Can you describe what kind of a problem you are having?
> On Apr 6, 2015, at 10:51 AM, Jim Christoff <[email protected]> wrote: > > That was not it > >> On Monday, April 6, 2015 at 9:05:16 AM UTC-4, Jim Christoff wrote: >> First, Julia is a great replacement for Octave and in most cases "C". I have >> used Octave for decades and Julia for a year or so. >> I still have problems making the conversions. Any assistance will be greatly >> appreciated. >> This is short example of the modified Octave code that I have tried to >> convert. I realize there is a difference >> in how Julia handles arrays, but I don't seem to be able to resolve this >> difference. >> The following code I use in Octave: >> >> function arburg1 (x, p) >> >> >> a = zeros(p) >> n = length(x) >> v = sum(x.*x) >> >> ## f and b are the forward and backward error sequences >> f = x[2:n] >> b = x[1:n-1] >> >> ## remaining stages i=2 to p >> for i=1:p >> >> ## get the i-th reflection coefficient >> g = 2 * sum(f.*b)/(sum(f.*f)+sum(b.*b)); >> >> >> ## generate next filter order >> if i==1 >> a = g >> else >> a = a-g*a[i-1:-1:1] **** this is my problem area***** >> endif >> >> ## keep track of the error >> v = v*(1-g^2); >> >> ## update the prediction error sequences >> oldf = f >> f = oldf[2:n-i]- g*b[2:n-i] >> b = b[1:n-i-1] - g*oldf[1:n-i-1] >> >> end >> a = -a[p:-1:1] >> >> end >> >> >> Thanks for any suggestions
