Sampling - HW7 - Revision history

Max.Woesner at 04:42, 2 December 2009

2009-12-02T04:42:01Z

← Older revision		Revision as of 21:42, 1 December 2009
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	<b>Solution</b><br>		<b>Solution</b><br>
	For frequency components <math> \frac{f_s}{2}<f<f_s\!</math>, our sampled signal <math> x(t)\!</math> in the frequency domain, or <math> X(f)\!</math>, is going to look like this.<br>		For frequency components <math> \frac{f_s}{2}<f<f_s\!</math>, our sampled signal <math> x(t)\!</math> in the frequency domain, or <math> X(f)\!</math>, is going to look like this.<br><br>
	[[Image:Sampling1a.jpg]]<br><br>		[[Image:Sampling1a.jpg]]<br><br>
	After sampling with frequency <math> f_s\!</math>, the signal is going to be shifted over by <math> \frac{1}{T}\!</math>, since <math> f_s = \frac{1}{T}\!</math>. It will look like this.<br>		After sampling with frequency <math> f_s\!</math>, the signal is going to be shifted over by <math> \frac{1}{T}\!</math>, since <math> f_s = \frac{1}{T}\!</math>. It will look like this.<br><br>
	[[Image:Sampling2.jpg]]<br><br>		[[Image:Sampling2.jpg]]<br><br>
	To recover the original signal <math> x(t)\!</math>, we need to a use bandpass filter to filter out the parts we don't want, as indicated by the red line in the figure below. <i>(Note: not to scale.)</i><br>		To recover the original signal <math> x(t)\!</math>, we need to a use bandpass filter to filter out the parts we don't want, as indicated by the red line in the figure below. <i>(Note: not to scale.)</i><br><br>
	[[Image:Sampling3.jpg]]<br><br>		[[Image:Sampling3.jpg]]<br><br>
	This leaves us with the original signal, as shown below.<br>		This leaves us with the original signal, as shown below.<br><br>
	[[Image:Sampling1a.jpg]]<br><br>		[[Image:Sampling1a.jpg]]<br><br>
	The transfer function of the bandpass filter that will accomplish this for us is <math>H(f)=\begin{cases} T, & \mbox{ } \frac{1}{2T}<\|f\|< \frac{1}{T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>		The transfer function of the bandpass filter that will accomplish this for us is <math>H(f)=\begin{cases} T, & \mbox{ } \frac{1}{2T}<\|f\|< \frac{1}{T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>
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	Also recall <math> sinc(\theta) = \frac{sin(\pi \theta)}{\pi \theta}\!</math>, so <br><br>		Also recall <math> sinc(\theta) = \frac{sin(\pi \theta)}{\pi \theta}\!</math>, so <br><br>
	<math> h(t) = \frac{2sin \Big(\frac{2 \pi t}{T} \Big)}{\frac{2 \pi t}{T}} \ - \ \frac{sin \Big(\frac{ \pi t}{T} \Big)}{\frac{ \pi t}{T}} = 2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\!</math><br><br>		<math> h(t) = \frac{2sin \Big(\frac{2 \pi t}{T} \Big)}{\frac{2 \pi t}{T}} \ - \ \frac{sin \Big(\frac{ \pi t}{T} \Big)}{\frac{ \pi t}{T}} = 2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\!</math><br><br>
	Now that we know <math> h(t) \!</math>, we can find <math> x(t) \!</math> by convolving the function for <math> x(t) \!</math> after sampling with <math> h(t) \!</math>.<br><br>		Now that we know <math> h(t) \!</math>, we can find <math> x(t) \!</math> by convolving the function for <math> x(t) \!</math> after sampling, or <math> \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) \!</math>, with <math> h(t) \!</math>.<br><br>
	<math> x(t) = \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) * \Bigg[2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\Bigg] = \sum_{n=-\infty}^\infty x(nT)\Bigg[2sinc \Bigg(\frac{2(t-nT)}{T} \Bigg) \ - \ sinc \Bigg(\frac{t-nT}{T} \Bigg)\Bigg]\!</math><br><br>		<math> x(t) = \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) * \Bigg[2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\Bigg] = \sum_{n=-\infty}^\infty x(nT)\Bigg[2sinc \Bigg(\frac{2(t-nT)}{T} \Bigg) \ - \ sinc \Bigg(\frac{t-nT}{T} \Bigg)\Bigg]\!</math><br><br>
	Or, if you prefer, <math> x(t) = \sum_{n=-\infty}^\infty x(nT)\Bigg[\frac{2sin\Big(\frac{2\pi (t-nT)}{T}\Big)}{\frac{2\pi (t-nT)}{T}} \ - \ \frac{sin\Big(\frac{\pi (t-nT)}{T}\Big)}{\frac{\pi (t-nT)}{T}}\Bigg]		Or, if you prefer, <math> x(t) = \sum_{n=-\infty}^\infty x(nT)\Bigg[\frac{2sin\Big(\frac{2\pi (t-nT)}{T}\Big)}{\frac{2\pi (t-nT)}{T}} \ - \ \frac{sin\Big(\frac{\pi (t-nT)}{T}\Big)}{\frac{\pi (t-nT)}{T}}\Bigg]\!</math><br><br>

