Power flow in a TE mode for a rectangular waveguide

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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.
 

Answers and Replies

  • #2
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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?
 
  • #3
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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##?
 

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