Quality factor (Q) of a waveguide resonator — correct definition in terms of stored energy and power loss Given ω0 as the resonant angular frequency, U as the time-average stored energy, and W_L as the average power loss, choose the correct expression for Q.

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:

  • Resonator operates in steady state at its resonance frequency ω0.
  • U is the average energy stored in electric and magnetic fields.
  • W_L is the average power lost due to wall loss, dielectric loss, radiation, etc.


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:

1) Start from Q = 2π * (Energy stored / Energy lost in one cycle).2) Energy lost in one cycle = W_L * T, with T = 2π / ω0.3) Substitute: Q = 2π * U / (W_L * 2π / ω0) = ω0 * U / W_L.4) Therefore, Q = ω0 * U / W_L.


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:

  • 2π * U / W_L: missing ω0; not dimensionally consistent with the standard steady-state definition unless implicitly paired with f0.
  • ω0 * U * W_L: multiplies instead of dividing by loss; incorrect units.
  • W_L / (ω0 * U): inverse of the correct relation.


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

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