DTFT of a sampled cosine Yijun ECE438 slecture - Rhea

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

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