3 - 8.6 The Fast Fourier Transform (FFT) 817 C OMPUTER...

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8.6 The Fast Fourier Transform (FFT) 817 COMPUTER EXAMPLE C8.3 Solve Example 8.10 using MATLAB. » 'P_0 4i N_O :.:: 32; '2:" rr_O/N_O; n :;:; (O:N_O-l); :1; » x_n [o~es(1,4) 0.5 zeros(1.23) 0.5 ones(1.3)]'; X_r fft (x_1J.); » H_Y [oCles(1,8) 0.5 zeros(l,lS) 0.5 ones(1,7))'; » Y _r }i_r~*X_Yi y_D = ifft{Y_r); »figure(1); subplot(2,2,1); ste:rr:(n,x_n,'k'); » xl.abel ('n'); y:iabel ('x_n'); axis ([0 31 .1 1.1)); » subplot(2,2,2); ste:rr:(r,real(X_r), 'k'); »xlabel('r'); ylabel('X_r'); axis([O 31 -2 8J); subplot(2,2,3); stem(Cl,real(y_n),'%'); »xlabel('Cl'}; ylabel('y_n'); axis([O 31 .11.1)); »subplot(2,2,4); stem(r,X_I.*H_r,'k'); »xlabel('r'); ylabel('Y_I X_RH_r'); axis([O 31 -2 8=); "," 30 n 0.8 0.6 ~ ~~illJk-;I~ 0 10 20 30 /1 Figure CS.3 ;.<- 8~----------------~ 6 4 2 ollllelll~,_,A'I,.,I11 8--- 6 r ~ 4 ><: II 2 ;..,' 0: -JJ,6 _2L_~_._.-.~~ o ]0. 20 30 r 8.6 THE FAST FOURIER TRANSFORM (FFT) The number of computations required in performing the DFT was dramatically reduced by an algorithm developed by Cooley and Tukey in 1965. 5 This algorithm, known as the fast Fourier transform (FFT), reduces the number of computations from something on the order of N6 to No log No. To compute one sampJe X, from Eq. (8.22a), we require complex multiplications and - 1 complex additions. To compute such values (XI for r = 0, 1 .... , - 1), we require a total of N(~ complex multiplications and No (No - 1) compJex additions. For a large No,
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818 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE these computations can be prohibitively time-consuming, even for a high-speed computer. TIl FFT algorithm is what made the use of Fourier transform accessible for digital signal processi e ng. How DOES FFT REDUCE NUMBER OF COMPUTATIONS? It is easy to understand the magic of FFT. The secret is in the linearity of the Fourier transform and also of the DFT. Because of linearity, we can compute the Fourier transform of a signal x(t) as a sum of the Fourier transforms of segments of of shorter duration. The same principle applies to computation of DFT. Consider a signal of length No = 16 samples. As seen earlier DFT computation of this sequence requires N~ = 256 multiplications and No(N o - 1) =: 240 additions. We can split this sequence in two shorter sequences, each of length 8. To compute DFT of each of these segments, we need 64 multiplications and 56 additions. Thus, we need a total of 128 multiplications and 112 additions. Suppose, we split the original sequence in fOllr segments of length 4 each. To compute the DFT of each segment, we require 16 multiplications
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3 - 8.6 The Fast Fourier Transform (FFT) 817 C OMPUTER...

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