Microwave Engineering - Example Problems

Problem 1

A transmission system using a $TE_{10}$ mode waveguide of dimensions $a = 5cm, b = 3cm$ is operating at 10GHz. The distance measured between two minimum power points is 1mm on a slotted line. Calculate the VSWR of the system.

Solution

Given that $f = 10GHz; a = 5cm; b = 3cm$

For $TE_{10}$ mode waveguide,

$$\lambda_c = 2a = 2 \times 5 = 10 cm$$

$$\lambda_0 = \frac{c}{f} = \frac{3\times10^{10}}{10\times10^9} = 3cm$$

$$d_2-d_1 = 1mm = 10^{-1}cm$$

We know

$$\lambda_g = \frac{\lambda_0}{1-({\lambda_0}/{\lambda_c})^2} = \frac{3}{\sqrt{1-({3}/{10})^2}} = 3.144cm$$

For double minimum method VSWR is given by

$$VSWR = \frac{\lambda_g}{\pi(d_2-d_1)} = \frac{3.144}{\pi(1\times10^{-1})} = 10.003 = 10$$

Hence, the VSWR value for the given transmission system is 10.

Problem 2

In a setup for measuring impedance of a reflectometer, what is the reflection coefficient when the outputs of two couplers are 2mw and 0.5mw respectively?

Solution

Given that

$$\frac{P_i}{100} = 2mw \quad and \quad \frac{P_r}{100} = 0.5mw$$

$$P_i = 2 \times 100mw = 200mw$$

$$P_r = 0.5 \times 100mw = 50mw$$

$$\rho = \sqrt{\frac{P_r}{P_i}} = \sqrt{\frac{50mw}{200mw}} = \sqrt{0.25} = 0.5$$

Hence, the reflection coefficient $\rho$ of the given set up is 0.5.

Problem 3

When two identical couplers are used in a waveguide to sample the incident power as 3mw and reflected power as 0.25mw, then find the value of $VSWR$.

Solution

We know that

$$\rho = \sqrt{\frac{P_r}{P_i}} = \sqrt{\frac{0.25}{3}} = \sqrt{0.0833} = 0.288$$

$$VSWR = S = \frac{1+\rho}{1-\rho} = \frac{1+0.288}{1-0.288} = \frac{1.288}{0.712} = 1.80$$

Hence, the $VSWR$ value for the above system is 1.80

Problem 4

Two identical 30dB directional couplers are used to sample incident and reflected power in a waveguide. The value of VSWR is 6 and the output of the coupler sampling incident power is 5mw. What is the value of the reflected power?

Solution

We know that

$$VSWR = S = \frac{1+\rho}{1-\rho} = 6$$

$$(1+\rho) = 6(1-\rho) = 6 - 6\rho$$

$$7\rho = 5$$

$$\rho = \frac{5}{7} = 0.174$$

To get the value of reflected power, we have

$$\rho = \sqrt{\frac{{P_r}/{10^3}}{{P_i}/{10^3}}} = \sqrt{\frac{P_r}{P_i}}$$

$$or \quad \rho^2 = \frac{P_r}{P_i}$$

$$P_r = \rho^2.P_i = (0.714)^2.5 = 0.510 \times 5 = 2.55$$

Hence, the reflected power in this waveguide is 2.55mW.

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