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Post task - Final activity Individual work

Luis Miguel Cortes Rodriguez Group 203058_27 1.061.718.214

UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA UNAD Escuela de Ciencias Básicas, Tecnología e Ingeniería Teoría Electromagnética y Ondas 2020 16-01

Introduction Apply the concepts of physics and electromagnetism to determine the parameters of the behavior of electromagnetic waves when they propagate in different media, using mathematical methods and the knowledge of the previous units, for the solution of the application exercises..

Application exercises: For the development of the following exercises, note that 27 corresponds to the group number and 106 to the first 3 digits of the identification number. 1. An electromagnetic wave of 𝑓 = 106 𝑀𝐻𝑧 is transmitted from the bottom of a ship to a receiver located at 𝑝 = 1 𝐾𝑚 depth. The wave is emitted with an advance angle of 𝑎 = (5 + 27)°. Determine the time it takes for the wave to reach the receiver. For development, follow the following steps: a. Calculate the tangent of losses 𝑇𝑎𝑛(𝛿). Table 1: Conductivity

𝛔 and electrical permittivity 𝛆𝐫 of some media.

Media

𝛔 [𝐒⁄𝐦]

𝛆𝐫 [ ]

5.80𝑥107

1

4

80

3. Vegetable soil

1.00𝑥10−2

14

4. Dry soil

1.00𝑥10−4

3

5. Sweet water

1.00𝑥10−3

80

1. Copper 2. Sea water

𝛆𝐨 = 8.8542x10

𝑇𝑎𝑛(𝛿) =

−12 2

𝐶 /N𝑚

2

𝜎 𝜎 4 𝑆⁄𝑚 = = = 𝟖, 𝟒𝟕𝟖𝟖 𝜔𝜀 2𝜋𝑓εr ε𝑜 2𝜋 ∗ 106𝑥106 𝐻𝑧 ∗ 80 ∗ 8.8542x10−12 𝐶 2 ⁄𝑁𝑚2 𝛿 = 𝑇𝑎𝑛−1 (8,4788) = 𝟖𝟑, 𝟐𝟕°

b. Classify the behavior of the medium. Table 2: Classification of propagation media. Media

Tan(δ)

δ [°]

1. Perfect dielectrics

(Not dissipative)

Tan(δ) = 0

δ = 0°

2. Good insulators

(Lost low dielectric)

0 < Tan(δ) 10

δ = 90°

c. Calculate the propagation parameters of the wave 𝛾, 𝛼 and 𝛽. Table 3: Propagation parameters in open media. Parameter

Not dissipative

Lost low dielectric

Dielectrics with losses

Good conductors



𝑗𝜔√𝜇𝜀

𝑗𝜔√𝜇𝜀

√𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀)

√𝑗𝜔𝜎𝜇𝑜



0

𝜎𝜂⁄2

𝑅𝑒()

√𝜋𝑓𝜎𝜇𝑜



𝜔√𝜇𝜀

𝜔√𝜇𝜀

𝐼𝑚()

√𝜋𝑓𝜎𝜇𝑜



√𝜇⁄𝜀

√𝜇⁄𝜀

√𝑗𝜔𝜇⁄(𝜎 + 𝑗𝜔𝜀)

√𝑗𝜔𝜇𝑜 ⁄𝜎

Propagation constant  (gamma). √𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀) √𝑗2𝜋 ∗ 106 ∗ 106 𝐻𝑧 ∗ 1 ∗ 1,2566 ∗ 10−12 𝑁/𝐴2 (4𝑆/𝑚 + 𝑗2𝜋 ∗ 106 ∗ 106 𝐻𝑧 ∗ 80 ∗ 8.8542 ∗ 10−12 𝐶 2 /𝑁𝑚2 )

 = 38.5749010252255 + 43.3918297856375𝑗 𝛂 = 38.5749 𝑁𝑝/𝑚 𝛃 = 43.3918 𝑅𝑎𝑑/𝑚

d. Calculate the propagation speed of the 𝑉𝑝 wave. 𝑉𝑝 =

𝑉𝑝 =

𝜔 𝛽

2𝜋 ∗ 106 ∗ 106 𝐻𝑧 = 1.535 ∗ 107 𝑚/𝑠 43.3918 𝑅𝑎𝑑/𝑚

e. Calculate the distance between the ship and the receiver 𝑑 .

