Class Wiki - User contributions [en]

https://fweb.wallawalla.edu/class-wiki/api.php?action=feedcontributions&feedformat=atom&user=Wonoje Class Wiki - User contributions [en] 2026-04-05T12:00:20Z User contributions MediaWiki 1.43.0 https://fweb.wallawalla.edu/class-wiki/index.php?title=2005-2006_Assignments&diff=2522 2005-2006 Assignments 2005-12-14T17:06:21Z

<p>Wonoje: /* HW #4 */</p> <hr /> <div>==Fall 2005 Homework Assignments==<br /> Assignments for this quarter will be listed here so that there is an easy place to look up the assignments. Below each assignment is the date that it was assigned.<br /> <br /> ===HW #1===<br /> Look at the Wiki & add your personal page. Spend two hours.<br /> <br /> - 9/26/05<br /> <br /> ===HW #2===<br /> Find the first three orthogonormal polynomials on <math>[-1,1]</math> (lowest order polynomials).<br /> <br /> - 9/30/05<br /> <br /> ===HW #3===<br /> 1) Work on the Wiki for two hours this week.<br /> <br /> 2) Find the output of a periodic function, <br /> <br /> <math>x(t) = \sum_{k = -\infty}^\infty \alpha_k e^{j \pi k \frac{t}{T}}</math><br /> <br /> to an RC filter with RC = T.<br /> <br /> - 10/3/05<br /> <br /> ===HW #4===<br /> Show how the real and imaginary parts of <math>\alpha_k</math> in the complex Fourier Series are related to the coefficients in the sine/cosine Fourier Series.<br /> <br /> 10/5/05<br /> <br /> ===HW #5===<br /> Individual Wiki pages on Fourier Transform.<br /> <br /> 10/13/05<br /> <br /> ===HW #6===<br /> 1) Find <math>\mathcal{F} [x(t) sin(2 \pi f_o t + \theta)</math>.<br /> <br /> 2) Convolve <math>u(t) - u(t-3)</math> with <math>cos(2 \pi t)[u(t-1)-u(t-2)]</math><br /> <br /> 10/14/05<br /> <br /> ===HW #7===<br /> 1) Find <math>\mathcal{F} [u(t) cos(2 \pi f_o t)]</math><br /> <br /> 2) Show <math>x(t)*\delta (t-t_o) = x(t-t_o)</math>.<br /> <br /> 10/19/05<br /> <br /> ===HW #8===<br /> There were two assignments that were labeled as HW #8.<br /> <br /> =====HW #8A=====<br /> Show that <br /> <br /> <math>\Phi_n (t) \equiv \frac{sin \left ( \frac{\pi (t - nT)}{T} \right)}{ \frac{\pi (t-nT)}{T} }</math><br /> <br /> form an orthogonal basis set. Tell me what functions these span.<br /> <br /> 10/21/05<br /> <br /> =====HW #8B=====<br /> Do a Wiki page on how a 2x oversampling CD Player works.<br /> <br /> 10/24/05<br /> <br /> ===HW #9===<br /> Handout on FIR filter.<br /> <br /> 11/2/05<br /> <br /> ===HW #10===<br /> Put something on FIR filters on the Wiki.<br /> <br /> 11/7/05<br /> <br /> ===HW #11===<br /> Work on the Wiki. Read someone else's contribution & fix or extend it a little. Continue with FIR and do a DFT section if you have time.<br /> <br /> 11/14/05<br /> <br /> ===HW #12===<br /> Come see me to discuss your Wiki contributions. I will give you suggestions for edits, etc.<br /> <br /> 11/30/05<br /> <br /> ===HW #13===<br /> Write a Wiki page on adaptive FIR filters. Spend at least 2 hours by sundown Friday.<br /> <br /> 11/30/05<br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=2005-2006_Assignments&diff=1495 2005-2006 Assignments 2005-12-09T07:09:12Z

