GATE 2020 EC Q.21: solution and answer

GATE 2020 · ECQ.21 · 1 mark · NAT

A transmission line of length $3 λ /4$ and having a characteristic impedance of $50 Ω$ is terminated with a load of $400 Ω$ . The impedance (rounded off to two decimal places) seen at the input end of the transmission line is ________ $Ω$ .

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Answer and solution

ANSWER

The impedance seen at the input end is $Z_{in} = Z_{0}^{2} / Z_{L} = 2500/400 = 6.25 Ω$ .

Official answer: 6.25Source: Official answer key, GATE 2020 EC (IIT Delhi)

The question gives no loss, so treat the line as lossless. Its input impedance is $Z_{in} = Z_{0} \frac{Z _{L} + j Z _{0} tan β l}{Z _{0} + j Z _{L} tan β l} .$
Electrical length: $β l = \frac{2 π}{λ} \cdot \frac{3 λ}{4} = \frac{3 π}{2}$ . Here $tan β l$ grows without bound, approaching $\pm \infty$ from either side.
Divide the numerator and the denominator by $tan β l$ and take the limit: $Z_{in} = Z_{0} \cdot \frac{j Z _{0}}{j Z _{L}} = \frac{Z _{0}^{2}}{Z _{L}} .$ This is the quarter-wave inverter relation. Because $Z_{in}$ repeats every $λ /2$ , a $3 λ /4$ line behaves exactly like a $λ /4$ line.
Substitute the values: $Z_{in} = \frac{( 50 ) ^{2}}{400} = \frac{2500}{400} = 6.25 Ω.$
Check: the load is real and $Z_{L} > Z_{0}$ , so it sits at a voltage maximum, where the impedance is $Z_{0} \cdot VSWR$ and $VSWR = 400/50 = 8$ . An odd number of quarter-waves away the line is at a voltage minimum, where the impedance is $Z_{0} / VSWR = 50/8 = 6.25 Ω$ . This agrees with step 4.

KEY IDEA

On a lossless line, any length that is an odd multiple of $λ /4$ ( $λ /4, 3 λ /4, 5 λ /4, \dots$ ) makes $tan β l \to \infty$ . The line then acts as an impedance inverter: $Z_{in} = Z_{0}^{2} / Z_{L}$ . Input impedance repeats every $λ /2$ , so only the length modulo $λ /2$ matters.

WATCH OUT FOR

One common slip is to treat $3 λ /4$ like a half-wave line and answer $Z_{in} = Z_{L} = 400 Ω$ . Only multiples of $λ /2$ repeat the load; $3 λ /4 = λ /2 + λ /4$ inverts it. Another is to invert the ratio and write $Z_{L}^{2} / Z_{0} = 3200 Ω$ . A quick sanity check catches both: a load larger than $Z_{0}$ must turn into an input impedance smaller than $Z_{0}$ after an odd number of quarter-waves.

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GATE 2020 EC Q.21: solution and answer

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