Telegrapher's Equation in Real Lines 2.3 2001/06/13 2005/07/18 14:15:22.819 GMT-5 Bill Wilson wlw@rice.edu CJ Ganier seejaie@rice.edu Bill Wilson wlw@rice.edu attenuation coefficient complex propagation constant telegrapher's equation Explains the behavior of "real lines" in terms of the telegrapher's equation. So far, the transmission lines we have looked at have been "ideal". That is they have been lossless and dispersionless. Lest you leave the course with a false idea of how things really work, we should go back to our model and try to get things adjusted just a bit. As you can probably imagine, a real transmission line is going to have some series resistance, associated with the real losses in the copper wire. There may also be some shunt conductance, if the insulating material holding the two conductors has some leakage current. We will need to include these effects along with the distributed inductance and capacitance which we have already talked about. Fixing up the model accordingly, we now draw a section of line Δ x long as shown in . Taking the same voltage loop and current sum that we did back in the discussion of transmission lines, we come up with the following version of the telegrapher's equations. x V x t R I x t L t I x t and x I x t G V x t C t V x t

Real Line Diagram A model for a line with losses.

Clearly, we would like to simplify things if we can. Let's again make a sinusoidal time excitation assumption, and let I x t and V x t become phasors. Since the time variation is now represented by a simple ω L the time derivatives become just ω . We have x V x R ω L I x and x I x G ω C V x The way to get a solution is, of course, just like we have always done. Take the derivative with respect to x of x 2 V x R ω L x I x and then plug in x 2 V x R ω L G ω C V x The obvious solution to this (See how easy this gets after you've done it once or twice) is V x V 0 ± γ x with γ R ω L G ω C This number, γ is called the complex propagation constant. Obviously, in general, it will have both a real and an imaginary part: γ α β and we have V x V 0 ± α β x Let's choose the minus sign in the exponent, and write the two terms as a product. V x V 0 α x β x We see we have something similar to what we had before, but with just a minor difference. The β x term is the propagating term which tells us how the phase angle of the phasor changes as we move along the line, and acts just like the β term we had before. Thus β 2 λ and ν p ω β The α is called the attenuation coefficient, and obviously, the α x term in causes the amplitude of the wave to decrease as it moves down the line. is a sketch of what a wave would look like if it is both propagating down the transmission line and also being attenuated. In a distance 1 α the amplitude of the propagating wave has fallen to -1 of the value it had when it started.

Wave Decay Sketch of a decaying wave on a transmission line.

Let's take the minus sign solution in and substitute back into x V x γ V 0 γ x R ω L I x From which we get I x γ R ω L V 0 γ x R ω L G ω C R ω L V x G ω C R ω L V x Thus we can say V x Z 0 I x where Z 0 R ω L G ω C R 0 X 0 In general, in order to find α, β, R 0 , and X 0 , we would have to find the square root given in and for specific values of R, L, G, and C. On the other hand, we could maybe come up with some reasonable approximations which might suffice for cases of real interest. Obviously, if a line is very lossy, we would not be very interested in using it, and so except in some very special cases where an extremely lossy line is unavoidable (usually having to do with signals at very high frequencies) we might see if we can find a low loss approximation.