<p>Wonoje: /* HW #8 */</p> <hr /> <div>==Fall 2005 Homework Assignments==<br /> Assignments for this quarter will be listed here so that there is an easy place to look up the assignments. Below each assignment is the date that it was assigned.<br /> <br /> ===HW #1===<br /> Look at the Wiki & add your personal page. Spend two hours.<br /> <br /> - 9/26/05<br /> <br /> ===HW #2===<br /> Find the first three orthogonormal polynomials on <math>[-1,1]</math> (lowest order polynomials).<br /> <br /> - 9/30/05<br /> <br /> ===HW #3===<br /> 1) Work on the Wiki for two hours this week.<br /> <br /> 2) Find the output of a periodic function, <br /> <br /> <math>x(t) = \sum_{k = -\infty}^\infty \alpha_k e^{j \pi k \frac{t}{T}}</math><br /> <br /> to an RC filter with RC = T.<br /> <br /> - 10/3/05<br /> <br /> ===HW #4===<br /> Show how the real and imaginary parts of \alpha_k in the complex Fourier Series are related to the coefficients in the sine/cosine Fourier Series.<br /> <br /> 10/5/05<br /> <br /> ===HW #5===<br /> Individual Wiki pages on Fourier Transform.<br /> <br /> 10/13/05<br /> <br /> ===HW #6===<br /> 1) Find <math>\mathcal{F} [x(t) sin(2 \pi f_o t + \theta)</math>.<br /> <br /> 2) Convolve <math>u(t) - u(t-3)</math> with <math>cos(2 \pi t)[u(t-1)-u(t-2)]</math><br /> <br /> 10/14/05<br /> <br /> ===HW #7===<br /> 1) Find <math>\mathcal{F} [u(t) cos(2 \pi f_o t)]</math><br /> <br /> 2) Show <math>x(t)*\delta (t-t_o) = x(t-t_o)</math>.<br /> <br /> 10/19/05<br /> <br /> ===HW #8===<br /> There were two assignments that were labeled as HW #8.<br /> <br /> =====HW #8A=====<br /> Show that <br /> <br /> <math>\Phi_n (t) \equiv \frac{sin \left ( \frac{\pi (t - nT)}{T} \right)}{ \frac{\pi (t-nT)}{T} }</math><br /> <br /> form an orthogonal basis set. Tell me what functions these span.<br /> <br /> 10/21/05<br /> <br /> =====HW #8B=====<br /> Do a Wiki page on how a 2x oversampling CD Player works.<br /> <br /> 10/24/05<br /> <br /> ===HW #9===<br /> Handout on FIR filter.<br /> <br /> 11/2/05<br /> <br /> ===HW #10===<br /> Put something on FIR filters on the Wiki.<br /> <br /> 11/7/05<br /> <br /> ===HW #11===<br /> Work on the Wiki. Read someone else's contribution & fix or extend it a little. Continue with FIR and do a DFT section if you have time.<br /> <br /> 11/14/05<br /> <br /> ===HW #12===<br /> Come see me to discuss your Wiki contributions. I will give you suggestions for edits, etc.<br /> <br /> 11/30/05<br /> <br /> ===HW #13===<br /> Write a Wiki page on adaptive FIR filters. Spend at least 2 hours by sundown Friday.<br /> <br /> 11/30/05<br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=2005-2006_Assignments&diff=1413 2005-2006 Assignments 2005-12-09T07:04:37Z

<p>Wonoje: </p> <hr /> <div>==Fall 2005 Homework Assignments==<br /> Assignments for this quarter will be listed here so that there is an easy place to look up the assignments. Below each assignment is the date that it was assigned.<br /> <br /> ===HW #1===<br /> Look at the Wiki & add your personal page. Spend two hours.<br /> <br /> - 9/26/05<br /> <br /> ===HW #2===<br /> Find the first three orthogonormal polynomials on <math>[-1,1]</math> (lowest order polynomials).<br /> <br /> - 9/30/05<br /> <br /> ===HW #3===<br /> 1) Work on the Wiki for two hours this week.<br /> <br /> 2) Find the output of a periodic function, <br /> <br /> <math>x(t) = \sum_{k = -\infty}^\infty \alpha_k e^{j \pi k \frac{t}{T}}</math><br /> <br /> to an RC filter with RC = T.<br /> <br /> - 10/3/05<br /> <br /> ===HW #4===<br /> Show how the real and imaginary parts of \alpha_k in the complex Fourier Series are related to the coefficients in the sine/cosine Fourier Series.<br /> <br /> 10/5/05<br /> <br /> ===HW #5===<br /> Individual Wiki pages on Fourier Transform.<br /> <br /> 10/13/05<br /> <br /> ===HW #6===<br /> 1) Find <math>\mathcal{F} [x(t) sin(2 \pi f_o t + \theta)</math>.<br /> <br /> 2) Convolve <math>u(t) - u(t-3)</math> with <math>cos(2 \pi t)[u(t-1)-u(t-2)]</math><br /> <br /> 10/14/05<br /> <br /> ===HW #7===<br /> 1) Find <math>\mathcal{F} [u(t) cos(2 \pi f_o t)]</math><br /> <br /> 2) Show <math>x(t)*\delta (t-t_o) = x(t-t_o)</math>.<br /> <br /> 10/19/05<br /> <br /> ===HW #8===<br /> Do a Wiki page on how a 2x oversampling CD Player works.<br /> <br /> 10/24/05<br /> <br /> ===HW #9===<br /> Handout on FIR filter.<br /> <br /> 11/2/05<br /> <br /> ===HW #10===<br /> Put something on FIR filters on the Wiki.<br /> <br /> 11/7/05<br /> <br /> ===HW #11===<br /> Work on the Wiki. Read someone else's contribution & fix or extend it a little. Continue with FIR and do a DFT section if you have time.<br /> <br /> 11/14/05<br /> <br /> ===HW #12===<br /> Come see me to discuss your Wiki contributions. I will give you suggestions for edits, etc.<br /> <br /> 11/30/05<br /> <br /> ===HW #13===<br /> Write a Wiki page on adaptive FIR filters. Spend at least 2 hours by sundown Friday.<br /> <br /> 11/30/05<br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=Signals_and_Systems&diff=2434 Signals and Systems 2005-12-09T06:48:04Z

