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An optical signal at a specific wavelength has lost 55% of its power after traversing 10 km of fiber. The attenuation will be _____ dB/km of this fiber

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\begin{array}{l} P_{R}=P_{T} (1-L o s s) \\ \frac{P_{R}}{P_{T}}=(1-\operatorname{Loss})----(1) \\ \text {Loss}_{d B}=10\log \left(\frac{P_{R}}{P_{T}}\right)----(2) \end{array}

\text{Substituting Eq. (1) into Eq. (2) we get the following;}

\begin{array}{l} \text {Loss}_{d B}=10\log (1-\operatorname{Loss})----(3) \\ \text {Attenuation}=\frac{{Loss}_{d B}}{\text {Distance}}----(4) \end{array}

\text{Substituting Eq. (3) into Eq. (4) we get the following;}

\begin{aligned} \text {Attenuation} &=\frac{10\log (1-\operatorname{Loss})}{\text {Dis } \tan \mathrm{ce}} \\ \text {Attenuation} &=\frac{10\log (1-0.55)}{10} \\ \text {Attenuatio} n &=-0.3467 \ \mathrm{dB} / \mathrm{km} \end{aligned}

\\\text{since the name Attenuation implies loss then we could take out the negative sign:} \\ \text{Attenuation }=0.3467 \ \mathrm{dB} / \mathrm{km}

 

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