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c
c-----------------------------------------------------------------------
c subroutine: iftsoh
c compute idft for real, symmetric, odd harmonic, n-point sequence
c using n/4-point fft
c symmetric sequence means x(m)=x(n-m), m=1,...,n/2-1
c odd harmonic means x(2*k)=0, all k, where x(k) is the dft of x(m)
c x(m) has the property x(m)=-x(n/2-m), m=0,1,...,n/4-1, x(n/4)=0
c note: index m is sequence index--not fortran index
c-----------------------------------------------------------------------
c
subroutine iftsoh(x, n, y)
dimension x(1), y(1)
c
c x = real array which on input contains the n/4 real points of
c the odd harmonics of the transform of the original time sequence
c i.e. the zero valued imaginary parts are not given nor are the
c zero valued even harmonics
c on output x contains the first n/4 points of the original input
c sequence (symmetrical)
c n = true size of input
c y = scratch array of size n/4+2
c
c
c handle n = 2 and n = 4 cases separately
c
if (n.gt.4) go to 10
c
c for n=2, 4 assume x(1)=x0, x(2)=-x0, compute idft directly
c
x(1) = x(1)/2.
return
c
c code for values of n which are multiples of 8
c
10 twopi = 8.*atan(1.0)
no2 = n/2
no4 = n/4
no8 = n/8
tpn = twopi/float(n)
c
c first compute x1=x(1) term directly
c use recursion on the sine cosine terms
c
cosd = cos(tpn*2.)
sind = sin(tpn*2.)
cosi = 2.*cos(tpn)
sini = 2.*sin(tpn)
x1 = 0.
do 20 i=1,no4
x1 = x1 + x(i)*cosi
temp = cosi*cosd - sini*sind
sini = cosi*sind + sini*cosd
cosi = temp
20 continue
x1 = x1/float(n)
c
c scramble original dft (x(k)) to give y(k)
c use recursion relation to give sin multipliers
c
cosi = cos(tpn)
sini = sin(tpn)
do 30 i=1,no8
ind = 2*i
ind1 = no4 + 1 - i
ak = (x(i)+x(ind1))/2.
bk = (x(i)-x(ind1))
y(ind-1) = ak
y(ind) = bk*sini
temp = cosi*cosd - sini*sind
sini = cosi*sind + sini*cosd
cosi = temp
30 continue
c
c the sequence y(k) is the odd harmonics dft output
c use subroutine iftohm to obtain y(m), the inverse transform
c
call iftohm(y, no2)
c
c form x(m) sequence from y(m) sequence
c use x1 initial condition on the recursion
c
x(1) = y(1)
x(2) = x1
if (no8.eq.1) return
do 40 i=2,no8
ind = 2*i
ind1 = no4 + 2 - i
t1 = (y(i)+y(ind1))/2.
x(ind-1) = (y(i)-y(ind1))/2.
x(ind) = t1 + x(ind-2)
40 continue
return
end
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