7.1 Problem Solving in Waveguide Systems

This final session is dedicated to applying the concepts we have learned to solve practical numerical problems. Working through these examples is a crucial step in reinforcing your understanding of the key formulas, definitions, and measurement techniques discussed in the previous modules. Each problem represents a common scenario encountered in a microwave laboratory or design environment.

Problem 1: VSWR from Slotted Line Measurements

• Problem Statement: A transmission system uses a TE₁₀ mode waveguide with dimensions a = 5cm and b = 3cm. It is operating at a frequency of 10 GHz. Using a slotted line with the double minimum method, the distance measured between the two half-power points is 1 mm. Calculate the VSWR of the system.
• Solution:
  1. Determine the Cut-off Wavelength (): First, we must determine the cut-off wavelength for the TE₁₀ mode, as this defines the lowest frequency that can propagate in this specific waveguide. For the dominant TE₁₀ mode, this depends only on the broad dimension ‘a’. \lambda_c = 2a = 2 \times 5cm = 10 \: cm
  2. Calculate the Free Space Wavelength (): This is the wavelength of the signal if it were propagating in a vacuum, determined by the operating frequency. \lambda_0 = \frac{c}{f} = \frac{3\times10^{10} \: cm/s}{10\times10^9 \: Hz} = 3 \: cm
  3. Calculate the Guide Wavelength (): Next, we calculate the guide wavelength, which is the effective wavelength of the signal inside the waveguide. It is always longer than the free-space wavelength. \lambda_g = \frac{\lambda_0}{\sqrt{1-(\frac{\lambda_0}{\lambda_c})^2}} = \frac{3}{\sqrt{1-(\frac{3}{10})^2}} = \frac{3}{\sqrt{1-0.09}} = 3.144 \: cm
  4. Calculate the VSWR: Finally, we apply the formula for the double minimum method, using the measured distance between the half-power points, (d_2 – d_1) = 1 \: mm = 0.1 \: cm. VSWR = \frac{\lambda_g}{\pi(d_2-d_1)} = \frac{3.144 \: cm}{\pi(0.1 \: cm)} = 10.003
• Final Answer: The VSWR of the system is approximately 10.

Problem 2: Reflection Coefficient from Reflectometer Readings

• Problem Statement: In a reflectometer setup, the output of the directional coupler sampling incident power is 2.0 mW, and the output of the coupler sampling reflected power is 0.5 mW. What is the reflection coefficient of the load?
• Solution:
  1. Analyze the Measurement: The problem provides the sampled powers from two identical directional couplers. Since the couplers are identical, the ratio of the true reflected power (P_r) to the true incident power (P_i) is the same as the ratio of their sampled outputs. Therefore, we can calculate the reflection coefficient directly from the output readings.
  2. Calculate the Reflection Coefficient (): \rho = \sqrt{\frac{P_r}{P_i}} = \sqrt{\frac{0.5 \: mW}{2.0 \: mW}} = \sqrt{0.25} = 0.5
• Final Answer: The reflection coefficient is 0.5.

Problem 3: VSWR from Incident and Reflected Power

• Problem Statement: Two identical couplers are used to measure the incident power as 3 mW and the reflected power as 0.25 mW. Find the value of the VSWR.
• Solution:
  1. First, calculate the reflection coefficient (): \rho = \sqrt{\frac{P_r}{P_i}} = \sqrt{\frac{0.25 \: mW}{3 \: mW}} = \sqrt{0.0833} \approx 0.288
  2. Second, use to find the VSWR (S): S = \frac{1+\rho}{1-\rho} = \frac{1+0.288}{1-0.288} = \frac{1.288}{0.712} \approx 1.809
• Final Answer: The VSWR value is approximately 1.81.

Problem 4: Reflected Power from VSWR and Incident Power

• Problem Statement: The measured VSWR in a waveguide is 6. Two identical 30 dB directional couplers are used, and the output of the coupler sampling the incident power is 5 mW. What is the value of the reflected power sampled by the second coupler?
• Solution:
  1. First, find the reflection coefficient () from the VSWR: We must rearrange the VSWR formula to solve for \rho. S = \frac{1+\rho}{1-\rho} \implies S(1-\rho) = 1+\rho \implies S – S\rho = 1+\rho S – 1 = \rho(S+1) \implies \rho = \frac{S-1}{S+1} Given S = 6: \rho = \frac{6-1}{6+1} = \frac{5}{7} \approx 0.714
  2. Then, use to find the reflected power: We know that \rho^2 = P_r / P_i. This relationship applies to both the true powers and the sampled powers from identical couplers. P_{r, sampled} = \rho^2 \times P_{i, sampled} = (0.714)^2 \times 5 \: mW P_{r, sampled} \approx 0.510 \times 5 \: mW = 2.55 \: mW
• Final Answer: The value of the sampled reflected power is 2.55 mW.

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This concludes our comprehensive lecture series on Microwave Engineering. We have journeyed from the fundamental principles of the microwave spectrum, through the detailed analysis of passive components like waveguides and couplers, and the intricate workings of active devices, both solid-state and vacuum tube. We concluded with the essential practical skills of microwave measurement and applied problem-solving. This field is a critical engine of modern technology, and I encourage you to continue exploring its many challenges and opportunities. The principles learned here are the foundation upon which the next generation of wireless and high-frequency systems will be built.