ISRO EC 2025 Q.19: solution and answer

ISRO 2025 · ECQ.19 · 1 mark · MCQ

A one wavelength long, 50 ohm line is terminated in Z_L = 30 - j40 ohms. Its input terminals are connected to the load end of a 100 ohm, quarter-wavelength line. What is the impedance seen at the input to the 100 ohm line?

(A)
30 + j40
(B)
60 + j40
(C)
120 + j160
(D)
None of the above

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Answer and solution

ANSWER

The impedance seen at the input to the $100 Ω$ line is $120 + j 160 Ω$ , option (c).

Official answer: CSource: ISRO's final answer key of 15 May 2026

A line exactly one wavelength long has $β l = 2 π$ , so $tan β l = 0$ and $Z_{in} = Z_{L}$ whatever its characteristic impedance. The $50 Ω$ line therefore presents $30 - j 40 Ω$ at its input terminals.
These terminals are connected to the load end of the $100 Ω$ quarter-wave line, so that line is terminated in $Z_{L}^{'} = 30 - j 40 Ω$ .
For a quarter-wave line, $β l = π /2$ and $Z_{in} = \frac{Z _{0}^{2}}{Z _{L}^{'}}$ , with $Z_{0} = 100 Ω$ .
$Z_{in} = \frac{10 0 ^{2}}{30 - j 40} = \frac{10000 ( 30 + j 40 )}{3 0 ^{2} + 4 0 ^{2}} = \frac{10000 ( 30 + j 40 )}{2500}$
$Z_{in} = 4 (30 + j 40) = 120 + j 160 Ω$
This is option (c). Option (a) is only the conjugate of the load impedance, and (b) does not follow from either transformation.

KEY IDEA

A line one wavelength long (or any whole number of half-wavelengths) repeats its load impedance, so only the quarter-wave section transforms it, by $Z_{in} = Z_{0}^{2} / Z_{L}$ .

WATCH OUT FOR

Using the $50 Ω$ value in the quarter-wave formula, or forgetting to rationalise $1/ (30 - j 40)$ by multiplying by the conjugate $30 + j 40$ , which leads to the conjugate-load answer (a).

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 6 ISRO Transmission lines questions, with notes

ISRO EC 2025 Q.19: solution and answer

Answer and solution

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