]
#*[[David Morgan's Octave Assignment]]
#*[[David Morgan's Fourier Series assignment]]
# [[Roth, Andrew]]
#*[[Andrew Roth's Fourier Series/Laplace Transform Project]]
#*[[Andrew's Octave Assignment]]
#*[[Example: LaTex format]]
#*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]
# [[Stirn, Jed]]
#*[[Jed's Octave Assignment]]
#*[[HW#8:Laplace Transforms/Fourier Series]]
# [[Stringer, Robert]]
#*[[Robert's Octave Assignment]]
#*[[Robert's HW #8]]
# [[Wooley, Andy]]
#*[[Andy's Octave Assignment]]
#*[[Andy's Fourier Series Project]]
# [[Zimmerly, Brian]]
#*[[Brian's Octave Assignment]]

Hildebrand, Kurt

2010-11-01T17:40:28Z

Kurt:

== Contact Information ==

'''Phone:''' (909)528-3558

'''Prefered E-mail:''' kurthildebrand@gmail.com

'''School E-mail:''' kurt.hildebrand@wallawalla.edu

== Latex Information ==

Wikibooks - LaTeX[[http://en.wikibooks.org/wiki/LaTeX]]

Kurt Hildebrand

2010-11-01T17:37:07Z

Kurt: moved Kurt Hildebrand to Hildebrand, Kurt

#REDIRECT [[Hildebrand, Kurt]]

Hildebrand, Kurt

2010-11-01T17:37:07Z

Kurt: moved Kurt Hildebrand to Hildebrand, Kurt

== Contact Information ==

'''Phone:''' (909)528-3558

'''Prefered E-mail:''' kurthildebrand@gmail.com

'''School E-mail:''' kurt.hildebrand@wallawalla.edu

== Useful Octave Templates ==

'''Plotting a function:'''

Kurt's Octave Assignment

2010-11-01T17:36:41Z

Kurt: Created page with 'Useful Octave Templates TODO: finish'

Useful Octave Templates

TODO: finish

Kurt's Assignment

2010-11-01T17:36:03Z

Kurt:

Common Synthesizer Waveforms

Square Wave

<math>x(t) = x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)</math>

<math>\begin{align}
a_0 &= \frac{1}{T}\int_0^T f(t) dt\\
&=\frac{1}{T}\int_0^{\frac{1}{2}T}H dt + \frac{1}{T}\int_{\frac{1}{2}T}^T 0 dt \\
&=\frac{1}{T}[Ht]
\end{align}</math>

TODO: finish

Kurt's Assignment

2010-11-01T17:35:47Z

Kurt: Created page with 'Common Synthesizer Waveforms Square Wave <math>x(t) = x(t+T) = a_0 + \sum_{n=1}^\infty a_n \cos(n\omega_0t) + b_n \sin(n\omega_0t)</math> <math>\begin{align} a_0 &= \f…'

Fall 2010

2010-11-01T17:04:12Z

Kurt:

Put links for your reports here.

Links to the posted reports are found under the publishing author's name.

Please number and sort Authors alphabetically.

0. [[Banton, Alex]]
*[[Alex's Octave Assignment]]
*[[Alex's Assignment #8]]

1. [[Bidwell, Kelvin]]
*[[Kelvin's Octave Assignment]]

2. [[Blaire, Matthew]]
*[[Matthew's Octave Assignment]]
*[[Matthew's Asgn #8]]

3. [[Boyd, Aaron]]
*[[ Aaron's Octave Assignment]]

4. [[Bryson, David]]
*[[David's Octave Assignment]]

5. [[Colls, David]]
*[[Colls Octave Assignment]]
*[[Fourier Series Assignment]]

6. [[Fullerton, Colby]]
*[[Colby's Octave Assignment]]

7. [[Hildebrand, Kurt]]
*[[Kurt's Octave Assignment]]
*[[Kurt's Assignment #8]]

8. [[Martinez, Jonathan]]
*[[Martinez's Octave Assignment]]
*[[Martinez's Fourier Assignment]]

9. [[Morgan, David]]
*[[David Morgan's Octave Assignment]]
*[[David Morgan's Fourier Series assignment]]

10. [[Roth, Andrew]]
*[[Andrew Roth's Fourier Series/Laplace Transform Project]]
*[[Andrew's Octave Assignment]]
*[[Example: LaTex format]]
*[[Chapter 22--Fourier Series: Fundamental Period, Frequency, and Angular Frequency]]

11. [[Stirn, Jed]]
*[[Jed's Octave Assignment]]
*[[HW#8:Laplace Transforms/Fourier Series]]

12. [[Stringer, Robert]]
*[[Robert's Octave Assignment]]

13. [[Wooley, Andy]]
*[[Andy's Octave Assignment]]
*[[Andy's Fourier Series Project]]

14. [[Zimmerly, Brian]]
*[[Brian's Octave Assignment]]

Hildebrand, Kurt

2010-10-04T17:34:05Z

Kurt:

Hildebrand, Kurt

2010-10-01T17:28:55Z

Kurt: Created page with '== Contact Information == '''Phone:''' (909)528-3558 '''Prefered E-mail:''' kurthildebrand@gmail.com '''School E-mail:''' kurt.hildebrand@wallawalla.edu'

== Contact Information ==

'''Phone:''' (909)528-3558

'''Prefered E-mail:''' kurthildebrand@gmail.com

'''School E-mail:''' kurt.hildebrand@wallawalla.edu