# 6.5: Parallel Plate Waveguide- TM Case, Electric Field


In Section 6.2, the parallel plate waveguide shown in Figure $$\PageIndex{1}$$ was introduced.

Figure $$\PageIndex{1}$$: TM component of the electric field in a parallel plate waveguide. (CC BY-SA 4.0; C. Wang)

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, $$\widetilde{\bf H} = \hat{\bf y}\widetilde{H}_y$$, with no component in the $$\hat{\bf x}$$ or $$\hat{\bf z}$$ directions. Following precisely the same reasoning employed in Section 6.2, we find the governing equation for the magnetic component of TM field is:

$\frac{\partial^2}{\partial x^2}\widetilde{H}_y + \frac{\partial^2}{\partial z^2}\widetilde{H}_y = - \beta^2 \widetilde{H}_y \label{m0177_eDH}$

The general solution to this partial differential equation is:

\begin{align} \widetilde{H}_y =&~~~~~e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right] \nonumber \\ &+e^{+jk_z z} \left[ C e^{-jk_x x} + D e^{+jk_x x} \right] \label{m0177_eGS}\end{align}

where $$A$$, $$B$$, $$C$$, and $$D$$ are complex-valued constants; and $$k_x$$ and $$k_z$$ 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 \ref{m0177_eGS} consists of two terms. The first term includes the factor $$e^{-jk_z z}$$, indicating a wave propagating in the $$+\hat{\bf z}$$ direction, and the second term includes the factor $$e^{+jk_z z}$$, indicating a wave propagating in the $$-\hat{\bf z}$$ direction. If we impose the restriction that sources exist only on the left ($$z<0$$) side of Figure $$\PageIndex{1}$$, and that there be no structure capable of wave scattering (in particular, reflection) on the right ($$z>0$$) side of Figure $$\PageIndex{1}$$, then there can be no wave components propagating in the $$-\hat{\bf z}$$ direction. In this case, $$C=D=0$$ and Equation \ref{m0177_eGS} simplifies to:

$\widetilde{H}_y = e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right] \label{m0177_eGS2}$

Before proceeding, let’s make sure that Equation \ref{m0177_eGS2} is actually a solution to Equation \ref{m0177_eDH}. As in the TE case, this check yields a constraint (in fact, the same constraint) on the as-yet undetermined parameters $$k_x$$ and $$k_z$$. First, note:

$\frac{\partial \widetilde{H}_y}{\partial x} = e^{-jk_z z} \left[-A e^{-jk_x x} + B e^{+jk_x x} \right]\left(jk_x\right) \label{m0177_edHydx}$

So:

\begin{align} \frac{\partial^2 \widetilde{H}_y}{\partial x^2} &= e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right]\left(-k_x^2\right) \nonumber \\ &= -k_x^2 \widetilde{H}_y\end{align} \nonumber

Next, note:

$\frac{\partial \widetilde{H}_y}{\partial z} = e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right]\left(-jk_z\right) \label{m0177_edHydz}$

So:

\begin{align} \frac{\partial^2 \widetilde{H}_y}{\partial z^2} &= e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right]\left(-k_z^2\right) \nonumber \\ &= -k_z^2 \widetilde{H}_y\end{align} \nonumber

Now summing these results:

$\frac{\partial^2 \widetilde{H}_y}{\partial x^2} + \frac{\partial^2 \widetilde{H}_y}{\partial z^2} = -\left( k_x^2 + k_z^2 \right) \widetilde{H}_y \label{m0177_eDE2}$

Comparing Equation \ref{m0177_eDE2} to Equation \ref{m0177_eDH}, we conclude that Equation \ref{m0177_eGS2} is a solution to Equation \ref{m0177_eDH} under the constraint that:

$\beta^2 = k_x^2 + k_z^2 \label{m0177_eBeta}$

This is precisely the same constraint identified in the TE case, and confirms that $$k_x$$ and $$k_y$$ are in fact the components of the propagation vector

