Lec 5 - Fall 2024 - EEE 314 Oblique Incidence II - Lecture Notes PDF

Lecture 5 Oblique Incidence II EEE 314 WAVES PROPAGATION 3rd year- Fall 2024 Electronics and Electrical Communications Assoc. Prof. Ahmed Farghal Dept. of Electrical Engineering,...

Lecture 5 Oblique Incidence II EEE 314 WAVES PROPAGATION 3rd year- Fall 2024 Electronics and Electrical Communications Assoc. Prof. Ahmed Farghal Dept. of Electrical Engineering, Faculty of Engineering, Sohag University 29/10/2024  Introduction The trick in deriving the components is to first get the propagation Lecture 5 vector k for incident, reflected, and transmitted waves. 𝑘𝑖 = 𝛽1 𝑘𝑟𝑧 = 𝛽1 cos 𝜃𝑟 𝑘𝑡 = 𝛽2 𝜃𝑟 𝑘𝑖𝑥 = 𝛽1 sin 𝜃𝑖 𝛽2 sin 𝜃𝑡 i 𝑘𝑟𝑥 = 𝛽1 sin 𝜃𝑟 t 𝑘𝑟 = 𝛽1 Dr. Ahmed Farghal 𝑘𝑖𝑧 = 𝛽1 cos 𝜃𝑖 𝑘𝑡𝑧 = 𝛽2 cos 𝜃𝑡 𝐤 𝑖 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝒂𝐤 𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 x 𝐤 𝑟 = 𝛽1 sin 𝜃𝑟 𝐚𝑥 − 𝛽1 cos 𝜃𝑟 𝐚𝑧 y z 𝒂𝐤 𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 Fall 2024 𝐤 𝑡 = 𝛽2 sin 𝜃𝑡 𝐚𝑥 + 𝛽2 cos 𝜃𝑡 𝐚𝑧 𝒂𝐤 𝑡 = 𝐚𝑥 sin 𝜃𝑡 + 𝐚𝑧 cos 𝜃𝑡 2  Introduction Once k is known, we define 𝐄 such that 𝐚𝑘 ∙ 𝐄 = 0 Lecture 5 General phasor form 𝐄 𝐫 = 𝐚𝐸 𝐸𝑜 𝑒 −𝑗𝐤⋅𝐫 position vector 𝐫 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 Dr. Ahmed Farghal wave number or propagation vector 𝐤 = 𝑘𝑥 𝐚𝑥 + 𝑘𝑦 𝐚𝑦 + 𝑘𝑧 𝐚𝑧 𝑘= 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = ω 𝜇𝜖 = 𝛽 𝐤 𝑖 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑖 + 𝛽1 𝑧 cos 𝜃𝑖 𝐤 𝑟 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑟 − 𝛽1 𝑧 cos 𝜃𝑟 𝐤 𝑡 ∙ 𝐫 = 𝛽2 𝑥 sin 𝜃𝑡 + 𝛽2 𝑧 cos 𝜃𝑡 Then 𝐇 is obtained from Fall 2024 1 𝐇 𝑥, 𝑧 = 𝐚𝑘 × 𝐄(𝑥, 𝑧) 𝜂 3  Parallel Polarization E𝑖 || plane of incidence and H𝑖 is ⊥ in 𝑦 direction Lecture 5 so that the incident wave propagates toward the interface. Medium 1 : 𝜀1 , 𝜇1 , 𝜎1 = 0 Medium 2: 𝜀2 , 𝜇2 , 𝜎2 = 0 𝐸𝑟 𝐤𝑟 Dr. Ahmed Farghal 𝐸𝑡 𝐻𝑟 kt r 𝜃𝑡 𝐻𝑡 z i 𝐤𝑖 𝐸𝑖 x Fall 2024 𝐻𝑖 y z 𝛽1 = 𝜔 𝜖1 𝜇1 𝛽2 = 𝜔 𝜖2 𝜇2 z=0 4  Parallel Polarization Lecture 5 −𝐚𝑧 sin 𝜃𝑖 𝜋 𝜋 − 𝜃𝑖 𝜃𝑖 − 𝜃𝑖 2 2 𝐚𝑥 cos 𝜃𝑖 𝐸𝑖 𝐸𝑖 𝜃𝑖 Dr. Ahmed Farghal 𝐄𝑖 , and 𝐇𝑖 in medium (1): 𝐤 𝑖 = 𝛽1 sin 𝜃𝑖 𝐚𝑥 + 𝛽1 cos 𝜃𝑖 𝐚𝑧 𝒂𝐤 𝑖 = 𝐚𝑥 sin 𝜃𝑖 + 𝐚𝑧 cos 𝜃𝑖 𝒂𝐤 𝑖 ∙ 𝐄𝑖 𝑥, 𝑧 = 0 𝐄𝑖 𝑥, 𝑧 = 𝐸𝑖𝑜 (𝐚𝑥 cos 𝜃𝑖 − 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖) 1 Fall 2024 𝐇𝑖 𝑥, 𝑧 = 𝒂𝐤 𝑖 × 𝐄𝑖 (𝑥, 𝑧) 𝜂1 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐇𝑖 𝑥, 𝑧 = 𝐚𝑦 𝑒 1 𝑖 𝑖 𝜂1 5  Parallel Polarization 1 𝒂𝐄 ∙ 𝐄 𝑥, 𝑧 = 0 Lecture 5 𝐇 𝑥, 𝑧 = 𝐚𝑘 × 𝐄(𝑥, 𝑧) 𝜂 𝐄𝑟 , and 𝐇𝑟 in medium (1) are: 𝐤 𝑟 = 𝛽1 sin 𝜃𝑟 𝐚𝑥 − 𝛽1 cos 𝜃𝑟 𝐚𝑧 𝒂𝐤 𝑟 = 𝐚𝑥 sin 𝜃𝑟 − 𝐚𝑧 cos 𝜃𝑟 𝐤 𝑟 ∙ 𝐫 = 𝛽1 𝑥 sin 𝜃𝑟 − 𝛽1 𝑧 cos 𝜃𝑟 Dr. Ahmed Farghal 𝐄𝑟 𝑥, 𝑧 = 𝐸𝑟𝑜 (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝐸𝑟𝑜 −𝑗𝛽 (𝑥 sin 𝜃 −𝑧 cos 𝜃 ) 𝐇𝑟 𝑥, 𝑧 = −𝐚𝑦 𝑒 1 𝑖 𝑖 𝜂1 𝐄𝑡 , and 𝐇𝑡 in medium (2): 𝐤 𝑡 = 𝛽2 sin 𝜃𝑡 𝐚𝑥 + 𝛽2 cos 𝜃𝑡 𝐚𝑧 𝒂𝐤 𝑡 = 𝐚𝑥 sin 𝜃𝑡 + 𝐚𝑧 cos 𝜃𝑡 𝐤 𝑡 ∙ 𝐫 = 𝛽2 𝑥 sin 𝜃𝑡 + 𝛽2 𝑧 cos 𝜃𝑡 Fall 2024 𝐄𝑡 𝑥, 𝑧 = 𝐸𝑡𝑜 (𝐚𝑥 cos 𝜃𝑡 − 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝐸𝑡𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐇𝑡 𝑥, 𝑧 = 𝐚𝑦 𝑒 2 𝑡 𝑡 𝜂2 6  Parallel Polarization Requiring that 𝜃𝑖 = 𝜃𝑟 and the tangential components of E and H Lecture 5 (𝑥 and 𝑦 components) be continuous @ boundary (𝑧 = 0), 𝐄𝑖 𝑥, 𝑧 = 𝐸𝑖𝑜 (𝐚𝑥 cos 𝜃𝑖 − 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 +𝑧 cos 𝜃𝑖) 𝐄𝑟 𝑥, 𝑧 = 𝐸𝑟𝑜 (𝐚𝑥 cos 𝜃𝑖 + 𝐚𝑧 sin 𝜃𝑖 ) 𝑒 −𝑗𝛽1 (𝑥 sin 𝜃𝑖 −𝑧 cos 𝜃𝑖 ) 𝐄𝑡 𝑥, 𝑧 = 𝐸𝑡𝑜 (𝐚𝑥 cos 𝜃𝑡 − 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) Dr. Ahmed Farghal 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐇𝑖 𝑥, 𝑧 = 𝐚𝑦 𝑒 1 𝑖 𝑖 𝜂1 𝐸𝑟𝑜 −𝑗𝛽 (𝑥 sin 𝜃 −𝑧 cos 𝜃 ) 𝐇𝑟 𝑥, 𝑧 = −𝐚𝑦 𝑒 1 𝑖 𝑖 𝜂1 𝐸𝑡𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐇𝑡 𝑥, 𝑧 = 𝐚𝑦 𝑒 2 𝑡 𝑡 𝜂2 Fall 2024 We obtain 1 𝐸𝑡𝑜 𝐸𝑖𝑜 + 𝐸𝑟𝑜 cos 𝜃𝑖 = 𝐸𝑡𝑜 cos 𝜃𝑡 & 𝐸𝑖𝑜 − 𝐸𝑟𝑜 = 𝜂1 𝜂2 01 02 7  Parallel Polarization Solving for 𝐸𝑟𝑜 and 𝐸𝑡𝑜 in terms of 𝐸𝑖𝑜 , we obtain Lecture 5 𝜂2 cos 𝜃𝑡 − 𝜂1 cos 