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Hard-coded four-step

This section presents an implementation of the four-step algorithm  [link] that leverages hard-coded sub-transforms to compute larger transforms. The implementation uses an implicit memory transpose (along with vector register transposes) and scales particularly well with VL. In contrast to the fully hard-coded implementation in theprevious section, the four-step implementation requires no new leaf primitives as VL increases, i.e., the code is much the same when V L > 1 as it is when V L = 1 .

The four-step algorithm

A transform of size N is decomposed into a two-dimensional array of size n 1 × n 2 where N = n 1 n 2 . Selecting n 1 = n 2 = N (or close) often obtains the best performance results  [link] . When either of the factors is larger than the other, it is the larger of the two factors that will determine performance, because the larger factor effectively brings the memory wall closer. The four steps of the algorithm are:

  1. Compute n 1 FFTs of length n 2 along the columns of the array;
  2. Multiply each element of the array with ω N i j , where i and j are the array coordinates;
  3. Transpose the array;
  4. Compute n 2 FFTs of length n 1 along the columns of the array.

For this out-of-place implementation, steps 2 and 3 are performed as part of step 1. Step 1 reads data from the input array and computes the FFTs, but before storing the data in the final pass, it is multiplied by the twiddle factors from step 2. After this, the data is stored to rows in the output array, and thus the transpose of step 3 is performed implicitly. Step 4 is then computed as usual: FFTs are computed along the columns of the output array.

This method of computing the four-step algorithm in two steps requires only minor modifications in order to support multiple vector lengths: with V L > 1 , multiple columns are read and computed in parallel without modification of the code, but before storing multiple columns of data to rows, a register transpose is required.

Vector length 1

When V L = 1 , three hard-coded FFTs are elaborated.

  1. FFT of length n 2 with stride n 1 × 2 for the first column of step 1;
  2. FFT of length n 2 with stride n 1 × 2 and twiddle multiplications on outputs – for all other columns of step 1;
  3. FFT of length n 1 with stride n 2 × 2 for columns in step 4.

In order to generate the code for the four-step sub-transforms, some minor modifications are made to the fully hard-coded code generator that was presented in the previous section.

The first FFT is used to handle the first column of step 1, where there are no twiddle factor multiplications because one of the array coordinates for step 2 is zero, and thus ω N 0 is unity. This FFT may be elaborated as in "Vector length 1" with the addition of a stride factor for the input address calculation. The second FFT is elaborated as per the first FFT, but with the addition of twiddle factor multiplications oneach register prior to the store operations. The third FFT is elaborated as per the first FFT, but with strided input and output addresses.

const SFFT_D __attribute__ ((aligned(32))) *LUT; const SFFT_D *pLUT;void sfft_dcf64_fs_x1_0(sfft_plan_t *p, const void *vin, void *vout){   const SFFT_D *in = vin;  SFFT_D *out = vout;   SFFT_R r0,r1,r2,r3,r4,r5,r6,r7;  L_4(in+0,in+64,in+32,in+96,&r0,&r1,&r2,&r3);   L_2(in+16,in+80,in+112,in+48,&r4,&r5,&r6,&r7);   K_0(&r0,&r2,&r4,&r6);   S_4(r0,r2,r4,r6,out+0,out+4,out+8,out+12);  K_N(VLIT2(0.7071,0.7071),VLIT2(0.7071,-0.7071),&r1,&r3,&r5,&r7);   S_4(r1,r3,r5,r7,out+2,out+6,out+10,out+14);} void sfft_dcf64_fs_x1_n(sfft_plan_t *p, const void *vin, void *vout){  const SFFT_D *in = vin;   SFFT_D *out = vout;  SFFT_R r0,r1,r2,r3,r4,r5,r6,r7;   L_4(in+0,in+64,in+32,in+96,&r0,&r1,&r2,&r3);   L_2(in+16,in+80,in+112,in+48,&r4,&r5,&r6,&r7);   K_0(&r0,&r2,&r4,&r6);   r2 = MUL(r2,LOAD(pLUT+4),LOAD(pLUT+6));  r4 = MUL(r4,LOAD(pLUT+12),LOAD(pLUT+14));   r6 = MUL(r6,LOAD(pLUT+20),LOAD(pLUT+22));  S_4(r0,r2,r4,r6,out+0,out+4,out+8,out+12);   K_N(VLIT2(0.7071,0.7071),VLIT2(0.7071,-0.7071),&r1,&r3,&r5,&r7);   r1 = MUL(r1,LOAD(pLUT+0),LOAD(pLUT+2));  r3 = MUL(r3,LOAD(pLUT+8),LOAD(pLUT+10));  r5 = MUL(r5,LOAD(pLUT+16),LOAD(pLUT+18));   r7 = MUL(r7,LOAD(pLUT+24),LOAD(pLUT+26));  S_4(r1,r3,r5,r7,out+2,out+6,out+10,out+14);   pLUT += 28;} void sfft_dcf64_fs_x2(sfft_plan_t *p, const void *vin, void *vout){  const SFFT_D *in = vin;   SFFT_D *out = vout;  SFFT_R r0,r1,r2,r3,r4,r5,r6,r7;   L_4(in+0,in+64,in+32,in+96,&r0,&r1,&r2,&r3);   L_2(in+16,in+80,in+112,in+48,&r4,&r5,&r6,&r7);   K_0(&r0,&r2,&r4,&r6);   S_4(r0,r2,r4,r6,out+0,out+32,out+64,out+96);  K_N(VLIT2(0.7071,0.7071),VLIT2(0.7071,-0.7071),&r1,&r3,&r5,&r7);   S_4(r1,r3,r5,r7,out+16,out+48,out+80,out+112);} void sfft_dcf64_fs(sfft_plan_t *p, const void *vin, void *vout) {  const SFFT_D *in = vin;   SFFT_D *out = vout;  pLUT =  LUT;   int i;  sfft_dcf64_fs_x1_0(p, in, out);   for(i=1;i<8;i++) sfft_dcf64_fs_x1_n(p, in+(i*2), out+(i*16));   for(i=0;i<8;i++) sfft_dcf64_fs_x2(p, out+(i*2), out+(i*2)); }
Hard-coded four-step VL-1 size-64 FFT

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
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/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
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?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
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Porter
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Yasmin
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Cesar
I'm interested in nanotube
Uday
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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 . 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
can nanotechnology change the direction of the face of the world
Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Computing the fast fourier transform on simd microprocessors. OpenStax CNX. Jul 15, 2012 Download for free at http://cnx.org/content/col11438/1.2
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