A distortionless line has a characteristic impedance [latex]Z_{o}[/latex] = 50[ Ω ] and a propagation constant γ = 0.01 + j 4.0[latex][m^{-1}][/latex] at frequency f = 100[MHz] . Find R , L , G , C , and [latex]v_{p}[/latex] .

From Eqs. (9-25c) and (9-25d), we have [latex]\beta = \omega \sqrt{LC} (9-25c)\\ Z_{o} = \sqrt{\frac{L (R / L + j \omega )}{ C ( G / C + j \omega )} } =\sqrt{\frac{L}{C} } \text{(characteristic impedance)} (9-25d) \\ \beta = \omega \sqrt{LC} = 2\pi \times 10^{8} \sqrt{LC} = 4.0 \\ Z_{o} = \sqrt{L / C} = 50[/latex] From the above equations we obtain L = 318[nH/m] C = 127[pF/m] From Eq. (9-25b) we obtain [latex]\alpha = G \sqrt{L / C} = R \sqrt{C/L} [/latex] (9-25b) [latex]\alpha = G \sqrt{L / C} = G Z_{o} = 0.01[/latex], and thus G = 200[ μ S/m] From the relation RC = GL we obtain [latex]R = \frac{GL}{C} = \frac{200 \times 10^{-6} \times 318 \times 10^{-9}}{127 \times 10^{-12}} = 0.50 [\Omega / m] [/latex] Phase velocity is therefore [latex]v_{p} = \frac{\omega }{\beta } = \frac{2 \pi \times 10^{8}}{4} = 1.57 \times 10^{8} [m / s] [/latex]

Question 9.2: A distortionless line has a characteristic impedance Zo = 50......

Introduction to Engineering Electromagnetics [1050204]

A distortionless line has a characteristic impedance $Z_{o}$ = 50[Ω] and a propagation constant γ = 0.01 + j 4.0 $[m^{-1}]$ at frequency f = 100[MHz] . Find R, L, G, C, and $v_{p}$ .

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From Eqs. (9-25c) and (9-25d), we have

\beta = \omega \sqrt{LC}                                                                      (9-25c)\\ Z_{o} = \sqrt{\frac{L (R / L + j \omega )}{ C ( G / C + j \omega )} } =\sqrt{\frac{L}{C} }                                   \text{(characteristic impedance)}                 (9-25d) \\ \beta = \omega \sqrt{LC} = 2\pi \times 10^{8} \sqrt{LC} = 4.0 \\ Z_{o} = \sqrt{L / C} = 50

From the above equations we obtain

L = 318[nH/m]

C = 127[pF/m]

From Eq. (9-25b) we obtain

$\alpha = G \sqrt{L / C} = R \sqrt{C/L}$ (9-25b)

$\alpha = G \sqrt{L / C} = G Z_{o} = 0.01$ , and thus

G = 200[μS/m]

From the relation RC = GL we obtain

R = \frac{GL}{C} = \frac{200 \times 10^{-6} \times 318 \times 10^{-9}}{127 \times 10^{-12}} = 0.50 [\Omega / m]

Phase velocity is therefore

v_{p} = \frac{\omega }{\beta } = \frac{2 \pi \times 10^{8}}{4} = 1.57 \times 10^{8} [m / s]

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Verified Answer:

(a) From Eq. (9-32) we obtain

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Question: 9.5

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Question: 9.11

A 50[Ω] -lossless line is terminated in a load impedance ZL = 17.5 – j 55 [Ω]. Find the location and length of the short-circuited stub to be connected in parallel for the impedance matching. ...

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Question: 9.10

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Question: 9.9

A lossless transmission line of Ro = 75[Ω] has a length of l = λ /8 . Locate the point of zL on the Smith chart, and find Γ and zin , if the line is terminated in (a) a short circuit, and (b) an open circuit. ...

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Question: 9.8

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Question: 9.7

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Question: 9.6

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Question: 9.4

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Question: 9.1

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Verified Answer:

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\...

Question 9.2: A distortionless line has a characteristic impedance Zo = 50......

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