Homework Eight - Revision history

Nicholas.Christman at 04:01, 9 November 2009

2009-11-09T04:01:57Z

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	[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>		[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>
	----		----
			'''''First and foremost, I'd like to give credit to [[Max Woesner]] for the edited class images.'''''

	Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>\scriptstyle x(t)</math> in the time domain ['''Figure 1'''] and <math>\scriptstyle X(f)</math> in the frequency domain ['''Figure 2'''].		Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>\scriptstyle x(t)</math> in the time domain ['''Figure 1'''] and <math>\scriptstyle X(f)</math> in the frequency domain ['''Figure 2'''].
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	[[Image:lowpass_G_f.jpg \|thumb\|center\|upright=2\|Figure 14: Low pass filter in the frequency domain]]		[[Image:lowpass_G_f.jpg \|thumb\|center\|upright=2\|Figure 14: Low pass filter in the frequency domain]]

	This is how digital ~~sampling~~ would ~~work~~ -- our final signal should now be the audio we desire.		This is how a CD player using digital filtering instead of oversampling would operate -- our final signal should now be the audio we desire.


			----
			--[[User:Nicholas.Christman\|Nicholas.Christman]] 04:01, 9 November 2009 (UTC)

Nicholas.Christman at 03:57, 9 November 2009

2009-11-09T03:57:10Z

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	At this point, we are ready to send the audio signal through a D/A converter -- which will essentially convolve the signal with a pulse function, say p(t). The pulse function is shown in the time domain ['''Figure 9'''] and the frequency domain ['''Figure 10''']		At this point, we are ready to send the audio signal through a D/A converter -- which will essentially convolve the signal with a pulse function, say p(t). The pulse function is shown in the time domain ['''Figure 9'''] and the frequency domain ['''Figure 10''']

	[[Image:p_t.jpg \|thumb\|center\|upright=2\|Figure 7: Pulse function in the time domain, p(t)]]		[[Image:p_t.jpg \|thumb\|center\|upright=2\|Figure 9: Pulse function in the time domain, p(t)]]
	[[Image:P_f.jpg \|thumb\|center\|upright=2\|Figure 8: Pulse function in the frequency domain, P(f)]]		[[Image:P_f.jpg \|thumb\|center\|upright=2\|Figure 10: Pulse function in the frequency domain, P(f)]]

	~~And the convolution of the pulse function with the above FIR filtered function in the time domain [~~''~~'Figure 11'''] and the frequency domain ['~~''~~Figure 12'''] is shown below~~		''Note: P(f) is Tsinc(fT)''

	[[Image:x_step.jpg \|thumb\|center\|upright=2\|Figure 7: FIR filtered function convolved with p(t) (time)]]		The convolution of the pulse function with the above FIR filtered function in the time domain ['''Figure 11'''] and the frequency domain ['''Figure 12'''] can be expressed as <math>\sum_{n=-\infty}^{\infty}x(nT)p(t-nT)</math> and <math>P(f)\left[ \sum_{n=-\infty}^{\infty}x(nT)e^{-j2 \pi fnT} \right]</math>, respectively.
	[[Image:x_frequency5.jpg \|thumb\|center\|upright=2\|Figure 8: FIR filtered function multiplied with P(f) (frequency)]]
			[[Image:x_step.jpg \|thumb\|center\|upright=2\|Figure 11: FIR filtered function convolved with p(t) (time)]]
			[[Image:x_frequency5.jpg \|thumb\|center\|upright=2\|Figure 12: FIR filtered function multiplied with P(f) (frequency)]]

			We are now very close to having the audio that we want -- now it would be good to run the signal through a low-pass filter to eliminate any high-frequency interference. As usual, we will convolve in time and multiply in frequency. The low pass filter in the time domain ['''Figure 13'''] and frequency domain ['''Figure 14'''] are shown below

			[[Image:lowpass_g_t.jpg \|thumb\|center\|upright=2\|Figure 13: Low pass filter in the time domain]]
			[[Image:lowpass_G_f.jpg \|thumb\|center\|upright=2\|Figure 14: Low pass filter in the frequency domain]]

			This is how digital sampling would work -- our final signal should now be the audio we desire.

