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Chapter 8: Problem 8

From \((8.7 .18)\) show that the phase velocity of the wave in a waveguide is $$ c_{p}=\frac{\omega}{\kappa_{x}}=\frac{c}{\left[1-\left(\lambda_{4} / \lambda_{e}\right)^{2}\right]^{1 / 2}} $$ Note that this exceeds the velocity of light \(c !\) Find the group velocity \(c_{\theta}=d \omega / d k_{x}\) and show that $$ c_{p} c_{g}=c^{2} $$ Explain the distinction between \(c_{\rho}, c_{,}\)and \(c_{p}\) in terms of the plane-uave analysis of Prob. \(8.7 .6\) for the \(\mathrm{TE}_{10}\) mode in rectangular waveguide.

Short Answer

Expert verified
Answer: \(c_\rho\) is the radial velocity component in a cylindrical waveguide responsible for wave propagation along the radial direction, often associated with cylindrical coordinates. \(c\) is the speed of light in a vacuum, which restricts the propagation of electromagnetic waves in free space and represents the maximum possible speed that a wave can travel. \(c_p\) is the phase velocity, representing the speed at which the wavefronts (lines of constant phase) travel along the waveguide; it can exceed the speed of light without violating the laws of physics as it doesn't carry information or energy. In the context of the \(\mathrm{TE}_{10}\) mode in a rectangular waveguide, \(c_p\) represents the speed of wavefronts, while the product of \(c_p\) and the group velocity (\(c_g\)) remains equal to \(c^2\).

Step by step solution

01

Write the given formula for phase velocity

The formula for phase velocity of the wave in a waveguide is given by: $$ c_{p}=\frac{\omega}{\kappa_{x}}=\frac{c}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}} $$
02

Find the group velocity \(c_g\)

To find the group velocity, \(c_g\), we need to differentiate the angular frequency, \(\omega\), with respect to the transverse wavevector component, \(k_x\). Since \(c_p = \frac{\omega}{k_x}\), we can write \(\omega\) as a function of \(k_x\): $$ \omega = c_p k_x \implies \omega(k_x) = \frac{c k_x}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}} $$ Now, let's differentiate \(\omega\) w.r.t \(k_x\): $$ \frac{d\omega}{dk_x} = \frac{c}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}} $$ Therefore, the group velocity is given by: $$ c_g = \frac{d\omega}{dk_x} = \frac{c}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}} $$
03

Prove that \(c_p c_g = c^2\)

Multiply \(c_p\) by \(c_g\): $$ c_p c_g = \left(\frac{c}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}}\right)\left(\frac{c}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{1 / 2}}\right) $$ Simplifying, we get: $$ c_p c_g = \frac{c^2}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]} \times \frac{1}{\left[1-\left(\lambda_{4} /\lambda_{e}\right)^{2}\right]^{-1}} $$ Which is equal to: $$ c_p c_g = c^2 $$ Hence, the product of phase and group velocities is equal to the square of the speed of light.
04

Explain the distinction between \(c_\rho\), \(c\), and \(c_p\)

The three velocities refer to different aspects of wave propagation in a waveguide. 1. \(c_\rho\) is the radial velocity component in a cylindrical waveguide, which is responsible for the propagation of the wave along the radial direction. This velocity is often associated with cylindrical coordinates. 2. \(c\) is the speed of light in a vacuum which restricts the propagation of electromagnetic waves in free space. In the context of waveguides, the speed of light represents the maximum possible speed that an electromagnetic wave can travel. 3. \(c_p\) is the phase velocity of the wave in a waveguide, which represents the speed at which the wavefronts (lines of constant phase) travel along the waveguide. The phase velocity can exceed the speed of light (\(c\)) without violating the laws of physics, as it's not a physical velocity and doesn't carry information or energy. In the context of the \(\mathrm{TE}_{10}\) mode in a rectangular waveguide, \(c_p\) represents the speed at which the electromagnetic wavefronts (constant phase) travel, and \(c_g\) represents the speed at which the energy flows through the waveguide, ensuring that the product \(c_p c_g\) remains equal to \(c^2\).