			Alternatively, a different bandpass filter can be used that will simplify the math we have to do, such as the one indicated by the red line in the figure below. <i>(Note: not to scale.)</i><br><br>
			[[Image:Sampling4.jpg]]<br><br>
			Leaving us with the signal shown below.<br><br>
			[[Image:Sampling5.jpg]]<br><br>
			The transfer function of this bandpass filter is <math>H(f)=\begin{cases} T, & \mbox{ } \|f\|< \frac{1}{2T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>
			To find the expression for <math> h(t)\!</math>, or the expression for the bandpass filter in the time domain, we can take the inverse Fourier transform of <math>H(f)\!</math>.<br><br>
			<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\|_{f=-\frac{1}{2T}}^{\frac{1}{2T}} = \frac{T}{j2}\Bigg[\frac{e^{j \pi t/T} - e^{-j \pi t/T}}{ \pi t}\Bigg]\!</math><br><br>
			Recall again that <math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math>, so <br><br>
			<math> h(t) = \frac{T}{j2}\Bigg[\frac{e^{j \pi t/T} - e^{-j \pi t/T}}{ \pi t}\Bigg] = \frac{T}{ \pi t}sin \Bigg(\frac{ \pi t}{T} \Bigg) \ = \frac{sin \Big(\frac{ \pi t}{T} \Big)}{\frac{ \pi t}{T}} \!</math><br><br>
			Also recall again that <math> sinc(\theta) = \frac{sin(\pi \theta)}{\pi \theta}\!</math>, so <br><br>
			<math> h(t) = \frac{sin \Big(\frac{ \pi t}{T} \bigg)}{\frac{ \pi t}{T}} = sinc \bigg(\frac{t}{T} \bigg) \!</math><br><br>
			Now that we know <math> h(t) \!</math>, we can find <math> x(t) \!</math> by convolving the function for <math> x(t) \!</math> after sampling, or <math> \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) \!</math>, with <math> h(t) \!</math>.<br><br>

			<math> x(t) = \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) * \Bigg[sinc \bigg(\frac{t}{T} \bigg)\Bigg] = \sum_{n=-\infty}^\infty x(nT)(sinc \Bigg(\frac{t-nT}{T} \Bigg)\!</math><br><br>
			Or, if you prefer, <math> x(t) = \sum_{n=-\infty}^\infty x(nT)\Bigg[\frac{sin\Big(\frac{\pi (t-nT)}{T}\Big)}{\frac{\pi (t-nT)}{T}}\Bigg]\!</math><br><br>

