{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Workshop10BounceP2

# Workshop10BounceP2 - V z,t = 0 = 1 3 V and V z,t = 0...

This preview shows page 1. Sign up to view the full content.

ECE 3030: Electromagnetic Fields and Waves Fall 2009 WORKSHOP 10 INSTRUCTOR NOTES: Bounce Diagrams, Waveguides Now we will use first method as outlined in the lectures. You might also compare this solution with that for Problem 13.33 given in Homework 10. V T ( z, t ) = V + ( z, t ) + V - ( z, t ) V - ( z = 0 , t ) = V + ( z = 0 , t L V + ( z = - ‘, t ) = V - ( z = - ‘, t S + V S Z 0 R L + Z 0 The initial values of the reflection coefficients are Γ S = 0 . 5 and Γ L = 0 . 5, and then after the switch closes Γ S = - 1. For the time before the switch closes, we have V T ( z, t ) = V + ( z, t ) + V - ( z, t ) = 1 V - ( z = 0 , t ) = 1 2 V + ( z = 0 , t ) V + ( z = - ‘, t ) = 1 2 V - ( z = - ‘, t ) + 2 50 150 + 50 Solving, we have V T ( z, t < 0) = 1 V, V + ( z, t < 0) = 2 / 3 V, and V - ( z, t < 0) = 1 / 3 V. When the switch closes, we still have
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: V-( z,t = 0 + ) = 1 / 3 V and V + ( z,t = 0 + ) changes to V + ( z,t = 0 + ) = V-( z,t = 0 + )Γ S = 1 / 3(-1) =-1 / 3 V leaving zero volts on the line. Once this wave reaches the load, V-( z,t = ‘/v + ) changes to V-( z,t = ‘/v + ) = V + ( z,t = ‘/v + )Γ L =-1 / 3(1 / 2) =-1 / 6 V putting-1 / 2 volt on the line. When this wave gets back to the source end, the new V + value is +1/2 V again canceling the voltage on the line. [The extra page provided 09/11/18.] CORNELL UNIVERSITY c ± WES SWARTZ (09/8/14) 10–2...
View Full Document

{[ snackBarMessage ]}