Lecture Notes: EEEC6430310 Electromagnetic Fields And Waves - Maxwell's Equation
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Lecture Notes: EEEC6430310 Electromagnetic Fields And Waves - Maxwell's Equation
- 1. EEEC6430310 ELECTROMAGNETIC FIELDS AND WAVES Maxwell’s Equation FACULTY OF ENGINEERING AND COMPUTER TECHNOLOGY BENG (HONS) IN ELECTRICALAND ELECTRONIC ENGINEERING Ravandran Muttiah BEng (Hons) MSc MIET
- 2. Maxwell’s Equation I. Maxwell’s Equation For Linear Media 𝛻 × 𝑬 = −𝜇 𝜕𝑯 𝜕𝑡 Faraday′ s Law 𝛻 × 𝑯 = 𝑱 + 𝜀 𝜕𝑬 𝜕𝑡 Ampere′ s Law 𝛻 ∙ 𝑬 = 𝜌f 𝜀 Gauss′ s Law 𝛻 ∙ 𝑯 = 0 Gauss′ s Law where, 𝛻 = vector differential (del) operator 𝑬 = electric field intensity 𝑯 = magnetic field intensity 𝑱 = electric current density 𝜌f = free electric charge density 1
- 3. 2 II. Pointing’s Theorem A. Power Flow And Electromagnetic Energy 𝛻 ∙ 𝑬 × 𝑯 = 𝑯 ∙ 𝛻 × 𝑬 − 𝑬 ∙ 𝛻 × 𝑯 = −𝜇𝑯 ∙ 𝜕𝑯 𝜕𝑡 − 𝑬 ∙ 𝑱 + 𝜀 𝜕𝑬 𝜕𝑡 = − 𝜇 2 𝜕 𝜕𝑡 𝑯 2 − 𝜀 2 𝜕 𝜕𝑡 𝑬 2 − 𝑬 ∙ 𝑱 𝛻 ∙ 𝑬 × 𝑯 + 𝜕 𝜕𝑡 1 2 𝜀 𝑬 2 + 1 2 𝜇 𝑯 2 = −𝑬 ∙ 𝑱
- 4. 3 V 𝛻 ∙ 𝑬 × 𝑯 d𝑉 = S 𝑬 × 𝑯 ∙ d𝑺 S 𝑬 × 𝑯 ∙ d𝒂 + d d𝑡 V 1 2 𝜀 𝑬 2 + 1 2 𝜇 𝑯 2 d𝑉 = − 𝑉 𝑬 ∙ 𝑱 d𝑉 𝑺 = 𝑬 × 𝑯 Poynting Vector Watt m2 𝑊 = V 1 2 𝜀 𝑬 2 + 1 2 𝜇 𝑯 2 d𝑉 Electromagnetic Stored Energy 𝑃d = V 𝑬 ∙ 𝑱 d𝑉 Power dissipated if 𝐽 ∙ 𝑬 > 0 e.g., 𝑱 = 𝜎𝑬 ⇒ 𝐽 ∙ 𝑬 = 𝜎 𝑬 𝟐 Power source if 𝐽 ∙ 𝑬 < 0 𝑃out = S 𝑬 × 𝑯 ∙ d𝒂 = S 𝑺 ∙ d𝒂 𝑃out + d𝑊 d𝑡 = −𝑃d 𝑤e = 1 2 𝜀 𝑬 2 Electric energy density in Joules m3 𝑤m = 1 2 𝜇 𝑯 2 Magnetic energy density in Joules m3
