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How did they get the values for E,Z and H? And why did they add the e exponential with -jkz?

How did they get the values for E,Z and H? And why did they add the e exponential with -jkz?

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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How did they get the values for E,Z and H? And why did they add the e exponential with -jkz?
Air Medium 1 (0,14) Metal block E- Medium 2 (02,82.42) H E, (Incident wave) H E. (Transmitted wave) H. (Reflected wave) := 0 E, = incident electric field H,= incident magnetic field for medium 1. E; = Epe-jkz For medium 1, wave impedance is Zair for medium 2, wave impedance is Zm z = Eiojkiz H Zair
Transcribed Image Text:Air Medium 1 (0,14) Metal block E- Medium 2 (02,82.42) H E, (Incident wave) H E. (Transmitted wave) H. (Reflected wave) := 0 E, = incident electric field H,= incident magnetic field for medium 1. E; = Epe-jkz For medium 1, wave impedance is Zair for medium 2, wave impedance is Zm z = Eiojkiz H Zair
E, = Eoe-jkz %3D H, = H1we¬ikrz %3D Incidence is normal, hence by boundary condition, at z=0, Htan 1 Htan 2 H = H, Eio-jkiz Zair E10 -jk2z Zm
Transcribed Image Text:E, = Eoe-jkz %3D H, = H1we¬ikrz %3D Incidence is normal, hence by boundary condition, at z=0, Htan 1 Htan 2 H = H, Eio-jkiz Zair E10 -jk2z Zm
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