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Student Harvey Meza Jacome Group 203058_20 Identification number 1091661053

UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA UNAD Escuela de Ciencias Básicas, Tecnología e Ingeniería Teoría Electromagnética y Ondas 2021 16-01

Exercises development Activity Application exercises: For the development of the following exercises, note that 𝐺𝐺 corresponds to the group number and 𝐶𝐶𝐶 to the last 3 digits of the identification number. 1. An electromagnetic wave of 𝑓 = 𝐶𝐶𝐶 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 + 𝐺𝐺)°. Determine the time it takes for the wave to reach the receiver.

Figure 1: wave propagation in open media. Image recovered from https://www.ee.co.za/article/new-economics-marineenvironmental-monitoring.html For development, follow the following steps: Data 𝑓 = 𝐶𝐶𝐶 𝑀𝐻𝑧 → 53 𝑀ℎ𝑧 𝑝 = 1𝐾𝑚 𝑎 = (5 + 𝐺𝐺)° → (5 + 20)° → 25° 𝜎= 4𝑠∕𝑚 𝜀𝑟 = 80 𝜀 = 𝜀𝑟 𝜀0 ; 𝜀0 = 8.8542 ∗ 10−12 𝐶 2 ∕ 𝑁𝑚2

a. Calculate the tangent of losses 𝑇𝑎𝑛(𝛿) = 𝜎/𝜔𝜀. Solution

We use the following formula 𝑇𝑎𝑛(𝛿) =

𝜎 𝜎 = 𝜔𝜀 2𝜋𝑓𝜀𝑟 𝜀0

We replace 𝑇𝑎𝑛(𝛿) =

2𝜋 ∗ 53 ∗

106 𝐻𝑧

4 ∗ 80 ∗ 8.8542 ∗ 10−12 𝐶 2 ∕ 𝑁𝑚2

𝑇𝑎𝑛(𝛿) = 17.557 (𝛿) = 𝑡𝑎𝑛 −117.557° (𝛿) = 86.73°

b. Classify the behavior of the medium. The values of the loss tangent are: 17.557, the values of the angle of the loss tangent are: 86.73° and this classifies me to seawater as a good conductor. c. Calculate the propagation parameters of the wave 𝛾, 𝛼 and 𝛽. Solution Data

𝜇 = 𝜇𝑟 ∗ 𝜇0 𝜇𝑟 = (𝑛𝑜𝑛 − 𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑚𝑒𝑑𝑖𝑎) 𝜇0 = 1.2566 ∗ 10−6𝑇𝑚 ∕ 𝐴 𝜔 = 2𝜋𝑓 𝑓 = 53𝑀𝐻𝑧 → 53 ∗ 106 𝐻𝑧 𝜎 = 4𝑠/𝑚 𝜀 = 𝜀𝑟 𝜀0 𝜀𝑟 = 80(𝑠𝑒𝑎 𝑤𝑎𝑡𝑒𝑟) 𝜀0 = 8.8542 ∗ 10−12𝐶 2 ∕ 𝑁𝑚2 Propagation constant 𝑦(𝑔𝑎𝑚𝑚𝑎) 𝑦 = √𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀) We replace 𝑦 = √𝑗(2𝜋𝑓)(𝜇𝑟 ∗ 𝜇0 )(𝜎 + 𝑗(2𝜋𝑓 )(𝜀𝑟 ∗ 𝜀0 )) 𝑦 = √𝑗(2𝜋 ∗ 53 ∗ 106 𝐻𝑧)(1.2566 ∗ 10−6 𝑇𝑚 ∕ 𝐴)(4𝑠/𝑚 + 𝑗(2𝜋 ∗ 53 ∗ 106 𝐻𝑧)(80 ∗ 8.8542 ∗ 10−12𝐶 2 ∕ 𝑁𝑚2 𝑦 ≈ 28.089 + 29.794𝑖 Now we calculate the attenuation constant 𝛼 (𝐴𝑙𝑝ℎ𝑎) 𝑦 = 28.089 + 29.794𝑖 𝛼 = 𝑅𝑒(𝑦) 𝛼 = 28.089𝑁𝑝/𝑚 Now we calculate the attenuation constant𝛽 (𝐵𝑒𝑡𝑎) 𝑦 = 28.089 + 29.794𝑖 𝛽 = 𝐼𝑚(𝑦)

