Skip to main content
Physics LibreTexts

6.7: General Relationships for Unidirectional Waves

  • Page ID
  • Analysis of electromagnetic waves in enclosed spaces – and in waveguides in particular – is quite difficult. The task is dramatically simplified if it can be assumed that the wave propagates in a single direction; i.e., is unidirectional. This does not necessarily entail loss of generality. For example: Within a straight waveguide, waves can travel either “forward” or “backward.” The principle of superposition allows one to consider these two unidirectional cases separately, and then to simply sum the results.

    In this section, the equations that relate the various components of a unidirectional wave are derived. The equations will be derived in Cartesian coordinates, anticipating application to rectangular waveguides. However, the underlying strategy is generally applicable.

    We begin with Maxwell’s curl equations:

    \[\begin{align} \nabla\times\widetilde{\bf E} &= -j\omega\mu\widetilde{\bf H} \label{m0222_eMCE1} \\ \nabla\times\widetilde{\bf H} &= +j\omega\epsilon\widetilde{\bf E} \label{m0222_eMCH1}\end{align}\]

    Let us consider Equation \ref{m0222_eMCH1} first. Solving for \(\widetilde{\bf E}\), we have:

    \[\widetilde{\bf E} = \frac{1}{j\omega\epsilon}\nabla\times\widetilde{\bf H} \label{m0222_eMCE2}\]

    This is actually three equations; that is, one each for the \(\hat{\bf x}\), \(\hat{\bf y}\), and \(\hat{\bf z}\) components of \(\widetilde{\bf E}\), respectively. To extract these equations, let us define the components as follows:

    \[\begin{align} \widetilde{\bf E} &= \hat{\bf x}\widetilde{E}_x + \hat{\bf y}\widetilde{E}_y + \hat{\bf z}\widetilde{E}_z \\ \widetilde{\bf H} &= \hat{\bf x}\widetilde{H}_x + \hat{\bf y}\widetilde{H}_y + \hat{\bf z}\widetilde{H}_z \end{align}\]

    Now applying the equation for curl in Cartesian coordinates (Equation 12.2.7 in Appendix 12.2)

    \[\nabla \times \mathbf{A}= \hat{\mathbf{x}}\left(\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}\right)
    +\hat{\mathbf{y}}\left(\frac{\partial A_{x}}{\partial z}-\frac{\partial A_{z}}{\partial x}\right)
    +\hat{\mathbf{z}}\left(\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}\right) \nonumber \]

    we find:

    \begin{align} \widetilde{E}_x &= \frac{1}{j\omega\epsilon}\left( \frac{\partial \widetilde{H}_z}{\partial y} - \frac{\partial \widetilde{H}_y}{\partial z} \right) \label{m0222_eEx} \\ \widetilde{E}_y &= \frac{1}{j\omega\epsilon}\left( \frac{\partial \widetilde{H}_x}{\partial z} - \frac{\partial \widetilde{H}_z}{\partial x} \right) \label{m0222_eEy} \\ \widetilde{E}_z &= \frac{1}{j\omega\epsilon}\left( \frac{\partial \widetilde{H}_y}{\partial x} - \frac{\partial \widetilde{H}_x}{\partial y} \right) \label{m0222_eEz}\end{align}

    Without loss of generality, we may assume that the single direction in which the wave is traveling is in the \(+\hat{\bf z}\) direction. If this is the case, then \(\widetilde{\bf H}\), and subsequently each component of \(\widetilde{\bf H}\), contains a factor of \(e^{-jk_z z}\) where \(k_z\) is the phase propagation constant in the direction of travel. The remaining factors are independent of \(z\); i.e., depend only on \(x\) and \(y\). With this in mind, we further decompose components of \(\widetilde{\bf H}\) as follows:

    \begin{align} \widetilde{H}_x &= \widetilde{h}_x(x,y) e^{-jk_z z} \label{m0222_eHx} \\ \widetilde{H}_y &= \widetilde{h}_y(x,y) e^{-jk_z z} \label{m0222_eHy} \\ \widetilde{H}_z &= \widetilde{h}_z(x,y) e^{-jk_z z} \label{m0222_eHz}\end{align}

    where \(\widetilde{h}_x(x,y)\), \(\widetilde{h}_y(x,y)\), and \(\widetilde{h}_z(x,y)\) represent the remaining factors. One advantage of this decomposition is that partial derivatives with respect to \(z\) reduce to algebraic operations; i.e.:

    \begin{align} \frac{\partial \widetilde{H}_x}{\partial z} &= -jk_z \widetilde{h}_x(x,y) e^{-jk_z z} = -jk_z \widetilde{H}_x \label{m0222_edzHx} \\ \frac{\partial \widetilde{H}_y}{\partial z} &= -jk_z \widetilde{h}_y(x,y) e^{-jk_z z} = -jk_z \widetilde{H}_y \label{m0222_edzHy} \\ \frac{\partial \widetilde{H}_z}{\partial z} &= -jk_z \widetilde{h}_z(x,y) e^{-jk_z z} = -jk_z \widetilde{H}_z \label{m0222_edzHz}\end{align}

    We now substitute Equations \ref{m0222_eHx} - \ref{m0222_eHz} into Equations \ref{m0222_eEx} - \ref{m0222_eEz} and then use Equations \ref{m0222_edzHx} - \ref{m0222_edzHz} to eliminate partial derivatives with respect to \(z\). This yields:

    \begin{align} \widetilde{E}_x &= \frac{1}{j\omega\epsilon}\left( \frac{\partial \widetilde{H}_z}{\partial y} + jk_z \widetilde{H}_y \right) \label{m0222_eEx2} \\ \widetilde{E}_y &= \frac{1}{j\omega\epsilon}\left( - jk_z \widetilde{H}_x - \frac{\partial \widetilde{H}_z}{\partial x} \right) \\ \widetilde{E}_z &= \frac{1}{j\omega\epsilon}\left( \frac{\partial \widetilde{H}_y}{\partial x} - \frac{\partial \widetilde{H}_x}{\partial y} \right) \label{m0222_eEz2}\end{align}

