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We call N the size of the problem. The rank of a problem is defined to be the rank of its size (i.e., the dimensionality of the DFT). Similarly, we call V the vector size of the problem, and the vector rank of a problem is correspondingly defined to be the rank of its vector size.Intuitively, the vector size can be interpreted as a set of “loops” wrapped around a single DFT, and we therefore refer to a singleI/O dimension of V as a vector loop . (Alternatively, one can view the problem as describing a DFT over a V -dimensional vector space.) The problem does not specify the order of execution of these loops, however, and thereforeFFTW is free to choose the fastest or most convenient order.

Dft problem examples

A more detailed discussion of the space of problems in FFTW can be found in [link] , but a simple understanding can be gained by examining a few examples demonstrating that the I/O tensorrepresentation is sufficiently general to cover many situations that arise in practice, including some that are not usually considered tobe instances of the DFT.

A single one-dimensional DFT of length n , with stride-1 input X and output Y , as in [link] , is denoted by the problem dft ( ( n , 1 , 1 ) , , X , Y ) (no loops: vector-rank zero).

As a more complicated example, suppose we have an n 1 × n 2 matrix X stored as n 1 consecutive blocks of contiguous length- n 2 rows (this is called row-major format). The in-place DFT of all the rows of this matrix would be denoted by the problem dft ( ( n 2 , 1 , 1 ) , ( n 1 , n 2 , n 2 ) , X , X ) : a length- n 1 loop of size- n 2 contiguous DFTs, where each iteration of the loop offsets its input/output data by a stride n 2 . Conversely, the in-place DFT of all the columns of this matrix would be denoted by dft ( ( n 1 , n 2 , n 2 ) , ( n 2 , 1 , 1 ) , X , X ) : compared to the previous example, N and V are swapped. In the latter case, each DFT operates on discontiguous data, andFFTW might well choose to interchange the loops: instead of performing a loop of DFTs computed individually, the subtransformsthemselves could act on n 2 -component vectors, as described in "The space of plans in FFTW" .

A size-1 DFT is simply a copy Y [ 0 ] = X [ 0 ] , and here this can also be denoted by N = (rank zero, a “zero-dimensional” DFT). This allows FFTW's problems to represent many kinds of copies andpermutations of the data within the same problem framework, which is convenient because these sorts of operations arise frequently inFFT algorithms. For example, to copy n consecutive numbers from I to O , one would use the rank-zero problem dft ( , ( n , 1 , 1 ) , I , O ) . More interestingly, the in-place transpose of an n 1 × n 2 matrix X stored in row-major format, as described above, is denoted by dft ( , ( n 1 , n 2 , 1 ) , ( n 2 , 1 , n 1 ) , X , X ) (rank zero, vector-rank two).

The space of plans in fftw

Here, we describe a subset of the possible plans considered by FFTW; while not exhaustive [link] , this subset is enough to illustrate the basic structure of FFTW and the necessity ofincluding the vector loop(s) in the problem definition to enable several interesting algorithms. The plans that we now describeusually perform some simple “atomic” operation, and it may not be apparent how these operations fit together to actually computeDFTs, or why certain operations are useful at all. We shall discuss those matters in "Discussion" .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Fast fourier transforms. OpenStax CNX. Nov 18, 2012 Download for free at http://cnx.org/content/col10550/1.22
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