MIT OpenCourseWare 6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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MIT OpenCourseWare http://ocw.mit.edu 6.013/ESD.013J Electromagnetics and Applications, Fall 2005 Please use the following citation format: Markus Zahn, 6.013/ESD.013J Electromagnetics and Applications, Fall 2005. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-Noncommercial-Share Alike. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms 6.013 - Electromagnetics and Applications Fall 2005 Lecture 8 - Transverse Electromagnetic Waves Prof. Markus Zahn October 4, 2005 I. Maxwell’s Equations for Linear Media ¯ ¯ = −µ ∂ H �×E ∂t ¯ ¯ = J¯ + � ∂ E �×H ∂t ¯ = ρf /� �·E ¯ =0 �·H II. Poynting’s Theorem A. Power Flow, Electromagnetic Energy � � � � � � ¯×H ¯ =H ¯ · �×E ¯ −E ¯· �×H ¯ �· E � ¯ ¯� ∂ H ∂ E ¯· ¯ · J¯ + � = −µH −E ∂t ∂t � � � � µ∂ � ∂ ¯ · J¯ |H̄ |2 − =− |Ē |2 − E 2 ∂t � 2�∂t � � ¯×H ¯ + ∂ 1 �|Ē |2 + 1 µ|H̄ |2 = −E ¯ · J¯ �· E ∂t 2 2 � � � � ¯ ¯ ¯×H ¯ · dS¯ � · E × H dV = E V S � � � � � d 1 ¯2 1 ¯ 2 ¯ ¯ ¯ · J¯ dV E × H · d¯ a+ �|E | + µ|H| dV = − E dt 2 2 S V V ¯×H ¯ S¯ = E � � � W = V 12 �|Ē |2 + 12 µ|H̄ |2 dV � ¯ · J¯ dV Pd = E Poynting Vector (watts / m2 ) Electromagnetic Stored Energy ¯>0 Power dissipated if J¯ · E V ¯ ⇒ J¯ · E ¯ = σ|Ē |2 e.g., J¯ = σE ¯<0 Power source if J¯ · E � ¯ ¯ dā = SE×H · Pout + dW dt = −Pd we = 21 �|Ē |2 wm = 12 µ|H̄ |2 Pout = � S S¯ · dā Electric energy density in Joules/m3 Magnetic energy density in Joules/m3 1 B. Power in Electric Circuits From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. Outside circuit elements � ¯ · d¯l ≈ 0, E ¯=0⇒E ¯ = −�Φ (Kirchoff’s Voltage Law �×E � k vk = 0 � ¯ = J¯ ⇒ � · J¯ = 0, ¯ ¯ �×H (Kirchoff’s S J · dS = 0 � current law k ik = 0 C � Pin = − ¯×H ¯ · dS¯ E S � =− � � ¯×H ¯ dV �· E V � � ¯ ·� ¯�−0 E ¯· �×H ¯ = −E ¯ · J¯ = �Φ · J¯ ¯×H ¯ =H �· E �� ×� E � � �0 � · J¯Φ = Φ� �� · J¯ + J¯ · (�Φ) � � � � ¯×H ¯ = J¯ · (�Φ) = � · ΦJ¯ �· E � � � � � � � ¯×H ¯ dV = − Pin = − �· E � · J¯Φ dV = − J¯Φ · dS¯ � � V V S On S, Φ = voltages on each wire, J¯ � is non-zero only on wires. � � � ¯ ·dS¯ = − N vk Pin = − S