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solution to homewrok 9

# solution to homewrok 9 - A PPP waveguide carrying the TE 1...

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0 50 100 150 200 250 300 0 50 100 150 200 kx kx kzTeflon 0 kzair 0 , note that the k vector moved away from the z-axis (the normal to the boundary) upon entering the air region At 9Ghz Gamma is 0.278 purely real and positive showing that the impedance in air is larger than in teflon. The angles of incidence are obtained from kz and kx Gamma 0.278 0.218 0.608 0.794i - = Gamma m ZTE1air m ZTE1Teflon m - ZTE1air m ZTE1Teflon m + = ZTE1Teflon m 377 2 kTeflon m kzTeflon m = ZTE1air m 377 k0 m kzair m = kzTeflon m kTeflon m ( 29 2 kx 2 - = kzair m k0 m ( 29 2 kx 2 - = kTeflon m k0 m 2 = k0 m 2 π freq m 3 10 8 = kx π a = For mode 1 m 0 2 .. = a 0.9 0.0254 = freq 9 10 9 12 10 9 6 10 9 = A PPP waveguide carrying the TE 1 mode Teflon Air

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At 12Ghz Gamma is 0.218 purely real and positive showing that the impedance in air is larger than in teflon. The angles of incidence are obtained from kz and kx both vectors are closer to the z-hjat direction but again the wave refracted away from the normal
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solution to homewrok 9 - A PPP waveguide carrying the TE 1...

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