From: Dorin Oprea (firstname.lastname@example.org)
Date: Fri Jun 23 2000 - 11:03:22 PDT
I believe the signal loop area is the key of this puzzle (no losses in
consideration). A lower dielectric permitivity allows getting the same
value within a smaller spacing between the conductor and the reference
(shield). Smaller distance between the two conductors means smaller loop
and therefore, less EM radiation = energy loss (Henry Ott) and also
Smaller L and smaller C means also faster transportation medium (coax
but L/C is still the same.
Hassan Ali wrote:
> I attended a presentation by a high-frequency (1GHz < f < 65GHz) coaxial
> cable vendor, and the presenter claimed that their cables use a material
> with a very low dielectric constant and therefore are ideal for high-speed
> application as they give rise to low capacitive loading. He gave a formula
> showing the capacitance (I think per unit length) decreasing as you decrease
> dielectric constant. This claim, however, perplexed me as I don't know how a
> cable's capacitance per unit length would give rise to a capacitive loading.
> All I know from my transmission line classes, a lossless transmission line
> with Z0 = sqrt(L/C) would transmit signals exactly the same way regardless
> of the value of the p.u.l. capacitance C as long as the ratio L/C is
> maintained. Am I missing something here?
> Hassan Ali <email@example.com>
> Equipment & Network Interconnect, Nortel Networks
> 2 Brewer Hunt Way, Kanata ON, K2K 2B5 Canada
> Tel: 613-765-1410 (ESN 395) Fax: 613-765-5512 (ESN 395)
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