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- Large dft modules: 11, 13, 16,
- Large dft modules: 11, 13, 16,
- Large dft modules: 11, 13, 16,
Length 19 module: 372 adds / 76 mpys
This module closely follows the traditional Winograd prime-length approach.
- Use the index map
to convert the DFT into a length 18 convolution plus a correction term for the DC component.
- Reduce the length 16 convolution modulo
and
.
- Use Nussbaumer's
convolution algorithm on r100-r108. This is a transposed tensor method, however it again uses an obscure reconstruction procedure. This algorithm computes nineteen intermediate quantities, r31-r319, which are then weighted against nineteen coefficients to produce t11-t119. This data is then partially reconstructed to yield the final result of the
convolution, t32-t310.
- In the course of the
convolution algorithm the
data is reduced modulo
and stored in r31. This quantity is added to
to patch up the DC term.
- An algebraic trick is used to compute the
convolution using the
algorithm. Suppose there exists a ring homomorphism H which maps elements of the ring of real polynomials modulo
into the ring of polynomials modulo
. Then
could be used on the
data, the resulting polynomial could be convolved in the modulo
domain using the existing procedure, and the output of that procedure could be mapped back through
into the modulo
domain. Such a homomorphism does exist, and moreover it happens to be its own inverse.
where
is a polynomial (in either
or
) may be formed from
by negating the sign on all odd-numbered coefficients, that is,
. The alternate negation of data values going into and coming out of the
convolution algorithm is accomplished without an increase in computing time by appropriate placement of negative signs. The nineteen intermediate values formed are r320-r338 which are then weighted by the (purely imaginary) coefficients to produce t120-t138. A partial reconstruction yields the
convolution result, t311-t319.
- The
convolution result is reconstructed from the
(real)
and
(imaginary) vectors and mapped back to the outputs using the
reverse of the input map.
- All coefficients were computed using the author's QR decomposition
linear equation solver and are accurate to at least 14 places.
Length 25 module: 420 adds / 132 mpys
This module is a common factor type module which uses length 5 convolutional DFT submodules. The length 5 submodules are implemented in a transposed tensor configuration using an index map
followed by a reduction modulo all the irreducible factors of
. The
convolution is implemented using Toom-Cook factors of
,
and
. The reconstruction matrix is exactly the transpose of the reduction procedure. The coefficients for the length 5 submodules were found using the author's QR procedure, and the twiddle factors were generated in a special FORTRAN program. The details of saving multiplies by scaling some of the prime length submodules in a common factor algorithm are discussed below in
[link] . This length 25 module has a total of 132 multiplies and 420 adds. Using Winograd's decomposition of the length 25 OFT into two length 5 DFT's and a length 20 convolution the best operation count generated by this author was 108 multiplies and 604 adds.
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?
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
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
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
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.
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
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?
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
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what are the types of wave
Maurice
fine, how about you?
Mohammed
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?
Who can show me the full solution in this problem?
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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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