This question was previously asked in

UPRVUNL AE EC 2016 Official Paper

Option 4 : 100 Ω

Hindi Subject Test 1

4474

10 Questions
10 Marks
10 Mins

__Concept__:

\({{Z}_{in}}={{Z}_{0}}\left( \frac{{{Z}_{L}}+j{{Z}_{0}}\tan \beta l}{{{Z}_{0}}+j{{Z}_{L}}\tan \beta l} \right)\)

\(\because \beta =\frac{2\pi }{\lambda }\), at a quarter \(\left( \frac{\lambda }{4} \right)\) distance away,

\({{Z}_{in}}={{Z}_{0}}\frac{\left( {{Z}_{L}}+j{{Z}_{0}}\tan \left( \frac{2\pi }{\lambda }.\frac{\lambda }{4} \right) \right)}{\left( {{Z}_{0}}+j{{Z}_{L}}\tan \left( \frac{2\pi }{\lambda }.\frac{\lambda }{4} \right) \right)}\)

\({{Z}_{in}}=\frac{Z_{0}^{2}}{{{Z}_{L}}}\)

__Calculation__:

Given, ZL = 200 Ω

Zin = 50 Ω

\({{Z}_{in}}=\frac{Z_{0}^{2}}{{{Z}_{L}}}\)

Putting on the respective values and solving for Z0, we get:

Z02 = Zin × ZL

Z02 = 50 × 200 = 10,000

Z0 = 100 Ω

So the quarter-wave transformer should have a characteristic impedance of 100 Ω.