Delivery Format - Step 2203058_24.docx

Step 2 To recognize the electrodynamic and waves applications

Group: 203058_24 Ricardo Cordoba. Alexis Pedroza. 67032716.

“UNIVERSIDAD NACIONAL ABIERTA Y A DISTANCIA – UNAD” School of Basic Sciences, Technology and Engineering Electromagnetic Theory and Waves 2019 16-01

Introduction The scope of the document is usually described in the introduction, and a brief explanation or summary of it is given. A reader reading the introduction should be able to get an idea about the content of the text, before beginning the reading itself. (5 to 10 lines).

Developed activity (consolidate) 1. Explain the concept of "loss tangent" applied to electromagnetic wave propagation media, and indicate how it is calculated. R//. The “loss tangent”: Also known as tan delta, is the relationship between the conduction current and the displacement through a specific medium, this relationship is constant and depends on the parameters of the medium and the frequency of the signal that apply to you The equation used to calculate it is: 𝜎 tan 𝛿 = 𝜔∗𝜖 2. According to the "tangent of losses", how can the means of propagation be classified? R//. Basic classification for media based on the “tangent of losses”:  Perfect dielectrics  Perfect drivers  Good insulators  Good drivers  Dissipative dielectrics 3. What concept does each of the means of propagation defined in the previous point have? R//. Concepts of each of the means of propagation:  Perfect dielectrics: they do not present conduction current, therefore, they do not have losses due to Joule effect. Tan 𝛿 = 0 ⟹ 𝛿 = 0  Perfect conductors: they do not present polarization current, therefore, they do not have capacitive effects or load accumulation. Tan 𝛿 → ∞ ⟹ 𝛿 =

𝜋 2

 Good insulators: they present conduction current and have losses due to Joule effect, but this effect is almost negligible compared to the capacitive effect, they are also called "low loss dielectrics". Tan 𝛿 → 0+ ⟹ 𝛿 → 0+  Good conductors: they present polarization current, therefore, they have capacitive or load accumulation effects, but the conduction current and Joule losses are much more 𝜋 significant. Tan 𝛿 → ∞ ⟹ 𝛿 → 2

 Dissipative dielectrics: they have both effects and none is negligible compared to the 𝜋 other. 0 < 𝛿 < 2

4. On what does the propagation velocity of an electromagnetic wave depend? R//. The phase velocity of electromagnetic waves depends only on the electromagnetic properties of the medium in which they propagate and not on the relative displacement between observers, which clearly violates the laws of mechanics known. This observation gave rise to the so-called special theory of relativity, whose fundamental statement was published, in the year 1905, by Albert Einstein. The phase velocity of an electromagnetic wave in a non-dissipative medium is: 1 𝑣𝑝 = √𝜇𝜀 The speed of electromagnetic waves in vacuum is a universal constant whose value turns out to be equal to the speed of light and is determined by: 1 𝐶𝑜 = = 3 × 108 𝑚⁄𝑠 √𝜇0 𝜀0 What is the relationship between the speed of propagation and the refractive index of a medium? R//. The refractive index of a substance is the ratio between the speed of light in vacuum and the phase velocity of an electromagnetic signal in a specific medium is represented by the letter n and is given by: 𝑐0 𝑛= 𝑣𝑝 Because the phase velocity in any medium is less than the speed of light in vacuum, the refractive index of a substance is always an amount greater than or equal to 1 In the case of perfect non-magnetic dielectrics, the refractive index remains: 𝑐0 𝑐0 𝑛= ⟹𝑛=𝑐 ⟹ 𝒏 = √𝜺𝒓 0 𝑣𝑝 ⁄ 𝜀 √ 𝑟

5. What is the penetration depth of an electromagnetic wave in a medium and how can I calculate it?

R//. The penetration depth of a wave 𝛿𝑝 is the inverse of the attenuation constant. Since the power of a signal is proportional to the square of its amplitude, when the signal drops to 36% of its value, its power drops to 13%. 𝛿𝑝 =

1 𝛼

Conclusiones Exercises (one per student) 1. Student name: A dissipative medium has the following parameters: 𝜀𝑟 = 3.5, 𝜇𝑟 = 2.2 and 𝜎 = 1.9 𝑆/𝑚 Find the wavelength and the amount of wavelengths that will penetrate a 10MHz signal. 2. Student name: In a medium with the following characteristics, 𝜀𝑟 = 2,5, 𝜇𝑟 = 1.3 and 𝜎 = 1.8𝑥10−3 𝑆/𝑚 find these parameters for a 1GHz signal: a. Loss tangent. b. Propagation constant. c. Phase velocity. d. Wavelength. e. Index of refraction. Explain the meaning of each found value.

3. Student name: An open medium has the following electromagnetic characteristics 𝜀𝑟 = 5.5, 𝜇𝑟 = 1.9 and 𝜎 = 14.6𝑥10−6 𝑆/𝑚 Find the power transmitted by a 200MHz signal with a maximum electric field of 127V/m and find the skin depth of the signal.

