Difficulty: Easy
Correct Answer: Q = ω0 * U / W_L
Explanation:
Introduction / Context:
The quality factor Q of a resonator quantifies how underdamped it is and how narrow its bandwidth is relative to its center frequency. For microwave cavities and waveguide resonators, Q directly relates to energy storage and dissipation, influencing selectivity, insertion loss, and phase noise in filters and oscillators. This item asks for the correct formula in terms of ω0, stored energy U, and power loss W_L.
Given Data / Assumptions:
Concept / Approach:
The standard definition for unloaded Q is Q = ω0 * (Energy stored per cycle / Energy lost per cycle). Since power is energy per unit time, and at steady state the rate of energy loss equals W_L, the expression simplifies to Q = ω0 * U / W_L. This connects directly to measurable quantities and underlies relations such as Q = f0 / Δf for a lightly loaded, weakly coupled resonator measured on a network analyzer.
Step-by-Step Solution:
Verification / Alternative check:
This form yields the familiar approximate relation Q ≈ f0 / (3 dB bandwidth) for lightly damped resonators when linked with coupling definitions and measured S-parameters.
Why Other Options Are Wrong:
Common Pitfalls:
Forgetting that W_L is power (J/s) while U is energy (J); mixing frequency f0 with angular frequency ω0.
Final Answer:
Q = ω0 * U / W_L
Discussion & Comments