$ \begin{aligned} \ &s_{12}=0.2 \angle-15^{\circ} \ &s_{21}=2.6 \angle 50^{\circ} \ &s_{22}=0.5 \angle-105^{\circ} \ \end{aligned} $ calculate and plot stability circles and choose $\Gamma_t$ for $\Gamma_{\text {in }}>>1$ Desigu load terminating network. (Io Marks)

Solution:

Stability checks -

$ \begin{aligned} \\ |\Delta| &=\left|s_{11} s_{22}-s_{12} s_{21}\right| \\\\ &=0.559\lt1 \\\\ k &=\frac{\left.|-| s_{11}\right|^2-\left|s_{22}\right|^2+|\Delta|^2}{2\left|s_{12} . s_{21}\right|}=0.2427\lt1 \\ \end{aligned}\\ $

As $k\lt1$ tiansistor is unstable $\&$ can be used to design oscillator.

Plotting of stability circles

$ \begin{aligned} \\ &C_S=\frac{\left(s_{11}-\Delta s_{22}^*\right)}{\left|s_{11}\right|^2-|\Delta|^2}=\frac{(0.9 L-150-0.559 \times 0.5\angle 105)}{0.9^2-0.559^2} \\\\ &\therefore C_S=\frac{1 \angle 134.4}{0.4975}=2 \angle 134^{\circ} \\ \end{aligned}\\ $

$ \begin{aligned} \\ &R s=\frac{\left|s_{12} s_{21}\right|}{\left|s_{11}\right|^2-|\Delta|^2}=\frac{A \text {bs }\left(0.2 L-15^{\circ} \times 2.6 L 50\right)}{0.4975} \\\\ &\therefore R_s=1.04\\ \end{aligned}\\ $

$ \begin{aligned} \\ C_L=\frac{(s_{22}-\Delta s_{11})}{|s_{22}|^2-|\Delta|^2} =\frac{(0.5 L-105^{\circ}-0.559 \times 0.9 L 150)}{0.5^2-0.559^2} \\\\ C_L =\frac{0.795 L 67}{0.062}=12.82 \angle 67^{\circ} \\\\ R_L=\frac{\left|s_{12} . s_{21}\right|}{\left|s_{22}\right|^2-|\Delta|^2}=\frac{A_{b s}\left(0.2 L-15^{\circ} \times 2.6 \angle 50^{\circ}\right)}{0.062}=8.38\\ \end{aligned}\\ $

For load stability chele $\left|s_{11}\right|=0.9\lt1$ indicates center of smith chait is representing stable point We also note that as center position and radius value of load stability cadres ale too lodge,

it becomes in-practicable to physically plot the output or load stability circle.

The large old loud stability circle covers the center of smith chait \& bence area inside old stability circle is stable. outside of this load stability circle the region is unstable.

we select $\Gamma_L=0.8 \angle 134^{\circ}$ in unstable region,

$ \begin{aligned} \\ &\Gamma_{1 N}=s_{11}+\frac{s_{12} s_{21} \Gamma_L}{1-s_{22} \Gamma_L} \\\\ &\therefore \Gamma_{1 N}=0.9 L-150^{\circ}+\left(\frac{0.2 L-15^{\circ} \times 2.6 \angle 50^{\circ} \times 0.8 \angle 134^{\circ}}{1-0.5 \angle-105^{\circ} \times 0.8 \angle 134^{\circ}}\right) \\\\ &\therefore \Gamma_{1 N}=1.48 L-159^{\circ} \therefore\left|\Gamma_{1 N}\right|=1.48\gt1 \\\\ &\therefore Z_{1 N}=z_0\left[\frac{1+\Gamma_{1 N}}{1-\Gamma_{1 N}}\right]^{\circ}=50\left[\frac{1+1.48 L-159^{\circ}}{1-1.48 L-159^{\circ}}\right]\\ \end{aligned}\\ $

$ \therefore z_{1 N}=50 \times 0.267 L-138^{\circ}=13.35 L-138^{\circ}=-9.92-j 8.9 \Omega \\ $

$\therefore$ Generator impedance

$ z_a=-\left[\frac{\operatorname{Rin}}{3} \pm j \times i n\right]=-\left[\frac{-9.92}{3}-j 8.9\right]=3.3+j 8.9 \Omega\\ $

Design of load terminating network:

Matching $\Gamma_L=0.8 L 134^{\circ}$ with center of smith chait.

Inserting a transmission line fiom $\Gamma_L$ to bring it on oxis $\therefore$ required length of transmission line

$ (0.25 \lambda-0.064 \lambda)=0.186 \lambda \\ $

$$ \begin{aligned} &\vec{R}_x=10 \\ &\therefore R_x=20 \times R_{\bar{x}}=50 \times 10=500 \Omega \end{aligned} $$ Now this $R x=500 \Omega$ we can match with $z_0=50 \Omega$ using a $\lambda 14$ transformer.

$ \begin{aligned} \\ &(0.25 \lambda-0.064 \lambda)=0.186 \lambda \\\\ &R_x=10 \\\\ &\therefore R_x=20 \times R_x=50 \times 10=500 \Omega \\ \end{aligned} \\ $

Now this $R_x=500 \Omega$ we can match with $z_0=50 \Omega$ using a $\lambda 14$ transformer.

$\therefore$ Required impedance of $\lambda / 4$ tiansformer

$ z(\lambda / 4)=\sqrt{z_0 \times R x} \\ $

$ \therefore z(\lambda 14)=\sqrt{50 \times 500}=158 \Omega$ $\therefore \vec{z}\left(\lambda(4)=\frac{158}{50}=3.16 \Omega\right.\\ $

$\therefore$ We take $\lambda / 4$ transmission line at this point with impedance o $158 \Omega$ will transform $500 \Omega$ to $50 \Omega$