Answer:
The answer is "150 [tex]\Omega[/tex]".
Explanation:
Its line length must be converted into wavelengths for the Smith chart to be used.
[tex]\to \beta = \frac{2 \pi }{\lambda } \\\\\to u_p = \frac{\omega }{\beta}\\\\\lambda =\frac{2 pi}{\beta} =\frac{2 \pi U_p}{\omega}=\frac{c}{\sqrt{\varepsilon_r f}}[/tex]
[tex]= \frac{3 \times 10^8}{ 2.25 \times (5 \times 10^9)}\\\\= \frac{3}{ 2.25 \times 5\times 10 }\\\\= \frac{3}{ 22.5 \times 5 }\\\\= \frac{3}{ 112.5}\\\\= 0.04 \ m.[/tex]
[tex]l =\frac{0.30 }{0.04 } \times \lambda[/tex]
[tex]= 7.5 \lambda[/tex]
Because it is an integrated half-wavelength amount,
[tex]Z_{in} = Z_L = 150 \ \Omega[/tex]