𝜃𝑖 𝐸𝑟𝑜 = 𝐸𝑖𝑜 , 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 2𝜂2 cos 𝜃𝑖 𝐸𝑡𝑜 = 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 Dr. Ahmed Farghal Γ and 𝜏 for || polarization: 𝐸𝑟𝑜 𝜂2 cos 𝜃𝑡 − 𝜂1 cos 𝜃𝑖 Γ|| = = or 𝐸𝑟𝑜 = Γ|| 𝐸𝑖𝑜 1 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 𝐸𝑡𝑜 2𝜂2 cos 𝜃𝑖 𝜏|| = = or 𝐸𝑡𝑜 = 𝜏|| 𝐸𝑖𝑜 2 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 Fall 2024 8  Parallel Polarization Lecture 5 The total fields in medium (1): 𝐸𝑟𝑜 = Γ|| 𝐸𝑖𝑜 𝐄1 𝑥, 𝑧 = 𝐚𝑥 𝐸𝑖𝑜 cos 𝜃𝑖 Γ|| 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 + 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 +𝐚𝑧 𝐸𝑖𝑜 sin 𝜃𝑖 Γ|| 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 − 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 Dr. Ahmed Farghal 𝐸𝑖𝑜 𝐇1 𝑥, 𝑧 = −𝐚𝑦 Γ|| 𝑒 𝑗𝛽1 𝑧 cos 𝜃𝑖 − 𝑒 −𝑗𝛽1 𝑧 cos 𝜃𝑖 𝑒 −𝑗𝛽1 𝑥 sin 𝜃𝑖 𝜂1 The fields in medium (2): 𝐸𝑡𝑜 = 𝜏|| 𝐸𝑖𝑜 𝐄𝑡 𝑥, 𝑧 = 𝜏|| 𝐸𝑖𝑜 𝐚𝑥 cos 𝜃𝑡 − 𝐚𝑧 sin 𝜃𝑡 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 Fall 2024 𝜏|| 𝐸𝑖𝑜 −𝑗𝛽 (𝑥 sin 𝜃 +𝑧 cos 𝜃 ) 𝐇𝑡 𝑥, 𝑧 = 𝐚𝑦 𝑒 2 𝑡 𝑡 𝜂2 9  Parallel Polarization 𝜂2 cos 𝜃𝑡 − 𝜂1 cos 𝜃𝑖 2𝜂2 cos 𝜃𝑖 Lecture 5 Γ|| = 𝜏|| = 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 cos 𝜃𝑡 It is easily shown that 1 + Γ|| = 𝜏|| cos 𝜃𝑖 When 𝜃𝑖 = 𝜃𝑡 = 0 (normal incidence), the equations reduce to Dr. Ahmed Farghal 𝜂2 − 𝜂1 2𝜂2 Γ|| = Γ = & 𝜏|| = 𝜏 = as expected, 𝜂2 + 𝜂1 𝜂2 + 𝜂1 Since 𝜃𝑖 & 𝜃𝑡 are related according to Snell’s law sin 𝜃𝑡 𝑛1 𝑣𝑝2 = = sin 𝜃𝑖 𝑛2 𝑣𝑝1 Γ|| and 𝜏|| can be written in terms of 𝜃𝑖 by substituting Fall 2024 2 cos 𝜃𝑡 = 1 − sin2 𝜃𝑡 = 1 − 𝑣𝑝2 Τ𝑣𝑝1 sin2 𝜃𝑖 Γ⊥ , 𝜏⊥ , Γ|| and 𝜏|| are known as Fresnel coefficients. 10  Brewster’s Angle Lecture 5 The reflection coefficients for ⊥ and || polarizations are 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 Γ⊥ = 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 𝜂2 cos 𝜃𝑡 − 𝜂1 cos 𝜃𝑖 Dr. Ahmed Farghal Γ|| = 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 Either reflection coefficient is zero because the numerator is the difference of two terms (if the numerator is zero); that is, Γ⊥ = 0 if 𝜂2 cos 𝜃𝑖 = 𝜂1 cos 𝜃𝑡 Γ|| = 0 if 𝜂2 cos 𝜃𝑡 = 𝜂1 cos 𝜃𝑖. Fall 2024 Under this condition, there is no reflection (𝐸𝑟𝑜 = 0). The incidence angle at which either condition is satisfied is called the Brewster angle 𝜃𝐵. 11  Brewster angle for ∥ Polarization Lecture 5 For || polarization, the Brewster angle is obtained by setting 𝜃𝑖 = 𝜃𝐵∥ when Γ|| = 0, that is 𝜂2 cos 𝜃𝑡 = 𝜂1 cos 𝜃𝐵∥ ⟹ 𝜂22 cos 2 𝜃𝑡 = 𝜂12 cos 2 𝜃𝐵∥ To find the angle 𝜃𝐵∥ @ which this is satisfied, we rewrite sin 𝜃𝑡 in terms of 𝜃𝐵∥ in Snell’s law. we have Dr. Ahmed Farghal 𝜀1 𝜇1 sin 𝜃𝑡 = sin 𝜃𝐵∥ 𝜀2 𝜇2 Using cos 𝜃𝑡 = 1 − sin2 𝜃𝑡 and cos 𝜃𝐵∥ = 1 − sin2 𝜃𝐵∥ 𝜀1 𝜇1 2 𝜂22 1− sin 𝜃𝐵∥ = 𝜂12 1 − sin2 𝜃𝐵∥ 𝜀2 𝜇2 Fall 2024 Now, using 𝜂1 = 𝜇1 Τℰ1 and 𝜂2 = 𝜇2 Τℰ2 , we get 1 − (𝜀1 𝜇2 /𝜀2 𝜇1 ) sin 𝜃𝐵∥ = 1 − (𝜀1 /𝜀2 )2 12  Brewster angle for ∥ Polarization This may also be written as Lecture 5 −1 1 − (𝜀1 𝜇2 /𝜀2 𝜇1 ) 𝜃𝐵∥ = sin 1 − (𝜀1 /𝜀2 )2 Thus, for any two materials except two materials of identical permittivity (𝜀1 = 𝜀2 ), there is a specific angle @ which there is no Dr. Ahmed Farghal reflected wave. For nonmagnetic media, (i.e., 𝜇1 = 𝜇2 = 𝜇0 ). The expression for Brewster’s angle is greatly simplified: 1 − (𝜀1 /𝜀2 ) ℰ2 1 sin 𝜃𝐵∥ = 2 = = 1 − (𝜀1 /𝜀2 ) ℰ1 + ℰ2 1 + ℰ1 Τℰ2 ℰ2 Τℰ1 𝜃𝐵∥ Fall 2024 ℰ2 ℰ 2 𝑛2 or 𝜃𝐵∥ = sin−1 or tan 𝜃𝐵∥ = = ℰ1 + ℰ2 ℰ1 𝑛1 1 ⟹ there is a Brewster angle for any combination of 𝜀1 and 𝜀2. 13  Brewster angle for ∥ Polarization Lecture 5 The importance of Brewster’s angle is twofold. 1. By proper choice of 𝜃𝑖 , the reflection from a material for a ∥ polarized wave can be canceled. 