RF toolkit capacitance calculation

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In RF toolkit/transmission line objects, the capacitance per unit length is calculated using the acosh function:
Most texts use this equation:
Are these expressions equal? If so, how are they equal?

Accepted Answer

David Goodmanson
David Goodmanson on 3 Oct 2020
Edited: David Goodmanson on 3 Oct 2020
Hi Derek,
The exact expression is indeed the acosh expression used in Matlab, and we would expect no less. The log expression is an approximation, and it's a bit of a holdover from the times when scientific software was not readily available. acosh was not so easy to calculate, but log tables were common. Leaving out the factor of pi*epsilon, the approximation goes as follows. Using the variable Cinv for 1/C, then
Cinv = acosh(D/2a)
cosh(Cinv) = D/2a
(exp(Cinv) + exp(-Cinv))/2 = D/2a
Now if D/2a is large, then Cinv is large and you can drop the exp(-Cinv) term compared to the exp(Cinv) term. Then
exp(Cinv) = D/a
Cinv = log(D/a)
The natural question is, how large is large?
Da = 2:.001:20; % D over a
Cinv1 = acosh(Da/2);
Cinv2 = log(Da);
The plot shows a pretty good approximation for D/a greater than about 5.
David Goodmanson
David Goodmanson on 12 Nov 2020
Edited: David Goodmanson on 17 Nov 2020
Hi Derek,
It's good to check these things out as you are doing. Looks like they are using radius instead of diameter because that doubles the resistance, and in this situation you have two wires carrying equal current, not just one wire.

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