DQ cannot be specified without knowing the cavity material and geometry
The relation Δω = ω₀/Q (or equivalently Δf = f₀/Q) connects bandwidth directly to Q. With f₀ = 10 GHz and Δf = 1 MHz, we need Q = f₀/Δf = 10×10⁹ / 1×10⁶ = 10,000. This illustrates how Q determines frequency selectivity: a higher Q means a narrower passband. The material and geometry determine whether Q = 10,000 is achievable, but the specification itself follows purely from the bandwidth requirement.
Question 2 Multiple Choice
What is the primary physical mechanism responsible for energy loss in a metal cavity resonator operating at microwave frequencies?
ARadiation leakage through imperfect seams in the cavity walls
BDielectric losses in the vacuum filling the cavity interior
COhmic dissipation as resonant-mode currents flow within the skin depth of the conducting walls
DThermal blackbody radiation from the heated cavity surfaces
The magnetic field of the resonant mode penetrates the conducting walls only to a depth equal to the skin depth δ = √(2/μσω). The oscillating field in this thin layer drives surface currents, which dissipate energy via Joule heating (I²R loss). Better conductors have thinner skin depths and smaller resistances, yielding less power loss per cycle and therefore higher Q. Dielectric losses are relevant when the cavity is not vacuum-filled, but for air- or vacuum-filled metal cavities, ohmic wall losses dominate.
Question 3 True / False
A cavity resonator with a higher Q factor produces a narrower resonance peak and responds efficiently only to signals within a correspondingly smaller frequency band.
TTrue
FFalse
Answer: True
This follows directly from Δω = ω₀/Q: the half-power bandwidth is inversely proportional to Q. A high-Q resonator stores energy efficiently for many oscillation cycles before dissipating it, which in the frequency domain corresponds to a sharp, narrow response centered at ω₀. This is why high-Q cavities are used as frequency discriminators in filters and oscillators — their narrow response means they can distinguish closely spaced frequencies.
Question 4 True / False
Increasing the electrical conductivity of a cavity's walls reduces its Q factor, because current flows more easily and dissipates energy faster.
TTrue
FFalse
Answer: False
This reasoning inverts the correct relationship. Higher conductivity reduces the skin depth (δ ∝ 1/√σ) and reduces wall resistance. Less resistance means less power dissipated per cycle by the surface currents, so Q = ω₀U/P_loss increases. The limiting case is a superconducting cavity (σ → ∞), where resistive wall losses approach zero and Q values of 10¹⁰ are achievable — roughly a million times higher than copper cavities.
Question 5 Short Answer
A superconducting cavity achieves Q ~ 10¹⁰ compared to Q ~ 10⁴ for a copper cavity at the same resonant frequency. What physical change accounts for this difference, and what engineering applications does it enable?
Think about your answer, then reveal below.
Model answer: In a normal metal, the skin depth δ = √(2/μσω) is small but nonzero, and the surface resistance R_s ∝ √(ω/σ) causes ohmic dissipation as resonant-mode currents flow in the walls. In a superconductor below its critical temperature, the DC resistance vanishes and the surface resistance drops by a factor of roughly 10⁶ compared to copper. This nearly eliminates wall loss, driving Q from ~10⁴ to ~10¹⁰. The resulting ultra-narrow bandwidth (Δf = f₀/Q ~ Hz at GHz frequencies) enables: (1) extremely low-phase-noise frequency standards and atomic clock cavities, (2) superconducting radio-frequency (SRF) cavities in particle accelerators that store enormous electromagnetic energy with minimal losses, and (3) cryogenic microwave resonators for quantum computing experiments requiring ultra-high coherence times.
The gain in Q comes entirely from eliminating ohmic loss — the stored energy U is nearly the same, but P_loss is reduced by ~10⁶. The tradeoff is the cryogenic infrastructure required to maintain superconductivity.