<p>Wonoje: /* Topics */</p> <hr /> <div>[http://www.wwc.edu/~frohro/ClassNotes/engr455index.htm Class notes for Signals & Systems]<br /> <br /> == Topics ==<br /> *[[Orthogonal functions]]<br /> *[[Fourier series]]<br /> *[[Fourier transform]]<br /> *[[Sampling]]<br /> *[[Discrete Fourier transform]]<br /> *[[Fourier series - by Ray Betz|Signals and Systems - by Ray Betz]]<br /> *[[FIR Filter Example]]<br /> *[[2005-2006 Assignments]]<br /> <br /> I couldn't figure out how to get to others Users pages easily so I decided to start posting them here, please add yours:<br /> <br /> [[User:Frohro|Rob Frohne]]<br /> <br /> ==2004-2005 contributors==<br /> <br /> [[User:Barnsa|Sam Barnes]]<br /> <br /> [[User:Santsh|Shawn Santana]]<br /> <br /> [[User:Goeari|Aric Goe]]<br /> <br /> [[User:Caswto|Todd Caswell]]<br /> <br /> [[User:Andeda|David Anderson]]<br /> <br /> [[User:Guenan|Anthony Guenterberg]]<br /> <br /> ==2005-2006 contributors==<br /> <br /> [[User:GabrielaV|Gabriela Valdivia]]<br /> <br /> [[User:SDiver|Raymond Betz]]<br /> <br /> [[User:chrijen|Jenni Christensen]]<br /> <br /> [[User:wonoje|Jeffrey Wonoprabowo ]]<br /> <br /> [[User:wilspa|Paul Wilson]]</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1470 FIRJEW 2005-12-06T22:52:31Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop. Because there is no feedback, the response of an FIR filter to an impulse is finite. <br /> <br /> The equation for an FIR filter would look like the following:<br /> <br /> <center><math>h(mT) = T \int_{-v}^v \hat H (f) e^{j 2 \pi f m t} df </math></center><br /> <br /> where the desired frequencies are in the range from <math>-v</math> to <math>v</math> and <math> \hat H (f) </math> is the desired response.<br /> <br /> Example: Desing an FIR low pass filter to pass between <math> - \frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.<br /> <br /> <math>\hat H (f) = \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math><br /> <br /> The FIR filter would then be:<br /> <br /> <math> h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}</math><br /> <br /> <math> = \frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m} = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math><br /> <br /> and the actual frequency response would then be:<br /> <br /> <math>\sum_{m=-M}^M h(mT) e^{- j 2 \pi f m T} = \sum_{m=-M}^M \frac{ sin \left ( \frac{\pi m}{2} \right) }{\pi m} e^{- j 2 \pi f m T} </math><br /> <br /> <br /> <br /> The frequency response plot of this particular FIR filter would look like this:<br /> <br /> [[Image:Frequency_Response_Pict.png]]<br /> <br /> <br /> <br /> <br /> <br /> ==Related Pages==<br /> [[AdaptiveFIRJEW | Adaptive FIR Filters]]<br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo<br /> <br /> Image of the frequency response plot taken from the FIR example page [[FIR_Filter_Example|here]].<br /> </small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=AdaptiveFIRJEW&diff=4078 AdaptiveFIRJEW 2005-12-06T22:51:59Z

<p>Wonoje: /* Related Pages */</p> <hr /> <div>==Adaptive FIR Filtes==<br /> <br /> <br /> ==Related Pages==<br /> [[FIRJEW | Finite Impulse Response Filters]]<br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=AdaptiveFIRJEW&diff=1410 AdaptiveFIRJEW 2005-12-06T22:51:31Z

<p>Wonoje: </p> <hr /> <div>==Adaptive FIR Filtes==<br /> <br /> <br /> ==Related Pages==<br /> [[FIRJEW | Finite Impulse Response Filters]]</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=User:Wonoje&diff=3829 User:Wonoje 2005-12-06T22:50:08Z

