https://fweb.wallawalla.edu/class-wiki/api.php?action=feedcontributions&feedformat=atom&user=Nicholas.ChristmanClass Wiki - User contributions [en]2026-06-07T18:27:29ZUser contributionsMediaWiki 1.43.0https://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_SSB&diff=9561Nick SSB2010-04-09T04:25:30Z<p>Nicholas.Christman: New page: ===<span style="font-variant: small-caps;">How It Works: Single Side Band (SSB) Modulation</span>===</p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Single Side Band (SSB) Modulation</span>===</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Engineering_Electronics&diff=9560Engineering Electronics2010-04-09T04:24:31Z<p>Nicholas.Christman: </p>
<hr />
<div>==Publish or Perish Game==<br />
*[[Electronics Score Pages]]<br />
*[[Rules]]<br />
*[[Conference Deadlines]]<br />
<br />
==Questions==<br />
<br />
==Links==<br />
*[http://www.dspguru.com/sites/dspguru//files/QuadSignals.pdf Quadrature Signals Explained]<br />
* Software Defined Radio Links<br />
**[http://people.wallawalla.edu/~Rob.Frohne/R2_DSP/9804x040.pdf R2 DSP (an early software defined radio using a dedicated DSP)]<br />
**[http://www.nonstopsystems.com/radio/frank_radio_sdr.htm Softrock and Theory]<br />
**[http://www.wb5rvz.com/sdr/ Softrock Build Instructions and Notes]<br />
**[http://groups.yahoo.com/group/softrock40/ Softrock Yahoo Interest Group]<br />
**[http://www.flex-radio.com/News.aspx?topic=publications This collection of Software Defined Radio publications is fantastic.]<br />
**[http://www.sdradio.eu/sdradio/ SDRadio]<br />
**[http://openhpsdr.org/ Open High Performance Software Defined Radio]<br />
<br />
==2010 Contributors==<br />
<br />
#[[Greg Fong|Fong, Greg]]<br />
#[[Ben Henry|Henry, Ben]]<br />
#[[Lau, Chris]]<br />
#[[Shepherd,Victor]]<br />
#[[Vier, Michael]]<br />
<br />
==2010 Articles==<br />
*[[Ideal vs. Nonideal Op Amps]]<br />
*[[Chapter 1]]<br />
*[[Chapter 2]]<br />
*[[Basic Op Amp circuits]]<br />
*[[Key Facts from Reading Chapter 1]]<br />
*[[Golden Rules]]<br />
*[[Integrator_Amplifier]] (by [[Ben Henry|Ben]])<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
<br />
*[[Chapter 3 Problems]] by [[Ben Henry|Ben]] <br />
* Disecting an Instrumentation Amplifier via [[Superposition]]<br />
*[[Reading from Chapter 4]]<br />
<br />
==Draft Articles awaiting review==<br />
*[[Feedback in Amplifiers]]<br />
<br />
<br />
==Contributing Articles==<br />
<br />
*[[Generalized Transmitter]] (in progress, Luke)<br />
*[[Generalized Receiver]] (in progress, Luke)<br />
*[[Electronics Receiver]] (in progress, Kevin)<br />
*[[Christman_GeneralizedReceiver|Generalized Receiver]] (Nick Christman)<br />
*[[Generalized Receiver Explanation]] (Jodi Hodge)<br />
*[[Eric's Generalized Receiver Explanation]] (Eric Clay)<br />
*[[Yet another Generalized Receiver Explanation]] (Joshua Sarris)</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Communications_Systems&diff=9559Communications Systems2010-04-09T04:24:07Z<p>Nicholas.Christman: </p>
<hr />
<div>ENGR 456<br />
====Articles====<br />
* [[AM vs FM Modulation]] (Jodi and Eric)<br />
* [[Nick_SSB | Single Side Band (SSB) Modulation]] (Nick Christman, in progress)<br />
<br />
====Links====<br />
*[http://people.wallawalla.edu/~Rob.Frohne/ClassHandouts/Communications/Jeff_McDow.pdf Jeff McDow's Senior Report on Morse Code Detection]</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Communications_Systems&diff=9558Communications Systems2010-04-09T04:23:28Z<p>Nicholas.Christman: </p>
<hr />
<div>ENGR 456<br />
====Articles====<br />
[[AM vs FM Modulation]] (Jodi and Eric)<br />
[[Nick_SSB | Single Side Band (SSB) Modulation]] (Nick Christman, in progress)<br />
<br />
====Links====<br />
*[http://people.wallawalla.edu/~Rob.Frohne/ClassHandouts/Communications/Jeff_McDow.pdf Jeff McDow's Senior Report on Morse Code Detection]</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9557Christman GeneralizedReceiver2010-04-09T04:21:55Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as ''baseband'', which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a ''bandpass'' signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
How is the data split and shifted you ask? Mathematically speaking, in the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{(\omega_c t)} + j\sin{(\omega_c t)}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{(\omega_c t)} + j\sin{(\omega_c t)})] = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
As you can also see from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>) which is at a frequency inaudible to human ears. Thus, in order to get the original communication data it is necessary to somehow bring the bandpass signal back down to its original baseband form -- and this is what the second quarter of electronics is all about.<br />
If you have not already discovered, to recover the original data <math>x(t)</math> and <math>y(t)</math> from the signal signal <math>v(t)</math> we must perform a task known as "mixing" and then run the results through low pass filters. Simply put, we need to multiply <math>v(t)</math> by <math>2 \cdot \cos{(\omega_c t)}</math> and <math>-2 \cdot \sin{(\omega_c t)}</math> (the "<math>2</math>" is not necessary, but it makes the math a little nicer in the end). To see this mathematically, recall that <math> v(t) = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)} </math>. Therefore,<br />
<br/><br />
<center><br />
<math>2\cos{(\omega_c t)} \cdot v(t) = 2cos{(\omega_c t)} \cdot [x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}] = 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)}</math><br />
</center><br />
<br/><br />
Using the trigonometric identities<br />
<br />
<center><br />
<math>\cos^{2}{\theta} = \frac{1+\cos{2 \theta}}{2}</math><br />
</center><br />
and<br />
<center><br />
<math>\cos{\theta}\sin{\theta} = \frac{1}{2}\sin{\theta}</math>,<br />
</center><br />
we can see that<br />
<center><br />
<br />
<math>\displaystyle 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)} = x(t) + x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)} </math>.<br />
<br />
</center><br />
<br />
<br/><br />
But when this result is ran through a low pass filter it is fairly obvious that <math>x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)}</math> is filtered out (because of the <math>2 \omega_c</math>), but <math>x(t)</math> is unaltered by the filter (because it is independent of frequency); therefore, we are left with <math>x(t)</math> as desired! (Similarly, <math>y(t)</math> can be obtained.)<br />
This is essentially how a generalized receiver functions: given a bandpass signal, <math>v(t)</math>, and we can use a mixer and low pass filters to obtain an original baseband signal, <math>m(t)</math> (in the form of <math>x(t)</math> and <math>y(t)</math>, of course).<br />
<br />
<br/><br />
----<br />
<br />
<br/><br />
Please feel free to fix any mistakes. Please list your name, date fixed, and short description below.<br />
<br />
* Name, Date, Description</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9556Christman GeneralizedReceiver2010-04-09T04:20:41Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as ''baseband'', which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a ''bandpass'' signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
How is the data split and shifted you ask? Mathematically speaking, in the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{(\omega_c t)} + j\sin{(\omega_c t)}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{(\omega_c t)} + j\sin{(\omega_c t)})] = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
As you can also see from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>) which is at a frequency inaudible to human ears. Thus, in order to get the original communication data it is necessary to somehow bring the bandpass signal back down to its original baseband form -- and this is what the second quarter of electronics is all about.<br />
If you have not already discovered, to recover the original data <math>x(t)</math> and <math>y(t)</math> from the signal signal <math>v(t)</math> we must perform a task known as "mixing" and then run the results through low pass filters. Simply put, we need to multiply <math>v(t)</math> by <math>2 \cdot \cos{(\omega_c t)}</math> and <math>-2 \cdot \sin{(\omega_c t)}</math> (the "<math>2</math>" is not necessary, but it makes the math a little nicer in the end). To see this mathematically, recall that <math> v(t) = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)} </math>. Therefore,<br />
<br />
<center><br />
<math>2\cos{(\omega_c t)} \cdot v(t) = 2cos{(\omega_c t)} \cdot [x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}] = 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)}</math><br />
</center><br />
<br />
Using the trigonometric identities<br />
<br />
<center><br />
<math>\cos^{2}{\theta} = \frac{1+\cos{2 \theta}}{2}</math><br />
</center><br />
and<br />
<center><br />
<math>\cos{\theta}\sin{\theta} = \frac{1}{2}\sin{\theta}</math>,<br />
</center><br />
we can see that<br />
<center><br />
<br />
<math>\displaystyle 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)} = x(t) + x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)} </math>.<br />
<br />
</center><br />
<br />
But when this result is ran through a low pass filter it is fairly obvious that <math>x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)}</math> is filtered out (because of the <math>2 \omega_c</math>), but <math>x(t)</math> is unaltered by the filter (because it is independent of frequency); therefore, we are left with <math>x(t)</math> as desired! (Similarly, <math>y(t)</math> can be obtained.)<br />
