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<font size="4">Downsampling </font>  
 
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We would like to determine what <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> looks like in order to find a way to reconstruct <span class="texhtml">''x''(''t'')</span>.
 
 
Since we have sampled at a rate <span class="texhtml">''f''<sub>''s''</sub> &gt; 2''f''<sub>''M''</sub></span>, the following inequalities hold:
 
 
<math> f_s > 2f_M \iff f_s - f_M > f_M \iff -f_s + f_M < f_M </math>.
 
 
Together, these inequalities, the graph of <span class="texhtml">''X''(''f'')</span>, and the expression for <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> in terms of <span class="texhtml">''X''(''f'')</span> imply that <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> will have the spectrum shown in the figure below.
 
 
[[Image:Xs1.png]]
 
 
Notice that the spectrum of the ideal sampling of a signal is an amplitude scaled periodic repetition of the original spectrum. Since <span class="texhtml">''x''(''t'')</span> is bandlimited and we have sampled at a rate <span class="texhtml">''f''<sub>''s''</sub> &gt; 2''f''<sub>''M''</sub></span>, the periodic repetitions of <span class="texhtml">''X''(''f'')</span> do not overlap.
 
 
All the information needed to reconstruct <span class="texhtml">''X''(''f'')</span> can be found in the portion of <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> that corresponds to <span class="texhtml">''X''(''f'')</span> (shown in red). Therefore we can use a simple lowpass filter with gain <math> \tfrac{1}{f_s} </math> and cutoff frequency <math> \tfrac{f_s}{2} </math> to recover <span class="texhtml">''X''(''f'')</span> from <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span>.
 
 
<br> <math>
 
X(f) = X_s(f)\left\{
 
\begin{array}{ll}
 
\frac{1}{f_s}, & |f| \le \frac{f_s}{2}\\
 
0, & \text{else}
 
\end{array}
 
\right. </math>
 
 
<math> \iff x(t) = x_s(t)*\text{sinc}(f_st) </math>
 
 
<math> \therefore </math> We can perfectly reconstruct <span class="texhtml">''x''(''t'')</span> from <span class="texhtml">''x''<sub>''s''</sub>(''t'')</span>.
 
 
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== Example  ==
 
 
Though the Nyquist theorem states that perfect reconstruction is possible if we satisfy the Nyquist condition <span class="texhtml">(''f''<sub>''s''</sub> &gt; 2''f''<sub>''M''</sub>)</span>, it is important to note that this condition is not necessary. The following example demonstrates how perfect reconstruction is sometimes possible even when undersampling.
 
 
Let the signal <span class="texhtml">''x''(''t'')</span> have a spectrum <span class="texhtml">''X''(''f'')</span> as seen in the figure below.
 
 
[[Image:X2.png]]
 
 
The Nyquist condition states that we should sample at a rate <span class="texhtml">''f''<sub>''s''</sub> &gt; 2(2''a'') = 4''a''</span>. Instead, let us sample at <span class="texhtml">''f''<sub>''s''</sub> = 2''a''</span>.
 
 
As before, we have <math> x_s(t) = x(t)p_{\frac{1}{f_s}}(t) </math> and <math> X_s(f) = f_s\sum_{k = -\infty}^\infty X(f-kf_s) </math>
 
 
<math> \implies X_s(f) = 2a\sum_{k = -\infty}^\infty X(f-2ka) </math>.
 
 
Therefore, <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> will have the spectrum shown in the figure below.
 
 
[[Image:Xs2.png]]
 
 
Notice that there is no aliasing in <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> even though <span class="texhtml">''f''<sub>''s''</sub> &lt; 4''a''</span>. In addition, the portion of <span class="texhtml">''X''<sub>''s''</sub>(''f'')</span> that corresponds to <span class="texhtml">''X''(''f'')</span> (shown in red) can be recovered using a bandpass filter with gain <math> \tfrac{1}{2a} </math> and cutoff frequencies <span class="texhtml">''a'' and 2''a''</span>.
 
