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Electromagnetic Power Density Introduction – Power is transmitted by an electromagnetic wave in the direction of propagation. – Poynting vector S ( x, y , z , t ) = E( x, y , z , t ) × H ( x, y , x, t ) V /m (W / m 2 ) A/ m • not only applied to uniform plane waves • can be used for any electromagnetic waves 1 Poynting Vector from Maxwell’s Equations It can be derived from Maxwell’s equation in the time domain that: ∂ − ∫ E × H ⋅ dS = ∫ E ⋅ Jdv + S V ∂t ε E 2 µH 2 ∫V 2 + 2 dv power loss the rate of increase of energies stored in the electric and magnetic fields S net power supplied to the enclosed surface V 2 Poynting Vector Net power supplied to the volume: Wout = ∫SE × H ⋅ dS Power density of the wave: P = E × H Therefore, Poynting vector P (in W/m2) represents the direction and density of power flow at a point. Example: If a plane wave propagating in a direction k̂ that makes an angle θ with the normal vector n̂ of an aperture P = ∫ S ⋅ nˆ dA S = SA cos θ 3 Example Find the Poynting vector on the surface of a long straight conducting wire (of radius b and conductivity σ) that carries a direct current I. Solution: Since we have a d-c situation, the current in the wire is uniformly distributed over its cross-sectional area. I J = zˆ 2 πb L J I E = = zˆ 2 σ σπb 4 Example On the surface of the wire, I ˆ H =φ 2πb P = E × H = zˆ × φˆ I2 2σπ b 2 3 = −rˆ I2 2σπ 2b 3 Integrating the Poynting vector over the wall of the wire I2 2πbL − ∫ P ⋅ ds = − ∫ P ⋅ rˆds =E × H = 2 3 s s 2σπ b L 2 2 =I = I R 2 σπb 5 Average Power Density In time-varying fields, it is more important to find the average power. We define the average Poynting vector for periodic signals as Pavg Pavg 1 = T ∫ T 0 Pdt It can be shown that the average power density can be expressed as Pavg 1 = Re(E( x, y, z ) × H ( x, y, z ) *) 2 Complex conjugate 6 Example: Plane Wave in a lossless Medium The general expression for the phasor electric field of a uniform plane wave with arbitrary polarization, traveling in the +z direction, is given by E( z ) = xˆE x ( z ) + yˆ E y ( z ) = (xˆE xo + yˆ E yo )e − jβz The magnetic field is 1 H ( z ) = zˆ × E( z ) η 1 = − ( yˆ E xo − xˆE yo )e − jβz η ( 2 1 2 ∴ Pav = zˆ E xo + E yo 2η ) 7 Example: Solar Power 8 Example: Solar Power 9 Example: Solar Power 10 Example: Solar Power 11 Example: Plane wave in a lossy medium The general expression for the phasor electric field of a uniform plane wave with arbitrary polarization, traveling in the +z direction, is given by E( z ) = xˆE x ( z ) + yˆ E y ( z ) = (xˆE xo + yˆ E yo )e −αz e − jβz and magnetic field is 1 H ( z ) = zˆ × E( z ) η 1 = − ( yˆ E xo − xˆE yo )e −αz e − jβz η 12 Example: Plane wave in a lossy medium The average power density is Pavg 1 = Re(E( x, y, z ) × H ( x, y, z ) *) 2 2 − 2αz 1 1 2 Re * = zˆ E xo + E yo e 2 η ) ( ( ) 2 − 2αz 1 2 = zˆ E xo + E yo e cos θη 2η if η = η e jθη Demonstration: D7.6 13 Reflection of Plane Waves at Normal Incidence Medium 1(σ1,ε1,µ1) x Medium 2 (σ2,ε2,µ2) Ei (incident wave) Er ak Et ak Hi Ht ak (transmitted wave) z Hr (reflected wave) y direction -y direction ak – propagation direction 14 Reflection of Plane Waves at Normal Incidence Incident wave: Ei = Eio e −γ 1 z H i = H io e Reflected wave: ax −γ 1 z ay = Eio η1 e −γ 1 z ay γ 1z E r = Ero e a x γ 1z H r = H ro e (−a y ) = − Transmitted wave: Et = Eto e −γ 2 z H t = H to e Ero η1 γ 1z e ay ax −γ 2 z ay = Eto η2 e −γ 2 z ay 15 Reflection of Plane Waves at Normal Incidence Boundary conditions at the interface (z=0): Tangential components of E and H are continuous in the absence of current sources at the interface, so that Eio + Ero = Eto H io + H ro = H to 1 η1 Reflection coefficient (Eio − Ero ) = Eto η2 Ero η 2 − η1 Γ= = Eio η 2 + η1 Eto 2η 2 = Transmission coefficient τ = Eio η1 + η 2 16 Reflection of Plane Waves at Normal Incidence Note that Γ and τ may be complex, and 1+ Γ = τ 0 ≤ Γ ≤1 Similar expressions may be derived for the magnetic field. In medium 1, a standing wave is formed due to the superimposition of the incident and reflected waves. Standing wave ratio can be defined as in transmission lines. 17 Example: Normal Incidence on a good conductor η 2 − η1 Γ= η 2 + η1 2η 2 τ= η1 + η 2 For good conductor,σ → ∞ jωµ η2 = →0 σ + jωε Γ = −1 τ =0 18