EMC Question of the Week: May 20, 2019
If the measured radiated emissions at a particular frequency is 44 dB(μV/m) and the limit is 38 dB(μV/m), then the product is over the limit by,
- 16%
- 200%
- 6 dB
- 6 dB(μV/m)
Answer
The correct answer is "c". 44 dB(μV/m) - 38 dB(μV/m) = 6 dB. 44 dB(μV/m) describes an electric field intensity that is 44 dB above 1 μV/m. 44 dB represents a field intensity ratio of about 158, therefore 44 dB(μV/m) is equivalent to approximately 158 μV/m. 38 dB(μV/m) is 79 μV/m. The product is over the limit by a factor of 158/79. Expressed in dB, this is 20*log(158/79) = 6 dB.
It is NOT correct to say the product is 6 dB(μV/m) over the limit. 6 dB(μV/m) is a field intensity, not a ratio. In fact, 6 dB(μV/m) corresponds to a field intensity of about 2 μV/m, which has nothing to do with the product's emissions or the limit.
16% and 200% are also clearly incorrect. If we wanted to express the relationship between the product emissions and the limit as a percentage, we might choose to say that the electric field intensity exceeds the limit by 100%. This can be confusing though. We have to make it clear that we are referring to the ratio of the field intensity and not the ratio of the power density. Also, while the emissions exceed the limit by 100%, the limit falls short of the emissions by 1 - 79/158 or 50%. Expressing ratios of electrical signal quantities in dB eliminates any confusion related to which quantity was the reference and whether we are referring to the signal amplitude or signal power.
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