𝛽 = 29.794𝑅𝑎𝑑/𝑚 d. Calculate the propagation speed of the 𝑉𝑝 wave. 𝑉𝑝 = 𝜔/𝛽 Solution 𝑉𝑝 =

𝜔 𝛽

We replace (2𝜋 ∗ 53 ∗ 106 𝐻𝑧) 𝑉𝑝 = 29.794𝑅𝑎𝑑/𝑚 𝑉𝑝 ≈ 11177043.071 𝑚/𝑠

e. Calculate the distance 𝑑 between the ship and the receiver. 𝑑 = 𝑃/𝑐𝑜𝑠(𝑎). Solution Data 𝑝 = 1𝐾𝑚 𝑎 = 25° We use the following formula 𝑑=

1 cos 𝑎

We replace 𝑑=

1 = 1.103 𝑘𝑚 cos(25°)

f. Based on 𝑉𝑝 and 𝑑 determine the time 𝑡 of the route. 𝑑 = 𝑡 ∗ 𝑉𝑝 Solution We use the following formula 𝑡 = 𝑑/𝑉𝑝 We replace

𝑡=

1.103 𝑘𝑚 11177043.071 𝑚/𝑠

𝑡 = 98.7182 ∗ 10−9 ≈ 98.718 𝑛𝑠

Verification of the results in GEOGEBRA

Interpretation: according to the concepts explored, explain the meaning of the value obtained for 𝑇𝑎𝑛(𝛿) and 𝑡. In the end we conclude that the loss value of the tangent is 17.557, which is a good conductor and that the time it takes to travel from one point to another is 98.718 𝑛𝑠.

2. From an airplane, which is ℎ1 = 1250 𝑚 high, a communication signal 𝑓 = 𝐶𝐶𝐶 𝑀𝐻𝑧 is emitted to a submarine that is ℎ2 = 800 𝑚 deep, the angle of incidence of the signal on the sea surface is 𝑎 = (5 + 𝐺𝐺)°. Determine the time it takes for the signal to reach the submarine. Note that 𝐶𝑜 = 3𝑥108 𝑚/𝑠.

Figure 2: wave propagation in bounded open media. Image recovered from https://byjus.com/physics/characteristics-of-soundwavesamplitude/ For development, follow the following steps: a. Calculate the distance between the plane and the point of incidence at sea 𝑑1 = ℎ1 /𝑠𝑒𝑛(𝑎). Solution Data 𝐶0 = 3 ∗ 108 𝑚/𝑠 ℎ1 = 1250𝑚 ℎ2 = 800𝑚 𝑓 = 𝐶𝐶𝐶 𝑀𝐻𝑧 → 53𝑀𝐻𝑧 𝑎 = (5 + 20)° → 25° We use the following formula 𝑑1 =

ℎ1 𝑠𝑒𝑛(𝑎)

We replace 𝑑1 =

1250𝑚 = 2957.751 𝑠𝑒𝑛(25°)

b. Calculate the velocity of propagation of the wave 𝑉𝑝1 in the air (𝑉𝑝1 = 𝐶𝑜/𝑛), where 𝑛 is the refractive index of air. Solution

Table indices de refraction Data Air: 1.000293 We use the following formula 𝑉𝑝1 =

