

 Large dft modules: 11, 13, 16,

A very efficient length N = 16 FFT module that can be use alone or with the PFA or the WFTA.
N=16 fft module
A FORTRAN implementation of a length16 FFT module to be
used in a Prime Factor Algorithm program.
C
CWFTA N=16C
116 R1 = X(I(1)) + X(I(9))R2 = X(I(1))  X(I(9))
S1 = Y(I(1)) + Y(I(9))S2 = Y(I(1))  Y(I(9))
R3 = X(I(2)) + X(I(10))R4 = X(I(2))  X(I(10))
S3 = Y(I(2)) + Y(I(10))S4 = Y(I(2))  Y(I(10))
R5 = X(I(3)) + X(I(11))R6 = X(I(3))  X(I(11))
S5 = Y(I(3)) + Y(I(11))S6 = Y(I(3))  Y(I(11))
R7 = X(I(4)) + X(I(12))R8 = X(I(4))  X(I(12))
S7 = Y(I(4)) + Y(I(12))S8 = Y(I(4))  Y(I(12))
R9 = X(I(5)) + X(I(13))R10= X(I(5))  X(I(13))
S9 = Y(I(5)) + Y(I(13))S10= Y(I(5))  Y(I(13))
R11 = X(I(6)) + X(I(14))R12 = X(I(6))  X(I(14))
S11 = Y(I(6)) + Y(I(14))S12 = Y(I(6))  Y(I(14))
R13 = X(I(7)) + X(I(15))R14 = X(I(7))  X(I(15))
S13 = Y(I(7)) + Y(I(15))S14 = Y(I(7))  Y(I(15))
R15 = X(I(8)) + X(I(16))R16 = X(I(8))  X(I(16))
S15 = Y(I(8)) + Y(I(16))S16 = Y(I(8))  Y(I(16))
T1 = R1 + R9T2 = R1  R9
U1 = S1 + S9U2 = S1  S9
T3 = R3 + R11T4 = R3  R11
U3 = S3 + S11U4 = S3  S11
T5 = R5 + R13T6 = R5  R13
U5 = S5 + S13U6 = S5  S13
T7 = R7 + R15T8 = R7  R15
U7 = S7 + S15U8 = S7  S15
T9 = C81 * (T4 + T8)T10= C81 * (T4  T8)
U9 = C81 * (U4 + U8)U10= C81 * (U4  U8)
R1 = T1 + T5R3 = T1  T5
S1 = U1 + U5S3 = U1  U5
R5 = T3 + T7R7 = T3  T7
S5 = U3 + U7S7 = U3  U7
R9 = T2 + T10R11= T2  T10
S9 = U2 + U10S11= U2  U10
R13 = T6 + T9R15 = T6  T9
S13 = U6 + U9S15 = U6  U9
T1 = R4 + R16T2 = R4  R16
U1 = S4 + S16U2 = S4  S16
T3 = C81 * (R6 + R14)T4 = C81 * (R6  R14)
U3 = C81 * (S6 + S14)U4 = C81 * (S6  S14)
T5 = R8 + R12T6 = R8  R12
U5 = S8 + S12U6 = S8  S12
T7 = C162 * (T2  T6)T8 = C163 * T2  T7
T9 = C164 * T6  T7T10 = R2 + T4
T11 = R2  T4R2 = T10 + T8
R4 = T10  T8R6 = T11 + T9
R8 = T11  T9U7 = C162 * (U2  U6)
U8 = C163 * U2  U7U9 = C164 * U6  U7
U10 = S2 + U4U11 = S2  U4
S2 = U10 + U8S4 = U10  U8
S6 = U11 + U9S8 = U11  U9
T7 = C165 * (T1 + T5)T8 = T7  C164 * T1
T9 = T7  C163 * T5T10 = R10 + T3
T11 = R10  T3R10 = T10 + T8
R12 = T10  T8R14 = T11 + T9
R16 = T11  T9U7 = C165 * (U1 + U5)
U8 = U7  C164 * U1U9 = U7  C163 * U5
U10 = S10 + U3U11 = S10  U3
S10 = U10 + U8S12 = U10  U8
S14 = U11 + U9S16 = U11  U9
CX(I( 1)) = R1 + R5
X(I( 9)) = R1  R5Y(I( 1)) = S1 + S5
Y(I( 9)) = S1  S5X(I( 2)) = R2 + S10
X(I(16)) = R2  S10Y(I( 2)) = S2  R10
Y(I(16)) = S2 + R10X(I( 3)) = R9 + S13
X(I(15)) = R9  S13Y(I( 3)) = S9  R13
Y(I(15)) = S9 + R13X(I( 4)) = R8  S16
X(I(14)) = R8 + S16Y(I( 4)) = S8 + R16
Y(I(14)) = S8  R16X(I( 5)) = R3 + S7
X(I(13)) = R3  S7Y(I( 5)) = S3  R7
Y(I(13)) = S3 + R7X(I( 6)) = R6 + S14
X(I(12)) = R6  S14Y(I( 6)) = S6  R14
Y(I(12)) = S6 + R14X(I( 7)) = R11  S15
X(I(11)) = R11 + S15Y(I( 7)) = S11 + R15
Y(I(11)) = S11  R15X(I( 8)) = R4  S12
X(I(10)) = R4 + S12Y(I( 8)) = S4 + R12
Y(I(10)) = S4  R12C
GOTO 20CFigure. Length16 FFT Module
Questions & Answers
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
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
J, combine like terms 7x4y
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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