This is essentially how a generalized receiver functions: given a bandpass signal, <math>v(t)</math>, and we can use a mixer and low pass filters to obtain an original baseband signal, <math>m(t)</math> (in the form of <math>x(t)</math> and <math>y(t)</math>, of course).<br />
----<br />
Please feel free to fix any mistakes. Please list your name, date fixed, and short description below.<br />
<br />
* Name, Date, Description</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9555Christman GeneralizedReceiver2010-04-09T04:18:38Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as ''baseband'', which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a ''bandpass'' signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
How is the data split and shifted you ask? Mathematically speaking, in the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{(\omega_c t)} + j\sin{(\omega_c t)}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{(\omega_c t)} + j\sin{(\omega_c t)})] = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
As you can also see from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>) which is at a frequency inaudible to human ears. Thus, in order to get the original communication data it is necessary to somehow bring the bandpass signal back down to its original baseband form -- and this is what the second quarter of electronics is all about.<br />
If you have not already discovered, to recover the original data <math>x(t)</math> and <math>y(t)</math> from the signal signal <math>v(t)</math> we must perform a task known as "mixing" and then run the results through low pass filters. Simply put, we need to multiply <math>v(t)</math> by <math>2 \cdot \cos{(\omega_c t)}</math> and <math>-2 \cdot \sin{(\omega_c t)}</math> (the "<math>2</math>" is not necessary, but it makes the math a little nicer in the end). To see this mathematically, recall that <math> v(t) = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)} </math>. Therefore,<br />
<br />
<center><br />
<math>2\cos{(\omega_c t)} \cdot v(t) = 2cos{(\omega_c t)} \cdot [x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}] = 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)}</math><br />
</center><br />
<br />
Using the trigonometric identities<br />
<br />
<center><br />
<math>\cos^{2}{\theta} = \frac{1+\cos{2 \theta}}{2}</math><br />
</center><br />
and<br />
<center><br />
<math>\cos{\theta}\sin{\theta} = \frac{1}{2}\sin{\theta}</math>,<br />
</center><br />
we can see that<br />
<center><br />
<br />
<math>\displaystyle 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)} = x(t) + x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)} </math>.<br />
<br />
</center><br />
<br />
But when this result is ran through a low pass filter it is fairly obvious that <math>x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)}</math> is filtered out (because of the <math>2 \omega_c</math>), but <math>x(t)</math> is unaltered by the filter (because it is independent of frequency); therefore, we are left with <math>x(t)</math> as desired! (Similarly, <math>y(t)</math> can be obtained.)<br />
This is essentially how a generalized receiver functions: given a bandpass signal, <math>v(t)</math>, and we can use a mixer and low pass filters to obtain an original baseband signal, <math>m(t)</math> (in the form of <math>x(t)</math> and <math>y(t)</math>, of course).</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9554Christman GeneralizedReceiver2010-04-09T04:17:57Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as ''baseband'', which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a ''bandpass'' signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
How is the data split and shifted you ask? Mathematically speaking, in the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{(\omega_c t)} + j\sin{(\omega_c t)}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{(\omega_c t)} + j\sin{(\omega_c t)})] = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>) which is at a frequency inaudible to human ears. Thus, in order to get the original communication data it is necessary to somehow bring the bandpass signal back down to its original baseband form -- and this is what the second quarter of electronics is all about.<br />
If you have not already discovered, to recover the original data <math>x(t)</math> and <math>y(t)</math> from the signal signal <math>v(t)</math> we must perform a task known as "mixing" and then run the results through low pass filters. Simply put, we need to multiply <math>v(t)</math> by <math>2 \cdot \cos{(\omega_c t)}</math> and <math>-2 \cdot \sin{(\omega_c t)}</math> (the "<math>2</math>" is not necessary, but it makes the math a little nicer in the end). To see this mathematically, recall that <math> v(t) = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)} </math>. Therefore,<br />
<br />
<center><br />
<math>2\cos{(\omega_c t)} \cdot v(t) = 2cos{(\omega_c t)} \cdot [x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}] = 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)}</math><br />
</center><br />
<br />
Using the trigonometric identities<br />
<br />
<center><br />
<math>\cos^{2}{\theta} = \frac{1+\cos{2 \theta}}{2}</math><br />
</center><br />
and<br />
<center><br />
<math>\cos{\theta}\sin{\theta} = \frac{1}{2}\sin{\theta}</math>,<br />
</center><br />
we can see that<br />
<center><br />
<br />
<math>\displaystyle 2x(t)\cos^{2}{(\omega_c t)} - 2y(t)\cos{(\omega_c t)}\sin{(\omega_c t)} = x(t) + x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)} </math>.<br />
<br />
</center><br />
<br />
But when this result is ran through a low pass filter it is fairly obvious that <math>x(t)\cos{(2 \omega_c t)} - y(t)\sin{(2 \omega_c t)}</math> is filtered out (because of the <math>2 \omega_c</math>), but <math>x(t)</math> is unaltered by the filter (because it is independent of frequency); therefore, we are left with <math>x(t)</math> as desired! (Similarly, <math>y(t)</math> can be obtained.)<br />
This is essentially how a generalized receiver functions: given a bandpass signal, <math>v(t)</math>, and we can use a mixer and low pass filters to obtain an original baseband signal, <math>m(t)</math> (in the form of <math>x(t)</math> and <math>y(t)</math>, of course).</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9553Christman GeneralizedReceiver2010-04-09T04:03:35Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as ''baseband'', which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a ''bandpass'' signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
How is the data split and shifted you ask? Mathematically speaking, in the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{(\omega_c t)} + j\sin{(\omega_c t)}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{(\omega_c t)} + j\sin{(\omega_c t)})] = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>) which is at a frequency inaudible to human ears. Thus, in order to get the original communication data it is necessary to somehow bring the bandpass signal back down to its original baseband form -- and this is what the second quarter of electronics is all about.<br />
If you have not already discovered, to recover the original data <math>x(t)</math> and <math>y(t)</math> from the signal signal <math>v(t)</math> we must perform a task known as "mixing" and then run the results through low pass filters. Simply put, we need to multiply <math>v(t)</math> by <math>\cos{(\omega_c t)}</math> and <math>-\sin{(\omega_c t)}</math>. To see this mathematically, recall that <math> v(t) = x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)} </math>. Therefore,<br />
<br />
<center><br />
<math>\cos{(\omega_c t)} \cdot v(t) = cos{(\omega_c t)} \cdot [x(t)\cos{(\omega_c t)} - y(t)\sin{(\omega_c t)}] = x(t)\cos^{2}{(\omega_c t)} - y(t)\cos{(\omega_c t)}\sin{(\omega_c t)}</math><br />
</center><br />
<br />
But when <math>\cos{(\omega_c t)}\sin{(\omega_c t)}</math> is ran through a low pass filter it is filtered out, whereas <math>\cos^{2}{(\omega_c t)}</math> is not; therefore, we are left with <math>x(t)</math> as desired! (Similarly, <math>y(t)</math> can be obtained.)</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9552Christman GeneralizedReceiver2010-04-09T03:36:34Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as \textit{baseband}, which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a \textit{bandpass} signal. This concept is illustrated below:<br />
<br/><br />
<br />
<center>[[Image:HW2_BandpassBaseband.png]]</center><br />
<br />
<br/><br />
Why is the data split and shifted you ask? In the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{\omega_{c}t} + j\sin{\omega_{c}t}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{\omega_{c}t} + j\sin{\omega_{c}t}) = x(t)\cos{\omega_{c}t} - y(t)\sin{\omega_{c}t}</math>.<br />
<br />
</center><br />
<br />
To visualize this process, observe the following figure:<br />
<br/><br />
This is why it is necessary to split <math>m(t)</math> into the the two signals <math>x(t) \text{ and } y(t)</math>. As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>).</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:HW2_BandpassBaseband.png&diff=9551File:HW2 BandpassBaseband.png2010-04-09T03:35:53Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9550Christman GeneralizedReceiver2010-04-09T03:25:11Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as \textit{baseband}, which consists of frequencies near D.C. (or <math>f_c = 0</math>). When you actually send the communication data, however, you want to send it via much higher frequency (one which is inaudible to humans) and this creates a \textit{bandpass} signal. This concept is illustrated below:<br />
<br />
