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EE 230: Optical Fiber Communication Lecture 17 From the movie Warriors of the Net System Considerations.

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Presentation on theme: "EE 230: Optical Fiber Communication Lecture 17 From the movie Warriors of the Net System Considerations."— Presentation transcript:

1 EE 230: Optical Fiber Communication Lecture 17 From the movie Warriors of the Net System Considerations

2 Basic Network Topologies

3 Bitrate Distance Graph for various point to point link technologies

4 System Design Determine wavelength, link distance, and bit-error rate Work out power budget Work out risetime budget Work out cost budget

5 Power Budget Steps Start with BER and bit rate, determine bandwidth B based on coding method Given BER, determine Q factor and S/N ratio B = 1/2  R L C gives the maximum load resistance R L based on B and C Based on R L and M, determine detector sensitivity (minimum power for S/N) Add system margin, typically 6 dB, to determine necessary power at receiver

6 Power budget steps, continued Add power penalties, if necessary, for extinction ratio, intensity noise (includes S/N degradation by amplifiers), timing jitter Add loss of fiber based on link distance Include loss contributions from connections and splices End up with required power of transmitter, or maximum length of fiber for a given transmitter power

7 Power budget example Imagine we want to set up a link operating at 1550 nm with a bit rate of 1 Gb/s using the RZ format and a BER of 10 -9. We want to use a PIN photodiode, which at this wavelength should be InGaAs. The R 0 for the diode is 0.9 A/W and its dark current is 4 nA.

8 Bandwidth required for bit rate For NRZ format, B=0.5 times bit rate For RZ format, B=bit rate For this example, the bandwidth B is equal to the bit rate, 10 9 /s.

9 Bandwidth limit C=2 pF for this photodiode. B = 1/2  R L C, so the load resistance R L must be (2  BC) -1 = 79.6 

10 Q Factor, S/N Ratio, and BER For our BER of 10 -9, Q=6 and S/N=144

11 Minimum signal power required Method 1 Signal to noise ratio is equal to where R 0 is responsivity, M is multiplier, i d is dark current, and P S is signal power. If we set this expression equal to the SNR of 144 we calculated, the required signal power P S turns out to be 6.09x10 -6 W = -22.2 dBm.

12 Noise Equivalent Power (NEP) Method 2 NEP is the signal power at which S/N=1 Units are W/Hz 1/2 In this case, M=1 and the dark current = 4 nA. The factor outside the radical is 1/R 0. We can thus determine the NEP by multiplying the above result by the square root of the bandwidth. The result is 5.1x10 -7 W, which equals -33.0 dBm.

13 NEP Method, continued Since Q is proportional to the square root of SNR, multiplying the NEP by that square root will give us the minimum acceptable power. In this case, SNR=144, so the minimum power is 12 times NEP = 6.12x10 -6 W = -22.1 dB, the same result as the other method.

14 Extinction ratio penalty Corrects for emission from transmitter during “0” bits Extinction ratio r ex =P 0 /P 1 If our extinction ratio is 0.1, the penalty is 0.87 dB.

15 Intensity noise penalty Corrects for power fluctuations during “1” bits r I =inverse of SNR of transmitter and amplifier output If the SNR of the amplified source is 13 dB = 20, then r I = 0.05 and we will have to raise the power at the receiver by δ I to keep the desired SNR. Since Q = 6, δ I = 0.41 dB.

16 Timing jitter penalty Parameter B  =fraction of bit period over which apparent clock time varies If our jitter represents 10% of the bit period, the power penalty is 0.34 dB

17 Fiber attenuation If the attenuation in the fiber is 0.2 dB/km and the link is 80 km long, the total loss in the fiber will be 16.0 dB

18 Example results Minimum power required for receiver: -22.1 dBm Safety margin: 6.0 dB Extinction ratio power penalty: 0.87 dB S/N power penalty: 0.41 dB Timing jitter power penalty: 0.34 dB Fiber loss over 80 km: 16.0 dB Total= minimum transmitter power= 1.52 dBm = 1.42 mW

19 Further steps Alternatively, previous data could be used with a fixed transmitter power to determine maximum length of a fiber link If power budget does not add up, one can replace PIN photodiode with APD add an EDFA to the link

20 Power Budget Example

21 Risetime Budget

22 Rise time budget components bit rate and coding format determine upper limit of rise time rise time of transmitter (from manufacturer; laser faster than LED) pulse spread due to dispersion rise time of receiver (from manufacturer; PIN faster than APD) Rise time components are combined by taking the square root of sums of squares

23 Upper limit for rise time For NRZ format, T r =0.70/B For RZ format, T r =0.35/B In this case, choose RZ format. T r must thus be less than or equal to 0.35/10 9 = 350 ps To add a safety margin of x%, multiply B by (1+x/100) before calculating rise time budget limit

24 Group Velocity Dispersion-based rise time Calculate from laser optical bandwidth if known, or from modulation rate: In this case, D=17 ps/nm-km, L=80 km, and  =0.016 nm, so t f =21.8 ps.

25 Modal dispersion rise time For multimode fiber, time spread due to modal dispersion is based on core index and fiber length L. For step-index fiber: For graded-index fiber:

26 Total rise time For this example, t MD =0, t TR =100 ps, t RC =0.5 ns, and t GVD = 21.8 ps as before. t r is therefore 510 ps, and the rise time budget does not meet the limit. Can use NRZ format Use faster detector or transmitter Use graded-index fiber for less dispersion

27 Computer Based Link Simulation Computer Simulation is often used to model opticla links to account for the complex interaction between components and nonlinear effects Commercial simulation tools are now available such as: Linksim from RSoft and the tools from VPI Systems Fiber-Optic Communication Systems-G. Agrawal

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