We will work with power ratios in this class. If you are confronted with an ampl

Question

We will work with power ratios in this class. If you are confronted with an amplitude ratio (e.g., the ratio of an output rms voltage to an input rms voltage), convert it first to a power ratio by an appropriate squaring operation. Example: Given transfer function H(omega), first evaluate and then take the log. Basic definition: Given power ratio P/P_0, we express it in dB using the formula Power ratio in dB = 10 log(P/P_0) Thus, if an amplifier increases the power from 4 mW to 30 W, the power ratio P/P_0 in dB equals 10 log(30/4 times 10^-3) = 10 log 7500 = 10 times 3.88 = 38.8 dB. (a) Confirm by direct calculation that a two-fold power amplification corresponds to a good approximation to 3 dB of gain. (b) The attenuation of a particular class of optical fiber is approximately 1 dB per km of fiber when used at the prescribed wavelength. Assume 1 mW of light power is input to the fiber. What is the output light power at the end of a 1 km fiber link? (d) What is the approximate output power of the link if the fiber is 2 km long?

Explanation / Answer

(a) P=2P0

power ratio in dB=10log(2P0/P0)=10log(2)=3 dB

So two-fold power amplification corresponds to a good approximation to 3 dB of gain.

(b) 10log(10-3/Po)=1

that is 10-3/Po=100.1=1.259

Po=10-3/1.259=0.794mW

(d) Given that the attenuation loss is 1 dB/km

For 2 km long fiber the attenuation loss is=2 dB/km

So 10log(10-3/Po)=2

that is 10-3/Po=100.1=1.585

Po=10-3/1.585=0.794mW=0.631 mW

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