The conductors of a coaxial transmission line are copper \(\left(\sigma_{c}=5.8 \times\right.\) \(\left.10^{7} \mathrm{~S} / \mathrm{m}\right)\), and the dielectric is polyethylene \(\left(\epsilon_{r}^{\prime}=2.26, \sigma / \omega \epsilon^{\prime}=0.0002\right) .\) If the inner radius of the outer conductor is \(4 \mathrm{~mm}\), find the radius of the inner conductor so that \((a) Z_{0}=50 \Omega ;(b) C=100 \mathrm{pF} / \mathrm{m} ;(c) L=0.2 \mu \mathrm{H} / \mathrm{m}\). A lossless line can be assumed.

Characteristic impedance (Z0) of a coaxial cable can be calculated as follows: \(Z_0 = \frac{1}{2\pi}\sqrt{\frac{\frac{\mu}{\epsilon^{\prime}}}{\sigma} }\ln{\frac{b}{a}}\) Where: - Z0: characteristic impedance - µ: permeability (for lossless line: µ = µ0) - σ: conductivity of the dielectric (σ = σ/ωε') - ε': relative permittivity - b: inner radius of the outer conductor - a: radius of the inner conductor From the given exercise: - \(\epsilon_r^{\prime} = 2.26\) - \(σ/ωε^{\prime} = 0.0002\) - Inner radius of the outer conductor: \(b = 4mm\) Now we can rearrange this equation for (a) to find the radius of the inner conductor for a given Z0: \(a = b\,e^{\frac{2\pi Z_0}{\sqrt{\frac{\mu_0}{\epsilon_r^{\prime}}\sigma}}}\)

The characteristic impedance is given as Z0 = 50 Ω. Plug this value and other given values into the equation for the characteristic impedance: \(a = 4\,e^{\frac{2\pi \times 50}{\sqrt{\frac{\mu_0}{2.26}\times 0.0002}}}\) Calculate the value of 'a' to find the radius of the inner conductor: \(a \approx 1.461\,\text{mm}\)

Capacitance (C) and inductance (L) per unit length of a coaxial cable can be calculated as follows: 1. Capacitance per unit length (C): \(C = 2\pi\epsilon_r^{\prime}\epsilon_0\frac{1}{\ln{\frac{b}{a}}}\) 2. Inductance per unit length (L): \(L = \frac{\mu_0}{2\pi}\ln{\frac{b}{a}}\)

We are given the following C and L values: - C = \(100\,\text{pF}\,/\,\text{m}\) - L = \(0.2\,\mu\,\text{H}\,/\,\text{m}\) Plugging these values into the capacitance and inductance formulas, we can calculate the inner conductor radius (a): 1. For given capacitance (C): \(a = b\,e^{\frac{1}{2\pi \epsilon_r^{\prime} \epsilon_0 C}} = 4\,e^{\frac{1}{2\pi\times 2.26\times 8.85\times 10^{-12}\times 100\times 10^{-12}}} \approx 2.038\,\text{mm}\) 2. For given inductance (L): \(a = b\,e^{\frac{2\pi L}{\mu_0}} = 4\,e^{\frac{2\pi\times 0.2\times 10^{-6}}{4\pi\times10^{-7}}} \approx 0.911\,\text{mm}\) Analysis of results: We found the following values for the inner conductor diameter for each parameter: - Z0 = 50 Ω: \(a \approx 1.461\,\text{mm}\) - C = 100 pF/m: \(a \approx 2.038\,\text{mm}\) - L = 0.2 µH/m: \(a \approx 0.911\,\text{mm}\) Notice that a different inner conductor diameter is required for each parameter, and therefore it cannot satisfy all three conditions simultaneously. Further analysis or optimization could be performed to find a compromise solution that comes close to satisfying all three conditions.