2. if a wave has an E which has components || and ⊥ to the plane of incidence and if the wave impinges on a material interface @ Dr. Ahmed Farghal Brewster’s angle, the reflection of the || polarized component is canceled but not that of the ⊥ polarized component. The reflected wave consists of ⊥ polarized component of the wave alone. Thus, for any general wave (polarized or unpolarized), the reflected wave @ Brewster angle of incidence is linearly polarized ⊥ to the plane of incidence. Brewster’s angle is also called a polarizing angle. Fall 2024 Because the wave at Brewster angle is not reflected, ⟹ it must be transmitted across the interface. Brewster angle may be called the angle of total transmission. 14  Brewster angle for ⊥ Polarization An angle of no reflection (Γ⊥ = 0 𝑜𝑟 𝐸𝑟 = 0) may also be defined Lecture 5 for ⊥ polarization by starting with Γ⊥. By replacing 𝜃𝑖 with 𝜃𝐵⊥ , we obtain 𝜂2 cos 𝜃𝐵⊥ = 𝜂1 cos 𝜃𝑡 Following steps identical to those for || polarization, the Brewster Dr. Ahmed Farghal angle is 1 − (𝜇1 𝜀2 /𝜇2 𝜀1 ) 𝜃𝐵⊥ = sin−1 1 − (𝜇1 /𝜇2 )2 Note that for nonmagnetic media (𝜇1 = 𝜇2 = 𝜇𝑜 ), sin 𝜃𝐵⊥ → ∞, so 𝜃𝐵⊥ does not exist because the sine of an angle is never > 1. Very few dielectrics have different permeabilities and almost Fall 2024 none have different permeabilities and the same permittivity. For this reason, the Brewster angle is most often associated with || polarization rather than ⊥ polarization. 15  Brewster angle for ⊥ Polarization Lecture 5 If 𝜇1 ≠ 𝜇2 , the condition can be satisfied and a Brewster angle exists. For materials with identical permittivities but different Dr. Ahmed Farghal permeabilities, the Brewster angle for ⊥ polarization is 𝜇2 sin 𝜃𝐵⊥ = if ℰ1 = ℰ2 𝜇1 + 𝜇2 𝜇2 tan 𝜃𝐵⊥ = 𝜇1 Fall 2024 Although this situation is theoretically possible, it rarely occurs in practice. 16  Total Reflection Lecture 5 If the wave propagates across an interface such that 𝜃𝑡 > 𝜃𝑖 , an ⇑ in 𝜃𝑖 leads to an angle @ which the refracted wave propagates @ 𝟗𝟎° to the normal. This angle is called a critical angle. Any further ⇑ in 𝜃𝑖 results in total reflection of the incident Dr. Ahmed Farghal wave. This condition occurs in lossless dielectrics if 𝜺𝟏 > 𝜺𝟐. For example, waves incident on the surface of Fall 2024 water from below satisfy this condition. 17  Total Reflection Total reflection phenomenon exists in either ⊥ or ∥ polarization. Lecture 5 Total reflection occurs when the reflection coefficient = unity. @ 𝜃𝑡 = 90° 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 𝜂2 cos 𝜃𝑡 − 𝜂1 cos 𝜃𝑖 Γ⊥ = = 1, Γ|| = = −1 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 𝜂2 cos 𝜃𝑡 + 𝜂1 cos 𝜃𝑖 Dr. Ahmed Farghal Transmitted wave Fall 2024 Incident Reflected wave wave 18  Total Reflection Lecture 5 Dr. Ahmed Farghal To define the critical angle, we again use Snell’s law : 𝜀1 𝜇1 sin 𝜃𝑡 = sin 𝜃𝑖 𝜀2 𝜇2 Substituting 𝜃𝑡 = 90° gives the critical angle: 𝜇2 𝜀2 Fall 2024 sin 𝜃𝑐 = for 𝜇2 𝜀2 < 𝜇1 𝜀1 𝜇1 𝜀1 The condition 𝜇2 𝜀2 < 𝜇1 𝜀1 is also necessary otherwise, sin 𝜃𝑐 would be > 1. 19  Fall 2024 Dr. Ahmed Farghal Lecture 5 Total Reflection 20  Total Reflection Now, suppose we ⇑ 𝜃𝑖 above 𝜃𝑐. Lecture 5 This leads to sin 𝜃𝑡 > 1; that is, sin 𝜃𝑡 = 1 for 𝜃𝑖 = 𝜃𝑐. 𝜀1 𝜇1 ∵ 𝜃𝑖 = 𝜃𝑐 < 90°, an ⇑ in 𝜃𝑖 ⇑ RHS of sin 𝜃𝑡 = sin 𝜃𝑖 above 1. 𝜀2 𝜇2 When substituting this condition in the reflection coefficients, it leads to complex values for the reflection coefficients. Dr. Ahmed Farghal The magnitude of the reflection coefficients remains = 𝟏, but they are no longer real values. Therefore, total reflection occurs for 𝜃𝑖 ≥ 𝜃𝑐 The condition for total reflection is 𝜇2 𝜀2 𝜃𝑖 ≥ 𝜃𝑐 = sin−1 for 𝜇2 𝜀2 ≤ 𝜇1 𝜀1 𝜇1 𝜀1 𝜀2 Fall 2024 𝜃𝑖 ≥ sin−1 for 𝜀2 ≤ 𝜀1 , 𝜇1 = 𝜇2 𝜀1 These relations are independent of polarization. 