Discussion
Why is a patch's resonant frequency lower than a simple half wavelength calculation predicts?
Answer: D
The fields extend a little beyond each end, so the effective length exceeds the physical one by roughly a substrate thickness at each edge and the resonance falls accordingly. Standard formulas add this length extension explicitly. Dielectric loss reduces the quality factor rather than the frequency, the ground plane is part of the resonator rather than an added inductance, and feed loading is a smaller secondary effect. Any resonator whose field escapes at its ends behaves as though it were longer than it is.
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