Why is the data split and shifted you ask? In the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{\omega_{c}t} + j\sin{\omega_{c}t}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{\omega_{c}t} + j\sin{\omega_{c}t}) = x(t)\cos{\omega_{c}t} - y(t)\sin{\omega_{c}t}</math>.<br />
<br />
</center><br />
<br />
To visualize this process, observe the following figure:<br />
<br/><br />
This is why it is necessary to split <math>m(t)</math> into the the two signals <math>x(t) \text{ and } y(t)</math>. As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>).</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9549Christman GeneralizedReceiver2010-04-09T03:23:19Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>, and shifted to a frequency <math>f_c</math> (and <math>-f_c</math> because of the notion of a complex conjugate). Note that the original wireless communication data that you want to send is in the form known as \textit{baseband}, which consists of frequencies near D.C.. When you actually send the communication data, however, you want to send it via much higher frequency (one inaudible to humans) and this creates a \textit{bandpass} signal. This concept is illustrated below:<br />
<br />
Why is the data split and shifted you ask? In the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{\omega_{c}t} + j\sin{\omega_{c}t}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{\omega_{c}t} + j\sin{\omega_{c}t}) = x(t)\cos{\omega_{c}t} - y(t)\sin{\omega_{c}t}</math>.<br />
<br />
</center><br />
<br />
To visualize this process, observe the following figure:<br />
<br/><br />
This is why it is necessary to split <math>m(t)</math> into the the two signals <math>x(t) \text{ and } y(t)</math>. As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. <br />
<br/><br />
<br />
The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br />
<center>[[Image:HW2_GenRec.jpg]]</center><br />
<br />
Receiving the original data from a wireless signal is very similar to that of transmitting the data (as described before). In short, the data that we are interested in (<math>m(t)</math>) is embedded within the carrying signal (<math>v(t)</math>).</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9513Christman GeneralizedReceiver2010-04-06T16:10:32Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>. Why is this you ask? In the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>,<br />
</center> <br />
<br />
<br/><br />
<br />
where <math>g(t) = x(t) + jy(t)</math> is the signal to be sent and <math>\scriptstyle e^{j\omega_{c}t} = \cos{\omega_{c}t} + j\sin{\omega_{c}t}</math> (Euler's identity) shifts the signal in the frequency domain by <math>\omega_{c} = 2 \pi f_{c}</math>.<br />
<br/><br />
The above formula can then be rewritten to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{\omega_{c}t} + j\sin{\omega_{c}t}) = x(t)\cos{\omega_{c}t} - y(t)\sin{\omega_{c}t}</math>.<br />
<br />
</center><br />
<br />
To visualize this process, observe the following figure:<br />
<br/><br />
This is why it is necessary to split <math>m(t)</math> into the the two signals <math>x(t) \text{ and } y(t)</math>. As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. The process of receiving the data is very similar, as is described below.<br />
<br/><br />
<center>[[Image:HW2_GenRec.jpg]]</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9470Christman GeneralizedReceiver2010-04-05T06:37:56Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
The process of transmitting data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:<br />
<br/><br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
As can be seen from the figure above, in order to transmit <math>m(t)</math>, one must first process the data using a baseband processor (usually accomplished with software). As a result of this process the original data will be split into two signals, <math>x(t)</math> and <math>y(t)</math>. Why is this you ask? In the world of "Communication Systems" a signal to be transmitted can be written as <br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>, where <math>\displaystyle g(t) = x(t) + jy(t)</math>.<br />
<br />
</center><br />
<br/><br />
Using Euler's identity, <math>\scriptstyle e^{j\omega_{c}t} = \cos{\omega_{c}t} + j\sin{\omega_{c}t}</math>, we are able to re-write the above equation and simplify so to obtain the following relationship:<br />
<br/><br />
<center><br />
<br />
<math>\displaystyle v(t) = Re[(x(t) + jy(t)) \cdot e^{j\omega_{c}t}] = Re[(x(t) + jy(t)) \cdot (\cos{\omega_{c}t} + j\sin{\omega_{c}t}) = x(t)\cos{\omega_{c}t} - y(t)\sin{\omega_{c}t}</math>.<br />
<br />
</center><br />
<br />
<br/><br />
<br />
This is why it is necessary to split <math>m(t)</math> into the the two signals <math>x(t) \text{ and } y(t)</math>. As you can also from the above relationship, in order to obtain the appropriate output signal <math>v(t)</math> one must multiply <math>x(t)</math> by a cosine function and <math>y(t)</math> by a negative sine function and then sum the results (this is illustrated in the figure above). In simplified terms and details, this is essentially how the data <math>m(t)</math> is transmitted. The process of receiving the data is very similar, as is described below.<br />
<br/><br />
<center>[[Image:HW2_GenRec.jpg]]</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9469Christman GeneralizedReceiver2010-04-05T05:39:47Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
<br />
<br />
<center>[[Image:HW2_GenTran.jpg]]</center><br />
<br/><br />
<center>[[Image:HW2_GenRec.jpg]]</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:HW2_GenTran.jpg&diff=9468File:HW2 GenTran.jpg2010-04-05T05:39:15Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:HW2_GenRec.jpg&diff=9467File:HW2 GenRec.jpg2010-04-05T05:38:34Z<p>Nicholas.Christman: uploaded a new version of "Image:HW2 GenRec.jpg"</p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:HW2_GenRec.jpg&diff=9466File:HW2 GenRec.jpg2010-04-05T05:21:39Z<p>Nicholas.Christman: uploaded a new version of "Image:HW2 GenRec.jpg"</p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9465Christman GeneralizedReceiver2010-04-05T05:17:19Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===<br />
<br/><br />
<br />
<br/><br />
<center>[[Image:HW2_GenRec.jpg]]</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:HW2_GenRec.jpg&diff=9464File:HW2 GenRec.jpg2010-04-05T05:15:23Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9463Christman GeneralizedReceiver2010-04-05T04:45:38Z<p>Nicholas.Christman: </p>
<hr />
<div>===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9462Christman GeneralizedReceiver2010-04-05T04:44:17Z<p>Nicholas.Christman: </p>
<hr />
<div>=<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>=</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Christman_GeneralizedReceiver&diff=9461Christman GeneralizedReceiver2010-04-05T04:42:28Z<p>Nicholas.Christman: New page: = How it Works: Generalized Receiver =</p>
<hr />
<div>= How it Works: Generalized Receiver =</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Engineering_Electronics&diff=9460Engineering Electronics2010-04-05T04:40:42Z<p>Nicholas.Christman: /* Contributing Articles */</p>
<hr />
<div>==Publish or Perish Game==<br />
*[[Electronics Score Pages]]<br />
*[[Rules]]<br />
*[[Conference Deadlines]]<br />
<br />
==Questions==<br />
<br />
==Links==<br />
*[http://www.dspguru.com/sites/dspguru//files/QuadSignals.pdf Quadrature Signals Explained]<br />
* Software Defined Radio Links<br />
**[http://people.wallawalla.edu/~Rob.Frohne/R2_DSP/9804x040.pdf R2 DSP (an early software defined radio using a dedicated DSP)]<br />
**[http://www.nonstopsystems.com/radio/frank_radio_sdr.htm Softrock and Theory]<br />
**[http://www.wb5rvz.com/sdr/ Softrock Build Instructions and Notes]<br />
**[http://groups.yahoo.com/group/softrock40/ Softrock Yahoo Interest Group]<br />
**[http://www.flex-radio.com/News.aspx?topic=publications This collection of Software Defined Radio publications is fantastic.]<br />
**[http://www.sdradio.eu/sdradio/ SDRadio]<br />
**[http://openhpsdr.org/ Open High Performance Software Defined Radio]<br />
<br />
==2010 Contributors==<br />
<br />
#[[Greg Fong|Fong, Greg]]<br />
#[[Ben Henry|Henry, Ben]]<br />
#[[Lau, Chris]]<br />
#[[Shepherd,Victor]]<br />
#[[Vier, Michael]]<br />
<br />
==2010 Articles==<br />
*[[Ideal vs. Nonideal Op Amps]]<br />
*[[Chapter 1]]<br />
*[[Chapter 2]]<br />
*[[Basic Op Amp circuits]]<br />
*[[Key Facts from Reading Chapter 1]]<br />
*[[Golden Rules]]<br />
*[[Integrator_Amplifier]] (by [[Ben Henry|Ben]])<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
<br />
*[[Chapter 3 Problems]] by [[Ben Henry|Ben]] <br />
* Disecting an Instrumentation Amplifier via [[Superposition]]<br />
*[[Reading from Chapter 4]]<br />
<br />
==Draft Articles awaiting review==<br />
*[[Feedback in Amplifiers]]<br />
<br />
<br />
==Contributing Articles==<br />
<br />
*[[Generalized Transmitter]] (in progress, Luke)<br />
*[[Generalized Receiver]] (in progress, Luke)<br />
*[[Electronics Receiver]] (in progress, Kevin)<br />
*[[Christman_GeneralizedReceiver|Generalized Receiver]] (in progress, Nick)</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431&diff=8758Nick ENGR4312010-01-27T01:08:10Z<p>Nicholas.Christman: /* Drafts */</p>
<hr />
<div>= ENGR431 Homework =<br />
<br />
'''Back to [[Nick Christman]]'''<br />
<br/><br />
<br/><br />
<br />
== Drafts ==<br />
----<br />
[[Transformer Example Problem]] (Aric Vyhmeister, Kevin Starkey, Nick Christman)<br />
<br />
== Published ==<br />
----<br />
[[Nick_ENGR431_P1 | Paper 1: ''Brief Introduction to Magnetostatics'']]<br />
<br />
== Points ==<br />
----</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=8757Electromechanical Energy Conversion2010-01-27T01:06:22Z<p>Nicholas.Christman: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
