Christman GeneralizedReceiver - Revision history

Nicholas.Christman at 04:21, 9 April 2010

2010-04-09T04:21:55Z

← Older revision		Revision as of 21:21, 8 April 2010
Line 39:		Line 39:
	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.		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.
	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,		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/>
	<center>		<center>
	<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>		<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>
	</center>		</center>
			<br/>
	Using the trigonometric identities		Using the trigonometric identities

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	</center>		</center>

			<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.)		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.)
	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).		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.		Please feel free to fix any mistakes. Please list your name, date fixed, and short description below.

	* Name, Date, Description		* Name, Date, Description

Nicholas.Christman at 04:20, 9 April 2010

2010-04-09T04:20:41Z

← Older revision		Revision as of 21:20, 8 April 2010
Line 62:		Line 62:
	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.)		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.)
	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).		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).
			----
			Please feel free to fix any mistakes. Please list your name, date fixed, and short description below.

			* Name, Date, Description

Nicholas.Christman at 04:18, 9 April 2010

2010-04-09T04:18:38Z

← Older revision		Revision as of 21:18, 8 April 2010
Line 30:		Line 30:

	<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.		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/>

Nicholas.Christman at 04:17, 9 April 2010

2010-04-09T04:17:57Z

← Older revision		Revision as of 21:17, 8 April 2010
Line 38:		Line 38:

	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.		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.
	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,		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,

	<center>		<center>
	<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>		<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>
	</center>		</center>

	~~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.)		Using the trigonometric identities

			<center>
			<math>\cos^{2}{\theta} = \frac{1+\cos{2 \theta}}{2}</math>
			</center>
			and
			<center>
			<math>\cos{\theta}\sin{\theta} = \frac{1}{2}\sin{\theta}</math>,
			</center>
			we can see that
			<center>

			<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>.

			</center>

			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.)
			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).

Nicholas.Christman at 04:03, 9 April 2010

2010-04-09T04:03:35Z

← Older revision		Revision as of 21:03, 8 April 2010
Line 5:		Line 5:
	<center>[[Image:HW2_GenTran.jpg]]</center>		<center>[[Image:HW2_GenTran.jpg]]</center>
	<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:		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/>

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	<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		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/>
	<center>		<center>
	<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\~~omega_{c}~~t}]</math>,		<math>\displaystyle v(t) = Re[g(t) \cdot e^{j(\omega_c t)}]</math>,
	</center>		</center>

	<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>.		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/>
	The above formula can then be rewritten to obtain the following relationship:		The above formula can then be rewritten to obtain the following relationship:
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	<center>		<center>

	<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>.		<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>.

	</center>		</center>

	~~To visualize this process, observe the following figure:~~
	<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.		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/>

	The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:		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:

	<center>[[Image:HW2_GenRec.jpg]]</center>		<center>[[Image:HW2_GenRec.jpg]]</center>

	~~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>).		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.
			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,

			<center>
			<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>
			</center>

			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.)

Nicholas.Christman at 03:36, 9 April 2010

2010-04-09T03:36:34Z

← Older revision		Revision as of 20:36, 8 April 2010
Line 6:		Line 6:
	<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:		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/>

			<center>[[Image:HW2_BandpassBaseband.png]]</center>

			<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		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/>

Nicholas.Christman at 03:25, 9 April 2010

2010-04-09T03:25:11Z

← Older revision		Revision as of 20:25, 8 April 2010
Line 5:		Line 5:
	<center>[[Image:HW2_GenTran.jpg]]</center>		<center>[[Image:HW2_GenTran.jpg]]</center>
	<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:		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:

	Why is the data split and shifted you ask? In the world of "Communication Systems" a signal to be transmitted can be written as		Why is the data split and shifted you ask? In the world of "Communication Systems" a signal to be transmitted can be written as

Nicholas.Christman at 03:23, 9 April 2010

2010-04-09T03:23:19Z

← Older revision		Revision as of 20:23, 8 April 2010
Line 5:		Line 5:
	<center>[[Image:HW2_GenTran.jpg]]</center>		<center>[[Image:HW2_GenTran.jpg]]</center>
	<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		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:

			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/>
	<center>		<center>
Line 25:		Line 27:
	To visualize this process, observe the following figure:		To visualize this process, observe the following figure:
	<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~~.		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/>

			The process of receiving data, <math>m(t)</math>, via a wireless signal, <math>v(t)</math>, is shown below:

	<center>[[Image:HW2_GenRec.jpg]]</center>		<center>[[Image:HW2_GenRec.jpg]]</center>

			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>).

Nicholas.Christman at 16:10, 6 April 2010

2010-04-06T16:10:32Z

← Older revision		Revision as of 09:10, 6 April 2010
Line 8:		Line 8:
	<br/>		<br/>
	<center>		<center>
			<math>\displaystyle v(t) = Re[g(t) \cdot e^{j\omega_{c}t}]</math>,
			</center>

	<~~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/>

	</~~center~~>		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/>
	~~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:		The above formula can then be rewritten to obtain the following relationship:
	<br/>		<br/>
	<center>		<center>

	<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>.		<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>.

	</center>		</center>

			To visualize this process, observe the following figure:
	<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.		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/>
	<center>[[Image:HW2_GenRec.jpg]]</center>		<center>[[Image:HW2_GenRec.jpg]]</center>

Nicholas.Christman at 06:37, 5 April 2010

2010-04-05T06:37:56Z

@@ Line 1: / Line 1: @@
 ===<span style="font-variant: small-caps;">How It Works: Generalized Receiver</span>===
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-<center>[[Image:HW2_GenTran.jpg]]</center>
+<br/>
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 <center>[[Image:HW2_GenRec.jpg]]</center>