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Re: A question on interval arithmetic


You need to average  (r1*r2)/(r1+r2) over the required interval.

Here is how I did it evaluating over the 2-dimensional tolerance region
around r1=r10 and r2=r20. I assume each resistor independently has a
tolerance +/-a which is uniformly distributed over an interval of length 2a,
which gives a probability density 1/(4a^2) over the 2-dimensional tolerance
region for the pair of resistors. I had to evaluate indefinite integrals and
then substitute in limits because definite integration seemed to lock
Mathematica up (I have version 5 for Windows).

Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)), r1]
Simplify[(% /. {r1 -> r10 + a}) - (% /. {r1 -> r10 - a})]
Integrate[%, r2]
Simplify[(% /. {r2 -> r20 + a}) - (% /. {r2 -> r20 - a})]

--
Steve Luttrell
West Malvern, UK

"Oliver Friedrich" <oliver.friedrich at tzm.de> wrote in message
news:blcq5l$p5f$1 at smc.vnet.net...
> Hallo,
>
> the resistance of 2 resistors in parallel is r1*r2/(r1+r2). Now I want to
> introduce tolerances in the resistors and ask for the range of resistance
> of the combination. One may think that e.g
>
> (r1*r2)/(r1+r2)/.{r1->Interval[{10,20}],r2->Interval[{20,40}]}
>
> would lead to the correct result, but there's a trap. If I replace the
> expressions by the intervals, Mathematica evaluates the new expression
assuming
> that all four intervals are independant from each other. And that's not
> correct. Taken either the minimum or the maximum from a certain interval ,
> Mathematica should stick to that, because it is nonsense to take Min[r1]
and Max
> [r1] within the same expression, r1 can have only one value at a time.
>
> How can I avoid this problem?
>
> Oliver Friedrich
>



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