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.

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.

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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