- Introduction to Optical Design
- Factors That Affect System Design
- Effect of Chromatic Dispersion on Transmission Length and Induced Power Penalty
- Design of a Point-to-Point Link Based on Q-Factor and OSNR
- Calculation of Q-Factor from OSNR
- Margin Requirements
- Design Using Chromatic Dispersion Compensation
- OSNR and Dispersion-Based Design
- Frequency Chirp
- Effects of FWM and XPM on Long-Haul Design
- PMD in Long-Haul Design
- Examples
- Summary
- References
Examples
The following case studies reinforce the principles discussed so far in this chapter.
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 Figure 4-16.
Figure 4-16 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:
OSNR_{stagei }= 1/(1/OSNR_{stage0} + NF.h.v.Δf /P_{in}) (1/OSNR_{stage0} = 0)
OSNR Stage 1
NF = 5 dB converting to linear = 3.166 (10^{NF 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)
P_{in} = input power at the amplifier
0 – 25 dB = –25 dB
OSNR_{stage1} = 28 dB
Output from the amplifier = –25 + 22 = –3 dB
OSNR Stage 2
OSNR_{stage2 }= 1/(1/OSNR_{stage1} + NF.h.v.Δf /P_{in})
Input at the second amplifier
= –3 – 25 = –28 dB
OSNR_{stage2} = 23 dB
Output from the amplifier = –28 + 22 = –6
OSNR Stage 3
OSNR_{stage3 }=1/(1/OSNR_{stage3} + NF.h.v.Δf /P_{in})
Input at the third amplifier = –6 – 25 = –31 dB
OSNR_{stage3} = 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 4-16.
Using Method 1: OSNR_{final }= 10 + 58 – 25 – 6 – 5 = 32
Method 2: gives OSNR_{final} = 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.
Case 2
Calculate the number of spans in this link, given P_{in} = 0 dB; OSNR_{final} = 20 dB total length = 300 km Bit rate = 5 Gbps; NF = 5 dB. (Assume fiber type is SMF ∝ = .2 dB/km)
Answer
Total loss over entire length = 300 × 0.2 = 60 dB
Attenuation per span = 60/N (assume N number of spans
OSNR_{Final} = P_{0} + 58 – –10logN – NF
20 = 0 + 58 – 60/N –10logN – 5
-33 = –60/N – 10logN
Rearranging, you get this:
60/N – 10logN = 33
Solving for N (for N = 2)
60/2 + 10log2 = 33
Therefore, the number of spans = 2.
Case 3
OSNR = 20 dB, dispersion of the fiber is 17 ps/nm-km, span loss = 22 dB.
What is the total length of the system? (NF of the amplifier = 4 dB, and dispersion tolerance is given as 1600 ps/nm, P_{in} = 10)
Answer
OSNR_{Final} = P0 + 58 – –10logN – NF
20 = 10 + 58 – 22 – 4 – 10logN
–22 = –10 logN
N = 158 spans
Therefore, total length = 158 * 22/0.2 = 17,280 km (theoretical limit)
But due to dispersion, max length = 1600 ps.nm/17 ps /nm.km = 94 km.
Case 4
Customer A wants to build a 200 km OC48 link to transport traffic. Design the link with the following parameters. (Assume SMF-28 fiber with α = .25 dB/km/ and 18 ps/nm/km as the dispersion characteristic.)
Receiver sensitivity: –18 dBm @BER=e^{-12}
Receiver overload: –10 dBm @BER=e^{-12}
Transmit power: +7 to +9 dBm
Dispersion tolerance: 1500 ps/nm
Dispersion penalty: 1.5 dB @ 1500 ps/nm
OSNR tolerance: 20 dB @ Resolution Bandwidth 0.1 nm
EDFA
Input power range: +3 to –25 dBm; Gain: 20 dB to 14 dB
Maximum output power: +17 dBm; Noise Figure: 5 dB
DCU
Loss 5 dB, dispersion compensation –1100 ps
Answer
The answer requires several steps. Refer to Figure 4-17.
