GATE 2018 EC Q.21: solution and answer

GATE 2018 · ECQ.21 · 1 mark · NAT

A lossy transmission line has resistance per unit length $R = 0.05 Ω/ m$ . The line is distortionless and has characteristic impedance of $50 Ω$ . The attenuation constant (in Np/m, correct to three decimal places) of the line is ______.

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Answer and solution

ANSWER

The attenuation constant is $α = R / Z_{0} = 0.05/50 = 0.001 Np/m$ .

Official answer: 0.001Source: Official answer key, GATE 2018 EC (IIT Guwahati)

Distortionless (Heaviside) condition: $\frac{R}{L} = \frac{G}{C}$ , which is the same as $R C = L G$ .
Write $R + j ω L = L (\frac{R}{L} + j ω)$ and $G + j ω C = C (\frac{G}{C} + j ω)$ . The brackets are equal under the distortionless condition, so the characteristic impedance is purely real: $Z_{0} = \frac{R + j ω L}{G + j ω C} = \frac{L}{C} = \frac{R}{G}$ .
The same factorisation gives the propagation constant: $γ = (R + j ω L) (G + j ω C) = L C (\frac{R}{L} + j ω) = R G + j ω L C$ . So $α = R G$ , which does not depend on frequency.
From $Z_{0} = R / G$ we get $G = \frac{R}{Z _{0}^{2}}$ . Substituting, $α = R \cdot \frac{R}{Z _{0}^{2}} = \frac{R}{Z _{0}}$ .
$α = \frac{R}{Z _{0}} = \frac{0.05}{50} = 0.001 Np/m$
Check: $G = \frac{R}{Z _{0}^{2}} = \frac{0.05}{2500} = 2 \times 1 0^{- 5} S/m$ , and $R G = 0.05 \times 2 \times 1 0^{- 5} = 1 0^{- 6} = 1 0^{- 3} Np/m$ .

KEY IDEA

On a distortionless line ( $R / L = G / C$ ), $Z_{0} = L / C = R / G$ is purely real and $α = R G$ . Eliminating $G$ gives $α = R / Z_{0}$ (equivalently $α = G Z_{0}$ ), so only $R$ and $Z_{0}$ are needed.

WATCH OUT FOR

Using the low-loss formula $α = \frac{R}{2 Z _{0}} + \frac{G Z _{0}}{2}$ but keeping only the $R$ term gives $0.0005$ . On a distortionless line the $G$ term is not negligible: $G Z_{0} = R / Z_{0}$ , so the two halves are equal and add up to $R / Z_{0} = 0.001$ . Another slip is to say $α$ cannot be found because $G$ , $L$ and $C$ are not given, when the distortionless condition fixes $G = R / Z_{0}^{2}$ .

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GATE 2018 EC Q.21: solution and answer

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