Rectangular wave guide, TE01 mode

[Phrak]

Rectangular wave guide, TE01 mode

I'm looking for the equations of the electric and magnetic fields over the volume of the rectangular guide. I've managed what appear to be the correct equations for the electric field,

E_y = sin\left(\frac{2\pi z}{\lambda_z}-\omega t\right) cos\left(\frac{2\pi x}{\lambda_x}\right)

I'm placing the guide cross section in the x-y plane. The wave propagates in the z direction. The electric field is in y direction.

I don't see anything obvious that hints at how to solve for the magnetic field.

 

[Born2bwire]

Just use Faraday's Law. The curl of the electric field is proportional to the time derivative of the magnetic field. Since it's a time-harmonic solution the time derivative is just a constant factor of i\omega (or whatever you choose as your time convention).

 

[Phrak]

Just use Faraday's Law. The curl of the electric field is proportional to the time derivative of the magnetic field. Since it's a time-harmonic solution the time derivative is just a constant factor of i\omega (or whatever you choose as your time convention).

Yeah, thanks. It should have been obvious and I do need the family of solutions. It only involves a single integral over time...

 

[Meir Achuz]

You have to use exp[i(kz-wt)] to get the correct magnetic field form Maxwell's equations.

 

[Born2bwire]

You have to use exp[i(kz-wt)] to get the correct magnetic field form Maxwell's equations.

Yeah, that's the easiest thing to do. Of course he should be able to work the problem the same way entirely in the real domain. Not as easy as converting back to the complex time-harmonic form I'll admit.

 

[Phrak]

I should explain the details.

Imagine a set of waves hitting a long straight beach as some particular angle other than straight-on. Now another set hits the beach at the complimentary angle. Now make the wave sets continuous sine waves. At some points on the beach, extending perpendicular to the beach, the wave amplitude will be nodal. This corresponds to the walls of the guide--or at least this is my premise, to be proved one way or the other.

The static solutions for the waves are

E_{y1}= sin\left(\frac{2\pi z}{\lambda_z}+\frac{2\pi x}{\lambda_x}\right)
and
E_{y2}= sin\left(\frac{2\pi z}{\lambda_z}-\frac{2\pi x}{\lambda_x}\right)

Using one of the many trig identities, these sum to

E_y= 2sin\left(\frac{2\pi z}{\lambda_z}\right)cos\left(\frac{2\pi x}{\lambda_x}\right)

Notice that at x=+\lambda_x/4 and x=-\lambda_x/4 the amplitude is zero (from the cosine term) for all z, so should correspond to the wave guide walls for TE₀₁, if my intuition serves me.

Including the time dependence,

E_y= 2sin \left( \frac{2\pi z}{\lambda_z} - \omega t \right) cos\left( \frac{2\pi x}{\lambda_x} \right) \ .

The alleged walls are still nodal at lambda/4. So far, so good. Visually what this looks like is a two dimensional checkerboard array of sine wave bumps and pits traveling in the z direction.

btw, I've didn't try using exponentials in defining E_y1 and E_y2, frankly because I didn't trust that they won't produce errors like inclusion of unwanted product terms. But, with some curiosity I tried to get the above trig identity using exponentials, and came up empty.

After taking the cross products of E, using Faraday, so far I get

\partial_x E_y = -\partial_t B_z=\frac{-4\pi}{\lambda_x}sin\left(\frac{2\pi z}{\lambda_z}-\omega t\right)sin\left(\frac{2\pi x}{\lambda_x}\right)

\partial_z E_y = \partial_t B_x=\frac{4\pi}{\lambda_z}cos\left(\frac{2\pi z}{\lambda_z}-\omega t\right)cos\left(\frac{2\pi x}{\lambda_x}\right)

if I haven't made a sign error or worse(??)

 

[Born2bwire]

I thought I wrote it out but I guess I accidentally removed it. Well then, if it's only traveling in the positive z direction then,

E_y = \Re \left\{ 2 E_0 \sin\left( \frac{\pi x}{a} \right) e^{i(k_zz-\omega t)} \right\} = 2 E_0 \sin\left( \frac{\pi x}{a} \right) \cos\left(k_zz-\omega t \right)

Yeah, so a quick check then, if this is a PEC rectangular waveguide then your equations are incorrect. Because the walls lie along the x and y axes, correct? Then the tangential electric fields should be zero along the walls and so E_y should be zero for x=0 and x=a. So the cosine factor is incorrect.

But I can't connect this to your description of waves reflecting off of a planar surface. You have only a single plane for reflection, how is there any guidance of the wave or boundary conditions similar to a rectangular waveguide?