21  Total Reflection Lecture 5 Transmitted wave has the general form 𝐄𝑡,⊥ 𝑥, 𝑧 = 𝐚𝑦 𝑡⊥ 𝐸𝑖𝑜⊥ exp 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) From Snell's law 𝜀1 𝜇1 sin 𝜃𝑡 = sin 𝜃𝑖 𝜀2 𝜇2 Dr. Ahmed Farghal 𝜀1 𝜇1 Recall that @ 𝜃𝑖 = 𝜃𝑐 ⇒ sin 𝜃𝑐 =1 𝜀2 𝜇2 𝜀1 𝜇1 If𝜀1 𝜇1 > 𝜀2 𝜇2 & 𝜃𝑖 > 𝜃𝑐 ⇒ sin 𝜃𝑡 = sin 𝜃𝑖 >1 𝜀2 𝜇2 cos 𝜃𝑡 = 1 − sin2 𝜃𝑡 = ±𝑗𝐴2 = ±𝑗 sin2 𝜃𝑡 − 1 +𝑗𝐴2 must be ignored (implies a wave with growing amplitude) Fall 2024 Taking cos 𝜃𝑡 = −𝑗𝐴2 ∴ 𝐄𝑡,⊥ 𝑥, 𝑧 = 𝐚𝑦 𝑡⊥ 𝐸𝑖𝑜⊥ 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧(𝑗𝐴2)) = 𝐚𝑦 𝑡⊥ 𝐸𝑖𝑜⊥ 𝑒 −𝛼2 𝑧 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 22  Total Reflection Lecture 5 𝛼2 = 𝛽2 𝐴2 = 𝛽2 sin2 𝜃𝑡 − 1 Traveling wave part 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 From Snell’s law: 𝛽1 sin 𝜃𝑖 = 𝛽2 sin 𝜃𝑡 = 𝑘𝑖𝑥 Dr. Ahmed Farghal ⇒ 𝑒 −𝑗𝛽2 𝑥 sin 𝜃𝑡 = 𝑒 −𝑗𝑘𝑖𝑥𝑥 ∴ 𝐄𝑡,⊥ 𝑥, 𝑧 = 𝐚𝑦 𝑡⊥ 𝐸𝑖𝑜⊥ 𝑒 −𝛼2𝑧 𝑒 −𝑗𝑘𝑖𝑥𝑥 where𝑘𝑖𝑥 = 𝛽1 sin 𝜃𝑖 is the wavevector of the incident wave along the 𝑥-axis, and 2 is an attenuation coefficient for the electric field penetrating into medium 2 The transmitted fields show propagation in the 𝑥 direction, along the interface, but exponential decay in the 𝑧 direction. Fall 2024 Such a field is known as a surface wave since it is tightly bound to the interface. 23  Total Reflection A surface wave is an example of a nonuniform plane wave, so Lecture 5 called because it has an amplitude variation in the z direction, apart from the propagation factor in the x direction. 𝑡⊥ 𝐸𝑖𝑜⊥ 𝐇𝑡 𝑥, 𝑧 = (−𝐚𝑥 cos 𝜃𝑡 + 𝐚𝑧 sin 𝜃𝑡 ) 𝑒 −𝑗𝛽2 (𝑥 sin 𝜃𝑡 +𝑧 cos 𝜃𝑡 ) 𝜂2 𝑡⊥ 𝐸𝑖𝑜⊥ Dr. Ahmed Farghal = (−𝐚𝑥 −𝑗𝐴2 + 𝐚𝑧 sin 𝜃𝑡 )𝑒 −𝛼2 𝑧 𝑒 −𝑗𝑘𝑖𝑥 𝑥 𝜂2 The complex Poynting vector for the surface wave fields 𝐸𝑖𝑜⊥ 2 𝑡⊥ 2 𝐒𝑡 = 𝐄𝑡 × 𝐇𝑡∗ = 𝐚𝑧 𝑗𝐴2 + 𝐚𝑥 sin 𝜃𝑡 𝑒 −2𝛼𝑧 𝜂2 This shows that no real power flow occurs in the 𝑧 direction. The real power flow in the 𝑥 direction is that of the surface wave Fall 2024 field, and it decays exponentially with distance into region 2. So even though no real power is transmitted into region 2, a nonzero field does exist there, in order to satisfy B.Cs at the interface. 24  Example Lecture 5 An EM wave travels in free space with the electric field component 𝐄𝑠 = 100 𝑒 𝑗(0.866𝑦+0.5𝑧) 𝐚𝑥 V/m Determine 𝐤 = 𝑘𝑖 𝐚𝑥 + 𝑘𝑖 𝐚𝑦 + 𝑘𝑧𝑖 𝐚𝑧 (a) 𝜔 and 𝜆 𝐫 = 𝑥𝐚𝑥 + 𝑦𝐚𝑦 + 𝑧𝐚𝑧 (b) The magnetic field component Dr. Ahmed Farghal 𝐤 ∙ 𝐫 = 𝑘𝑥 𝑥 + 𝑘𝑦 𝑦 + 𝑘𝑧 𝑧 (c) The time average power in the wave (a) Comparing the given E with 𝐄 = 𝐸𝑜 𝑒 𝑗𝐤⋅𝐫 𝐚𝐸 = 𝐸𝑜 𝑒 𝑗(𝑘𝑥 𝑥+𝑘𝑦 𝑦+𝑘𝑧 𝑧) 𝐚𝑥 it is clear that 𝑘𝑥 = 0, 𝑘𝑦 = 0.866, 𝑘𝑧 = 0.5 Thus, Fall 2024 𝑘= 𝑘𝑥2 + 𝑘𝑦2 + 𝑘𝑧2 = 0 2 + 0.866 2 + 0.5 2 =1 25  But in free space, Example 𝜔 2𝜋 Lecture 5 𝑘 = 𝛽 = 𝜔 𝜇𝑜 𝜀𝑜 = = 𝑐 𝜆 Hence, 8 2𝜋 𝜔 = 𝑘𝑐 = 3 × 10 radΤs , 𝜆= = 2𝜋 = 6.283 m 𝑘 (b) The corresponding magnetic field is given by Dr. Ahmed Farghal 𝐄 0.866𝐚𝑦 + 0.5𝐚𝑧 100 𝑒 𝑗(0.866𝑦+0.5𝑧) 𝐚𝑥 𝐇 = 𝐚𝑘 × = × 𝜂1 2 0.866 + 0.5 2 120𝜋 100 𝑒 𝑗(0.866𝑦+0.5𝑧) 𝐇 = 0.5𝐚𝑦 − 0.866𝐚𝑧 × 120𝜋 𝑗(0.866𝑦+0.5𝑧) 𝐇 = 132.63𝐚𝑦 − 229.7𝐚𝑧 𝑒 mA/m (c) The time-average power is 𝐄𝑠 = 100 𝑒 𝑗(0.866𝑦+0.5𝑧) 𝐚𝑥 1 𝐸 2 𝑜 𝐒 = Re 𝐄 × 𝐇 ∗ = 𝐚𝑘 Fall 2024 2 2𝜂 100 2 = 0.866𝐚𝑦 + 0.5𝐚𝑧 = 11.49𝐚𝑦 + 6.631𝐚𝑧 WΤm2 2(120𝜋) 26  Example A uniform plane wave in air with Lecture 5 𝐄𝑖 = 8 cos 𝜔𝑡 − 4𝑥 − 3𝑧 𝐚𝑦 V/m is incident on a dielectric slab (𝑧 ≥ 0) with 𝜇𝑟 = 1.0, 𝜀𝑟 = 2.5, 𝜎 = 0. Find (a) The polarization of the wave (b) The angle of incidence Dr. Ahmed Farghal (c) The reflected E field (d) The transmitted H field (a) From 𝐄𝑖 , it is evident that the propagation vector is 2 2 𝜔 𝐤 𝑖 = 4𝐚𝑥 + 3𝐚𝑧 → 𝑘𝑖 = 4 + 3 = 5 = 𝜔 𝜇𝑜 𝜀𝑜 = 𝑐 𝜔 = 5𝑐 = 15 × 108 rad/s A unit vector normal to the interface (𝑧 = 0) is 𝐚𝑧. Fall 2024 The plane containing k and 𝐚𝑧 is 𝑦 = constant, which is the 𝑥𝑧-plane, the plane of incidence. Since 𝐄𝑖 is normal to this plane, we have perpendicular polarization. 27  Example Lecture 5 (b) The propagation vectors are illustrated in Figure where it is clear that Dr. Ahmed Farghal 𝐤𝑖 4𝐚𝑥 + 3𝐚𝑧 4𝐚𝑥 + 3𝐚𝑧 𝐚𝑘𝑖 = = = 𝐤𝑖 𝑘𝑖 5 𝑘𝑖𝑥 4 4 tan 𝜃𝑖 = = → 𝜃𝑖 = tan−1 = 53.13° 𝑘𝑖𝑧 3 5 Fall 2024 Alternatively, we can obtain 𝜃𝑖 from the fact that 𝜃𝑖 is the angle between k and 𝐚𝑛 ; that is 4𝐚𝑥 + 3𝐚𝑧 3 cos 𝜃𝑖 = 𝐚𝑘𝑖 ∙ 𝐚𝑛 = ∙ 𝐚𝑧 = → 𝜃𝑖 = 53.13° 5 5 28  Example Let Lecture 5 𝐄𝑟 = 𝐸𝑟𝑜 cos 𝜔𝑡 − 𝐤 𝑟. 