JΦ J¯ · dS¯ = N k=1 vk ik k=1 S � �� � −ik C. Complex Poynting’s Theorem (Sinusoidal Steady State, ejωt ) � � � � ˆ (r̄)ejωt = 1 Ē ˆ (r̄)ejωt + E ˆ¯ ∗ (r̄)e−jωt Ē(r̄, t) = Re Ē 2 � � 1� � ˆ (r̄)ejωt = ˆ¯ (r̄)ejωt + H ˆ¯ ∗ (r̄)e−jωt H̄ (r̄, t) = Re H̄ H 2� �� � The real part of a complex number is one-half of the sum of the number and its complex conjugate 2 Maxwell’s Equations in Sinusoidal Steady State ˆ¯ (r̄) = −jωµH ˆ¯ (r̄) �×E ˆ¯ (r̄) = J¯(r̄) + jω�Ē ˆ (r̄) �×H ˆ¯ (r̄) = ρ̂ (r̄)/� �·E f ˆ ¯ (r̄) = 0 �·H S̄(r̄, t) = Ē(r̄, t) × H̄ (r̄, t) � � � 1�ˆ ˆ¯ (r̄)ejωt + H ˆ¯ ∗ (r̄)e−jωt ˆ ∗ (r̄)e−jωt × H ¯ (r̄)ejωt + Ē E = 4 � 1�ˆ ˆ ˆ¯ ∗ (r̄) × H ˆ¯ (r̄) + Ē ˆ (r̄) × H̄ ˆ ∗ (r̄) + E ˆ¯ ∗ (r̄) × H ˆ¯ ∗ (r̄)e−2jωt ¯ (r̄) × H ¯ (r̄)e2jωt + E = E 4 � � � 1 � ˆ∗ ˆ¯ ˆ¯ (r̄) × H ˆ¯ ∗ (r̄) ¯ = ¯ (r̄) × H(r̄) S E +E 4 � � � � 1 ˆ ˆ ˆ¯ ∗ (r̄) × H ˆ¯ (r̄) ¯ (r̄) × H ¯ ∗ (r̄) = 1 Re E = Re E 2 2 (A complex number plus its complex conjugate is twice the real part of that number.) 1ˆ ˆ (r̄)∗ Ē (r̄) × H̄ 2 � 1 � ¯ˆ ∗ ˆ ∗ (r̄) ¯ˆ (r̄) − E ¯ˆ (r̄) · � × H ¯ H (r̄) · � × E 2 � � �� 1 � ˆ¯ ∗ � ¯ˆ ¯ˆ ¯ˆ ∗ (r̄) = H (r̄) −jωµH(r̄) − E(r̄) · J¯ˆ∗ (r̄) − jω�E 2 � 1� ˆ¯ (r̄)|2 + jω�|E ˆ¯ (r̄)|2 − 1 Ē ˆ (r̄) · Jˆ¯∗ (r̄) = −jωµ|H 2 2 Sˆ¯ = � � 1 ¯ˆ ∗ ˆ ˆ ¯ ¯ � · S = � · E (r̄) × H (r̄) = 2 1 ˆ¯ 1 ˆ¯ �wm � = µ|H (r̄)|2 , �we � = �|E (r̄)|2 4 4 1ˆ P̂d = Ē (r̄) · J¯ˆ∗ (r̄) 2 � · S¯ˆ + 2jω [�wm � − �we �] = −P̂d III. Transverse Electromagnetic Waves (ρf = 0, J¯ = 0) A. Wave equation ¯ ∂H ∂t ¯ ∂E ¯ = −µ �×E ¯ =� �×H ¯=0 � ·E ¯ =0 �·H 3 ∂t � ¯� � � ∂ � ∂ ∂E ¯ ¯ � × H = −µ � × � × E = −µ � ∂t ∂t ∂t � � 2 ¯ � � � ¯ =� � ¯� 0− �2 E ¯ = −�µ ∂ E �× �×E �� ·E ∂t2 � Wave equation ¯= �2 E where c = √1 �µ ¯ 1 ∂2E c2 ∂t2 is the speed of the electromagnetic wave. In free space µ = µ0 = 4π × 10−7 henries/m and � = �0 ≈ c0 = √�10 µ0 ≈ 3 × 108 m/s. 10−9 36π farads/m, which leads to Similarly 2 ¯ � � � ¯ = −�µ ∂ H ¯ =� � ¯� 0− �2 H ¯ =� ∂ �×E �× �×H �� ·H 2 � � � � ∂t ∂t ¯ ∂2H 1 ¯ = 1 ,c = √ �2 H 2 2 c ∂t �µ B. Plane waves x Kx = Re[K0 exp (jω t)] Ex- = Re[Ex-(z)exp (j ωt)] ε, µ ε, µ Ex+ = Re[Ex+ (z)exp (jω t)] Hy+ = Re[Hy+ (z)exp (jω t)] Hy- = Re[Hy-(z)exp (jω t)] y z Image by MIT OpenCourseWare. � Ex (z, t) = Re Êx (z)ejωt ω2 d2 Êx = − Êx dz 2 c2 d2 Êx + k 2 Êx = 0 dz 2 4 � where we have ω2 = ω 2 �µ c2 2π k= is the wavenumber, λ is the wavelength λ ω k = ± ⇒ ω = kc c 2π ω = 2πf = c ⇒ fλ = c λ k2 = Êx = A1 ejkz + A2 e−jkz � � Ex = Re A1 ej(ωt+kz) + A2 ej(ωt−kz) � �� � � �� � traveling wave traveling wave in the −z direction in the +z direction For the wave in the −z direction we have: ωt + kz = constant ωdt + kdz = 0 dz ω = − = −c dt k For the wave in the +z direction we have: ωt − kz = constant ωdt − kdz = 0 dz ω = = +c dt k � Êx+ e−jkz Êx (z) = Êx− e+jkz z>0 z<0 ¯ ¯ = −µ ∂H ⇒ dÊx = −jωµĤy ⇒ Ĥy = − 1 dÊx �×E ∂t dz jωµ dz ¯ ¯ = � ∂E ⇒ dĤy = −jω�Êx �×H ∂t dz � k Êx+ e−jkz z>0 Ĥy = ωµ k +jkz − ωµ Êx− e z<0 � � √ ω �µ k � µ � = = ,η = is the wave impedance ωµ ωµ µ � � � Eˆ x+ −jkz z>0 η e Ĥy = Êx− +jkz − η e z<0 5 Now we look at the boundary conditions: Ex (z = 0+ , t) = Ex (z = 0− , t) ⇒ Êx+ = Êx− ˆx+ −Êx− − E ˆ0 Hy (z = 0− , t) − Hy (z = 0+ , t) = Kx (z = 0, t) ⇒ =K η Êx+ = Êx− = − Êx (z) = Ĥy (z) = � 0 −jkz − ηK̂ 2 e ˆ 0 jkz − ηK 2 e � − K̂20 e−jkz K̂0 jkz 2 e 1 ¯ˆ ˆ∗ = Ŝ = E × H̄ 2 ηK̂0 2 z>0 z<0 z>0 z<0 � ηK 2 0 8 īz ηK 2 − 8 0 īz z>0 z<0 (K̂0 real) � ηK0 � � ˆx (z)ejωt = − 2 cos(ωt − kz) z > 0 Ex (z, t) = Re E 0 − ηK 2 cos(ωt + kz) z < 0 � � � − K20 cos(ωt − kz) z > 0 jωt Hy (z, t) = Re Ĥy (z)e = + K20 cos(ωt + kz) z < 0 � ηK 2 2 0 z>0 4 cos (ωt − kz) Sz = Ex Hy = 2 ηK0 2 − 4 cos (ωt + kz) z < 0 � ηK 2 0 z>0 8 �Sz � = ηK02 − 8 z<0 C. Normal Incidence Onto a Perfect Conductor � � Ēi (z, t) = Re Êi ej(ωt−kz) īx � � Êi j(ωt−kz) H̄i (z, t) = Re e īy η � � Reflected Fields: Ēr (z, t) = Re Êr ej(ωt+kz) īx � � Êr j(ωt+kz) H̄r (z, t) = Re − e īy η � µ √ k = ω �µ, η = � Incident Fields: The boundary conditions require that Ex (z = 0, t) = Ex,i (z = 0, t) + Ex,r (z = 0, t) = 0 Êi + Êr = 0 ⇒ Êr = −Êi 6 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. For Êi = Ei real we have: � � � � Ex (z, t) = Ex,i (z, t) + Ex,r (z, t) = Re Êi e−jkz − e+jkz ejωt = 2Ei sin(kz) sin(ωt) � � � Êi � −jkz Hy (z, t) = Hy,i (z, t) + Hy,r (z, t) = Re e + e+jkz ejωt η = Kz (z = 0, t) = Hy (z = 0, t) = 2Ei cos(kz) cos(ωt) η 2Ei cos(ωt) η Radiation pressure in free space (µ = µ0 , � = �0 ) Forcez �� 1¯ ¯ = 1 µ0 Kx Hy |z=0 īz = 1 µ0 Hy2 (z = 0)īz = K × µ0 H � Area z=0 2 2 2 2µ0 Ei2 cos2 (ωt)īz = η02 2µ0 2 = E cos2 (ωt)īz µ0 /�0 i = 2�0 Ei2 cos2 (ωt)īz 7 From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. IV. Normal Incidence Onto a Dielectric From Electromagnetic Field Theory: A Problem Solving Approach, by Markus Zahn, 1987. Used with permission. 8 � � √ Ēi (z, t) = Re Êi ej(ωt−k1 z) īx , k1 = ω �1 µ1 � � � Êi j(ωt−k1 z) H̄i (z, t) = Re e īy , η1 = µ1 /�1 η1 � � Ēr (z, t) = Re Êr ej(ωt+k1 z) īx � � ˆr −E j(ωt+k1 z) H̄r (z, t) = Re e īy η1 � � √ Ēt (z, t) = Re Êt ej(ωt−k2 z) īx , k2 = ω �2 µ2 � � � Êt j(ωt−k2 z) H̄t (z, t) = Re e īy , η2 = µ2 /�2 η2 Ex (z = 0− ) = Ex (z = 0+ ) ⇒ Êi + Êr = Êt Hy (z = 0− ) = Hy (z = 0+ ) ⇒ Êi − Êr Êt = η1 η2 R≡ T ≡ Êr Êi Êt = η2 − η1 is the Reflection coefficient η1 + η2 = 2η2 is the Transmission coefficient η1 + η2 Êi 1+R=T � � 1 Re Êx (z)Ĥy∗ (z) 2 �� �� �� 1 = Re Êi e−jk1 z + Êr ejk1 z Êi∗ e+jk1 z − Êr∗ e−jk1 z 2η1 � 1 � = |Êi |2 − |Êr |2 2η1 � � 1 + Re Êr Êi∗ e2jk1 z − Êr∗ Êi e−2jk1 z 2η1 � �� � �Sz,i � = pure imaginary � 1 � = |Êi |2 − |Êr |2 2η1 � |Êi |2 � = 1 − R2 2η1 ˆi |2 T 2 ˆi |2 (1 − R2 ) |E |E 1 �Sz,t � = |Êt |2 = = = �Sz,i � 2η2 2η2 2η1 ¯ V. Lossy Dielectrics - J¯ = σE ¯ = J¯ + � ∂ E¯ = σE ¯ + � ∂ E¯ Ampere’s Law: � × H ∂t ∂t For ejωt fields � � σ ˆ ˆ ˆ¯ ¯ ¯ � × H = (jω� + σ)E = jω� 1 + E jω� 9 � � σ Define complex permittivity by �ˆ = � 1 + jω� . Then complex amplitude solutions are the same as real amplitude solutions if we replace � by �ˆ: � � √ “ µ ” k = ω �ˆµ, η = µ�ˆ = σ � 1+ jω� � � � σ k = ω �µ 1 + jω� A. Low loss limit: σ ω� �1 � σ √ k = ω �µ 1 + jω� � � 1 σ √ ≈ ω �µ 1 + 2 jω� � jσ µ √ ≈ ω �µ − 2 � ση √ ≈ ω �µ − j 2 √ −jkz −jω �µz −j(−j ση )z 2 e =e e = e−jω √ �µz ση e�−��2 �z slow exponential decay B. Large loss limit: σ ω� �1 √ � σ k = ω �µ 1 + jω� � � σ � −j ≈ ω ��µ ω�� � �� � 1−j √ 2 � ωµσ (1 − j) 2 � 1−j 2 ≈ δ≡ is the skin depth ωµσ δ ≈ e−jkz = e−j(1−j)z/δ = e−jz/δ �e−z/δ �� � fast exponential decay 10
