# Power flow in a TE mode for a rectangular waveguide

• PhDeezNutz
In summary, the magnetic field is greater in magnitude than the electric field within the skin depth of the guide walls.
PhDeezNutz
Homework Statement
Derive an integral expression for power flow in a TE rectangular guide

The correct answer is supposed to be

$$P_{flow} = \frac{1}{2} \frac{\mu \omega k_z}{\gamma^2} \int H_z H_z^* \,dA$$
Relevant Equations
The transverse directions are the x and y directions whereas the longitudinal direction is z

Here are some general formulas, capital Z is impedance

$$\frac{dP_{flow}}{dA} = \frac{1}{2} Re \{\vec{E} \times \vec{H^*} \} \cdot \hat{z}$$

$$\vec{H_t^*} = \pm \frac{1}{Z} \hat{z} \times \vec{E_t^*} \Rightarrow \vec{E_t^*} = \mp Z \left[ \hat{z} \times \vec{H_t^*} \right]$$

Specifically for TE modes

$$\vec{H_t} = \pm \frac{i k_z}{\gamma^2} \nabla_t H_z$$

$$Z_{TE} = \frac{\mu \omega}{k_z}$$

Just for completeness I will mention that TE modes are such that

1) $E_z = 0$ everywhere inside

2) $\frac{\partial B_z}{\partial n} = 0$ at the walls of the guide (x=0,a)(y=0,b)
$$\frac{dP_{flow}}{dA} = \frac{1}{2} Re \{\vec{E} \times \vec{H^*} \} \cdot \hat{z}$$

since ##E_z = 0## everywhere ##\vec{E} = \vec{E_t}##$$\frac{dP_{flow}}{dA} = \frac{1}{2} Re \{\vec{E_t} \times \vec{H^*} \} \cdot \hat{z}$$

$$\frac{1}{2} Re \left\{\mp Z_{TE} \left[ \left(\hat{z} \times \vec{H_t} \right) \times \vec{H^*}\right] \cdot \hat{z} \right\}$$

Using a vector identity for triple cross products we get

$$\frac{Z_{TE}}{2} Re \left\{\left[ \left(\hat{z} \cdot \vec{H^*} \right)\vec{H_t} - \left( \vec{H_t} \cdot \vec{H^*}\right)\hat{z}\right] \cdot \hat{z} \right\}$$

the first term directly above cancels to zero because ##\vec{H_t} \cdot \hat{z} = 0## so we have

$$-\frac{Z_{TE}}{2} Re \left\{ \vec{H_t} \cdot \vec{H_t^*} \right\}$$

Using the formula for ##H_t##and ##Z_{TE}## we get

$$\frac{dP_{flow}}{dA} = \frac{\mu \omega k_z}{2 \gamma^4} Re \left\{ \left( \nabla_t H_z\right) \cdot \left( \nabla_t H_z\right)^*\right\}$$

Now to integrate over a cross section of the guide for total Power

I'm going to invoke this vector identity ##\left( \nabla_t H_z\right) \cdot \left( \nabla_t H_z \right)^* = \nabla_t \cdot \left(H_z \left( \nabla_t H_z\right)^* \right) - H_z \left( \nabla_t^2 H_z \right)^*##$$P = \int \frac{dP_{flow}}{dA} \, dA = \frac{\mu \omega k_z}{2 \gamma^4} \int Re\left\{ \nabla_t \cdot \left(H_z \left( \nabla_t H_z\right)^*\right) \right\} \, dA - \frac{\mu \omega k_z}{2 \gamma^4} \int Re\left\{ H_z \left( \nabla_t^2 H_z\right)\right\}\,dA$$

Using the wave equation for a guide ##\nabla_t^2 H_z = - \gamma^2 H_z## we have$$P = \int \frac{dP_{flow}}{dA} \, dA = \frac{\mu \omega k_z}{2 \gamma^4} \int Re\left\{ \nabla_t \cdot \left(H_z \left( \nabla_t H_z\right)^*\right) \right\} \, dA + \frac{\mu \omega k_z}{\gamma^2} \int H_zH_z^* \,dA$$

If only I could get the first term in the expression directly above to vanish I will have the right answer. Loosely speaking the integral of a derivative (of a function) throughout a region is equal to the integral of the original function around the boundary. ##H_z = 0## on the boundary so it stands to reason that the first integral vanishes. However that's a little bit too handwaving for me and I'd like to hammer out the details if possible.

