# 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 Commented Nov 5, 2017 at 7:45
• @SolarMike that I know its 90 degree but how that came? Commented Nov 5, 2017 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 $90^{\circ}$

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