𝐫 𝐚𝑦 From Figure 𝐤 𝑟 = 𝑘𝑟𝑥 𝐚𝑥 − 𝑘𝑟𝑧 𝐚𝑧 where 𝑘𝑟𝑥 = 𝑘𝑟 sin 𝜃𝑟 , 𝑘𝑟𝑧 = 𝑘𝑟 cos 𝜃𝑟 Dr. Ahmed Farghal But 𝜃𝑟 = 𝜃𝑖 and 𝑘𝑟 = 𝑘𝑖 = 5 because both 𝑘𝑟 and 𝑘𝑖 are in the same medium. ⟹ 𝐤 𝑟 = 4𝐚𝑥 − 3𝐚𝑧 To find 𝐸𝑟𝑜 , we need Γ⊥ ⟹ we need 𝜃𝑡. From Snell’s law 𝑣𝑝2 𝑐 𝜇𝑟1 𝜀𝑟1 sin 53.13° sin 𝜃𝑡 = sin 𝜃𝑖 = sin 𝜃𝑖 = 𝑣𝑝1 𝑐 𝜇𝑟2 𝜀𝑟2 2.5 Fall 2024 ⇒ 𝜃𝑡 = 30.39° 𝐸𝑟𝑜 𝜂2 cos 𝜃𝑖 − 𝜂1 cos 𝜃𝑡 Γ⊥ = = 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 29  Example Lecture 5 𝜇𝑜 𝜇𝑟2 377 Where 𝜂1 = 𝜂𝑜 = 377 Ω, 𝜂2 = = = 238.4 Ω 𝜀𝑜 𝜀𝑟2 2.5 238.4 cos 53.13° − 377 cos 30.39° Γ⊥ = = −0.389 238.4 cos 53.13° + 377 cos 30.39° Hence, Dr. Ahmed Farghal 𝐸𝑟𝑜 = Γ⊥ 𝐸𝑖𝑜 = −0.389 8 = −3.112 And 𝐄𝑟 = −3.112 cos 15 × 108 𝑡 − 4𝑥 + 3𝑧 𝐚𝑦 V/m (d) Similarly, let the transmitted electric field be 𝐄𝑡 = 𝐸𝑡𝑜 cos 𝜔𝑡 − 𝐤 𝑡. 𝐫 𝐚𝑦 Where Fall 2024 𝜔 15 × 108 𝑘𝑡 = 𝛽2 = 𝜔 𝜇2 𝜀2 = 𝜇𝑟2 𝜀2𝑟 = 8 1 × 2.5 = 7.906 𝑐 3 × 10 30  Example Lecture 5 From figure 𝑘𝑡𝑥 = 𝑘𝑡 sin 𝜃𝑡 = 7.906 sin 30.39° = 4 𝑘𝑡𝑧 = 𝑘𝑡 cos 𝜃𝑡 = 7.906 cos 30.39° = 6.819 Or 𝐤 𝑡 = 4𝐚𝑥 + 6.819 𝐚𝑧 Dr. Ahmed Farghal Notice that 𝑘𝑖𝑥 = 𝑘𝑟𝑥 = 𝑘𝑡𝑥 as expected. 𝐸𝑡𝑜 2𝜂2 cos 𝜃𝑖 𝜏⊥ = = 𝐸𝑖𝑜 𝜂2 cos 𝜃𝑖 + 𝜂1 cos 𝜃𝑡 2 × 238.4 cos 53.13° = = 0.611 238.4 cos 53.13° + 377 cos 30.39° The same result could be obtained from the relation 𝜏⊥ = 1 + Γ⊥. Hence, Fall 2024 𝐸𝑡𝑜 = 𝜏⊥ 𝐸𝑖𝑜 = 0.611 × 8 = 4.888 𝐄𝑡 = 4.888 cos 15 × 108 𝑡 − 4𝑥 − 6.819𝑧 𝐚𝑦 V/m 31  Example Lecture 5 From 𝐄𝑡 , 𝐇𝑡 is easily obtained as 𝐄𝑡 𝐇𝑡 = 𝐚𝐤 𝑡 × 𝜂2 4𝐚𝑥 + 6.819 𝐚𝑧 4.888 cos 15 × 108 𝑡 − 4𝑥 − 6.819𝑧 𝐚𝑦 = × 42 + 6.819 2 238.4 Dr. Ahmed Farghal 𝐇𝑡 = −17.69𝐚𝑦 + 10.37𝐚𝑧 cos 15 × 108 𝑡 − 4𝑥 − 6.819𝑧 mA/m Fall 2024 32  Fall 2024 Dr. Ahmed Farghal Lecture 5 APPLICATION NOTE 33  MICROWAVES At the moment, there are three means for carrying thousands of channels Lecture 5 over long distances: (a) microwave links, (b) coaxial cables, and (c) fiber optic. Microwaves are EM waves whose frequencies range from approximately 300 MHz to 300 GHz. Dr. Ahmed Farghal For comparison, signal from an AM radio station is about 1 MHz, while that from an FM station is about 100 MHz. You may be familiar with microwave appliances such as the microwave oven, which operates at 2.4 GHz, the satellite TV receiver, which operates @ ≈ 4 GHz, and the police radar, which works at about 22 GHz. Features that make microwaves attractive for communications 1. wide available bandwidths (capacities to carry information) Fall 2024 a 1% bandwidth, for example, provides more absolute frequency range at MW frequencies than that at HF. 2. directive (line of sight (LOS)) properties of short wavelengths. 34  MICROWAVES Since the amount of information that can be transmitted is limited Lecture 5 by the available bandwidth, the microwave spectrum provides more communication channels than the radio and TV bands. 𝑆 𝐶 = 𝐵 log 2 1 + 𝑁 With the ever-increasing demand for channel allocation, Dr. Ahmed Farghal microwave communications has become more common. A microwave system normally consists of a transmitter (including a microwave oscillator, waveguides, and a transmitting antenna) a receiver subsystem (including a receiving antenna, transmission line or waveguide, microwave amplifiers, and a receiver). Fall 2024 A microwave network is usually an interconnection of various microwave components and devices. There are several microwave components and variations of these components. 