CDPlayerJEW - Revision history

Wonoje: /* How a CD Player Works */

2005-12-06T19:22:33Z

How a CD Player Works

← Older revision		Revision as of 12:22, 6 December 2005
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	Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:		Then, the discrete signal is convolved by the D/A converter with a function <math>p(t)</math>:

	<center><math>p(t) = u(t+\frac{T}{2}) - u(t-\frac{T}{2})</math></center>		<center><math>p(t) = u \left (t+\frac{T}{2} \right) - u \left (t-\frac{T}{2} \right)</math></center>

Wonoje: /* 2x Oversampling */

2005-12-06T00:55:32Z

2x Oversampling

← Older revision		Revision as of 17:55, 5 December 2005
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	This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.		This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.



	<small>- Principle author of this page: Jeffrey Wonoprabowo		<small>- Principle author of this page: Jeffrey Wonoprabowo

	- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>		- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>

Wonoje: /* 2x Oversampling */

2005-12-06T00:55:20Z

2x Oversampling

← Older revision		Revision as of 17:55, 5 December 2005
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	</math></center>		</math></center>


			<math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.

			This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.

	<small>- Principle author of this page: Jeffrey Wonoprabowo		<small>- Principle author of this page: Jeffrey Wonoprabowo

	- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>		- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>

	~~</math>~~

	~~<math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.~~

	This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.

Wonoje: /* \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m */

2005-12-06T00:54:44Z

\hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m

← Older revision		Revision as of 17:54, 5 December 2005
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	<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>		<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>

	= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)		<center><math>= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)
	</math></center>		</math></center>

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	</math>		</math>

			<math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.

			This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.

Wonoje: /* 2x Oversampling */

2005-12-06T00:54:04Z

2x Oversampling

← Older revision		Revision as of 17:54, 5 December 2005
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	Mathematically, the equation is:		Mathematically, the equation is:
	<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>		<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>


	~~<math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.~~

	This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.

	= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)		= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)

Wonoje: /* 2x Oversampling */

2005-12-06T00:53:19Z

2x Oversampling

← Older revision		Revision as of 17:53, 5 December 2005
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	<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>		<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>

	<~~center~~><math>
			<math>h(t)</math> effectively fills in the gaps by taking the average of two adjacent data points and placing a data point, whose value is the average of it's two adjacent points, half way in between.

			This process can be extended to cover 4, 8, 16 or higher oversampling. The result is a smoother stepped function (in the time domain) and a frequency function that looks more and more like the original signal.

	= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)		= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)
	</math></center>		</math></center>

Wonoje: /* 2x Oversampling */

2005-12-06T00:48:00Z

2x Oversampling

← Older revision		Revision as of 17:48, 5 December 2005
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	In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.		In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.

	<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{n}) \delta (t-\frac{mT}{n}) </math></center>		<center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{n} \right) \delta \left (t-\frac{mT}{n} \right) </math></center>

	where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,		where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,

	<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>		<center><math>h(t) = \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t-\frac{mT}{2} \right) </math></center>

	Graphically, this looks like this:		Graphically, this looks like this:
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	Mathematically, the equation is:		Mathematically, the equation is:
	<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t - \frac{mT}{2}) = \hat y (t) = \sum_{l=-\infty}^\infty \left \sum_{m=-M}^M x \left \frac{l-m}{2} T \right h(\frac{mT}{2}) \right \delta (t - \frac{lT}{2})		<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h \left (\frac{mT}{2} \right) \delta \left (t - \frac{mT}{2} \right) </math></center>

			<center><math>
			= \hat y (t) = \sum_{l=-\infty}^\infty \left ( \sum_{m=-M}^M x \left (\frac{l-m}{2} T \right) h \left(\frac{mT}{2} \right) \right) \delta \left (t - \frac{lT}{2} \right)
			</math></center>

Wonoje: /* 2x Oversampling */

2005-12-06T00:34:35Z

2x Oversampling

← Older revision		Revision as of 17:34, 5 December 2005
Line 44:		Line 44:
	<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>		<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>

			Graphically, this looks like this:
	[[Image:2x01.jpg\|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg\|Illustration of 2x Oversampling pt.2]]		[[Image:2x01.jpg\|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg\|Illustration of 2x Oversampling pt.2]]


			Mathematically, the equation is:
			<center><math>x(t)h(t) = \sum_{n=-\infty}^\infty x(nT) \delta (t-nT) \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t - \frac{mT}{2}) = \hat y (t) = \sum_{l=-\infty}^\infty \left \sum_{m=-M}^M x \left \frac{l-m}{2} T \right h(\frac{mT}{2}) \right \delta (t - \frac{lT}{2})


	<small>- Principle author of this page: Jeffrey Wonoprabowo		<small>- Principle author of this page: Jeffrey Wonoprabowo

	- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>		- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>

			</math>

Wonoje: /* 2x Oversampling */

2005-12-06T00:27:25Z

2x Oversampling

← Older revision		Revision as of 17:27, 5 December 2005
Line 44:		Line 44:
	<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>		<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>

	[[Image:2x01.jpg\|Illustration of 2x Oversampling pt.1]][[Image:2x02~~.jpg~~.jpg\|Illustration of 2x Oversampling pt.2]]		[[Image:2x01.jpg\|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg\|Illustration of 2x Oversampling pt.2]]

	<small>- Principle author of this page: Jeffrey Wonoprabowo		<small>- Principle author of this page: Jeffrey Wonoprabowo

	- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>		- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>

Wonoje: /* 2x Oversampling */

2005-12-06T00:26:52Z

2x Oversampling

← Older revision		Revision as of 17:26, 5 December 2005
Line 38:		Line 38:
	In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.		In order to perform oversampling, an extra step must be inserted into the process described above. The discretized time signal must be convolved with a function <math>h(t)</math>. The resulting convolution will then be convolved with the pulse function <math>p(t)</math>. This is the only change in the process.

			<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{n}) \delta (t-\frac{mT}{n}) </math></center>

			where <math>n</math> is the number of times oversampling desired. In the case of 2x oversampling, <math>n=2</math> and so,

			<center><math>h(t) = \sum_{m=-M}^M h(\frac{mT}{2}) \delta (t-\frac{mT}{2}) </math></center>

			[[Image:2x01.jpg\|Illustration of 2x Oversampling pt.1]][[Image:2x02.jpg.jpg\|Illustration of 2x Oversampling pt.2]]

	<small>- Principle author of this page: Jeffrey Wonoprabowo		<small>- Principle author of this page: Jeffrey Wonoprabowo

	- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>		- Image of CD Player Diagram and Convolution by Aric Goe or Todd Caswell (not sure which since both of them had it on their pages)</small>