GATE 2026 EC – Question 52
A complex load (in Ω) is represented as $\Gamma_L=0.5\angle30^\circ$ 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)?
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Correct answer: (B) $17.7-j11.8$
Explanation
The diametrically opposite point has $\Gamma=0.5\angle210^\circ=-0.433-j0.25$. Then $Z=50(1+\Gamma)/(1-\Gamma)=50(0.567-j0.25)/(1.433+j0.25)\approx17.7-j11.8\ \Omega$.