Solution:

calculation of input impedance of diode:

$ \begin{aligned} \\ &z_{\text {in }}=z_0\left[\frac{1+\Gamma_{1 N}}{1-\Gamma_{1 N}}\right]=50 \times\left[\frac{1+1.25 \angle 40^{\circ}}{1-1.25 L 40^{\circ}}\right]=50 \times 2.5 \angle 40^{\circ} \\\\ &\therefore \pi_{\text {in }}=125 \angle 40^{\circ}=95.75+180.34 \Omega \\ \end{aligned} \\ $

calculation of cinerator impedance:

$ \begin{aligned} \\ &z_G=-z_{1 N}=-\left[\frac{R_{1 N}}{3} \pm j x_{1 N}\right]=-\left[\frac{95.75}{3}+j 80.34\right] \\\\ &\therefore z_G=-\frac{95.75}{3}-j 80.34=-31.9-j 80.34 \Omega \\\\ &\therefore z_G \approx-32-j 80 \Omega \\ \end{aligned}\\ $

Matching circuit desigu for $z_a$

$\therefore$ plot $\bar{z} g=\frac{32-j 80 \Omega}{50}=0.64-j 1.6 \Omega$

using single parallel s.c. stub locating $\bar{y}_g$ at $0.082 \lambda$ at design point 1 WTa $=0.187 \lambda$

$\therefore$ location of stub $=d=0.187 \lambda-0.082 \lambda=0.105 \lambda$

at $1^{\prime} \bar{y}_1^{\prime}=1+j 2$

$\therefore$ locating $-j 2$ we get $W T G=0.324 \lambda$

$\therefore$ lengtu of stab $=0.324 \lambda-0.25 \lambda=0.074 \lambda$