GATE 2023 EC Q.54: solution and answer

GATE 2023 · ECQ.54 · 2 marks · MSQ

The following circuit(s) representing a lumped element equivalent of an infinitesimal section of a transmission line is/are

GATE 2023 EC question 54: Each option is a two-port with a top (signal) wire and a bottom (return) wire, input terminals on the left and output… — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
Figure — series $L Δ z$ and $R Δ z$ ; shunt $C Δ z /2$ only at the input side and shunt $G Δ z /2$ only at the output side
(B)
Figure — shunt $G Δ z ∥ C Δ z$ at the input, followed by series $L Δ z$ and $R Δ z$
(C)
Figure — shunt $G Δ z /2 ∥ C Δ z /2$ at the input, series $L Δ z$ and $R Δ z$ , shunt $G Δ z /2 ∥ C Δ z /2$ at the output
(D)
Figure — series $R Δ z /2$ and $L Δ z /2$ , shunt $G Δ z ∥ C Δ z$ at the middle, series $L Δ z /2$ and $R Δ z /2$

Answer and solution

ANSWER

(B), (C) and (D). Each one carries the full series impedance $(R + j ω L) Δ z$ and the full shunt admittance $(G + j ω C) Δ z$ , as an L-, $π$ - and T-section respectively. Circuit (A) has only half the shunt admittance, $\frac{1}{2} (G + j ω C) Δ z$ .

Official answer: B, C, DSource: Official answer key, GATE 2023 EC (IIT Kanpur)

A section of length $Δ z$ must carry the full per-unit-length parameters: a total series impedance of $(R + j ω L) Δ z$ and a total shunt admittance of $(G + j ω C) Δ z$ .
Apply KVL and KCL to any such section, divide by $Δ z$ and let $Δ z \to 0$ . This gives the telegrapher's equations $- \frac{\partial V}{\partial z} = (R + j ω L) I, - \frac{\partial I}{\partial z} = (G + j ω C) V$ How the elements are split or placed (L-, T- or $π$ -section) changes $Δ V$ and $Δ I$ only by terms of order $(Δ z)^{2}$ . After dividing by $Δ z$ those terms are of order $Δ z$ , so they vanish in the limit. The only thing that matters is that the totals are right.
(A): the series arm is correct ( $L Δ z$ , $R Δ z$ ), but the only shunt elements are one $C Δ z /2$ and one $G Δ z /2$ . The total shunt admittance is $(\frac{G}{2} + j ω \frac{C}{2}) Δ z$ , half the required value. It would give $- \frac{\partial I}{\partial z} = \frac{1}{2} (G + j ω C) V$ , which is wrong. Not valid.
(B): shunt $G Δ z ∥ C Δ z$ , then series $L Δ z$ and $R Δ z$ . The totals are $(R + j ω L) Δ z$ and $(G + j ω C) Δ z$ . This is the standard L-section. Valid.
(C): series $L Δ z$ and $R Δ z$ , with $G Δ z /2 ∥ C Δ z /2$ at each end. The total shunt admittance is $2 \times \frac{( G + j ω C ) Δ z}{2} = (G + j ω C) Δ z$ . This is a symmetric $π$ -section. Valid.
(D): series $\frac{R Δ z}{2} + j ω \frac{L Δ z}{2}$ on each side, giving a total of $(R + j ω L) Δ z$ , with $G Δ z ∥ C Δ z$ in the middle. This is a symmetric T-section. Valid.
The correct options are (B), (C) and (D).

KEY IDEA

A lumped model of a length $Δ z$ of line is valid if its series elements add up to $(R + j ω L) Δ z$ and its shunt elements add up to $(G + j ω C) Δ z$ . The layout (L, T or $π$ , split or unsplit) only changes terms of order $(Δ z)^{2}$ , which disappear when you derive the telegrapher's equations.

WATCH OUT FOR

Accepting (A) because it contains all four symbols $L$ , $R$ , $C$ , $G$ and looks like a $π$ -section. It has only one $C Δ z /2$ and one $G Δ z /2$ , so the shunt admittance is halved. The opposite mistake is rejecting (B) because it is not symmetric. An unsymmetric L-section still gives the correct telegrapher's equations in the limit $Δ z \to 0$ .

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 13 Transmission lines questions, with notes

GATE 2023 EC Q.54: solution and answer

Answer and solution

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