Question

At an operating frequency of 5 GHz, a 50 lossless coaxial line with insulating material having a relative permittivity r = 2.25 is terminated in an antenna with an impedance ZL = 150 . Use the Smith chart to find Zin. The line length is 30 cm.

Answer 1

Answer:

The answer is "150 $\Omega$".

Explanation:

Its line length must be converted into wavelengths for the Smith chart to be used.

$\to \beta = \frac{2 \pi }{\lambda } \\\\\to u_p = \frac{\omega }{\beta}\\\\\lambda =\frac{2 pi}{\beta} =\frac{2 \pi U_p}{\omega}=\frac{c}{\sqrt{\varepsilon_r f}}$

                        $= \frac{3 \times 10^8}{ 2.25 \times (5 \times 10^9)}\\\\= \frac{3}{ 2.25 \times 5\times 10 }\\\\= \frac{3}{ 22.5 \times 5 }\\\\= \frac{3}{ 112.5}\\\\= 0.04 \ m.$

$l =\frac{0.30 }{0.04 } \times \lambda$

  $= 7.5 \lambda$

Because it is an integrated half-wavelength amount,

$Z_{in} = Z_L = 150 \ \Omega$