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Questions: (write with your own words) 1. What are the modes of propagation of electromagnetic waves? 2. What does it mean to polarize an electromagnetic wave? 3. What is the phenomenon of total reflection of an electromagnetic wave? 4. What is the phenomenon of total refraction of an electromagnetic wave? 5. What is the purpose of Snell's Law in the study of the propagation of waves?

1. An electromagnetic wave of 𝑓 = 22 𝑀𝐻𝑧 and 𝑃1+ = 200𝑚𝑊/𝑚2, incident from the air (𝜂1 = 120𝜋 𝛺 ), perpendicular to an infinite wall with an intrinsic impedance 𝜂2 = 105 𝛺. Calculate the reflected power 𝑃1− and the transmitted power 𝑃2+ to the wall.

Figure 1: Propagation of “normal wave” in infinite medium. 𝛤= 𝛤=

𝑛2 − 𝑛1 𝑛2 + 𝑛1

105 − 120𝜋 = −0.56 = < 180° 105 + 120𝜋

Reflectance:

𝑅 = |𝛤|2 𝑅 = |0.56|2 = 0.313 = 31.3% Transmittance:

𝑇 = 1 − 𝑅 = 68.7%

Reflected power:

𝑃1− = 31.3% 𝑥200

𝑚𝑊 = 62.6𝑚𝑊/𝑚2 2 𝑚

Transmitted Power:

𝑃2+ = 68.7% 𝑥200

𝑚𝑊 = 137.4𝑚𝑊/𝑚2 𝑚2

2. An electromagnetic wave of 𝑓 = 22𝑀 𝐻𝑧 and 𝑃1+ = 200𝑚𝑊/𝑚2 , coming from a wave generator located 30𝑐𝑚 from the wall, which impinges from the air (𝜂1 = 120𝜋 𝛺) perpendicularly on a wall with an intrinsic impedance 𝜂2 = 105 𝛺 and 10𝑐𝑚 thick. The wall is made of a non-magnetic and non-dissipative material. On the other side of the wall is a receiver located 20cm away.

Figure 2: Propagation of “normal wave” in finite medium.

a. Calculate the coefficient of reflection and transmission seen by the generator. b. Determine in [%] and [𝑚𝑊/𝑚2 ] the power that is transmitted to the receiver.

120𝜋 2 𝐸𝑟 = ( ) = 12.89 105𝛺 Wall phase constant: 𝛽=

(2𝜋) ∗ (22𝑥106 ) ∗ √12.89 3 ∗ 108

𝛽 = 1.654 rad/m Input impedance: 𝑛𝑖𝑛 = 𝑛1

𝑛2 + 𝑗𝑛1 𝑇𝑎𝑛(𝛽𝑥 ) 𝑛1 + 𝑗𝑛2 𝑇𝑎𝑛(𝛽𝑥 )

rad

𝑛𝑖𝑛

(120𝜋105 𝛺) + 𝑗(105 𝛺)𝑇𝑎𝑛(1.654 m ∗ 0.1) = (105 𝛺) rad (105 𝛺) + 𝑗(120𝜋𝛺)𝑇𝑎𝑛(1.546 m ∗ 0.1)

𝑛𝑖𝑛 = 238.3366 − 168.620𝑖

𝛤1 =

𝛤1 =

𝑛𝑖𝑛 − 𝑛𝑎𝑖𝑟𝑒 𝑛𝑖𝑛 + 𝑛𝑎𝑖𝑟𝑒

(238.3366 − 168.620𝑖) − (120𝜋 𝛺) = −0.1397 − 0.3123𝑖 = 0.3421 (65.9°) (238.3366 − 168.620𝑖) + (120𝜋 𝛺)

𝜏1 = 1 + 𝛤1 𝜏1 = 1 + (−0.1397 − 0.3123𝑖) = 0.86025 − 0.3123𝑖

𝑆𝑊𝑅 =

1 + |𝛤1 | 1 + |0.3421| = 1 − |𝛤1 | 1 − |0.3421| 𝑆𝑊𝑅 = 2.0399

𝑇1 = 1 − |𝛤1 |2 = 0.8829 = 88.29% 𝑅1 = 1 − 𝑇1 𝑅1 = 1 − (0.8829) = 0.117 = 11.7% + 𝑃1− = 𝑅1 ∗ 𝐸𝑦1 = (0.117)200 = 23.4 𝑚𝑊⁄𝑚2