<p>Wonoje: /* My Wiki Pages */</p> <hr /> <div>==My Wiki Pages==<br /> [[FourierTransformsJW | HW #5: Fourier Transforms]]<br /> <br /> [[CDPlayerJEW | HW #8: How a CD Player works]]<br /> <br /> [[FIRJEW | HW #10: FIR Filters]]<br /> <br /> [[DFTJEW | HW #11: Discrete Fourier Transform]]<br /> <br /> [[AdaptiveFIRJEW | HW #13: Adaptive FIR Filters]]<br /> <br /> ==Jeffrey Wonoprabowo==<br /> Making a page about myself is the first assignment for the class; although it isn't exactly the easiest thing to do as I never know what to write for these things.<br /> <br /> This is my fourth and final year at Walla Walla College. When I first came my concentration was in computer engineering. However, towards the end of my second year, I switched to Bioengineering and am currently applying to medical schools.<br /> <br /> At this point I am very sad because I cannot tell you how a CD player works... All I can do is write about myself. So below you'll find a modified picture of the one I put up on my hall in Meske; and the only reason I put one up is cause the RAs were asked to do so... I did not just put up a poster of myself up for fun.<br /> <br /> [[Image:JeffSS.jpg]]</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=DFTJEW&diff=4070 DFTJEW 2005-12-06T22:36:24Z

<p>Wonoje: /* Discrete Fourier Transform */</p> <hr /> <div>==Discrete Fourier Transform==<br /> The [[FourierTransformsJW|Fourier Transform]] is a powerful tool to convert a continuous function from the time domain into the frequency domain. The Fourier transform, however, is an integral transform; it is done by integration. This cannot be done with a discrete function. The Discrete Fourier Transform (DFT) allows us to transform a discrete function from the time domain into the frequency domain.<br /> <br /> If <math>x(n)</math> is a discrete function in time, then the DFT of <math>x(n)</math> would be:<br /> <br /> <math>\mbox{DFT}[x(n)] \equiv X(m) \equiv \sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n m}{N} }</math><br /> <br /> <br /> When dealing with a continuous function, an inverse Fourier Transform can be used to go back from the frequency domain into the time domain. The discete analog of the inverse Fourier Transform is the IDFT (Inverse Discrete Fourier Transform) which is defined below:<br /> <br /> <math> \mbox{IDFT}[X(m)] \equiv x(k) \equiv \frac{1}{N} \sum_{m=0}^{N-1} X(m) e^{j \frac{2 \pi k m}{N} } </math><br /> <br /> One thing to note about the DFT is that it assumes periodicity. Let us, for example, consider a DFT on the interval of <math>-m </math> to <math> +m</math>. The DFT will expect the data at +<math>m</math> to be equal to the data at -<math>m</math>.<br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=DFTJEW&diff=1408 DFTJEW 2005-12-06T22:30:38Z

<p>Wonoje: /* Discrete Fourier Transform */</p> <hr /> <div>==Discrete Fourier Transform==<br /> The [[FourierTransformsJW|Fourier Transform]] is a powerful tool to convert a continuous function from the time domain into the frequency domain. The Fourier transform, however, is an integral transform; it is done by integration. This cannot be done with a discrete function. The Discrete Fourier Transform (DFT) allows us to transform a discrete function from the time domain into the frequency domain.<br /> <br /> If <math>x(n)</math> is a discrete function in time, then the DFT of <math>x(n)</math> would be:<br /> <br /> <math>\mbox{DFT}[x(n)] \equiv X(m) \equiv \sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n m}{N} }</math><br /> <br /> <br /> When dealing with a continuous function, an inverse Fourier Transform can be used to go back from the frequency domain into the time domain. The discete analog of the inverse Fourier Transform is the IDFT (Inverse Discrete Fourier Transform) which is defined below:<br /> <br /> <math> \mbox{IDFT}[X(m)] \equiv x(k) \equiv \frac{1}{N} \sum_{m=0}^{N-1} X(m) e^{j \frac{2 \pi k m}{N} } </math><br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=DFTJEW&diff=1407 DFTJEW 2005-12-06T22:13:19Z

<p>Wonoje: /* Discrete Fourier Transform */</p> <hr /> <div>==Discrete Fourier Transform==<br /> The [[FourierTransformsJW|Fourier Transform]] is a powerful tool to convert a continuous function from the time domain into the frequency domain. The Fourier transform, however, is an integral transform; it is done by integration. This cannot be done with a discrete function. The Discrete Fourier Transform (DFT) allows us to transform a discrete function from the time domain into the frequency domain.<br /> <br /> If <math>x(n)</math> is a discrete function in time, then the DFT of <math>x(n)</math> would be:<br /> <br /> <math>\mbox{DFT}[x(n)] \equiv X(m) \equiv \sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n m}{N} }</math><br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=DFTJEW&diff=1406 DFTJEW 2005-12-06T22:11:24Z

