[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: unwrap modulo 2pi
I just don't see that you can do this at all. Modulo operator discards
information about the number of multiples of B from A, leaving only the
remainder of the division operation. For example, if:
c = [2., 4., 6., 1.71681, 3.71681, 5.71681]
b = 2*!PI
; Each of the following A will produce the given
; C through "C=A mod B" :
a=[2.,4.,6.,8.,10.,12.]
a=[2.,4.,6.,8.,10.,12.] + b
a=[2.,4.,6.,8.,10.,12.] +2*b
a=[2.,4.,6.,8.,10.,12.] & a[3:5] = a[3:5] + 3*b
Unless I am missing something, I see no way how a unique solution can be
obtained from the *remainder of division* and *divisor*. But I studied
arithmetics a long time ago :-)
Cheers,
Pavel
graham_wilson@my-deja.com wrote:
>
> My appologies for not being explicit enough...
>
> IDL> a=[2,4,6,8,10,12]
> IDL> a=[2.,4.,6.,8.,10.,12.]
> IDL> b=2*!PI
> IDL> c=a mod b
> IDL> print, c
> 2.00000 4.00000 6.00000 1.71681 3.71681 5.71681
>
> What I mean by 'unwraping' is: Given I know 'c' and 'b' how do I
> explicitly find a?
>
> GW
>
> In article <3A817F92.1DBCDCCA@noaa.gov>,
> "Pavel A. Romashkin" <pavel.romashkin@noaa.gov> wrote:
> > I am sorry, but I am still a little behind here, please bear with me.
> > Must be the lack of coffe. What is the "unwrapping of the function
> > mod(x,y)" ? I might think of a solution if I knew what I am looking
> at.
> >
> > Cheers,
> > Pavel
>
> Sent via Deja.com
> http://www.deja.com/