enter image description here

In the solution they have given like this

enter image description here

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

Can anyone help me Please?

  • \$\begingroup\$ 2 pi / 4 is 90degrees \$\endgroup\$
    – Solar Mike
    Commented Nov 5, 2017 at 7:45
  • 1
    \$\begingroup\$ @SolarMike that I know its 90 degree but how that came? \$\endgroup\$
    – Rohit
    Commented Nov 5, 2017 at 7:47

1 Answer 1


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}\$

finally the answer becomes



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.