6.5: Parallel Plate Waveguide- TM Case, Electric Field
( \newcommand{\kernel}{\mathrm{null}\,}\)
In Section 6.2, the parallel plate waveguide shown in Figure 6.5.1 was introduced.

At the end of that section, we decomposed the problem into its TE and TM components. In this section, we find the TM component of the fields in the waveguide.
“Transverse magnetic” means the magnetic field vector is perpendicular to the plane of interest, and is therefore parallel to the conducting surfaces. Thus, ˜H=ˆy˜Hy, with no component in the ˆx or ˆz directions. Following precisely the same reasoning employed in Section 6.2, we find the governing equation for the magnetic component of TM field is:
∂2∂x2˜Hy+∂2∂z2˜Hy=−β2˜Hy
The general solution to this partial differential equation is:
˜Hy= e−jkzz[Ae−jkxx+Be+jkxx]+e+jkzz[Ce−jkxx+De+jkxx]
where A, B, C, and D are complex-valued constants; and kx and kz are real-valued constants. We have assigned variable names to these constants with advance knowledge of their physical interpretation; however, at this moment they remain simply unknown constants whose values must be determined by enforcement of boundary conditions.
Note that Equation 6.5.2 consists of two terms. The first term includes the factor e−jkzz, indicating a wave propagating in the +ˆz direction, and the second term includes the factor e+jkzz, indicating a wave propagating in the −ˆz direction. If we impose the restriction that sources exist only on the left (z<0) side of Figure 6.5.1, and that there be no structure capable of wave scattering (in particular, reflection) on the right (z>0) side of Figure 6.5.1, then there can be no wave components propagating in the −ˆz direction. In this case, C=D=0 and Equation 6.5.2 simplifies to:
˜Hy=e−jkzz[Ae−jkxx+Be+jkxx]
Before proceeding, let’s make sure that Equation ??? is actually a solution to Equation ???. As in the TE case, this check yields a constraint (in fact, the same constraint) on the as-yet undetermined parameters kx and kz. First, note:
∂˜Hy∂x=e−jkzz[−Ae−jkxx+Be+jkxx](jkx)
So:
∂2˜Hy∂x2=e−jkzz[Ae−jkxx+Be+jkxx](−k2x)=−k2x˜Hy
Next, note:
∂˜Hy∂z=e−jkzz[Ae−jkxx+Be+jkxx](−jkz)
So:
∂2˜Hy∂z2=e−jkzz[Ae−jkxx+Be+jkxx](−k2z)=−k2z˜Hy
Now summing these results:
∂2˜Hy∂x2+∂2˜Hy∂z2=−(k2x+k2z)˜Hy
Comparing Equation ??? to Equation ???, we conclude that Equation ??? is a solution to Equation ??? under the constraint that:
β2=k2x+k2z
This is precisely the same constraint identified in the TE case, and confirms that kx and ky are in fact the components of the propagation vector
k≜βˆk=ˆxkx+ˆyky+ˆzkz
where ˆk is the unit vector pointing in the direction of propagation, and ky=0 in this particular problem.
Our objective in this section is to determine the electric field component of the TM field. The electric field may be obtained from the magnetic field using Ampere’s law:
∇טH=jωϵ˜E
Thus:
˜E=1jωϵ ∇טH=1jωϵ ∇×(ˆy˜Hy)
The relevant form of the curl operator is
∇×A=ˆx(∂Az∂y−∂Ay∂z)+ˆy(∂Ax∂z−∂Az∂x)+ˆz(∂Ay∂x−∂Ax∂y)
from Appendix 12.2. Although the complete expression consists of 6 terms, all but 2 of these terms are zero because the ˆx and ˆz components of ˜H are zero. The two remaining terms are −ˆx∂˜Hy/∂z and +ˆz∂˜Hy/∂x. Thus:
˜E=1jωϵ (−ˆx∂˜Hy∂z+ˆz∂˜Hy∂x)
We may further develop this expression using Equations ??? and ???. We find the ˆx component of ˜E is:
˜Ex=kzωϵ e−jkzz[Ae−jkxx+Be+jkxx]
and the ˆz component of ˜E is:
˜Ez=kxωϵ e−jkzz[−Ae−jkxx+Be+jkxx]
The solution has now been reduced to the problem of finding the constants A, B, and either kx or kz. This is accomplished by enforcing the relevant boundary conditions. In general, the component of the electric field which is tangent to a perfectly-conducting surface is zero. Applied to the present (TM) case, this means ˜Ez(x=0)=0 and ˜Ez(x=a)=0. Referring to Equation ???, the boundary condition at x=0 means