[http://people.wallawalla.edu/~rob.frohne/ClassNotes/engr431index.htm Class Notes]<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
* If it's been approved by the reviewers, move it to the articles section<br />
<br />
==Announcements==<br />
<br />
If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.<br />
<br />
==Article Suggestions==<br />
(Please remove these when you complete the article.)<br />
# Rewrite the notes for the wiki.<br />
# Draw and explain the effect of the non-linear B-H curve on current waveforms for a voltage excited inductor.<br />
# Explain how to measure the B-H curve experimentally.<br />
# If the B-H curve was traced out more quickly in the experiment above, would the curve look different? If so why?<br />
# Show how to calculate the core losses of a nonlinear inductor using its i-v curve.<br />
# Explore transformers with more than one secondary winding.<br />
# What is the input impedance of an idea transformer with two secondaries, one with N2 turns and one with N<sub>3</sub> turns, each with a different load resistor on attached.<br />
# How do the mutual impedances relate to the turns ratios in transformers with more than one secondary?<br />
# Develop a circuit model for a non-ideal transformer with multiple secondaries.<br />
# Develop the theory of autotransformers.<br />
# Explore how leakage flux affects the inductance of an inductor. What if that flux is then recovered and the effect accounted for by mutual inductance? Does the result agree with the simple calculation of inductance without leakage?<br />
# Describe the coupling factor, k, used in Spice simulators and other circuit simulators. Relate it to the leakage, magnetizing, and mutual inductances. <br />
# Derive the <math>Y/\Delta</math> transformations.<br />
# Explore the voltage regulation <math>(V_{full ~load} - V_{no ~load} ) \over {V_{full ~load} } </math>x 100% as a function of the power factor angle on the load of a transformer. (You will note some surprising results in some cases.)<br />
# Describe the open circuit and short circuit test as applied to transformers.<br />
# Explore how much flux is in the core of a loaded ideal transformer.<br />
# Calculate and compare how much power can be delivered with three phase circuits as compared to a single phase circuits. Assume that the same amount of copper is available for the wire of both systems.<br />
# And if you don't understand any of the above, no surprise.<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Example: Ampere's Law]] (Tyler Anderson)<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Faraday's Law]]<br />
* [[Eddy Current]]<br />
* [[Example Problems of Magnetic Circuits]]<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
<br />
* [[Example: Ideal Transformer Exercise]] (John Hawkins)<br />
* [[Reference Terms and Units]] (Amy Crosby)<br />
* [[Ideal Transformer Example|Example: Ideal Transformer]]<br />
* [[Problem Set 1]](Jodi Hodge)<br />
* [[ANOTHER IDEAL TRANSFORMER!!!!!!!!!]]<br />
* [[Example: Magnetic Field]] (Amy Crosby) <br />
* [[Example: Metal Cart]] (Amy Crosby)<br />
* [[Class Notes]](Tyler Anderson)<br />
* [[The Class Notes]] ([[Kirk Betz]])<br />
* [[Transformer Example Problem]] (Aric Vyhmeister, Kevin Starkey, Nick Christman)<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Gauss Meters]] (Tyler Anderson)<br />
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)<br />
* [[Magnetic Flux]] (Jason Osborne)<br />
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]] (Chris Lau)<br />
* [[An Ideal Transformer Example]] (Chris Lau)<br />
* [[Magnetic Circuit]] (John Hawkins)<br />
* [[AC Motors]]<br />
* [[Example Problem - Toroid]] ([[Kirk Betz]])<br />
* [[Transformer_example_problem|Ideal Transformer Example]] (Tim Rasmussen)<br />
* [[AC vs. DC]] (Wesley Brown)</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_Example_Problem&diff=8756Transformer Example Problem2010-01-27T01:05:06Z<p>Nicholas.Christman: </p>
<hr />
<div>'''[[Kevin Starkey EMEC]], [[Nick Christman]], [[Aric Vyhmeister]]'''<br />
<br />
<br />
<u>'''Problem:'''</u> <br />
<br />
A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance, <math>i_{m}(t)</math>.<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
<u>'''Solution:'''</u><br />
<br />
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div><br />
<br/><br />
<br />
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>.<br />
</div><br />
<br />
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>.<br />
</div><br />
<br />
Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky."</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Transformer_Example_Problem&diff=8755Transformer Example Problem2010-01-27T01:04:34Z<p>Nicholas.Christman: New page: ==Transformer Example== '''Kevin Starkey EMEC, Nick Christman, Aric Vyhmeister''' <u>'''Problem:'''</u> A step down transformer has a winding of <math> N_1 = 10 \text{ tur...</p>
<hr />
<div>==Transformer Example==<br />
<br />
'''[[Kevin Starkey EMEC]], [[Nick Christman]], [[Aric Vyhmeister]]'''<br />
<br />
<br />
<u>'''Problem:'''</u> <br />
<br />
A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance, <math>i_{m}(t)</math>.<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
<u>'''Solution:'''</u><br />
<br />
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div><br />
<br/><br />
<br />
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>.<br />
</div><br />
<br />
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>.<br />
</div><br />
<br />
Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky."</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Kevin_Starkey_EMEC&diff=8754Kevin Starkey EMEC2010-01-27T01:04:14Z<p>Nicholas.Christman: </p>
<hr />
<div>=My Articles=<br />
<br />
Article#1-[[Example Problems of Magnetic Circuits]]<br />
<br />
Article#2 - [[Transformer Example Problem]]<br />
<br />
Article#2-[[Example Problems with Transformers]]</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8753Example Problems with Transformers2010-01-27T01:03:17Z<p>Nicholas.Christman: </p>
<hr />
<div>==Transformer Example==<br />
<br />
'''[[Kevin Starkey EMEC]], [[Nick Christman]], [[Aric Vyhmeister]]'''<br />
<br />
<br />
<u>'''Problem:'''</u> <br />
<br />
A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance, <math>i_{m}(t)</math>.<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
<u>'''Solution:'''</u><br />
<br />
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div><br />
<br/><br />
<br />
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>.<br />
</div><br />
<br />
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>.<br />
</div><br />
<br />
Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky."</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8752Example Problems with Transformers2010-01-27T01:02:24Z<p>Nicholas.Christman: </p>
<hr />
<div>==Transformer Example==<br />
<br />
'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''<br />
<br />
<br />
<u>'''Problem:'''</u> <br />
<br />
A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance, <math>i_{m}(t)</math>.<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
<u>'''Solution:'''</u><br />
<br />
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div><br />
<br/><br />
<br />
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>.<br />
</div><br />
<br />
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>.<br />
</div><br />
<br />
Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky."</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8751Example Problems with Transformers2010-01-27T01:01:08Z<p>Nicholas.Christman: </p>
<hr />
<div>==Transformer Example==<br />
<br />
'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''<br />
<br />
<br />
<u>'''Problem:'''</u> <br />
<br />
A step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. (a) If the input voltage is 1200V, what is the resulting output voltage? (b) If the input and output currents are <math>5A \mbox{ and } 20A</math>, respectively, what is the current loss due to the leakage inductance (<math>i_{m}(t)</math>.<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
<u>'''Solution:'''</u><br />
<br />
a) By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div><br />
<br />
b) From the figure above, we need to find the true value of current (i.e. the current that is not lost by leakage inductance). To accomplish this we will use equation 5-40 (Mohan 5-23) to obtain the value of <math>i_{1}^{'}</math>. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{1}^{'} = \left( \frac{N_2}{N_1} \right) i_{2} = \left( \frac{2}{10} \right) 20A = 4A</math>.<br />
</div><br />
<br />
Now from equation 5-42 (Mohan 5-23) we can obtain the current loss due to the leakage inductance, which is<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math>i_{m} = i_{1} - i_{1}^{'} = 5A - 4A = 1A</math>.<br />
</div><br />
<br />
Therefore, the current loss in our transformer is <math>1A</math>, which means this specific transformer is very "leaky."</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8749Example Problems with Transformers2010-01-27T00:37:08Z<p>Nicholas.Christman: </p>
<hr />
<div>==Problems 1-3==<br />
<br />
'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''<br />
<br />
<br />
'''Problem 1.''' An ''ideal'' step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. If the input voltage is 1200V, what is the resulting output voltage?<br />
<br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center><br />
<br />
<br />
'''Solution'''<br />
<br />
By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information provided we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \left( \frac{N_2}{N_1} \right) e_1 = \left( \frac{2}{10} \right) 1200V = 240V </math><br />
