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Question

Find α\alpha and Z0Z_{0} of a distortionless line whose R=2Ω/mR^{\prime}=2 \Omega / \mathrm{m} and G=2×104S/mG^{\prime}=2 \times 10^{-4} \mathrm{S} / \mathrm{m}.

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Answered 2 years ago
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From Problem 2.13, we know that

α=RG.\alpha = \sqrt{R'G'}.

With R=2  Ω/R' = 2\;\Omega/m and G=2×104G' = 2\times10^{-4} S/m, we find

α=2×2×104=0.02  (Np/m).\alpha = \sqrt{2\times2\times10^{-4}} = 0.02\;\mathrm{(Np/m)}.

Again from Problem 2.13, we realize that

Z0=LC=LGCG=RCCG=RG,Z_0 = \sqrt{\frac{L'}{C'}} = \sqrt{\frac{L'G'}{CG'}} = \sqrt{\frac{R'C'}{C'G'}} = \sqrt{\frac{R'}{G'}},

where we used RC=LGR'C' = L'G' (the distortionless-line condition). Thus

Z0=22×104=100  (Ω).Z_0 = \sqrt{\frac{2}{2\times10^{-4}}} = 100\;(\Omega).

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