𝐶𝑜𝑠(𝛼) =

𝑑=

𝑝 𝑝 = 𝑑= 𝑑 𝐶𝑜𝑠(𝛼)

1000𝑚 = 𝟏𝟏𝟕𝟗 𝒎 𝐶𝑜𝑠(32°)

f. Based on 𝑉𝑝 and 𝑑 determine the time 𝑡 of the route. 𝑋 =𝑡∗𝑣

𝑡=

=

𝑡=

𝑋 𝑣

1179𝑚 = 7.680 ∗ 10−5 𝑠 1.535 ∗ 107 𝑚/𝑠

Figure 1: wave propagation in open media. Image recovered from https://www.ee.co.za/article/new-economics-marine-environmentalmonitoring.html

2. From an airplane, which is 1250 𝑚 high, a communication signal 𝑓 = 106 𝑀𝐻𝑧 is emitted to a submarine that is 800 𝑚 deep, the angle of incidence of the signal on the sea surface is 𝑎 = (5 + 27)°. Determine the time it takes for the signal to reach the submarine. For development, follow the following steps: a. Calculate the distance between the plane and the point of incidence at sea 𝑑1 .

𝑆𝑒𝑛(𝛼) =

𝑑1 =

ℎ1 ℎ1 = 𝑑1 = 𝑑1 𝑆𝑒𝑛(𝛼)

1250𝑚 = 𝟐𝟑𝟓𝟖. 𝟖𝟓𝒎 𝑆𝑒𝑛(32°)

b. Calculate the velocity of propagation of the wave 𝑉𝑝1 in the air (𝑉𝑝 = 𝐶𝑜/𝑛) 𝑛=

𝐶𝑜 3 ∗ 108 𝑚/𝑠 = = 𝟐. 𝟗𝟗𝟏𝟑𝟐𝟓 ∗ 𝟏𝟎𝟖 𝒎/𝒔 𝑉𝑝 1,00029

c. Using Snell's Law, calculate the angle of refraction of the wave in the sea. 90 − 32° = 58° 𝑛=

𝐶𝑜 3 ∗ 108 𝑚/𝑠 = = 𝟏𝟗, 𝟓𝟒𝟑 𝒎/𝒔 𝒔𝒆𝒂 𝒓𝒆𝒇𝒓𝒂𝒄𝒕𝒊𝒐𝒏 𝒊𝒏𝒅𝒆𝒙 𝑉𝑝 1,535 ∗ 107 𝑚/𝑠 𝑛1 𝑠𝑒𝑛(𝜃1 ) = 𝑛2 𝑠𝑒𝑛(θ2 ) 1.00029 ∗ 𝑠𝑒𝑛(58°) = 19.543 ∗ 𝑠𝑒𝑛(θ2 ) n1 θ2 = 𝑠𝑒𝑛−1 ( 𝑠𝑒𝑛(θ1 )) 𝑛2 1.00029 θ2 = 𝑠𝑒𝑛−1 ( 𝑠𝑒𝑛(58°)) = 𝟐, 𝟒𝟗° 19.543

d. Calculate the distance between the point of incidence in the sea and the submarine. 𝐶𝑜𝑠(𝜃2 ) =

𝑑2 =

ℎ2 ℎ2 = 𝑑2 = 𝑑2 𝐶𝑜𝑠(𝜃2 )

800 𝑚 = 𝟖𝟎𝟎. 𝟕𝟓𝒎 𝐶𝑜𝑠(2,49°)

e. Calculate the propagation speed of the wave 𝑉𝑝2 at sea (it is 𝑉𝑝 from exercise 1). 𝑉𝑝 =

𝑉𝑝 =

𝜔 𝛽

2𝜋 ∗ 106 ∗ 106 𝐻𝑧 = 1.535 ∗ 107 𝑚/𝑠 43.3918 𝑅𝑎𝑑/𝑚

f. Based on 𝑉𝑝1 and 𝑑1 determine the time 𝑡1 of the first path (𝑡1 = 𝑑1 /𝑉𝑝1 ). 𝑋 =𝑡∗𝑣

𝑡=

=

𝑡=

𝑋 𝑣

2358.85𝑚 = 𝟕. 𝟖𝟖𝟓𝟔 ∗ 𝟏𝟎−𝟔 𝒔 2.991325 ∗ 108 𝑚/𝑠

g. Based on 𝑉𝑝2 and 𝑑2 determine the time 𝑡2 of the second path. 𝑋 =𝑡∗𝑣

𝑡=

=

𝑡=

𝑋 𝑣

800.75𝑚 = 𝟓. 𝟐𝟏𝟕𝟐 ∗ 𝟏𝟎−𝟓 𝒔 2.991325 ∗ 108 𝑚/𝑠

h. Calculate the total time of the route 𝑡 = 𝑡1 + 𝑡2 . 7.8856 ∗ 10−6 𝑠 + 5.2172 ∗ 10−5 𝑠 = 𝟔. 𝟎𝟎𝟓𝟕 ∗ 𝟏𝟎−𝟓

Figure 2: wave propagation in bounded open media. Image recovered from https://byjus.com/physics/characteristics-of-sound-wavesamplitude/

3. A lossless transmission line has a characteristic impedance of 𝑍0 = 75Ω, a length of 𝐿 = 5𝑚 and is excited by a signal of 𝑓 = 500 𝑀𝐻𝑧. The line is connected to an antenna with load impedance 𝑍𝐿 = (45 + 𝑗45)Ω. Taking into account that 𝑉𝑝 =

