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Details of the generation of very efficient length 11, 13, 17, 19, and 25 FFTs using the techniques of Winograd. Originally, Technical Report number 8105 from the EE department of Rice University in 1981, by H. W. Johnson and C. S. Burrus

Introduction

This report describes three large DFT modules (17,19,25) which were developed by the first author, Howard Johnson, in June of 1981, and two previously undocumented modules (11,13) which were originally generated at Stanford in 1978 [link] .

The length 17 and 19 modules were created in the style of Winograd's convolutional DFT programs with strict adherence to three additional module development principles. First, as much code as possible was automatically generated. This included use of FORTRAN programs to generate the input and output mapping statements and the multiplication statements, and heavy use of EDIT commands to copy redundant sections of code. The code for imaginary data manipulation was copied directly from a working listing of code for the real part. All discussion below therefore centers on producing code only for the real part of the input data array. Even the EDIT commands for copying sections of code and substituting variable names were themselves listed in a command file. In this way, the programmer was prevented from introducing occasional typographical errors which are the bane of the DFT module debugger. Errors which did occur tended to be very large and obvious. Test routines were written to test particularly difficult sections of code before they were inserted into the DFT module (such as the modulo z 8 + 1 convolution subsection).

Once the reduction, or PRE-WEAVE, section was written, the reconstruction, or POST-WEAVE, section was arranged to be the transpose of the reduction equations, according to the method of 'transposing the tensor' [link] . Although the problem of minimizing the number of additions in a module is not necessarily solved by transposing the tensor, due to the inordinate difficulty of finding suitable substitutions which would abate the addition count, and the high probability of error involved in making such substitutions, it was decided to use this method. This method also provides a convenient way to check the correctness of the reconstruction procedure by computing the matrices of the reduction and reconstruction subroutines and testing to see that they are indeed a transpose pair.

Intrinsic to the method of transposing the tensor is the fact that the matrix B used to compute the algorithm's multiplication coefficients from the Nth roots of unity is generally more complicated than either the reduction matrix or its transpose, the reconstruction matrix. This result is a consequence of B having been generated from Toom-Cook polynomial reconstruction procedures and also CRT polynomial reconstructions, which are both known to be more complicated than their associated reduction procedures. The problem of finding B in order to compute a set of multipliers may be neatly circumvented by directly solving a set of linear equations to find a coefficient vector which makes the algorithm work. The details of this trick are not reported here, but may be found in [link] . Suffice to say that given working FORTRAN subroutines for the reduction and reconstruction procedures, a FORTRAN program exists which will solve for the correct coefficients.

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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cm
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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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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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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?
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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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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, Large dft modules: 11, 13, 16, 17, 19, and 25. revised ece technical report 8105. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10569/1.7
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