# Step 1_Ejercicio_Practico_1_MarioDeLosRios.docx

Description
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 Sucr
Categories
Published

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
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 srcinal signal has an electric field of 120Vrms/m. Find the losses in watts when the signal travels 10m in the medium.    = 9 ,    = 1.6  and  = 1.210 −  /   Para un medio con las mismas características electromagnéticas que el tercer problema, encuentre las pérdidas por unidad de longitud para una señal de 500MHz. Si la señal srcinal tiene un campo eléctrico de 120 Vrms / m. Encuentra las pérdidas en vatios cuando la señal viaja 10 m en el medio. Medium Non dissipative Lossless dielectric Dielectric with losses Good conductor    ≈ 0   ≤ 0.1   0.1 <  < 10   ≥ 10       √     √    √   +    √         0   2      √        √    √        √                   +           Solución: First we define the constants  =          = 136  10 − /    =    Magnetic permeability  =       Electric permittivity  =          We apply the data:  =  = 1.2∗ 10 5 2 ∗ ∗5 ∗10   ∗9∗ 136 ∗ 10 −    = 0.000048   It is a lossless dielectric  =      =   1.6 ∗ 4  10 7 9 ∗ 136 ∗ 10 9  = 158.95    = 2    =  1.210 5   158.952 = 0.0009537   % = 1 − 100%   % = (1 −.∗ )100% = 1.88%    =   /    =     = 120    = 120  158.95 = 90.59      Solución: First we define the constants  =          = 136  10 − /    =    Magnetic permeability  =       Electric permittivity

Apr 16, 2018

#### 1-s2.0-S2092678216303545-main.docx

Apr 16, 2018
Search
Similar documents

View more...
Tags

Related Search