Jackson Problem 8.19 (c)

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Discuss the modifications that occur if the guide, instead of running off to infinity in both direction is terminated with a perfectly conducting surface at $ z = L$. what values of L will maximize the power flow for a fixed current $ I_0$? What is the radiation resistance of the probe (defined as the ration of power flow to one-half the square of the current at the base of the probe) at maximum?

Placing a perfectly conducting surface at $ z = L$ will cause the right moving wave to be perfectly reflected at this surface. Then the wave flowing out the left side will be a linear superposition of the left moving wave generated by the probe and the reflected wave. Since the parallel component to the electric field at the surface of a perfect conductor must vanish, the reflected wave will come back $ 180^0$ out of phase. We can write the left moving wave as

$\displaystyle \vec{E}^{(-)}$ $\displaystyle = A^{(-)}_{10}\vec{E}_{t,10} e^{-\imath kz}$ (16)

and it is easily seen that the reflected wave must be


$\displaystyle \vec{E}^{(refl)}$ $\displaystyle = -A^{(+)}_{10}\vec{E}_{t,10} e^{\imath k(2L-z)}$ (17)

so that at $ z = L$ there is no electric field. Therefore, for $ Z < 0$


$\displaystyle \vec{E}$ $\displaystyle = \vec{E}^{(-)} + \vec{E}^{(refl)} = A_{10}(1 - e^{2\imath k L})\vec{E}_{t,10} e^{-\imath kz}$ (18)

The maximum amplitude case occurs when there is constructive interference


$\displaystyle kL$ $\displaystyle = (n+ \ensuremath{\frac{1}{2}})\pi$ (19)

since the amplitude doubles, the power is then increased by a factor of four. For this maximum power case, the radiation resistance is given by


$\displaystyle P^{(\pm)}$ $\displaystyle = \frac{4\mu c^2 I^2}{\omega k a b} \sin^2 \left( \frac{\pi X}{a} \right) \sin^4 \left( \frac{\omega h}{2c} \right)$ (20)
  $\displaystyle = \ensuremath{\frac{1}{2}}I^2_0 R_{rad}$ (21)

where $ R_{rad}$ is the radiation resistance.

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