Homework Seven - Revision history

Nicholas.Christman at 01:17, 30 November 2009

2009-11-30T01:17:40Z

← Older revision		Revision as of 18:17, 29 November 2009
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	So to find the transfer function in the time-domain we simply need to find the inverse Fourier Transform of H(f):		So to find the transfer function in the time-domain we simply need to find the inverse Fourier Transform of H(f):

	<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{2T}}^{\frac{1}{2T}}Te^{j2 \pi ft} \,df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{-\frac{1}{2T}}^{-\frac{1}{2T}} = \frac{-T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] </math>		<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{2T}}^{\frac{1}{2T}}Te^{j2 \pi ft} \,df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{-\frac{1}{2T}}^{\frac{1}{2T}} = \frac{T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] </math>

	By Euler's Identiy, however, we can see that		By Euler's Identiy, however, we can see that
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	<math> \frac{-T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] = \frac{-T}{\pi t} sin \left( \frac{ \pi t}{T} \right) </math>		<math> \frac{T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] = \frac{T}{\pi t} sin \left( \frac{ \pi t}{T} \right) </math>
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	~~Therefore~~,		Recall that <math>\textstyle sinc(\theta ) = \frac{sin(\pi \theta )}{\pi \theta }</math>. Thus,
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	<math> h(~~t) =~~ \frac{-T}{\pi t} ~~sin~~ \left( \frac{ \pi t}{T} \right) </math>		<math> \frac{sin \left( \frac{ \pi t}{T} \right)}{\frac{\pi t}{T}} = sinc\left( \frac{t}{T} \right) \Longrightarrow h(t) = sinc\left( \frac{t}{T} \right)</math>
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			We learned in class that <math> \textstyle x'(t) = \sum_{n=-\infty }^{\infty}x(nT)\delta (t-nT)</math> and, thus, we must convolve <math>\textstyle x'(t) \mbox{ with } h(t)</math> to find the function of <math>\textstyle x(t) </math> desired:
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			<math> x(t) = x'(t)h(t) = \sum_{n=-\infty }^{\infty}x(nT)\delta (t-nT)\left[ sinc\left( \frac{t}{T} \right) \right]
			= \sum_{n=-\infty }^{\infty}x(nT) \mbox{ } sinc\left( \frac{t-nT}{T} \right)
			</math>
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			Therefore, the original signal (in terms of <math> \scriptstyle x(nT)</math>) is
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			<math> x(t)= \sum_{n=-\infty }^{\infty}x(nT) \mbox{ } sinc\left( \frac{t-nT}{T} \right) = \sum_{n=-\infty }^{\infty}\frac{x(nT) \mbox{ } sin \left( \frac{\pi (t-nT)}{T} \right)}{\frac{\pi (t-nT)}{T}}</math>

Nicholas.Christman at 20:46, 29 November 2009

2009-11-29T20:46:56Z

← Older revision		Revision as of 13:46, 29 November 2009
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	So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5 & 6).		So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5 & 6).
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			Our next task is to find an expression of x(t) in terms of x(nT). To accomplish this, recall from class that the transfer function of the lowpass filter is
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			<math>H(f)=\begin{cases} T, & \mbox{ } \|f\| < \frac{1}{2T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>

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			So to find the transfer function in the time-domain we simply need to find the inverse Fourier Transform of H(f):

			<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{2T}}^{\frac{1}{2T}}Te^{j2 \pi ft} \,df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{-\frac{1}{2T}}^{-\frac{1}{2T}} = \frac{-T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] </math>

			By Euler's Identiy, however, we can see that
			<br/>
			<br/>
			<math> \frac{-T}{\pi t}\left[ \frac{e^{j \pi t/T}-e^{-j \pi t/T}}{j2}\right] = \frac{-T}{\pi t} sin \left( \frac{ \pi t}{T} \right) </math>
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			Therefore,
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			<math> h(t) = \frac{-T}{\pi t} sin \left( \frac{ \pi t}{T} \right) </math>
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Nicholas.Christman at 20:02, 29 November 2009

2009-11-29T20:02:35Z

← Older revision		Revision as of 13:02, 29 November 2009
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	[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 6: Original signal obtained using a lowpass filter.]]		[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 6: Original signal obtained using a lowpass filter.]]

	So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).		So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5 & 6).

Nicholas.Christman at 20:02, 29 November 2009

2009-11-29T20:02:02Z

← Older revision		Revision as of 13:02, 29 November 2009
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	[[Image:signal_sampled.jpg \|thumb\|center\|upright=2\|Figure 4: Signal being sampled at a rate of 1/T.]]		[[Image:signal_sampled.jpg \|thumb\|center\|upright=2\|Figure 4: Signal being sampled at a rate of 1/T.]]

	In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:		In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal and filter out any high frequency components:

	[[Image:signal_lowpass.jpg \|thumb\|center\|upright=2\|Figure 5: The use of a lowpass filter to obtain the original signal as shown in Figure 1.]]		[[Image:signal_lowpass.jpg \|thumb\|center\|upright=2\|Figure 5: The use of a lowpass filter to obtain the original signal as shown in Figure 1.]]

			[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 6: Original signal obtained using a lowpass filter.]]

	So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).		So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).

Nicholas.Christman at 19:57, 29 November 2009

2009-11-29T19:57:06Z

← Older revision		Revision as of 12:57, 29 November 2009
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	Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:		Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:

	[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]		[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 1: Original signal for which we want to analyze.]]

