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FAQ-How to calculate OSNR value of a WDM link

Publication Date:  2012-07-25 Views:  66 Downloads:  1
Issue Description
Many of us having doubt what is OSNR and how we calculate it?

Alarm Information
None

Handling Process
First we will see what is OSNR?---OSNR is Optical Signal to Noise Ratio and just like any Telecom System, it is one of the most important parameters. When the signal is amplified by the Optical Amplifier, like EDFA, its OSNR is reduced, and this is the primary reason why the number of OAs is limited in a network. One mitigation is to use RAMAN Amplifier but it also has some intrinsic noise, though it is less than that of EDFA.
OSNR value is specified by the manufacturer just as is the Receive Sensitivity. The design value of OSNR is typically 20 dB or more. For measurement of OSNR different instruments are used and they use either In-Band or Out-of-Band techniques...

For Calculation of OSNR we will take one example.....



 
 
Case 1
Design a 4 ×25 span WDM link with an optical amplifier gain of 22 dB and NF equal to 5 dB.
Calculate the final OSNR if the input power is 0 dB. Calculate the signal power at the receiver.
Will this system work if the receiver sensitivity is a minimum of –25 dB?
Will the system work if the input power is 10 dB?
The answer is shown in below figure
Answer to the Problem
You can use one of the two methods to determine the final OSNR.

Method 1


Method 2

OSNR stage by stage analysis using the formula:
OSNRstagei = 1/(1/OSNRstage0 + NF.h.v.Δf /Pin) (1/OSNRstage0 = 0)
OSNR Stage 1
NF = 5 dB converting to linear = 3.166 (10NF dB/10)
h = Plank's constant = 6.6260E – 34
v = frequency of light 1.9350E + 14
Δf = bandwidth (measuring the NF) = 12.5 KHz (.1 nm)
                              Pin = input power at the amplifier
                      0 – 25 dB = –25 dB
             OSNRstage1 = 28 dB
     Output from the amplifier = –25 + 22 = –3 dB
OSNR Stage 2
OSNRstage2 = 1/(1/OSNRstage1 + NF.h.v.Δf /Pin)
Input at the second amplifier
= –3 – 25 = –28 dB
OSNRstage2 = 23 dB
Output from the amplifier = –28 + 22 = –6
OSNR Stage 3
OSNRstage3 =1/(1/OSNRstage3 + NF.h.v.Δf /Pin)
Input at the third amplifier = –6 – 25 = –31 dB
OSNRstage3 = 20 dB
Output from the amplifier = –31 + 22 = –9 dB
Power at the receiver
= –9 – 25 = –34 dB
If the receiver sensitivity is –25, the system will not work.
The solution is to (a) increase the gain of the amplifier (b) increase the input power of the transmitter.
The same solution for 10dB input power is shown in Figure
Using Method 1: OSNRfinal = 10 + 58 – 25 – 6 – 5 = 32
Method 2: gives OSNRfinal = 29 dB
The difference in value is due to the approximation made in the parameters.
Final power at the receiver
= 10(Tr) – 25 (loss1) + 22 (gain1) – 25 (loss2) + 22 (gain 2) – 25 (loss 3) + 22 (gain 3) – 25 (loss) = –24 dB
The receiver sensitivity is given as –25 dB, so the system will work.

 

 



Root Cause
None

Suggestions
None

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