Talk:Heaviside condition

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The article says that "the condition is satisfied when the ratio of conductance G (in siemens) to capacitance C (in farads) equals the ratio of resistance R (in ohms) to inductance L (in henries)"

     G/C = R/L

But, G=1/R, therefore

G/C = R/L <=> 1/RC = R/L <=> C = L

So, it appears that resistance is actually irrelevant. Am I missing something? Nikola 18:22, 15 November 2005 (UTC)

Of course I am. It's hard to work with slashes, so actually L/C=R^2. Resistance is relevant. Nikola 05:05, 22 January 2006 (UTC)

Also, G is not 1/R. R represents the series resistance per unit length; it's (approximately, ignoring G) what you'd get if you short-circuited one end of a length of cable, and measured the other end with an ohmmeter. G represents the shunt resistance; it's (approximately, ignoring R) what you'd get if you open-circuited the cable and repeated your measurement. They just happen to be expressible in the same units (ohms, or 1/ohms). It makes the math work out nicer to express the series resistance in ohms, and the shunt resistance in 1/ohms. —The preceding unsigned comment was added by 66.30.10.35 (talk • contribs) 17:01, 3 December 2006 (UTC).
R is the series resistance component in the equivalent circuit model of a non-ideal transmission line, and G is the shunt conductance in the same model. Physically, the R and the G arise due to completely different reasons: R is due to the small, yet finite, resistivity in the non-ideal metal conductor(s) of the line, whereas G is due to the small, yet again finite, conductivity of the non-ideal dielectric material between the metal conductor(s) of the line. Therefore, in a lossless transmission line, where the metal is a perfect conductor and the dielectric is a perfect insulator, the metal exhibits no resistivity and the dielectric exhibits no conductivity. Hence, the characteristic impedance of the lossless line reduces to Z0 = sqrt(L/C).
To look at it another way, the non-ideal metals and dielectrics that are used in transmission lines (which is practically the case) are the root cause of signal distortion, but by implementing Heaviside's condition distortionless transmission lines can be designed even with lossy materials.
NOTE: I keep using the terms "lossy" and "lossless." "Lossy" simply means the transmission line has some R and/or G associated with it, and since any resistance when none is wanted (or conversely any conductance when non is wanted) results in loss of some of the energy of a signal, a signal propagating through a "lossy" transmission line will lose some energy. On the other hand a "lossless," i.e., perfect transmission line will cause no loss of signal energy.
Idunno271828 (talk) 09:25, 21 May 2008 (UTC)