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329sp09hw11sol - ECE-329 Spring 2009 Homework 11 Solution 1...

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ECE-329 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 = 50 Ω , 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
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ECE-329 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 0 1 2 3 3 2 1 0 1/2 z t [ μ s] 1/6 Γ CS = 0 Γ CL = - 1 3 0 1 2 3 3 2 1 0 1/100 z -1/300 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 ) . b) According to the previous result, the impulse voltage response at z = l 2 is h v ( t ) = 1 2 δ ( t - 0 . 5 μ s ) + 1 6 δ ( t - 1 . 5 μ s ) .
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