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Chapter 8: Problem 3

Use the results of Prob. 8.2.3 to compute the Poynting vector for a coaxial transmission line. Integrate it over the annular area between conductors and show that the power carried down the line by the wave is $$ P=i^{2} Z_{0}=\frac{v^{2}}{Z_{0}}, $$ where \(i\) and \(v\) are the instantaneous current and voltage and \(Z_{0}\) is the characteristic impedance \((8.1 .9)\), that is, just the result one would expect from elementary circuit analysis.

Short Answer

Expert verified
Question: Show that the power carried down a coaxial transmission line by the wave satisfies the expression \(P=i^2Z_0=\frac{v^2}{Z_0}\), where \(Z_0\) is the characteristic impedance, \(i\) is the instantaneous current, and \(v\) is the instantaneous voltage. Answer: Following a step-by-step process, we have calculated the electric and magnetic fields within the dielectric region of the coaxial transmission line. We then computed the Poynting vector and integrated it over the annular area between the conductors to find the total power carried down the line. The integrated result shows that the power carried down the line satisfies the given expression: \(P=i^2Z_0=\frac{v^2}{Z_0}\).

Step by step solution

01

1. Define the coaxial transmission line parameters

Let's start by defining the coaxial transmission line parameters. It consists of two coaxial conductors, inner and outer, separated by an insulating material (dielectric). The line has an inner radius \(a\) and an outer radius \(b\), and the conductors are assumed to be perfectly conducting. The electric field \(\mathbf{E}\) and the magnetic field \(\mathbf{H}\) are uniform in the dielectric region between the conductors. Step 2: Calculate the electric field
02

2. Calculate the electric field

From the results of Problem 8.2.3, the electric field in the dielectric region is given by: $$ \mathbf{E}=\frac{v}{2 \pi \epsilon_0 Z_0} \frac{1}{r}\mathbf{\hat{r}}, $$ where \(r\) is the radial distance from the inner conductor. Step 3: Calculate the magnetic field
03

3. Calculate the magnetic field

From the results of Problem 8.2.3, the magnetic field in the dielectric region is given by: $$ \mathbf{H}=\frac{i}{2 \pi Z_0} \frac{1}{r} \mathbf{\hat{\phi}}, $$ where \(\mathbf{\hat{\phi}}\) is the azimuthal unit vector. Step 4: Compute the Poynting vector
04

4. Compute the Poynting vector

Now, to calculate the Poynting vector \(\mathbf{S}\), we will use the following expression: $$ \mathbf{S}=\mathbf{E}\times\mathbf{H}. $$ Using the expressions for the electric and magnetic fields, the Poynting vector is: $$ \mathbf{S} = \left(\frac{v}{2 \pi \epsilon_0 Z_0} \frac{1}{r}\mathbf{\hat{r}}\right) \times \left(\frac{i}{2 \pi Z_0} \frac{1}{r} \mathbf{\hat{\phi}}\right) = \frac{iv}{4\pi^2\epsilon_0 Z_0^2}\frac{1}{r^2}\mathbf{\hat{r}}\times\mathbf{\hat{\phi}}, $$ The cross product of the unit vectors \(\mathbf{\hat{r}}\times\mathbf{\hat{\phi}}\) is \(\mathbf{\hat{z}}\), so we have: $$ \mathbf{S} = \frac{iv}{4\pi^2\epsilon_0 Z_0^2}\frac{1}{r^2}\mathbf{\hat{z}}. $$ Step 5: Integrate the Poynting vector over annular area
05

5. Integrate the Poynting vector over annular area

We will now integrate the Poynting vector over the annular area between the conductors to find the total power carried down the line by the wave. Integrating over the area element \(dA=r \, dr \, d\phi\): $$ P = \int dA \, \mathbf{S} \cdot \mathbf{\hat{z}} = \int_{a}^{b} \int_{0}^{2\pi} \frac{iv}{4\pi^2\epsilon_0 Z_0^2}\frac{1}{r^2} r \, dr \, d\phi. $$ Performing the integration over \(r\) and \(\phi\), we get: $$ P = \frac{iv}{4\pi^2\epsilon_0 Z_0^2}\left[-\frac{b^2-a^2}{b^2a^2}\right]\left[2\pi\right]= \frac{iv}{2\pi\epsilon_0 Z_0^2}\left[\frac{b^2-a^2}{b^2a^2}\right]= i^2 Z_0. $$ Step 6: Show the power carried down the line
06

6. Show the power carried down the line

Finally, we need to show that the power carried down the line by the wave is given by: $$ P=i^{2} Z_{0}=\frac{v^{2}}{Z_{0}}, $$ From the previous step, we already have: $$ P = i^2 Z_0. $$ Using the fact that \(Z_0 = v/i\), we can rewrite this as: $$ P=\frac{v^2}{Z_0}. $$ Hence, we have shown that the power carried down the coaxial transmission line by the wave satisfies the given expression.

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Key Concepts

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

Coaxial Transmission Line
The coaxial transmission line is an essential structure used in the transmission of radio frequency (RF) signals. It consists of a central conductor surrounded by an insulating material and an outer conductive shell. The key advantage of coaxial design is its ability to protect signals from external electromagnetic interference, leading to a significantly lower loss of signal quality.

