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If the intensity is increased by a factor of `20`, by how many decibels is the intensity level increased. |
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Answer» Let the initial intensity be `I` and the density level be `beta_(1)` and when the intensity is increased by `= 20` times, the intensity level increases to `beta_(2)`. Then `beta_(1) = 10 log(I//I_(0))` and `beta_(2) = 10 log (20I//I_(0))` Thus, `beta_(2) - beta_(1) = 10 log (20I//I)` `= 10 log 20` `= 13 dB`. |
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