${\bf k} \triangleq \beta\hat{\bf k} = \hat{\bf x}k_x + \hat{\bf y}k_y + \hat{\bf z}k_z \nonumber$

where $$\hat{\bf k}$$ is the unit vector pointing in the direction of propagation, and $$k_y=0$$ 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:

$\nabla \times \widetilde{\bf H} = j\omega\epsilon \widetilde{\bf E} \nonumber$

Thus:

\begin{align} \widetilde{\bf E} &= \frac{1}{j\omega\epsilon}~\nabla \times \widetilde{\bf H} \nonumber \\ &= \frac{1}{j\omega\epsilon}~\nabla \times \left(\hat{\bf y}\widetilde{H}_y\right) \end{align} \nonumber

The relevant form of the curl operator is

\begin{aligned}
\nabla \times \mathbf{A}=& \hat{\mathbf{x}}\left(\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}\right) \\
&+\hat{\mathbf{y}}\left(\frac{\partial A_{x}}{\partial z}-\frac{\partial A_{z}}{\partial x}\right) \\
&+\hat{\mathbf{z}}\left(\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}\right)
\end{aligned}

from Appendix 12.2. Although the complete expression consists of 6 terms, all but 2 of these terms are zero because the $$\hat{\bf x}$$ and $$\hat{\bf z}$$ components of $$\widetilde{\bf H}$$ are zero. The two remaining terms are $$-\hat{\bf x}\partial \widetilde{H}_y/\partial z$$ and $$+\hat{\bf z}\partial \widetilde{H}_y/\partial x$$. Thus:

$\widetilde{\bf E} = \frac{1}{j\omega\epsilon}~\left( -\hat{\bf x}\frac{\partial\widetilde{H}_y}{\partial z} +\hat{\bf z}\frac{\partial\widetilde{H}_y}{\partial x} \right) \label{m0177_eE}$

We may further develop this expression using Equations \ref{m0177_edHydx} and \ref{m0177_edHydz}. We find the $$\hat{\bf x}$$ component of $$\widetilde{\bf E}$$ is:

$\widetilde{E}_x = \frac{k_z}{\omega\epsilon}~e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right] \label{m0177_eEx}$

and the $$\hat{\bf z}$$ component of $$\widetilde{\bf E}$$ is:

$\widetilde{E}_z = \frac{k_x}{\omega\epsilon}~e^{-jk_z z} \left[ -A e^{-jk_x x} + B e^{+jk_x x} \right] \label{m0177_eEz}$

The solution has now been reduced to the problem of finding the constants $$A$$, $$B$$, and either $$k_x$$ or $$k_z$$. 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 $$\widetilde{E}_z\left(x=0\right) = 0$$ and $$\widetilde{E}_z\left(x=a\right) = 0$$. Referring to Equation \ref{m0177_eEz}, the boundary condition at $$x=0$$ means

$\frac{k_x}{\omega\epsilon}~e^{-jk_z z} \left[ -A \left(1\right) + B \left(1\right) \right] = 0 \nonumber$

The factor $$e^{-jk_z z}$$ always has unit magnitude, and so cannot be zero. We could require $$k_x$$ to be zero, but this is unnecessarily restrictive. Instead, we require $$A=B$$ and we may rewrite Equation \ref{m0177_eEz} as follows:

$\widetilde{E}_z = \frac{B k_x}{\omega\epsilon}~e^{-jk_z z} \left[ e^{+jk_x x} - e^{-jk_x x} \right] \nonumber$

This expression is simplified using a trigonometric identity:

$\sin{k_x a} = \frac{1}{2j}\left[ e^{+jk_x a} - e^{-jk_x a} \right] \nonumber$

Thus:

$\widetilde{E}_z = \frac{j2B k_x}{\omega\epsilon} e^{-jk_z z} \sin k_x x \nonumber$

