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450fall09hw13_Sol

# 450fall09hw13_Sol - ECE 450 Fall 2009 Homework 13 Solutions...

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ECE 450 Fall 2009 Homework 13 Solutions Due: Tuesday, Dec 1 st , 2009 1. Given the assumptions, model the waveguide/air region as a transmission line (TL) problem wherein the effective load at the end of the waveguide (the air region) is of impedance 0 L Z η = and the characteristic impedance of the waveguide is of 10 0 TE Z η = -- the intrinsic impedance of the TE 10 mode in the waveguide. We can relate the reflection and transmission problem of the waveguide/air interface to the reflection and transmission from the mismatch between the TL line impedance and the load (See TL model in the figure below). That is, as in the lecture notes, one can use guide impedance of waveguides to set up TL models for waveguide circuits. Γ 10 0 TE Z η = 0 L Z η = In the figure, Γ is the well known reflection coefficient given by 10 10 0 0 0 0 TE L Z Z Z Z η L TE η η η = + + Γ = . Your previous experience with TLs should tell you that the percentage of incident power delivered to the load, which equates to the percentage of time average incident power that gets radiated out of the waveguide opening, is 2 1 trans inc P P = − Γ . The formulation above can be proved via power conservation and utilizing plane-wave reflection and transmission-like properties. Power conservation states that power must be 1

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conserved across the cross section of the waveguide, i.e., the sum of the incident of the reflected time average power ( P inc + P ref ) must be equal to the time average power transmitted, radiated out ( P trans ). Æ In a waveguide ( for non-lossy medium ): * Write the incident field (TE m0 modes) 0 0 ˆ ˆ ( , ) cos( ) sin( ) z z jk z jk z z x inc x x x z x e z H k x e k = + H ± jk H k x 0 0 sin( ˆ ( , ) ) z jk inc x j H k x z y e k z x x ωμ = − E ± * Write the reflected field mode (note the sign changes for ˆ z propagation) 0 0 ˆ ˆ ( , ) cos( ) sin( ) z z jk z jk z z x ref x jk H k x x z x e z H k x e k + + Γ = − Γ H ± x 0 0 sin( ˆ ( , ) ) z jk inc x j H k x z y k z x x e ωμ Γ = − E ± + * Time average Poynting Vector in the guide { } 2 aginary Part" 1 1 Re Re 2 2 1 2 ˆ sin ( ) z inc inc z x x nc x x i k H z k x k k H z k x k ωμ ωμ 0 0 * 2 2 2 0 0 2 2 ˆ sin ( ) "Im × = + = = E H S ± ± { } 2 2 "Imaginary P 1 1 Re Re art" ˆ sin ( 2 ) 2 1 2 z ref ref x z x x ref k H z k x k k H z k x k ωμ ωμ 0 0 * 2 2 2 2 0 0 2 2 ˆ sin ( ) x Γ × = + Γ = = − S E H ± ± , where the minus sign in the above indicates that ref S is flowing in the minus ˆ z direction for the reflected wave. 2
* Time average power in the guide, say at the interface ( ) ( ) ( ) ( ) 0 2 2 0 0 2 2 2 2 2 2 0 0 0 0 2 0 0 2 2 0 ˆ ˆ ˆ 1 sin ( ) 1 si 1 2 1 ( ) 1 2 2 2 n z x x a z z x x x guide inc ref S a y x b x z k H z k x P z k k H k H ab k x dx k ds dx dy b k ωμ ωμ ωμ = = = − Γ − Γ = − Γ = + = = S S i i As in the lecture notes, define 0 0 0 0 0 x c E H E k k λ η λ = η , with 0 0 k ωμ η = one can show that ( ) 2 2 0 0 2 2 1 , where 2 1 1 2 TE TE g c uide E ab f f P η η η − Γ = = .

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