@David 3 dB of optical loss is equivalent to 6 dB of electrical loss.
With optical the power is directly proportional to the current in the led or laser (the voltage remains almost constant), and the same happens in the photodetector at the receiver in that the current is directly related to the incoming optical power. Therefore half the optical power (-3dB) results in half the current which results in half the voltage at the output of the transimedance stage.
But electrically half the voltage and current is -6dB since electrical power is related to the square of the voltage or current. With optical there is no square in the relationship, the power is linearly related to the current.
I used to get a chuckle years ago when all the FDDI hype was for that newfangled optical fiber and the magazine articles compared fiber to copper and breathlessly announced "...only 3dB per kilometer..." when in reality it was 6dB/Km once converted back into electrical terms.
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