Nicholas.Christman at 03:42, 9 November 2009

2009-11-09T03:42:21Z

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	[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 7: Original function, x(nT), convolved with the FIR filter, g(t)]]		[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 7: Original function, x(nT), convolved with the FIR filter, g(t)]]
	[[Image:x_frequency4a.jpg \|thumb\|center\|upright=2\|Figure 8: Original function multiplied by the FIR filter in frequency domain]]		[[Image:x_frequency4a.jpg \|thumb\|center\|upright=2\|Figure 8: Original function multiplied by the FIR filter in frequency domain]]

			At this point, we are ready to send the audio signal through a D/A converter -- which will essentially convolve the signal with a pulse function, say p(t). The pulse function is shown in the time domain ['''Figure 9'''] and the frequency domain ['''Figure 10''']

			[[Image:p_t.jpg \|thumb\|center\|upright=2\|Figure 7: Pulse function in the time domain, p(t)]]
			[[Image:P_f.jpg \|thumb\|center\|upright=2\|Figure 8: Pulse function in the frequency domain, P(f)]]

			And the convolution of the pulse function with the above FIR filtered function in the time domain ['''Figure 11'''] and the frequency domain ['''Figure 12'''] is shown below

			[[Image:x_step.jpg \|thumb\|center\|upright=2\|Figure 7: FIR filtered function convolved with p(t) (time)]]
			[[Image:x_frequency5.jpg \|thumb\|center\|upright=2\|Figure 8: FIR filtered function multiplied with P(f) (frequency)]]

Nicholas.Christman at 03:30, 9 November 2009

2009-11-09T03:30:46Z

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	[[Image:x_frequency2a.jpg \|thumb\|center\|upright=2\|Figure 4: x(nT) written as impulses for sampling purposes in the frequency domain]]		[[Image:x_frequency2a.jpg \|thumb\|center\|upright=2\|Figure 4: x(nT) written as impulses for sampling purposes in the frequency domain]]

	As one may suspect, in order to get a more accurate function you must take more data points -- in other words, a higher sampling rate (i.e. 2x, 8x, etc. oversampling) leads to a more accurate collection of data. For this case, we are asked to looked a digital sampling, or 1x oversampling. In order to accomplish this, we will convolve our function from above with a Finite Impulse Response filter (FIR filter). Let's call it g(t and define it as <math>\sum_{-M}^{M}g(\frac{mT}{p})\delta (t-\frac{mT}{p}) </math> where <math>\scriptstyle p</math> is the oversampling rate. In our case <math>\scriptstyle p = 1</math>. Because we are dealing with digital sampling, the convolution will result in the original function ['''Figure 5'''].		As one may suspect, in order to get a more accurate function you must take more data points -- in other words, a higher sampling rate (i.e. 2x, 8x, etc. oversampling) leads to a more accurate collection of data. For this case, we are asked to looked a digital sampling, or 1x oversampling. In order to accomplish this, we will convolve our function from above with a Finite Impulse Response filter (FIR filter). Let's call the FIR g(t) and define it as <math>\sum_{-M}^{M}g(\frac{mT}{p})\delta (t-\frac{mT}{p}) </math> where <math>\scriptstyle p</math> is the oversampling rate. In our case <math>\scriptstyle p = 1</math>. Because we are dealing with digital sampling, the convolution will result in the original function ['''Figure 5''']. It is fairly easy to see that in the frequency domain, the FIR filter will be expressed as <math> G(f) = \sum_{-M}^{M}g(mT)e^{-j2 \pi mT}</math> ['''Figure 6'''].

	[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 5: x(nT) ~~after~~ the ~~convolution mentioned above~~]]		[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 5: FIR filter in time domain, g(t)]]
			[[Image:x_frequency3a.jpg \|thumb\|center\|upright=2\|Figure 6: FIR filter in frequency domain, G(f)]]

			As we know, convolution in the time domain results in multiplication in the frequency domain. The results of the convolution and multiplication of the before mentioned FIR filter (time, frequency) are shown below:

			[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 7: Original function, x(nT), convolved with the FIR filter, g(t)]]
			[[Image:x_frequency4a.jpg \|thumb\|center\|upright=2\|Figure 8: Original function multiplied by the FIR filter in frequency domain]]

Nicholas.Christman at 03:17, 9 November 2009

2009-11-09T03:17:36Z

← Older revision		Revision as of 20:17, 8 November 2009
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	----		----

	Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain ('''Figure 1''') and <math>X(f)</math> in the frequency domain ('''Figure 2''').		Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>\scriptstyle x(t)</math> in the time domain ['''Figure 1'''] and <math>\scriptstyle X(f)</math> in the frequency domain ['''Figure 2'''].