𝛤2 =

𝑛𝑎𝑖𝑟𝑒 − 𝑛𝑝𝑎𝑟𝑒𝑑 𝑛𝑎𝑖𝑟𝑒 + 𝑛𝑝𝑎𝑟𝑒𝑑

𝛤2 =

(120𝜋 𝛺) − 112 = 0.5419 (120𝜋 𝛺) + 112

𝑇2 = 1 − |𝛤2 |2 = 0.7063 = 70.63% 𝑅1 = 1 − 𝑇2 = 0.2937 = 29.37%

𝑃2− = 𝑅1 ∗ 200 𝑚𝑊⁄𝑚2 𝑃2− = (0.2937) ∗ 200 𝑚𝑊⁄𝑚2 = 58.7 𝑚𝑊⁄𝑚2 𝑇𝑡 = 𝑇1 ∗ 𝑇2 𝑇𝑡 = (0.8829) ∗ (0.7063) = 0.62359 = 62.3% 𝑃3+ = 𝑇𝑡 ∗ 200 𝑚𝑊⁄𝑚2 𝑃3+ = (0.62359) ∗ 200 𝑚𝑊⁄𝑚2 = 124.718 𝑚𝑊⁄𝑚2 3. An electromagnetic wave propagates through several media as shown in the graph.

Figure 3: Propagation of “oblique wave” in finite media.

Initially the ray travels through the ice layer striking the air layer at point B, forming an angle of 𝜃𝑎 = 44,9°. Using Snell's Law, calculate step by step the total path of the wave to determine the value of "d". Note that each layer is 105 𝑚𝑚 thick and that at point C there is a total refractive effect, so it is necessary to identify which material is in layer 3 by calculating its refractive index.

𝜃1 = 90° − 𝜃𝑎 = 45.1° 𝑛1 sin(𝜃1 ) = 𝑛2 sin(𝜃2 ) 𝑛1 𝜃2 = 𝑠𝑖𝑛−1 ( 𝑠𝑖𝑛(𝜃1 )) 𝑛2 𝜃2 = 𝑠𝑖𝑛−1 (

1.31 𝑠𝑖𝑛(45.1°)) 1,0002926

𝜃2 = 68.071° 𝜃2 = 𝜃𝑏 𝐷𝑎, 𝑏 = 105𝑚𝑚 ∗ tan(68.071°) 𝐷𝑎, 𝑏 = 260.814𝑚𝑚 𝜃𝑏 + 𝜃𝑐 = 90° 𝜃𝑐 = 90° − 𝜃𝑏 𝜃𝑐 = 90° − 68.071° 𝜃𝑐 = 21.929° 𝑛3 sin(𝜃3 ) = 𝑛2 sin(𝜃2 ) 𝑛3 = 𝑛3 =

𝑛2 sin(𝜃2 ) sin(𝜃3 )

(1.0002926) sin(68.071°) sin(21.929°) 𝑛3 = 2.4846

𝐷𝑏, 𝑐 = 105𝑚𝑚 ∗ tan(21.929°) = 42.271𝑚𝑚 𝑛3 𝜃4 = 𝑠𝑖𝑛−1 ( sin(𝜃𝑐 )) 𝑛4 𝜃4 = 𝑠𝑖𝑛−1 (

2.4846 sin(21.929°)) 2.42

𝜃4 = 22.546°° 𝐷𝑐, 𝑑 = 105𝑚𝑚 ∗ tan(22.546°) 𝐷𝑐, 𝑑 = 43.59𝑚𝑚 𝐷 = 𝐷𝑎, 𝑏 + 𝐷𝑏, 𝑐 + 𝐷𝑐, 𝑑 𝐷 = (260.814𝑚𝑚) + (42.271𝑚𝑚) + (43.59𝑚𝑚) 𝐷 = 346.675𝑚𝑚