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103 lines
2.2 KiB
Fortran
103 lines
2.2 KiB
Fortran
subroutine lorentzian(y,npts,a)
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! Input: y(npts); assume x(i)=i, i=1,npts
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! Output: a(5)
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! a(1) = baseline
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! a(2) = amplitude
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! a(3) = x0
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! a(4) = width
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! a(5) = chisqr
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real y(npts)
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real a(5)
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real deltaa(4)
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a=0.
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df=12000.0/8192.0 !df = 1.465 Hz
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width=0.
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ipk=0
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ymax=-1.e30
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do i=1,npts
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if(y(i).gt.ymax) then
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ymax=y(i)
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ipk=i
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endif
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! write(50,3001) i,i*df,y(i)
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!3001 format(i6,2f12.3)
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enddo
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! base=(sum(y(ipk-149:ipk-50)) + sum(y(ipk+51:ipk+150)))/200.0
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base=(sum(y(1:20)) + sum(y(npts-19:npts)))/40.0
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stest=ymax - 0.5*(ymax-base)
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ssum=y(ipk)
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do i=1,50
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if(ipk+i.gt.npts) exit
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if(y(ipk+i).lt.stest) exit
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ssum=ssum + y(ipk+i)
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enddo
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do i=1,50
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if(ipk-i.lt.1) exit
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if(y(ipk-i).lt.stest) exit
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ssum=ssum + y(ipk-i)
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enddo
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ww=ssum/y(ipk)
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width=2
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t=ww*ww - 5.67
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if(t.gt.0.0) width=sqrt(t)
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a(1)=base
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a(2)=ymax-base
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a(3)=ipk
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a(4)=width
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! Now find Lorentzian parameters
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deltaa(1)=0.1
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deltaa(2)=0.1
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deltaa(3)=1.0
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deltaa(4)=1.0
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nterms=4
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! Start the iteration
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chisqr=0.
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chisqr0=1.e6
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do iter=1,5
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do j=1,nterms
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chisq1=fchisq0(y,npts,a)
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fn=0.
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delta=deltaa(j)
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10 a(j)=a(j)+delta
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chisq2=fchisq0(y,npts,a)
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if(chisq2.eq.chisq1) go to 10
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if(chisq2.gt.chisq1) then
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delta=-delta !Reverse direction
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a(j)=a(j)+delta
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tmp=chisq1
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chisq1=chisq2
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chisq2=tmp
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endif
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20 fn=fn+1.0
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a(j)=a(j)+delta
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chisq3=fchisq0(y,npts,a)
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if(chisq3.lt.chisq2) then
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chisq1=chisq2
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chisq2=chisq3
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go to 20
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endif
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! Find minimum of parabola defined by last three points
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delta=delta*(1./(1.+(chisq1-chisq2)/(chisq3-chisq2))+0.5)
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a(j)=a(j)-delta
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deltaa(j)=deltaa(j)*fn/3.
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! write(*,4000) iter,j,a,chisq2
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!4000 format(i1,i2,4f10.4,f11.3)
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enddo
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chisqr=fchisq0(y,npts,a)
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! write(*,4000) 0,0,a,chisqr
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if(chisqr/chisqr0.gt.0.99) exit
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chisqr0=chisqr
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enddo
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a(5)=chisqr
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return
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end subroutine lorentzian
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