The characteristic impedance $Z_0$ of a parallel-wire transmission line can be calculated using the formula:

$Z_0 = \frac{120}{\sqrt{\varepsilon_r}} \ln\left(\frac{d}{r}\right)$

Where:

• $\varepsilon_r$ is the relative permittivity of the medium (assume $\varepsilon_r = 1$ for air),
• $d$ is the center-to-center spacing between the wires,
• $r$ is the radius of each wire.

Given:

• Wire spacing $d = 1.27 \, \text{cm}$,
• Diameter of the wire $0.21 \, \text{cm}$, hence $r = 0.21/2 = 0.105 \, \text{cm}$.

Now calculate:

$Z_0 = 120 \ln\left(\frac{d}{r}\right)$

Substitute the values:

$Z_0 = 120 \ln\left(\frac{1.27}{0.105}\right)$

Step 1: Calculate $\frac{d}{r}$

$\frac{d}{r} = \frac{1.27}{0.105} \approx 12.095$

Step 2: Find the natural logarithm

$\ln(12.095) \approx 2.492$

Step 3: Calculate $Z_0$

$Z_0 = 120 \cdot 2.492 \approx 299.04 \, \Omega$

Final Answer:

The characteristic impedance is approximately 299 Ω.

Would you like more details or have any questions? Here are some follow-ups to expand your understanding:

1. How does changing the wire spacing affect the characteristic impedance?
2. What happens if the transmission line is in a medium other than air?
3. Can this formula be used for coaxial cables, and how would it differ?
4. How does the frequency of operation affect the characteristic impedance in real-world scenarios?
5. Why is $\varepsilon_r$ assumed to be 1 in this calculation?

Tip: Always ensure unit consistency when applying formulas in electrical engineering.