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=Homework 5, [[ECE438]], Fall 2011, [[user:mboutin|Prof. Boutin]]=
 
=Homework 5, [[ECE438]], Fall 2011, [[user:mboutin|Prof. Boutin]]=
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==Question 1==
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Diagram of "decimation by two" FFT computing 8-pt DFT.
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[[Image:HW5Q1fig.jpg]]
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where <math>W_N^k = e^{-j2\pi k/N},\ N=8</math>
  
 
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==Question 2==
 
==Question 2==
  
Diagram of 8-pt FFT.
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Diagram of "radix-2" FFT computing 8-pt DFT.
  
 
[[Image:HW5Q2fig1.jpg]]
 
[[Image:HW5Q2fig1.jpg]]

Revision as of 16:01, 23 October 2011

Homework 5, ECE438, Fall 2011, Prof. Boutin


Question 1

Diagram of "decimation by two" FFT computing 8-pt DFT.

File:HW5Q1fig.jpg

where $ W_N^k = e^{-j2\pi k/N},\ N=8 $


Question 2

Diagram of "radix-2" FFT computing 8-pt DFT.

HW5Q2fig1.jpg

Recall the definition of DFT:

$ X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N} $

In this question N=8

If we use summation formula to compute DFT, for each k, we need N times complex multiplications and N-1 times complex additions.

In total, we need N*N=64 times of complex multiplications and N*(N-1)=56 times of complex additions.

In decimation-in-time FFT algorithm, we keep on decimating the number of points by 2 until we get 2 points DFT. At most, we can decimate $ v=log_2 N $ times. As a result, we get v levels of DFT. Except for the first level (2-pt FFT), which only needs N times complex additions, for the rest of levels, we need N/2 times of complex multiplications and N times of complex additions.

In total, we need $ \frac{N}{2}(log_2 N -1)=8 $ times of complex multiplications and $ Nlog_2 N=24 $ times of complex additions.

(Note: when $ N $ is large, $ log_2 N -1 \approx log_2 N $. So the number of multiplications becomes $ \frac{N}{2}log_2 N $.)


Question 3


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