35  Common Microwave Components Coaxial cables: are TLs for interconnecting microwave components. Lecture 5 Resonators: are usually cavities in which EM waves are stored. Waveguide sections: may be straight, curved, or twisted. Antennas: transmit or receive EM waves efficiently Terminators: are designed to absorb the input power and therefore act Dr. Ahmed Farghal as one-port networks. Attenuators: are designed to absorb some of the EM power passing through the device, thereby ↓ the power level of the microwave signal. Directional couplers: consist of two waveguides and a mechanism for coupling signals between them. Isolators: allow energy flow in only one direction. Circulators: are designed to establish various entry/exit points where Fall 2024 power can be either fed or extracted. Filters: suppress unwanted signals and/or separate signals of different frequencies. 36  1- Telecommunications Lecture 5 Telecommunications (the transmission of analog or digital information from one point to another) This is the largest application of microwave frequencies. Microwaves propagate along a straight line like light rays and are not bent by the ionosphere as are signals of LF. Dr. Ahmed Farghal Consequently, communication links between (and among) satellites and terrestrial stations are possible. Communication satellite is a microwave relay station that is used to link two or more ground-based transmitters and receivers. The satellite receives signals at one frequency, repeats or amplifies them, and transmits at another frequency. Fall 2024 Two common modes of operation for satellite communication: ▪ point-to-point link in Figure (a) ▪ multiple links between one ground-based transmitter and several ground-based receivers in Figure (b). 37  Fall 2024 Dr. Ahmed Farghal Lecture 5 1- Telecommunications 38  2- Radar systems Radar systems provided the major stimulant for the development Lecture 5 of microwave technology because they give better resolution for radar instruments at higher frequencies. Only the microwave region of the spectrum could provide the required resolution with antennas of reasonable size. Dr. Ahmed Farghal The ability to focus a radiated wave sharply is what makes microwaves so useful in radar applications. Radar is used to detect aircraft, guide supersonic missiles, observe and track weather patterns, and control flight traffic at airports. Fall 2024 It is also used in burglar alarms, garage-door openers, and police speed detectors. 39  3- Heating Microwave energy is more easily directed, controlled, and Lecture 5 concentrated than low-frequency EM waves. Also, various atomic and molecular resonances occur @ microwave frequencies, creating diverse application areas in basic science, remote sensing, and heating methods. Dr. Ahmed Farghal The heating properties of microwave power are useful in a wide variety of commercial and industrial applications. The microwave oven is a typical example. When the magnetron oscillates, microwave energy is extracted from the resonant cavities. The reflections from the stationary walls and the motion of the stirring fan cause the microwave energy to be well distributed. Fall 2024 Thus the microwave enables the cooking process to be fast and even. Microwave heating properties also are used in physical diathermy and in drying potato chips, paper, cloth, and so on. 40  Fall 2024 Dr. Ahmed Farghal Lecture 5 3- Heating 41  4- 60 GHz TECHNOLOGY The next-generation wireless technology—namely, 60 GHz—can provide Lecture 5 wireless connectivity for short distances between electronic devices at speeds in the multigiga byte per second range. @ such a high frequency, the wavelength is nearly 5 mm. Although millimeter wave (mmWave) technology has been known for a long time, it was used initially only for military applications. Dr. Ahmed Farghal With the strides in process technologies and low-cost integration solutions, academia, industry, and standardization bodies also turned to mmWave technology. mmWave refers to EM spectrum that spans 30 to 300 GHz, corresponding to wavelengths from 10 mm down to 1 mm. At these smaller wavelengths, the data rates are expected to be 40 to 100 times faster than current