</div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8748Example Problems with Transformers2010-01-27T00:34:39Z<p>Nicholas.Christman: /* Problems 1-3 */</p>
<hr />
<div>==Problems 1-3==<br />
<br />
'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''<br />
<br />
<br />
'''Problem 1.''' An ''ideal'' step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. If the input voltage is 1200V, what is the resulting output voltage?<br />
<br />
'''Solution'''<br />
<br />
By modifying equation 5-39 (Mohan 5-22) we can obtain an equation for the output voltage. That is, <br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \frac{N_2}{N_1}e_1 </math>.<br />
</div><br />
<br />
<br/><br />
<br />
With the information above we can now determine the output voltage:<br />
<br/><br />
<br />
<div style="text-align:center"><br />
<math> e_2 = \frac{2}{10}1200 = 240V </math><br />
</div><br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_Problems_with_Transformers&diff=8432Example Problems with Transformers2010-01-20T15:45:54Z<p>Nicholas.Christman: </p>
<hr />
<div>==Problems 1-3==<br />
<br />
'''Kevin Starkey, Nick Christman, Aric Vyhmeister'''<br />
<br />
<br />
'''Problem 1.''' An ''ideal'' step down transformer has a winding of <math> N_1 = 10 \text{ turns and } N_2 = 2 </math> turns. If the input voltage is 1200V, what is the resulting output voltage?<br />
<br />
'''Solution''' Using the equation <math> e_2 = \frac{N_2}{N_1}e_1 </math> we get <br><br />
<br />
<math> e_2 = \frac{2}{10}1200 = 240V </math><br />
<br />
<center><br />
{|<br />
|[[Image:IdealTransformer1-nka.jpg|thumb|center|upright=2|Figure 1: Model for an ideal transformer.]]<br />
|[[Image:IdealTransformer2-nka.jpg|thumb|center|upright=2|Figure 2: Magnetic circuit of an ideal transformer.]]<br />
|}<br />
</center></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:IdealTransformer2-nka.jpg&diff=8431File:IdealTransformer2-nka.jpg2010-01-20T15:39:09Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:IdealTransformer1-nka.jpg&diff=8430File:IdealTransformer1-nka.jpg2010-01-20T15:34:02Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:Ideal_transformer1_nka.jpg&diff=8429File:Ideal transformer1 nka.jpg2010-01-20T15:32:06Z<p>Nicholas.Christman: </p>
<hr />
<div></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=8058Electromechanical Energy Conversion2010-01-16T05:53:11Z<p>Nicholas.Christman: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
==Articles==<br />
*[[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]]<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
* If it's been approved by the reviewers, move it to the articles section<br />
<br />
==Announcements==<br />
<br />
If anyone wants to write the derivation of Ampere's Law you can put it on my (Wesley Brown) [[Ampere's Law]] page and be a co-author.<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Gauss Meters]]<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[AC vs. DC]]<br />
* [[AC Motors]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Example problems of magnetic circuits]]<br />
* [[Magnetic Circuit]]<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
* [[An Ideal Transformer Example]]<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Nick_ENGR431_P1|Magnetostatics]] (Nick Christman)<br />
* [[Magnetic Flux]] (Jason Osborne)</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electromechanical_Energy_Conversion&diff=7705Electromechanical Energy Conversion2010-01-11T19:33:56Z<p>Nicholas.Christman: </p>
<hr />
<div>[[Rules]]<br />
<br />
[[Class Roster]]<br />
<br />
[[Points]]<br />
<br />
==Articles==<br />
None published to date<br />
<br />
==Questions==<br />
<br />
What do we do when we are finished with the draft and ready to publish?<br />
<br />
==Draft Articles==<br />
These articles are not ready for reading and error checking. They are listed so people will not simultaneously write about similar topics.<br />
* [[Ferromagnetism]]<br />
* [[Magnetic Circuits]]<br />
* [[Gauss Meters]]<br />
* [[Ampere's Law]]<br />
* [[DC Motor]]<br />
* [[AC vs. DC]]<br />
* [[An Application of Electromechanical Energy Conversion: Hybrid Electric Vehicles]]<br />
* [[AC Motors]]<br />
* [[Fringing]]<br />
* [[Electrostatics]]<br />
* [[Example problems of magnetic circuits]]<br />
* [[Magnetic Circuit]]<br />
* [[Ohm's Law and Reluctance]]<br />
* [[Magnetic Flux]]<br />
<br />
==Reviewed Articles==<br />
These articles have been reviewed and submitted.<br />
* [[Nick_ENGR431_P1|Magnetostatics (Nick Christman)]]</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431&diff=7499Nick ENGR4312010-01-11T03:48:54Z<p>Nicholas.Christman: </p>
<hr />
<div>= ENGR431 Homework =<br />
<br />
'''Back to [[Nick Christman]]'''<br />
<br/><br />
<br/><br />
<br />
== Drafts ==<br />
----<br />
[[Nick_ENGR431_P1 | Paper 1]]<br />
<br />
== Published ==<br />
----<br />
[[Nick_ENGR431_P1 | Paper 1: ''Brief Introduction to Magnetostatics'']]<br />
<br />
== Points ==<br />
----</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431_P1&diff=7498Nick ENGR431 P12010-01-11T03:45:57Z<p>Nicholas.Christman: </p>
<hr />
<div>== Nick Christman: Paper 1 - Magnetostatics ==<br />
<br />
Back to [[Nick ENGR431 | Nick's EMEC Wiki]]<br />
<br />
Back to [[Electromechanical Energy Conversion | Class EMEC Wiki]]<br />
<br/><br />
<br/><br />
<br />
<br />
<div style="text-align: center; font-size:15pt;">'''Brief Introduction to Magnetostatics'''</div><br />
<div style="text-align: center; font-size:13pt;">'''Nick Christman'''</div><br />
<br />
<br/><br />
<p><br />
According to several sources, there is an ancient story about a Cretan shepherd named Magnes who accidentally discovered a naturally occurring magnetic material known as lodestone (or loadstone). As the legend goes, Magnes was herding his sheep one day when the iron nails in his shoes and the iron tip on his staff became unusually attracted to a large, dark stone – lodestones contain a mineral, now known as magnetite (<math>Fe_{3}O_{4}</math>), that consists of naturally occurring magnetic properties<ref>Jezek, How Magnets Work</ref>. The first record of using magnets is somewhat of a debatable topic because magnetic properties were first recorded by Greek philosophers possibly as early as the 7th century BC and the Chinese have written records dating circa 4th century BC. In either case, the awareness of magnetic properties has been around for quite some time and the study of magnetism continues to be important and popular portion of research topics.<br />
</p><br />
<br/><br />
<p><br />
Magnetostatics, the study of static magnetic fields, is only a small portion of the overall study of magnetic properties. As just stated, magnetostatics implies that the magnetic field is static – that is, the flow of current that creates the magnetic field is steady or direct current (DC)<ref>Wikipedia, Magnetostatics</ref> – and this allows scientists to make very accurate approximations of how magnetic fields act. In this document, the theory and application behind magnetostatics will be addressed.<br />
</p><br />
<br/><br />
<p><br />
It is said that currents in opposite directions repel while currents in the same directions attract. To demonstrate this, take a common example that is illustrated in David Griffiths ''Introduction to Electrodynamics'', 3rd edition: “[Imagine] two wires hang from the ceiling, a few centimeters apart; when I turn on a current so that it passes up one wire and back down the other, the wires jump apart. Moreover, I could hook up my demonstration so as to make the current flow up both wires; in this case they are found to attract” <ref>Griffiths, p. 203</ref>(Griffiths 203). This demonstration is illustrated in Figure 1. Through experimentation it has been proven that the forces of attraction in this system are not due to electrostatics, but instead are the result of magnetic forces – ''in addition to an electric field, a moving charge also generates a magnetic field''<ref>Griffiths, p. 203</ref> (Griffiths 203). In fact, if one were to allow a current to flow down a straight wire, then a magnetic field circling the wire will be formed – recall the right-hand-rule which states that if you grasp the wire with your right hand (thumb pointing in the direction of current) then your fingers curl in the direction of the magnetic field. Place two wires in close proximity to each other, such as those mentioned in the demonstration above, and there will be a magnetic force between the two wires – hence, in one case the two wires repel and in the other they attract.<br />
</p><br />
[[Image:Paper1_img1.png |thumb|center|upright=2.5|Figure 1: Illustration of (A) currents in opposite directions repel and (B) currents in the same directions attract. Image created by Nick Christman.]]<br />
<p><br />
This concept of magnetic force is an interesting, yet complicating topic that is often covered in series of undergraduate and graduate level courses. Generally speaking, however, there is one law that encapsulates all of magnetostatics, which is Lorentz Force Law<br />
<br />
<div style="text-align:center;"><br />
<math>\textstyle F = Q(v \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
<br />
where Q is the charge for which the magnetic field is acting upon, B is the magnetic field, and v is the velocity of the charge Q. (Note that this force, F, is due to the magnetic field acting on Q and that there are no external electric fields present. In the presence of an external electric field, there would be an additional term.) Furthermore, Lorentz Force Law can be expanded in order to account for a current carrying wire rather than a point charge. In the case of a current carrying wire, Lorentz Force Law becomes<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = \int (I \times B)dl = \int I(dl \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
