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Unformatted text preview: 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 e iωt + γx and i ( x,t ) = I e iωt + γx , when put into either of the above two ’equations of motion’, give γ 2 = LC (- ω 2 ) + ( RC + GL )( iω ) + 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 )( iω ) + RG we get α 2- k 2 = RG- LCω 2 and 2 αk = ( RC + GL ) ω . Thus we obtain for k....
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