OTN Amplifier Nominal Input Power for 10 of 80 Wavelengths
Short answer
D Calculate the 10-wavelength nominal amplifier input as 4 - 10log80 + 10log10, giving about -5.0 dBm.
Question
Key Takeaways
It tests dB power summation across wavelengths and the trap of applying 10log(N) in the wrong direction or using the wrong endpoint of the -16 to +4 dBm range.
This OTN optical-power question tests converting an amplifier's full 80-wavelength input range to a 10-wavelength multiplexed value using 10log(N) scaling. The correct expression is D: 4 - 10log80 + 10log10.
Common mistake: The most plausible wrong answer is C, which starts from 4 dBm but adds 10log80 while subtracting 10log10, incorrectly predicting higher total power when only 10 rather than 80 wavelengths are present.
Expert Analysis
What the range represents
The amplifier specification of -16 dBm to 4 dBm is a total input-power window for the fully loaded 80-wavelength OTN configuration, not a per-wavelength value. Because the question asks for the nominal multiplexed-wavelength input power when only 10 wavelengths pass, the 80-channel total must be normalized to one channel and then re-scaled to ten channels. Noise is explicitly excluded, so the calculation is pure passive power summation.
The dB scaling step
For equal-power WDM channels, total power is P_total = P_channel + 10log(N). Taking the upper/nominal endpoint, 4 dBm for N=80, gives P_channel = 4 - 10log80 dBm. For N=10, P_total(10) = P_channel + 10log10 = 4 - 10log80 + 10log10 dBm. Numerically that is 4 - 19.03 + 10, approximately -5.03 dBm. The lower end of the range would similarly become -16 - 10log80 + 10log10, approximately -25.03 dBm, but the options are built around the 4 dBm nominal endpoint.
Eliminating the options
Option A uses 16 and adds 10log80 while subtracting 10log10, which would raise power when the wavelength count falls; that is the opposite direction. Option B uses 16 instead of 4 and therefore starts from the wrong endpoint. Option C has the right 4 dBm endpoint but applies the scaling backwards, predicting about 13 dBm for 10 wavelengths, higher than the full 80-wavelength limit. Option D preserves the correct order: subtract the full-configuration channel-count term, then add the active-channel term. D is the only expression consistent with both the 80-to-10 conversion and the stated nominal upper limit.
Exam Strategy
For optical-power scaling questions, first decide whether the stated dBm value is per wavelength or total multiplexed power. Convert to per-channel power by subtracting 10log(N_full), then rebuild the total for the active count by adding 10log(N_active). Keep signs straight: fewer active wavelengths must lower total power if per-channel power is unchanged. Use the endpoint identified as nominal, and reject options that increase total power when N drops.
Frequently Asked Questions
Why does the correct expression start from 4 dBm rather than -16 dBm?
Why subtract 10log80 and then add 10log10?
Does ignoring noise affect the power calculation?
Community Votes
100% of anonymous learners picked answer D. Votes are pick records left by other test-takers — they are not the verified answer.
Discussion 1 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 1
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ethassan 👍 1 Answered DThe correct formula for the nominal multiplexed-wavelength input optical power of the amplifier with 10 active channels is 4 - 10log80 + 10log10
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