Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
dwi
Posts:
15
Registered:
10/18/12
|
|
Re: sum of exponentials
Posted:
Nov 23, 2012 10:16 AM
|
|
"dwi" wrote in message <k8o2mg$pdp$1@newscl01ah.mathworks.com>...
> "Roger Stafford" wrote in message <k8lqar$4po$1@newscl01ah.mathworks.com>...
> > "dwi" wrote in message <k8l6sb$2fj$1@newscl01ah.mathworks.com>...
> > > I have a matrix whose data are interrupted by sequences of zeros. I need every time that there's a zero value to substitute it with a sum of exponentials using the previous data, eg:
> > > x=[x1 x2 x3 0 0 0 x7 0 0]
> > > When i find the first zero in the element x4 I want:
> > > x4=(x3*e^(-1)+x2*e^(-2)+x1*e^(-3))/(e^(-1)+e^(-2)+e^(-3));
> > > However, when I find the second zero value I need to calculate the same expression but without using the previous recalculated values. That is,
> > > x4=x5=x6
> > > and for x8 I will use only the values in x7,x3,x2,x1 and then x8=x9 etc
> > > And all this for a 180000-length data.
> > > Any ideas on how to do this?
> > > Thanks in advance
> > - - - - - - - - -
> > a = 0; b = 0;
> > for k = 1:length(x)
> > if x(k) ~= 0
> > a = x(k) + a*e^(-1);
> > b = 1 + b*e^(-1);
> > else
> > x(k) = a/b;
> > end
> > end
> >
> > Roger Stafford
> Thank you for your answer but if I use this the index of the exponent never changes.I want it to decrease as k increases, and i don't know hot to calculate a sum like this.
A correction:
x8=(x7*e^(-1))/e^(-1)
ie without using x1,x2,x3 but only the immediate previous non-zero values
|
|
|
|