	[[Image:x_continuous3.jpg \|thumb\|center\|upright=2\|Figure 1: Audio data as a function of time, x(t)]]		[[Image:x_continuous3.jpg \|thumb\|center\|upright=2\|Figure 1: Audio data as a function of time, x(t)]]
	[[Image:x_frequency1b.jpg \|thumb\|center\|upright=2\|Figure 2: Audio data as a function of frequency, X(f)]]		[[Image:x_frequency1b.jpg \|thumb\|center\|upright=2\|Figure 2: Audio data as a function of frequency, X(f)]]

	When one wants to store the data of <math>x(t)</math> (i.e. reading the data onto a CD in this case) an infinite number of data points is not ideal -- in fact, it is impossible. Therefore, we must sample the data at the (typical) rate of <math>f_{s} = \frac{1}{T}</math>. This will give us a periodic function which we will call <math>x(nT) \mbox{, where } n \in \mathbb{Z} \mbox{ and } T</math> is the sampling period. It is common to allow a sampling rate of twice the frequency at which a human can hear (i.e. 2 x 22 kHz) -- that means, <math>f_{s} = 44</math> kHz.<br/>		When one wants to store the data of <math>\scriptstyle x(t)</math> (i.e. reading the data onto a CD in this case) an infinite number of data points is not ideal -- in fact, it is impossible. Therefore, we must sample the data at the (typical) rate of <math>\scriptstyle f_{s} = \frac{1}{T}</math>. This will give us a periodic function which we will call <math>\scriptstyle x(nT) \mbox{, where } n \in \mathbb{Z} \mbox{ and } T</math> is the sampling period. It is common to allow a sampling rate of twice the frequency at which a human can hear (i.e. 2 x 22 kHz) -- that means, <math>\scriptstyle f_{s} = 44</math> kHz.<br/>

	Recall from class that in order to sample the original function, <math>x(nT)</math>, we need to use a delta function. In other words, the continuous function of <math>x(nT)</math> can be written as <math>\sum_{-\infty}^{\infty}x(nT) \delta (t-nT_{0})</math> ('''Figure 3''')		Recall from class that in order to sample the original function, <math>\scriptstyle x(nT)</math>, we need to use a delta function. In other words, the continuous function of <math>\scriptstyle x(nT)</math> can be written as <math> \sum_{n=-\infty}^{\infty}x(nT) \delta (t-nT_{0})</math> ['''Figure 3''']. As one may expect, in the frequency domain we should get the same plot as in '''Figure 2''', but it should repeat. In the frequency domain, this can be expressed as <math>\frac{1}{T}\sum_{n=-\infty}^{\infty}X(f-\frac{n}{T})</math> ['''Figure 4'''].

	[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 3: x(nT) written as impulses for sampling purposes]]		[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 3: x(nT) written as impulses for sampling purposes]]
			[[Image:x_frequency2a.jpg \|thumb\|center\|upright=2\|Figure 4: x(nT) written as impulses for sampling purposes in the frequency domain]]

			As one may suspect, in order to get a more accurate function you must take more data points -- in other words, a higher sampling rate (i.e. 2x, 8x, etc. oversampling) leads to a more accurate collection of data. For this case, we are asked to looked a digital sampling, or 1x oversampling. In order to accomplish this, we will convolve our function from above with a Finite Impulse Response filter (FIR filter). Let's call it g(t and define it as <math>\sum_{-M}^{M}g(\frac{mT}{p})\delta (t-\frac{mT}{p}) </math> where <math>\scriptstyle p</math> is the oversampling rate. In our case <math>\scriptstyle p = 1</math>. Because we are dealing with digital sampling, the convolution will result in the original function ['''Figure 5'''].

			[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 5: x(nT) after the convolution mentioned above]]

Nicholas.Christman at 01:51, 9 November 2009

2009-11-09T01:51:27Z

← Older revision		Revision as of 18:51, 8 November 2009
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	----		----

	Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain and <math>X(f)</math> in the frequency domain~~, as follows~~		Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain ('''Figure 1''') and <math>X(f)</math> in the frequency domain ('''Figure 2''').