- 5. 4 Figure 1: The circuit power into an N terminal network 𝑘=1 𝑁 𝑉𝑘𝐼𝑘 equals the electromagnetic power flow into the surface surrounding the network, − s 𝑬 × 𝑯 · d𝑺 B. Power In Electric Circuits E H 𝑆 = 𝐸 × 𝐻 𝑉1 𝑉2 𝑉3 𝑉𝑁−1 𝑉𝑁 𝐼1 𝐼2 𝐼3 𝐼𝑁−1 𝐼𝑁 𝑆
- 6. 5 Outside circuit elements C 𝑬 ∙ d𝒍 ≈ 0, 𝛻 × 𝑬 = 0 ⇒ 𝑬 = −𝛻Ф (Kirchoff’s Voltage Law 𝑘 𝑣𝑘 = 0) 𝛻 × 𝑯 = 𝑱 ⇒ 𝛻 ∙ 𝑱 = 0, S 𝑱 ∙ d𝑺 = 0 (Kirchoff’s Current Law 𝑘 𝑖𝑘 = 0) 𝑃in = − S 𝑬 × 𝑯 ∙ d𝑺 = − V 𝛻 ∙ 𝑬 × 𝑯 d𝑉 𝛻 ∙ 𝑬 × 𝑯 = 𝑯 ∙ 𝛻 × 𝑬 − 𝑬 ∙ 𝛻 × 𝑯 = −𝑬 ∙ 𝑱 = 𝛻Ф ∙ 𝑱 𝛻 ∙ 𝑱Ф = Ф𝛻 ∙ 𝑱 + 𝑱 ∙ 𝛻Ф 𝛻 ∙ 𝑬 × 𝑯 = 𝑱 ∙ 𝛻Ф = 𝛻 ∙ Ф𝑱 𝑃in = − V 𝛻 ∙ 𝑬 × 𝑯 d𝑉 = − V 𝛻 ∙ 𝑱Ф d𝑉 = − S 𝑱Ф ∙ d𝑺 On 𝑆, Ф = voltages on each wire, 𝑱 is non-zero only on wires. 𝑃in = − S 𝑱Ф ∙ d𝑺 = − 𝑘=1 𝑁 𝑣𝑘 S 𝑱 ∙ d𝑺 = 𝑘=1 𝑁 𝑣𝑘𝑖𝑘 0 0 −𝑖𝑘
- 7. C. Complex Poynting’s Theorem (Sinusoidal Steady State, ej𝜔𝑡 ) 6 𝑬 𝒓, 𝑡 = Re 𝑬 𝒓 ej𝜔𝑡 = 1 2 𝑬∗ 𝒓 ej𝜔𝑡 + 𝑬∗ 𝒓 e−j𝜔𝑡 𝑯 𝒓, 𝑡 = Re 𝑯 𝒓 ej𝜔𝑡 = 1 2 𝑯∗ 𝒓 ej𝜔𝑡 + 𝑯∗ 𝒓 e−j𝜔𝑡 The real part of a complex number is one-half of the sum of the number and its complex conjugate
- 8. 7 Maxwell’s Equations In Sinusoidal Steady State 𝛻 × 𝑬 𝒓 = −j𝜔𝜇𝑯 𝒓 𝛻 × 𝑯 𝒓 = 𝑱 𝒓 + j𝜔𝜀 𝒓 𝛻 ∙ 𝑬 𝒓 = 𝝆f 𝒓 𝜀 𝛻 ∙ 𝑯 𝒓 = 0
- 9. 8 𝑺 𝒓, 𝑡 = 𝑬 𝒓, 𝑡 × 𝑯 𝒓, 𝑡 = 1 4 𝑬 𝒓 ej𝜔𝑡 + 𝑬 𝒓 e−j𝜔𝑡 × 𝑯 𝒓 ej𝜔𝑡 + 𝑯∗ 𝒓 e−j𝜔𝑡 = 1 4 𝑬 𝒓 × 𝑯 𝒓 e2j𝜔𝑡 +𝑬∗ 𝒓 × 𝑯 𝒓 + 𝑬 𝒓 × 𝑯∗ 𝒓 + 𝑬∗ 𝒓 × 𝑯∗ 𝒓 e−2j𝜔𝑡 𝑺 = 1 4 𝑬∗ 𝒓 × 𝑯 𝒓 + 𝑬 𝒓 × 𝑯∗ 𝒓 = 1 2 Re 𝑬 𝒓 × 𝑯∗ 𝒓 = 1 2 Re 𝑬∗ 𝒓 × 𝑯 𝒓 (A complex number plus its complex conjugate is twice the real part of that number)
