# Reflection coefficient at the load when reflection coefficient at the mid point of transmission line given

In the solution they have given like this

Here at the end they have calculated the angle that did't get why they have added it by $2\pi/4$.?

• 2 pi / 4 is 90degrees – Solar Mike Nov 5 '17 at 7:45
• @SolarMike that I know its 90 degree but how that came? – Rohit Nov 5 '17 at 7:47

The General formula of Reflection coefficient at the line is

$$\Gamma_l(l)=\mid\Gamma_L\mid e^{j(\theta_\Gamma-2\beta l)}$$

at the mid point that is at $l/2$ the reflection coefficient is $$\Gamma_l(l/2)=\mid\Gamma_L\mid e^{j(\theta_\Gamma-\beta l)}$$

Now $\Gamma_l(l/2)$ same as the $\Gamma'$ in the solution manual that is $$\Gamma_l(l/2)=0.4685\angle-38.666^{\circ}$$

Now I need to calculate the reflection coefficient at the load that is $\mid\Gamma_L\mid e^{j(\theta_\Gamma)}$=$\Gamma_L$

$$\Gamma_L=\Gamma_l(l/2)e^{\beta l}$$

Now calculate $\beta l=\frac{2\pi\times300\times10^6\times4.2}{0.8\times3\times 10^8}=\frac{21 \pi}{2}=1890^{\circ}$ $$\Gamma_L=\Gamma_l(l/2)e^{\beta l}=0.4685\angle-38.666e\times e^{j1890^{\circ}}$$ $$=0.4685\angle-38.666\times [cos(1890^{\circ})+jsin(1890)^{\circ}]$$ $$=0.4685\angle-38.666\times(0+j)$$ where j is nothing but $90^{\circ}$

$$\boxed{0.4685\angle-38.666^{\circ}+90^{\circ}=0.4685\angle51.54^{\circ}}$$