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Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. 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—
    1. (0.01 – j 0.02) mho
    2. (0.02 – j 0.01) mho
    3. (0.04 – j 0.02) mho
    4. (0.01 – j 0) mho
Correct Option: A

When

l =
λ
8

For short circuit stub
Zin = Z0
ZL + j Z0 tan βI
Z0 + j ZL tan βI

= Z0
0 + j 50 tan π/4
Z0 + j 0 tan π/4

=
Z0[j 50]
∵ 2π/λ , λ/8 = π/4
Z0ZL = 0

= j 50
or
Yin =
1
= - 0.02
50

when
I = λ/2 then βl = 2π/ λ · λ/2 = π
Zin2 = Z0
ZL + j Z0 tan π
Z0 + j ZL tan π

= Z0
ZL + j Z0 0
Z0 + j ZL 0

ZL = 100
or
Yin2 = 1/100 = 0.01
so resultant admittance Y = Yin1 + Yin2 = 0.01 – j 0.02 mho.



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