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[[Category:ECE438Fall2014Boutin]]
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[[Category:ECE438]]
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[[Category:signal processing]] 
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[[Category:Discrete-time_Fourier_transform]]
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<center>
 
<font size="4">DTFT of a Cosine Signal Sampled Above and Below the Nyquist Frequency </font>  
 
<font size="4">DTFT of a Cosine Signal Sampled Above and Below the Nyquist Frequency </font>  
  
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----
 
----
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 +
 +
<font size="4">Outline</font>
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<font size="3">
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<ul>
 +
<li> Introduction
 +
<li> Sampling Above Nyquist Frequency </li>
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<li> Sampling Below Nyquist Frequency </li>
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<li> Conclusion </li>
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</ul>
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</font>
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 +
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----
 +
 +
 +
<font size="4">Introduction</font>
 +
  
 
<font size = '3'>
 
<font size = '3'>
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----
 
  
 
Lets look at a pure tone frequency F4 = 349Hz  
 
Lets look at a pure tone frequency F4 = 349Hz  
  
We will represent this tone as a cosine signal, <math>cos*(2\pi349t)</math>  
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We will represent this tone as a cosine signal, <math>cos(2\pi349t)</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>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})
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<font size="4">Sampling <u>Above</u> Nyquist Frequency</font>
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 +
For this signal <math>f_{s} > 2f_{m} = 2(349)Hz = 698Hz</math> or else aliasing will occur. We will choose a sampling frequency of <math>f_{s} = 1/T_{1} = 1000Hz</math>.
 +
 +
 
 +
<math>
 +
\begin{align}
 +
x_{1}(n) &= x(nT_{1}) \\
 +
 
 +
&= cos(2\pi349nT_{1}) \\
 +
 
 +
&= cos(\frac{2\pi349n}{1000}) \\
 +
 
 +
&= \frac{1}{2}(e^{\frac{-j2\pi349n}{1000}} + e^{\frac{j2\pi349n}{1000}}) \\
 +
 
 +
\\
 +
\\
 +
\end{align}
 +
</math>
 +
 
 +
<math>Note\ that:\ 0 < |\pm2\pi\frac{349}{1000}| < \pi </math>
 +
 
 +
This means the original signal can be properly represented when sampled at <math>f_{s} = 1000Hz.</math>
 +
 
 +
Using the discrete-time Fourier transform pair for cosine:
 +
 
 +
<math>x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) </math>
 +
 
 +
<math>cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) </math>
 +
 
 +
<math>
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\begin{align}
 +
X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{1000} +2\pi k) + \delta(\omega + 2\pi\frac{349}{1000} + 2\pi k)) \\
 +
 
 +
&= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{1000}) + \delta(\omega + 2\pi\frac{349}{1000})) \\
 +
 
 +
 
 +
&= rep_{2\pi}\frac{1000}{2} ((\delta(\frac{1000}{2\pi}\omega - 349) + \delta(\frac{1000}{2\pi}\omega + 349)) \\
 +
\end{align}
 +
</math>
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 +
Plot of <math>X_{1}(\omega)</math>:
 +
 
 +
https://www.projectrhea.org/rhea/images/4/4f/X1wplot.png
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Repeats every <math>2\pi</math>
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 +
The frequency content of the original signal lies within the -<math>\pi</math> to <math>\pi </math> band. Therefore the signal can be properly reconstructed.
 +
 
 +
 
 +
 
 +
----
 +
 
 +
 
 +
<font size="4">Sampling <u>Below</u> Nyquist Frequency</font>
 +
 +
For this signal <math>f_{s} > 2f_{m} = 2(349)Hz = 698Hz</math> or else aliasing will occur. We will choose a sampling frequency that does not satisfy this condition this time. Let's use <math> f_{s} = 1/T_{2} = 500Hz</math>.
 +
 +
 
 +
<math>
 +
\begin{align}
 +
x_{2}(n) &= x(nT_{2}) \\
 +
 
 +
&= cos(2\pi349nT_{2}) \\
 +
 
 +
&= cos(\frac{2\pi349n}{500}) \\
 +
 
 +
&= \frac{1}{2}(e^{\frac{-j2\pi349n}{500}} + e^{\frac{j2\pi349n}{500}}) \\
 +
 
 +
\\
 +
\\
 +
\end{align}
 +
</math>
 +
 
 +
<math>Note\ that:\ \pi < |\pm2\pi\frac{349}{500}| < 2\pi </math>
 +
 
 +
This means the original signal <u>cannot</u> be properly represented when sampled at <math>f_{s} = 500Hz.</math>
 +
 
 +
Using the discrete-time Fourier transform pair for cosine again:
 +
 
 +
<math>x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) </math>
 +
 
 +
<math>cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) </math>
 +
 
 +
<math>
 +
\begin{align}
 +
X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{500} +2\pi k) + \delta(\omega + 2\pi\frac{349}{500} + 2\pi k)) \\
 +
 
 +
&= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{500}) + \delta(\omega + 2\pi\frac{349}{500})) \\
 +
 
 +
&= rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 349) + \delta(\frac{500}{2\pi}\omega + 349)) \\
 +
\end{align}
 +
</math>
 +
 
 +
Use periodicity property of cosine to shift from  -<math>2\pi</math> to <math>2\pi </math> to -<math>\pi</math> to <math>\pi </math> band.
 +
 