	[[Image:x_continuous3.jpg \|thumb\|~~left~~\|upright=2\|Audio data as a function of time, x(t)]]		[[Image:x_continuous3.jpg \|thumb\|center\|upright=2\|Figure 1: Audio data as a function of time, x(t)]]
	[[Image:x_frequency1b.jpg \|thumb\|~~left~~\|upright=2\|Audio data as a function of frequency, X(f)]]<br>\		[[Image:x_frequency1b.jpg \|thumb\|center\|upright=2\|Figure 2: Audio data as a function of frequency, X(f)]]

			When one wants to store the data of <math>x(t)</math> (i.e. reading the data onto a CD in this case) an infinite number of data points is not ideal -- in fact, it is impossible. Therefore, we must sample the data at the (typical) rate of <math>f_{s} = \frac{1}{T}</math>. This will give us a periodic function which we will call <math>x(nT) \mbox{, where } n \in \mathbb{Z} \mbox{ and } T</math> is the sampling period. It is common to allow a sampling rate of twice the frequency at which a human can hear (i.e. 2 x 22 kHz) -- that means, <math>f_{s} = 44</math> kHz.<br/>

			Recall from class that in order to sample the original function, <math>x(nT)</math>, we need to use a delta function. In other words, the continuous function of <math>x(nT)</math> can be written as <math>\sum_{-\infty}^{\infty}x(nT) \delta (t-nT_{0})</math> ('''Figure 3''')

			[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 3: x(nT) written as impulses for sampling purposes]]

Nicholas.Christman at 01:06, 9 November 2009

2009-11-09T01:06:03Z

← Older revision		Revision as of 18:06, 8 November 2009
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	Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain and <math>X(f)</math> in the frequency domain, as follows		Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain and <math>X(f)</math> in the frequency domain, as follows

	[[Image:x_continuous3.jpg]] ~~{time} --------> {frequency}~~ [[Image:x_frequency1b.jpg]]<br>		[[Image:x_continuous3.jpg \|thumb\|left\|upright=2\|Audio data as a function of time, x(t)]]
			[[Image:x_frequency1b.jpg \|thumb\|left\|upright=2\|Audio data as a function of frequency, X(f)]]<br>\

Nicholas.Christman at 00:57, 9 November 2009

2009-11-09T00:57:13Z

← Older revision		Revision as of 17:57, 8 November 2009
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	[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>		[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>
	----		----

			Recall that when the audio for a music CD is produced it has an infinite amount of data points which can be expressed as <math>x(t)</math> in the time domain and <math>X(f)</math> in the frequency domain, as follows

			[[Image:x_continuous3.jpg]] {time} --------> {frequency} [[Image:x_frequency1b.jpg]]<br>

Nicholas.Christman: New page: '''How does a CD player work with no oversampling, but digital filtering (1x oversampling)?'''
Nick Christman
----

2009-11-01T00:56:03Z

New page: '''How does a CD player work with no oversampling, but digital filtering (1x oversampling)?''' Nick Christman ----

New page

'''How does a CD player work with no oversampling, but digital filtering (1x oversampling)?'''

 

[[Nick Christman|Nick Christman]] 
----

← Older revision		Revision as of 20:42, 8 November 2009
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	[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 7: Original function, x(nT), convolved with the FIR filter, g(t)]]		[[Image:x_discrete_a.jpg \|thumb\|center\|upright=2\|Figure 7: Original function, x(nT), convolved with the FIR filter, g(t)]]
	[[Image:x_frequency4a.jpg \|thumb\|center\|upright=2\|Figure 8: Original function multiplied by the FIR filter in frequency domain]]		[[Image:x_frequency4a.jpg \|thumb\|center\|upright=2\|Figure 8: Original function multiplied by the FIR filter in frequency domain]]

			At this point, we are ready to send the audio signal through a D/A converter -- which will essentially convolve the signal with a pulse function, say p(t). The pulse function is shown in the time domain ['''Figure 9'''] and the frequency domain ['''Figure 10''']

			[[Image:p_t.jpg \|thumb\|center\|upright=2\|Figure 7: Pulse function in the time domain, p(t)]]
			[[Image:P_f.jpg \|thumb\|center\|upright=2\|Figure 8: Pulse function in the frequency domain, P(f)]]

			And the convolution of the pulse function with the above FIR filtered function in the time domain ['''Figure 11'''] and the frequency domain ['''Figure 12'''] is shown below

			[[Image:x_step.jpg \|thumb\|center\|upright=2\|Figure 7: FIR filtered function convolved with p(t) (time)]]
			[[Image:x_frequency5.jpg \|thumb\|center\|upright=2\|Figure 8: FIR filtered function multiplied with P(f) (frequency)]]