Mumbai University > Electronics and Telecomm > Sem 7 > Microwave & Radar Engineering

Marks: 10M

Given,

$Z_L=200-j100 Ω \\ Z_0=100 Ω \\ F=500 MHz$

$\mathcal{Z}_L=\frac{Z_L}{Z_0} =\frac{200-j100}{100} \\ \mathcal{Z}_L=2-j \Omega \\ \therefore \mathcal{Z}=1-j1.2 \Omega \space \dots \text{(Refer Smith chart 1)} \\ \mathcal{Z}=y=1 \\ \mathcal{Z}_L \rightarrow \mathcal{Z} \rightarrow \mathcal{Z} = 1$

On Y circle downwards (Shunt Capacitor C),

$∆y=y_a-y_L \\ y_L=0.4+j0.2 \\ y_a=0.4+j0.5\\ ∆y=j0.3$

For shunt capacitor,

$jw C Z_0=j0.3 \\ j×2π ×500 ×10^6×C×100=j0.3 \\ C=0.9549 pF$

On Z circle upwards (Series Inductor L),

$∆z=z-z_a \\ ∆z=j1.2$

For Series Inductor L,

$\frac{jwL}{Z_0} =j1.2 \\ \frac{j×2π ×500 ×10^6×L}{100}=j1.2 \\ L=38.1971 nH$

Circuit -

Smith Chart: