GATE 2026 EC Q.52: solution and answer

GATE 2026 · ECQ.52 · 2 marks · MCQ

A complex load (in $Ω$ ) is represented as $Γ_{L} = 0.5∠3 0^{o}$ on the Smith chart.

A co-axial cable with a characteristic impedance of $50 Ω$ is connected to the load.

The new input impedance of the load now moves to a diametrically opposite point on the same $Γ$ circle on the Smith chart.

Which option is the nearest input impedance of the cable connected load (in $Ω$ )?

(A)
$20.7 - j 5.1$
(B)
$17.7 - j 11.8$
(C)
$97.5 - j 65.0$
(D)
$97.5 + j 65.0$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

(B) $17.7 - j 11.8 Ω$ . The input reflection coefficient is $Γ_{in} = - Γ_{L} = 0.5∠ - 15 0^{\circ}$ , which gives $z_{in} = 0.354 - j 0.236$ and so $Z_{in} = 50 z_{in} \approx 17.7 - j 11.8 Ω$ .

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Diametrically opposite point. On a constant- $∣Γ∣$ circle, the diametrically opposite point has the same magnitude and a phase $18 0^{\circ}$ away. So $Γ_{in} = - Γ_{L} = 0.5∠ (3 0^{\circ} + 18 0^{\circ}) = 0.5∠21 0^{\circ} = 0.5∠ - 15 0^{\circ}$ . One full turn of the Smith chart is $λ /2$ of line, so a half turn means the lossless cable is $λ /4$ long.
Rectangular form. $Γ_{in} = 0.5 [cos (- 15 0^{\circ}) + j sin (- 15 0^{\circ})] = 0.5 (- 0.866 - j 0.5) = - 0.433 - j 0.25$ .
Normalised input impedance from $z = \frac{1 + Γ}{1 - Γ}$ : $z_{in} = \frac{0.567 - j 0.25}{1.433 + j 0.25} = \frac{( 0.567 - j 0.25 ) ( 1.433 - j 0.25 )}{1.43 3 ^{2} + 0.2 5 ^{2}} = \frac{0.75 - j 0.5}{2.116} = 0.354 - j 0.236$
De-normalise with $Z_{0} = 50 Ω$ : $Z_{in} = 50 z_{in} \approx 17.7 - j 11.8 Ω$ .
Cross-check with the quarter-wave rule $Z_{in} = Z_{0}^{2} / Z_{L}$ . The load is $z_{L} = \frac{1 + Γ _{L}}{1 - Γ _{L}} = \frac{1.433 + j 0.25}{0.567 - j 0.25} = \frac{0.75 + j 0.5}{0.384} = 1.953 + j 1.302$ , so $Z_{L} \approx 97.7 + j 65.1 Ω$ . Then $Z_{in} = \frac{2500}{97.7 + j 65.1} = \frac{2500 ( 97.7 - j 65.1 )}{97. 7 ^{2} + 65. 1 ^{2}} = \frac{2500 ( 97.7 - j 65.1 )}{13775} \approx 17.7 - j 11.8 Ω,$ which agrees. (Equivalently, $Γ \to - Γ$ turns $z$ into $1/ z$ : $z_{in} = 1/ z_{L}$ .)
The nearest option is (B), $17.7 - j 11.8 Ω$ .

KEY IDEA

A diametrically opposite point on the same $∣Γ∣$ circle means $Γ_{in} = - Γ_{L}$ . That is a $18 0^{\circ}$ turn of the chart, which is a $λ /4$ lossless line, so the normalised impedance inverts: $z_{in} = 1/ z_{L}$ , or $Z_{in} = Z_{0}^{2} / Z_{L}$ .

WATCH OUT FOR

Option (D), $97.5 + j 65.0 Ω$ , is the load impedance itself. You get it by converting $Γ_{L}$ and forgetting to move to the opposite point. Option (C) is its conjugate, which you get by reflecting the point across the real axis (flipping the sign of the phase) instead of turning it through $18 0^{\circ}$ . A further slip is adding $18 0^{\circ}$ and then taking $cos 21 0^{\circ}$ and $sin 21 0^{\circ}$ with the wrong signs. Both parts of $Γ_{in}$ are negative.

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 13 Transmission lines questions, with notes

GATE 2026 EC Q.52: solution and answer

Answer and solution

More Transmission lines questions