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In the classic book by Collin (Foundations for Microwave Engineering, 2nd ed.), the author postulates on p. 278 that, given a waveguide structure that stretches along the $z$ axis with open ends (see figure attached), and given a current source $J$ inside the region captured by the two transverse planes at $z=z_{1}$ and $z=z_{2}$ in the wvaeguide, then the electromagnetic fields generated to the right ($E^{+},H^{+}$) and to the left ($E^{-},H^{-}$) may be given as expansions of the orthogonal waveguide natural modes (eigenfunctions) as:

$$ \boldsymbol{E}^{+}=\sum_{n}C^{+}_{n} (\boldsymbol{e_{n}}+\boldsymbol{e_{zn}})e^{-i\beta_{n}z}\ \ \ ; \ \ \ z>z_{2} $$ $$ \boldsymbol{H}^{+}=\sum_{n}C^{+}_{n} (\boldsymbol{h_{n}}+\boldsymbol{h_{zn}})e^{-i\beta_{n}z}\ \ \ ; \ \ \ z>z_{2} $$ $$ \boldsymbol{E}^{-}=\sum_{n}C^{-}_{n} (\boldsymbol{e_{n}}-\boldsymbol{e_{zn}})e^{i\beta_{n}z}\ \ \ ; \ \ \ z<z_{1} $$ $$ \boldsymbol{H}^{-}=\sum_{n}C^{-}_{n} (-\boldsymbol{h_{n}}+\boldsymbol{h_{zn}})e^{i\beta_{n}z}\ \ \ ; \ \ \ z<z_{1} $$

where $C$ are amplitude coefficients, $e_{n},h_{n}$ are the transverse components of such modes, while the $e_{zn},h_{zn}$ are their longitudinal components. Note that the author uses symbol $j$ instead of $i$ for $\sqrt{-1}$ in the image attached. Bold symbols are vectors.

I understand the expansion itself in terms of the modes, but I am not sure why did the author insert a minus sign ($-$) in front of $e_{zn}$ and $h_{n}$ in the last two equations? I assume that explaining one would lead to the other (from curl relations).

Could it be because he assumed, for example, that the $e_{zn}$ mode is origially defined to be along the $+z$-direction, and by looking into the opposite direction for $E^{-}$ we then see this reversed? And shouldn't inverting $e^{-i\beta_{n}z}$ to $e^{i\beta_{n}z}$ be the only mathematical operation made in moving from one direction to another?

Copyrights to Collin's cited book

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The direction of the longitudinal field component $\textbf{e}_{zn}$ is parallel with the $z$ axis. When the wave is moving in the same direction its propagation factor is $e^{-j\beta_n z}$. This is denoted by the $+$ sign as superscript such as $\textbf{E}^+$. The reflected wave denoted by $\textbf{E}^-$ is moving backwards, i.e., anti-parallel with the $z$ axis therefore its longitudinal component must have the opposite sign to that of the forward wave, hence the term $-\textbf{e}_{zn}$ in (4.107c). The sign of the transversal component $\textbf{e}_n$ of the E-field is not affected by the reflection along the z-axis.

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  • $\begingroup$ Why would one assume that travelling in the negative direction should affect the longitudinal vector arrow to be also pointing to negative z ? A vector pointing in positive z direction can still travel in the opposite direction. If there was a common boundary between the two oppositely travelling waves on which the longitudinal components must vanish (e.g. if z1=z2 and a perfect magnetic conductor boundary existed there), then I could appreciate that we take one as negative so they would cancel there to meet the condition. But there is no such clear boundary in our case here.Hence the question $\endgroup$ – user135626 Oct 18 '17 at 0:34
  • $\begingroup$ You probably meant to say "perfect electric boundary" not magnetic, but at any rate the particular section 4.12 in Collin is just about that: antenna probe next to a metal wall (in practice a quarter wave length from a back-short.). But this is not the only reason for the sign change, that sign change is derived in equations (3.58a-f) on page 98, and implied specifically in (e) and (f). $\endgroup$ – hyportnex Oct 18 '17 at 12:38
  • $\begingroup$ 1/1: No, I meant magnetic wall across the guide at the source region, so that the component Ez perpendicular to it would be cancelled (logitudinal=0) there. If it were electric wall, the transverse one (tangent to wall) would be zero. (2) I dont concur about sign change being there due to topic of sec 4.12: Collin explicitly derives these "general" equations on p.278 before applying them to the coupled probe guide problem later on p.279. Perhaps you hit the nail in the head by referring to eqns 3.58, but only if we think of them outside the source region for divergence to be zero and... $\endgroup$ – user135626 Oct 19 '17 at 15:36
  • $\begingroup$ 2/2: ... assumed that source holds same $\nabla_{t}\cdot \boldsymbol{e}$ behaviour to left and right sides of the source region. Then the change in sign between $e^{\pm i\beta z}$ will force eqn (3.58f) to change sign for $e_{z}$ to hold, which respectively forces transverse $h_{t}$ to flip sign due to (3.58c). There will be double sign flips in (3.58e) so it won't be affected, and no sign flips at all in (3.58a). If we look at it this way, I think it works (one loses his head with Collin sometimes). If you wish to update your answer, I'd be happy to "tick" answer. Thanks for the insight. $\endgroup$ – user135626 Oct 19 '17 at 15:37

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