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![]() Cecil Moore wrote: Reg Edwards wrote: To clear away all misconceptions and confusion, try returning to square one and begin again with a clean slate. dV/dz = -(R+j*Omega*L)*I dI/dz = -(G+j*Omega*C)*V Those are essentially the differential equations used by Ramo and Whinnery to develop the following exponential equations for load matched transmission lines (no reflections): V = V+(e^-az)(e^-jbz) I = V+(e^-az)(e^-jbz)/Z0 where 'a' is the attenuation factor. So why does an attenuated voltage drop while a current, attenuated by exactly the same percentage, doesn't drop? Depends on whether 'a' is in series or in shunt. 73, ac6xg |
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