<p>Wonoje: /* Discrete Fourier Transform */</p> <hr /> <div>==Discrete Fourier Transform==<br /> The [[FourierTransformsJW|Fourier Transform]] is a powerful tool to convert a continuous function from the time domain into the frequency domain. The Fourier transform, however, is an integral transform; it is done by integration. This cannot be done with a discrete function. The Discrete Fourier Transform (DFT) allows us to transform a discrete function from the time domain into the frequency domain.<br /> <br /> Let <math>x(n)</math> be a discretized function in time.<br /> <br /> Then the DFT of <math>x(n)</math> would be:<br /> <br /> <math>\mbox{DFT}[x(n)] \equiv X(m) \equiv \sum_{n=0}^{N-1} x(n) e^{-j \frac{2 \pi n m}{N} }</math><br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=DFTJEW&diff=1405 DFTJEW 2005-12-06T22:03:07Z

<p>Wonoje: /* Discrete Fourier Transform */</p> <hr /> <div>==Discrete Fourier Transform==<br /> The Fourier Transform is a powerful tool to convert a continuous function from the time domain into the frequency domain. The Fourier transform, however, is an integral transform; it is done by integration. This cannot be done with a discrete function. The Discrete Fourier Transform (DFT) allows us to transform a discrete function from the time domain into the frequency domain.<br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1411 FIRJEW 2005-12-06T21:57:58Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop. Because there is no feedback, the response of an FIR filter to an impulse is finite. <br /> <br /> The equation for an FIR filter would look like the following:<br /> <br /> <center><math>h(mT) = T \int_{-v}^v \hat H (f) e^{j 2 \pi f m t} df </math></center><br /> <br /> where the desired frequencies are in the range from <math>-v</math> to <math>v</math> and <math> \hat H (f) </math> is the desired response.<br /> <br /> Example: Desing an FIR low pass filter to pass between <math> - \frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.<br /> <br /> <math>\hat H (f) = \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math><br /> <br /> The FIR filter would then be:<br /> <br /> <math> h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}</math><br /> <br /> <math> = \frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m} = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math><br /> <br /> and the actual frequency response would then be:<br /> <br /> <math>\sum_{m=-M}^M h(mT) e^{- j 2 \pi f m T} = \sum_{m=-M}^M \frac{ sin \left ( \frac{\pi m}{2} \right) }{\pi m} e^{- j 2 \pi f m T} </math><br /> <br /> <br /> <br /> The frequency response plot of this particular FIR filter would look like this:<br /> <br /> [[Image:Frequency_Response_Pict.png]]<br /> <br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo<br /> <br /> Image of the frequency response plot taken from the FIR example page [[FIR_Filter_Example|here]].<br /> </small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1403 FIRJEW 2005-12-06T21:54:47Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop. Because there is no feedback, the response of an FIR filter to an impulse is finite. <br /> <br /> The equation for an FIR filter would look like the following:<br /> <br /> <center><math>h(mT) = T \int_{-v}^v \hat H (f) e^{j 2 \pi f m t} df </math></center><br /> <br /> where the desired frequencies are in the range from <math>-v</math> to <math>v</math> and <math> \hat H (f) </math> is the desired response.<br /> <br /> Example: Desing an FIR low pass filter to pass between <math> - \frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.<br /> <br /> <math>\hat H (f) = \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math><br /> <br /> The FIR filter would then be:<br /> <br /> <math> h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}</math><br /> <br /> <math> = \frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m} = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math><br /> <br /> and the actual frequency response would then be:<br /> <br /> <math>\sum_{m=-M}^M h(mT) e^{- j 2 \pi f m T} = \sum_{m=-M}^M \frac{ sin \left ( \frac{\pi m}{2} \right) }{\pi m} e^{- j 2 \pi f m T} </math><br /> <br /> <br /> <br /> The frequency response plot of this particular FIR filter would look like this:<br /> <br /> [[Image:Frequency_Response_Pict.png]]<br /> <br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo<br /> <br /> Image of the frequency response plot taken from the FIR example page [[FIR_Filter_Example|here]].</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1402 FIRJEW 2005-12-06T21:48:23Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop. Because there is no feedback, the response of an FIR filter to an impulse is finite. <br /> <br /> The equation for an FIR filter would look like the following:<br /> <br /> <center><math>h(mT) = T \int_{-v}^v \hat H (f) e^{j 2 \pi f m t} df </math></center><br /> <br /> where the desired frequencies are in the range from <math>-v</math> to <math>v</math> and <math> \hat H (f) </math> is the desired response.<br /> <br /> Example: Desing an FIR low pass filter to pass between <math> - \frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.<br /> <br /> <math>\hat H (f) = \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math><br /> <br /> The FIR filter would then be:<br /> <br /> <math> h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}</math><br /> <br /> <math> = \frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m} = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math><br /> <br /> and the actual frequency response would then be:<br /> <br /> <math>\sum_{m=-M}^M h(mT) e^{- j 2 \pi f m T} = \sum_{m=-M}^M \frac{ sin \left ( \frac{\pi m}{2} \right) }{\pi m} e^{- j 2 \pi f m T} </math><br /> <br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1401 FIRJEW 2005-12-06T21:41:59Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop. Because there is no feedback, the response of an FIR filter to an impulse is finite. <br /> <br /> The equation for an FIR filter would look like the following:<br /> <br /> <center><math>h(mT) = T \int_{-v}^v \hat H (f) e^{j 2 \pi f m t} df </math></center><br /> <br /> where the desired frequencies are in the range from <math>-v</math> to <math>v</math> and <math> \hat H (f) </math> is the desired response.<br /> <br /> Example: Desing an FIR low pass filter to pass between <math> - \frac{1}{4T} < f < \frac{1}{4T} </math> and reject the rest.<br /> <br /> <math>\hat H (f) = \begin{cases} 1, & |f| \le \frac{1}{4T} \\ 0, & else \end{cases} </math><br /> <br /> The FIR filter would then be:<br /> <br /> <math> h(mT) = T \int_{\frac{-1}{4T}}^{\frac{1}{4T}} 1 e^{j 2 \pi f m t} df = T\frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m T}</math><br /> <br /> <math> = \frac{e^{j 2 \pi \frac{1}{4T} m T} - e^{j 2 \pi \frac{-1}{4T} m T}}{j 2 \pi m} = \left ( \frac{1}{2} \right) \frac{sin \left ( \frac{\pi m}{2} \right)} {\left(\frac{\pi m}{2}\right)}</math><br /> <br /> <br /> <br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1399 FIRJEW 2005-12-06T21:01:54Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback loop.<br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1397 FIRJEW 2005-12-06T20:55:08Z

