Question

Find \(R, L, C\), and \(G\) for a two-wire transmission line in polyethylene at \(f=800 \mathrm{MHz}\). Assume copper conductors of radius \(0.50 \mathrm{~mm}\) and separation \(0.80 \mathrm{~cm}\). Use \(\epsilon_{r}=2.26\) and \(\sigma /\left(\omega \epsilon^{\prime}\right)=4.0 \times 10^{-4}\).

Short answer

Question: Calculate the transmission line parameters (Resistance, Inductance, Capacitance, and Conductance per unit length) for a two-wire transmission line, given the following information: - Frequency: \(800\) MHz - Radius of copper conductors: \(0.50\) mm - Separation between conductors: \(0.80\) cm - Relative permittivity of polyethylene: \(2.26\) - Loss tangent value: \(\frac{\sigma}{\omega \epsilon'} = 4.0 \times 10^{-4}\) Answer: The transmission line parameters are as follows: 1. Resistance per unit length (R): \(0.00683 \, \Omega/m\) 2. Inductance per unit length (L): \(3.65 \times 10^{-7} \, H/m\) 3. Capacitance per unit length (C): \(2.37\times 10^{-11} \, F/m\) 4. Conductance per unit length (G): \(4.63\times 10^{-12} \, S/m\)

Step-by-step solution

Calculate the Resistance per unit length (R)

First, we need to find the conductivity of copper, which is \(\sigma_c = 5.8 \times 10^7 S/m\). Now, we'll use the given radius (\(r = 0.50 \times 10^{-3}\) m) to calculate the resistance per unit length using the above formula: $$ R = \frac{1}{\pi (0.50\times10^{-3})^2 (5.8 \times 10^7)} = 0.00683 \, \Omega/m $$

Calculate the Inductance per unit length (L)

To find the inductance per unit length, we'll use the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \, H/m\). We plug in the conductor separation \(d = 0.80 \times 10^{-2}\) m and conductor radius \(r = 0.50 \times 10^{-3}\) m in the formula: $$ L = \frac{4\pi \times 10^{-7}}{\pi} \ln\frac{0.80 \times 10^{-2}}{0.50 \times 10^{-3}} = 3.65 \times 10^{-7}\,H/m $$

Calculate the Capacitance per unit length (C)

For the capacitance per unit length, we'll use the permittivity of free space, \(\epsilon_0 = 8.854 \times 10^{-12} \, F/m\), as well as the given relative permittivity for polyethylene, \(\epsilon_r = 2.26\). Plug these values, along with \(d\) and \(r\), into the formula: $$ C = \frac{\pi (8.854\times10^{-12})(2.26)}{\ln\frac{0.80 \times 10^{-2}}{0.50 \times 10^{-3}}} = 2.37\times 10^{-11} \, F/m $$

Calculate the Conductance per unit length (G)

Finally, let's find the conductance per unit length. We need the angular frequency \(\omega\), which is \(2\pi f\), with the given frequency \(f = 800\times10^6\) Hz: $$ \omega = 2\pi (800\times10^6) = 5.027\times10^9 \, rad/s $$ Now, using the given formula and values: $$ G = \frac{\pi(5.027\times10^9)(8.854\times10^{-12})(2.26)(4.0 \times 10^{-4})}{\ln\frac{0.80 \times 10^{-2}}{0.50 \times 10^{-3}}} = 4.63\times 10^{-12}\,S/m $$ To summarize our findings: 1. Resistance per unit length (R): \(0.00683 \, \Omega/m\) 2. Inductance per unit length (L): \(3.65 \times 10^{-7} \, H/m\) 3. Capacitance per unit length (C): \(2.37\times 10^{-11} \, F/m\) 4. Conductance per unit length (G): \(4.63\times 10^{-12} \, S/m\)

Key concepts

Resistance per unit length

Resistance per unit length is a crucial parameter in understanding how much opposition a conductor presents to the flow of electric current. This is especially important in transmission lines where energy efficiency is paramount. For a two-wire transmission line with copper conductors, like in the exercise presented, resistance is calculated using the conductivity of the material, the radius of the conductors, and the cross-sectional area. Copper, being a highly conductive material, has a well-defined conductivity, \( \sigma_c = 5.8 \times 10^7 \ S/m \). By plugging this into the formula for resistance:\[ R = \frac{1}{\pi r^2 \sigma_c} \]we arrive at the value that shows how much resistance the current will encounter per meter of the transmission line. Understanding this helps in designing efficient electrical circuits and systems.

Inductance per unit length

Inductance per unit length represents the tendency of a conductor to oppose a change in the current flowing through it. This property plays a vital role in how transmission lines transmit signals and energy. In the calculation, the key elements include the permeability of free space, \( \mu_0 = 4\pi \times 10^{-7} \ H/m \), along with the physical layout of the transmission line, such as the separation distance and radius of the conductors. Using the formula:\[ L = \frac{\mu_0}{\pi} \ln\frac{d}{r} \]where \( d \) is the center-to-center distance between the wires, we determine how inductive effects will influence the transmission line's behavior. These effects are crucial for signal integrity in communication lines.

Capacitance per unit length

Capacitance per unit length defines the ability of a transmission line to store charge, which is pivotal in alternating current scenarios. Here, permittivity plays a central role, especially the permittivity of free space, \( \epsilon_0 = 8.854 \times 10^{-12} \ F/m \), and the relative permittivity of the insulating material (polyethylene in this case) with \( \epsilon_r = 2.26 \). The formula used is:\[ C = \frac{\pi \epsilon_0 \epsilon_r}{\ln\frac{d}{r}} \]This calculation shows how much electrical energy the line can store and release, impacting how effectively it can transmit signals over distance. A higher capacitance can lead to energy losses or signal distortions if not properly managed.

Conductance per unit length

Conductance per unit length is linked to the leakage of current through the insulating material between the conductors, an important factor for signal quality. This parameter directly relates to how well the dielectric material insulates and is characterized under certain conditions by relative permittivity, frequency, and other specific transmission line parameters. The angular frequency \( \omega = 2\pi f \), where \(f = 800\times10^6\) Hz, is involved in the computation:\[ G = \frac{\pi \omega \epsilon_0 \epsilon_r \sigma_d}{\ln\frac{d}{r}} \]where \( \sigma_d = 4.0 \times 10^{-4} \) accounts for dielectric losses. Conductance per unit length indicates how much unwanted attenuation or distortion might affect electrical signals in high-frequency applications.

Polyethylene dielectric properties

Polyethylene is a commonly used dielectric material in transmission lines due to its excellent insulating properties. Its relative permittivity, \( \epsilon_r = 2.26 \),makes it suitable for high-frequency applications like the problem presented. This property dictates how well it can store electrical energy relative to air or vacuum. The low value of dielectric loss factor represented by \( \sigma/\left(\omega \epsilon^{\prime}\right)=4.0 \times 10^{-4} \)allows polyethylene to efficiently insulate without significantly degrading the signal over distance. In practical terms, this means less energy is lost as heat, and signals can travel farther with less distortion, making it a great choice for many telecommunications and data transport applications.