Question

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

• A. (0.02 + j0) mho
• B. (0.02 - j0.01) mho
• C. (0.01 - j0.02)mho
• D. (0.04 - j0.02) mho

Answer 1

The correct option is C (0.01 - j0.02)mho

Ans : (a)

$\beta =\frac{2\pi }{\lambda }$

$For{Y}_d\beta d=\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi$

$Z_d=\frac{Z_0[Z_L+j{Z}_{0}tan\beta d]}{Z_0+j{Z}_{L}tan\beta d}$

$Z_d=\frac{50[100+j50tan\pi ]}{(50+j100tan\pi )}=100$

$Y_d=\frac{1}{Z_d}=\frac{1}{100}=0.01$

$For{Y}_s\beta d=\frac{2\pi }{\lambda }\times \frac{\lambda }{8}=\frac{\pi }{4},Z_L\to 0$

$Z_s=\frac{Z_0[Z_L+j{Z}_{0}tan\frac{\pi }{4}]}{[Z_0+j{Z}_{L}tan\frac{\pi }{4}]}=j{Z}_0$

$Y_s=\frac{1}{Z_s}=\frac{1}{j{Z}_0}=-0.02j$

$\therefore Y=Y_d+Y_s=0.01-0.02j$