I have designed one cavity to excite for the first mode TE110 at 2.9 GHz. When i have etched the circular ring slot of radius 18 mm on the top plane the resonant frequency is shifted at 5.1 GHz. While increasing the radius the resonant frequency shifts towards higher side.
If radius of ring slot is increased, capacitance among elements on top metallic layer will decrease. Consequently, resonance frequency will shift towards higher side.
Well,if you understand its physics and identify its equivalent circuit in R,L,C form then your query will be solved.
The slot size determines the coupling so a minor change in its size would result in non optimum excitation due to impedance mismatch. Impedance mismatch means that there there has been a phase shift in impedance from its zero value at resonance. From the equation Frequency x Time = Phase, frequency is expected to change as phase changes.
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