Now following up with $$\widetilde{E}_x$$, beginning from Equation \ref{m0177_eEx}:

\begin{align} \widetilde{E}_x &= \frac{k_z}{\omega\epsilon}~e^{-jk_z z} \left[ A e^{-jk_x x} + B e^{+jk_x x} \right] \nonumber \\ &= \frac{B k_z}{\omega\epsilon}~e^{-jk_z z} \left[ e^{-jk_x x} + e^{+jk_x x} \right] \nonumber \\ &= \frac{2B k_z}{\omega\epsilon}~e^{-jk_z z} \cos k_x x\end{align} \nonumber

For convenience we define the following complex-valued constant:

$E_{x0}\triangleq \frac{2B k_z}{\omega\epsilon} \nonumber$

This yields the following simpler expression:

$\widetilde{E}_x = E_{x0}~e^{-jk_z z} \cos k_x x \nonumber$

Now let us apply the boundary condition at $$x=a$$ to $$\widetilde{E}_{z}$$:

$\frac{j2B k_z}{\omega\epsilon} e^{-jk_z z} \sin k_x a = 0 \nonumber$

Requiring $$B=0$$ or $$k_z=0$$ yields only trivial solutions, therefore, it must be true that

$\sin{k_x a} = 0 \nonumber$

This in turn requires that

$k_x a = m \pi \nonumber$

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 $$k_x=0$$, and the associated field turned out to be identically zero. In the present (TM) case, $$m=0$$ also yields $$k_x=0$$, but the associated field is not necessarily zero. That is, for $$m=0$$, $$\widetilde{E}_z=0$$ but $$\widetilde{E}_x$$ 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 $$k_x$$. 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 $$k_z$$ for mode $$m$$ is obtained using Equation \ref{m0177_eBeta} as follows:

\begin{align} k_z &= \sqrt{\beta^2-k_x^2} \nonumber \\ & =\sqrt{\beta^2-\left(\frac{m\pi}{a}\right)^2}\end{align} \nonumber

Since $$k_z$$ is specified to be real-valued, we require:

$\beta^2-\left(\frac{m\pi}{a}\right)^2 > 0 \nonumber$

This constrains $$\beta$$; specifically:

$\beta > \frac{m\pi}{a} \nonumber$

Recall that $$\beta=\omega\sqrt{\mu\epsilon}$$ and $$\omega=2\pi f$$ where $$f$$ is frequency. Solving for $$f$$, we find:

$f > \frac{m}{2a\sqrt{\mu\epsilon}} \nonumber$

Thus, each mode exists only above a certain frequency, which is different for each mode. This cutoff frequency $$f_c$$ for mode $$m$$ is given by

$\boxed{ f_c^{(m)} \triangleq \frac{m}{2a\sqrt{\mu\epsilon}} } \label{m0177_efcm}$

At frequencies below the cutoff frequency for mode $$m$$, modes $$0$$ through $$m-1$$ exhibit imaginary-valued $$k_z$$ 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 $$\PageIndex{1}$$, we find that the electric field component of the TM field is given by:

$\boxed{ \widetilde{\bf E} = \sum_{m=0}^{\infty} \left[ \hat{\bf x} \widetilde{E}_x^{(m)} + \hat{\bf z} \widetilde{E}_z^{(m)} \right] } \label{m0177_eEsum}$

where

$\boxed{ \widetilde{E}_x^{(m)} \triangleq \begin{cases} 0, & f<f_c^{(m)} \\ E_{x0}^{(m)} e^{-jk_z^{(m)} z} \cos k_x^{(m)} x, & f\ge f_c^{(m)} \end{cases} } \nonumber$

and

$\boxed{ \widetilde{E}_z^{(m)} \triangleq \begin{cases} 0, & f<f_c^{(m)} \\ j\frac{k_x^{(m)}}{k_z^{(m)}}E_{x0}^{(m)} e^{-jk_z^{(m)} z} \sin k_x^{(m)} x, & f\ge f_c^{(m)} \end{cases} } \nonumber$

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.

This page titled 6.5: Parallel Plate Waveguide- TM Case, Electric Field is shared under a CC BY-SA license and was authored, remixed, and/or curated by Steven W. Ellingson (Virginia Tech Libraries' Open Education Initiative) .