Q factor of a resonator

Sacademy August 11, 2023 1 min read

Q factor of a resonator

Q = ω_r × (Energy stored in a cavity resonator / Average energy loss)

Here ω_r is angular resonance frequency of a resonator.
The stored energy (U_stored) in a resonator is obtained by integrating energy density over the volume of the resonator.

$U_{stored}=\frac{1}{2}\int_{V}^{}E^{2}dV=\frac{1}{2}\mu\int_{V}^{}H^{2}dV$

The energy loss (U_lost) is obtained by integrating the energy density over the surface of resonator.

$U_{lost}=\frac{1}{2}R_{S}\int_{S}^{}|H_{t}|^{2}dS$

$And \;\;\;\;\; U_{stored}=\frac{1}{2}\mu\int_{V}^{}H^{2}dV$

$\therefore\;\;\;\;\;Q=\omega _{r}\times\frac{\frac{1}{2}\mu\int_{V}^{}H^{2}dV}{\frac{1}{2}R_{S}\int_{S}^{}|H_{t}|^{2}dS}$

$or\;\;\;\;\;Q=\frac{\omega _{r}\mu\int_{V}^{}H^{2}dV}{R_{S}\int_{S}^{}|H_{t}|^{2}dS}$

Since the peak value of the magnetic field intensity is related to its tangential and normal component and it is given by

H² = |H_t|² + |H_n|²

|H_t|² ≈ 2 |H|² average over the volume

$\therefore \;\;\;\;\;Q=\frac{\omega _{r}\mu\int_{V}^{}H^{2}dV}{R_{S}\int_{S}^{}2|H|^{2}dS}$

or Q = (ω_rμ × volume) / (2 R_s × surface area)

$\therefore \;\;\;\;\;Q=\frac{\omega^{2}\mu^{2}}{4 R_{S}\pi^{2}}a^{3}\times\frac{1}{3}$

$or\;\;\;\;\;Q=\frac{1}{12}\frac{\omega^{2}\mu^{2}a^{3}}{R_{S}\pi^{2}}$

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