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The Q-factor of a waveguide resonator is given by (ω0 is resonant frequency, U is energy storage and ωL is power loss)1. \(Q = \frac{{{\omega _0}U}}{{{\omega _L}}}\)2. \(Q = \frac{{{\omega _0}{\omega _L}}}{U}\)3. Q = ω0UωL4. \(Q = \frac{{U{\omega _L}}}{{{\omega _0}}}\) |
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Answer» Correct Answer - Option 1 : \(Q = \frac{{{\omega _0}U}}{{{\omega _L}}}\) Concept: The quality factor is in general a measure of the ability of a resonator to store energy in relation to time-average power dissipation. Specifically, the Q of a resonator is defined as: \(Q=2\pi\frac{Maximum \ energy \ stored}{Energy \ dissipated \ per \ cycle}\) In the case of resonators, the Quality factor is given by: \(Q=\frac{ω_0W}{P_l}\) ---(1) Where, ω0 = Resonant frequency W = Total Energy Pl = Power loss in the cavity. Calculation: Given: Pl = ωL ω0 = resonant frequency W = U Putting the above values in equation (1) we get the Quality factor as: \(Q = \frac{{{\omega _0}U}}{{{\omega _L}}}\) Hence option (1) is the correct answer.
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