To determine the attenuation constant $\alpha$ for a transmission line, we use the following general formula for the attenuation constant of a transmission line:

$\alpha = \text{Re} \left( \sqrt{(R' + j \omega L')(G' + j \omega C')} \right)$

Where:

• $R'$ is the resistance per unit length (Ω/m)
• $L'$ is the inductance per unit length (H/m)
• $G'$ is the conductance per unit length (S/m)
• $C'$ is the capacitance per unit length (F/m)
• $\omega$ is the angular frequency $\omega = 2 \pi f$, where $f$ is the frequency (Hz)

We are given the following values:

• $R' = 0.100 \, \Omega/\text{m}$
• $L' = 25 \, \mu\text{H/m} = 25 \times 10^{-6} \, \text{H/m}$
• $G' = 0 \, \text{S/m}$
• $C' = 25 \, \text{nF/m} = 25 \times 10^{-9} \, \text{F/m}$
• $f = 80 \, \text{kHz} = 80 \times 10^3 \, \text{Hz}$

Step 1: Calculate the angular frequency $\omega$

$\omega = 2 \pi f = 2 \pi \times 80 \times 10^3 \, \text{Hz} = 502,654.82 \, \text{rad/s}$

Step 2: Plug in the values into the formula

Since $G' = 0$, the formula simplifies to: $\alpha = \text{Re} \left( \sqrt{(R' + j \omega L') j \omega C'} \right)$

Let's compute the attenuation constant $\alpha$ step by step.The attenuation constant $\alpha$ for the given two-wire air line at 80 kHz is approximately $0.00158 \, \text{Np/m}$.

Would you like further details or have any additional questions?

Here are some related questions for deeper understanding:

1. How does the attenuation constant change with frequency?
2. What effect does increasing the resistance per unit length $R'$ have on the attenuation constant?
3. How does the capacitance per unit length $C'$ affect the transmission line’s behavior at higher frequencies?
4. What is the physical meaning of the attenuation constant in transmission lines?
5. How would the presence of conductance $G'$ affect the calculation of $\alpha$?

Tip: The attenuation constant $\alpha$ determines how much signal power is lost per unit length due to resistance and dielectric losses in the line.