# Estimating loss-limited transmission distance

• wu_weidong

## Homework Statement

A high-speed optical data communication system is composed of a transmitter, an unamplified transmission fiber link, and a receiver. The optical transmitter generates a 10-Gb/s non-return-to-zero (NRZ) signal using a 1550-nm laser diode (linewidth=2 MHz) followed by a chirp-free Mach-Zehnder modulator. The signal has a very high extinction ratio and the fiber launch power is 0 dBm. After transmission over single-mode fiber (loss: 0.2 dB/km, chromatic dispersion: 17 ps/nm/km), the signal is detected by a PIN receiver. The responsivity of the PIN detector is 0.8 A/W and the receiver has a load resistance of 50 W. The bandwidth of the receiver is 8 GHz. Assume that the noise temperature is 300 K, and the target bit-error rate (BER) is 10^-9.

## The Attempt at a Solution

First of all, even though I'm given the dispersion of 17 ps/nm/km, the "loss-limited" term refers only to the attenuation fiber loss, is this correct?

I've done the following, and I hope to know if I'm correct.

For BER of 10-9, Q-factor = 6.0.

For high extinction ratio, SNR ≈ 4Q2 = 144.

Power at transmitter, Ptrans = 0dBm = 1mW

SNR = (R2RL Prec2) / (4kBTΔf)
144 = (0.82×50×Prec2) / (4×1.38×10-23×300×8×109)
Prec = 2.44×10-5W

α = 0.2 = -(10/L)log (Prec / Ptrans)
L = 80.6 km ← loss-limited transmission distance

I think it is OK, still thinking. Is Q the peak to average noise voltage?

Q is the Quality factor, as described here.

## Homework Statement

A high-speed optical data communication system is composed of a transmitter, an unamplified transmission fiber link, and a receiver. The optical transmitter generates a 10-Gb/s non-return-to-zero (NRZ) signal using a 1550-nm laser diode (linewidth=2 MHz) followed by a chirp-free Mach-Zehnder modulator. The signal has a very high extinction ratio and the fiber launch power is 0 dBm. After transmission over single-mode fiber (loss: 0.2 dB/km, chromatic dispersion: 17 ps/nm/km), the signal is detected by a PIN receiver. The responsivity of the PIN detector is 0.8 A/W and the receiver has a load resistance of 50 W. The bandwidth of the receiver is 8 GHz. Assume that the noise temperature is 300 K, and the target bit-error rate (BER) is 10^-9.

## The Attempt at a Solution

First of all, even though I'm given the dispersion of 17 ps/nm/km, the "loss-limited" term refers only to the attenuation fiber loss, is this correct?
Yes but the fiber length is not given so I would ignore this and the dispersion rate also.
Second, what is the problem statement?
I would just compute the noise over the receiver bandwidth assuming it all comes from the 50 ohm load resistor, & compare to the signal strength at the photodiode.
Seems like a flaky problem set; additionally, the receiver noise would probably be dominated by the amplifier and tghe laser, not the 50 ohm, but ...
Don't know about "Q" & forgot how to relate noise to BER. Sorry.

Yes but the fiber length is not given so I would ignore this and the dispersion rate also.
Second, what is the problem statement?
I would just compute the noise over the receiver bandwidth assuming it all comes from the 50 ohm load resistor, & compare to the signal strength at the photodiode.
Seems like a flaky problem set; additionally, the receiver noise would probably be dominated by the amplifier and tghe laser, not the 50 ohm, but ...
Don't know about "Q" & forgot how to relate noise to BER. Sorry.
Agree. Taking it a bit further using dB for simplicity:-
Boltzmann's Constant can be expressed as -228.6 dBW/K/Hz/
Noise power in 50 ohms at 300K = kTB
Pn = -228.6 + 10 log 300 + 10 log 8x10^9
Pn = -72.3dBW
Required signal power, Pr, in 50 ohm load = Pn + 20 log Q [Q seems to be the signal to RMS noise voltage for 10^-9 BER]
For BER 10^-9, Q = 6 [see web link quoted]
Signal to noise ratio for BER 10^-9 = 20 log 6 = 15.6dB
Pr = -72.3 + 15.6 = -56.7 dBW
Pr = 10^-5.67 W
Resistor Current
I ^2 x 50 = 10^-5.67
I = 2 x 10 ^-4 Amp
Optical Power required at diode 0.8A/W
Po = 2 x 10^-4 / 0.8 (W)
Po = 2.5 x 10^-4 (W)
Po = - 6 dBm
TX power
Pt = 0 dBm
Max path loss = 6 dB
Fibre attenuation = 0.2dB/km
Max fibre length = 6/0.2 = 30km
Sounds plausible??

Agree. Taking it a bit further using dB for simplicity:-
Boltzmann's Constant can be expressed as -228.6 dBW/K/Hz/
Noise power in 50 ohms at 300K = kTB
Pn = -228.6 + 10 log 300 + 10 log 8x10^9
Pn = -72.3dBW
Required signal power, Pr, in 50 ohm load = Pn + 20 log Q [Q seems to be the signal to RMS noise voltage for 10^-9 BER]
For BER 10^-9, Q = 6 [see web link quoted]
Signal to noise ratio for BER 10^-9 = 20 log 6 = 15.6dB
Pr = -72.3 + 15.6 = -56.7 dBW
Pr = 10^-5.67 W
Resistor Current
I ^2 x 50 = 10^-5.67
I = 2 x 10 ^-4 Amp
Optical Power required at diode 0.8A/W
Po = 2 x 10^-4 / 0.8 (W)
Po = 2.5 x 10^-4 (W)
Po = - 6 dBm
TX power
Pt = 0 dBm
Max path loss = 6 dB
Fibre attenuation = 0.2dB/km
Max fibre length = 6/0.2 = 30km
Sounds plausible??
You're way ahead of me I'm afraid. My fiber background is with optical fiber sensors which do not involve digital data so I don't know how to relate BER to noise. Also, since the problem statement seems to ask for the permissible length of fiber I'm afraid dispersion and loss are material in coming up with the answer. Sorry, this is a pretty specific problem for us here at pf ...

You're way ahead of me I'm afraid. My fiber background is with optical fiber sensors which do not involve digital data so I don't know how to relate BER to noise. Also, since the problem statement seems to ask for the permissible length of fiber I'm afraid dispersion and loss are material in coming up with the answer. Sorry, this is a pretty specific problem for us here at pf ...
I notice you have a factor of 4 with the S/N calculation. I am guessing you are correct here as my definition of Q might not include a half amplitude term. I have taken Q = signal/RMS noise. I have 15dB S/N and you have 144 = 21.5 dB. Allowing for this, I think we agree, and your calculation is ten times shorter!