Demo10Transient_1 - ECE 3030: Electromagnetic Fields and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/16/2010 for the course ECE 3030 at Cornell University (Engineering School).

Page1 / 6

Demo10Transient_1 - ECE 3030: Electromagnetic Fields and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online