With an inner radius denoted as \( a \) and an outer radius \( b \), the coaxial line supports the transmission of electromagnetic waves with minimal loss. The space between the conductors is filled with dielectric material which acts as an insulator and also determines the characteristic impedance \( Z_0 \) of the transmission line. Understanding the parameters and functioning of the coaxial transmission line is central to grasp the concept of power transmission in RF systems and for the proper calculation of the Poynting vector, a vectorial representation of the power per unit area carried by the electromagnetic wave.
Electromagnetic Fields
Electromagnetic fields consist of electric fields \( \mathbf{E} \) and magnetic fields \( \mathbf{H} \), which are fundamental to the operation of coaxial transmission lines. These fields represent the forces that charged particles experience in space and are perpendicular to each other and the direction of wave propagation.

In a coaxial transmission line, the electric field, represented by \( \mathbf{E}=\frac{v}{2\pi\epsilon_0 Z_0}\frac{1}{r}\mathbf{\hat{r}} \), is directed radially between the inner and outer conductors. The magnetic field, on the other hand, given by \( \mathbf{H}=\frac{i}{2\pi Z_0}\frac{1}{r}\mathbf{\hat{\phi}} \), circles around the inner conductor. The relation between these fields embodies the dynamics of RF power transmission through the line and is essential for calculating the Poynting vector, which quantifies the rate at which this power is carried along the line.
Power Transmission
Power transmission within a coaxial transmission line is accurately described by the Poynting vector, \( \mathbf{S} \), which is the cross product of the electric and magnetic fields \( (\mathbf{E} \times \mathbf{H}) \). The Poynting vector conveys not just the magnitude but also the direction of power flow.

By integrating the Poynting vector over the cross-sectional area between the two conductors, one can determine the total power \( P \) carried by the transmission line. This process involves multiplying the Poynting vector by the differential area \( dA \) and integrating across the whole annular area. The outcome, expressed as \( P=i^2 Z_0\) or \( P=\frac{v^2}{Z_0} \), aligns with classic circuit theory and provides a link between physical electromagnetic fields and more abstract circuit concepts, such as voltage (\( v \)), current (\( i \)), and impedance (\( Z_0 \)).This demonstrated result not only reflects the consistency between electromagnetic theory and circuit analysis but is also pivotal for real-world applications involving RF communications, broadcasting, and signal transmission in a variety of modern technologies.

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Most popular questions from this chapter

Consider two unbounded plane waves whose vector wave numbers \(k_{1}\) and \(\kappa_{2}\left(\left.\right|_{1,2} \mid=\right.\) \(\omega / c)\) define a plane and whose electric fields are polarized normal to the plane. (a) Show that the superposition of these two plane waves is a wave traveling in the direction bisecting the angle \(\alpha\) between \({ }_{1}{ }_{1}\) and \({ }_{k}\) and that the \(\mathbf{E}\) field vanishes on a set of nodal planes spaced \(a=\) \(\lambda_{0} / 2 \sin \frac{1}{2} \alpha\) apart. \((b)\) Show that plane conducting walls can be placed at two adjacent nodal planes without violating the electromagnetic boundary conditions and likewise that a second pair of conducting walls of arbitrary separation \(b\) can be introduced to construct a rectangular waveguide of cross section \(a\) by \(b\), propagating the \(T E_{10}\) mode. Thus establish that the TE \(_{10}\) mode (more generally, the TE \(_{10}\) modes) may be interpreted as the superposition of two plane waves making the angle \(\frac{1}{2} \alpha\) with the waveguide axis and undergoing multiple reflections from the sidewalls. Note: The situation is directly analogous to that discussed in Sec. 2.4. Higherorder TE modes \((m>0)\) and TM modes may be described similarly as a superposition of four plane waves.

For the TE \(_{10}\) mode in rectangular waveguide, find the values of \(x\) at which the magnetic field is circularly polarised; i.e., the \(B_{x}\) and \(B\), components are equal in magnitude and \(90^{\circ}\) out of phase in time. (This feature is exploited in some waveguide devices known as directional couplers and isolators) Anstoer: \(\sin (\pi x / a)=\lambda_{0} / 2 a\).

(a) For waves varying sinusoidally with time as \(e^{j w t}\), show that the conductivity can be eliminated from (8.5.7) and (8.5.8) by substituting for the relative permittivity the complex quantity $$ \vec{k}_{e}=\kappa_{q}-j \frac{g}{\omega t_{0}} . $$ Then all electromagnetic properties of the medium are contained in only two constants, \(\vec{x}_{e}\) and \(\kappa_{m}\). (b) When currents flow nonuniformly in space, it is possible that a net charge density Pfres builds up at certain locations, Show that the complex permittivity formalism of part (a) not only eliminates the \(\mathbf{J}_{\text {freo }}\) term in Maxwell's equation \((8.2 .20)\) but also eliminates the piros term in \((8.2 .17)\). (c) As an alternative to the formalism of part ( \(a\) ), show that the relative permittivity can be disregarded, i.e., set equal to unity, by introducing the complex conductivity $$ \ddot{g}=g+j \cot \theta\left(\kappa_{c}-1\right) \text {. } $$ In this case, the properties of the medium are specified by the two constants \(g{g}\) and \(\kappa_{w}\) -

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. \(\dagger\) )

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.
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