- 10. 9 𝑺 = 1 2 Re 𝑬 𝒓 × 𝑯∗ 𝒓 𝛻 ∙ 𝑺 = 𝛻 ∙ 1 2 𝑬 𝒓 × 𝑯∗ 𝒓 = 1 2 𝑯∗ 𝒓 ∙ 𝛻 × 𝑬 𝒓 − 𝑬 𝒓 ∙ 𝛻 × 𝑯∗ 𝒓 = 1 2 𝑯∗ 𝒓 −j𝜔𝜇𝑯 𝒓 − 𝑬 𝒓 ∙ 𝑱∗ 𝒓 − j𝜔𝜀𝑬∗ 𝒓 = 1 2 −j𝜔𝜇 𝑯 𝒓 2 + j𝜔𝜀 𝑬 𝒓 2 − 1 2 𝑬 𝒓 ∙ 𝑱∗ 𝒓 𝑤m = 1 4 𝜇 𝑯 𝒓 2 , 𝑤e = 1 4 𝜀 𝑬 𝒓 2 𝑷d = 1 2 𝑬 𝒓 ∙ 𝑱∗ 𝒓 𝛻 ∙ 𝑺 + 2j𝜔 𝑤m − 𝑤e = −𝑷d
- 11. III. Transverse Electromagnetic Waves (𝑱 = 0, 𝜌f = 0) 10 The great success of Maxwell’s equations lies partly in the simple prediction of electromagnetic waves and their simple characterization of materials in terms of conductivity 𝜎, permittivity 𝜀, and permeability 𝜇. In vacuum we find 𝜎 = 0, 𝜀 = 𝜀o and 𝜇 = 𝜇o. Therefore, 𝑱 = 𝜎𝑬 = 0 and 𝜌f = 0. A. Wave Equation 𝛻 × 𝑬 = −𝜇 𝜕𝑯 𝜕𝑡 𝛻 × 𝑯 = 𝜀 𝜕𝑬 𝜕𝑡 𝛻 ∙ 𝑬 = 0 𝛻 ∙ 𝑯 = 0 In contrast the nano-structure of media can be quite complex and requires quantum mechanics and for its full explanation. Fortunately, simple classical approximations to atoms and molecules suffice to understand the origin of 𝜎, 𝜀 and 𝜇.
- 12. 11 𝛻 × 𝛻 × 𝑬 = −𝜇 𝜕 𝜕𝑡 𝛻 × 𝐻 = −𝜇 𝜕 𝜕𝑡 𝜀 𝜕𝑬 𝜕𝑡 𝛻 × 𝛻 × 𝑬 = 𝛻 𝛻 ∙ 𝑬 − 𝛻2 𝑬 = −𝜀𝜇 𝜕2 𝑬 𝜕𝑡2 Wave equation, 𝛻2 𝑬 = 1 𝑐2 𝜕2 𝑬 𝜕𝑡2 where, 𝑐 = 1 𝜀𝜇 is the velocity of the electromagnetic wave. In free space (i.e. vacuum), 𝜇 = 𝜇o = 4π × 10−7 henries m and 𝜀 = 𝜀o ≈ 10−9 36π farads m , which leads to 𝑐o = 1 𝜀o𝜇o ≈ 3 × 108 m s . Similarly, 𝛻 × 𝛻 × 𝑯 = 𝛻 𝛻 ∙ 𝑯 − 𝛻2 𝑯 = 𝜀 𝜕 𝜕𝑡 𝛻 × 𝑬 = −𝜀𝜇 𝜕2 𝑯 𝜕𝑡2 𝛻2 𝑯 = 1 𝑐2 𝜕2 𝑯 𝜕𝑡2 , 𝑐 = 1 𝜀𝜇 0 0
