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== Introduction == | == 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 | 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(2pi*440t)</math>. | the cosine signal is <math>x(t)=cos(2pi*440t)</math>. | ||
Line 35: | Line 35: | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
+ | |||
since <math>\frac{2\pi*440}{1000}</math> is between -\pi and \pi, so for <math>\omega \isin [-\pi,\pi]</math> | since <math>\frac{2\pi*440}{1000}</math> is between -\pi and \pi, so for <math>\omega \isin [-\pi,\pi]</math> | ||
+ | |||
+ | <math> \begin{align} \\ | ||
+ | \mathcal{X_1}(\omega) & =\frac{1}{2}[2\pi\dleta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)] \\ | ||
+ | & =cos(\frac{2\pi440*n}{1000}) \\ | ||
+ | & =\frac{1}{2}(e^{\frac{j2\pi440*n}{1000}}+e^{\frac{-j2\pi440*n}{1000}})\\ | ||
+ | \end{align} | ||
+ | </math> | ||
== Sampling rate below Nyquist rate == | == Sampling rate below Nyquist rate == | ||
Revision as of 12:53, 27 September 2014
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(2pi*440t) $.
Sampling rate above Nyquist rate
The Nyquist sampling rate $ fs=2fM=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\dleta(\frac{1000}{2\pi}\omega-440)+\delta(\frac{1000}{2\pi}\omega+440)] \\ & =cos(\frac{2\pi440*n}{1000}) \\ & =\frac{1}{2}(e^{\frac{j2\pi440*n}{1000}}+e^{\frac{-j2\pi440*n}{1000}})\\ \end{align} $
Sampling rate below Nyquist rate
Conclusion
References
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
Questions and comments
If you have any questions, comments, etc. please post them on this page.