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|
program test
parameter(M=2**2)
integer i,j
complex A(2*m),B(2*m),C(2*m),D(2*m)
do i=1,2*M
A(i)=cmplx(1.,2.*i)
B(i)=cmplx(2.,3.*i)
enddo
call conv(A,B,M,C)
print 20,(C(i),i=1,2*M)
20 format('C=',8(F16.8,1X))
call fft1d(A,200,2*M)
call fft1d(B,200,2*M)
do i=1,2*M
D(i)=A(i)*B(i)
enddo
call ifft1d(D,200,2*M)
print 30,(D(i),i=1,2*m)
30 format('S=',8(F16.8,1X))
end
! subroutine pour calculer A*B
subroutine conv(A,B,M,C)
integer i,j,l1,l2
complex A(2*M),B(2*M),C(2*M)
do i=1,2*M
C(i)=(0.,0.)
do l1=1,2*M
C(i)=C(i)+A(i-l1)*B(l1)
enddo
enddo
return
end
! subroutine pour calcule le produit element par element
subroutine wise(AA,BB,MM,CC)
integer i,j,l1,l2
complex AA(2*MM),BB(2*MM),CC(2*MM)
do l1=1,2*MM
CC(i)=AA(i)*BB(i)
enddo
return
end
subroutine ifft1d(z, m, n)
complex z(*)
complex zt,wt,w(2**15)
double precision p
integer n,m,m2
save w, m2
data m2 /-1/
if (m .eq. 0) then
return
endif
C Pass back sin/cos table if not previously initialized.
if (m .ne. m2) then
p = 6.283185307179586d0/n
do k = 1,n/2
C Coefficients for ifft.
w(k) = cmplx(cos(p*(k-1)),sin(p*(k-1)))
C Coefficients for fft.
w(k+n/2) = conjg(w(k))
end do
m2 = m
end if
C Unit stride.
ks = 1
nlast = n
n2 = n
kn2 = ks*n
ie = 1
do k = 1,m
n1 = n2
kn1 = kn2
n2 = n2/2
kn2 = kn2/2
ia = 1
kj = 1
do j = 1,n2
wt = w(ia)
ia = ia + ie
ki = kj
do i = j,n,n1
kl = ki + kn2
zt = z(ki) - z(kl)
z(ki) = z(ki) + z(kl)
z(kl) = wt*zt
ki = ki+kn1
end do
kj = kj+ks
end do
ie = ie+ie
end do
ki = 1
do i = 1,n
z(ki) = z(ki)/float(n)
ki = ki+ks
end do
j = 1
n1 = n - 1
ki = 1
kj = 1
do i=1,n1
if(i.lt.j) then
zt = z(kj)
z(kj) = z(ki)
z(ki) = zt
endif
k = n/2
do while (k.lt.j)
j = j - k
k = k/2
end do
j = j + k
kj = ks*(j-1) + 1
ki = ki + ks
end do
return
end
subroutine fft1d(z,m,n)
complex z(*)
complex zt,wt,w(2**15)
double precision p
integer n,m,m2
save w, m2
data m2 /-1/
if (m .eq. 0) then
return
endif
C Pass back sin/cos table if not previously initialized.
if (m .ne. m2) then
p = 6.283185307179586d0/n
do k = 1,n/2
C Coefficients for ifft.
w(k) = cmplx(cos(p*(k-1)),sin(p*(k-1)))
C Coefficients for fft.
w(k+n/2) = conjg(w(k))
enddo
m2 = m
endif
C Unit stride.
ks = 1
n2 = n
kn2 = ks*n
ie = 1
C Use second half of table.
ia1 = n/2 + 1
do k = 1,m
n1 = n2
kn1 = kn2
n2 = n2/2
kn2 = kn2/2
ia= ia1
kj = 1
do j = 1,n2
wt = w(ia)
ia = ia + ie
ki = kj
do i = j,n,n1
kl = ki + kn2
zt = z(ki) - z(kl)
z(ki) = z(ki) + z(kl)
z(kl) = wt*zt
ki = ki+kn1
end do
kj = kj+ks
end do
ie = ie+ie
end do
j = 1
n1 = n - 1
ki = 1
kj = 1
do i=1,n1
if(i.lt.j) then
zt = z(kj)
z(kj) = z(ki)
z(ki) = zt
endif
k = n/2
do while (k.lt.j)
j = j - k
k = k/2
end do
j = j + k
kj = ks*(j-1) + 1
ki = ki + ks
end do
return
end |
Problème convolution discrète
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