Question

Each conductor of a two-wire transmission line has a radius of $0.5\mathrm{mm}$ their center-to-center separation is $0.8\mathrm{cm}$. Let $f=150\mathrm{MHz}$, and assume $\sigma$ and ${\sigma }_c$ are zero. Find the dielectric constant of the insulating medium if (a) $Z_0=300\mathrm{\Omega };$ (b) $C=20\mathrm{pF}/\mathrm{m};$ ( $c$ ) $v_p=2.6\times {10}^8\mathrm{m}/\mathrm{s}$.

Short answer

Question: Calculate the dielectric constant of the insulating medium under the given conditions. Answer: The dielectric constant of the insulating medium under the given conditions is as follows: - For given impedance (300Ω): εr ≈ 2.05 - For given capacitance (20pF/m): εr ≈ 2.65 - For given propagation velocity (2.6x10^8 m/s): εr ≈ 1.33

Step-by-step solution

Interpret the given data

We are given the conductor radius, $r=0.5\mathrm{mm}$, the separation between conductors, $d=0.8\mathrm{cm}$, and frequency, $f=150\mathrm{MHz}$. Next, we will convert the values into SI units: - $r=0.5\times {10}^{-3}\mathrm{m}$ - $d=0.8\times {10}^{-2}\mathrm{m}$ - $f=150\times {10}^6\mathrm{Hz}$

Calculate the characteristic impedance ($Z_0$) formula

The formula for characteristic impedance ($Z_0$) of a two-wire transmission line is: $$Z_0=120\pi \sqrt{\frac{{ϵ}_r}{2}}\ln (\frac{D}{r}),$$ where ${ϵ}_r$ is the relative permittivity (dielectric constant) of the insulating medium, $D$ is the center-to-center separation, and $r$ is the radius of the conductor.

Calculate the dielectric constant with given characteristic impedance ($Z_0$)

In part (a), we are given $Z_0=300\mathrm{\Omega }$. Substituting the given values in the impedance formula, we can solve for the dielectric constant (${ϵ}_r$). $$300=120\pi \sqrt{\frac{{ϵ}_r}{2}}\ln (\frac{0.8\times {10}^{-2}}{0.5\times {10}^{-3}})$$ Solving for ${ϵ}_r$: $${ϵ}_r=\frac{2(300/120\pi {)}^2}{{\ln }^2(\frac{0.8\times {10}^{-2}}{0.5\times {10}^{-3}})}\approx 2.05$$

Calculate the capacitance ($C$) formula

The formula for capacitance per unit length ($C$) of a two-wire transmission line is: $$C=\frac{\pi {ϵ}_0{ϵ}_r}{\ln (\frac{D}{r})},$$ where ${ϵ}_0$ is the vacuum permittivity, approximately $8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}$.

Calculate the dielectric constant with given capacitance ($C$)

In part (b), we are given $C=20\mathrm{pF}/\mathrm{m}$. Substituting the given values in the capacitance formula, we can solve for the dielectric constant (${ϵ}_r$). $$20\times {10}^{-12}=\frac{\pi \times 8.85\times {10}^{-12}{ϵ}_r}{\ln (\frac{0.8\times {10}^{-2}}{0.5\times {10}^{-3}})}$$ Solving for ${ϵ}_r$: $${ϵ}_r=\frac{20\times {10}^{-12}\ln (\frac{0.8\times {10}^{-2}}{0.5\times {10}^{-3}})}{\pi \times 8.85\times {10}^{-12}}\approx 2.65$$

Calculate the propagation velocity ($v_p$) formula

The formula for propagation velocity ($v_p$) of a two-wire transmission line is: $$v_p=\frac{c}{\sqrt{{ϵ}_r}},$$ where $c$ is the speed of light, approximately $3\times {10}^8\mathrm{m}/\mathrm{s}$.

Calculate the dielectric constant with given propagation velocity ($v_p$)

In part (c), we are given $v_p=2.6\times {10}^8\mathrm{m}/\mathrm{s}$. Substituting the given value in the propagation velocity formula, we can solve for the dielectric constant (${ϵ}_r$). $$2.6\times {10}^8=\frac{3\times {10}^8}{\sqrt{{ϵ}_r}}$$ Solving for ${ϵ}_r$: $${ϵ}_r={\left(\frac{3\times {10}^8}{2.6\times {10}^8}\right)}^2\approx 1.33$$ The dielectric constant of the insulating medium under the given conditions is as described: - For given impedance ($300\mathrm{\Omega }$): ${ϵ}_r\approx 2.05$ - For given capacitance ($20\mathrm{pF}/\mathrm{m}$): ${ϵ}_r\approx 2.65$ - For given propagation velocity ($2.6\times {10}^8\mathrm{m}/\mathrm{s}$): ${ϵ}_r\approx 1.33$

Key concepts

Characteristic Impedance

Characteristic impedance, denoted as $Z_0$, is a fundamental property of transmission lines, reflecting how the line behaves when transmitting signals. It is the ratio of voltage to current at any point on the line when the load is perfectly matched, meaning no signal gets reflected.

In our two-wire transmission line scenario, $Z_0$ is calculated using the specific geometry of the conductors and the properties of the insulating medium between them. The formula for $Z_0$ involves the relative permittivity (dielectric constant) of the insulating medium, ${ϵ}_r$, the conductor separation, $D$, and the radius of the conductors, $r$. Accurate computation of this impedance is crucial for ensuring signal integrity and minimizing reflections.

Two-Wire Transmission Line

A two-wire transmission line is a simple type of line used for transmitting electrical signals over distances. It consists of two parallel conductors with a consistent separation distance. This design is instrumental in RF and microwave circuits since it supports high-frequency signal transmission.

The dimensions of the conductors and the choice of insulation material, which determines the dielectric constant, are critical in designing such lines. Adjustments to these parameters affect the line's characteristic impedance, capacitance, and propagation velocity, thereby influencing the quality and speed of signal transmission.

Propagation Velocity

Propagation velocity, $v_p$, is the speed at which an electromagnetic wave travels through a transmission line. It provides valuable insight into the time it takes for a signal to traverse the line's length.

This velocity is inversely proportional to the square root of the dielectric constant, ${ϵ}_r$, of the insulating medium. Indeed, for two-wire transmission lines, the choice of insulating material significantly impacts the propagation speed, with a lower ${ϵ}_r$ facilitating faster signal travel. Knowing $v_p$ helps predict signal timing and phase, both crucial for the synchronization of electronic systems.

Capacitance

Capacitance is a measure of a system's ability to store an electric charge. For a two-wire transmission line, capacitance per unit length, $C$, captures the system's ability to store electrical energy between the two conductors.

The capacitance depends on the dielectric constant of the insulating material between conductors and the geometry of the conductors themselves. Accurate calculation of capacitance is necessary for designing circuits because it affects the transmission line's ability to maintain signal integrity over a range of frequencies and enables the prediction of possible signal distortions.