Consider a length of Teflon-filled, copper K-band rectangular waveguide having dimensions a = 1.07 cm and b=0.43 cm. Find the cutoff frequencies of the first five propagating modes. If the operating frequency is 15 GHz, find the attenuation due to dielectric and conductor losses.

Question

Consider a length of Teflon-filled, copper K-band rectangular wave guide having dimensions a = 1.07 cm and b=0.43 cm. Find the cutoff frequencies of the first five propagating modes. If the operating frequency is 15 GHz, find the attenuation due to dielectric and conductor losses.

Answer

Cutoff Frequencies of the First Five Propagating Modes:

For a rectangular waveguide filled with Teflon (dielectric constant ε_r = 2.08), the cutoff frequencies of the first five propagating modes can be calculated using the following equations:

TE Modes (Transverse Electric):

  • TE_mn: f_mn = c / sqrt(ε_r) * sqrt[(m/a)^2 + (n/b)^2]

TM Modes (Transverse Magnetic):

  • TM_mn: f_mn = c / a * b * sqrt(ε_r) * sqrt[(m/a)^2 + (n/b)^2]

where:

  • c is the speed of light in free space (3 x 10^8 m/s)
  • a and b are the dimensions of the waveguide (a = 1.07 cm, b = 0.43 cm)
  • m and n are integers representing the mode order (m for the longer dimension, n for the shorter dimension)

Calculation:

TE Modes:

  1. TE_10: f_10 = (3 x 10^8 m/s) / sqrt(2.08) * sqrt[(1/1.07 cm)^2 + (0/0.43 cm)^2] = 6.98 GHz
  2. TE_01: f_01 = (3 x 10^8 m/s) / (1.07 cm * 0.43 cm) * sqrt(2.08) * sqrt[(0/1.07 cm)^2 + (1/0.43 cm)^2] = 16.25 GHz
  3. TE_20: f_20 = (3 x 10^8 m/s) / sqrt(2.08) * sqrt[(2/1.07 cm)^2 + (0/0.43 cm)^2] = 13.96 GHz
  4. TE_11: f_11 = (3 x 10^8 m/s) / (1.07 cm * 0.43 cm) * sqrt(2.08) * sqrt[(1/1.07 cm)^2 + (1/0.43 cm)^2] = 17.68 GHz
  5. TE_02: f_02 = (3 x 10^8 m/s) / (1.07 cm * 0.43 cm) * sqrt(2.08) * sqrt[(0/1.07 cm)^2 + (2/0.43 cm)^2] = 32.50 GHz

TM Modes:

  • TM_mn: not propagating in this case due to higher cutoff frequencies compared to TE modes for the given waveguide dimensions and material.

Attenuation at 15 GHz:

There are two main types of losses affecting waveguides:

  1. Dielectric Loss: Due to the dissipation of energy within the dielectric material (Teflon).
  2. Conductor Loss: Due to the finite conductivity of the wave guide walls (copper).

Calculating Attenuation:

The total attenuation (α_total) is the sum of the dielectric attenuation (α_d) and the conductor attenuation (α_c):

α_total = α_d + α_c

Dielectric Attenuation:

  • α_d = tan(δ_d) * πf / c * sqrt(ε_r)

where:

  • δ_d is the loss tangent of the dielectric material (0.0004 for Teflon)

Conductor Attenuation:

  • α_c = 8.686 * σ / (a * b * sqrt(f))

where:

  • σ is the conductivity of the waveguide material (5.8 x 10^7 S/m for copper)

Calculation:

  • α_d = tan(0.0004) * π * 15 GHz / (3 x 10^8 m/s) * sqrt(2.08) ≈ 0.004 dB/m
  • α_c = 8.686 * 5.8 x 10^7 S/m / (1.07 cm * 0.43 cm * sqrt(15 GHz)) ≈ 0.075 dB/m

Therefore, the total attenuation at 15 GHz is:

α_total ≈ 0.004 dB/m + 0.075 dB/m ≈ 0.079 dB/m

Leave a Comment