Figure 4-17 Case 4 Answer
Step 1 |
The total distance is 200 km; therefore, total loss is 200 * .25 = 50 dB. You need amplifiers to reach that distances. |
Step 2 |
Total dispersion is 200 km * 18 ps.nm / km = 3600 ps.nm |
You need to use dispersion compensators because the given dispersion tolerance is 1500 ps.nm. The maximum distance for this system without DCU is 1500/18 = 83.33 km given DCU = –1100 ps. If you strategically place 3 DCU, you get (3600 – 3 * 1100) = 300 ps, which is well within the limit. The DCU has a 5 dB passthrough loss, so it is best to place it before preamplifiers.
Analysis of the problem
Transmit power = + 7 dB
Stage 1:
loss = –10 (link loss) – 6 (DCU loss) – 1.5 (margin) = –17.5 dB
P_{in} = Power at the end of stage 1 = 7 – 17.5 = –10.5 dB
Stage 1 amplifier gain: 20
Power output from the amplifier: (power input + gain) – 10.5 dB + 20 = 9.5 dB
OSNR calculation
OSNR_{stagei }=1/(1/OSNR_{0} + NF.h.v.Δf /P_{in}) (for stage 1 1/OSNR_{0} = 0)
OSNR_{1 = }Pin/NF.h.v. Δf
Substituting the values Pin = –10.5 dB = 8.9125E – 05 W (convert to watts)
NF = 5 dB converting to linear = 3.166 (10^{NF dB/10})
h = planks constant = 6.6260E – 34
v = frequency of light 1.9350E + 14
Δf = bandwidth (measuring the NF) = 12.5 KHz (.1 nm)
Substituting, you get OSNR = 42 dB
Stage 2:
Loss = –20 (link loss – 6 (DCU loss) – 1.5 (margin) = –27.5 dB
Power at the end of stage 2
P_{in }= 9.5 – 20 – 6 – 1.5 (margin) = –18 dB
Stage 2 amplifier gain: 20
Power output from amplifier: –18 + 20 = 2 dB
OSNR_{stage2} = 1/(1/OSNR_{1} + NF.h.v.Δf /P_{in})
OSNR1 is the OSNR of stage 1 = 42 dB and P_{in} = –18 dB
OSNR_{stage 2 }= 34 dB
Stage 3 loss:
–20(link loss) – 6 (DCU loss) – 1.5 (margin) = –27.5 dB (loss)
Power at the end of stage 3
P_{in}= 2 – 27.5 = –25.5 dB
Stage 3 amplifier gain 20 dB
Power output from the amplifier: = –25.5 + 20 = –5.5 dB
OSNR
OSNR_{stage3} = 1/(1/OSNR_{stage2} + NF.h.v.Df /P_{in})
OSNR = 26
Case 5
Calculate the composite power at the output of a DWDM 8 channel multiplexor (shown in Figure 4-18) if the input power is 0 dB (insertion loss = 5 dB).
Figure 4-18 Calculate the Composite Power
P_{composite} = P_{channel} + 10logN – Insertion loss (where N is the number of channels)
= 0 + 10log8 – 5 dB = 0 + 9 – 5 dB
= 4 dB
Exercises
The following exercises are left for you to solve. Use previous examples as guidelines.
Calculate the dispersion and the receiver power of an optical link that is 90 km in length. The transmit power = +8 dB, the receiver sensitivity = –18 dB, and the dispersion tolerance = 1500 ps/nm (SMF ∝ = .25 dB/km 18 ps/nm-km). Is it dispersion limited? Design the link for proper network operation if you need to use EDFA 15 dB (gain) and DCU –1100 ps and 5 dB loss).
Design a 4 × 25 span WDM link with optical amplifier gain = 18 dB and NF = 6 dB.
Calculate power in milliwatts if the input power were 0 dBm.
Calculate the composite power at the output of a DWDM 16-channel mux if the input power is 4 dB. If the input is plugged into an amplifier, how much attenuation do you need (given NF = 5 dB, input range = 0–25 dB, gain = 22)
Calculate final OSNR if the input power is 0 dB. Calculate signal power at the receiver.
(The receiver sensitivity is –25 dB at BER 10^{-15}). Does the system work?
Now calculate the final OSNR by replacing the following:
— The transmitter with input power to +7 dB
— The amplifier gain to 22 dB
Calculate the power in dBm if the input power were 12 mW.
Hint: XmW power is 10 * log_{10}(x) in dBm_{ }
Y dBm power is 10^{Y/10} in mW