    Applying the same procedure to the curl equation for \(\widetilde{\bf H}\) (Equation \ref{m0222_eMCE1}), one obtains:

    \begin{align} \widetilde{H}_x &= \frac{1}{-j\omega\mu}\left( \frac{\partial \widetilde{E}_z}{\partial y} + jk_z \widetilde{E}_y \right) \label{m0222_eHx2} \\ \widetilde{H}_y &= \frac{1}{-j\omega\mu}\left( - jk_z \widetilde{E}_x - \frac{\partial \widetilde{E}_z}{\partial x} \right) \label{m0222_eHy2} \\ \widetilde{H}_z &= \frac{1}{-j\omega\mu}\left( \frac{\partial \widetilde{E}_y}{\partial x} - \frac{\partial \widetilde{E}_x}{\partial y} \right) \label{m0222_eHz2}\end{align}

    Equations \ref{m0222_eEx2} - \ref{m0222_eHz2} constitute a set of simultaneous equations that represent Maxwell’s curl equations in the special case of a unidirectional (specifically, a \(+\hat{\bf z}\)-traveling) wave. With just a little bit of algebraic manipulation of these equations, it is possible to obtain expressions for the \(\hat{\bf x}\) and \(\hat{\bf y}\) components of \(\widetilde{\bf E}\) and \(\widetilde{\bf H}\) which depend only on the \(\hat{\bf z}\) components of \(\widetilde{\bf E}\) and \(\widetilde{\bf H}\). Here they are:1

    \begin{align} \widetilde{E}_x &= \frac{-j}{k_{\rho}^2}\left( +k_z \frac{\partial \widetilde{E}_z}{\partial x} + \omega\mu \frac{\partial \widetilde{H}_z}{\partial y} \right) \label{m0222_eExu} \\ \widetilde{E}_y &= \frac{+j}{k_{\rho}^2}\left( -k_z \frac{\partial \widetilde{E}_z}{\partial y} + \omega\mu \frac{\partial \widetilde{H}_z}{\partial x} \right) \label{m0222_eEyu} \\ \widetilde{H}_x &= \frac{+j}{k_{\rho}^2}\left( \omega\epsilon \frac{\partial \widetilde{E}_z}{\partial y} - k_z \frac{\partial \widetilde{H}_z}{\partial x} \right) \label{m0222_eHxu} \\ \widetilde{H}_y &= \frac{-j}{k_{\rho}^2}\left( \omega\epsilon \frac{\partial \widetilde{E}_z}{\partial x} + k_z \frac{\partial \widetilde{H}_z}{\partial y} \right) \label{m0222_eHyu} \end{align}


    \[k_{\rho}^2 \triangleq \beta^2 - k_z^2\]

    Why define a parameter called “\(k_{\rho}\)”? Note from the definition that \(\beta^2 = k_{\rho}^2 + k_z^2\). Further, note that this equation is an expression of the Pythagorean theorem, which relates the lengths of the sides of a right triangle. Since \(\beta\) is the “overall” phase propagation constant, and \(k_z\) is the phase propagation constant for propagation in the \(\hat{\bf z}\) direction in the waveguide, \(k_{\rho}\) must be associated with variation in fields in directions perpendicular to \(\hat{\bf z}\). In the cylindrical coordinate system, this is the \(\hat{\bf \rho}\) direction, hence the subscript “\(\rho\).” We shall see later that \(k_{\rho}\) plays a special role in determining the structure of fields within the waveguide, and this provides additional motivation to identify this quantity explicitly in the field equations.

    Summarizing: If you know the wave is unidirectional, then knowledge of the components of \(\widetilde{\bf E}\) and \(\widetilde{\bf H}\) in the direction of propagation is sufficient to determine each of the remaining components of \(\widetilde{\bf E}\) and \(\widetilde{\bf H}\). There is one catch, however: For this to yield sensible results, \(k_{\rho}\) may not be zero. So, ironically, these expressions don’t work in the case of a uniform plane wave, since such a wave would have \(\widetilde{E}_z=\widetilde{H}_z=0\) and \(k_z=\beta\) (so \(k_{\rho}=0\)), yielding values of zero divided by zero for each remaining field component. The same issue arises for any other transverse electromagnetic (TEM) wave. For non-TEM waves – and, in particular, unidirectional waves in waveguides – either \(\widetilde{E}_z\) or \(\widetilde{H}_z\) must be non-zero and \(k_{\rho}\) will be non-zero. In this case, Equations \ref{m0222_eExu} - \ref{m0222_eHyu} are both usable and useful since they allow determination of all field components given just the \({\bf z}\) components.

    In fact, we may further exploit this simplicity by taking one additional step: Decomposition of the unidirectional wave into transverse electric (TE) and transverse magnetic (TM) components. In this case, TE means simply that \(\widetilde{E}_z=0\), and TM means simply that \(\widetilde{H}_z=0\). For a wave which is either TE or TM, Equations \ref{m0222_eExu} - \ref{m0222_eHyu} reduce to one term each.

    1. Students are encouraged to derive these for themselves – only algebra is required. Tip to get started: To get \(\widetilde{E}_x\), begin with Equation \ref{m0222_eEx2}, eliminate \(\widetilde{H}_y\) using Equation \ref{m0222_eHy2}, and then solve for \(\widetilde{E}_x\).