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A parallel-plate guide is to be constructed for operation in the TEM mode only over the frequency range \(0

Short Answer

Expert verified
Based on the given information about the frequency range, dielectric between the plates, and operating mode, we used the formulas for propagation constant, impedance, capacitance, and inductance to determine the maximum allowable plate separation (d̄) for this parallel-plate waveguide. The final result is approximately 1.59 x 10^(-4) meters.

Step by step solution

01

Write the given information

We're given the following information: - Frequency range: 0 < f < 3 GHz - Dielectric between plates: Teflon (ε_r' = 2.1) Our objective is to calculate the maximum allowable plate separation, d̄.
02

Calculate the propagation constant γ using the frequency of 3 GHz

Since the cutoff frequency for the TEM mode is zero, we'll use the maximum frequency of the range, which is 3 GHz. We have to find the propagation constant (γ) using the formula: γ = √(ω²με - β²) Where ω is the angular frequency, μ is the permeability, ε is the permittivity, and β is the phase constant. We can simplify this formula as follows: γ = √((2πf)²με - β²) Now, we need to find ε and μ, knowing that the dielectric between the plates is Teflon with a relative permittivity of ε_r' = 2.1. The permeability μ can be assumed to be equal to the permeability of free space (μ_0), since Teflon is a non-magnetic material: μ = μ₀ = 4π10^(-7) H/m The permittivity (ε) is given by: ε = ε₀ε_r' Where ε₀ is the permittivity of free space, ε₀ = 8.854 x 10^(-12) F/m. Now, we can calculate the permittivity of Teflon: ε = (8.854 x 10^(-12) F/m)(2.1) = 18.6 x 10^(-12) F/m
03

Calculate the propagation constant γ

Next, we will substitute the values found in step 2 for μ, ε and f in the formula: γ = √((2π(3 x 10^9))^²(4π10^(-7) H/m)(18.6 x 10^(-12) F/m) - β²) We'll simplify the above equation: γ = √((2π(3 x 10^9))^²(4π10^(-7) H/m)(18.6 x 10^(-12) F/m)) Now we can solve for γ: γ ≈ 1064.5 Np/m
04

Calculate the maximum allowable plate separation

Now that we have found the propagation constant γ, we can determine the maximum allowable plate separation, d̄, using the formula for impedance Z and associated equations: Z = R + jX = (ωμ/γ) + j(ωε/γ) For the TEM mode, the imaginary part (jX) should be zero, and thus the impedance Z becomes: Z = ωμ/γ From the equations of capacitance and inductance per unit length to the distance d between the parallel plates: L = μ/d C = εd Where L is the inductance per unit length and C is the capacitance per unit length. We can write the impedance equation as: Z = ωL/ωC = L/C Now, we can substitute L and C: Z = (μ/d)/(εd) To find the maximum allowable plate separation d̄, when the imaginary part (jX) is zero: d̄ = μ/Zε Now, we can substitute the values found for μ, ε, and Z: d̄ = (4π10^(-7) H/m)/(1064.5 Np/m)(18.6 x 10^(-12) F/m) Finally, we can calculate the maximum allowable plate separation d̄: d̄ ≈ 1.59 x 10^(-4) m So, the maximum allowable plate separation for this parallel-plate waveguide is approximately 1.59 x 10^(-4) meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

TEM Mode
The Transverse Electromagnetic (TEM) mode is a type of wave propagation where the electric and magnetic fields are both perpendicular to the direction of wave travel; this is also true for the direction of the wave's energy flow. The TEM mode is of special significance as it is the only mode that can propagate at any frequency, including DC (0 Hz). The existence of TEM mode is dependent on having two conductors, which, in the case of our exercise, are the parallel plates of the waveguide.