𝐶𝑜 𝑛

We replace . Air 𝑉𝑝1 =

3 ∗ 108 𝑚/𝑠 = 2.9991 ∗ 108 1.000293

Seawater 𝑛𝑠𝑒𝑎 = 𝑛𝑠𝑒𝑎

𝐶𝑜 𝑉𝑝1

3 ∗ 108 𝑚/𝑠 = = 1.0003 2.9991 ∗ 108 𝑉𝑝2 =

𝐶𝑜 𝑛𝑠𝑒𝑎

We replace 𝑉𝑝2 =

3 ∗ 108 𝑚/𝑠 = 2.99 ∗ 108 𝑚/𝑠 1.0003

c. Using Snell's Law, calculate the angle of refraction of the wave in the sea. Solution 𝑎𝑖𝑟 = 1.000293 𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 = 1.0003 For the angle of incidence, we apply 𝜃𝑎 − 𝜃𝑎´ = 90 𝜃𝑎´ = 90 − 𝜃𝑎 𝜃𝑎´ = 90 − 25° = 65° We find the refractive angle of point B sin 𝜃𝑏 =

1.000293 𝑠𝑖𝑛𝜃𝑎 𝑛2

We replace sin 𝜃𝑏 =

1.000293 sin 65° 1.0003

𝜃𝑏 = 𝑠𝑖𝑛 −1

1.000293 sin 65° 1.0003

𝜃𝑏 = 𝑠𝑖𝑛 −1

1.000293 (0.826) 1.0003

𝜃𝑏 = 55.689°

d. Calculate the distance between the point of incidence in the sea and the submarine. Solution We use the following formula 𝑑2 =

ℎ2 cos 𝑏

We replace 𝑑2 =

800𝑚 = 1419.233 𝑚 cos(55.689°)

e. Calculate the propagation speed of the wave 𝑉𝑝2 at sea (take 𝑉𝑝 from exercise 1). Solution We use the following formula 𝑉𝑝 =

𝐶0 𝑛𝑠𝑒𝑎

We replace 𝑛𝑠𝑒𝑎 =

3 ∗ 108 𝑚/𝑠 = 1.0003 2.991 ∗ 108 𝑚/𝑠

3 ∗ 108 𝑚/𝑠 = 2.99 ∗ 108 𝑚/𝑠 𝑉𝑝2 = 1.0003

f. Based on 𝑉𝑝1 and 𝑑1 determine the time 𝑡1 of the first path (𝑡1 = 𝑑1 /𝑉𝑝1). Solution We use the following formula 𝑡1 =

𝑑1 𝑉𝑝1

We replace values.

𝑡1 =

2957.751𝑚 = 9.862 ∗ 10−6 𝑠 2.991 ∗ 108 𝑚/𝑠

g. Based on 𝑉𝑝2 and 𝑑2 determine the time 𝑡2 of the second path (𝑡2 = 𝑑2 /𝑉𝑝2). Solution We use the following formula 𝑡2 =

𝑑2 𝑉𝑝2

We replace 𝑡2 =

1419.233 𝑚 = 4.73 ∗ 10−6𝑠 2.99𝑥 108 𝑚/𝑠

h. Calculate the total time of the route 𝑡 = 𝑡1 + 𝑡2 . Solution 𝑡 = 𝑡1 + 𝑡2 We replace 𝑡 = 9.862 ∗ 10−6 𝑠 + 4.73 ∗ 10−6 𝑠 = 1.459 ∗ 10−5 𝑠

Verification of the results in GEOGEBRA

Interpretation: according to the concepts explored, explain the meaning of the value obtained for 𝑉𝑝1 , 𝑉𝑝2 and 𝑡. We observe the speed of propagation of the wave in each medium, and the speed of the wave in the air compared to the speed of the wave in the sea is very similar and there is not much difference. The time it takes for the wave to travel from the plane to the submarine at sea is 1.459 ∗ 10−5 𝑠

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: Solution Data 𝑍𝐿 = (45 + 𝑗45)Ω 𝑓 = 500𝑀𝐻𝑧 𝑍0 = 75Ω 𝐿 = 5𝑚 𝑉𝑝 = 3 ∗ 107 𝑚/𝑠 a. Wavelength 𝜆. We use the following 𝜆=

