Demo10Transient_1

# Demo10Transient_1 - ECE 3030: Electromagnetic Fields and...

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ECE 3030: Electromagnetic Fields and Waves Fall 2009 DEMO 10 INSTRUCTOR NOTES: Transmission Line Transients and Dielectric Slabs Reminder: Date: Tuesday 11/3 1. Problem 13.40 The switch in the circuit shown in Figure 1 is closed at t = 0. a. Find the reﬂection coeﬃcients at both ends. Ω = 50 o Z V 75 0 = z " - = z z Ω 25 Short circuit Figure 1: Transmission line with a shorted load. b. What is the ﬁnal voltage and current in the load after a long time? c. Construct the appropriate bounce diagram and answer table following the models shown below in Figures 2 and 1. d. Plot the current at the load as a function of time in units of the one-way bounce time. Solution: This is a problem with transients on a short circuited transmis- sion line showing how the current builds up through the short. a. The reﬂection coeﬃcients are Γ S = - 25 - 50 25 + 50 = - 0 . 333 and Γ L = 0 - 50 0 + 50 = - 1 b. Final states: V ( t = ) = 0 V and I ( t = ) = 75 / 50 = 1 . 5 A c. Fill out the table and add the numbers to the bounce diagram and ﬁll out the table. The bounce diagram can either show the total voltages and total currents after each wave, or show the component voltages and currents of each wave as given in the table. d. Plot of the current through the load. CORNELL UNIVERSITY c ± WES SWARTZ (09/10/21; Rev.09/11/16) 10–1

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ECE 3030: Electromagnetic Fields and Waves Fall 2009 DEMO 10 INSTRUCTOR NOTES: Transmission Line Transients and Dielectric Slabs Source Voltage wave component Current wave component Load Time Γ S = - 1 3 at z = ‘/ 2 z = ‘/ 2 Γ L = - 1 t = < 0 V S = 0 V T = 0 V I T = 0 A I L = 0 A t = 0 + V S = 50 V 1+ = 0 V I 1+ = 0 A I L = 0 A t = + 2 v V S = 50 V 1+ = 75 · 50 / (25 + 50) = 50 V I 1+ = 50 / 50 = 1 A I L = 0 A t = 3 + 2 v V S = 50 V 1 - = 50 · ( - 1) = - 50 V I 1 - = - ( - 50)(50) = 1 A I L = 2 A t = 5 + 2 v V S = +16 . 667 V 2+ = - 50 / ( - 3) = +16 . 667 I 2+ = - 0 . 333 A I L = 2 A t = 7 + 2 v V S = +16 . 667 V 2 - = 16 . 667( - 1) = - 16 . 667 I 2 - = +0 . 333 A I L = 2 . 667 A t = 9 + 2 v V S = 5 . 556 V 3+ = 16 . 667 / 3 = 5 . 556 I 3+ = 0 . 111 A
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## This note was uploaded on 01/16/2010 for the course ECE 3030 at Cornell University (Engineering School).

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Demo10Transient_1 - ECE 3030: Electromagnetic Fields and...

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