mirror of https://github.com/saitohirga/WSJT-X.git
85 lines
2.2 KiB
Fortran
85 lines
2.2 KiB
Fortran
subroutine sync(y1,y2,y3,y4,npts,jpk,baud,bauderr)
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C Input data are in the y# arrays: detected sigs in four tone-channels,
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C before decimation by NSPD.
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parameter (NSPDMAX=200)
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real y1(npts)
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real y2(npts)
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real y3(npts)
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real y4(npts)
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real zf(NSPDMAX)
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complex csum
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integer nsum(NSPDMAX)
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real z(65538) !Ready for FSK110
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complex cz(0:32768)
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equivalence (z,cz)
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data twopi/6.283185307/
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NSPD=25
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do i=1,NSPD
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zf(i)=0.0
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nsum(i)=0
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enddo
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do i=1,npts
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a1=max(y1(i),y2(i),y3(i),y4(i)) !Find the largest one
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if(a1.eq.y1(i)) then !Now find 2nd largest
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a2=max(y2(i),y3(i),y4(i))
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else if(a1.eq.y2(i)) then
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a2=max(y1(i),y3(i),y4(i))
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else if(a1.eq.y3(i)) then
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a2=max(y1(i),y2(i),y4(i))
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else
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a2=max(y1(i),y2(i),y3(i))
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endif
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z(i)=1.e-6*(a1-a2) !Subtract 2nd from 1st
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j=mod(i-1,NSPD)+1
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zf(j)=zf(j)+z(i)
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nsum(j)=nsum(j)+1
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enddo
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n=log(float(npts))/log(2.0)
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nfft=2**(n+1)
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call zero(z(npts+1),nfft-npts)
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call xfft(z,nfft)
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C Now find the apparent baud rate.
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df=11025.0/nfft
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zmax=0.
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ia=391.0/df !Was 341/df
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ib=491.0/df !Was 541/df
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do i=ia,ib
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z(i)=real(cz(i))**2 + imag(cz(i))**2
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if(z(i).gt.zmax) then
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zmax=z(i)
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baud=df*i
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endif
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enddo
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C Find phase of signal at 441 Hz.
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csum=0.
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do j=1,NSPD
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pha=j*twopi/NSPD
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csum=csum+zf(j)*cmplx(cos(pha),-sin(pha))
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enddo
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pha=-atan2(imag(csum),real(csum))
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jpk=nint(NSPD*pha/twopi)
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if(jpk.lt.1) jpk=jpk+NSPD
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C The following is nearly equivalent to the above. I don't know which
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C (if either) is better.
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c zfmax=-1.e30
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c do j=1,NSPD
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c if(zf(j).gt.zfmax) then
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c zfmax=zf(j)
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c jpk2=j
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c endif
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c enddo
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bauderr=(baud-11025.0/NSPD)/df !Baud rate error, in bins
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return
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end
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