Math Problem Statement

A rectangular waveguide is measuring a=4.5 cm,

b=3cm and 9GHz signal propagated in it. Calculate

guide wavelength, phase and group velocities for

dominant mode.

Solution

To solve for the guide wavelength, phase velocity, and group velocity of the rectangular waveguide, we need to follow these steps for the dominant mode (TE₁₀ mode).

Given:

  • a=4.5cm=0.045ma = 4.5 \, \text{cm} = 0.045 \, \text{m} (broad dimension)
  • b=3cm=0.03mb = 3 \, \text{cm} = 0.03 \, \text{m} (narrow dimension)
  • Frequency, f=9GHz=9×109Hzf = 9 \, \text{GHz} = 9 \times 10^9 \, \text{Hz}

The dominant mode in a rectangular waveguide is the TE₁₀ mode, and the waveguide cutoff frequency for the TE₁₀ mode is given by:

fc=c2af_c = \frac{c}{2a} where c=3×108m/sc = 3 \times 10^8 \, \text{m/s} is the speed of light.

Step 1: Calculate the cutoff frequency

fc=3×1082×0.045=3.33GHzf_c = \frac{3 \times 10^8}{2 \times 0.045} = 3.33 \, \text{GHz}

Step 2: Guide wavelength λg\lambda_g

The guide wavelength is given by:

λg=λ01(fcf)2\lambda_g = \frac{\lambda_0}{\sqrt{1 - \left(\frac{f_c}{f}\right)^2}} where λ0\lambda_0 is the wavelength in free space:

λ0=cf=3×1089×109=0.0333m=3.33cm\lambda_0 = \frac{c}{f} = \frac{3 \times 10^8}{9 \times 10^9} = 0.0333 \, \text{m} = 3.33 \, \text{cm}

Now, calculate the guide wavelength:

λg=0.03331(3.339)2=0.033310.137=0.03330.933=0.0357m=3.57cm\lambda_g = \frac{0.0333}{\sqrt{1 - \left(\frac{3.33}{9}\right)^2}} = \frac{0.0333}{\sqrt{1 - 0.137}} = \frac{0.0333}{0.933} = 0.0357 \, \text{m} = 3.57 \, \text{cm}

Step 3: Phase velocity vpv_p

The phase velocity is given by:

vp=c1(fcf)2v_p = \frac{c}{\sqrt{1 - \left(\frac{f_c}{f}\right)^2}}

Substituting values:

vp=3×1081(3.339)2=3×1080.933=3.215×108m/sv_p = \frac{3 \times 10^8}{\sqrt{1 - \left(\frac{3.33}{9}\right)^2}} = \frac{3 \times 10^8}{0.933} = 3.215 \times 10^8 \, \text{m/s}

Step 4: Group velocity vgv_g

The group velocity is given by:

vg=c1(fcf)2v_g = c \sqrt{1 - \left(\frac{f_c}{f}\right)^2}

Substituting values:

vg=3×108×1(3.339)2=3×108×0.933=2.799×108m/sv_g = 3 \times 10^8 \times \sqrt{1 - \left(\frac{3.33}{9}\right)^2} = 3 \times 10^8 \times 0.933 = 2.799 \times 10^8 \, \text{m/s}

Summary of results:

  • Guide wavelength λg=3.57cm\lambda_g = 3.57 \, \text{cm}
  • Phase velocity vp=3.215×108m/sv_p = 3.215 \times 10^8 \, \text{m/s}
  • Group velocity vg=2.799×108m/sv_g = 2.799 \times 10^8 \, \text{m/s}

Would you like more details on any of these steps?

5 Related Questions:

  1. How would the guide wavelength change if the frequency increased?
  2. What happens to the group velocity as the operating frequency approaches the cutoff frequency?
  3. How do higher modes (e.g., TE₁₁, TE₂₀) affect the wave propagation in a rectangular waveguide?
  4. Can you calculate the cutoff frequency for the TE₁₁ mode in the same waveguide?
  5. How would the results change if the waveguide had a different broad dimension?

Tip:

In waveguides, the phase velocity always exceeds the speed of light, but the signal information travels at the group velocity, which remains below the speed of light.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Electromagnetic Waves
Wave Propagation
Waveguide Theory

Formulas

Cutoff frequency: f_c = c / (2a)
Guide wavelength: λ_g = λ_0 / sqrt(1 - (f_c/f)^2)
Phase velocity: v_p = c / sqrt(1 - (f_c/f)^2)
Group velocity: v_g = c * sqrt(1 - (f_c/f)^2)

Theorems

Waveguide Modes
Electromagnetic Propagation

Suitable Grade Level

Undergraduate