

 Large dft modules: 11, 13, 16,

The length 25 module does not follow the traditional Winograd approach. This module is an inline code version of a commonfactor 5x5 DFT. Each length 5 DFT is a primelength convolutional module. The output unscrambling is included in the assignment statements at the end of the program. Some of the length 5 modules used in this program are implemented as scaled versions of conventional length 5 modules in order to save some multiplies by 1/4. The scaling factors are then compensated for by adjusting the twiddle factors. This module has three multiply sections, one for the row DFT's with a data expansion factor of 6/5, one for the twiddle factors (expansion=33/25) and on for the column DFT's (expansion=6/5).
Modules for lengths 11 and 13 are very similar in spirit to the length 19 and 17 modules. Derivations are presented for both the 11 and 13 length modules which are consistent with the listings, although these interpretations may not agree with the original intentions of the designer
[link] they are correct in the sense that the algorithms could have been derived in the stated manner. Both the modules are of prime length and they are implemented in Winograd's convolutional style.
FORTRAN listings for all five modules are included with this report in a subroutine form suitable for use in Burrus' PFA program
[link] . Addition and multiplication counts given are for complex input data.
17 module: 314 adds / 70 mpys
This module closely follows the traditional Winograd primelength approach.
 Use the index map
$\overline{x}\left(n\right)=x(<{3}^{n}{>}_{mod17})$ to convert the DFT into a length 16 convolution, plus a correction term for the DC component.
 Reduce the length 16 convolution modulo all the irreducible factors of
${z}^{16}1$ . (Irreducible over the rationals).
$$\begin{array}{ccc}\hfill mod{z}^{8}+1& :& r108r115\hfill \\ \hfill mod{z}^{8}1& :& r100r107\hfill \end{array}$$
From
${z}^{8}1$ data
$$\begin{array}{ccc}\hfill mod{z}^{4}+1& :& r31r34\hfill \\ \hfill mod{z}^{4}1& :& r200r203\hfill \end{array}$$
From
${z}^{4}1$ data
$$\begin{array}{ccc}\hfill mod{z}^{2}+1& :& r35r36\hfill \\ \hfill mod{z}^{2}1& :& r204r205\hfill \end{array}$$
From
${z}^{2}1$ data
$$\begin{array}{ccc}\hfill modz+1& :& r38\hfill \\ \hfill modz1& :& r37\hfill \end{array}$$
 Reduce the convolution modulo
${z}^{2}+1$ using ToomCook factors of
$z$ ,
$1/z$ and
$z+1$ . This creates variables r35, r36, and r314.
 Reduce the modulo
${z}^{4}+1$ convolution with an iterated ToomCook reduction using the factors
$z$ ,
$1/z$ and
$z1$ for the first step, and the factors
$z$ ,
$1/z$ and
$z+1$ for the second step. The first step produces r310 and r39, and the second step computes r313, r312 and r311. This is exactly the reduction procedure used in Nussbaumer's
${z}^{4}+1$ convolution algorithm.
 Patch up the DC term by adding the
$z1$ reduction result to
$x\left(i\right(1\left)\right)$ .
 Use Nussbaumer's
${z}^{8}+1$ convolution algorithm
[link] on r108r115. This is the only exception to the strict use of transposing the tensor, as his algorithm saves two additions by computing the transposed reconstruction procedure in an obscure fashion. The result, however, is an exact calculation of the transpose. This reduction computes twentyone values, r315r335, which must be weighted by coefficients to produce the reconstructed
${z}^{8}+1$ output, t115t135.
 Weight the variables r31r39, r310r314 by coefficients to produce t11t19, t110t114.
 The reconstruction procedure for the
${z}^{8}1$ terms is a straightforward transpose of the reduction procedure.
 The
${z}^{16}1$ convolution result is reconstructed from the
${z}^{8}1$ (real)
and
${z}^{8}+1$ (imaginary) vectors and mapped back to the outputs using the
reverse of the input map.
 All coefficients were computed using the author's QR decompositionlinear equation solver and are accurate to at least 14 places.
Questions & Answers
find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
sure. what is your question?
ninjadapaul
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X6)^2
so it's 20 divided by X6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
is it a question of log
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
if A not equal to 0 and order of A is n prove that adj (adj A = A
rolling four fair dice and getting an even number an all four dice
Differences Between Laspeyres and Paasche Indices
No. 7x 4y is simplified from 4x + (3y + 3x) 7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)1/7 (x1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is system testing?
AMJAD
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field .
1Electronicsmanufacturad IC ,RAM,MRAM,solar panel etc
2Helth and MedicalNanomedicine,Drug Dilivery for cancer treatment etc
3 Atomobile MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
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
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^15
Prasenjit
can nanotechnology change the direction of the face of the world
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.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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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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