WSJT-X/stdecode.f
Joe Taylor 2c17544f3f initial import
git-svn-id: svn+ssh://svn.code.sf.net/p/wsjt/wsjt/WSJT/trunk@1 ab8295b8-cf94-4d9e-aec4-7959e3be5d79
2005-12-22 16:40:53 +00:00

107 lines
3.3 KiB
Fortran

subroutine stdecode(s2,nchan,nz,sigma,dtbuf,df,stlim0,
+ DFTolerance,cfile6,pick,istart)
C Search for and decode single-tone messages.
real s2(nchan,nz)
integer DFTolerance
logical pick
character cfile6*6,msg3*3
character*90 line
common/ccom/nline,tping(100),line(100)
NSPD=25 !Change if FSK110 is implemented
LTone=2
NBaud=11025/NSPD
stlim=stlim0
if(pick) stlim=stlim0-1.0
iwidth=1
ts0=-1.0
dt=1.0/11025.0
C In each time slice, find largest peak between LTone*NBaud-DFTolerance and
C (LTone+3)*NBaud+DFTolerance.
ia=(LTone*NBaud-DFTolerance)/df - 5.0
ib=((LTone+3)*NBaud+DFTolerance)/df - 4.0
do j=1,nz
smax=0.
do i=ia,ib !Get the spectral peak
if(s2(i,j).gt.smax) then
smax=s2(i,j)
ipk=i
endif
enddo
peak=dB(smax/sigma) - 14.0 !Empirical
C constant should be dB(43/2500) = -17.6 dB?
if(peak.gt.stlim) then
C To minimize false ST decodings from QRN and MT pings, find the
C second best peak (excluding points around the first peak).
smax2=0.
do i=ia,ib
if((abs(i-ipk).gt.iwidth) .and. s2(i,j).gt.smax2) then
smax2=s2(i,j)
ipk2=i
endif
enddo
C Larger values of ratlim make it more likely to report ST decodings.
ratlim=0.18
if(stlim.lt.-2.5) ratlim=0.20
if(stlim.lt.-3.5) ratlim=0.22
if(stlim.lt.-4.5) ratlim=0.24
if(pick) ratlim=0.27 !Fine tuning here...
if(smax2/smax.gt.ratlim) goto 20
call peakup(s2(ipk-1,j),s2(ipk,j),s2(ipk+1,j),dx)
freq=(ipk+5+dx)*df
tstart=j*dtbuf + dt*(istart-1)
mswidth=20
nwidth=0
nstrength=0
n=nint(freq/NBaud)
noffset=freq-n*NBaud
if((noffset.lt.-DFTolerance) .or.
+ (noffset.gt.DFTolerance)) goto 20
C The numbers 2 and 5 depend on Ltone:
if(n.lt.2 .or. n.gt.5) goto 20
C OK, this detection has survived all tests. Save it for output
C (uness perhaps it is redundant).
if(n.eq.LTone) msg3='R26'
if(n.eq.LTone+1) msg3='R27'
if(n.eq.LTone+2) msg3='RRR'
if(n.eq.LTone+3) msg3='73'
C Now check for redundant detections. (Not sure, now, why I chose
C the time span 0.11 s.)
if(tstart-ts0.gt.0.11) then
peak0=0. !If time diff>0.11s, start fresh
else
if(peak.le.peak0) goto 20
nline=nline-1 !Delete previous, this one's better
peak0=peak !Save best peak
endif
C OK, we want to output this one. Save the information.
if(nline.le.99) nline=nline+1
ts0=tstart
tping(nline)=tstart
nst=(int(smax/smax2)-4)/2 + 1
if(nst.lt.1) nst=1
if(nst.gt.3) nst=3
write(line(nline),1050) cfile6,tstart,mswidth,int(peak),
+ nwidth,nstrength,noffset,msg3,nst
1050 format(a6,f5.1,i5,i3,1x,2i1,i5,1x,a3,40x,i3)
endif
20 continue
enddo
return
end