sol6 - PHYS 218 SOLUTION TO ASSIGNMENT 6 16 Mar 2007 By...

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PHYS 218 - SOLUTION TO ASSIGNMENT 6 16 Mar 2007 By Chung Koo Kim e-mail : [email protected] 1. Telegrapher’s Equation (a) Taking the derivative of the first equation with respect to x and using the second equation gives: (subscript taken out for convenience) 2 v ∂x 2 = - L 2 i ∂x∂t - R ∂i ∂x = - L ∂t - C ∂v ∂t - Gv - R - C ∂v ∂t - Gv = LC 2 v ∂t 2 + ( RC + GL ) ∂v ∂t + RGv (b) Similar steps as in (a) gives identical equation, with the current taking the place of voltage: 2 i ∂x 2 = LC 2 i ∂t 2 + ( RC + GL ) ∂i ∂t + RGi (c) The trial forms of solution v ( x, t ) = V 0 e iωt + γx and i ( x, t ) = I 0 e iωt + γx , when put into either of the above two ’equations of motion’, give γ 2 = LC ( - ω 2 ) + ( RC + GL )( ) + RG (d)-(e) As the general form we use γ = ± ( α + ik ). Then from γ 2 = ( α 2 - k 2 ) + 2 iαk = LC ( - ω 2 ) + ( RC + GL )( ) + RG we get α 2 - k 2 = RG - LCω 2 and 2 αk = ( RC + GL ) ω . Thus we obtain for k k 2 = ( LCω 2 - RG ) ± ( LCω 2 - RG ) 2 + ω 2 ( RC + GL ) 2 2 LCω 2 ± LCω 2 2 ( R ωL, G ωC -→ RG ω ( RC + GL ) ω 2 LC ) Taking the non-trivial solution (non-zero k) only, k = ω
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