Approfondimenti sulle guide d'onda en

From "Physics, Waves, Music": a website about the physics of waves and sound, acoustics of musical instruments, musical scales, harmony and music.

AI-translated from Italian. Original page: Approfondimenti_sulle_guide_d'onda

Description of the ideal system

Imagine a pair of metal plates as in the figure.

Suppose we want to send electromagnetic radiation of wavelength λ into the space between the two plates.

To simplify the calculations, suppose the radiation is linearly polarised along the y direction.

Suppose the spacing between the plates is a, while in the y and z directions the system extends to infinity (in practice it is enough for the plates to be much larger than λ along y and z).

Recall that the magnitude k of the wave vector k is related to the wavelength of the incoming wave by

λ=2πk.

In general, that is, for a wave propagating in three dimensions,

k=kx2+ky2+kz2,

where kx, ky, kz are the Cartesian components of k. When the wave enters the guide, however, not all values of kx, ky, kz are allowed, because of the reflecting nature of the walls of the guide.

Introducing the boundary conditions

Since the electric field tangent to the surface of a conductor is always zero, along the plates, that is, for x=0 and x=a, we must have

Ey(0,y,z)=Ey(a,y,z)=0

and, since we have chosen to deal with radiation polarised in the y direction and the electric field is always perpendicular to the wave vector,

ky=0.

The wave is free in the y and z directions, while in the x direction it bounces back and forth between the walls, which forces this component of the wave vector to a discrete set of values, exactly as happens for the frequencies of a vibrating string:

kx=nπan=1,2,….

The mathematical expression of the wave propagating between the plates must therefore be that of a standing wave in the x direction and a travelling wave in the z direction

Ey(x,y,z,t)=E0cos⁡(ωt−kzz)sin⁡(kxx).

For sound waves the boundary conditions are different: the rigid walls require the air velocity perpendicular to the walls to be zero. For this reason the standing wave in the x direction contains cos⁡(kxx) instead of sin⁡(kxx). The allowed values are still kx=nπ/a, but now n = 0 is also possible: it is the plane wave that travels along the guide without bouncing off the walls. The same happens for an electromagnetic wave polarised in the x direction, perpendicular to the plates.

Propagation speed

Let c be the speed at which the wave propagates freely in the medium between the plates. For example, for a sound wave c is the speed of sound in the medium; for an electromagnetic wave, c is the speed of light in the medium.

The speed at which the wave moves along the guide (that is, in the z direction) is lower than c, because the wave, while still propagating in the medium at speed c, follows a zig-zag path in the guide, bouncing back and forth between the walls. The effective propagation speed in the z direction, which is the group velocity of the wave in the guide, is
vg=BD‾AD‾c=kzkc<c.
On the other hand, the drawing shows that in one period T the wavefront, which travels a distance λ in the direction of motion of the wave, covers a length L along z. This means that the wavefront moves along the guide with speed
vf=Lλc=kkzc.

This is, by definition, the phase velocity of the wave in the guide. Note that, since L>λ,

vf>c.

In other words, for electromagnetic waves in a waveguide, the phase velocity can exceed the speed of light in vacuum. This does not contradict the theory of relativity, because the phase velocity only describes the motion of the wavefronts of an infinitely long wave: energy and signals travel at the group velocity, which is always less than c. The two velocities are related by

vgvf=c2.
  • The App Onde 2D en can be used to visualise the propagation of waves in a guide in real time, and to observe directly the various effects described on this page.
  • The page virtual experiments on waveguides offers guided tours to help you use the app

Dispersion relation

In free space, that is, in the medium without the guide, the relation between ω and k, called the dispersion relation, is

ω=kc,

where c is the speed of the waves in the medium. Note that the relation is linear.

For a wave propagating in the guide, it becomes

ω=c(nπa)2+kz2.

This relation is particularly interesting for several reasons:

  • First of all, there are different dispersion curves for the different modes of the guide. Indeed, each mode is identified by a different value of n.
  • For each n the dispersion is no longer linear.
  • The dispersion approaches the linear relation of free space only for large values of kz, that is (since k is fixed) the more nearly parallel the original direction of the wave is to the z axis.
  • Conversely, for kz small compared with k the dispersion departs from the linear case, and in particular, at the same frequency, the values of kz in the guide are smaller than the wavenumber k=ω/c in free space. As we have already seen, this means a lower group velocity.
  • For each mode there is a specific cutoff frequency (in the figure the cutoff frequency corresponds to the intersection of the dispersion curves with the frequency axis: we have chosen the units so that it equals 1 for the first mode, 2 for the second, and so on). As is easy to show by imposing the condition kz=0, for the n-th mode it is
fcutoff=n⋅c2a
  • If the waves have a frequency below the cutoff frequency, there is no wave vector kz that can propagate them.

Denoting by fn=nc/(2a) the cutoff frequency of mode n, the group and phase velocities of this mode can be written as functions of the frequency f of the wave:

vg=c1−(fnf)2,vf=c1−(fn/f)2.

Approaching the cutoff frequency, the group velocity tends to zero (the wave bounces between the plates almost perpendicularly) and the phase velocity becomes infinite.

Check your understanding

Problem. Two metal plates are 2 cm apart with air between them. Calculate the cutoff frequency of the first mode for electromagnetic waves polarised parallel to the plates. Solution. f1=c/(2a) = 3·108/0.04 Hz = 7.5·109 Hz = 7.5 GHz: these are microwaves, with a wavelength in vacuum of 4 cm.

Problem. A 10 GHz wave is sent into the guide of the previous problem in the first mode. Calculate the group velocity and the phase velocity. Solution. f1/f = 0.75, so √(1 − 0.75²) ≈ 0.661. The group velocity is about 0.66 c ≈ 1.98·108 m/s and the phase velocity about 1.51 c ≈ 4.5·108 m/s, greater than the speed of light. Their product is exactly c2.

Problem. A 20 GHz wave is sent into the same guide. Which modes can propagate? Solution. The cutoff frequencies are 7.5 GHz, 15 GHz, 22.5 GHz, … The modes with a cutoff frequency below 20 GHz propagate, that is n = 1 and n = 2, each with its own group velocity: a signal that excites both arrives as two copies shifted in time. This is why in telecommunications waveguides are used at frequencies where only one mode propagates.


"Physics, Waves, Music": a website about the physics of waves, acoustics of musical instruments, musical scales, harmony and music.