  
 
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<font size="3">To summarize, the Nyquist theorem states that any bandlimited signal can be perfectly reconstructed from its sampling if sampled at a rate greater than twice its bandwidth <span class="texhtml">(''f''<sub>''s''</sub> &gt; 2''f''<sub>''M''</sub>)</span>. However, the Nyquist condition is not necessary for perfect reconstruction as shown in the example above. </font>  
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<font size="3">To summarize, when signal is downsampled its&nbsp; . </font>  
  
 
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Revision as of 20:31, 9 October 2014


Downsampling

A slecture by ECE student Yerkebulan Yeshmukhanbetov

Partly based on the ECE438 Fall 2014 lecture material of Prof. Mireille Boutin.








































Outline

  1. Introduction
  2. Definition of Downsampling
  3. Derivation of DTFT of downsampled signal
  4. Example
  5. Conclusion
  6. References

Introduction

This slecture provides definition of downsampling, derives DTFT  downsampled signal and demonstrates it in a frequency domain. Also, it explains process of decimation and why it needs a low-pass filter.


Definition of Downsampling

Downsampling is an operation which involves throwing away samples from discrete-time signal. Let  x[n] be a digital-time signal shown below:

GRAPH

 then y[n] will be produced by downsampling x [n]  by factor D = 3. So, y [n] = x[Dn].

GRAPH

As seen in above graph, y [n] is obtained by throwing away some samples from x [n]. So, y [n] is a downsampled signal from

x [n].


Derivation of DTFT of downsampled signal

Let x (t) be a continuous time signal. Then x1 [n] = x (T1n) and  x2 [n] = x (T2n). And ratio of sampling periods would be

D = T2/T1,   which is an integer greater than 1. From these equations we obtain realtionship between x1 [n] and x2 [n].

$ \begin{align} x_2 [n] = x(T_2 n) = x(DT_1 n) = x_1 [nD] \end{align} $

Below we derive Discrete-Time Fourier Transform of x2 [n] in terms of DTFT of x1 [n].


$ \begin{align} &\mathcal{X}_2(\omega)= \mathcal{F}(x_2 [n]) = \mathcal{F}(x_1 [Dn])\\ &= \sum_{n = -\infty}^\infty x_1[Dn] e^{-j \omega n} = \sum_{m = -\infty}^\infty x_1[m] e^{-j \omega {\frac{m}{D}}}\\ &= \sum_{n = -\infty}^\infty s_D[m]* x_1 [m] e^{-j \omega {\frac{m}{D}}}\\ \end{align} $


where

$ s_D [m]=\left\{ \begin{array}{ll} 1,& \text{ if } n \text{ is a multiple of } 4,\\ 0, & \text{ else}. \end{array}\right. = {\frac{1}{D}} \sum_{k = -\infty}^{D-1} e^{jk {\frac{2 \pi}{D} m}} $


$ \begin{align} &\mathcal{X}_2(\omega)= \sum_{m = -\infty}^\infty {\frac{1}{D}} \sum_{k = -\infty}^{D-1} e^{jk {\frac{2 \pi}{D} m}} x_1[m] e^{-j \omega {\frac{m}{D}}}\\ &= {\frac{1}{D}} \sum_{k = -\infty}^{D-1} \sum_{m = -\infty}^\infty x_1[m] e^{-jm ({\frac{\omega - 2 \pi k}{D}})} = \\ &= {\frac{1}{D}} \sum_{k = -\infty}^{D-1} \mathcal{X}_1 ({\frac{\omega - 2 \pi k}{D}}) \\ \end{align} $


Example


Let's take a look  at  an original signal X1 (w) and  X2 (w) which is obtained after downsampling X1(w) by factor D = 2 in a frequency domain.

Downsamplegraph.jpg

Downsampler is a part of a decimator which also has a low-pass filter to  prevent aliasing.  LPF reduces signal components which has  frequencies higher than cutoff frequency, which can be found from graphs shown above.

                             $ \begin{align} & D 2 \pi T_1 f_{max} < \pi\\ & {\frac{T_2}{T_1}} 2\pi T_1 f_{max} < \pi \\ & 2\pi T_2f_{max} < \pi \\ &f_{max} < {\frac{1}{2T_2}} \end{align} $




Conclusion

To summarize, when signal is downsampled its  .


References

[1] John G. Proakis, Dimitris G. Manolakis, "Digital Signal Processing with Principles, Algorithms, and Applications" 4th Edition,2006




Questions and comments

If you have any questions, comments, etc. please post them on this page.


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