𝑉𝑝 𝑓

We replace 𝜆=

3 ∗ 107 𝑚/𝑠 = 0.06𝑚 500𝑀𝐻𝑧

b. Electrical length ℓ. We use the following ℓ=

𝐿 𝜆

We replace ℓ=

5𝑚 = 83.333 𝐿𝑎𝑚𝑏𝑑𝑎 0.06𝑚

c. Input impedance 𝑍𝑖𝑛 . We use the following 𝑍𝑖𝑛 = 𝑍0

𝑍𝐿 + 𝑗𝑍0 𝑡𝑎𝑛(2𝜋ℓ) 𝑍0 + 𝑗𝑍𝐿 tan(2𝜋ℓ)

We replace 𝑍𝑖𝑛 = 75Ω

(45Ω + 𝑗45Ω) + 𝑗75Ω tan(2Ω ∗ 83.333𝑚) 75Ω + 𝑗(45Ω + 𝑗75Ω) tan(2Ω ∗ 83.333𝑚 𝑍𝑖𝑛 = 34.361 − 24.124𝑗 𝑂ℎ𝑚

d. Reflection coefficient Γ (magnitude and phase).

We use the following Γ=

𝑍𝐿 − 𝑍0 𝑍𝐿 + 𝑍0

We replace Γ=

(45Ω + 𝑗45Ω) − 75Ω (45Ω + 𝑗45Ω) + 75Ω

Γ = −0.0958 + 0.410j Ohm We went from rectangular to a polar Γ = 0.421 ∢103.134° Ohm

e. VSWR.

We use the following 𝑉𝑆𝑊𝑅 =

1 + |Γ| 1 − |Γ|

We replace 𝑉𝑆𝑊𝑅 =

1 + |−0.0958 + 0.410j | 1 − |−0.0958 + 0.410j| 𝑉𝑆𝑊𝑅 = 2.46

Verification of the results in GEOGEBRA

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

Transmission line.

Software Smith 4.1

Verification results in software Γ = 0.422 ∢102.707° 𝑉𝑆𝑊𝑅 = 2.46 𝑍𝑖𝑛 = 34.419 − 24.298𝑗 𝑂ℎ𝑛

Interpretation: according to the concepts explored, explain the meaning of the value obtained for 𝑍𝑖𝑛 , Γ and 𝑉𝑆𝑊𝑅 . We conclude that the magnitude of the input impedance of the line is equal to 34.361 − 24.124𝑗 𝑂ℎ𝑚, this indicates the relationship that exists in the different components that are connected in the current line, and the propagation coefficient is 0.421 ∢103.134° Ohm, the VSWR value of 2.46. Which comparing them with the results obtained in the software are very similar.

References

Chen, W. (2005). The Electrical Engineering Handbook. Boston: Academic Press, (pp.

525-551).

Recovered

from http://bibliotecavirtual.unad.edu.co:2048/login?url=http://search.e bscohost.com/login.aspx?direct=true&db=nlebk&AN=117152&lang=es&s ite=ehost-live&ebv=EB&ppid=pp_525 Joines, W., Bernhard, J., & Palmer, W. (2012). Microwave Transmission Line Circuits.

Boston:

Artech

House,

(pp.

23-68).

Recovered

from http://bibliotecavirtual.unad.edu.co:2051/login.aspx?direct=true&d b=nlebk&AN=753581&lang=es&site=eds-live&ebv=EB&ppid=pp_23 Hierauf, S. (2011). Understanding Signal Integrity. Boston: Artech House, Inc. Chapter

6,

7,

11.

Recovered

from http://bibliotecavirtual.unad.edu.co:2051/login.aspx?direct=true&d b=nlebk&AN=345692&lang=es&site=eds-live&ebv=EB&ppid=pp_49 Impedance

Matching

Communication

Networks. Circuits,

(2001).

Radio-Frequency

(pp.

146-188).

&

Microwave Recovered

from http://bibliotecavirtual.unad.edu.co:2051/login.aspx?direct=true&d b=aci&AN=14528229&lang=es&site=eds-live...