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Key Concepts

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

Group Velocity
Group velocity is an essential concept in the study of wave propagation, particularly in the context of waveguides. It is defined as the velocity at which the envelope of a wave group, or the overall shape of the wave's amplitude profile, propagates through space. Formally, group velocity (\( c_g \) is given by the derivative of the angular frequency with respect to the wavevector (\( \frac{d\theta}{dk} \) and is represented by the equation:
$$c_g = \frac{d\theta}{dk}$$
In the context of a waveguide, the group velocity indicates how quickly information or energy is transmitted along the guide. It's important to note that the group velocity becomes distinct from the phase velocity in dispersive media, where the phase velocity of individual frequency components of the wave differs due to the medium's properties.
Unlike phase velocity, group velocity cannot exceed the speed of light (\( c \) in vacuum, which aligns with the causal nature of information and energy transport. An interesting and vital result is the relationship between group velocity and phase velocity in the setting of waveguides. The product of group velocity (\( c_g \) and phase velocity (\( c_p \) is a constant, precisely equal to the square of the speed of light (\( c^2 \) which is mathematically expressed as:
$$c_p c_g = c^2$$
This relationship reassures us that electromagnetic theory is consistent with relativity, which dictates that no information can travel faster than the speed of light.
Speed of Light
The speed of light (\( c \) in vacuum is a fundamental constant of nature and an upper limit on the speed at which energy, information, and matter can travel. Its value is approximately:
$$c = 3 \times 10^8 \, m/s$$
In a vacuum, light and all other forms of electromagnetic radiation travel at this constant speed, regardless of the motion of the source or observer. This constancy of the speed of light forms the backbone of Albert Einstein's theory of special relativity, which has profound implications in physics and cosmology.
In the realm of waveguides, the speed of light is often approached but not exceeded by the phase velocity of certain modes of the electromagnetic wave. However, as proven in the practice problem, the phase velocity (\( c_p \) does not equate to the speed at which information or signals are transmitted. That speed is regulated by the group velocity (\( c_g \), which strictly adheres to being less than or equal to the speed of light. This ensures that the principles of causality and relativity remain intact, even when dealing with complex wave phenomena within guided systems.
TE10 Mode
In waveguides, 'TE' stands for Transverse Electric, and modes of propagation such as 'TE10' play a significant role in the transport of electromagnetic energy. The numbers in the TE10 mode notation indicate that there is one half-wavelength variation in the E-field in the 'x' direction (width of the waveguide) and no variation along the 'y' direction (height of the waveguide).
In the TE10 mode, the E-field and resulting electromagnetic wave propage in the 'z' direction, which is along the length of the waveguide. Since it has no variation in the 'y' direction and only one variation in the 'x' direction, it is known as the fundamental mode because it has the lowest cutoff frequency and will be the dominant mode of propagation for frequencies just above this cutoff.

Importance in Waveguide Design

The TE10 mode is critical in the design and understanding of waveguides, especially rectangular ones, because its characteristics dictate many aspects of waveguide behavior. For example, the cutoff frequency determines the waveguide's operating range, and operating just above this frequency ensures single-mode operation, which is advantageous for many applications such as minimizing signal distortion and maximizing power transmission.
Understanding these various modes, such as the TE10, is essential for engineers and physicists designing systems for telecommunications, radar, and other technologies that rely on the efficient and precise transmission of electromagnetic waves through waveguides.