<p>Wonoje: /* Finite Impulse Response Filters */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> A Finite Impurse Response filter (aka FIR filter) is a type of filter often used in digital signal processing and has no feedback looop.<br /> <br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=4066 CDPlayerJEW 2005-12-06T19:22:33Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u \left (t+\frac{T}{2} \right) - u \left (t-\frac{T}{2} \right)</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <center><math>= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.<br /> <br /> This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.<br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1395 CDPlayerJEW 2005-12-06T00:55:32Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <center><math>= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.<br /> <br /> This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.<br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1391 CDPlayerJEW 2005-12-06T00:55:20Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <center><math>= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.<br /> <br /> This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.<br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1390 CDPlayerJEW 2005-12-06T00:54:44Z

<p>Wonoje: /* \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <center><math>= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small><br /> <br /> </math><br /> <br /> <math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.<br /> <br /> This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1389 CDPlayerJEW 2005-12-06T00:54:04Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> = \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1388 CDPlayerJEW 2005-12-06T00:53:19Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <br /> <math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.<br /> <br /> This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.<br /> <br /> = \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1387 CDPlayerJEW 2005-12-06T00:48:00Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center><br /> <br /> <center><math><br /> = \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)<br /> </math></center><br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1386 CDPlayerJEW 2005-12-06T00:34:35Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{n}) \delta (t-\frac{mT}{n}) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center><br /> <br /> Graphically, this looks like this:<br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <br /> Mathematically, the equation is:<br /> <center><math>x(t)*h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t - \frac{mT}{2}) = \hat y (t) = \sum_{l=-\infty}^\infty \left \sum_{m=-M}^M x \left \frac{l-m}{2} T \right h(\frac{mT}{2}) \right \delta (t - \frac{lT}{2})<br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1385 CDPlayerJEW 2005-12-06T00:27:25Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{n}) \delta (t-\frac{mT}{n}) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center><br /> <br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1384 CDPlayerJEW 2005-12-06T00:26:52Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{n}) \delta (t-\frac{mT}{n}) </math></center><br /> <br /> where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,<br /> <br /> <center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center><br /> <br /> [[Image:2x01.jpg|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg.jpg|Illustration of 2x Oversampling pt.2]]<br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=File:2x02.jpg&diff=4077 File:2x02.jpg 2005-12-06T00:25:46Z

<p>Wonoje: </p> <hr /> <div></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=File:2x01.jpg&diff=4076 File:2x01.jpg 2005-12-06T00:25:32Z

<p>Wonoje: </p> <hr /> <div></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1383 CDPlayerJEW 2005-12-06T00:18:56Z

