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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

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
I'm not sure why it wrote it the other way
I got X =-6
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
Idrissa Reply
im all ears I need to learn
right! what he said ⤴⤴⤴
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
I'm not good at math so would you help me
what is the problem that i will help you to self with?
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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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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