In this chapter, let us have some fun by solving a few numerical problems related to microwaves.

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,

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

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

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

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

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

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

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

$$(1+rho) = 6(1-rho) = 6 – 6rho$$

$$7rho = 5$$

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

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.