GATE 2023 EC Q.52: solution and answer

GATE 2023 · ECQ.52 · 2 marks · MSQ

The standing wave ratio on a 50 $Ω$ lossless transmission line terminated in an unknown load impedance is found to be 2.0. The distance between successive voltage minima is 30 cm and the first minimum is located at 10 cm from the load. $Z_{L}$ can be replaced by an equivalent length $l_{m}$ and terminating resistance $R_{m}$ of the same line. The value of $R_{m}$ and $l_{m}$ , respectively, are

GATE 2023 EC question 52: Two transmission lines, one drawn above the other. — Figure from the GATE 2023 EC paper. It is part of the question.

(A)
$R_{m} = 100 Ω$ , $l_{m} = 20$ cm
(B)
$R_{m} = 25 Ω$ , $l_{m} = 20$ cm
(C)
$R_{m} = 100 Ω$ , $l_{m} = 5$ cm
(D)
$R_{m} = 25 Ω$ , $l_{m} = 5$ cm

Answer and solution

ANSWER

B and C: $R_{m} = 25 Ω$ with $l_{m} = 20$ cm, or $R_{m} = 100 Ω$ with $l_{m} = 5$ cm. Both reproduce $Z_{L} = \frac{50 ( 8 - j 3 3 )}{7} \approx (57.14 - j 37.12) Ω$ at the load plane.

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

Successive voltage minima are $λ /2$ apart, so $λ /2 = 30 cm$ , which gives $λ = 60 cm$ and $β = \frac{2 π}{60} rad/cm$ . A voltage maximum sits midway between two minima, $λ /4 = 15 cm$ from each.
On a lossless line the impedance is purely resistive at the extremes of the standing wave. At a voltage minimum the voltage is smallest and the current largest, so $Z = \frac{V _{min}}{I _{ma x}} = Z_{0} \frac{1 - ∣Γ∣}{1 + ∣Γ∣} = \frac{Z _{0}}{S} = \frac{50}{2} = 25 Ω$ . At a voltage maximum, $Z = Z_{0} S = 50 \times 2 = 100 Ω$ .
On a lossless line, $∣Γ (d) ∣ = \frac{S - 1}{S + 1}$ stays the same everywhere, and the phase of $Γ (d) = Γ_{L} e^{- j 2 β d}$ changes steadily with distance. So the impedance at every point is fixed by $S$ and the position of one minimum. Carry the same pattern on past the load, as in the lower figure. Minima repeat every $30 cm$ , so the real minimum $10 cm$ on the generator side of the load has a copy $30 - 10 = 20 cm$ beyond the load. The maximum between these two minima lies $20 - 15 = 5 cm$ beyond the load.
Now end the extended line at either point with the resistance the line shows there. A $25 Ω$ resistor gives $Γ = \frac{25 - 50}{25 + 50} = - \frac{1}{3}$ : $S = 2$ with a minimum at the resistor. Placed $20 cm$ beyond the load, it puts minima $20 cm$ beyond the load and $10 cm$ before it, which is exactly the measured pattern. A $100 Ω$ resistor gives $Γ = + \frac{1}{3}$ : $S = 2$ with a maximum at the resistor. Placed $5 cm$ beyond the load, it puts the next minimum $15 cm$ nearer the generator, at $10 cm$ before the load, which is again the measured pattern. The same pattern means the same impedance at $z = 0$ , namely $Z_{L}$ . That gives two equivalents: $R_{m} = 25 Ω$ with $l_{m} = 20 cm$ , and $R_{m} = 100 Ω$ with $l_{m} = 5 cm$ .
Check by finding $Z_{L}$ directly. $∣ Γ_{L} ∣ = \frac{S - 1}{S + 1} = \frac{1}{3}$ . At a voltage minimum $θ_{Γ} - 2 β d_{min} = - π$ , where $2 β d_{min} = 2 \cdot \frac{2 π}{60} \cdot 10 = \frac{2 π}{3}$ , so $θ_{Γ} = - \frac{π}{3}$ and $Γ_{L} = \frac{1}{6} - j \frac{3}{6}$ . Then $Z_{L} = 50 \frac{1 + Γ _{L}}{1 - Γ _{L}} = 50 \frac{7 - j 3}{5 + j 3} = \frac{50 ( 8 - j 3 3 )}{7} \approx (57.14 - j 37.12) Ω.$
The input impedance of a length $l$ of line ending in $R$ is $Z_{in} = Z_{0} \frac{R + j Z _{0} tan β l}{Z _{0} + j R tan β l}$ . For $R_{m} = 25 Ω$ and $l_{m} = 20 cm$ : $β l_{m} = 12 0^{\circ}$ and $tan β l_{m} = - 3$ , so $Z_{in} = 50 \frac{25 - j 50 3}{50 - j 25 3} = 50 \frac{1 - j 2 3}{2 - j 3} = \frac{50 ( 8 - j 3 3 )}{7} .$ For $R_{m} = 100 Ω$ and $l_{m} = 5 cm$ : $β l_{m} = 3 0^{\circ}$ and $tan β l_{m} = \frac{1}{3}$ , so $Z_{in} = 50 \frac{100 + j \frac{50}{3}}{50 + j \frac{100}{3}} = 50 \frac{2 3 + j}{3 + j 2} = \frac{50 ( 8 - j 3 3 )}{7} .$ Both equal $Z_{L}$ exactly.
Options A and D pair each resistance with the wrong length. $100 Ω$ at $20 cm$ and $25 Ω$ at $5 cm$ both give $Z_{in} = \frac{50 ( 8 + j 3 3 )}{13} \approx (30.77 + j 19.99) Ω \neq = Z_{L}$ . Each puts a maximum where the measured pattern has a minimum, or the other way round. The correct options are B and C.

KEY IDEA

On a lossless line the impedance at any point depends only on the SWR and where the minima are. It is purely resistive at the extremes of the standing wave: $Z_{0} / S$ at a voltage minimum and $Z_{0} S$ at a voltage maximum. So any load can be replaced by extending the line to a minimum or a maximum of its own standing-wave pattern and ending it there in that resistance. Because minima repeat every $λ /2$ , the minimum 10 cm towards the generator has a copy 20 cm beyond the load, and there is a maximum 5 cm beyond the load.

WATCH OUT FOR

Using the 10 cm distance itself as $l_{m}$ , or measuring the extension towards the generator instead of past the load. A second trap is pairing $Z_{0} / S = 25 Ω$ with the position of a maximum or $Z_{0} S = 100 Ω$ with a minimum (options A and D). Both give $(30.77 + j 19.99) Ω$ , not $Z_{L}$ . Finally, this is an MSQ: marking only one of B and C loses the marks, because both equivalents are valid.

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GATE 2023 EC Q.52: solution and answer

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