Step 1 Ejercicio Practico 1 MarioDeLosRios

NATIONAL UNIVERSITY OPEN AND DISTANCE (UNAD)

Electromagnetic theory and waves

Unit 1 - Step 2

Activity:

Recognize the electrodynamic and waves applications

Group: 203058_3

Produced by:

Mario Jose De Los Rios Torres Code: 1102851098

Presented to: Wilmer Hernan Gutierrez

Course tutor

Place and Date: Corozal Sucre March 07, 2018

Choose one of the following problems, solve it and share the solution in the forum. Perform a critical analysis on the group members’ contributions and reply this in the forum.

4. For a medium with the same electromagnetic characteristics than the third problem, find the losses per length unit for a 500MHz signal. If the original signal has an electric field of 120Vrms/m. Find the losses in watts when the signal travels 10m in the medium.

 = 9  = 1.6  = 1.210− / ,

 and

Para un medio con las mismas características electromagnéticas que el tercer problema, encuentre las pérdidas por unidad de longitud para una señal de 50 0MHz. Si la señal original tiene un campo eléctrico de 120 Vrms / m. Encuentra las pérdidas en vatios c uando la señal viaja 10 m en el medio.

Medium

 

Non dissipative

≈0

      √   0  √   

Lossless dielectric

≤ 0.1

Dielectric with losses

0.1 <  < 10 ≥ 10

 √  √  +  √ √   2√    √      +    

Solución: First we define the constants

Magnetic permeability

Good conductor

 =    = 361   10−/  =   = 

Electric permittivity

 =   

We apply the data:

5    1. 2 ∗ 10  =  = 2 ∗ ∗5 ∗10 ∗ 9∗ 361  ∗ 10−  = 0.000048

It is a lossless dielectric

 =   7 1. 6 ∗ 4  10  =  9 ∗ 361  ∗ 109 = 158.95

 = 2 5  158.95    1. 2 10  = 2 = 0.0009537 % = 1−100% % = (1−.∗)100% = 1.88%  =  / =  = 120  120  = 158.95 = 90.59

Solución: First we define the constants

Magnetic permeability

Electric permittivity

 =    = 361   10−/  =   = 

 =    We apply the data:

 =   =  = 2∗  ∗2.5 ∗ 10 ∗4 80∗ 361  ∗ 10−  = 0.36

It is a dielectric with losses

 = √  +   = 