Notes 12 Spring 2005

# Notes 12 Spring 2005 - Notes 12 Spring 2005 Power...

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Notes 12 Spring 2005.doc Power Considerations As a notational convenience we will represent the magnitude of the time-averaged Poynting vector as P instead of S r and call it simply the power density of the wave. The power density of an incident wave is 1 {} η = η = = × = + + + + + + + 1 2 10 x 1 * 10 x 10 x * 10 y 10 x * 1 y 1 x i 1 E 2 1 E E 2 1 H E 2 1 H E 2 1 P The power density of its reflected wave is i 2 1 2 10 x 2 1 * 10 x 10 x * 10 y 10 x * 1 y 1 x r P 1 E 2 1 E E 2 1 H E 2 1 H E 2 1 P Γ = η Γ = η Γ Γ = = × = + + + Then, since energy must be conserved, we note that the power that is not reflected must be transmitted, i.e., i 2 i 2 i r i t P 1 P P P P P Γ = Γ = = Summarizing, we find the reflected and transmitted powers in terms of the incident power: Incident power density: i Reflected power density: i 2 r P P Γ = Transmitted power density: i 2 t P 1 P Γ = We see that with knowledge of the magnitude of the incident wave’s electric field and the constitutive parameters of the two media on either side of a boundary surface, for normal incidence we can calculate the reflected and transmitted field values and power values! Let us look at the interesting limiting case of having the incident region 1 be a perfect dielectric and have a transmitted region 2 be a perfect conductor ( σ ). ε ′ ω + σ = 2 2 eff 2 Multiplied by on top and bottom. ω j 0 j j j 1 2 eff 2 2 eff 2 2 2 2 2 2 2 1 = ε′ ω + σ ωµ = ε′ ω σ ε′ µ = ε µ = η and then ° = = η + η η η = Γ 180 1 1 1 2 1 2 From this we determine that τ and so E . 0 1 = Γ + = 0 E 10 x 20 x = τ = + + Also from this we determine that E. + + = Γ = 10 x 10 x 10 x E E 1 1 Recall that for any complex number c , jb a + = () 2 2 2 c b a jb a jb a = + = + = cc

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Notes 12 Spring 2005.doc This last line shows that the magnitudes of the incident and reflected electric fields are equal (for this special case) which is another way of saying that all of the energy is reflected . This agrees with what we said earlier; that no time varying fields exist in the perfect conductor (skin depth is zero). In fact we can show from the power equations that i 2 i 2 r P 1 P P = = Γ = ( ) ( ) 0 P 1 1 P 1 P i 2 i 2 t = = Γ = Our chart becomes (for this special case): Region 1 is a perfect dielectric and Region 2 is a Perfect conductor Electric Field Magnetic Field Incident z jk 10 x 1 x 1 e E E + + = z jk 10 x 1 1 y 1 e E 1 H + + η = Reflected z jk 10 x z jk 10 x 1 x 1 1 e E e E E + + = Γ = z jk 10 x 1 z jk 10 x 1 1 y 1 1 e E 1 e E 1 H + + η = Γ η = Transmitted 0 E 2 x = + 0 H 2 y = + To represent the total phasor E-field in region 1 we write as α as before and note that since region 1 is a perfect dielectric 1 jk = γ 1 1 j β + 0 1 = α 2 : () + β β + β + β + + β = = = + = 10 x 1 z j z j 10 x z j 10 x z j 10 x 1 x 1 x 1 x E z sin 2 j e e
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Notes 12 Spring 2005 - Notes 12 Spring 2005 Power...

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