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\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)} $
Sampling rate below Nyquist rate
Conclusion
References
[1].Mireille Boutin, "ECE438 Digital Signal Processing with Applications," Purdue University August 26,2009
Questions and comments
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