The propagation constant of a lossy transmission line is (2 + ๐‘—5) m−1 and its characteristic impedance is (50 + ๐‘—0) Ω at ๐œ” = 106 rad s−1. The values of the line constants L, C,R, G are, respectively,

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  1. L = 200 μH/m, C = 0.1 μF/m, R = 50 Ω/m, G = 0.02 S/m
  2. L = 250 μH/m, C = 0.1 μF/m, R = 100 Ω/m, G = 0.04 S/m
  3. L = 200 μH/m, C = 0.2 μF/m, R = 100 Ω/m, G = 0.02 S/m
  4. L = 250 μH/m, C = 0.2 μF/m, R = 50 Ω/m, G = 0.04 S/m

Answer (Detailed Solution Below)

Option 2 : L = 250 μH/m, C = 0.1 μF/m, R = 100 Ω/m, G = 0.04 S/m
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Detailed Solution

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Concept:

If the characteristic impedance is purely real and the propagation constant is complex, the transmission line is distortionless and for this line  \({\rm{\alpha }} = \sqrt {{\rm{RG}}}\) and \({\rm{\beta }} = {\rm{\omega }}\sqrt {{\rm{LC}}}\). and \({{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}\)

Application: 

As \({\rm{\alpha }} = \sqrt {{\rm{RG}}}\) and \({{\rm{Z}}_{\rm{o}}} = \sqrt {\frac{{\rm{R}}}{{\rm{G}}}}\)

\(\begin{array}{l} \Rightarrow {\rm{\alpha }}{{\rm{Z}}_{\rm{o}}} = {\rm{R}}\\ {{\rm{Z}}_{\rm{o}}}{\rm{R}} = 2 \times 50 = 100{\rm{\;\Omega }}/{\rm{m}} \end{array}\)

\({\rm{G}} = \frac{{\rm{R}}}{{{\rm{Z}}_0^2}} = \frac{{100}}{{2500}} = 0.04\frac{{\rm{S}}}{{\rm{m}}}\) . We see only option b fits these two requirements. Confirming \({\rm{\beta }}\) from option b we see,

\(\begin{array}{l} {\rm{\omega }}\sqrt {{\rm{LC}}} = {10^6}\sqrt {250 \times {{10}^{ - 6}} \times 0.1 \times {{10}^{ - 6}}} \\ = {10^6} \times {10^{ - 6}} \times 5\\ = 5 = {\rm{\beta }} \end{array}\)

Thus, option b is true
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