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Using the relation \(\langle S\rangle=\frac{1}{2} \operatorname{Re}\left\\{\mathbf{E}_{s} \times \mathbf{H}_{s}^{*}\right\\}\) and Eqs. (106) through (108), show that the average power density in the \(\mathrm{TE}_{10}\) mode in a rectangular waveguide is given by $$ \langle S\rangle=\frac{\beta_{10}}{2 \omega \mu} E_{0}^{2} \sin ^{2}\left(\kappa_{10} x\right) \mathbf{a}_{z} \mathrm{~W} / \mathrm{m}^{2} $$

Short Answer

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Based on the step-by-step solution provided, create a short answer question and answer below: Question: Derive the expression for the average power density 鉄潙嗏煩 in the TE10 mode in a rectangular waveguide using the provided field components and relations. Answer: To calculate the average power density 鉄潙嗏煩 in the TE10 mode, follow these steps: 1. Write down the expression for 饾惛s. In the TE10 mode, 饾惛s is given by -jE鈧sin(魏鈧佲個x)e^(-j尾鈧佲個z)饾悮岬. 2. Write down the expression for 饾惢s*. The complex conjugate of 饾惢s in the TE10 mode is given by -j(魏鈧佲個/蠅渭)E鈧cos(魏鈧佲個x)e^(j尾鈧佲個z)饾悮鈧 + (尾鈧佲個/蠅渭)E鈧sin(魏鈧佲個x)e^(j尾鈧佲個z)饾悮饾懅. 3. Compute the cross product 饾惛s 脳 饾惢s*: The cross product simplifies to (尾鈧佲個/蠅渭)E鈧虏sin虏(魏鈧佲個x)饾悮饾懅. 4. Calculate the average power density 鉄潙嗏煩 using the provided relation: The average power density is given by 鉄潙嗏煩 = (尾鈧佲個/2蠅渭)E鈧虏sin虏(魏鈧佲個x)饾悮饾懅 W/m虏.

Step by step solution

01

Write down the expression for E饾憼#

Using Eqs. (106) and (107), we have the electric field components for TE10 mode as follows: $$ \mathbf{E}_{s}=-j E_{0} \sin (\kappa_{10} x) e^{-j \beta_{10} z} \mathbf{a}_{y} $$ Here, \(E_0\) is the electric field amplitude, \(\kappa_{10}\) is the wave number in the x direction, and \(\beta_{10}\) is the propagation constant in the z direction.
02

Write down the expression for H饾憼*#

Now, using Eq. (108), we have the following magnetic field components for TE10 mode: $$ \mathbf{H}_{s}=\frac{j \kappa_{10}}{\omega \mu} E_{0} \cos (\kappa_{10} x) e^{-j \beta_{10} z} \mathbf{a}_{x} - \frac{\beta_{10}}{\omega \mu} E_{0} \sin (\kappa_{10} x) e^{-j \beta_{10} z} \mathbf{a}_{z} $$ Taking the complex conjugate of H饾憼 gives H饾憼*: $$ \mathbf{H}_{s}^*=-\frac{j \kappa_{10}}{\omega \mu} E_{0} \cos (\kappa_{10} x) e^{j \beta_{10} z} \mathbf{a}_{x} + \frac{\beta_{10}}{\omega \mu} E_{0} \sin (\kappa_{10} x) e^{j \beta_{10} z} \mathbf{a}_{z} $$
03

Compute the cross product E饾憼 脳 H饾憼*#

Now, we need to compute the cross product: $$ \mathbf{E}_{s} \times \mathbf{H}_{s}^*=-j E_{0} \sin (\kappa_{10} x) e^{-j \beta_{10} z} \mathbf{a}_{y} \times \left(-\frac{j \kappa_{10}}{\omega \mu} E_{0} \cos (\kappa_{10} x) e^{j \beta_{10} z} \mathbf{a}_{x} + \frac{\beta_{10}}{\omega \mu} E_{0} \sin (\kappa_{10} x) e^{j \beta_{10} z} \mathbf{a}_{z}\right) $$ The cross product simplifies to: $$ \mathbf{E}_{s} \times \mathbf{H}_{s}^* = \frac{\beta_{10}}{\omega \mu} E_{0}^2 \sin^2(\kappa_{10} x) \mathbf{a}_z $$
04