wireless technologies for local-area networks. Fall 2024 42  4- 60 GHz TECHNOLOGY There is a large bandwidth (up to 7 GHz) available worldwide. Lecture 5 The 60 GHz technology is less restricted in terms of power limits. @ 60 GHz, the path loss is higher, but higher transmitting power overcomes this, especially when the operation is restricted to indoor environments. The effective interference levels for 60 GHz are less severe than those systems located in the congested 2–2.5 and 5–5.8 GHz bands. Dr. Ahmed Farghal In addition, higher frequency reuse can also be achieved in indoor environments, allowing a very high throughput. The compact size of the 60 GHz radio band permits the use of multiple antenna arrays, which can be conveniently integrated into consumer electronic products. Operators at these bands are exempt from license fees. Narrow beamwidth is possible. For example, this technology can be applied to an in-flight entertainment distribution system without causing interference with Fall 2024 flight controls or navigation equipment. Oxygen absorption does not pose a problem when a 60 GHz system is used between satellites. 43  5- Stealth Aircraft There are two methods of avoiding detection by radar. Lecture 5 1. ensure that the reflection coefficient of the aircraft, as a whole, is as nearly as possible close to zero. If there is no energy reflected back from the aircraft, there is no energy reaching the antenna of the radar and the aircraft is “transparent” to the incoming wave. To do so, the aircraft is coated with materials which have the same intrinsic impedance as air but which also absorb (dissipate) energy. Dr. Ahmed Farghal The latter is required because if it were not for this, the wave would propagate through the coating and reflect off the metallic surfaces of the aircraft. 𝜇 𝜇0 Materials appropriate for this purpose are those for which =. 𝜀 𝜀0 The required ratio is usually obtained by varying the permeability by adding ferromagnetic powders. The absorption of energy must be over a wide enough spectrum to avoid detection by shifting frequencies of the radar system (shifting frequencies is the simplest way to detect “undetectable” aircraft). Fall 2024 In most cases, radar-absorbing materials are used only where necessary (such as engine intakes, wing tips and edges, etc.) to reduce rather than eliminate the aircraft radar visibility. 44  5- Stealth Aircraft 2. A second method of avoiding detection is to reflect the incoming waves Lecture 5 but to deflect these in directions away from the antenna. In this method, no energy is absorbed, but little is reflected back to the antenna. Aircraft of this type will have sharp angles, as shown in Figure. The sharper the corners, the less energy will be reflected. However, the flat surfaces employed are quite visible if viewed Dr. Ahmed Farghal from a steep enough angle. In the example in Figure, the aircraft is visible from underneath or even from above, but these are not normal angles of observation. Typically a radar installation will try to detect aircraft at low angles, possibly from the front or side. For these angles, the bottom flat surface in Figure is not detectable. Fall 2024 45  6- Scattering of Waves The reflection of waves by any material, including perfect dielectrics, is Lecture 5 due to the fact that the reflection coefficient is almost always ≠ 𝟎. For example, the atmosphere has a permittivity different than free space and the permittivity differs from place to place depending on atmospheric pressure and weather conditions. Dr. Ahmed Farghal Similarly, any substance in the atmosphere (e.g., dust, cloud, an airplane, rain or snow, or a pressure) will have permittivities that differ from that of air. These variations may be harmful to communication in that some energy is reflected in various directions (scattered) rather than serving a useful purpose, whereas in some cases, this scattering is quite useful. Because of scattering, many of the effects mentioned above can be Fall 2024 detected by measuring the reflectivity of the materials or conditions present. This is extremely important in weather prediction and remote sensing of the environment. 