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[[Category:ECE438]] | [[Category:ECE438]] | ||
[[Category:signal processing]] | [[Category:signal processing]] | ||
+ | [[Category:Discrete-time_Fourier_transform]] | ||
<center><font size= 5> | <center><font size= 5> | ||
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---- | ---- | ||
== outline == | == outline == | ||
− | <font size = | + | <font size = 3> |
*Introduction | *Introduction | ||
*Sampling rate above Nyquist rate | *Sampling rate above Nyquist rate | ||
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*Conclusion | *Conclusion | ||
*References | *References | ||
− | + | ---- | |
+ | ---- | ||
== Introduction == | == Introduction == | ||
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slecture, I will talk about how does the discrete-time Fourier transform of the sampling of this signal look like. Suppose | slecture, I will talk about how does the discrete-time Fourier transform of the sampling of this signal look like. Suppose | ||
the cosine signal is <math>x(t)=cos(2\pi 440 t)</math>. | the cosine signal is <math>x(t)=cos(2\pi 440 t)</math>. | ||
− | + | ---- | |
+ | ---- | ||
== Sampling rate above Nyquist rate == | == Sampling rate above Nyquist rate == | ||
The Nyquist sampling rate <math>f_s=2f_M=880</math>,so we pick a sample frequency 1000 which is above the Nyquist rate. | The Nyquist sampling rate <math>f_s=2f_M=880</math>,so we pick a sample frequency 1000 which is above the Nyquist rate. | ||
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</math> | </math> | ||
− | + | Since <math>\frac{2\pi 440}{1000}</math> is between <math>-\pi </math> and <math>\pi</math>, so for <math>\omega \isin [-\pi,\pi]</math> | |
<math> \begin{align} \\ | <math> \begin{align} \\ | ||
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</math> | </math> | ||
− | + | For all w, | |
<math>\mathcal{X}_1(\omega) = rep_{2\pi}{\frac{1000}{2}[\delta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)]}</math> | <math>\mathcal{X}_1(\omega) = rep_{2\pi}{\frac{1000}{2}[\delta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)]}</math> | ||
− | The graph of <math>\mathcal{X}_1(\omega)</math> | + | The graph of <math>\mathcal{X}_1(\omega)</math> is |
[[Image:Xd1(w).jpg]] | [[Image:Xd1(w).jpg]] | ||
+ | ---- | ||
+ | ---- | ||
== Sampling rate below Nyquist rate == | == Sampling rate below Nyquist rate == | ||
we pick a sample frequency 800 which is below the Nyquist rate. | we pick a sample frequency 800 which is below the Nyquist rate. | ||
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</math> | </math> | ||
− | + | Since <math>\frac{2\pi*360}{800}</math> is between <math>-\pi</math> and <math>\pi</math>, so for <math>\omega \isin [-\pi,\pi]</math> | |
<math> \begin{align} \\ | <math> \begin{align} \\ | ||
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</math> | </math> | ||
− | + | For all w, | |
<math>\mathcal{X}_2(\omega) = rep_{2\pi}{\frac{800}{2}[\delta(\frac{800}{2\pi}\omega-360)+\delta(\frac{800}{2\pi}\omega+360)]}</math> | <math>\mathcal{X}_2(\omega) = rep_{2\pi}{\frac{800}{2}[\delta(\frac{800}{2\pi}\omega-360)+\delta(\frac{800}{2\pi}\omega+360)]}</math> | ||
− | The graph of <math>\mathcal{X}_2(\omega)</math> | + | The graph of <math>\mathcal{X}_2(\omega)</math> is |
[[Image:Xd2(w).jpg]] | [[Image:Xd2(w).jpg]] | ||
+ | ---- | ||
+ | ---- | ||
== Conclusion == | == Conclusion == | ||
+ | The discrete Fourier transform of a sampled cosine signal with sample frequency f has the following properties: | ||
+ | |||
+ | compared with the CTFT of cosine signal | ||
+ | |||
+ | if ''f'' is above the Nyquist rate | ||
+ | 1. <math>\mathcal{X}(\omega)</math> is a rep with period <math>2\pi</math>. | ||
+ | |||
+ | 2. <math>\mathcal{X}(\omega)</math> re-scales by <math>\frac{2\pi}{f}</math>. | ||
+ | |||
+ | 3. <math>\mathcal{X}(\omega)</math> has amplitude of ''f''. | ||
+ | |||
+ | if ''f'' is below the Nyquist rate | ||
+ | |||
+ | 1. <math>\mathcal{X}(\omega)</math> is a rep with period <math>2\pi</math>. | ||
+ | |||
+ | 2. <math>\mathcal{X}(\omega)</math> re-scales by <math>\frac{2\pi(f-f_0)}{f}</math>. | ||
+ | |||
+ | 3. <math>\mathcal{X}(\omega)</math> has amplitude of ''f''. | ||
+ | |||
+ | ---- | ||
+ | ---- | ||
== References == | == References == | ||
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009 | [1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009 | ||
+ | |||
+ | [2].Steve Eddins,"Discrete-time Fourier transform (DTFT)," Matlab Central December 31, 2009 | ||
---- | ---- | ||
---- | ---- | ||
− | ==[[ | + | </font size> |
+ | ==[[Slecture_DescreteTime_Fourier_transform_ECE438_review|Questions and comments]]== | ||
− | If you have any questions, comments, etc. please post them on [[ | + | If you have any questions, comments, etc. please post them on [[Slecture_DescreteTime_Fourier_transform_ECE438_review|this page]]. |
---- | ---- | ||
− | [[ | + | [[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]] |
Latest revision as of 20:05, 16 March 2015
Discrete-time Fourier transform (DTFT) of a sampled cosine
A slecture by ECE student Yijun Han
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
Contents
outline
- Introduction
- Sampling rate above Nyquist rate
- Sampling rate below Nyquist rate
- Conclusion
- References
Introduction
Consider a CT cosine signal (a pure frequency), and sample that signal with a rate above or below Nyquist rate. In this slecture, I will talk about how does the discrete-time Fourier transform of the sampling of this signal look like. Suppose the cosine signal is $ x(t)=cos(2\pi 440 t) $.