- 13. 12 x z y 𝜀1, 𝜇1 𝐸𝑥− = Re 𝑬𝑥− 𝑧 ej𝜔𝑡 𝐻𝑦− = Re 𝑯y− 𝑧 ej𝜔𝑡 𝐻𝑦+ = Re 𝑯y+ 𝑧 ej𝜔𝑡 𝐸𝑥+ = Re 𝑬𝑥+ 𝑧 ej𝜔𝑡 𝐾𝑥+ = Re 𝑲0 𝑧 ej𝜔𝑡 𝜀2, 𝜇2 B. Plane Waves 𝐸𝑥 𝑧, 𝑡 = Re 𝑬𝑥 𝑧 ej𝜔𝑡 d2 𝑬𝑥 d𝑧2 = − 𝜔2 𝑐2 𝑬𝑥 d2𝑬𝑥 d𝑧2 + 𝑘2𝑬𝑥 = 0
- 14. 13 where we have, 𝑘2 = 𝜔2 𝑐2 = 𝜔2 𝜀𝜇 𝑘 = 2π 𝜆 is the wavenumber, 𝜆 is the wavelength 𝑘 = ± 𝜔 𝑐 ⇒ 𝜔 = 𝑘𝑐 𝜔 = 2π𝑓 = 2π 𝜆 𝑐 ⇒ 𝑓𝜆 = 𝑐 𝑬𝑥 = 𝐴1ej𝑘𝑧 + 𝐴2e−j𝑘𝑧 𝐸𝑥 = Re 𝐴1ej 𝜔𝑡+𝑘𝑧 + 𝐴2ej 𝜔𝑡−𝑘𝑧 For the wave in the −𝑧 direction we have: 𝜔𝑡 + 𝑘𝑧 = constant, 𝜔d𝑡 + 𝑘d𝑧 = 0 d𝑧 d𝑡 = − 𝜔 𝑘 = −𝑐 travelling wave in the − 𝑧 direction travelling wave in the +𝑧 direction
- 15. 14 For the wave in the +𝑧 direction we have: 𝜔𝑡 − 𝑧 = constant, 𝜔d𝑡 − 𝑘d𝑧 = 0, d𝑧 d𝑡 = 𝜔 𝑘 = +𝑐 𝑬𝑥 𝑧 = 𝑬𝑥+e−j𝑘𝑧 𝑧 > 0 𝑬𝑥−ej𝑘𝑧 𝑧 < 0 𝛻 × 𝑬 = −𝜇 𝜕𝑯 𝜕𝑡 ⇒ d𝑬𝑥 d𝑧 = −j𝜔𝜇𝑯𝑦 ⇒ 𝑯𝑦 = − 1 j𝜔𝜇 d𝑬𝑥 d𝑧 𝛻 × 𝑯 = 𝜀 𝜕𝑬 𝜕𝑡 ⇒ d𝑯𝑦 d𝑧 = −j𝜔𝜀𝑬𝑥 𝑯𝑦 = 𝑘 𝜔𝜇 𝑬𝑥+e−j𝑘𝑧 𝑧 > 0 − 𝑘 𝜔𝜇 𝑬𝑥−ej𝑘𝑧 𝑧 < 0 𝑘 𝜔𝜇 = 𝜔 𝜀𝜇 𝜔𝜇 = 𝜀 𝜇 , 𝜂 = 𝜇 𝜀 is the wave impedance 𝑯𝑦 = 𝑬𝑥+ 𝜂 e−j𝑘𝑧 𝑧 > 0 − 𝑬𝑥− 𝜂 ej𝑘𝑧 𝑧 < 0
- 16. 15 Now we look at the boundary conditions: 𝑬𝑥 𝑧 = 0+, 𝑡 = 𝑬𝑥 𝑧 = 0−, 𝑡 ⇒ 𝑬𝑥+ = 𝑬𝑥− 𝑯𝑦 𝑧 = 0−, 𝑡 − 𝑯𝑦 𝑧 = 0+, 𝑡 = 𝑲𝑥 𝑧 = 0, 𝑡 ⇒ −𝑬𝑥− − 𝑬𝑥+ 𝜂 = 𝑲0 𝑬𝑥+ = 𝑬𝑥− = − 𝜂𝑲0 2 𝑬𝑥 𝑧 = − 𝜂𝑲0 2 e−j𝑘𝑧 𝑧 > 0 − 𝜂𝑲0 2 ej𝑘𝑧 𝑧 < 0 𝑯𝑦 𝑧 = − 𝑲0 2 e−j𝑘𝑧 𝑧 > 0 𝑲0 2 ej𝑘𝑧 𝑧 < 0 𝑺 = 1 2 𝑬 × 𝑯∗ = 𝜂𝐾0 2 8 𝒊𝑧 𝑧 > 0 − 𝜂𝐾0 2 8 𝒊𝑧 𝑧 < 0 (𝑲0 real)
- 17. 16 𝐸𝑥 𝑧, 𝑡 = Re 𝑬𝑥 𝑧 ej𝜔𝑡 = − 𝜂𝑲0 2 cos 𝜔𝑡 − 𝑘𝑧 𝑧 > 0 − 𝜂𝑲0 2 cos 𝜔𝑡 + 𝑘𝑧 𝑧 < 0 𝐻𝑦 𝑧, 𝑡 = Re 𝑯𝑦 𝑧 ej𝜔𝑡 = − 𝑲0 2 cos 𝜔𝑡 − 𝑘𝑧 𝑧 > 0 𝑲0 2 cos 𝜔𝑡 + 𝑘𝑧 𝑧 < 0 𝑆𝑧 = 𝐸𝑥𝐻𝑦 = 𝜂𝐾0 2 4 cos2 𝜔𝑡 − 𝑘𝑧 𝑧 > 0 − 𝜂𝐾0 2 4 cos2 𝜔𝑡 + 𝑘𝑧 𝑧 < 0 𝑆𝑧 = 𝜂𝐾0 2 8 𝑧 > 0 − 𝜂𝐾0 2 8 𝑧 < 0