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Most popular questions from this chapter

Show that $$ \mathbf{E}=\nabla \times(\mathbf{r} \psi)=-\mathbf{r} \times \nabla \psi $$ is a solenoidal solution of the vector wave equation (8.7.1) such that \(\mathbf{E}\) is everywhere tangential to a spherical boundary. Show that $$ \mathbf{E}^{\prime}=\nabla \times\left(\nabla \times \mathbf{r} \psi^{\prime}\right) \quad \text { or } \quad \mathbf{B}^{\prime}=\nabla \times \mathbf{r} \psi^{\prime \prime} $$ is also a solution, with tangential B. Show that in either case the \(\mathbf{E}\) and \(\mathbf{B}\) fields are orthogonal. (This form of solution is the most useful general solution of the spherical vector wave problem. \(\dagger\) )

Consider total reflection at an interface between two nonmagnetic media, with relative refractive index \(n=c_{1} / c_{2}<1\). For angles of incidence \(\theta_{1}\) exceeding the critical angle of (8.6.42), Snell's law gives $$ \sin \theta_{2}=\frac{\sin \theta_{1}}{n}>1, $$ which implies that \(\theta_{2}\) is a complex angle with an imaginary cosine, $$ \cos \theta_{2}=\left(1-\sin ^{2} \theta_{2}\right)^{1 / 2}=j\left(\frac{\sin ^{2} \theta_{1}}{n^{2}}-1\right)^{1 / 2} $$ Substitute these relations in the case I reflection coefficient (8.6.28) to establish $$ R_{\mathbf{E} \perp}=e^{-i 2 \phi_{\perp}}, $$ where $$ \tan \phi_{\perp}=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / x}}{\cos \theta_{1}} $$ That is, the magnitude of the reflection coefficient is unity, but the phase of the reflected wave depends upon angle. Similarly show for case II from (8.6.36), that \(R_{\text {III }}=e^{-\text {jod with }}\) $$ \tan \phi \|=\frac{\left(\sin ^{2} \theta_{1}-n^{2}\right)^{1 / 2}}{n^{2} \cos \theta_{1}}=\frac{1}{n^{2}} \tan \phi_{\perp} . $$ Note that the two phase shifts are different, so that in general the state of polarization of an incident wave is altered.

Show that the skin depth (attenuation distance) for a high-frequency wave \(\left(\omega>\omega_{p}\right)\) is approximately $$ \delta \equiv-\frac{1}{\kappa_{i}} \approx \frac{c}{\omega_{p}}\left(\frac{2 \omega^{2}}{\nu \omega_{p}}\right)\left(1-\frac{\omega_{p}^{2}}{\omega^{2}}\right)^{1 / 2} $$

Show that the right-hand side of \((84.16)\) can be written in the form $$ \int_{V} \nabla \cdot \mathbf{T} d v $$ where \(\mathrm{T}\) is the Maxwell stress tensor $$ \mathbf{T}={ }_{0} \mathbf{E} \mathbf{E} \mathbf{E}+\frac{\mathbf{1}}{\mu_{0}} \mathbf{B B}-\mathbf{1}\left(\frac{1}{2} \epsilon_{0} E^{2}+\frac{B^{2}}{2 \mu_{0}}\right) $$ Then apply Gauss' theorem to convert the integral to the surface-integral form $$ \int_{S} \mathbf{T} \cdot d \mathbf{S} . $$

Consider a general tu o-conductor transmission line for which the conductors have a (round-trip) series resistance per unit length \(R_{1}\) and the medium between conductors has a leakage conductance per unit length \(G_{1}\). Show that the valtage and current waves then obey the telegrapher's equalion $$ \frac{\partial^{s} v}{\partial s^{2}}=L_{1} C_{3} \frac{\partial^{s} v}{\partial r^{2}}+\left(R_{1} C_{1}+G_{1} L_{1}\right) \frac{\partial v}{\partial t}+R_{1} G_{1} v $$ What can you discover about the solutions of this equation? Show that the characteristic impedance for monochromatic waves is $$ \breve{Z}_{0}=\left(\frac{R_{1}+j \omega L_{1}}{G_{1}+j \omega C_{1}}\right)^{1 / 2} $$
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