You are given an incident and a reflected electric in region 1. E inc = 400e (-j

Question

You are given an incident and a reflected electric in region 1.

Einc = 400e(-j*5*pi*y) az v/m and Eref = 80e(j*5*pi*y) az v/m

a) Find the operating frequency

b) What is the intrinsic impedance of the material in region 2. Can you find it?

c) Find the incident, reflected and transmitter average pointing vectors

You are given an incident and a reflected electric in region 1. Einc = 400e^(-j*5*pi*y) az v/m and Eref = 80e^(j*5*pi*y) az v/m a) Find the operating frequency b) What is the intrinsic impedance of the material in region 2. Can you find it? c) Find the incident, reflected and transmitter average pointing vectors

Explanation / Answer

Given:-

Incident wave is represented by

Einc = Eie(-j*B*y)az v/m => Einc = 400e(-j*5*pi*y) az v/m

and

Reflected wave is represented by

Eref = -Ere(j*B*y)az v/m   =>   Eref = 80e(j*5*pi*y) az v/m

a) From the above two equations:- B = 5*pi rad/m

=> Wavelength (w)= 2*pi / B

=> w = 2*pi / 5*pi

=> w =  0.4 m

frequency (f) = c/w

=> f = 3*108 m/s/ 0.4 m

=> f = 7.8 *108 m

b) The intrinsic impedance of the wave is defined as the ratio of the incident and a reflected electric field amplitude

Intrinsic impedance of the material (Z0) = Ei/Er

Z0 = 400/ 80

Z0 = 50

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