<p>Wonoje: /* 2x Oversampling */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.<br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1382 CDPlayerJEW 2005-12-06T00:09:52Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> It turns out that the frequency graph shown above is very similar to the frequency graph of the original sound signal that was recorded. So, the signal is then sent to the speaker(s) for playback.<br /> <br /> ==2x Oversampling==<br /> Some feel that the sound quality of the playback is increased through oversampling. Oversampling requires the signal processor to "fill in" the gaps in the discretized signal thus forming a smoother stepped graph. As noted above, the stepped graph is the result of convolving the discretized sound signal with a pulse function.<br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1381 CDPlayerJEW 2005-12-05T23:57:28Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) * p(t) = \sum_{n=-\infty}^\infty x(nT)p(t-nT) </math></center><br /> <br /> This convolution, which is the convolution of the discretized signal and <math>p(t)</math>, a pulse function, will yield a graph that is no longer discrete, but is stepped. The following is an example:<br /> <br /> <center>[[Image:DAOutput.jpg|Convolved Graph]]</center><br /> <br /> In the frequency domain, the above stepped signal will look like the following:<br /> <br /> <center>[[Image:TnD.jpg|Convolution (Time Domain) and it's frequency domain counterpart]]</center><br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=File:TnD.jpg&diff=4075 File:TnD.jpg 2005-12-05T23:56:38Z

<p>Wonoje: CD Player Assignment: Stepped function (time domain) and its frequency domain counterpart.</p> <hr /> <div>CD Player Assignment: Stepped function (time domain) and its frequency domain counterpart.</div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1380 CDPlayerJEW 2005-12-05T23:32:29Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:<br /> <br /> <center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center><br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image Player Diagram by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1379 CDPlayerJEW 2005-12-05T23:28:56Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> This discrete signal can be represented mathematically by:<br /> <br /> <center><math>\sum_{n=-\infty}^\infty x(nT) \delta (t-nT) </math><br /> </center><br /> <br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image Player Diagram by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1378 CDPlayerJEW 2005-12-05T23:15:04Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the interval between samples.<br /> <br /> <br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image Player Diagram by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=CDPlayerJEW&diff=1377 CDPlayerJEW 2005-12-05T23:14:08Z

<p>Wonoje: /* How a CD Player Works */</p> <hr /> <div>==How a CD Player Works==<br /> A CD player reads a dicrete set of data off a CD. In short, a CD player takes this data and sends it through a digitla to analog converter, then through a low pass filter, and finally is output through speakers. A simple diagram illustrates this below.<br /> <br /> <br /> <center><br /> [[Image:CDplayerdiagram.jpg|Simple Block Diagram of Signal Pathway]]<br /> </center><br /> <br /> <br /> When an audio CD is recorded, the music has an infinite amount of data points and can be represented as a continuous function of time <math> x(t) </math>. Because a medium, such as a CD, has a finite amount of space, it will not be able to hold <math> x(t) </math> since it has an infinite amount of data. Instead, the music is sampled at intervals to create a discrete function of time <math> x(nT) </math> where <math> n </math> is an integer and <math> T </math> is the period.<br /> <br /> <br /> <br /> <small>- Principle author of this page: Jeffrey Wonoprabowo<br /> <br /> - Image Player Diagram by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FIRJEW&diff=1396 FIRJEW 2005-12-05T03:58:13Z

<p>Wonoje: /* Finite Impulse Response */</p> <hr /> <div>==Finite Impulse Response Filters==<br /> <br /> <small>Principle author: Jeffrey Wonoprabowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=4067 FourierTransformsJW 2005-12-05T03:57:15Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Given: <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> Given: <math>X(f) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t}df </math><br /> <br /> <math>X(f-f_o) = \int_{-\infty}^\infty X(f) e^{j 2 \pi (f-f_o) t}df = \mathcal{F}[e^{j 2 \pi f_o t}x(t)]</math><br /> <br /> =====Modulation=====<br /> <math>\mathcal{F}[cos(2 \pi f_o t)x(t)] = \int_{\infty}^\infty x(t)cos(2 \pi f_o t) e^{-j 2 \pi f t} dt</math><br /> <br /> <math>= \int_{\infty}^\infty \frac{e^{j 2 \pi f_o t} + e^{-j 2 \pi f_o t}}{2} x(t) e^{j 2 \pi f t} dt </math><br /> <br /> <math> = \frac{1}{2} \int_{\infty}^\infty x(t) e^{-j 2 \pi (f-f_o) t} dt + \frac{1}{2} \int_{\infty}^\infty x(t)e^{-j 2 \pi (f+f_o) t} dt</math><br /> <br /> <math> = \frac{1}{2}X(f-f_o) + \frac{1}{2}X(f+f_o)</math><br /> <br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1374 FourierTransformsJW 2005-12-05T03:52:30Z

