Question 6.8: SHAPE PERTURBATION OF A RECTANGULAR CAVITY A thin screw of r...
SHAPE PERTURBATION OF A RECTANGULAR CAVITY A thin screw of radius\ r_{0} extends a distance \ \ell through the center of the top wall of a rectangular cavity operating in the \ TE_{101} mode, as shown in Figure 6.26. If the cavity is air filled, use
(6.107)\ \frac{\omega -\omega _{0}}{\omega _{0}} \simeq \frac{\int_{\Delta V }^{}{\left(\mu \left|\bar{H_{0}} \right|^{2}-\epsilon \left|\bar{E_{0}} \right|^{2} \right)}dv }{ \int_{ V_{0} }^{}{\left(\mu \left|\bar{H_{0}} \right|^{2}+\epsilon \left|\bar{E_{0}} \right|^{2} \right)}dv}
to derive an expression for the change in resonant frequency from the unperturbed cavity.
From(6.42a)\ E_{y}=E_{0}\sin \frac{\pi x}{a} \sin \frac{\ell \pi z}{d}–
(6.42c)\ H_{z}=\frac{j\pi E_{0}}{k\eta a} \cos \frac{\pi x}{a} \sin \frac{\ell \pi z}{d}
\ E_{y}=A\sin \frac{\pi x}{a} \sin \frac{\pi z}{d}\ H_{x}=\frac{-jA}{Z_{TE}} \sin \frac{\pi x}{a} \cos \frac{\pi z}{d}
\ H_{z}=\frac{j\pi A}{k\eta a} \cos \frac{\pi x}{a} \sin \frac{ \pi z}{d}
If the screw is thin, we can assume that the fields are constant over the cross
section of the screw and can be represented by the fields at x = a/2, z = d/2:
\ H_{x}\left(x=\frac{a}{2},y,z=\frac{d}{2} \right) =0
\ H_{z}\left(x=\frac{a}{2},y,z=\frac{d}{2} \right) =0
Then the numerator of (6.107) can be evaluated as
where\ \Delta V=\pi \ell r^{2}_{0} is the volume of the screw. The denominator of (6.107) is,from (6.43),
\ \int_{\Delta V}^{}{\left(\mu \left|\bar{H_{0}} \right|^{2}-\epsilon \left|\bar{E_{0} } \right| ^{2} \right) } dv=\frac{adb\epsilon _{0}A^{2}}{2} =\frac{V_{0}\epsilon _{0}A^{2}}{2} ,where \ V_{0}=adb is the volume of the unperturbed cavity. Then (6.107) gives
\ \frac{\omega -\omega _{0}}{\omega _{0}} =\frac{-2\ell \pi r^{2}_{0} }{adb} =\frac{-2\Delta V}{V_{0}},
which indicates a lowering of the resonant frequency.