where <math>I</math>, the current, is represented by a vector pointing in the same direction as <math>dl</math>. In most cases, however, current is treated as constant; thus, Lorentz Force Law can be written as<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = I \int (dl \times B) </math><ref>Griffiths, p. 209</ref><br />
</div><br />
</p><br />
<br/><br />
<p><br />
Overall, magnetostatics has been derived from a compilation of different sources and it is used to more simply model the properties of magnetism. In both the constant and non-constant current cases presented above, it is clear that the Lorentz Force Law is an important aspect of magnetostatics that can be expanded to account for nearly every study of magnetism. This short overview of magnetostatics is important when one desires to develop and understand magnetic circuits and, ultimately, transformers.<br />
</p><br />
<br/><br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''References''':</div><br />
<br />
<br />
Griffiths, David J. "Introduction to Electrodynamics." Upper Saddle River, N.J: Prentice Hall, 1999.<br />
<br />
Jezek, Geno. "History of Magnets." How Magnets Work. Web. 9 Jan. 2010. <http://www.howmagnetswork.com/history.html>.<br />
<br />
Mohan, Ned. Electric Drives An Integrative Approach. Minneapolis: Mnpere, 2004.<br />
<br />
Wikepdia. "Magnetostatics." Wikipedia, the free encyclopedia. Web. 10 Jan. 2010. <http://en.wikipedia.org/wiki/Magnetostatics>.<br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''In-text Citations''':</div><br />
<references/><br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Statistics''':</div><br />
<br />
Word Count: 845 (content, equations, work cited, and title)<br />
<br />
Image Count: 1<br />
<br />
Potential Points: TBD<br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Reviewers''':</div><br />
<br />
* Kevin Starkey<br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Readers''':</div><br />
<br />
*</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431_P1&diff=7497Nick ENGR431 P12010-01-11T03:40:37Z<p>Nicholas.Christman: </p>
<hr />
<div>== Nick Christman: Paper 1 - Magnetostatics ==<br />
<br />
Back to [[Nick ENGR431 | Nick's EMEC Wiki]]<br />
<br />
Back to [[Electromechanical Energy Conversion | Class EMEC Wiki]]<br />
<br/><br />
<br/><br />
<br />
<br />
<div style="text-align: center; font-size:15pt;">'''Brief Introduction to Magnetostatics'''</div><br />
<div style="text-align: center; font-size:13pt;">'''Nick Christman'''</div><br />
<br />
<br/><br />
<p><br />
According to several sources, there is an ancient story about a Cretan shepherd named Magnes who accidentally discovered a naturally occurring magnetic material known as lodestone (or loadstone). As the legend goes, Magnes was herding his sheep one day when the iron nails in his shoes and the iron tip on his staff became unusually attracted to a large, dark stone – lodestones contain a mineral, now known as magnetite (<math>Fe_{3}O_{4}</math>), that consists of naturally occurring magnetic properties<ref>Jezek, How Magnets Work</ref>. The first record of using magnets is somewhat of a debatable topic because magnetic properties were first recorded by Greek philosophers possibly as early as the 7th century BC and the Chinese have written records dating circa 4th century BC. In either case, the awareness of magnetic properties has been around for quite some time and the study of magnetism continues to be important and popular portion of research topics.<br />
</p><br />
<br/><br />
<p><br />
Magnetostatics, the study of static magnetic fields, is only a small portion of the overall study of magnetic properties. As just stated, magnetostatics implies that the magnetic field is static – that is, the current creating the magnetic field is steady or direct current (DC)<ref>Wikipedia, Magnetostatics</ref> – and this allows scientists to make very accurate approximations of how magnetic fields act. In this document, the theory and application behind magnetostatics will be addressed.<br />
</p><br />
<br/><br />
<p><br />
It is said that currents in opposite directions repel while currents in the same directions attract. To demonstrate this, take a common example that is illustrated in David Griffiths ''Introduction to Electrodynamics'', 3rd edition: “[Imagine] two wires hang from the ceiling, a few centimeters apart; when I turn on a current so that it passes up one wire and back down the other, the wires jump apart. Moreover, I could hook up my demonstration so as to make the current flow up both wires; in this case they are found to attract” <ref>Griffiths, p. 203</ref>(Griffiths 203). This demonstration is illustrated in Figure 1. Through experimentation it has been proven that the forces of attraction in this system are not due to electrostatics, but instead are the result of magnetic forces – ''in addition to an electric field, a moving charge also generates a magnetic field''<ref>Griffiths, p. 203</ref> (Griffiths 203). In fact, if one were to allow a current to flow down a straight wire, then a magnetic field circling the wire will be formed – recall the right-hand-rule which states that if you grasp the wire with your right hand (thumb pointing in the direction of current) then your fingers curl in the direction of the magnetic field. Place two wires in close proximity to each other, such as those mentioned in the demonstration above, and there will be a magnetic force between the two wires – hence, in one case the two wires repel and in the other they attract.<br />
</p><br />
[[Image:Paper1_img1.png |thumb|center|upright=2.5|Figure 1: Illustration of (A) currents in opposite directions repel and (B) currents in the same directions attract. Image created by Nick Christman.]]<br />
<p><br />
This concept of magnetic force is an interesting, yet complicating topic that is often covered in series of undergraduate and graduate level courses. Generally speaking, however, there is one law that encapsulates all of magnetostatics, which is Lorentz Force Law<br />
<br />
<div style="text-align:center;"><br />
<math>\textstyle F = Q(v \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
<br />
where Q is the charge for which the magnetic field is acting upon, B is the magnetic field, and v is the velocity of the charge Q. (Note that this force, F, is due to the magnetic field acting on Q and that there are no external electric fields present. In the presence of an external electric field, there would be an additional term.) Furthermore, Lorentz Force Law can be expanded in order to account for a current carrying wire rather than a point charge. In the case of a current carrying wire, Lorentz Force Law becomes<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = \int (I \times B)dl = \int I(dl \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
where <math>I</math>, the current, is represented by a vector pointing in the same direction as <math>dl</math>. In most cases, however, current is treated as constant; thus, Lorentz Force Law can be written as<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = I \int (dl \times B) </math><ref>Griffiths, p. 209</ref><br />
</div><br />
</p><br />
<br/><br />
<p><br />
Overall, magnetostatics has been derived from a compilation of different sources and it is used to more simply model the properties of magnetism. In both the constant and non-constant current cases presented above, it is clear that the Lorentz Force Law is an important aspect of magnetostatics that can be expanded to account for nearly every study of magnetism. This short overview of magnetostatics is important when one desires to develop and understand magnetic circuits and, ultimately, transformers.<br />
</p><br />
<br/><br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''References''':</div><br />
<br />
<br />
Griffiths, David J. "Introduction to Electrodynamics." Upper Saddle River, N.J: Prentice Hall, 1999.<br />
<br />
Jezek, Geno. "History of Magnets." How Magnets Work. Web. 9 Jan. 2010. <http://www.howmagnetswork.com/history.html>.<br />
<br />
Mohan, Ned. Electric Drives An Integrative Approach. Minneapolis: Mnpere, 2004.<br />
<br />
Wikepdia. "Magnetostatics." Wikipedia, the free encyclopedia. Web. 10 Jan. 2010. <http://en.wikipedia.org/wiki/Magnetostatics>.<br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''In-text Citations''':</div><br />
<references/><br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Statistics''':</div><br />
<br />
Word Count: 845 (content, equations, work cited, and title)<br />
<br />
Image Count: 1<br />
<br />
Potential Points: TBD<br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Reviewers''':</div><br />
<br />
* Kevin Starkey<br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Readers''':</div><br />
<br />
*</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431_P1&diff=7496Nick ENGR431 P12010-01-11T03:40:06Z<p>Nicholas.Christman: </p>
<hr />
<div>== Nick Christman: Paper 1 - Magnetostatics ==<br />
<br />
Back to [[Nick ENGR431 | Nick's EMEC Wiki]]<br />
<br />
Back to [[Electromechanical Energy Conversion | Class EMEC Wiki]]<br />
<br/><br />
<br/><br />
<br />
<br />
<div style="text-align: center; font-size:15pt;">'''Brief Introduction to Magnetostatics'''</div><br />
<div style="text-align: center; font-size:13pt;">'''Nick Christman'''</div><br />
<br />
<br/><br />
<p><br />