Sampling rate above Nyquist rate
The Nyquist sampling rate $ f_s=2f_M=880 $,so we pick a sample frequency 1000 which is above the Nyquist rate.
$ \begin{align} \\ x_1[n] & =x(\frac{n}{1000}) \\ & =cos(\frac{2\pi440 n}{1000}) \\ & =\frac{1}{2}(e^{\frac{j2\pi440 n}{1000}}+e^{\frac{-j2\pi440 n}{1000}})\\ \end{align} $
Since $ \frac{2\pi 440}{1000} $ is between $ -\pi $ and $ \pi $, so for $ \omega \isin [-\pi,\pi] $
$ \begin{align} \\ \mathcal{X}_1(\omega) & =\frac{1}{2}[2\pi\delta(\omega-2\pi\frac{440}{1000})+2\pi\delta(\omega+2\pi\frac{440}{1000})] \\ & = \frac{1000}{2}[\delta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)] \end{align} $
For all w,
$ \mathcal{X}_1(\omega) = rep_{2\pi}{\frac{1000}{2}[\delta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)]} $
The graph of $ \mathcal{X}_1(\omega) $ is
Sampling rate below Nyquist rate
we pick a sample frequency 800 which is below the Nyquist rate.
$ \begin{align} \\ x_2[n] & =x(\frac{n}{800}) \\ & =cos(\frac{2\pi440 n}{800}) \\ & =cos(\frac{2\pi440 n}{800}-2\pi n) \\ & =cos(2\pi n \frac{-360}{800}) \\ & =cos(\frac{2\pi360 n}{800}) \\ & =\frac{1}{2}(e^{\frac{j2\pi360*n}{800}}+e^{\frac{-j2\pi360*n}{800}})\\ \end{align} $
Since $ \frac{2\pi*360}{800} $ is between $ -\pi $ and $ \pi $, so for $ \omega \isin [-\pi,\pi] $
$ \begin{align} \\ \mathcal{X}_2(\omega) & =\frac{1}{2}[2\pi\delta(\omega-2\pi\frac{360}{800})+2\pi\delta(\omega+2\pi\frac{360}{800})] \\ & = \frac{800}{2}[\delta(\frac{800}{2\pi}\omega-360)+\delta(\frac{800}{2\pi}\omega+360)] \end{align} $
For all w,
$ \mathcal{X}_2(\omega) = rep_{2\pi}{\frac{800}{2}[\delta(\frac{800}{2\pi}\omega-360)+\delta(\frac{800}{2\pi}\omega+360)]} $
The graph of $ \mathcal{X}_2(\omega) $ is
Conclusion
The discrete Fourier transform of a sampled cosine signal with sample frequency f has the following properties:
compared with the CTFT of cosine signal
if f is above the Nyquist rate
1. $ \mathcal{X}(\omega) $ is a rep with period $ 2\pi $.
2. $ \mathcal{X}(\omega) $ re-scales by $ \frac{2\pi}{f} $.
3. $ \mathcal{X}(\omega) $ has amplitude of f.
if f is below the Nyquist rate
1. $ \mathcal{X}(\omega) $ is a rep with period $ 2\pi $.
2. $ \mathcal{X}(\omega) $ re-scales by $ \frac{2\pi(f-f_0)}{f} $.
3. $ \mathcal{X}(\omega) $ has amplitude of f.
References
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
[2].Steve Eddins,"Discrete-time Fourier transform (DTFT)," Matlab Central December 31, 2009
Questions and comments
If you have any questions, comments, etc. please post them on this page.