	In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:		In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:

	[[Image:signal_split.jpg \|thumb\|center\|upright=2\|Figure 1:~~Original signal for which we want to analyze~~.]]		[[Image:signal_split.jpg \|thumb\|center\|upright=2\|Figure 2: Signal separated into positive and negative aspects.]]

	To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>, we get the following signal, <math> \textstyle X(f) </math>:		To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>, we get the following signal, <math> \textstyle X(f) </math>:

	[[Image:signal_export.jpg \|thumb\|center\|upright=2\|Figure 1:~~Original~~ signal ~~for which we want~~ to ~~analyze~~.]]		[[Image:signal_export.jpg \|thumb\|center\|upright=2\|Figure 3: Preparing signal to be sent and sampled.]]

	Finally, we need to sample the signal at a rate of <math>\textstyle f_s = \frac{1}{T} </math>. Theoretically, this leads to the following frequency plot:		Finally, we need to sample the signal at a rate of <math>\textstyle f_s = \frac{1}{T} </math>. Theoretically, this leads to the following frequency plot:

	[[Image:signal_sampled.jpg \|thumb\|center\|upright=2\|Figure 1~~:Original signal for which we want to analyze~~.]]		[[Image:signal_sampled.jpg \|thumb\|center\|upright=2\|Figure 4: Signal being sampled at a rate of 1/T.]]

	In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:		In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:

	[[Image:signal_lowpass.jpg \|thumb\|center\|upright=2\|Figure 1:~~Original~~ signal ~~for which we want to analyze~~.]]		[[Image:signal_lowpass.jpg \|thumb\|center\|upright=2\|Figure 5: The use of a lowpass filter to obtain the original signal as shown in Figure 1.]]

	So what have we accomplished? We have taken a signal (Figure 1), prepared it to be ~~analyzed and~~ sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).		So what have we accomplished? We have taken a signal (Figure 1), prepared it to be sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).

Nicholas.Christman at 19:53, 29 November 2009

2009-11-29T19:53:27Z

← Older revision		Revision as of 12:53, 29 November 2009
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	Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:		Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:

	[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 1: ~~Audio data as a function of time, x(t)~~]]		[[Image:original.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]

	In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:		In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:

			[[Image:signal_split.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]

	To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>, we get the following signal, <math> \textstyle X(f) </math>:		To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>, we get the following signal, <math> \textstyle X(f) </math>:

			[[Image:signal_export.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]

	Finally, we need to sample the signal at a rate of <math>\textstyle f_s = \frac{1}{T} </math>. Theoretically, this leads to the following frequency plot:		Finally, we need to sample the signal at a rate of <math>\textstyle f_s = \frac{1}{T} </math>. Theoretically, this leads to the following frequency plot:

			[[Image:signal_sampled.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]

	In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:		In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:

			[[Image:signal_lowpass.jpg \|thumb\|center\|upright=2\|Figure 1:Original signal for which we want to analyze.]]

	So what have we accomplished? We have taken a signal (Figure 1), prepared it to be analyzed and sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).		So what have we accomplished? We have taken a signal (Figure 1), prepared it to be analyzed and sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).

Nicholas.Christman at 19:50, 29 November 2009

2009-11-29T19:50:01Z

← Older revision		Revision as of 12:50, 29 November 2009
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	Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:		Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:

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

	In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:		In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:

Nicholas.Christman at 19:08, 29 November 2009

2009-11-29T19:08:16Z

← Older revision		Revision as of 12:08, 29 November 2009
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	[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>		[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br/>
	----		----

			Imagine we have an original signal that has the following frequency plot, centered about the "zero frequency" origin:


			In order to analyze this signal we must look at both the positive and negative frequency aspects -- therefore, we will split the signal into two parts, positive and negative:

			To "send the signal" we must essentially move it to a higher frequency and, remembering that x(t) has frequency components for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>, we get the following signal, <math> \textstyle X(f) </math>:

			Finally, we need to sample the signal at a rate of <math>\textstyle f_s = \frac{1}{T} </math>. Theoretically, this leads to the following frequency plot:


			In order to get the original signal, we simply need to create a lowpass filter that will essentially encompass the the original (desired) signal:


			So what have we accomplished? We have taken a signal (Figure 1), prepared it to be analyzed and sampled (Figure 2 & 3), sampled it at a sampling rate of <math>\textstyle f_s = \frac{1}{T} </math> (Figure 4), and used a lowpass filter to collect the original (now sampled) signal (Figure 5).

Nicholas.Christman at 00:46, 1 November 2009

2009-11-01T00:46:46Z

← Older revision		Revision as of 17:46, 31 October 2009
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	'''Figure out what happens if your sampled signal, x(t), has frequency components only for <math>\textstyle \frac{f_{s~~}}{2~~} < f < f_{s}</math>. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nt).'''		'''Figure out what happens if your sampled signal, x(t), has frequency components only for <math>\textstyle \frac{1}{2}f_{s} < f < f_{s}</math>. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nt).'''

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Nicholas.Christman at 00:46, 1 November 2009

2009-11-01T00:46:14Z

← Older revision		Revision as of 17:46, 31 October 2009
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	'''Figure out what happens if your sampled signal, x(t), has frequency components only for <math>\frac{f_{s}}{2} < f < f_{s}</math>. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nt).'''		'''Figure out what happens if your sampled signal, x(t), has frequency components only for <math>\textstyle \frac{f_{s}}{2} < f < f_{s}</math>. Can you recover the original signal from it? If so, find the expression for x(t) in terms of x(nt).'''

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