According to several sources, there is an ancient story about a Cretan shepherd named Magnes who accidentally discovered a naturally occurring magnetic material known as lodestone (or loadstone). As the legend goes, Magnes was herding his sheep one day when the iron nails in his shoes and the iron tip on his staff became unusually attracted to a large, dark stone – lodestones contain a mineral, now known as magnetite (<math>Fe_{3}O_{4}</math>), that consists of naturally occurring magnetic properties<ref>Jezek, How Magnets Work</ref>. The first record of using magnets is somewhat of a debatable topic because magnetic properties were first recorded by Greek philosophers possibly as early as the 7th century BC and the Chinese have written records dating circa 4th century BC. In either case, the awareness of magnetic properties has been around for quite some time and the study of magnetism continues to be important and popular portion of research topics.<br />
</p><br />
<br/><br />
<p><br />
Magnetostatics, the study of static magnetic fields, is only a small portion of the overall study of magnetic properties. As just stated, magnetostatics implies that the magnetic field is static – that is, the current creating the magnetic field is steady or direct current (DC)<ref>Wikipedia, Magnetostatics</ref> – and this allows scientists to make very accurate approximations of how magnetic fields act. In this document, the theory and application behind magnetostatics will be addressed.<br />
</p><br />
<br/><br />
<p><br />
It is said that currents in opposite directions repel while currents in the same directions attract. To demonstrate this, take a common example that is illustrated in David Griffiths ''Introduction to Electrodynamics'', 3rd edition: “[Imagine] two wires hang from the ceiling, a few centimeters apart; when I turn on a current so that it passes up one wire and back down the other, the wires jump apart. Moreover, I could hook up my demonstration so as to make the current flow up both wires; in this case they are found to attract” <ref>Griffiths, p. 203</ref>(Griffiths 203). This demonstration is illustrated in Figure 1. Through experimentation it has been proven that the forces of attraction in this system are not due to electrostatics, but instead are the result of magnetic forces – ''in addition to an electric field, a moving charge also generates a magnetic field''<ref>Griffiths, p. 203</ref> (Griffiths 203). In fact, if one were to allow a current to flow down a straight wire, then a magnetic field circling the wire will be formed – recall the right-hand-rule which states that if you grasp the wire with your right hand (thumb pointing in the direction of current) then your fingers curl in the direction of the magnetic field. Place two wires in close proximity to each other, such as those mentioned in the demonstration above, and there will be a magnetic force between the two wires – hence, in one case the two wires repel and in the other they attract.<br />
</p><br />
[[Image:Paper1_img1.png |thumb|center|upright=2.5|Figure 1: Illustration of (A) currents in opposite directions repel and (B) currents in the same directions attract. Image created by Nick Christman.]]<br />
<p><br />
This concept of magnetic force is an interesting, yet complicating topic that is often covered in series of undergraduate and graduate level courses. Generally speaking, however, there is one law that encapsulates all of magnetostatics, which is Lorentz Force Law<br />
<br />
<div style="text-align:center;"><br />
<math>\textstyle F = Q(v \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
<br />
where Q is the charge for which the magnetic field is acting upon, B is the magnetic field, and v is the velocity of the charge Q. (Note that this force, F, is due to the magnetic field acting on Q and that there are no external electric fields present. In the presence of an external electric field, there would be an additional term.) Furthermore, Lorentz Force Law can be expanded in order to account for a current carrying wire rather than a point charge. In the case of a current carrying wire, Lorentz Force Law becomes<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = \int (I \times B)dl = \int I(dl \times B) </math><ref>Griffiths, p. 204</ref><br />
</div> <br />
where <math>I</math>, the current, is represented by a vector pointing in the same direction as <math>dl</math>. In most cases, however, current is treated as constant; thus, Lorentz Force Law can be written as<br />
<br />
<div style="text-align:center;"><br />
<math> \textstyle F = I \int (dl \times B) </math><ref>Griffiths, p. 209</ref><br />
</div><br />
</p><br />
<br/><br />
<p><br />
Overall, magnetostatics has been derived from a compilation of different sources and it is used to more simply model the properties of magnetism. In both the constant and non-constant current cases presented above, it is clear that the Lorentz Force Law is an important aspect of magnetostatics that can be expanded to account for nearly every study of magnetism. This short overview of magnetostatics is important when one desires to develop and understand magnetic circuits and, ultimately, transformers.<br />
</p><br />
<br/><br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''References''':</div><br />
<br />
<br />
Griffiths, David J. "Introduction to Electrodynamics." Upper Saddle River, N.J: Prentice Hall, 1999.<br />
<br />
Jezek, Geno. "History of Magnets." How Magnets Work. Web. 9 Jan. 2010. <http://www.howmagnetswork.com/history.html>.<br />
<br />
Mohan, Ned. Electric Drives An Integrative Approach. Minneapolis: Mnpere, 2004.<br />
<br />
Unknown. "Magnetostatics." Wikipedia, the free encyclopedia. Web. 10 Jan. 2010. <http://en.wikipedia.org/wiki/Magnetostatics>.<br />
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<div style="text-align: center; font-size:13pt;">'''In-text Citations''':</div><br />
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<div style="text-align: center; font-size:13pt;">'''Statistics''':</div><br />
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Word Count: 845 (content, equations, work cited, and title)<br />
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Image Count: 1<br />
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Potential Points: TBD<br />
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<div style="text-align: center; font-size:13pt;">'''Reviewers''':</div><br />
<br />
* Kevin Starkey<br />
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<div style="text-align: center; font-size:13pt;">'''Readers''':</div><br />
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*</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Nick_ENGR431_P1&diff=7495Nick ENGR431 P12010-01-11T03:35:57Z<p>Nicholas.Christman: </p>
<hr />
<div>== Nick Christman: Paper 1 - Magnetostatics ==<br />
<br />
Back to [[Nick ENGR431 | Nick's EMEC Wiki]]<br />
<br />
Back to [[Electromechanical Energy Conversion | Class EMEC Wiki]]<br />
<br/><br />
<br/><br />
<br />
<br />
<div style="text-align: center; font-size:15pt;">'''Brief Introduction to Magnetostatics'''</div><br />
<div style="text-align: center; font-size:13pt;">'''Nick Christman'''</div><br />
<br />
<br/><br />
<p><br />
According to several sources, there is an ancient story about a Cretan shepherd named Magnes who accidentally discovered a naturally occurring magnetic material known as lodestone (or loadstone). As the legend goes, Magnes was herding his sheep one day when the iron nails in his shoes and the iron tip on his staff became unusually attracted to a large, dark stone – lodestones contain a mineral, now known as magnetite (<math>Fe_{3}O_{4}</math>), that consists of naturally occurring magnetic properties<ref>Jezek, How Magnets Work</ref>. The first record of using magnets is somewhat of a debatable topic because magnetic properties were first recorded by Greek philosophers possibly as early as the 7th century BC and the Chinese have written records dating circa 4th century BC. In either case, the awareness of magnetic properties has been around for quite some time and the study of magnetism continues to be important and popular portion of research topics.<br />
</p><br />
<br/><br />
<p><br />
Magnetostatics, the study of static magnetic fields, is only a small portion of the overall study of magnetic properties. As just stated, magnetostatics implies that the magnetic field is static – that is, the current creating the magnetic field is steady or direct current (DC)<ref>Wikipedia, Magnetostatics</ref> – and this allows scientists to make very accurate approximations of how magnetic fields act. In this document, the theory and application behind magnetostatics will be addressed.<br />
</p><br />
<br/><br />
<p><br />
It is said that currents in opposite directions repel while currents in the same directions attract. To demonstrate this, take a common example that is illustrated in David Griffiths ''Introduction to Electrodynamics'', 3rd edition: “[Imagine] two wires hang from the ceiling, a few centimeters apart; when I turn on a current so that it passes up one wire and back down the other, the wires jump apart. Moreover, I could hook up my demonstration so as to make the current flow up both wires; in this case they are found to attract” <ref>Griffiths, p. 203</ref>(Griffiths 203). This demonstration is illustrated in Figure 1. Through experimentation it has been proven that the forces of attraction in this system are not due to electrostatics, but instead are the result of magnetic forces – ''in addition to an electric field, a moving charge also generates a magnetic field''<ref>Griffiths, p. 203</ref> (Griffiths 203). In fact, if one were to allow a current to flow down a straight wire, then a magnetic field circling the wire will be formed – recall the right-hand-rule which states that if you grasp the wire with your right hand (thumb pointing in the direction of current) then your fingers curl in the direction of the magnetic field. Place two wires in close proximity to each other, such as those mentioned in the demonstration above, and there will be a magnetic force between the two wires – hence, in one case the two wires repel and in the other they attract.<br />