kxωϵ e−jkzz[−A(1)+B(1)]=0
The factor e−jkzz always has unit magnitude, and so cannot be zero. We could require kx to be zero, but this is unnecessarily restrictive. Instead, we require A=B and we may rewrite Equation ??? as follows:
˜Ez=Bkxωϵ e−jkzz[e+jkxx−e−jkxx]
This expression is simplified using a trigonometric identity:
sinkxa=12j[e+jkxa−e−jkxa]
Thus:
˜Ez=j2Bkxωϵe−jkzzsinkxx
Now following up with ˜Ex, beginning from Equation ???:
˜Ex=kzωϵ e−jkzz[Ae−jkxx+Be+jkxx]=Bkzωϵ e−jkzz[e−jkxx+e+jkxx]=2Bkzωϵ e−jkzzcoskxx
For convenience we define the following complex-valued constant:
Ex0≜2Bkzωϵ
This yields the following simpler expression:
˜Ex=Ex0 e−jkzzcoskxx
Now let us apply the boundary condition at x=a to ˜Ez:
j2Bkzωϵe−jkzzsinkxa=0
Requiring B=0 or kz=0 yields only trivial solutions, therefore, it must be true that
sinkxa=0
This in turn requires that
kxa=mπ
where m is an integer. Note that this is precisely the same relationship that we identified in the TE case. There is an important difference, however. In the TE case, m=0 was not of interest because this yields kx=0, and the associated field turned out to be identically zero. In the present (TM) case, m=0 also yields kx=0, but the associated field is not necessarily zero. That is, for m=0, ˜Ez=0 but ˜Ex is not necessarily zero. Therefore, m=0 as well as m=1, m=2, and so on are of interest in the TM case.
At this point, we have uncovered a family of solutions with m=0,1,2,.... Each solution is referred to as a mode, and is associated with a particular value of kx. In the discussion that follows, we shall find that the consequences are identical to those identified in the TE case, except that m=0 is now also allowed. Continuing: The value of kz for mode m is obtained using Equation ??? as follows:
kz=√β2−k2x=√β2−(mπa)2
Since kz is specified to be real-valued, we require:
β2−(mπa)2>0
This constrains β; specifically:
β>mπa
Recall that β=ω√μϵ and ω=2πf where f is frequency. Solving for f, we find:
f>m2a√μϵ
Thus, each mode exists only above a certain frequency, which is different for each mode. This cutoff frequency fc for mode m is given by
f(m)c≜m2a√μϵ
At frequencies below the cutoff frequency for mode m, modes 0 through m−1 exhibit imaginary-valued kz and therefore do no propagate. Also, note that the cutoff frequency for m=0 is zero, and so this mode is always able to propagate. That is, the m=0 mode may exist for any a>0 and any f>0. Once again, this is a remarkable difference from the TE case, for which m=0 is not available.
Let us now summarize the solution. With respect to Figure 6.5.1, we find that the electric field component of the TM field is given by:
˜E=∞∑m=0[ˆx˜E(m)x+ˆz˜E(m)z]
where
˜E(m)x≜{0,f<f(m)cE(m)x0e−jk(m)zzcosk(m)xx,f≥f(m)c
and
˜E(m)z≜{0,f<f(m)cjk(m)xk(m)zE(m)x0e−jk(m)zzsink(m)xx,f≥f(m)c
where m enumerates modes (m=0,1,2,...) and
\boxed{ k_z^{(m)} \triangleq \sqrt{\beta^2-\left[k_x^{(m)}\right]^2} } \nonumber
\boxed{ k_x^{(m)} \triangleq m\pi/a } \label{m0177_ekxma}
Finally, the coefficients E_{x0}^{(m)} depend on sources and/or boundary conditions to the left of the region of interest.
For the scenario depicted in Figure \PageIndex{1}, the electric field component of the TM solution is given by Equation \ref{m0177_eEsum} with modal components determined as indicated by Equations \ref{m0177_efcm}-\ref{m0177_ekxma}. This solution presumes all sources lie to the left of the region of interest, with no additional sources or boundary conditions to the right of the region of interest.
The m=0 mode, commonly referred to as the “TM_0” mode, is of particular importance in the analysis of microstrip transmission line, and is addressed in Section 6.6.