c This routine calculates the resulting convolution between 2 filters

      SUBROUTINE convol_new(af1,aw1,kmax1,af2,aw2,kmax2,af3,aw3,kmax)
      INTEGER i,kmax1,kmax2,kmax
      REAL af1,aw1,af2,aw2,af3,aw3,ddd,ddf
      DIMENSION af1(4000),aw1(4000),af2(4000),aw2(4000)
      DIMENSION af3(4000),aw3(4000),ddd(4000),ddf(4000)

c... Filters have the same delta lambda value
c... We take the first filter as a reference and we rebin the 2ond one
c... kmax is the maximum number of wavelengts per filter

      DO i=1,kmax1
         ddd(i)=aw1(i)
         p1=0.0
         DO j=2,kmax2
            IF (ddd(i).ge.aw2(j-1).and.ddd(i).lt.aw2(j)) THEN
               p2=((ddd(i)-aw2(j-1))/(aw2(j)-aw2(j-1)))
     .              *(af2(j)-af2(j-1))
               p1=af2(j-1)+p2
            END IF
         END DO
         ddf(i)=p1*af1(i)
      END DO

c... Maximum number of points (kmax)

      kmax=1
      ntest=1
      DO i=2,kmax1
         IF (ddf(i).gt.0.0.and.ntest.eq.1) THEN
            ntest=2
            aw3(kmax)=ddd(i-1)
            af3(kmax)=ddf(i-1)
         END IF
         IF (ddf(i).gt.0.0.and.ntest.gt.1) THEN
            kmax=kmax+1
            aw3(kmax)=ddd(i)
            af3(kmax)=ddf(i)
         END IF
         IF (ddf(i).le.0.0.and.ntest.gt.1) THEN
            kmax=kmax+1
            ntest=1
            aw3(kmax)=ddd(i)
            af3(kmax)=ddf(i)
         END IF
      END DO
      
      amin=1.e15
      amax=0.0
      DO i=1,kmax
         IF (af3(i).lt.amin) amin=af3(i)
         IF (af3(i).gt.amax) amax=af3(i)
      END DO

      RETURN
      END