Line 6: | Line 6: | ||
Diagram of "decimation by two" FFT computing 8-pt DFT. | Diagram of "decimation by two" FFT computing 8-pt DFT. | ||
− | [[Image: | + | [[Image:HW5Q1fig1.jpg]] |
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> | ||
+ | |||
+ | |||
---- | ---- |
Revision as of 16:02, 23 October 2011
Homework 5, ECE438, Fall 2011, Prof. Boutin
Question 1
Diagram of "decimation by two" FFT computing 8-pt DFT.
where $ W_N^k = e^{-j2\pi k/N},\ N=8 $
Question 2
Diagram of "radix-2" FFT computing 8-pt DFT.
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
Back to Homework 5
Back to ECE 438 Fall 2011