</p><br />
[[Image:Paper1_img1.png |thumb|center|upright=2.5|Figure 1: Illustration of (A) currents in opposite directions repel and (B) currents in the same directions attract. Image created by Nick Christman.]]<br />
<p><br />
This concept of magnetic force is an interesting, yet complicating topic that is often covered in series of undergraduate and graduate level courses. Generally speaking, however, there is one law that encapsulates all of magnetostatics, which is Lorentz Force Law<br />
<br />
<div style="text-align:center;"><math>\textstyle F = Q(v \times B) </math></div><br />
<br />
where Q is the charge for which the magnetic field is acting upon, B is the magnetic field, and v is the velocity of the charge Q. (Note that this force, F, is due to the magnetic field acting on Q and that there are no external electric fields present. In the presence of an external electric field, there would be an additional term.) Furthermore, Lorentz Force Law can be expanded in order to account for a current carrying wire rather than a point charge. In the case of a current carrying wire, Lorentz Force Law becomes<br />
<br />
<div style="text-align:center;"><math> \textstyle F = \int (I \times B)dl = \int I(dl \times B) </math></div><br />
<br />
where <math>I</math>, the current, is represented by a vector pointing in the same direction as <math>dl</math>. In most cases, however, current is treated as constant; thus, Lorentz Force Law can be written as<br />
<br />
<div style="text-align:center;"><math> \textstyle F = I \int (dl \times B) </math></div><br />
</p><br />
<br/><br />
<p><br />
Overall, magnetostatics has been derived from a compilation of different sources and it is used to more simply model the properties of magnetism. In both the constant and non-constant current cases presented above, it is clear that the Lorentz Force Law is an important aspect of magnetostatics that can be expanded to account for nearly every study of magnetism. This short overview of magnetostatics is important when one desires to develop and understand magnetic circuits and, ultimately, transformers.<br />
</p><br />
<br/><br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''References''':</div><br />
<br />
<br />
Griffiths, David J. "Introduction to Electrodynamics." Upper Saddle River, N.J: Prentice Hall, 1999.<br />
<br />
Jezek, Geno. "History of Magnets." How Magnets Work. Web. 9 Jan. 2010. <http://www.howmagnetswork.com/history.html>.<br />
<br />
Mohan, Ned. Electric Drives An Integrative Approach. Minneapolis: Mnpere, 2004.<br />
<br />
Unknown. "Magnetostatics." Wikipedia, the free encyclopedia. Web. 10 Jan. 2010. <http://en.wikipedia.org/wiki/Magnetostatics>.<br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''In-text Citations''':</div><br />
<references/><br />
<br />
<br/><br />
<br />
----<br />
<br />
<div style="text-align: center; font-size:13pt;">'''Statistics''':</div><br />
<br />
Word Count: 845 (content, equations, work cited, and title)<br />
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Image Count: 1<br />
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Potential Points: TBD<br />
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<div style="text-align: center; font-size:13pt;">'''Reviewers''':</div><br />
<br />
* Kevin Starkey<br />
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----<br />
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<div style="text-align: center; font-size:13pt;">'''Readers''':</div><br />
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*</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_problems_of_magnetic_circuits&diff=7480Example problems of magnetic circuits2010-01-11T02:56:10Z<p>Nicholas.Christman: </p>
<hr />
<div>Given:<br />
<br />
A copper core with susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math><br />
<br />
length of core L = 1 m<br />
<br />
Gap length g = .01 m<br />
<br />
cross sectional area A = .1 m<br />
<br />
current I = 10A<br />
<br />
N = 5 turns<br />
<br />
<br />
<br />
Find: <math> B_g </math><br />
<br />
Solution:<br />
First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br><br />
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br><br />
<br />
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br><br />
<br />
<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br><br />
Similarly to get the reluctance of the gap <br><br />
<br />
<math>R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} = \frac {.01}{4 \times \pi \times 10^{-7} (\sqrt{.1} + .01)^2} = 74.8 \times 10^{3} </math> <br> <br><br />
<br />
Now Using <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br><br />
Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^2} = .789 \times 10^{-9}</math><br />
<br />
<br />
===Reviewers:===<br />
----<br />
[[Nick Christman]]:<br />
* I would change "Given: (...)[list] A copper core with..." to "Given a copper core with: [list]" to make it a little more consistent or even take all the information you have and make it into a complete sentence/paragraph.<br />
* This looks strange to me, <math>\mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6})</math> maybe make it <math>\mu = 4 \pi \times 10^{-7}(1-9.7 \times 10^{-6})</math> or <math>\mu = 4 \pi \times 10^{-7}(1+(-9.7 \times 10^{-6}))</math><br />
* This sentence is kind of strange, "Now with this, the length and cross sectional area of the core we can solve for reluctance..." Maybe make it, "With the permeability, length, and cross sectional area of the copper core we can now solve for the reluctance..." Something like that might flow a little better."<br />
* Below that, I think you need a comma after "Similarly."<br />
* You might want to add some more words to the last two lines... Instead of saying "Now using..." say something like, "Recall that the equation for the magnetic field of the gap is..." or something to that effect. <br />
* Lastly, you should think of some sort of conclusion... what exactly does this mean?</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_problems_of_magnetic_circuits&diff=7474Example problems of magnetic circuits2010-01-11T02:47:56Z<p>Nicholas.Christman: </p>
<hr />
<div>Given:<br />
<br />
A copper core with susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math><br />
<br />
length of core L = 1 m<br />
<br />
Gap length g = .01 m<br />
<br />
cross sectional area A = .1 m<br />
<br />
current I = 10A<br />
<br />
N = 5 turns<br />
<br />
<br />
<br />
Find: <math> B_g </math><br />
<br />
Solution:<br />
First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br><br />
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br><br />
<br />
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br><br />
<br />
<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br><br />
Similarly to get the reluctance of the gap <br><br />
<br />
<math>R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} = \frac {.01}{4 \times \pi \times 10^{-7} (\sqrt{.1} + .01)^2} = 74.8 \times 10^{3} </math> <br> <br><br />
<br />
Now Using <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br><br />
Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^2} = .789 \times 10^{-9}</math><br />
<br />
<br />
===Reviewers:===<br />
----<br />
[[Nick Christman]]:<br />
* I would change "Given: ... A copper core with" to "Given a copper core with:" to make it a little more consistent or even take all the information you have and make it into a complete sentence/paragraph.<br />
* This looks strange to me, <math>\mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6})</math> maybe make it <math>\mu = 4 \pi \times 10^{-7}(1-9.7 \times 10^{-6})</math> or <math>\mu = 4 \pi \times 10^{-7}(1+(-9.7 \times 10^{-6}))</math></div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Example_problems_of_magnetic_circuits&diff=7467Example problems of magnetic circuits2010-01-11T02:42:23Z<p>Nicholas.Christman: </p>
<hr />
<div>Given:<br />
<br />
A copper core with susceptibility <math> \chi_m = -9.7 \times 10^{-6} </math><br />
<br />
length of core L = 1 m<br />
<br />
Gap length g = .01 m<br />
<br />
cross sectional area A = .1 m<br />
<br />
current I = 10A<br />
<br />
N = 5 turns<br />
<br />
<br />
<br />
Find: <math> B_g </math><br />
<br />
Solution:<br />
First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br><br />
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br><br />
<br />
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br><br />
<br />
<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br><br />
Similarly to get the reluctance of the gap <br><br />
<br />
<math>R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} = \frac {.01}{4 \times \pi \times 10^{-7} (\sqrt{.1} + .01)^2} = 74.8 \times 10^{3} </math> <br> <br><br />
<br />
Now Using <math> B_g = \frac{NI}{(R_g R_c)((\sqrt{A} + g)^2} </math> <br><br />
Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^2} = .789 \times 10^{-9}<br />
<br />
==Reviewers==<br />
----<br />
[[Nick Christman]]:</div>Nicholas.Christmanhttps://fweb.wallawalla.edu/class-wiki/index.php?title=File:Paper1_img1.png&diff=7423File:Paper1 img1.png2010-01-11T01:02:32Z<p>Nicholas.Christman: uploaded a new version of "Image:Paper1 img1.png": Added more graphics to better indicate repulsion and attraction.</p>
<hr />
<div>Image for magnetostatics illustration.</div>Nicholas.Christman