Re: A question on interval arithmetic
- To: mathgroup at smc.vnet.net
- Subject: [mg43760] Re: A question on interval arithmetic
- From: "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk>
- Date: Fri, 3 Oct 2003 02:28:46 -0400 (EDT)
- References: <blcq5l$p5f$1@smc.vnet.net> <blgjf4$i1d$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Groan! I didn't read the question properly. My suggestion computed the expected value (mean) of the parallel resistance. I see now that all you wanted was the "range of resistance of the combination", like everyone else noticed first time around. Apologies for the confusion. -- Steve Luttrell West Malvern, UK "Steve Luttrell" <luttrell at _removemefirst_westmal.demon.co.uk> wrote in message news:blgjf4$i1d$1 at smc.vnet.net... > You need to average (r1*r2)/(r1+r2) over the required interval. > > Here is how I did it evaluating over the 2-dimensional tolerance region > around r1=r10 and r2=r20. I assume each resistor independently has a > tolerance +/-a which is uniformly distributed over an interval of length 2a, > which gives a probability density 1/(4a^2) over the 2-dimensional tolerance > region for the pair of resistors. I had to evaluate indefinite integrals and > then substitute in limits because definite integration seemed to lock > Mathematica up (I have version 5 for Windows). > > Integrate[((r1*r2)/(r1 + r2))*(1/(4*a^2)), r1] > Simplify[(% /. {r1 -> r10 + a}) - (% /. {r1 -> r10 - a})] > Integrate[%, r2] > Simplify[(% /. {r2 -> r20 + a}) - (% /. {r2 -> r20 - a})] > > -- > Steve Luttrell > West Malvern, UK > > "Oliver Friedrich" <oliver.friedrich at tzm.de> wrote in message > news:blcq5l$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 > > > >