| Line 22: | Line 22: | ||
For this signal <math>f_{s} > 2f_{m} = 2(349)Hz = 698Hz</math>, so we will choose a sampling frequency of <math>1000Hz = 1/T_{1}</math> | For this signal <math>f_{s} > 2f_{m} = 2(349)Hz = 698Hz</math>, so we will choose a sampling frequency of <math>1000Hz = 1/T_{1}</math> | ||
| − | \begin{align} | + | |
| − | + | <math> | |
| − | + | \begin{align} | |
| + | x_{1}(n) = x(nT_{1}) = cos(2\pi349nT_{1}) = cos(2\pi349n/1000) \\ | ||
&= frac{1}{2}(e^{-j2\pi349n/1000} + e^{j2\pi349n/1000}) | &= frac{1}{2}(e^{-j2\pi349n/1000} + e^{j2\pi349n/1000}) | ||
Revision as of 18:08, 1 October 2014
DTFT of a Cosine Signal Sampled Above and Below the Nyquist Frequency
A slecture by ECE student Andrew Pawling
Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.
In this slecture we will look at an example that illustrates the Nyquist condition. When a signal is sampled, frequencies above half the sampling rate cannot be properly represented and result in aliasing.
Lets look at a pure tone frequency F4 = 349Hz
We will represent this tone as a cosine signal, $ cos*(2\pi349t) $
For this signal $ f_{s} > 2f_{m} = 2(349)Hz = 698Hz $, so we will choose a sampling frequency of $ 1000Hz = 1/T_{1} $
$ \begin{align} x_{1}(n) = x(nT_{1}) = cos(2\pi349nT_{1}) = cos(2\pi349n/1000) \\ &= frac{1}{2}(e^{-j2\pi349n/1000} + e^{j2\pi349n/1000}) $
