# Large dft modules: 11, 13, 16, 17, 19, and 25  (Page 2/5)

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The length 25 module does not follow the traditional Winograd approach. This module is an in-line code version of a common-factor 5x5 DFT. Each length 5 DFT is a prime-length 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 prime-length approach.

1. Use the index map $\overline{x}\left(n\right)=x\left(<{3}^{n}{>}_{mod17}\right)$ to convert the DFT into a length 16 convolution, plus a correction term for the DC component.
2. Reduce the length 16 convolution modulo all the irreducible factors of ${z}^{16}-1$ . (Irreducible over the rationals).
$\begin{array}{ccc}\hfill mod{z}^{8}+1& :& r108-r115\hfill \\ \hfill mod{z}^{8}-1& :& r100-r107\hfill \end{array}$
From ${z}^{8}-1$ data
$\begin{array}{ccc}\hfill mod{z}^{4}+1& :& r31-r34\hfill \\ \hfill mod{z}^{4}-1& :& r200-r203\hfill \end{array}$
From ${z}^{4}-1$ data
$\begin{array}{ccc}\hfill mod{z}^{2}+1& :& r35-r36\hfill \\ \hfill mod{z}^{2}-1& :& r204-r205\hfill \end{array}$
From ${z}^{2}-1$ data
$\begin{array}{ccc}\hfill modz+1& :& r38\hfill \\ \hfill modz-1& :& r37\hfill \end{array}$
3. Reduce the convolution modulo ${z}^{2}+1$ using Toom-Cook factors of $z$ , $1/z$ and $z+1$ . This creates variables r35, r36, and r314.
4. Reduce the modulo ${z}^{4}+1$ convolution with an iterated Toom-Cook reduction using the factors $z$ , $1/z$ and $z-1$ 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.
5. Patch up the DC term by adding the $z-1$ reduction result to $x\left(i\left(1\right)\right)$ .
6. Use Nussbaumer's ${z}^{8}+1$ convolution algorithm [link] on r108-r115. 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 twenty-one values, r315-r335, which must be weighted by coefficients to produce the reconstructed ${z}^{8}+1$ output, t115-t135.
7. Weight the variables r31-r39, r310-r314 by coefficients to produce t11-t19, t110-t114.
8. The reconstruction procedure for the ${z}^{8}-1$ terms is a straightforward transpose of the reduction procedure.
9. 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.
10. 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
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salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
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ninjadapaul
20/(×-6^2)
Salomon
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I don't understand what the A with approx sign and the boxed x mean
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it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
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I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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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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