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In the solution they have given like this

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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?

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  • \$\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

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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

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

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