Now, suppose we combine length p and q modules with Good's prime factor algorithm (not using twiddles). The following scaling procedure will work:
Assume the input data has been appropriately loaded into a
$pxq$ data array
Scale the non-DC outputs of the length
$p$ module and apply the modified module to all columns of the data array.
Now all the rows are scaled by
$(p-1)$ except the zeroeth row, corresponding to the DC outputs of the length
$p$ modules. Apply a normal length
$q$ module to the zeroeth row. Modify the length
$q$ module to scale by
$1/(p-1)$ and apply the modified version to all the other rows. The DFT is now complete.
As an example, consider the 3x7 DFT. In the length 3 module scaling the non-DC outputs trades one multiply for one add. When the scaled DFT is constructed, the modified length 3 module is used 7 times. But two rows must be scaled by modified length 7 modules, which brings the total multiply savings to 5 at a cost of 7 adds. This looks like a nice tradeoff. The total number of multiplies in a normal 3x7 PFA is 38.
These ideas can be expanded to multidimensional cases, although it quickly becomes difficult to keep track of which rows and columns need to be counter-scaled.
Length 11 module: 168 adds / 40 mpys
Use the index map
$\overline{x}\left(n\right)=x(<{8}^{n}{>}_{mod11})$ to convert the DFT into a length 10 convolution, plus a correction term for the DC components.
Reduce the length 10 convolution modulo all the irreducible factors of
${z}^{10}-1$
Patch up the DC terms by adding the
$z-1$ reduction result to
$X\left(I\right(1\left)\right)$ and store the result in AMO.
The
${z}^{5}-1$ convolution proceeds in four steps. First, do the irreducible factor reductions, then reduce further with an iterated Toom-Cook procedure, weight all remaining variables, and apply the transpose of the complete reduction stage to the weighted results. The first Toom-Cook reduction uses the factors
$z$ ,
$1/z$ and
$z+1$ on the vectors AM4,AM3 and AM7,AM6 which generates the new vector AM4-AM7,AM3-AM6. Each of the original two vectors is then individually reduced using factors of
$z$ ,
$1/z$ and
$z+1$ , while the new vector is reduced by
$A$ ,
$1/z$ and
$z-1$ . This procedure generates nine variables: AM4,AM3,AM5; AM7,AM6,AM8; S7,S8,AM11. (The expressions for S6 and S8 contain the variables of interest).
The nine variables from 4) are weighted along with T13.
An exact transpose of the reduction algorithm is applied to the weighted variables (and AMO).
The result S16,S15,S18,S17,S19 is the real part of the answer and is mapped back to the output using the map
$\overline{x}\left(n\right)=x(<{8}^{n+1}>mod11$ . This is an unusual map, but it is perfectly acceptable.
A in the length 19 transform the
${z}^{5}+1$ convolution is computed with a variation of the
${z}^{5}-1$ algorithm. First the inputs T6,-T8,-T7,-T1O,T9 are alternately negated, then the
${z}^{5}-1$ algorithm is applied
The second stage of the Toom-Cook reductions uses the factors z, liz and z+l for all three length two vectors. Also, the DC patch is not used here. and the outputs alternately negated.
The result S21,S20,S23,S22,S24, representing the imaginary part of the answer, is mapped back to the output using the map
$\overline{x}\left(n\right)=x(<{8}^{n+1}>mod11)$ .
In both this algorithm and the length 13 DFT plus and minus signs have been freely altered to force all constants to be positive. Also, many shortcut computations were used to save adds, obscuring in some places the logical flow of the algorithm.
All coefficients were computed using the author's QR decomposition linear equation solver and are accurate to at least 14 places.
Length 13 module: 188 adds / 40 mpys
Use the index map
$\overline{x}\left(n\right)=x(<{2}^{n}{>}_{mod13})$ to convert the DFT into a length 12 convolution, plus a correction term for the DC components.
Reduce the length 12 convolution modulo all the irreducible factors of
${z}^{12}-1$
Patch up the DC terms by adding the
$z-1$ reduction result to
$X\left(I\right(1\left)\right)$ and store the result in AMO.
The
${z}^{2}-z+1$ and
${z}^{2}+z+1$ convolutions are reduced using Toom-cook factors of
$z$ ,
$1/z$ and
$z+1$ in one case and
$z$ ,
$1/z$ and
$z-1$ in the other case, and then all the reduced quantities are weighted by constants generating new variables:
from
${z}^{2}-z+1$
The original
$modz+1$ reduction quantity is weighted and passed, along with AMO and the above six variables, to a reconstruction procedure which first combines the
$z-1$ and
${z}^{2}+z+1$ data to compute the convolution mod
${z}^{3}-1$ (CC4,CC5,CC6), and then combines the
$z+1$ and
${z}^{2}-z+1$ data to compute the convolution mod
${z}^{3}+1$ (CC1,CC2,CC3). These two vectors are combined to compute the complete
${z}^{6}-1$ output, which appears in permuted form in CC15 through CC20.
The
${z}^{2}+1$ vector is decomposed with Toom-Cook factors of
$z$ ,
$1/z$ and
$z+1$ yielding A17,A16 and the implicit term (A16+A17).
The
${z}^{4}-{z}^{2}+1$ vector is decomposed with a double iterated Toom-Cook scheme. First the vector is broken into two length two pieces: A27,A26 and A31,A30. Then the vectors are reduced by the factors of
$z$ ,
$1/z$ and
$z+1$ operating on whole vectors to produce a set of three length two vectors:
Ā27,A26A31,A30
A29,A28 = (A27+A31), (A26+A30)These vectors are not calculated in a straightforward manner. Each length two vector is further reduced, in the second iteration, by the factors
$z$ ,
$1/z$ and
$z+1$ to create three new implicit variables
$(A27+A26)$ ,
$(A31+A30)$ and
$(A29+A28)$ .
The nine variables from
[link] and the three variables from
[link] are weighted by constants and the
$mod{z}^{6}+1$ reconstruction proceeds in an ad-hoc fashion which closely resembles a transposed tensor method, but has some differences. The add count for the reconstruction would have been the same if the transposed tensor method had been applied. The
${z}^{6}+1$ result appears in permuted form in variables CC21 through CC26.
The final result is reconstructed from the
${z}^{6}-1$ and
${z}^{6}-1$ vectors. The DC term,
$x\left(i\right(1\left)\right)$ is set equal' to AMO.
All coefficients were computed using the author's QR decomposition linear
equation solver and are accurate to at least 14 places.
Questions & Answers
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,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
Hello
Uday
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
how did you get the value of 2000N.What calculations are needed to arrive at it
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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