3𝑥107 𝑚/𝑠, calculate: a. Wavelength 𝜆. 6

𝑓 = 500 𝑀𝐻𝑧 = 500 ∗ 10 𝐻𝑧 7 𝑉𝑝 = 3𝑥10 𝑚/𝑠 𝑉𝑝 = 𝜆𝑓

𝜆=

𝑉𝑝

𝑓

𝜆=

3 ∗ 107 𝑚/𝑠

500 ∗ 106 𝐻𝑧

= 𝟎. 𝟎𝟔 𝒎

b. Electrical length ℓ. ℓ=

𝐿 5𝑚 = = 𝟖𝟑. 𝟑 𝒍𝒂𝒎𝒃𝒅𝒂 𝜆 0.06 𝑚

c. Input impedance 𝑍𝑖𝑛 . 𝑍𝑖𝑛 = 𝑍0

𝑍𝑖𝑛 = 75Ω

𝑍𝑙 + 𝑖𝑍0 tan(2𝜋𝜆) 𝑍0 + 𝑖𝑍𝑙 tan(2𝜋𝜆)

(45 + 𝑗45Ω) + 𝑗(75Ω)tan(2𝜋 ∗ 83.3) 75Ω + 𝑗(45 + 𝑗45Ω)tan(2𝜋 ∗ 83.3)

𝒁𝒊𝒏 = 𝟑𝟒. 𝟒𝟔𝟎𝟒𝟗𝟏𝟏𝟏𝟕𝟎𝟔𝟔𝟕 + 𝟐𝟒. 𝟒𝟐𝟐𝟗𝟒𝟐𝟗𝟒𝟓𝟗𝟐𝟒𝒊 = 𝟒𝟐. 𝟐 < 𝟑𝟓. 𝟑𝟐° d. Reflection coefficient Γ (magnitude and phase). Γ=

Γ=

𝑍𝑙 − 𝑍0 𝑍𝑙 + 𝑍0

45 + 𝑗45Ω − 75Ω 45 + 𝑗45Ω + 75Ω

Γ = −0.0958904109589 + 0.4109589041096𝑗 = 0.422 < 103.13°

e. VSWR 𝑉𝑆𝑊𝑅 =

𝑉𝑆𝑊𝑅 =

1 + |Γ| 1 − |Γ|

1 + |0.422| = 𝟐. 𝟒𝟔𝟎𝟐 1 − |0.422|

f. Check the results c, d and e with the Smith 4.1 software.

g. Using the Smith 4.1 software, get an input impedance that is only real.

Figure 3: Transmission line.

Conclusions Using mathematical methods, it was possible to characterize and analyze an electromagnetic wave when it propagates freely through some medium, it was possible to observe how it behaves when it travels through the medium. Also, it can be concluded that when passing from one medium to another an electromagnetic wave generates loss of power, simply for changing material and it is possible that part of it will be reflected or refracted.

Too it is important to keep in mind that in reflection the incident angle is equal to the reflected angle, this indicates that the reflected wave when it obliquely hits does not change the speed nor its frequency, the only thing that changes is its direction. I learned that In electromagnetic waves, speed does not depend on power, it depends on the medium; the fastest is in a vacuum, where the refractive index is 1, but when it begins to rise when passing from one medium to another, the OEM is slower. The higher the refractive index, the slower it travels with respect to vacuum. Bibliography Quesada, M., & Maroto, J. (2014). Plane electromagnetic waves. Electromagnetic waves in free space. From Maxwell's Equations to Free and Guided Electromagnetic Waves: An Introduction for First-year Undergraduates. New York: Nova Science Publishers, Inc. (pp. 49-60). Recovered from http://bibliotecavirtual.unad.edu.co/login?url=http://search.ebscohost.com/login .aspx?direct=true&db=nlebk&AN=746851&lang=es&site=edslive&scope=site&ebv=EB&ppid=pp_49 Alternate bibliography is shared in Spanish for the exploration of the topics of unit 2 of the course. Paz, A. (2013). Ondas en medios abiertos acotados. Capítulo 7. Electromagnetismo para Ingeniería electrónica. pp.249-287 https://drive.google.com/drive/folders/1rzanqW4xM53fGC-eNT8AlMb34gunumBu Chen, W. (2005). The Electrical Engineering Handbook. Boston: Academic Press. (pp. 525-537). Recovered from https://bibliotecavirtual.unad.edu.co/login?url=http://search.ebscohost.com/login.asp x?direct=true&db=nlebk&AN=117152&lang=es&site=eds-live&scope=site Gutiérrez, W. (2017). The Smith http://hdl.handle.net/10596/13141

Chart

basics.

[Video].

Recovered

from