Chapter 13, Problem 13.22P

Consider the TE₁₁ mode in a rectangular guide having width a and height b. Using Eqs. (96d) and (96e), along with methods described in Chapter 2, Section 2.6, show that the equation for the streamlines for this field (at z = 0) is $y=b\pi \cos -1[C\cos \left(\pi x/a\right)]\left(0<x<a,0<y,b\right)$

where C is a constant that identifies a given streamline.

To determine

To prove:

The equation for the streamlines for transverse electric field $E_{xs}$ and $E_{ys}$ for $T{E}_{11}$ mode in a rectangular is

   $y=\frac{b}{\pi }{\cos }^{-1}\left[\frac{C}{\cos \left(\pi x/a\right)}\right]$

Explanation of Solution

Given:

Width = a

Hight = b

   $\begin{array}{l}{E}_{xs}=j\omega \mu \frac{k_p}{k_{mp}{}^2}A\cos \left(k_{m}x\right)\sin \left(k_{p}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\\ E_{ys}=-j\omega \mu \frac{k_p}{k_{mp}{}^2}A\sin \left(k_{m}x\right)\cos \left(k_{p}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\end{array}$

Mode = $T{E}_{11}$

   $\begin{array}{l}T{E}_{11}=T{E}_{mp}\\ m=1\\ p=1\end{array}$

   $\begin{array}{l}0\le x\le a\\ 0\le y\le b\end{array}$

Concept used:

Formula to calculate equation for the streamlines is shown below:

   $\frac{dy}{dx}=\frac{E_y}{E_x}$

Calculation:

Although,

   $\begin{array}{l}{E}_{xs}=j\omega \mu \frac{k_p}{k_{mp}{}^2}A\cos \left(k_{m}x\right)\sin \left(k_{p}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\\ E_{ys}=-j\omega \mu \frac{k_p}{k_{mp}{}^2}A\sin \left(k_{m}x\right)\cos \left(k_{p}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\end{array}$

   $\begin{array}{l}\because k_m=\frac{m\pi }{a}\\ k_p=\frac{p\pi }{b}\end{array}$

Therefore,

   $\begin{array}{l}{k}_m=\frac{\pi }{a}\\ k_p=\frac{\pi }{b}\end{array}$

Plugging value of k_m and k_p in transverse electric field equation,

   $\begin{array}{l}{E}_{xs}=j\omega \mu \frac{k_p}{k_{mp}{}^2}A\cos \left(\frac{\pi }{a}x\right)\sin \left(\frac{\pi }{b}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\\ E_{ys}=-j\omega \mu \frac{k_m}{k_{mp}{}^2}A\sin \left(\frac{\pi }{a}x\right)\cos \left(\frac{\pi }{b}y\right)e^{-j{\beta }_{mp}z}\text{V/m}\end{array}$

Since,

   $\frac{dy}{dx}=\frac{E_y}{E_x}$

Hence,

   $\begin{array}{l}\frac{dy}{dx}=\frac{j\omega \mu \frac{k_p}{k_{mp}{}^2}A\sin \left(\frac{\pi }{a}x\right)\cos \left(\frac{\pi }{b}y\right)e^{-j{\beta }_{mp}z}}{-j\omega \mu \frac{k_p}{k_{mp}{}^2}A\cos \left(\frac{\pi }{a}x\right)\sin \left(\frac{\pi }{b}y\right)e^{-j{\beta }_{mp}z}}\\ \frac{dy}{dx}=\frac{b\sin \left(\frac{\pi }{a}x\right)\cos \left(\frac{\pi }{b}y\right)}{-a\cos \left(\frac{\pi }{a}x\right)\sin \left(\frac{\pi }{b}y\right)}\\ \frac{dy}{dx}=-\frac{b\tan \left(\frac{\pi }{a}x\right)}{a\tan \left(\frac{\pi }{b}y\right)}\end{array}$

   $\begin{array}{l}\tan \left(\frac{\pi }{b}y\right)dy=-\frac{b}{a}\tan \left(\frac{\pi }{a}x\right)dx\\ \text{Integrating}\text{both}\text{sides}\text{with}\text{limits,}\\ {\displaystyle \int \tan \left(\frac{\pi }{b}y\right)dy}=-\frac{b}{a}{\displaystyle \int \tan \left(\frac{\pi }{a}x\right)dx}\\ \left[-\frac{b}{\pi }\ln \cos \frac{\pi }{b}y+c_1\right]=-\frac{b}{a}\left[\frac{a}{\pi }\ln \cos \frac{\pi }{a}x+c_2\right]\end{array}$

For boundary condition of x and y final equivalent of above integral is shown below:

   $\begin{array}{l}\cos \left(\frac{y\pi }{b}\right)\cdot \cos \left(\frac{x\pi }{a}\right)=C\left(C=\text{constant}\right)\\ y=\frac{b}{\pi }{\cos }^{-1}\left[\frac{C}{\cos \left(\pi x/a\right)}\right]\end{array}$

Conclusion:

Hence,the streamlines for transverse electric field $E_{xs}$ and $E_{ys}$ for $T{E}_{11}$ mode in a rectangular is

   $y=\frac{b}{\pi }{\cos }^{-1}\left[\frac{C}{\cos \left(\pi x/a\right)}\right]$