## 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:**

- 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
- 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
- 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
- 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
- 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:

**Dielectric Loss:**Due to the dissipation of energy within the dielectric material (Teflon).**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