# Half sphere in waveguide

So, the cutoff freq, $$f_c$$ of a rectangular waveguide is following:

$$f_{c}=\frac{1}{2\pi\sqrt{\mu \epsilon }}\sqrt{(\frac{\pi}{a})^{2}}.$$

For TE10 mode, the equation is a little weird as I am dropping some terms unnecessary for TE10 mode.

If I put a dielectric half sphere, with $$\epsilon$$ say 10, into the waveguide. It will decrease the cutoff frequency as this effectively decrease

$$\frac{1}{2\pi\sqrt{\mu \epsilon }}$$

part of the equation.

Now, what happen if I put a perfect electric conductor half sphere into the waveguide? It seems this also will decrease the cutoff frequency as this seems make

$$\frac{\pi}{a}$$

become larger.

But this is kinda counter-intuitive to me. What would happen to cutoff frequency if a PEC half sphere is put into a rectangular waveguide?

The textbook I am using only discuss the dielectric case but not conductor case.

Normally, when you restrict the size of a passage through which a wave can propagate you increase the cutoff frequency because you need "roughly" a wavelength's worth of free space to fit a wave inside. A metallic piece inside the waveguide restricts the wave because there is no propagation inside an ideal metal. This is unlike in a dielectric where the wavelength decreases proportionally to $$1/\sqrt{\epsilon}$$, and in fact you have now more space available relative to the propagating wavelength that would be without the dielectric, thus placing a dielectric slab along the guide will lower the cutoff frequency. This of course assumes that the length of the restriction you place in the waveguide is at least a wavelength's long otherwise there is no meaning to "propagation".
The effect of placing a hemispherical metal piece inside the waveguide would have no effect on the cutoff frequency because it is not long enough to have one. In other words, instead of changing the cutoff frequency, or the guide wavelength, the hemispherical metallic restriction will act as a *reactive* discontinuity, an inductive or capacitive, or a much more complicated circuit combination of inductive and capacitive elements. Which reactive term is dominant will depend on the size and placement of object on the wall (wide or narrow). For example a *small* hemispherical indentation in the middle of the wide wall is *capacitive* but on the narrow wall is *inductive* both assuming $$TE_{10}$$ mode propagation. If the mode, or the location, or the size of the metallic object is different then the equivalent reactive circuit is different too.