DESIGN OF A RECTANGULAR CAVITY RESONATOR A rectangular waveguide cavity is made from a piece of copper WR-187 H-band waveguide, with a = 4.755 cm and b = 2.215 cm. The cavity is filled with polyethylene (εr = 2.25, tan δ = 0.0004). If resonance is to occur at f = 5 GHz, find the required length, d,

Question

DESIGN OF A RECTANGULAR CAVITY RESONATOR

A rectangular waveguide cavity is made from a piece of copper WR-187 H-band waveguide, with a = 4.755 cm and b = 2.215 cm. The cavity is filled with polyethylene [latex]\ \left( \epsilon _{r}=2.25, an \delta =0.0004 ight)[/latex]. If resonance is to occur at f = 5 GHz, find the required length, d, and the resulting unloaded Q for the [latex]\ell= 1 [/latex] and [latex]\ell = 2[/latex] resonant modes.

Accepted Answer

The wave number k is

[latex]\ k=\frac{2\pi f\sqrt{\epsilon _{r}}}{c}=157.08m^{-1}. [/latex]

From (6.40) [latex]\ f_{mn\ell}=\frac{ck_{mn\ell}}{2\pi \sqrt{\mu _{r}\epsilon _{r}} }=\frac{c}{2\pi \sqrt{\mu _{r}\epsilon _{r}}} \sqrt{\left(\frac{m\pi }{a} ight)^{2}+\left(\frac{n\pi }{b} ight) ^{2}+\left(\frac{\ell\pi }{d} ight)^{2} } [/latex]the required length for resonance can be found as (m = 1, n = 0)

[latex]\ d=\frac{\ell\pi}{\sqrt{k^{2}-\left(\pi/a ight)^{2}}},[/latex]

for [latex]\ell= 1 [/latex],[latex]\ d=\frac{\pi}{\sqrt{\left(157.08 ight)^{2}-\left(\pi/0.04755 ight)^{2}}}=2.20cm[/latex]

for [latex]\ell= 2 [/latex],[latex]\ d=2\left(2.20 ight)=4.40cm.[/latex]

From Examples 6.1, the surface resistivity of copper at 5 GHz is[latex]\ R_{s}=1.84 imes 10^{-2}\Omega [/latex] . The intrinsic impedance is

[latex]\ \eta =\frac{377}{\sqrt{\epsilon_{r}} }=251.3\Omega [/latex]

for [latex]\ell= 1 [/latex],[latex]\ Q_{c} = 8,403,[/latex]

for [latex]\ell = 2[/latex],[latex]\ Q_{c} = 11,898.[/latex]

From (6.48) the Q due to dielectric loss only is, for both [latex]\ell = 1[/latex] and [latex]\ell= 2 [/latex],

[latex]\ Q_{d}=\frac{1}{ an \delta } =\frac{1}{0.0004}=2500.[/latex]

Then total unloaded Qs are, from (6.49)

for [latex]\ell = 1[/latex],[latex]\ Q_{0}=\left(\frac{1}{8403}+\frac{1}{2500} ight) ^{-1}=1927[/latex]

for [latex]\ell = 2[/latex],[latex]\ Q_{0}=\left(\frac{1}{11,898}+\frac{1}{2500} ight) ^{-1}=2065.[/latex]

Note that the dielectric loss has the dominant effect on the Q; higher Q could
be obtained using an air-filled cavity. These results can be compared to those
of Examples 6.1 and 6.2, which used similar types of materials at the same
frequency.

Question 6.3: DESIGN OF A RECTANGULAR CAVITY RESONATOR A rectangular waveg...

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