<p>Wonoje: /* Modulation */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Given: <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> Given: <math>X(f) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t}df </math><br /> <br /> <math>X(f-f_o) = \int_{-\infty}^\infty X(f) e^{j 2 \pi (f-f_o) t}df = \mathcal{F}[e^{j 2 \pi f_o t}x(t)]</math><br /> <br /> =====Modulation=====<br /> <math>\mathcal{F}[cos(2 \pi f_o t)x(t)] = \int_{\infty}^\infty x(t)cos(2 \pi f_o t) e^{-j 2 \pi f t} dt</math><br /> <br /> <math>= \int_{\infty}^\infty \frac{e^{j 2 \pi f_o t} + e^{-j 2 \pi f_o t}}{2} x(t) e^{j 2 \pi f t} dt </math><br /> <br /> <math> = \frac{1}{2} \int_{\infty}^\infty x(t) e^{-j 2 \pi (f-f_o) t} dt + \frac{1}{2} \int_{\infty}^\infty x(t)e^{-j 2 \pi (f+f_o) t} dt</math><br /> <br /> <math> = \frac{1}{2}X(f-f_o) + \frac{1}{2}X(f+f_o)</math><br /> <br /> =====Time Scaling=====<br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1373 FourierTransformsJW 2005-12-05T03:44:28Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Given: <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> Given: <math>X(f) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t}df </math><br /> <br /> <math>X(f-f_o) = \int_{-\infty}^\infty X(f) e^{j 2 \pi (f-f_o) t}df = \mathcal{F}[e^{j 2 \pi f_o t}x(t)]</math><br /> <br /> =====Modulation=====<br /> <math>\mathcal{F}[cos(2 \pi f_o t)x(t)] = \int_{\infty}^\infty x(t)cos(2 \pi f_o t) e^{j 2 \pi f t} dt</math><br /> <br /> =====Time Scaling=====<br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1372 FourierTransformsJW 2005-12-05T03:37:20Z

<p>Wonoje: /* Frequency Shift */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Let <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> <math>X(f) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t}df </math><br /> <br /> <math>X(f-f_o) = \int_{-\infty}^\infty X(f) e^{j 2 \pi (f-f_o) t}df </math><br /> <br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1371 FourierTransformsJW 2005-12-05T03:36:50Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Let <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> <math>X(f) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t}df </math<br /> <br /> <math>X(f-f_o) = \int_{-\infty}^\infty X(f) e^{j 2 \pi (f-f_o) t}df </math><br /> <br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1370 FourierTransformsJW 2005-12-05T03:33:27Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Let <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> =====Time Shift=====<br /> <math>x(t-t_o) = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f(t-t_o)} df</math><br /> <br /> <math> = \int_{-\infty}^\infty e^{-j 2 \pi f t_o} X(f) e^{j 2 \pi f t} df</math><br /> <br /> <math> = \mathcal{F}^{-1}[e^{-j 2 \pi f t_o}X(f)] </math><br /> <br /> =====Frequency Shift =====<br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1369 FourierTransformsJW 2005-12-05T03:28:06Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Let <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> =====Differentiation=====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]</math><br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1368 FourierTransformsJW 2005-12-05T03:27:31Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> Let <math>x(t) = \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df</math><br /> <br /> ====Differentiation====<br /> <math>\frac{dx}{dt} = \int_{-\infty}^\infty j 2 \pi X(f) e^{j 2 \pi f t} df = \mathcal{F}^{-1}[j 2 \pi f X(f)]<br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small><br /> <br /> </math></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1367 FourierTransformsJW 2005-12-05T03:21:55Z

<p>Wonoje: /* Fourier Transform */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <br /> ===Some Properties of the Fourier Transform===<br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje https://fweb.wallawalla.edu/class-wiki/index.php?title=FourierTransformsJW&diff=1366 FourierTransformsJW 2005-12-05T03:16:05Z

<p>Wonoje: /* Fourier Transform Defined */</p> <hr /> <div>==Fourier Transform==<br /> ===Introduction===<br /> A Fourier series allows a periodic function to be represented as the sum of sine and/or cosine waves. This is very useful, because functions in the time domain can be expressed in the frequency domain. The frequency domain can, at times, be easier to work with. This is where the Fourier transfrom comes in. It allows for a nonperidic function of time to be converted (or transformed) into a function of frequency. <br /> <br /> This conversion is analgous to the conversion from cartesian coordinates to polar or spherical coordinates. The location of the point does not change; only the directions for how to get there.<br /> <br /> ===Fourier Transform Defined===<br /> The Fourier Transform of a function <math>x(t)</math> can be defined as:<br /> <br /> <math> \mathcal{F}[x(t)] = \int_{-\infty}^\infty x(t) e^{-j 2 \pi f t} dt = X(f) </math><br /> <br /> then the inverse Fourier transform of a function <math>X(f)</math> will be:<br /> <br /> <math> \mathcal{F}^{-1}[X(f)] = \int_{-\infty}^\infty X(f) e^{j 2 \pi f t} df = x(t) </math><br /> <br /> where <math>x(t)</math> is a function of time and <math>X(f)</math> is the Fourier transform of <math>x(t)</math> and is it's Fourier Transform.<br /> <br /> <small>Principle author of this page: Jeffrey Wonoprbowo</small></div>

Wonoje