Jackson Problem 8.19 (a)

The Cantenna - A Physics 632 Web Project

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Calculate the amplitudes for excitation of both TE and TM modes for all $ (m,n)$ and show how the amplitudes depend on $ m$ and $ n$ for $ m,n \gg 1$ for a fixed frequency $ \omega$.

For TE modes, we know that

$\displaystyle H_z$ $\displaystyle = H_0 \cos \left( \frac{m\pi x}{a} \right) \cos \left( \frac{n\pi y}{b} \right)$ (1)
$\displaystyle E_t$ $\displaystyle = \frac{\imath \mu_{0} \omega}{\gamma^2_{mn}} \hat{z} \times \vec{\nabla}_tH_z$ (2)
$\displaystyle \Rightarrow E_y$ $\displaystyle = -\frac{\imath \mu_{0} \omega}{\gamma^2_{mn}}\frac{m \pi}{a} H_0 \sin \left( \frac{m\pi x}{a} \right) \sin \left( \frac{n\pi y}{b} \right)$ (3)

where


$\displaystyle k^2_{mn}$ $\displaystyle = \frac{\omega^2}{c^2}-\gamma^2_{mn},$    and $\displaystyle \gamma^2_{mn} = \pi^2 \left( \frac{m^2}{a^2}+\frac{n^2}{b^2}\right)$    

and when normalized


$\displaystyle H_0$ $\displaystyle = \frac{2\imath \gamma_{mn}}{\mu_{0} \omega \sqrt{ab}}$    

except when $ m,n=0$ where $ a,b\rightarrow2a,2b$ as the constant mode averages to $ 1$, not $ 1/2$.

For TM modes, we know that

$\displaystyle E_z$ $\displaystyle = E_0 \sin \left( \frac{m\pi x}{a} \right) \sin \left( \frac{n\pi y}{b} \right)$ (4)
$\displaystyle E_t$ $\displaystyle = \frac{\imath k_{mn}}{\gamma^2_{mn}} \vec{\nabla}_tE_z$ (5)
$\displaystyle \Rightarrow E_y$ $\displaystyle = -\frac{\imath k_{mn}}{\gamma^2_{mn}}\frac{n \pi}{b} E_0 \sin \left( \frac{m\pi x}{a} \right) \sin \left( \frac{n\pi y}{b} \right)$ (6)

and when normalized


$\displaystyle E_0$ $\displaystyle = \frac{2\imath \gamma_{mn}}{k_{mn}\sqrt{ab}}$      

and as $ m,n \not= 0$

The excitation amplitudes are

$\displaystyle A^{(\pm)}_{mn}$ $\displaystyle = -\frac{Z_{mn}}{2}I_0\int_V \vec{J} \cdot \vec{E}^{(\mp)}_{mn} d^3x$ (7)

where $ Z_{mn} = k_{mn}/ \epsilon_0\omega$ for TM modes and $ Z_{mn} = \mu_0\omega / k_{mn}$ for TE modes. The source current density may be written as:


$\displaystyle \vec{J}$ $\displaystyle = \hat{y} I_0 \sin \left[ \frac{\omega}{c}(h-y)\right] \delta(x-X) \delta(z) \Theta(h-y)$ (8)

Hence


$\displaystyle A^{(\pm)}_{mn}$ $\displaystyle = -\frac{Z_{mn}}{2}I_0\int^h_0 \sin \left[ \frac{\omega}{c}(h-y))\right] E^{(\mp)}_{y,mn}(X,y,0)dy$ (9)

Taking the integral gives


$\displaystyle A^{(\pm)}_{mn}$ $\displaystyle = \beta_{mn}^{(TM,TE)} \sin \left( \frac{m\pi X}{a}\right) \left[... ...s\left[ \frac{n \pi h}{b} \right]- \cos \left[ \frac{\omega h}{c}\right]\right)$ (10)

where $ \beta^{TM(TE)}_{mn}$ is all the constants out front for the TM(TE) mode. For $ m,n \gg 1$ with fixed $ \omega$, we see


$\displaystyle A^{(\pm)}_{mn}$ $\displaystyle \sim \frac{1}{n}$    (TM) (11)

since at large m,n the wave number is imaginary as all TM modes are cutoff modes. Likewise


$\displaystyle A^{(\pm)}_{mn}$ $\displaystyle \sim \frac{m}{n^2} \frac{1}{(m/a)^2 + (n/b)^2} \sim \frac{1}{N^3}$    (TE) (12)

as all the modes are cutoff for $ m,n \gg 1$. The propagating $ TE_{10}$ mode has amplitude


$\displaystyle A^{(\pm)}_{10}$ $\displaystyle = \beta^{(TE)}_{10} \sin \left( \frac{\pi X}{a}\right) \left[1- \cos \left( \frac{\omega h}{c}\right) \right]$    
  $\displaystyle = \beta^{(TE)}_{10} \sin \left( \frac{\pi X}{a}\right) \sin^2 \left( \frac{\omega h}{2c}\right)$ (13)

where we have made sure to use the normalization appropriate for an $ n=0$ mode in $ \beta^{(TE)}_{10}$.

Solution to part (b)