Max.Woesner at 04:48, 10 November 2009

2009-11-10T04:48:59Z

← Older revision		Revision as of 21:48, 9 November 2009
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	This leaves us with the original signal, as shown below.<br>		This leaves us with the original signal, as shown below.<br>
	[[Image:Sampling1a.jpg]]<br><br>		[[Image:Sampling1a.jpg]]<br><br>
	The transfer function of the bandpass filter that will accomplish this for us is <math>H(f)=\begin{cases} T, & \mbox{ }- \frac{1}{2T}<\|f\|< \frac{1}{T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>		The transfer function of the bandpass filter that will accomplish this for us is <math>H(f)=\begin{cases} T, & \mbox{ } \frac{1}{2T}<\|f\|< \frac{1}{T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>
	To find the expression for <math> h(t)\!</math>, or the expression for the bandpass filter in the time domain, we can take the inverse Fourier transform of <math>H(f)\!</math>.<br><br>		To find the expression for <math> h(t)\!</math>, or the expression for the bandpass filter in the time domain, we can take the inverse Fourier transform of <math>H(f)\!</math>.<br><br>
	<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{T}}^{-\frac{1}{2T}}Te^{j2 \pi ft}df\ + \int_{\frac{1}{2T}}^{\frac{1}{T}}Te^{j2 \pi ft}df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=-\frac{1}{T}}^{-\frac{1}{2T}} + \ \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=\frac{1}{2T}}^{\frac{1}{T}}\!</math><br><br>		<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{T}}^{-\frac{1}{2T}}Te^{j2 \pi ft}df\ + \int_{\frac{1}{2T}}^{\frac{1}{T}}Te^{j2 \pi ft}df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=-\frac{1}{T}}^{-\frac{1}{2T}} + \ \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=\frac{1}{2T}}^{\frac{1}{T}}\!</math><br><br>

Max.Woesner at 06:05, 29 October 2009

2009-10-29T06:05:31Z

← Older revision		Revision as of 23:05, 28 October 2009
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	=== Homework #7 - Sampling ===		=== Homework #7 - Sampling ===

			<br><b>Problem Statement</b><br>
	Figure out what happens if your sampled signal, <math> x(t)\!</math>, 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 <math> x(t)\!</math> in terms of <math> x(nT)\!</math>.<br>		Figure out what happens if your sampled signal, <math> x(t)\!</math>, 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 <math> x(t)\!</math> in terms of <math> x(nT)\!</math>.<br>