- 18. 17 C. Normal Incidence Onto A Perfect Conductor Incident Fields: 𝑬i 𝑧, 𝑡 = Re 𝑬iej 𝜔𝑡−𝑘𝑧 𝒊𝑥 𝑯i 𝑧, 𝑡 = Re 𝑬i 𝜂 ej 𝜔𝑡−𝑘𝑧 𝒊𝑦 Reflected Fields: 𝑬r 𝑧, 𝑡 = Re 𝑬rej 𝜔𝑡+𝑘𝑧 𝒊𝑥 𝑯r 𝑧, 𝑡 = Re − 𝑬r 𝜂 ej 𝜔𝑡+𝑘𝑧 𝒊𝑦 𝑘 = 𝜔 𝜀𝜇, 𝜂 = 𝜇 𝜀 The boundary conditions require that, 𝐸𝑥 𝑧 = 0, 𝑡 = 𝐸𝑥, i 𝑧 = 0, 𝑡 + 𝐸𝑥, r 𝑧 = 0, 𝑡 = 0 𝑬i + 𝑬r = 0 ⇒ 𝑬r = −𝑬i
- 19. 18 Figure 2: Time varying electromagnetic phenomena differ only in the scaling of time (frequency) and size (wavelength). In linear dielectric media the frequency and wavelength are related as 𝑓λ = 𝑐 (𝜔 = 𝑘𝑐), where 𝑐 = 1 𝜀𝜇 is the velocity of light in the medium. sin 𝑘z 𝜆 = 2π 𝑘 𝑇 = 2π 𝜔 2π 𝑘 2π 𝜔 𝑧 𝑡 π 𝜔 π 𝑘 −1 −1 1 1 sin 𝜔𝑡 0 102 104 106 108 1010 1012 1014 1016 1018 1020 Circuit Theory Microwaves f (Hz) Visible Light Ultraviolet X-Rays Gamma Rays Power Infrared (Heat) Radio and Television AM FM 3 × 106 λ meters 3 × 104 3 × 102 3 3 × 10−2 3 × 10−4 3 × 10−6 3 × 10−8 3 × 10−10 3 × 10−12
- 20. 19 For 𝑬i = 𝐸i real we have: 𝐸𝑥 𝑧, 𝑡 = 𝐸𝑥, i 𝑧, 𝑡 + 𝐸𝑥, r 𝑧, 𝑡 = Re 𝑬i e−j𝑘𝑧 − ej𝑘𝑧 ej𝜔𝑡 = 2𝐸i sin 𝑘𝑧 sin 𝜔𝑡 𝐻𝑦 𝑧, 𝑡 = 𝐻𝑦, i 𝑧, 𝑡 + 𝐻𝑦, r 𝑧, 𝑡 = Re 𝑬i 𝜂 e−j𝑘𝑧 − ej𝑘𝑧 ej𝜔𝑡 = 2𝐸i 𝜂 cos 𝑘𝑧 cos 𝜔𝑡 𝐾𝑧 𝑧 = 0, 𝑡 = 𝐻𝑦 𝑧 = 0, 𝑡 = 2𝐸i 𝜂 cos 𝜔𝑡 Radiation pressure in free space 𝜇 = 𝜇o, 𝜀 = 𝜀o Force𝑧 Area 𝑧=0 = 1 2 𝑲 × 𝜇o𝑯 = 1 2 𝜇o𝐾𝑥𝐻𝑦 𝑧=0 𝒊𝑧 = 1 2 𝜇o𝐻𝑦 2 𝑧 = 0 𝒊𝑧 = 2𝜇o𝐸i 2 𝜂o 2 cos2 𝜔𝑡 𝒊𝑧 = 2𝜇o 𝜂o 𝜀o 𝐸i 2 cos2 𝜔𝑡 𝒊𝑧 = 2𝜀o 𝐸i 2 cos2 𝜔𝑡 𝒊𝑧
