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(a) Show that the Npoint DFT of x ( n ) has only odd harmonics. that is.
X(k) =0 k even
(b) Show how to find the Npoint DFT of x ( n ) by finding the N/2point DFT of an appropriately modified sequence. (a) The Npoint DFT of x(n) is
Because x(n) = x(n
+ N/2), if k is even, each term in the sum is zero, and X(k) = 0 fork = 0 , 2 , 4 , . . .. (b) In the first stage of a decimationinfrequency FFT algorithm, we separately evaluate the evenindex and oddindex san~ples X(k). If X(k) has only odd harmonics, the even samples are zero, and we need only evaluate of the odd samples. From Eq. (7.4) we see that the odd samples are given by With x ( n ) = x(n
+ N/2) this becomes
which is the N/2point DFT of the sequence y(n) = 2WE;x(n). Therefore, to find the Npoint DFT of x(n), we multiply the first N/2 points of x(n) by 2W;. and then compute the N/2point DFT of y(nh The N/2point DFT of x(n) is then given by
THE FAST FOURIER TRANSFORM
[CHAP. 7
FFT Algorithms for Composite N
When the number of points in the D F T is a power of 4, w e can use a radix2 FFT algorithm. However, when N = 4", it is more efficient to use a radix4 FFT algorithm. (a) Derive the r a d i x 4 decimationintime FFT algorithm when N = 4". (b) Draw the structure for the butterfly in the radix4 FFT, and compare the number of complex multiplies and adds with a radix4 F F T to a radix2 FFT. (a) To derive a decimationintime radix4 FFT. let NI = N/4 and N2 = 4. and define the index maps
We then express X ( k ) using the decomposition given in Eq. (7.7) with NI = N/4 and N2 = 4, The inner summation.
is the N/4point DFT of the sequence x(4nI
+ n2),and the outer summation is a 4point DFT, Since W4 = j, these 4point transforms have the form
for kl = 0. 1 , 2 . 3 , and n2 = 0, I , . . . (N/4) 1. If N2 = N/4 is divisible by 4, then the process is repeated. In this way, we generate v = Iog, N stages with N/4 butterflies in each stage. (b) The 4point butterflies in the radix4 FFT perform operations of the form
CHAP. 71 With
THE FAST FOURIER TRANSFORM
Since multiplications by ij only requires interchanging real and imaginary parts and possibly changing a sign bit, then each 4point butterfly only requires 3 complex multiplications. With v = log, N stages, and N/4 butterflies per stage, the number of complex multiplies for a DFT of length N = 4" is N 3N 3 .  log, N =  log, N 4 8 For a radix:! decimationintime FFT, on the other hand, the number of multiplications is N log, N 2 Therefore, the number of multiplications in a radix4 FFT is
& times the number in a radix2 FFT.
Suppose that we would like to find the Npoint DFT of a sequence where N is a power of 3, N = 3". ( a ) Develop a radix3 decimationintime FFT algorithm, and draw the corresponding flowgraph for N =9. ( h ) How many multiplications are required for a radix3 FFT (c) Can the computations be performed in place (a) A radix3 decimationintime FFT may be derived in exactly the same way as a radix2 FFT. First, x ( n ) is decimated by a factor of 3 to form three sequences of length Nj3: THE FAST FOURIER TRANSFORM Expressing the Npoint DFT in terms of these sequences, we have
[CHAP. 7
Since w$' = W k 3 ,then
Note that the first term is the N/3point DFT o f f (n), the second is W i times the N/3point DFT of g(n), and the third is w;' times the N 13point DFT of h(n), We may continue decimating by factors of 3 until we are left with only 3point DFTs. The flowgraph for a 9point decimationintime FFT is shown in Fig. 7 1 I . Only one of the 3point butterflies is shown in the second stage in order to allow for the labeling of the branches. The complete flowgraph is formed by replicating this 3point butterfly up by one node, and down by one node, and changing the branch multiplies to their appropriate values. 0 X(8) W" J Fig. 711. Flowgraph for a9point decimationintime FFT (only one butterfly in the second stage is shown).

