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A short circuited stub is shunt connected to a T.L. as shown. If Zo = 50 Ω, the admittance Y seen at the function of stub and the T.L. is—
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- (0.01 – j 0.02) mho
- (0.02 – j 0.01) mho
- (0.04 – j 0.02) mho
- (0.01 – j 0) mho
Correct Option: A
When
l = | |
8 |
For short circuit stub
Zin = Z0 | ![]() | ![]() | |
Z0 + j ZL tan βI |
= Z0 | ![]() | ![]() | |
Z0 + j 0 tan π/4 |
= | ![]() | ![]() | ||
Z0 | ZL = 0 |
= j 50
or
Yin = | = - 0.02 | |
√50 |
when
I = λ/2 then βl = 2π/ λ · λ/2 = π
Zin2 = Z0 | ![]() | ![]() | |
Z0 + j ZL tan π |
= Z0 | ![]() | ![]() | |
Z0 + j ZL 0 |
ZL = 100
or
Yin2 = 1/100 = 0.01
so resultant admittance Y = Yin1 + Yin2 = 0.01 – j 0.02 mho.