- 21. 20 Figure 3: A uniform plane wave normally incident upon a perfect conductor has zero electric field at the conducting surface thus requiring a reflected wave. The source of this reflected wave is the surface current at 𝑧 = 0, which equals the magnetic field there. The total electric and magnetic fields are 90° out of phase in time and space. x z y 𝐸i = Re 𝑬iej 𝜔𝑡−𝑘𝑧 𝒊𝑥 𝐻i = Re 𝑬i 𝑍o ej 𝜔𝑡−𝑘𝑧 𝒊𝑦 𝑘 = 𝜔 𝑐 𝒊𝑧 𝐻r = Re − 𝑬r 𝑍o ej 𝜔𝑡+𝑘𝑧 𝒊𝑦 𝐸r = Re 𝑬rej 𝜔𝑡+𝑘𝑧 𝒊𝑥 𝑘r = − 𝜔 c1 𝒊𝑧 𝜀o, 𝜇o 𝜂o = 𝑍o= 𝜇o 𝜀o 𝐻𝑦 𝑠, 𝑡 = 2𝑬i 𝑍o cos 𝑘𝑠 cos 𝜔𝑡 𝐸𝑥 𝑠, 𝑡 = 2𝑬isin 𝑘𝑠 sin 𝜔𝑡 𝜎 → ∞
- 22. 21 Figure 4: A uniform plane wave normally incident upon a dielectric interface separating two different materials has part of its power reflected and part transmitted. IV. Normal Incidence Onto A Dielectric x z y 𝜀1, 𝜇1 𝜂1 = 𝑍1 = 𝜇1 𝜀1 , 𝑐1 = 1 𝜀1𝜇1 𝜀2, 𝜇2 𝑍2 = 𝜇2 𝜀2 , 𝑐2 = 1 𝜀2𝜇2 𝐸i = Re 𝑬iej 𝜔𝑡−𝑘1𝑧 𝒊𝑥 𝐸t = Re 𝑬tej 𝜔𝑡−𝑘2𝑧 𝒊𝑥 𝐻t = Re 𝑬t 𝑍2 ej 𝜔𝑡−𝑘2𝑧 𝒊𝑦 𝐻r = Re − 𝑬r 𝑍1 ej 𝜔𝑡+𝑘1𝑧 𝒊𝑦 𝐻i = Re 𝑬i 𝑍1 ej 𝜔𝑡−𝑘1𝑧 𝒊𝑦 𝐸r = Re 𝑬rej 𝜔𝑡+𝑘1𝑧 𝒊𝑥 𝑘i = 𝑘1𝒊𝑧 = 𝜔 𝑐1 𝒊𝑧 𝑘r = −𝑘1𝒊𝑧 = −𝜔 𝑐1 𝒊𝑧 𝑘t = 𝑘2𝒊𝑧 = 𝜔 𝑐2 𝒊𝑧
- 23. 22 𝑬i 𝑧, 𝑡 = Re 𝑬iej 𝜔𝑡−𝑘1𝑧 𝑖𝑥 , 𝑘1 = 𝜔 𝜀1𝜇1 𝑯i 𝑧, 𝑡 = Re 𝑯i 𝜂1 ej 𝜔𝑡−𝑘1𝑧 𝑖𝑦 , 𝜂1 = 𝜔 𝜇1 𝜀1 𝑬r 𝑧, 𝑡 = Re 𝑬rej 𝜔𝑡+𝑘1𝑧 𝑖𝑥 𝑯r 𝑧, 𝑡 = Re − 𝑬r 𝜂1 ej 𝜔𝑡+𝑘1𝑧 𝑖𝑦 𝑬t 𝑧, 𝑡 = Re 𝑬tej 𝜔𝑡−𝑘2𝑧 𝑖𝑥 , 𝑘2 = 𝜔 𝜀2𝜇2 𝑯t 𝑧, 𝑡 = Re 𝑬t 𝜂2 ej 𝜔𝑡−𝑘2𝑧 𝑖𝑦 , 𝜂2 = 𝜔 𝜇2 𝜀2
