GATE EC 2003 Electromagnetics Solution Q 9

GATE EC 2003 Electromagnetics Solution Q 9

A short-circuited stub is shunt connected to a transmission line as shown in the figure. If Z_0=50\;\Omega , the admittance Y seen at the junction of the stub and the transmission line is _.

GATE EC 2003 Electromagnetics Solution img2 - Grad Plus

(a) (0.01 – j0.02)mho

(b) (0.02 – j0.01)mho

(c) (0.04 – j0.02)mho

(d) (0.02 + j0)mho

Ans :- (a)

Explanation

The equivalent admittance seen at the junction is the addition of the admittance of the λ/2 line terminated with ZL and admittance of short circuited λ/8 line.

i.e. Y = Yd+Ys

For Yd, i.e. admittance seen because of λ/2 line terminated with ZL

\beta d=\frac{2\pi}\lambda\times\frac\lambda2=\pi Z_d=\frac{Z_0\left[Z_L+jZ_0\;\tan\beta d\right]}{Z_0+jZ_L\tan\beta d} Z_d=\frac{50\left[100+j50\tan\;\pi\right]}{\left(50+j100\tan\;\pi\right)}=100 Y_d=\frac1{Z_d}=\frac1{100}=0.01

For Ys, admittance of short circuited λ/8 line

\beta d=\frac{2\pi}\lambda\times\frac\lambda8=\frac\pi4 Z_s=\frac{Z_0\left[Z_L+jZ_0\;\tan{\displaystyle\frac\pi4}\right]}{\left[Z_0+jZ_L\;\tan{\displaystyle\frac\pi4}\right]}=jZ_0 Y_s=\frac1{Z_s}=\frac1{jZ_0}=-0.02j \therefore Y=Y_d+Y_s=0.01-0.02j

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