For the TEM mode to exist, there should not be any component of the electric or magnetic field in the direction of propagation, which makes the wave impedance purely real. This is why in step 4 of the solution, the imaginary part of the wave impedance is set to zero. In a parallel-plate waveguide, the TEM mode allows us to calculate the maximum allowable plate separation, ensuring that higher modes do not propagate and that the waveguide operates efficiently within the specified frequency range.
Dielectric Permittivity
Dielectric permittivity is a fundamental material property that quantifies how an electric field affects, and is affected by, a dielectric medium. It is a measure of the dielectric's ability to store electrical energy and is denoted by the symbol \( \epsilon \). In more technical terms, it is the ratio of the electric flux density produced in the material to the electric field strength. It affects the capacitance of a region of space and the speed at which electromagnetic waves travel through a material.

In our exercise, teflon is used as a dielectric material between the plates with a relative permittivity \( \epsilon_r' = 2.1 \). This value is crucial in determining the waveguide's characteristics. We use the product of the dielectric permittivity of free space \( \epsilon_0 \) and the relative permittivity \( \epsilon_r' \) to find the actual permittivity \( \epsilon \) of teflon, which is used in calculating the propagation constant and, subsequently, the maximum plate separation.
Propagation Constant
The propagation constant \( \gamma \) is a complex quantity that characterizes the change in amplitude and phase of an electromagnetic wave as it propagates through a medium. It is composed of two parts: the attenuation constant \( \alpha \) and the phase constant \( \beta \), with \( \gamma = \alpha + j\beta \). The attenuation constant represents the exponential decay of the wave's amplitude, while the phase constant corresponds to the phase shift per unit length.

In the context of the exercise, the propagation constant plays a pivotal role. Since we're dealing with the TEM mode, the cutoff frequency is zero, so the propagation constant can be calculated using only the maximum frequency of 3 GHz. It relates the waveguide's physical parameters, such as the permittivity \( \epsilon \) and permeability \( \mu \) of the material, to the frequency of operation. The real part of the propagation constant should result in a value that guarantees the desired impedance for the TEM mode, which ultimately allows us to determine the maximum allowable separation between the plates to solely support the TEM mode without any attenuation.

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Most popular questions from this chapter

In a symmetric slab waveguide, \(n_{1}=1.50, n_{2}=1.45\), and \(d=10 \mu \mathrm{m}\). (a) What is the phase velocity of the \(m=1 \mathrm{TE}\) or \(\mathrm{TM}\) mode at cutoff? (b) How will your part ( \(a\) ) result change for higher-order modes (if at all)?

Show that the group dispersion parameter, \(d^{2} \beta / d \omega^{2}\), for a given mode in a parallel-plate or rectangular waveguide is given by $$ \frac{d^{2} \beta}{d \omega^{2}}=-\frac{n}{\omega c}\left(\frac{\omega_{c}}{\omega}\right)^{2}\left[1-\left(\frac{\omega_{c}}{\omega}\right)^{2}\right]^{-3 / 2} $$ where \(\omega_{c}\) is the radian cutoff frequency for the mode in question [note that the first derivative form was already found, resulting in Eq. (57)].

A step index optical fiber is known to be single mode at wavelengths \(\lambda>1.2 \mu \mathrm{m}\). Another fiber is to be fabricated from the same materials, but it is to be single mode at wavelengths \(\lambda>0.63 \mu \mathrm{m} .\) By what percentage must the core radius of the new fiber differ from the old one, and should it be larger or smaller?

The cutoff frequency of the \(m=1 \mathrm{TE}\) and TM modes in an air-filled parallel-plate guide is known to be \(f_{c 1}=7.5 \mathrm{GHz}\). The guide is used at wavelength \(\lambda=1.5 \mathrm{~cm}\). Find the group velocity of the \(m=2 \mathrm{TE}\) and \(\mathrm{TM}\) modes.

Two microstrip lines are fabricated end-to-end on a \(2-\mathrm{mm}\) -thick wafer of lithium niobate \(\left(\epsilon_{r}^{\prime}=4.8\right)\). Line 1 is of \(4 \mathrm{~mm}\) width; line 2 (unfortunately) has been fabricated with a \(5 \mathrm{~mm}\) width. Determine the power loss in \(\mathrm{dB}\) for waves transmitted through the junction.

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