Max Schmitt wrote:
>
> > impedance of a coax line having a symmetrically placed infinitely
> > thin strip as its centre conductor?
>
> Z = 138 * log ( 2D/b ) ohm
>
> where D = inside diameter of outer circular conductor
>
> b = width of thin metal strip
>
> condition ; D/b > 2
Please note that Max means the 10-based logarithm here, and also that
the factor sqrt(mu_r/eps_r) for material other than air is missing.
The expression in my previous post, when using D and b, becomes:
Zline = Z0 Sqrt[mur/epsr] * EllipticK[1-m] / (4 EllipticK[m])
with: m = 4 / (b/D+D/b)^2
It can be series-expanded to give (for air:)
Zline = Z0/(2Pi) * [ log(2D/b) - 1/8 (b/D)^4 + O((b/D)^6) ]
= 138.0595 * log_10(2D/b) - 7.49481 * (b/D)^4 + ...
Keeping only the firtst term, as Max does, will indeed be good enough
for b/D<0.5, since the error is then about 0.5 percent. For strips
that come closer to the wall, however, the exact expression should be
preferred, since the second term increases as a 4th power!
In a case like this, the main purpose of the exact solution is to see
what the error in the simplified expression is (and that is important
information) and not to do the actual calculation.
Although, on the other hand, if you have a math-package which will
compute the elliptic integral stuff, then it's nice just for once to
be able to use it. :-)
Cheers,
Jos
-- Dr. Jozef R. Bergervoet Electromagnetism and EMC Philips Research Laboratories, Eindhoven, The Netherlands Building WS01 FAX: +31-40-2742224 E-mail: bergervo_at_natlab.research.philips.com Phone: +31-40-2742403Received on Tue Apr 25 2000 - 05:10:05 EDT
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