Mumbai University > Electronics and telecommunication > Sem 7 > Microwave and Radar Engineering

Marks: 10 Marks

Year: Dec 2015

$Z_L = 200 –j100Ω$

$$\overline{Z_L}=\dfrac{Z_L}{Z_0} = \dfrac{200-j100}{100}=2-j1$$

Select path A-B-C

Which will give shunt capacitor with series resistance

For path A to B

$\overline{Z_A}=2-j1 \ \ \ \ \overline{Z_B} =1-j1.2 \ \ \ \ \overline{Z_C}=1+j0 \\ \overline{Y_A}=0.4+0.2j \ \ \ \ \overline{Y_B} =0.4+0.5j \ \ \ \ \overline{Y_C}=1+j0$

$∴\overline{ΔY}$ $=\overline{Y_B}-\overline{Y_A} \\ =(0.4+0.5j)-(0.4+0.2j)$

$\overline{ΔY}=0.3j \\ ∴ΔY=\dfrac{0.3j}{Z_0} =\dfrac{0.3j}{100}=0.003j \\ ∴j2πfc=0.003j \\ c=\dfrac{0.003}{2π×500MHz} \\ c=0.9549 pf$

Now for path B-C

$\overline{ΔZ}$ $=\overline{Z_c}-\overline{Z_B} \\ =(1+j0)-(1-j1.2) \\ =1.2j$

$ΔZ$ $=\overline{ΔZ} \times100 \\ =120 j$

Compare with

$j2πfL=120j \\ L= \dfrac{120}{2π×500 MHz} \\ L=38.19 nH$