- 24. 23 𝐸𝑥 𝑧 = 0− = 𝐸𝑥 𝑧 = 0+ ⇒ 𝑬i + 𝑬r = 𝑬t 𝐻𝑦 𝑧 = 0− = 𝐻𝑦 𝑧 = 0+ ⇒ 𝑬i − 𝑬r 𝜂1 = 𝑬t 𝜂2 𝑅 = 𝐸r 𝐸i = 𝜂2 − 𝜂1 𝜂1 + 𝜂2 is the Reflection coefficient 𝑇 = 𝐸t 𝐸i = 2𝜂2 𝜂1 + 𝜂2 is the Transmission coefficient 1 + 𝑅 = 𝑇
- 25. 24 𝑆𝑧, i = 1 2 Re 𝑬𝑥 𝑧 𝑯𝑦 ∗ 𝑧 = 1 2𝜂1 Re 𝑬ie−j𝑘1𝑧 + 𝑬rej𝑘1𝑧 𝑬i ∗ ej𝑘1𝑧 − 𝑬r ∗ e−j𝑘1𝑧 = 1 2𝜂1 𝑬i 2 − 𝑬r 2 + 1 2𝜂1 Re 𝑬r𝑬i ∗ e2j𝑘1𝑧 − 𝑬r ∗ 𝑬ie−2j𝑘1𝑧 = 1 2𝜂1 𝑬i 2 − 𝑬r 2 = 𝑬i 2 2𝜂1 1 − 𝑅2 𝑆𝑧, t = 1 2𝜂2 𝑬t 2 = 𝑬i 2 𝑇2 2𝜂2 = 𝑬i 2 1 − 𝑅2 2𝜂2 = 𝑆𝑧, i pure imaginary
- 26. 25 V. Lossy Dielectrics, 𝑱 = 𝜎𝑬 Ampere′ s Law: 𝛻 × 𝑯 = 𝑱 + 𝜀 𝜕𝑬 𝜕𝑡 = 𝜎𝑬 + 𝜀 𝜕𝑬 𝜕𝑡 For ej𝜔𝑡 fields, 𝛻 × 𝑯 = j𝜔𝜀 + 𝜎 𝑬 = j𝜔𝜀 1 + 𝜎 j𝜔𝜀 𝑬 where, 𝜎 = conductivity of a medium ( Siemens m )
- 27. 26 Define complex permittivity by 𝜺 = 𝜀 1 + 𝜎 j𝜔𝜀 . Then complex amplitude solutions are the same as real amplitude solutions if we replace 𝜀 by 𝜺: 𝑘 = 𝜔 𝜺𝜇 𝜂 = 𝜇 𝜺 = 𝜇 𝜀 1 + 𝜎 j𝜔𝜀 𝑘 = 𝜔 𝜀𝜇 1 + 𝜎 j𝜔𝜀
- 28. 27 A. Low Loss Limit: 𝜎 𝜔𝜀 ≪ 1 𝑘 = 𝜔 𝜀𝜇 1 + 𝜎 j𝜔𝜀 ≈ 𝜔 𝜀𝜇 1 + 1 2 𝜎 j𝜔𝜀 ≈ 𝜔 𝜀𝜇 − j𝜎 2 𝜇 𝜀 ≈ 𝜔 𝜀𝜇 − j 𝜎𝜂 2 e−j𝑘𝑧 = e−j𝜔 𝜀𝜇𝑧 e −j −j 𝜎𝜂 2 𝑧 = e−j𝜔 𝜀𝜇𝑧 e− 𝜎𝜂 2 𝑧 slow exponential decay
- 29. 28 B. Large Loss Limit: 𝜎 𝜔𝜀 ≫ 1 𝑘 = 𝜔 𝜀𝜇 1 + 𝜎 j𝜔𝜀 ≈ 𝜔 𝜀𝜇 𝜎 𝜔𝜀 −j we know that, 𝜔 𝜔 = 𝜔 ≈ 𝜔𝜇𝜎 2 1 − j ≈ 1 − j 𝛿 𝛿 = 2 𝜔𝜇𝜎 is the skin depth e−j𝑘𝑧 = e −j 1−j 𝑧 𝛿 = e −j 𝑧 𝛿 e − 𝑧 𝛿 fast exponential decay 1 − j 2
- 30. (1) Markus Zahn, Electromagnetics and Applications, Massachusetts Institute of Technology: MIT Open Course Ware, 2005. References 29