What is the worst case acceptance value for attenuation when testing 2000 m (6562 ft) of 50/125 multimode fiber at a wavelength of 1300 nm when the fiber has 2 connector pairs and 3 splices?
What is the worst case acceptance value for attenuation when testing 2000 m (6562 ft) of 50/125 multimode fiber at a wavelength of 1300 nm when the fiber has 2 connector pairs and 3 splices?
To determine the worst-case acceptance value for attenuation, we need to consider the maximum cable attenuation, connector pairs, and splices. The typical maximum attenuation for 50/125 multimode fiber at 1300 nm is about 1.5 dB/km. For a 2000 m (2 km) length of fiber, the attenuation would be 2 km * 1.5 dB/km = 3.0 dB. Additionally, we have to account for the two connector pairs and three splices. Each connector pair typically adds 0.75 dB of loss, and each splice adds 0.3 dB. Therefore, the total loss from connectors is 2 * 0.75 dB = 1.5 dB, and the total loss from splices is 3 * 0.3 dB = 0.9 dB. Summing these values gives us a total worst-case attenuation of 3.0 dB (fiber attenuation) + 1.5 dB (connectors) + 0.9 dB (splices) = 5.4 dB. Thus, the worst-case acceptance value for attenuation is 5.4 dB.
Maximum cable attenuation coefficient for 50/125 multimode fiber at a wavelength of 1300 nm is 1.5 dB/km. So (Maximum channel attenuation – [# connector pairs x 0.75 dB] – [# splices x 0.3 dB])=-1.5dB/km x 2km- 2 x 0.75dB - 3 x 0.3 dB = -5.4dB.
Can someone help me with this one?
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