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the cosine signal is <math>x(t)=cos(2pi*440t)</math>. | the cosine signal is <math>x(t)=cos(2pi*440t)</math>. | ||
== Sampling rate above Nyquist rate == | == Sampling rate above Nyquist rate == | ||
− | The Nyquist sampling rate <math>fs=2fM=880 | + | The Nyquist sampling rate <math>fs=2fM=880</math>,so we pick a sample frequency 1000 which is above the Nyquist rate. |
<math>x1[n]=x(n/1000)</math> | <math>x1[n]=x(n/1000)</math> | ||
+ | <math>x1[n]=cos(2pi*440*n/1000)</math> | ||
+ | <math>x1[n]=1/2(exp(j2pi440*n/1000)+exp(-j2pi440*n/1000))</math> | ||
== Sampling rate below Nyquist rate == | == Sampling rate below Nyquist rate == | ||
Revision as of 12:24, 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.
$ x1[n]=x(n/1000) $ $ x1[n]=cos(2pi*440*n/1000) $ $ x1[n]=1/2(exp(j2pi440*n/1000)+exp(-j2pi440*n/1000)) $
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.