 

[Phrak]

I thought I wrote it out but I guess I accidentally removed it. Well then, if it's only traveling in the positive z direction then,

E_y = \Re \left\{ 2 E_0 \sin\left( \frac{\pi x}{a} \right) e^{i(k_zz-\omega t)} \right\} = 2 E_0 \sin\left( \frac{\pi x}{a} \right) \cos\left(k_zz-\omega t \right)

Yeah, so a quick check then, if this is a PEC rectangular waveguide then your equations are incorrect. Because the walls lie along the x and y axes, correct? Then the tangential electric fields should be zero along the walls and so E_y should be zero for x=0 and x=a. So the cosine factor is incorrect.

But I can't connect this to your description of waves reflecting off of a planar surface. You have only a single plane for reflection, how is there any guidance of the wave or boundary conditions similar to a rectangular waveguide?

Your origin is different, taking sine to cosine, so my solution appears at first glance to be OK. I'll look closer at yours in the morning. I have zero electric field strength at the walls. I'm taking the walls to be perpendicular to the x-z plane. The origin is half way between the two walls.

I could have done better with the waves-on-beach visual, it seems. The waves do not reflect off the beach. There are just two sets of waves hitting the beach (the x-y plane) at two angles, +theta and -theta.

 

[Born2bwire]

Your origin is different, taking sine to cosine, so my solution appears at first glance to be OK. I'll look closer at yours in the morning. I have zero electric field strength at the walls. I'm taking the walls to be perpendicular to the x-z plane. The origin is half way between the two walls.

I could have done better with the waves-on-beach visual, it seems. The waves do not reflect off the beach. There are just two sets of waves hitting the beach (the x-y plane) at two angles, +theta and -theta.

Ok then. Yeah, the sine of x term is behaving as the superposition of two traveling waves at are propagating in the +/- x direction. We can also include a traveling wave in the -z direction and this would split the cos(k_zz-\omega t) into something like sine(k_zz)cos(\omega t).

 

[Meir Achuz]

Phrak: You are doing it in the most complicated way.
Don't you have a textbook that shows how to do it using
the boundary conditions and Maxwell's equations?

 

[Meir Achuz]

For a TE mode in an AXB wave guide, the easiest thing to get first is
H_z=H_0\cos(\pi x/A)\cos(\pi y/B)exp[i(kz-\omega t)],
which just follows from the BC conditions on H.
The other components follow using Maxwell's equations.

 

[Phrak]

For a TE mode in an AXB wave guide, the easiest thing to get first is
H_z=H_0\cos(\pi x/A)\cos(\pi y/B)exp[i(kz-\omega t)],
which just follows from the BC conditions on H.
The other components follow using Maxwell's equations.

What boundary conditions have you imposed to require H_z take this form? I take it that z is the axis of the guide.

 

[Meir Achuz]

The boundary condition on H in a wave guide is that the normal derivative of its tangential component is zero. This is a little tricky to derive from the usual boundary condition that the normal component must be zero and Maxwell's curl H equation. It is easier to get the cos cos form by realizing that the functions of x and y can only be sin or cos. Since the tangential component of H doesn't vanish at the boundary, it must be cos.

 

[Phrak]

The boundary condition on H in a wave guide is that the normal derivative of its tangential component is zero. This is a little tricky to derive from the usual boundary condition that the normal component must be zero and Maxwell's curl H equation. It is easier to get the cos cos form by realizing that the functions of x and y can only be sin or cos. Since the tangential component of H doesn't vanish at the boundary, it must be cos.

This clarifies things greatly. So that if I'm looking at the curl of H it should be zero tangential to the surface. Or, it seems, constrained by dE_tangential/dt+J=0.

 

[Phrak]

I've managed to identify 5 boundary conditions pertaining to a conducting waveguide surface.

1) E_tangential=0, current must be finite for an ideally zero resistivity waveguide surface.
2) E_normaldA=dq/ε, electric field terminates on charge.

3) J_normal=0, current does not leave the surface.
4) J_tangential = (∇×B)_tangential - ∂E_tangential/∂t

5) ∇·J = -∂ρ/∂t, the charge continuity condition. Currents can't converging on some area element in the floor and ceiling of the waveguide without the right amount of charge accumulating.

Are any of these wrong, or did I miss any?

 

[Born2bwire]

There is a bit of redundancy here. First, since the tangential electric field is zero then you can just simplify the fourth boundary condition. In addition, we can make use of vector calculus via Stokes' Law and such to remove the derivatives. So instead you can say that:

\mathbf{E}_{1tan} = \mathbf{E}_{2tan} = 0

\mathbf{E}_{1n} = \frac{\rho_s}{\epsilon_1}

\mathbf{E}_{2n} = 0

\mathbf{H}_{1n} = \mathbf{H}_{2n} = 0

\hat{n} \times \mathbf{H}_{1tan} = \mathbf{J}_s

\mathbf{H}_{2tan} = 0

Basically all that is needed is that the tangential electric field is continuous (and thus zero) and that the normal H field is continuous (and also zero). From these two boundary conditions we can derive the full set of equations. If we have the boundary lying in the x-y plane, then we can fully describe the TE and TM modes by their tangential electric and magnetic components respectively. Then we use Faraday's Law to find the missing magnetic and electric field components.

