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Now for the fun. Because N 2 L , each of the half-length transforms can be reduced to two quarter-length transforms, each of these to twoeighth-length ones, etc. This decomposition continues until we are left with length-2 transforms. This transform is quitesimple, involving only additions. Thus, the first stage of the FFT has N 2 length-2 transforms (see the bottom part of [link] ). Pairs of these transforms are combined by adding one to the other multiplied by a complexexponential. Each pair requires 4 additions and 2 multiplications, giving a total number of computations equaling 6 · N 4 3 N 2 . This number of computations does not change from stage to stage.Because the number of stages, the number of times the length can be divided by two, equals 2 logbase --> N , the number of arithmetic operations equals 3 N 2 2 logbase --> N , which makes the complexity of the FFT O N 2 logbase --> N .

Length-8 dft decomposition

The initial decomposition of a length-8 DFT into the terms using even- and odd-indexed inputs marks the first phase ofdeveloping the FFT algorithm. When these half-length transforms are successively decomposed, we are left with the diagram shownin the bottom panel that depicts the length-8 FFT computation.

Doing an example will make computational savings more obvious. Let's look at the detailsof a length-8 DFT. As shown on [link] , we first decompose the DFT into two length-4 DFTs, with the outputs added and subtracted together in pairs.Considering [link] as the frequency index goes from 0 through 7, we recycle values fromthe length-4 DFTs into the final calculation because of the periodicity of the DFT output. Examining how pairs of outputsare collected together, we create the basic computational element known as a butterfly ( [link] ).


The basic computational element of the fast Fourier transform is the butterfly. It takes two complex numbers, representedby a and b , and forms the quantities shown. Each butterfly requires onecomplex multiplication and two complex additions.
By considering together the computations involving common output frequencies from the two half-length DFTs, we see that the twocomplex multiplies are related to each other, and we can reduce our computational work even further. By further decomposing thelength-4 DFTs into two length-2 DFTs and combining their outputs, we arrive at the diagram summarizing the length-8 fastFourier transform ( [link] ). Although most of the complex multiplies are quite simple(multiplying by 2 means swapping real and imaginary parts and changing their signs), let's count those forpurposes of evaluating the complexity as full complex multiplies. We have N 2 4 complex multiplies and N 8 complex additions for each stage and 2 logbase --> N 3 stages, making the number of basic computations 3 N 2 2 logbase --> N as predicted.

Note that the ordering of the input sequence in the two parts of [link] aren't quite the same. Why not? How is the ordering determined?

The upper panel has not used the FFT algorithm to compute the length-4 DFTs while the lower one has. The ordering isdetermined by the algorithm.

Other "fast" algorithms were discovered, all of which make use of how many common factors the transformlength N has. In number theory, the number of prime factors a given integer has measures how composite it is. The numbers 16 and 81 are highly composite (equaling 2 4 and 3 4 respectively), the number 18 is less so ( 2 1 · 3 2 ), and 17 not at all (it's prime). In over thirty years of Fourier transform algorithm development, the originalCooley-Tukey algorithm is far and away the most frequently used. It is so computationally efficient that power-of-twotransform lengths are frequently used regardless of what the actual length of the data.

Suppose the length of the signal were 500 ? How would you compute the spectrum of this signal using the Cooley-Tukeyalgorithm? What would the length N of the transform be?

The transform can have any greater than or equal to the actual duration of the signal. We simply“pad” the signal with zero-valued samples until a computationally advantageous signal length results. Recallthat the FFT is an algorithm to compute the DFT . Extending the length of the signal this way merely means weare sampling the frequency axis more finely than required. To use the Cooley-Tukey algorithm, the length of theresulting zero-padded signal can be 512, 1024, etc. samples long.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
on number 2 question How did you got 2x +2
combine like terms. x + x + 2 is same as 2x + 2
Q2 x+(x+2)+(x+4)=60 3x+6=60 3x+6-6=60-6 3x=54 3x/3=54/3 x=18 :. The numbers are 18,20 and 22
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Need help solving this problem (2/7)^-2
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A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Discrete-time fourier analysis. OpenStax CNX. Sep 20, 2008 Download for free at http://cnx.org/content/col10579/1.1
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