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where <math>W_N^k = e^{-j2\pi k/N},\ N=8</math>
 
where <math>W_N^k = e^{-j2\pi k/N},\ N=8</math>
  
 +
Recall the definition of DFT:
 +
 +
<math>X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N},\ k=0,...,N-1</math>.
 +
 +
For each k, we need N times complex multiplications and N-1 times complex additions.
 +
 +
In total, we need <math>N^2</math> times of complex multiplications and <math>N^2-N</math> times of complex additions.
 +
 +
In this algorithm, the DFT is computed in two steps. For the first step, two N/2-pt DFT are computed with <math>2\cdot (\frac{N}{2})^2</math> multiplications and <math>2((\frac{N}{2})^2-\frac{N}{2})</math>. For the second step, <math>\frac{N}{2}</math> multiplications and <math>N</math> additions are needed.
 +
 +
When <math>N=8</math>, the total numbers of complex operations are
 +
 +
multiplications: <math>2\cdot (\frac{N}{2})^2 + \frac{N}{2} = 36</math>
  
 +
additions: <math>2((\frac{N}{2})^2-\frac{N}{2}) + N = 32</math>
  
 
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Recall the definition of DFT:  
 
Recall the definition of DFT:  
  
<math>X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N}</math>
+
<math>X[k]=\sum_{n=0}^{N-1} x[n]e^{-j2\pi k/N},\ k=0,...,N-1</math>
  
 
In this question N=8
 
In this question N=8

Revision as of 16:17, 23 October 2011

Homework 5, ECE438, Fall 2011, Prof. Boutin


Question 1

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

HW5Q1fig1.jpg

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

Recall the definition of DFT:

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

For each k, we need N times complex multiplications and N-1 times complex additions.

In total, we need $ N^2 $ times of complex multiplications and $ N^2-N $ times of complex additions.

In this algorithm, the DFT is computed in two steps. For the first step, two N/2-pt DFT are computed with $ 2\cdot (\frac{N}{2})^2 $ multiplications and $ 2((\frac{N}{2})^2-\frac{N}{2}) $. For the second step, $ \frac{N}{2} $ multiplications and $ N $ additions are needed.

When $ N=8 $, the total numbers of complex operations are

multiplications: $ 2\cdot (\frac{N}{2})^2 + \frac{N}{2} = 36 $

additions: $ 2((\frac{N}{2})^2-\frac{N}{2}) + N = 32 $


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},\ k=0,...,N-1 $

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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