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Unformatted text preview: Derivation of the Telegraph Equation Model an infinitesmal piece of telegraph wire as an electrical circuit which consists of a resistor of resistance R dx and a coil of inductance L dx. If i(x, t) is the current through the wire, the voltage across i the resistor is iR dx while that across the coil is t L dx. Denoting by u(x, t) the voltage at position x and time t, we have that the change in voltage between the ends of the piece of wire is du = -iRdx -
i t Ldx Suppose further that current can escape from the wire to ground, either through a resistor of conductance G dx or through a capacitor of capacitance C dx. The amount that escapes through the resistor is uG dx. C dx 1/(G dx) x L dx R dx x + dx Because the charge on the capacitor is q = uC dx, the amount of current that escapes through the capacitor is qt = ut C dx. In total di = -uGdx - ut Cdx Dividing by dx and taking the limit dx 0 we get the differential equations ux + Ri + Lit = 0 Cut + Gu + ix = 0 (1) (2) Solving (2) for ix gives ix = -Cut - Gu. Substituting this and its consequence ixt = -Cutt - Gut into x (1), which is uxx + Rix + Lixt = 0, gives uxx + R(-Cut - Gu) + L(-Cutt - Gut ) = 0 Dividing by LC and moving some terms to the other side of the equation gives
1 LC uxx = utt + R L + G C ut + GR LC u Renaming some constants, we get the telegraph equation utt + ( + )ut + u = c2 uxx where c2 =
1 LC = G C = R L January 3, 2007 Derivation of the Telegraph Equation ...
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This note was uploaded on 01/27/2011 for the course MATH 267 taught by Professor Yilmaz during the Spring '10 term at The University of British Columbia.
- Spring '10