How does a signal that reads 20 dB over S9 compare to one that reads S9 on a receiver, assuming a properly calibrated S meter?

Explication

Decibels of power relate to a ratio by dB = 10 × log₁₀(P₂/P₁), so 20 dB corresponds to a power ratio of 10^20/10 = 10² = 100. A reading of 20 dB over S9 therefore represents a signal 100 times more powerful than one reading S9.