55 Waveguides II
In the previous lecture, we established that there were three different possible modes for traveling waves in a hollow cylindrical (not in general circular) conducting waveguide, following (more or less) the presentation in Griffiths. Here we’ll investigate further, more closely following Jackson.
1. We repeat the equations from the previous lecture, but now with dielectric filling, so replacing with
2. For the given frequency we label the free space wave number
TM (transverse magnetic) solutions: zero longitudinal (parallel to the guide) magnetic field.
That is, everywhere in the waveguide, at the conducting surface (as always).
We found for TM waves, the longitudinal electric field satisfies
and the transverse electric field magnetic field
TE (transverse electric) solutions: everywhere in the waveguide, at the surface.
the longitudinal magnetic field satisfies
and the transverse fields are
From these equations, and using the transverse magnetic and electric fields satisfy
(for ) where the wave impedance, is given by (remember we’ve switched from to )
Recall that (lecture 46) is the impedance for a free space plane wave. (See the zigzag picture given below for some insight into the other term.)
The impedance is important when waveguides are put together, or waveguides are connected to sources or sinks: the impedances need to match, otherwise there will be reflection of the wave and consequent inefficiency.
The units are ohms, typical coax cable is around 75 ohms for TV signals.
Solving for the Modes: TEM
Let's first look for TEM modes. Recall now our earlier discussion of the coaxial cable: that was clearly a TEM mode. To get more quantitative, since for a TEM mode, the now (only two-dimensional!) electric field satisfies so we can write
with the two-dimensional potential satisfying
For the usual coaxial cable, we’d take polar coordinates. In fact, this is a familiar electrostatics problem, we know the surfaces are perfect conductors, the solution for the coaxial case is a radial electric field, the potential goes as the field as The magnetic field is in circles, of strength and of course both have the factor
But what about TEM modes in just a hollow waveguide? This same analysis shows that the transverse electric field is again like an electrostatic field, but this time with a potential constant over the whole boundarythe unique possible field is identically zero: so there cannot be a TEM mode in a single hollow conducting pipe! Essentially, TEM modes need two surfaces, and the transverse electric field look like the electrostatic field from opposite charges on the two (conducting) surfaces. For example, for two parallel cylinders (or wires) see the diagrams in lecture 15.
Notation! We’ll follow Jackson closely in this section, so we replace both and with
For TM waves, we have identically zero and nonzero
which we write as
cannot be negative since we need wave-type solutions on the transverse plane, to satisfy at all boundaries: negative would give exponentials. Given the boundary condition, the differential equation will have a set of eigenvalues and for each one there will be waves having -direction wavenumber
Notice this means there is a cutoff (minimum) frequency for each family,
above which the -direction wavenumber will be
At high frequencies, the waves move close to the speed of light in the dielectric.
Near the threshold the group velocity
is very slow, going to zero. (On the other hand, the phase velocity, , goes to infinity at threshold. From note that )
*Optional Aside: The Zigzag Picture for a Rectangular Pipe
To reduce clutter, we’ll just take vacuum in this section.
We just mention this picture (first suggested in 1903) as a slightly different way of imagining the wave going down the guide.
The fields have ordinary traveling wave behavior along the tube, but standing wave behavior in the perpendicular directions, to satisfy the boundary conditions.
For a particular mode, a way to visualize this is to imagine a wave aimed into the tube at an angle so that it bounces up and down in a zigzag path, reflected in turn by the top and bottom of a rectangular waveguide, and it will bounce off the sides too, we won’t try to draw that, the math is straightforward, as you’ll see. We could also add in a wave aimed in at the same angle downwards, as shown here. (The relative phase of the two waves determines if the mode is TE or TM.)
The wave will interfere with itself destructively unless the parallel segments of the wave are in phase with each other. This means the angle can only have certain values, as discussed below.
These bouncing waves are ordinary light waves, with perpendicular to each other and to as usual.
Taking to be at an angle to the direction, the in is given by For the lowest mode, we need the wavelengths in the perpendicular directions to be twice the tube dimensions, that is, to give simple standing waves in the perpendicular directions, eliminating destructive interference.
Now the zigzagging wave is just light in a vacuum (we’re assuming no dielectric in the waveguide), so and so generalizing in the obvious way to all standing waves in the perpendicular directions,
for positive integers
Rectangular Waveguide: Details
For TE waves, there is no electric field pointing down the pipe, but a nonzero ,
and, now following Jackson, we write with
and boundary conditions:
The solutions are (check thesenote the origin is at bottom left in the diagram)
Then from we have
For the lowest frequency mode is
and the TE1,0 fields are (from Jackson):
Exercise: Check that these follow from
For a given frequency of wave sent down the pipe, the -direction wave number will be
is the cutoff frequency: if the incoming the wave number is imaginary, meaning the wave exponentially decays in the pipe. (What about the Poynting vector?)
Exercise: try sketching the fields in one wavelength for this 1,0 mode, taking . This is in fact the mode most widely used commercially: for there is no TM mode until the smaller of twice cutoff or times cutoff, so no mixing problems. For the rectangular pipe, as we are about to show, in general there are TE and TM modes at the same frequency.
For the TM modes, has to go to zero at the walls, so , and the lowest mode will be 1,1, a higher cutoff frequency than the lowest TE mode, but for a general TM mode the value of and hence the frequency is the same as for the corresponding TE. (This is in contrast to, for example, a waveguide with a circular cross-section, or indeed almost any cross-section other than rectangular.)
Exercise: Examine carefully this representation of field lines for two modes in a rectangular waveguide, and answer the following:
(a) Identify the modes allegedly represented.
(b) There are some definite errors here! Re-sketch the diagrams correctly.
(c) Sketch the charge density and current density in the walls for this field configuration.
I found this online years ago, but it has been taken down. I suspect the authors are not anxious to have their work attributed.
Energy loss and Attenuation
For real walls, as opposed to perfect conductors, the wave penetrated into the walls, where Joule heating takes place, and energy is drained from the wave. We've found the formula for loss per unit area of wall surface in terms of the parallel magnetic field at the wall. We also know how to find the power being delivered down the guide by the wave. The rate of power loss is proportional to the square of the field, as is the power delivered, so the power going down the guide will decrease exponentially, in Jackson's notation
so the attenuation coefficient (for mode ) is
where the power drain is found by integrating around the circumference to find the loss rate per unit length of pipe:
This is straightforward given the waveguide shape and the mode. Consult Jackson for details of the calculation. He ends with an estimate, we'll just look at it and interpret the terms. The attenuation coefficient is
For ordinary waveguide shapes, it turns out that are numbers of order unity, and for TM (it comes from the component of magnetic field in the -direction for TE waves). The term encodes the relative surface area to volume, the "geometric factor". The crucial physics is that at high frequencies the is the dominating factor, it comes directly from the Joule heating, so energy loss increases with frequency in this regime. But it also increases going to low frequencies (meaning just above cutoff) : the blows upthat’s the group velocity going to zero, so little power is being delivered down the pipe, but the Joule heating is still happening. Evidently the attenuation has a minimum value somewhere.
For TM modes the attenuation has a simple frequency dependence, and the minimum loss rate is at
Exercise: Check the last statement.
(Jackson 8.6: Perturbation of Boundary Conditions: we'll skip this).