如何使用c语言实现hilbert变换

#include <math.h>

#include <stdio.h>

#define N 8

void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il)

void main()

{

double xr[N],xi[N],Yr[N],Yi[N],l=0,il=0

int i,j,n=N,k=3

for(i=0i<Ni++)

{

xr[i]=i

xi[i]=0

}

printf("陪昌脊------FFT------\n")

l=0

kkfft(xr,xi,n,k,Yr,Yi,l,il)

for(i=0i<Ni++)

{

printf("%-11lf + j* %-11lf\n",Yr[i],Yi[i])

}

printf("-----DFFT-------\n")

l=1

kkfft(Yr,Yi,n,k,xr,xi,l,il)

for(i=0i<芦渗Ni++)

{

printf("%-11lf + j* %-11lf\n",xr[i],xi[i])

}

getch()

}

void kkfft(double pr[], double pi[], int n, int k, double fr[], double fi[], int l, int il)

{

int it,m,is,i,j,nv,l0

double p,q,s,vr,vi,poddr,poddi

for (it=0it<=n-1it++)

{

m = it

is = 0

for(i=0i<=k-1i++)

{

j = m/2

is = 2*is+(m-2*j)

m = j

}

fr[it] = pr[is]

fi[it] = pi[is]

}

pr[0] = 1.0

pi[0] = 0.0

p = 6.283185306/(1.0*n)

pr[1] = cos(p)

pi[1] = -sin(p)

if (l!=0)

pi[1]=-pi[1]

for (i=2i<=n-1i++)

{

p = pr[i-1]*pr[1]

q = pi[i-1]*pi[1]

s = (pr[i-1]+pi[i-1])*(pr[1]+pi[1])

pr[i] = p-q

pi[i] = s-p-q

}

for (it=0it<=n-2it=it+2)

{

vr = fr[it]

vi = fi[it]

fr[it] = vr+fr[it+1]

fi[it] = vi+fi[it+1]

fr[it+1] = vr-fr[it+1]

fi[it+1] = vi-fi[it+1]

}

m = n/2

nv = 2

for (l0=k-2l0>=0l0--)

{

m = m/2

nv = 2*nv

for(it=0it<=(m-1)*nvit=it+nv)

for (j=0j<迅迅=(nv/2)-1j++)

{

p = pr[m*j]*fr[it+j+nv/2]

q = pi[m*j]*fi[it+j+nv/2]

s = pr[m*j]+pi[m*j]

s = s*(fr[it+j+nv/2]+fi[it+j+nv/2])

poddr = p-q

poddi = s-p-q

fr[it+j+nv/2] = fr[it+j]-poddr

fi[it+j+nv/2] = fi[it+j]-poddi

fr[it+j] = fr[it+j]+poddr

fi[it+j] = fi[it+j]+poddi

}

}

/*逆傅立叶变换*/

if(l!=0)

{

for(i=0i<=n-1i++)

{

fr[i] = fr[i]/(1.0*n)

fi[i] = fi[i]/(1.0*n)

}

}

/*是否计算模和相角*/

if(il!=0)

{

for(i=0i<=n-1i++)

{

pr[i] = sqrt(fr[i]*fr[i]+fi[i]*fi[i])

if(fabs(fr[i])<0.000001*fabs(fi[i]))

{

if ((fi[i]*fr[i])>0)

pi[i] = 90.0

else

pi[i] = -90.0

}

else

pi[i] = atan(fi[i]/fr[i])*360.0/6.283185306

}

}

return

}

function [x,y] = hilbert(n)

if n<薯陪=0

  x=0

  y=0

else

  局散[xo,yo]=hilbert(n-1)

  x=.5*[-.5+yo -.5+xo .5+xo  .5-yo]

  y=.5*[-.5+xo  .5+yo 数腊蠢.5+yo -.5-xo]

end

#include ‘“搭晌stdio.h”

void main()

{

int count=0,i,j,m,n

int a[]

scanf("%d",&a[count])

while(a[count]!=0)

{

printf("%d\n",a[count])

count++

scanf("%d"知姿锋册巧,&a[count])

}

printf("%d\n",a[count])

for(i=0i<counti++)

{

j=0

for(m=5m<a[i]m=m+4)

for(n=5n<a[i]n=m+4)

if(a[i]==m*n)

j++

printf("%d %d\n",a[i],j)

}

}

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