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Re: sum of exponentials
Posted:
Nov 24, 2012 12:28 PM
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"dwi" wrote in message <k8qlsv$ktj$1@newscl01ah.mathworks.com>...
> Ok, I understand now how this works. But still, you said the result will be
> x(3)+x(2)*e^(-1)+x(1)*e^(-2))/(1+e^(-1)+e^(-2));
> while I want
> (x(3)*e^(-1)+x(2)*e^(-2)+x(1)*e^(-3))/(e^(-1)+e^(-2)+e^(-3));
> Also, how would your code change if I had e^(-1/20), e^(-2/20), e^(-3/20) etc?
- - - - - - - -
The two expressions
(x(3)*e^(-1)+x(2)*e^(-2)+x(1)*e^(-3))/(e^(-1)+e^(-2)+e^(-3))
and
(x(3)+x(2)*e^(-1)+x(1)*e^(-2))/(1+e^(-1)+e^(-2))
are identically equal. Just divide the numerator and denominator of the first expression by e^(-1) to get the second expression. What you want and what this code produces are the same thing.
As to your second question, just the two lines
a = x(k) + a*f*e^(-1);
b = 1 + b*f*e^(-1);
would need to be changed to:
a = x(k) + a*f*e^(-1/20);
b = 1 + b*f*e^(-1/20);
Roger Stafford
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