Calculate the average power density 饾憜 using the provided relation #

Now, we can use the provided relation to calculate the average power density: $$ \langle S\rangle=\frac{1}{2} \operatorname{Re}\left\{\mathbf{E}_{s} \times \mathbf{H}_{s}^{*}\right\} = \frac{1}{2} \operatorname{Re} \left\{\frac{\beta_{10}}{\omega \mu} E_{0}^2 \sin^2(\kappa_{10} x) \mathbf{a}_z\right\} $$ Since there is no imaginary part, the real part is equal to the whole expression, which leaves us with the desired expression for the average power density in the TE10 mode in a rectangular waveguide: $$ \langle S\rangle=\frac{\beta_{10}}{2\omega \mu} E_{0}^2 \sin^2(\kappa_{10} x) \mathbf{a}_z \mathrm{~W} / \mathrm{m}^{2} $$

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Key Concepts

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

TE10 Mode
When studying the concept of waveguides, it's crucial to understand the various modes of electromagnetic wave propagation within them. The TE10 mode is the fundamental transverse electric mode in a rectangular waveguide. This mode is characterized by having no electric field component in the direction of wave propagation, known as the z-direction in a standard coordinate system.

The 'TE' signifies 'Transverse Electric,' indicating that the electric field lines are purely transverse to the direction of wave propagation, with no longitudinal electric field component. The '10' denotes that there is one half-wave variation of the electric field in the 'b' dimension (the wider side of the waveguide) and zero half-wave variations in the 'a' dimension (the narrow side).

The TE10 mode is significant as it has the lowest cutoff frequency among all the modes that can propagate in a rectangular waveguide. The frequency must be greater than this cutoff frequency for wave propagation to occur. This unique feature of TE10 makes it highly efficient for microwave frequency ranges, leading to its widespread use in various applications such as satellite communications and radar systems.
Electromagnetic Wave Propagation
Electromagnetic wave propagation within a waveguide involves complex interactions between electric and magnetic fields. In a rectangular waveguide operating in the TE10 mode, the wave propagates in the z-direction, and the electric and magnetic fields are confined within the waveguide walls.

As a wave travels, its shape and intensity can be described mathematically by field components. These components depend on the waveguide's physical dimensions and the frequency of the electromagnetic wave. The propagation constant \(\beta_{10}\) identifies how the phase of the wave changes along the waveguide, while the wave number \(\kappa_{10}\) describes the variation across the waveguide's width.

Understanding the fundamentals of wave propagation aids in determining how waves behave under different circumstances and the implications for power distribution and signal strength. Moreover, this knowledge is essential for designing waveguides that are optimal for certain frequencies and applications, ensuring maximum efficiency and minimal signal loss.
Poynting Vector Calculation
The Poynting vector, represented by \(\mathbf{S}\), is a vital concept in electromagnetics that provides the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field.

To calculate the average power density, which is the average energy flow per unit area, we use the Poynting vector. The equation \(\langle S\rangle=\frac{1}{2}\operatorname{Re}\{\mathbf{E}_{s} \times \mathbf{H}_{s}^{*}\}\) is critical for this calculation. In this context, \(\mathbf{E}_{s}\) refers to the electric field, and \(\mathbf{H}_{s}^{*}\) is the complex conjugate of the magnetic field.

To obtain the average power density in the TE10 mode of a rectangular waveguide, one takes the real part of the cross product of these two field components. This real part calculation assumes time-harmonic fields and results in the average power density over a wave cycle. The Poynting vector highlights an essential aspect of electromagnetic wave propagation, indicating how much power is actually being transmitted through a given area, and is a fundamental tool for engineers and scientists working with waveguides and other electromagnetic systems.

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Most popular questions from this chapter

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.

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?

Pertinent dimensions for the transmission line shown in Figure \(13.2\) are \(b=\) \(3 \mathrm{~mm}\) and \(d=0.2 \mathrm{~mm}\). The conductors and the dielectric are nonmagnetic. (a) If the characteristic impedance of the line is \(15 \Omega\), find \(\epsilon_{r}^{\prime}\). Assume a low-loss dielectric. ( \(b\) ) Assume copper conductors and operation at \(2 \times 10^{8}\) \(\mathrm{rad} / \mathrm{s}\). If \(R C=G L\), determine the loss tangent of the dielectric.

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 \(\sigma /\left(\omega \epsilon^{\prime}\right)=4.0 \times 10^{-4}\).

A rectangular waveguide has dimensions \(a=6 \mathrm{~cm}\) and \(b=4 \mathrm{~cm} .(a)\) Over what range of frequencies will the guide operate single mode? \((b)\) Over what frequency range will the guide support both \(\mathrm{TE}_{10}\) and \(\mathrm{TE}_{01}\) modes and no others?

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