# 0.6 Winograd's short dft algorithms

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The operation of discrete convolution defined by

$y\left(n\right)=\sum _{k}h\left(n-k\right)\phantom{\rule{4pt}{0ex}}x\left(k\right)$

is called a bilinear operation because, for a fixed $h\left(n\right)$ , $y\left(n\right)$ is a linear function of $x\left(n\right)$ and for a fixed $x\left(n\right)$ it is a linear function of $h\left(n\right)$ . The operation of cyclic convolution is the same but with all indices evaluated modulo $N$ .

Recall from Polynomial Description of Signals: Equation 3 that length-N cyclic convolution of $x\left(n\right)$ and $h\left(n\right)$ can be represented by polynomial multiplication

$Y\left(s\right)=X\left(s\right)\phantom{\rule{4pt}{0ex}}H\left(s\right)\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\text{mod}\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}\left({s}^{N}-1\right)$

This bilinear operation of [link] and [link] can also be expressed in terms of linear matrix operators and a simpler bilinearoperator denoted by $o$ which may be only a simple element-by-element multiplication of the two vectors [link] , [link] , [link] . This matrix formulation is

$Y=C\left[AXoBH\right]$

where $X$ , $H$ and $Y$ are length-N vectors with elements of $x\left(n\right)$ , $h\left(n\right)$ and $y\left(n\right)$ respectively. The matrices $A$ and $B$ have dimension $M$ x $N$ , and $C$ is $N$ x $M$ with $M\ge N$ . The elements of $A$ , $B$ , and $C$ are constrained to be simple; typically small integers or rational numbers. It will be thesematrix operators that do the equivalent of the residue reduction on the polynomials in [link] .

In order to derive a useful algorithm of the form [link] to calculate [link] , consider the polynomial formulation [link] again. To use the residue reduction scheme, the modulus is factored into relatively prime factors. Fortunately the factoringof this particular polynomial, ${s}^{N}-1$ , has been extensively studied and it has considerable structure. When factored over the rationals,which means that the only coefficients allowed are rational numbers, the factors are called cyclotomic polynomials [link] , [link] , [link] . The most interesting property for our purposes is that most of the coefficients of cyclotomic polynomialsare zero and the others are plus or minus unity for degrees up to over one hundred. This means the residue reduction will generallyrequire no multiplications.

The operations of reducing $X\left(s\right)$ and $H\left(s\right)$ in [link] are carried out by the matrices $A$ and $B$ in [link] . The convolution of the residue polynomials is carried out by the $o$ operator and the recombination by the CRT is done by the $C$ matrix. More details are in [link] , [link] , [link] , [link] , [link] but the important fact is the $A$ and $B$ matrices usually contain only zero and plus or minus unity entries and the $C$ matrix only contains rational numbers. The only general multiplications are those represented by $o$ . Indeed, in the theoretical results from computational complexity theory,these real or complex multiplications are usually the only ones counted. In practical algorithms, the rational multiplicationsrepresented by $C$ could be a limiting factor.

what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
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Rafiq
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LITNING
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write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
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what king of growth are you checking .?
Renato
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Kyle
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why?
what school?
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biomolecules are e building blocks of every organics and inorganic materials.
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research.net
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sciencedirect big data base
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Introduction about quantum dots in nanotechnology
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absolutely yes
Daniel
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