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Unformatted text preview: ECE329 Spring 2009 Homework 11 — Solution April 22, 2009 1. The voltage v ( z,t ) and current i ( z,t ) along a transmission line satisfy the set of partial differential equations (PDE’s) ∂v ∂z = L ∂i ∂t and ∂i ∂z = C ∂v ∂t . Combining both equations, we can find the following 2nd order PDE in terms of the voltage v ( z,t ) , ∂ 2 v ∂z 2 = LC ∂ 2 v ∂t 2 . a) Replacing v ( z,t ) = cos( ωt βz ) in the previous equation we find that ∂ 2 v ∂z 2 = β 2 cos( ωt βz ) and LC ∂ 2 v ∂t 2 =LC ω 2 cos( ωt βz ) . The equality holds if and only if β = ω √ LC . b) To find the corresponding current, we first replace v ( z,t ) in ∂i ∂t = 1 L ∂v ∂z and get ∂i ∂t = β L sin( ωt βz ) . Then, integrating over t , we obtain i ( z,t ) = β ω L cos( ωt βz ) . 2. Let us consider a T.L. with characteristic impedance Z o = 50Ω , length l = 300 m , and propagation velocity v p = c = 3 × 10 8 m / s . A voltage source f ( t ) with internal resistance R S = Z o is connected at z = 0 and a load resistance R L = 2 Z o is placed at z = l as shown in the following diagram. R S = 50 Ω R L = 100 Ω f ( t ) z = 0 m z = 300 m Z o = 50 Ω v p = c a) For f ( t ) = δ ( t ) , the initial voltage at the input of the T.L. is given by the voltage divider Z o R S + Z o δ ( t ) = 1 2 δ ( t ) . In addition, the voltage reflection coefficients at the source and at the load are respectively Γ V S = R S Z o R S + Z o = 50 50 50 + 50 = 0 Γ V L = R L Z o R L + Z o = 100 50 100 + 50 = 1 3 . 1 ECE329 Spring 2009 The corresponding current reflection coefficients are Γ CS = Γ V S = 0 Γ CL = Γ V L = 1 3 . Using these information, we can build the following "bounce diagrams" for the voltage v ( z,t ) and current i ( z,t ) on the transmission line. z = 0 m z = 300 m Γ V S = 0 Γ V L = 1 3 1 2 3 3 2 1 1/2 z t [ μ s] 1/6 Γ CS = 0 Γ CL = 1 3 1 2 3 3 2 1 1/100 z1/300 z = 0 m z = 300 m (a) (b) Figure 1: (a) Voltage and (b) current bounce diagrams for voltage source f ( t ) = δ ( t ) . The corresponding expressions for v ( z,t ) and i ( z,t ) are v ( z,t ) = 1 2 δ ( t z c ) + 1 6 δ ( t + z c 2 μ s ) i ( z,t ) = 1 100 δ ( t z c ) 1 300 δ ( t + z c 2 μ s ) ....
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This note was uploaded on 11/16/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 FRANKE
 Electromagnet, Volt

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