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

Marks: 5M

Given,

$Z_L=10-j10 Ω \\ Z_0=50 Ω \\ F=1 GHz$

$\mathcal{Z}_L=\frac{Z_L}{Z_0} =\frac{10-j10}{50} \\ \mathcal{Z}_L=0.2-0.2j Ω \\ ∴ \mathcal{Z}_a=0.2-j0.4 Ω …\text{(Refer Smith chart 2)} \\ \mathcal{Z}=y=1 \\ y_a=1+j2 Ω$

On Z Circle Downwards (Series Capacitor C),

$∆\mathcal{z}=\mathcal{z}-\mathcal{z}_a \\ ∆\mathcal{z}=-j0.2$

For Series Capacitor C,

$\frac{1}{jwC Z_0}=-j0.2 \\ \frac{1}{j×2π×1×10^9×C×50}=-j0.2 \\ C=15.91 pF$

On Y Circle Upwards (Shunt Inductor L),

$∆y=y-y_a \\ ∴∆y=-j2$

For Series Inductor L,

$\frac{Z_0}{jwL}=-j2 \\ \frac{50}{j×2π×1×10^9×L}=-j2 \\ L=3.97 nH$

Circuit

Smith Chart: