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Topic: sum of exponentials
Replies: 9   Last Post: Nov 24, 2012 7:16 PM

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Roger Stafford

Posts: 5,892
Registered: 12/7/04
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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