Jackson invokes Green's identities and then arbitrarily reduces them to two dimensions...which doesn't sit well with me.

As always, any help, guidance, insight etc is appreciated.

I think I got it

I mis-spoke towards the end of my last post. It's not the z-component of ##H_z## that vanishes at the boundary but rather it's normal derivative (i.e. the gradient?) (I presume the walls of the waveguide are an equipotential?)

Apparently the divergence theorem in 2D (for two dimensional vector fields and two dimensional del operator) is

$$\int \nabla \cdot \vec{F} \, dA = \oint \vec{F} \cdot \, d\ell$$

In this case ##nabla_t## is a two dimensional del operator and subsequently ##(\nabla_t H_z)*## and ##H_z (\nabla_t H_z)*## are two dimensional vector fields

so

$$\frac{\mu \omega k_z}{2 \gamma^4} \int Re\left\{ \nabla_t \cdot \left(H_z \left( \nabla_t H_z\right)^*\right) \right\} \, dA = \int H_z \left(\nabla_t H_z \right)^* \cdot \, d \ell$$

$$\left(\nabla_t H_z \right)_{boundary} = 0$$ on the boundary so the integral that I needed to vanish indeed vanishes.

Does this argument hold up?

This doesn't warrant a new thread so I'll just ask it here. I'm considering the fields within the skin depth ##\delta## of the guide walls. Supposedly the magnetic field is greater in magnitude than the electric field.

my expression relating the fields within the guide walls (i.e. the conductor) is

$$\vec{E_c} = \left(\frac{\mu \omega}{2\sigma}\right)^{\frac{1}{2}} \left( 1 - i \right) \left(\hat{n} \times \vec{H_{\parallel}} \right) e^{-\frac{n}{\delta}}e^{\frac{n}{\delta}} = \left(\frac{\mu \omega}{2\sigma}\right)^{\frac{1}{2}} \left( 1 - i \right) \vec{H_c}$$

Where ##\sigma## is the conductivity. ##\omega## is the angular frequency. ##\mu## of course is the magnetic permeability.

I can't for the life of me justify why ##\left(\frac{\mu \omega}{2\sigma}\right)^{\frac{1}{2}} \lt 1## categorically. On one hand ##\sigma## is really big so that should minimize the term. But for what reason is ##2 \sigma \gt \mu \omega##?

## 1. What is a TE mode in a rectangular waveguide?

A TE (transverse electric) mode in a rectangular waveguide is an electromagnetic wave that propagates through the waveguide with its electric field perpendicular to the direction of propagation. This mode is characterized by a zero electric field component in the direction of the width of the waveguide.

## 2. How does power flow in a TE mode for a rectangular waveguide?

The power in a TE mode for a rectangular waveguide flows through the waveguide in the form of an electromagnetic field. As the wave propagates through the waveguide, the electric and magnetic fields interact to transfer energy from one end of the waveguide to the other.

## 3. What factors affect the power flow in a TE mode for a rectangular waveguide?

The power flow in a TE mode for a rectangular waveguide is affected by several factors, including the dimensions of the waveguide, the frequency of the wave, and the material properties of the waveguide. The power flow is also influenced by any obstacles or imperfections in the waveguide that may cause reflections or losses.

## 4. How is the power flow in a TE mode for a rectangular waveguide calculated?

The power flow in a TE mode for a rectangular waveguide can be calculated using Maxwell's equations and the boundary conditions at the walls of the waveguide. This involves solving for the electric and magnetic fields within the waveguide and then calculating the power using the Poynting vector, which describes the direction and magnitude of energy flow in an electromagnetic field.

## 5. What are some practical applications of understanding power flow in a TE mode for a rectangular waveguide?

Understanding power flow in a TE mode for a rectangular waveguide is essential for the design and operation of many microwave and optical devices, such as antennas, filters, and couplers. This knowledge is also important for optimizing the performance of communication systems, radar systems, and other technologies that use waveguides to transmit and receive electromagnetic signals.

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