Electromagnetic waves carry both energy and momentum. It should therefore be possible to transmit energy contained in the wave from one place to another. For low frequencies (typically, less than 1 MHz), this is done by parallel transmission lines or coaxial cables. However, for higher frequencies, such as microwave frequencies, we need special conduits such a hollow metal tubes or optical fibers.
We will first explain the basic concepts of guiding waves by taking the simple case of a pair of parallel, infinite metal plates with a separation d between them. The wave is made to propagate in the hollow region between the plates, which we take to be empty space. We have seen that the electric field is primarily confined to the surface, penetrating a small “skin depth” which becomes smaller with increasing frequency. As a result we take the electric field to vanish at the surface of the guide.
Let us rewrite the Maxwell’s equations for the case where there are free charges or currents.
where we have time variation to be
Taking the curl of the these equations and substituting from the other equations, we get
For propagation between the plannes, σ = 0, so that we have,
We will return to thiese equations a little while later.
Consider the original curl equations and write them in component form (with σ = 0)
Parallely, the derivatives of electric field components satisf,
The solutions of these set of equations can be classified into three distinct groups. The direction of propagation being along the z direction, we term this direction as the longitudinal direction and a direction perpendicular to it (i.e. x and y directions) as the transverse direction. The distinct solutions are grouped as
It is of course possible for a solution not to belong to any of these distinct categories in which case it would be a "mixed mode" solution.
Since the wave propagates along the z direction, the z dependence of the field is specified,
where, the complex factor Thus is equivalent to multiplication by - Further, since the plates are of infinite extent in y direction, there is no field variation in this direction so that we can replace the derivative by zero. Using these, we can rewrite the equations above as
The wave equation (1) takes the form,
We will discuss in detail the TE solution and leave the TM solution as an exercise.
TE- Mode :
In this case Ez = 0. From Eqn. (2) , we get Hy = constant, which we can choose to be zero. This in turn implies, from Eqn. (3), Ex = 0. We will first solve for Ey using Eqn. (4).
Define k2 = ŷ2 + ω2∈μ. We have,
the solution of which are well known to be
(complete solution will be obtained by multiplying this with
We now insert the boundary condition, Ey = 0 on both the plates, i.e. at x= 0 and at x= d.
The former gives B= 0, so that Ey = A sin Kx. The latter condition restricts the values that k can take to where n = 1,2, … .( n cannot take the value zero because that would make the field identically zero. )
Thus we have,
where A = Ey0 is the maximum value of the field. We can now use this expression in Eqn. (3) to obtain the magnetic field components
Modes are named by specifying the value that n takes. As we have seen the lowest mode is n = 1. This is termed as TE10 mode, the meaning of the second index will be clear when we discuss rectangular waveguides but for the present case it remains zero for all values of n. The following figure gives the electric field profile for n=1 and n=2 for a fixed z.
If we look along the direction of propagation, for the mode, thefield lines crowdat the centre of the guide, where the field strength is strongest.
The corresponding magnetic fields for are shown below:
Is transmission in this manner always possible? We have,
Propagating solution implies that Thus we requires,
If the frequency is less than this, the wave attenuates. The phase velocity for the propagating solution is given by
As frequency decreases and approaches the critical value, it becomes infinite. For very large frequencies, the velocity in vacuum approaches that of light
We will not work out the TM mode algebra. In this case . The non-zero field components are
Note that in TM case , unlike in the case of TE modes, we can have m=0 here because the solutions are in terms of cosine functions. In this case we have,
which gives the ratio which is the intrinsic impedance we have seen to characterize propagation of wave in a uniform medium.
Solutions to Tutorial Assignments
so that the cutoff frequency is given by
If the operating frequency is 3 GHz, the propagation constant is given by
The phase velocity is given by
2. The amplitude of linear current density on the plates is equal to the tangential component of the magnetic field on the planes,
Power loss per unit length of conductor
We take . The loss for TE10 is (in J/m)
Self Assessment Questions
Solutions to Self Assessment Questions
1. The propagation constant is given by
For propagation to take place 22 > 5n, so that n<5. This implies 4 TE modes, 4 TM modes and one TEM mode, giving a total of 9 modes.
2. The cutoff frequency is given by The wavelength in the x direction for m. If the operating frequency is 2.5 × 109Hz., the propagation vector is given by (using n=2)
The fields are as follows :
To determine λz, we use the fact that it is equal to the distance over which the phase of the propagating wave changes by 2π. Thus
so that λz= 3/20m.
If the operating frequency is 1.2 GHz, the wave attenuates, and we have,
so that the attenuation distance is
3. The fields are given by
Power transmitted per unit area is
Power transmitted in z direction through an area of unit width,