 +
<math>X_{1}(\omega) = rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 151) + \delta(\frac{500}{2\pi}\omega + 151)) </math>
 +
 
 +
Plot of <math>X_{2}(\omega)</math>:
 +
 
 +
https://www.projectrhea.org/rhea/images/0/00/X2wplot.png
 +
 
 +
As you can see the original signal frequency exists outside of the -<math>\pi</math> to <math>\pi </math> band.
 +
The frequencies that exists within this band are copies of the original signal that are 'rep'ed. Therefore these frequencies do not represent the frequency content of the the original signal, resulting in aliasing.
 +
 
 +
 
 +
----
 +
 
 +
<font size ="4">
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Conclusion
 +
 
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<font size="3">
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When sampling above the Nyquist frequency the signal can be accurately represented but when sampling below the signal will become aliased.
 +
 
 +
 
 +
 
 +
[[DTFTCosinePawlingQuestions| Questions and Comments]]
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----
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[[2014_Fall_ECE_438_Boutin_digital_signal_processing_slectures|Back to ECE438 slectures, Fall 2014]]

Latest revision as of 20:04, 16 March 2015


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.




Outline

  • Introduction
  • Sampling Above Nyquist Frequency
  • Sampling Below Nyquist Frequency
  • Conclusion




Introduction


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) $




Sampling Above Nyquist Frequency

For this signal $ f_{s} > 2f_{m} = 2(349)Hz = 698Hz $ or else aliasing will occur. We will choose a sampling frequency of $ f_{s} = 1/T_{1} = 1000Hz $.


$ \begin{align} x_{1}(n) &= x(nT_{1}) \\ &= cos(2\pi349nT_{1}) \\ &= cos(\frac{2\pi349n}{1000}) \\ &= \frac{1}{2}(e^{\frac{-j2\pi349n}{1000}} + e^{\frac{j2\pi349n}{1000}}) \\ \\ \\ \end{align} $

$ Note\ that:\ 0 < |\pm2\pi\frac{349}{1000}| < \pi $

This means the original signal can be properly represented when sampled at $ f_{s} = 1000Hz. $

Using the discrete-time Fourier transform pair for cosine:

$ x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) $

$ cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) $

$ \begin{align} X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{1000} +2\pi k) + \delta(\omega + 2\pi\frac{349}{1000} + 2\pi k)) \\ &= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{1000}) + \delta(\omega + 2\pi\frac{349}{1000})) \\ &= rep_{2\pi}\frac{1000}{2} ((\delta(\frac{1000}{2\pi}\omega - 349) + \delta(\frac{1000}{2\pi}\omega + 349)) \\ \end{align} $

Plot of $ X_{1}(\omega) $:

X1wplot.png Repeats every $ 2\pi $

The frequency content of the original signal lies within the -$ \pi $ to $ \pi $ band. Therefore the signal can be properly reconstructed.




Sampling Below Nyquist Frequency

For this signal $ f_{s} > 2f_{m} = 2(349)Hz = 698Hz $ or else aliasing will occur. We will choose a sampling frequency that does not satisfy this condition this time. Let's use $ f_{s} = 1/T_{2} = 500Hz $.


$ \begin{align} x_{2}(n) &= x(nT_{2}) \\ &= cos(2\pi349nT_{2}) \\ &= cos(\frac{2\pi349n}{500}) \\ &= \frac{1}{2}(e^{\frac{-j2\pi349n}{500}} + e^{\frac{j2\pi349n}{500}}) \\ \\ \\ \end{align} $

$ Note\ that:\ \pi < |\pm2\pi\frac{349}{500}| < 2\pi $

This means the original signal cannot be properly represented when sampled at $ f_{s} = 500Hz. $

Using the discrete-time Fourier transform pair for cosine again:

$ x[n] \ \ \ \ \ \ \ \ \ \ -----> \ \ \ \ X(\omega) $

$ cos(\omega_{0}n) \ \ \ \ -----> \ \ \ \ \pi\sum_{k=-\infty}^\infty (\delta(\omega -\omega_{0}+2\pi k) + \delta(\omega + \omega_{0}+2\pi k)) $

$ \begin{align} X_{1}(\omega) &= \pi \sum_{k=-\infty}^\infty(\delta(\omega - 2\pi\frac{349}{500} +2\pi k) + \delta(\omega + 2\pi\frac{349}{500} + 2\pi k)) \\ &= rep_{2\pi}\pi (\delta(\omega - 2\pi\frac{349}{500}) + \delta(\omega + 2\pi\frac{349}{500})) \\ &= rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 349) + \delta(\frac{500}{2\pi}\omega + 349)) \\ \end{align} $

Use periodicity property of cosine to shift from -$ 2\pi $ to $ 2\pi $ to -$ \pi $ to $ \pi $ band.

$ X_{1}(\omega) = rep_{2\pi}\frac{500}{2} ((\delta(\frac{500}{2\pi}\omega - 151) + \delta(\frac{500}{2\pi}\omega + 151)) $

Plot of $ X_{2}(\omega) $:

X2wplot.png

As you can see the original signal frequency exists outside of the -$ \pi $ to $ \pi $ band. The frequencies that exists within this band are copies of the original signal that are 'rep'ed. Therefore these frequencies do not represent the frequency content of the the original signal, resulting in aliasing.



Conclusion

When sampling above the Nyquist frequency the signal can be accurately represented but when sampling below the signal will become aliased.


Questions and Comments


Back to ECE438 slectures, Fall 2014

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