Question 3-64J5

Harmonic attenuation is calculated by comparing the power of the fundamental signal to the power of the harmonic, expressed in decibels (dB). The formula for attenuation in dB is $10 \log_{10} (P_{initial} / P_{final})$, where $P_{initial}$ is the reference power (in this case, the fundamental signal's power) and $P_{final}$ is the power of the attenuated harmonic. Given: Primary signal output ($P_{initial}$) = 500 watts 2nd harmonic output ($P_{final}$) = 0.5 watt Attenuation = $10 \log_{10} (500 \text{ W} / 0.5 \text{ W})$ Attenuation = $10 \log_{10} (1000)$ Since $\log_{10} (1000) = 3$, Attenuation = $10 \times 3 = 30 \text{ dB}$. This means the 2nd harmonic is 30 dB below the primary signal level. Therefore, option B is correct. Options A, C, and D would correspond to different power ratios, not the 1000:1 ratio observed here.