46  6- Scattering of Waves Other applications include communication such as the Lecture 5 tropospheric scattering method shown in Figure. In this method, the transmitter sends a rather narrow beam upward into the troposphere. The waves are scattered and some of the scattered waves are then reflected back into the receiver. Dr. Ahmed Farghal With this method there is no need for a reflector to reflect the waves back into the receiver; use is made of the natural reflections that occur in the troposphere. Fall 2024 47  Scattering of Waves Another simple use of scattering is in microwave testing of Lecture 5 lossless dielectrics, shown in Figure b. A microwave beam illuminates the test sample. Some of the waves propagate through the material and some are reflected back into the transmitter. Dr. Ahmed Farghal However, none will be coupled to the upper receiver. If, however, there are inclusions, defects, etc., in the material, these will scatter waves in many directions, some of which will be received in the upper receiver. This reception is then an indication of the defects or foreign materials in the test sample. Fall 2024 48  Microwave Reflectors Many antennas rely on reflectors to guide the beam in specific Lecture 5 directions. A typical parabolic (dish) antenna is shown in Figure. It consists of a parabolic dish very much like the surface in a car’s headlights. The antenna itself (also called a feed) is a small horn located in the focal point of the reflector (similar to the bulb in the Dr. Ahmed Farghal headlight). The feed radiates toward the reflector and the reflector then reflects the beam into the direction required. These antennas are highly useful because they transmit energy in narrow beams in the required direction. They are common in satellites and other communication systems. Fall 2024 The common thread in these applications is the need for good reflectivity. In antenna reflectors, this is obtained by use of highly conductive, polished materials. 49  Radomes and Dielectric Windows It is often necessary to transmit EM waves from one area into Lecture 5 another through a physical barrier. Radome. Whenever an antenna must be physically separated from the environment, its energy can only be transmitted and received through this barrier. For example, in an airplane, the radar antenna must be located Dr. Ahmed Farghal within the body of the airplane for aerodynamic purposes. A window is then provided to allow transmission and reception. On ships, the antenna must be protected from the environment by a cover. Fall 2024 50  Radomes and Dielectric Windows Lecture 5 A radome may be designed in two ways. 1. One is to choose permittivities and permeabilities such that the intrinsic impedance of the radome material equals that of the surrounding domain (air or free space). For this purpose, the material must be a lossless dielectric with Dr. Ahmed Farghal 𝜇 𝜇 material properties such that = 0. 𝜀 𝜀0 This method has the distinct advantage that the design is independent of frequency. 2. The second method is to choose a perfect dielectric and design its thickness such that there are no reflections at the required frequency. In practical applications, it is often required to switch frequencies, Fall 2024 and the design of the radome must be such that it is transparent at all required frequencies, a design which is often difficult to achieve. 51  Have a nice day. Dr. Ahmed Farghal 52