			<b>Solution</b><br>
			For frequency components <math> \frac{f_s}{2}<f<f_s\!</math>, our sampled signal <math> x(t)\!</math> in the frequency domain, or <math> X(f)\!</math>, is going to look like this.<br>
			[[Image:Sampling1a.jpg]]<br><br>
			After sampling with frequency <math> f_s\!</math>, the signal is going to be shifted over by <math> \frac{1}{T}\!</math>, since <math> f_s = \frac{1}{T}\!</math>. It will look like this.<br>
			[[Image:Sampling2.jpg]]<br><br>
			To recover the original signal <math> x(t)\!</math>, we need to a use bandpass filter to filter out the parts we don't want, as indicated by the red line in the figure below. <i>(Note: not to scale.)</i><br>
			[[Image:Sampling3.jpg]]<br><br>
			This leaves us with the original signal, as shown below.<br>
			[[Image:Sampling1a.jpg]]<br><br>
			The transfer function of the bandpass filter that will accomplish this for us is <math>H(f)=\begin{cases} T, & \mbox{ }- \frac{1}{2T}<\|f\|< \frac{1}{T} \\ 0, & \mbox{ else } \end{cases}\!</math><br><br>
			To find the expression for <math> h(t)\!</math>, or the expression for the bandpass filter in the time domain, we can take the inverse Fourier transform of <math>H(f)\!</math>.<br><br>
			<math> h(t) = \mathcal{F}^{-1}[H(f)] = \int_{-\frac{1}{T}}^{-\frac{1}{2T}}Te^{j2 \pi ft}df\ + \int_{\frac{1}{2T}}^{\frac{1}{T}}Te^{j2 \pi ft}df = \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=-\frac{1}{T}}^{-\frac{1}{2T}} + \ \frac{Te^{j2 \pi ft}}{j2 \pi t} \Bigg\|_{f=\frac{1}{2T}}^{\frac{1}{T}}\!</math><br><br>
			<math> h(t) = T \frac{e^{-j \pi t/T} - e^{-j2 \pi t/T}}{j2( \pi t)} \ + \ T \frac{e^{j2 \pi t/T} - e^{j \pi t/T}}{j2( \pi t)} = \frac{T}{j2}\Bigg[\frac{e^{-j \pi t/T} - e^{-j2 \pi t/T}+e^{j2 \pi t/T} - e^{j \pi t/T}}{ \pi t}\Bigg]\!</math><br><br>
			<math> h(t) = \frac{T}{j2}\Bigg[\frac{- e^{j \pi t/T} + e^{-j \pi t/T} + e^{j2 \pi t/T} - e^{-j2 \pi t/T}}{ \pi t}\Bigg] = \frac{T}{j2}\Bigg[\frac{e^{j2 \pi t/T} - e^{-j2 \pi t/T}}{ \pi t}\Bigg] - \ \frac{T}{j2}\Bigg[\frac{e^{j \pi t/T} - e^{-j \pi t/T}}{ \pi t}\Bigg] \!</math><br><br>
			Recall <math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math>, so <br><br>
			<math> h(t) = \frac{T}{j2}\Bigg[\frac{e^{j2 \pi t/T} - e^{-j2 \pi t/T}}{ \pi t}\Bigg] - \ \frac{T}{j2}\Bigg[\frac{e^{j \pi t/T} - e^{-j \pi t/T}}{ \pi t}\Bigg] =\frac{T}{ \pi t}sin \Bigg(\frac{2 \pi t}{T} \Bigg) \ - \ \frac{T}{ \pi t}sin \Bigg(\frac{ \pi t}{T} \Bigg) = \frac{2sin \Big(\frac{2 \pi t}{T} \Big)}{\frac{2 \pi t}{T}} \ - \ \frac{sin \Big(\frac{ \pi t}{T} \Big)}{\frac{ \pi t}{T}} \!</math><br><br>
			Also recall <math> sinc(\theta) = \frac{sin(\pi \theta)}{\pi \theta}\!</math>, so <br><br>
			<math> h(t) = \frac{2sin \Big(\frac{2 \pi t}{T} \Big)}{\frac{2 \pi t}{T}} \ - \ \frac{sin \Big(\frac{ \pi t}{T} \Big)}{\frac{ \pi t}{T}} = 2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\!</math><br><br>
			Now that we know <math> h(t) \!</math>, we can find <math> x(t) \!</math> by convolving the function for <math> x(t) \!</math> after sampling with <math> h(t) \!</math>.<br><br>
			<math> x(t) = \sum_{n=-\infty}^\infty x(nT)\delta(t-nT) * \Bigg[2sinc \Bigg(\frac{2t}{T} \Bigg) - sinc \Bigg(\frac{t}{T} \Bigg)\Bigg] = \sum_{n=-\infty}^\infty x(nT)\Bigg[2sinc \Bigg(\frac{2(t-nT)}{T} \Bigg) \ - \ sinc \Bigg(\frac{t-nT}{T} \Bigg)\Bigg]\!</math><br><br>
			Or, if you prefer, <math> x(t) = \sum_{n=-\infty}^\infty x(nT)\Bigg[\frac{2sin\Big(\frac{2\pi (t-nT)}{T}\Big)}{\frac{2\pi (t-nT)}{T}} \ - \ \frac{sin\Big(\frac{\pi (t-nT)}{T}\Big)}{\frac{\pi (t-nT)}{T}}\Bigg]

Max.Woesner at 01:42, 23 October 2009

2009-10-23T01:42:25Z

← Older revision		Revision as of 18:42, 22 October 2009
Line 4:		Line 4:
	=== Homework #7 - Sampling ===		=== Homework #7 - Sampling ===

	~~<br>~~
	Figure out what happens if your sampled signal, <math> x(t)\!</math>, 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 <math> x(t)\!</math> in terms of <math> x(nT)\!</math>.<br>		Figure out what happens if your sampled signal, <math> x(t)\!</math>, 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 <math> x(t)\!</math> in terms of <math> x(nT)\!</math>.<br>

Max.Woesner: New page: == Max Woesner == Back to my Home Page === Homework #7 - Sampling ===
Figure out what happens if your sampled signal, $x(t)\!$ , has frequency components ...

2009-10-23T01:21:18Z

New page: == Max Woesner == Back to my Home Page === Homework #7 - Sampling === <br> Figure out what happens if your sampled signal, <math> x(t)\!</math>, has frequency components ...

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