4. Student name: Ricardo Cordoba. For a medium with the same electromagnetic characteristics than the third problem, find the losses per length unit for a 400MHz signal. If the original signal has an electric field of 120Vrms/m. Find the losses in watts when the signal travels 20m in the medium. Solution: As the statement tells me that the characteristics of the medium are the same as the previous problem, we must bear in mind that: 𝜀𝑟 = 5.5, 𝜇𝑟 = 1.9 and 𝜎 = 14.6𝑥10−6 𝑆/𝑚 The first thing I do is calculate the tangent of losses to discriminate the medium like this: 1

Knowing that: 𝜔 = 2𝜋 ∗ 4 × 108 𝑟𝑎𝑑/𝑠𝑒𝑔 y 𝜖 = 5.5 ∗ 36𝜋 ∗ 10−9 𝐹 ⁄𝑚 so: 𝜎 14.6𝑥10−6 𝑆/𝑚 𝑡𝑎𝑛 𝛿 = ⟹ 𝑡𝑎𝑛 𝛿 = 1 𝜔∗𝜖 2𝜋 ∗ 4 × 108 𝑟𝑎𝑑/𝑠𝑒𝑔 ∗ 5.5 ∗ 36𝜋 × 10−9 𝐹 ⁄𝑚 𝒕𝒂𝒏 𝜹 = 𝟏. 𝟐𝟏 × 𝟏𝟎−𝟓

The medium is a low loss dielectric so the intrinsic impedance is given by: 𝜇 4𝜋 × 10−7 𝜂=√ ⟹𝜂=√ 1 𝜀 5.5 ∗ 36𝜋 × 10−9 𝜼 = 𝟏𝟔𝟎. 𝟕𝟒𝟗𝟓𝜴

The attenuation constant is: 𝑆 14.6𝑥10−6 𝑚 ∗ 160.7495𝛺 𝜎𝜂 𝛼= ⟹𝛼= 2 2 𝜶 = 𝟎. 𝟎𝟎𝟏𝟏𝟕𝟑 ⟹ 𝜶 = 𝟏. 𝟏𝟕𝟑 × 𝟏𝟎−𝟑 𝑵𝒑⁄𝒎

As the losses in units of length are requested, 𝑥 = 1𝑚 %𝑃𝑒𝑟𝑑𝑖𝑑𝑎𝑠 = 1 − 𝑒 −2𝛼𝑥 −3

%𝑃𝑒𝑟𝑑𝑖𝑑𝑎𝑠 = 1 − 𝑒 −2∗1.173×10

%𝑷𝒆𝒓𝒅𝒊𝒅𝒂𝒔 = 𝟎. 𝟎𝟎𝟐𝟑𝟒𝟑 = 𝟎. 𝟐% From the above we can deduce that in this medium for each meter traveled, 0.2% of the power is lost. Therefore if a 20m run is made: %𝑃𝑒𝑟𝑑𝑖𝑑𝑎𝑠 = 1 − 𝑒 −2𝛼𝑥 −3

%𝑃𝑒𝑟𝑑𝑖𝑑𝑎𝑠 = 1 − 𝑒 −2∗20∗1.173×10

%𝑷𝒆𝒓𝒅𝒊𝒅𝒂𝒔 = 𝟎. 𝟎𝟒𝟓𝟖𝟑 = 𝟒. 𝟓𝟖𝟑%

Now the initial power is: 𝑃0 =

|120 𝑉𝑅𝑀𝑆 ⁄𝑚|2 ⟹ 𝑷𝟎 = 𝟖𝟗. 𝟓𝟖 𝑾⁄𝒎𝟐 |160.7495|𝛺

The losses therefore are: 𝑃𝑒𝑟𝑑𝑖𝑑𝑎𝑠 = 89.58 𝑊 ⁄𝑚2 ∗ 0.04583 𝑷𝒆𝒓𝒅𝒊𝒅𝒂𝒔 = 𝟒. 𝟏𝟎𝟓𝟒 𝑾⁄𝒎𝟐

5. Student name: Alexis Pedroza

For a 400MHz signal, traveling in seawater find the attenuation per length unit. How long does the signal have to travel, in order to have an attenuation greater than 3dB? Conclusions (one per student) Conclusion 1: Student name Conclusion 2: Student name

21/5000 (one per student)

Conclusion 3: Student name Conclusion 4: Student name Conclusion 5: Student name ... The conclusions should be written with their own words and should focus on the concepts explored, learned, discovered and practiced in the development of the activity, it is suggested to present a conclusion by topic, the result of learning obtained as evidence of conceptual assimilation. To obtain a good writing it is suggested to read the written several times, correcting and adjusting the text until obtaining a clear and coherent postulate. Avoid superficiality and simplicity.

References (one per student) Bibliography 1: Ricardo Cordoba Paz. P, A., (2013). Electrodinámicas y Ondas. Electromagnetismo para ingeniería electrónica (pp. 196-199). Cali, Colombia: Editorial. Sello Editorial Javeriano. Retrieved from: http://www.academia.edu/15312004/Electromagnetismo Bibliography 2: Student name Bibliography 3: Student name Bibliography 4: Student name Bibliography 5: Student name … Examples: Ebook. Surname, A. (Year). Title of the book. (pp. xx-xx). Country: Editorial. Retrieved from http: // ... Chapter of an electronic book. Surname, A., and Surname, B. (Year). Title of the chapter or the entry. Title of the book (pp. Xx-xx). City, Country: Editorial. Retrieved from http: // ... Internet video. Surname, N. (Year). Title of the video Video server [Video]. Retrieved from http: // ...