Using the integral method to measure the OSNR value of a 100G signal, the integrated value of the signal power + noise power in the channel interval read from the OSA is 9.68uW, the integrated value of the noise power in the channel interval is 29.58nW, and the signal center frequency is 0.1nm. The integrated power within is 7.395nW, then the OSNR value is?
A . 10log [(9.68-0.02958)/0.007395]
B. 10log [(9.68+0.02958)/0.007395]
C. 10log [(9.68-0.007395)/0.02958]
D. 10log [(9.68+0.007395)/0.02958]
Answer: A
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