We also can just find the H_z field for the TE mode and the E_z field for the TM mode and characterize the problem from those two fields alone (again since we can find the other fields trivially).

 

[Phrak]

I don't know what the subscripts 1 and 2 mean.

 

[Meir Achuz]

In a wave guide with perfectly conducting walls, another BC can be derived: ({\hat n}\cdot\nabla)H_{\rm tangential}=0
That is useful in finding H_z for TE.

 

[Born2bwire]

I don't know what the subscripts 1 and 2 mean.

1 is air, 2 is the PEC. Or like Meir Achuz states above there is another set of equivalent boundary conditions. We have a Neumann boundary condition for the normal H field (EDIT: I seem to recall that it should be a Neumann boundary conditon for the tangential field, not the normal. I'll have to check my references when I get back to the office. Either way Meir Achuz's BC is equivalent for PEC since the gradient operator will just pull out the wave vector and we can remove that since the RHS is zero). This can be found more explicitly by looking at the vector wave equation for a plane wave in a planar geometry. Of course it applies to any geometric boundary but using planar boundaries makes it easier to demonstrate. Like I said though, I removed the derivatives by making use of Stoke's Law and such like that. What you do is you take a volume integral of the curl of the field. You can remove the operator by changing from the volume to the surface integral. You then take the limit of the integrating volume to lie along the surface of the boundary. Here is a simple set of slides discussing this.

http://www.amanogawa.com/archive/docs/EM5.pdf

Amanogawa also has sets of slides discussing the rectangular waveguide and so forth (from the main page: Electromagnetic Waves -> Instructional Material). In Chapter 2 of Chew's "Waves and Fields in Inhomogeneous Media" is a derivation of the boundary conditions that results in Meier's BC. I'm sure that Balanis' text would include that too but I cannot recall to what detail his discussion is.

 

[Phrak]

There is a bit of redundancy here. First, since the tangential electric field is zero then you can just simplify the fourth boundary condition. In addition, we can make use of vector calculus via Stokes' Law and such to remove the derivatives. So instead you can say that:

\mathbf{E}_{1tan} = \mathbf{E}_{2tan} = 0

\mathbf{E}_{1n} = \frac{\rho_s}{\epsilon_1}

\mathbf{E}_{2n} = 0

\mathbf{H}_{1n} = \mathbf{H}_{2n} = 0

\hat{n} \times \mathbf{H}_{1tan} = \mathbf{J}_s

\mathbf{H}_{2tan} = 0

I'm assuming subscript 1 refers to the floor and ceiling of the waveguide and subscript 2 to the side walls when the electric field is vertically polarized. But I don't see is how \mathbf{H}_{1n} is constrained to zero.

 

[Phrak]

In a wave guide with perfectly conducting walls, another BC can be derived: ({\hat n}\cdot\nabla)H_{\rm tangential}=0
That is useful in finding H_z for TE.

Did you mean something different??. ({\hat n}\cdot\nabla) looks like a vector dotted into a vector--that is, a scalar--acting on H_tan.

 

[Born2bwire]

I'm assuming subscript 1 refers to the floor and ceiling of the waveguide and subscript 2 to the side walls when the electric field is vertically polarized. But I don't see is how \mathbf{H}_{1n} is constrained to zero.

No. Read my post above yours. I stated that 1 is the fields on the surface of the boundary in air, 2 is the fields on the surface of the boundary in the PEC wall of the waveguide. Take a look at the linked slides to see a derivation.

The general boundary conditions stipulate that the tangential E fields are continuous and the normal B fields are continuous. The normal E and tangential H fields are discontinous by the induced charge and currents. So, in a PEC the fields are zero which means that the tangential E and normal B (or H in this case by the fact that the permeability is non-zero) must both be zero on either side of the interface.

 

[Phrak]

No. Read my post above yours. I stated that 1 is the fields on the surface of the boundary in air, 2 is the fields on the surface of the boundary in the PEC wall of the waveguide. Take a look at the linked slides to see a derivation.

The general boundary conditions stipulate that the tangential E fields are continuous and the normal B fields are continuous. The normal E and tangential H fields are discontinous by the induced charge and currents. So, in a PEC the fields are zero which means that the tangential E and normal B (or H in this case by the fact that the permeability is non-zero) must both be zero on either side of the interface.

Thanks. I wasn't sure. I re-read your post and looked over the slides.

It seems that invoking PEC complicates things more than required. I didn't know what it mean. I'm going to consider a finite, and small wall thickness. I don't think the answer will deviate from a lossless waveguide--zero resistivity.

 

[Phrak]

Born2bwire. If you were to recommend one and only one text for waveguides, what would it be?