Topics covered in this week: • Special case-1 of reflection: 𝛼1

1 | P a g e Umm Al-Qura Universtiy, Makkah

Department of Electrical Engineering Electromagnetics (2) - (8022320)

Term 1: 2021/2022

Solution Homework 7 Dr. Waheed Ahmad Younis

Do not submit this homework. It will be included in the midterm on Monday, 25 Oct 2021.

Topics covered in this week:

• Special case-1 of reflection: 𝛼₁ = 0, 𝜂₂ = 0 𝐻𝑒𝑛𝑐𝑒 Γ = −1

Net field in media-1: 𝐸_𝑥1 = 2𝐸_𝑥10⁺ sin 𝜔𝑡 sin 𝛽₁𝑧 Field is zero at 𝑧 = 𝑛^𝜆1

2

𝐻_𝑦1 = 2^𝐸𝑥10+

𝜂1 cos 𝜔𝑡 cos 𝛽₁𝑧 Field is zero at 𝑧 = (¹

2+ 𝑛)^𝜆1

2

𝑃_𝑎𝑣(𝑧) = 0 These are standing waves.

• Special case-2 of reflection: 𝛼₁ = 0, 𝜂₁ = 𝜂₂ 𝐻𝑒𝑛𝑐𝑒 Γ = 0 Net field in media-1: 𝐸_𝑥1 = 𝐸_𝑥10⁺ cos(𝜔𝑡 − 𝛽₁𝑧)

𝐻_𝑦1 = ^𝐸𝑥10+

𝜂1 cos(𝜔𝑡 − 𝛽₁𝑧) These are travelling waves.

• General case of reflection: 𝛼₁ = 0 𝑎𝑛𝑑 Γ = |Γ|𝑒^𝑗𝜙 Incident wave: 𝐸_𝑥𝑠1⁺ = 𝐸_𝑥10⁺ 𝑒^{−𝑗𝛽1𝑧}

Reflected wave: 𝐸_𝑥𝑠1⁻ = |Γ|𝑒^𝑗𝜙𝐸_𝑥10⁺ 𝑒^{𝑗𝛽1𝑧} = |Γ|𝐸_𝑥10⁺ 𝑒^{𝑗(𝛽1𝑧+𝜙)}

Net field in media-1: 𝐸_𝑥𝑠1 = 𝐸_𝑥𝑠1⁺ + 𝐸_𝑥𝑠1⁻ = 𝐸_𝑥10⁺ 𝑒^{−𝑗𝛽1𝑧}+ |Γ|𝐸_𝑥10⁺ 𝑒^{𝑗(𝛽1𝑧+𝜙)}

= 𝐸_𝑥10⁺ [𝑒^{−𝑗𝛽1𝑧}+ |Γ|𝑒^{𝑗(𝛽1𝑧+𝜙)}]

Maximum value of field: 𝐸_{𝑥𝑠1,𝑚𝑎𝑥} = 𝐸_𝑥10⁺ [1 + |Γ|] At −𝛽₁𝑧_𝑚𝑎𝑥 =^𝜙

2 + 𝑛𝜋 Minimum value of field: 𝐸_{𝑥𝑠1,𝑚𝑖𝑛} = 𝐸_𝑥10⁺ [1 − |Γ|] At −𝛽₁𝑧_𝑚𝑖𝑛 =^𝜙

2 + 𝑛𝜋 +^𝜋

2

Standing wave ratio = 𝑆𝑊𝑅 =^{𝐸𝑥𝑠1,𝑚𝑎𝑥}

𝐸_{𝑥𝑠1,𝑚𝑖𝑛} = ^1+|Γ|

1−|Γ|

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Q1. Consider a standing wave in a lossless region 𝐸⃗ = 200 sin 10⁹𝑡 sin 20𝑧 𝑎⃗⃗⃗⃗ V/m _𝑥

If the amplitude of associated magnetic field is 1 A/m, find (a) 𝜇 𝑎𝑛𝑑 𝜖 for the medium, (b) 𝐻⃗⃗ , (c) 𝐸⃗⃗⃗⃗ _𝑠. Solution:

a. 𝜂 = ^𝐸

𝐻= √^𝜇_𝜖 = 200 ⇒ ^𝜇

𝜖 = 4 × 10⁴ Also 𝛽 = 𝜔√𝜇𝜖 = 20 ⇒ 𝜇𝜖 = (₁₀²⁰9)² = 4 × 10⁻¹⁶

Hence (^𝜇

𝜖) (𝜇𝜖) = (4 × 10⁴)(4 × 10⁻¹⁶) = 16 × 10⁻¹² ⇒ 𝜇 = 4 𝜇𝐻 𝑚⁄ Also 𝜇𝜖 = 4 × 10⁻¹⁶ ⇒ 𝜖 = ^4×10−16

𝜇 =^4×10−16

4×10⁻⁶ = 10⁻¹⁰ = 100 𝑝𝐹/𝑚 b. 𝐻⃗⃗ = cos 10⁹𝑡 cos 20𝑧 𝑎⃗⃗⃗⃗ A/m _𝑦

c. 𝐸⃗ = 200 sin 10⁹𝑡 sin 20𝑧 𝑎⃗⃗⃗⃗ = 200 sin 20𝑧 𝑅𝑒[−𝑗𝑒_𝑥 ^𝑗109𝑡] 𝑎⃗⃗⃗⃗ _𝑥 𝐸_𝑠

⃗⃗⃗⃗ = −𝑗200 sin 20𝑧 𝑎⃗⃗⃗⃗ V/m _𝑥

2 | P a g e Q2. Consider two perfect dielectrics:

𝜇₁ = 𝜇₀, 𝜖₁ = 3.6𝜋 𝑝𝐹 𝑚⁄ , 𝑎𝑛𝑑 𝜎₁ = 0.

𝜇₂ = 𝜇₀, 𝜖₂ = 14.4𝜋 𝑝𝐹 𝑚⁄ , 𝑎𝑛𝑑 𝜎₂ = 0 Find the SWR in dielectric-1.

Solution:

𝜂₁ = √^𝜇_𝜖¹

1 = √ ^{4𝜋×10−7}

3.6𝜋×10⁻¹² = 333.33 Ω 𝜂₂ = √^𝜇_𝜖²

2 = √ ^{4𝜋×10−7}

14.4𝜋×10⁻¹² = 166.67 Ω Γ =^{𝜂2−𝜂1}

𝜂2+𝜂1= 166.67−333.33

166.67+333.33= −0.333 𝑆𝑊𝑅 =^1+|Γ|

1−|Γ|= ^1+0.333

1−0.333= 2

Q3. In region A (x<0), the incident and reflected electric fields are given as 𝐸_𝑠𝐴⁺

⃗⃗⃗⃗⃗⃗ = 1000 𝑒^{−𝑗200𝑥} 𝑎⃗⃗⃗⃗ V/m _𝑦 𝐸_𝑠𝐴⁻

⃗⃗⃗⃗⃗⃗ = 141.3 𝑒𝑗(200𝑥+106°) 𝑎⃗⃗⃗⃗ V/m _𝑦

Find the reflection co-efficient and standing wave ratio.

Solution:

Γ =^{𝐸𝑠𝐴0−}

𝐸_𝑠𝐴0⁺ = ^{141.3 𝑒𝑗(106.08°)}

1000 = 0.1413𝑒^𝑗(106°)= 0.1413∠106°

𝑆𝑊𝑅 =^1+|Γ|

1−|Γ|= ^1+0.1413

1−0.1413= 1.33