Question 6 of 6: Power and Rise-Time Budget for a 1550 nm Link

Question 6: Power and Rise-Time Budget for a 1550 nm Link (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A repeaterless 75 km single-mode link at 1550 nm carrying 250 Mb/s in RZ format, with the component data collected below.

Find. The optical power budget and system margin with the resulting photocurrent; the maximum system rise time and a receiver bandwidth that meets it; whether NRZ would also work; and the highest rate at which the laser could be directly modulated.

Approach. Convert the launched power to dBm, subtract every loss and penalty in decibels to reach the power at the detector, and compare with the sensitivity to get the margin; then treat rise times as independent contributions adding in quadrature against the format-dependent budget $0.35/B$ for RZ.

1. Express the launched power logarithmically. $$P_t = 5\ \text{mW} \quad\Longrightarrow\quad 10\log_{10}\frac{5\ \text{mW}}{1\ \text{mW}} = +6.99\ \text{dBm}.$$
2. Total the losses over 75 km. Fibre attenuation contributes $0.25 \times 75 = 18.75$ dB and the splices, at one per kilometre, contribute $0.2 \times 75 = 15.0$ dB — note the splices cost more than the glass itself on this route, which is a strong argument for longer drum lengths. Adding the fixed terms, $$\begin{aligned} \text{input coupling} &= 1.0\ \text{dB}\\ \text{fibre attenuation} &= 18.75\ \text{dB}\\ \text{splices} &= 15.0\ \text{dB}\\ \text{output coupling} &= 0.5\ \text{dB}\\ \text{excess noise penalty} &= 2.5\ \text{dB}\\[2pt] \hline \text{total} &= 37.75\ \text{dB} \end{aligned}$$
3. Power at the detector and the system margin. $$P_r = 6.99 - 37.75 = \boxed{\;-30.76\ \text{dBm}\;}$$ and against the quoted sensitivity of $-36$ dBm the link retains $$\text{margin} = -30.76 - (-36.0) = \boxed{\;5.24\ \text{dB}\;}$$ A margin of 5 dB is a sound but not generous allowance for ageing, repair splices and temperature drift; typical practice asks for 3–6 dB, so the design is acceptable with little room for added repairs.
4. Detector photocurrent. Converting $-30.76$ dBm back to watts, $$P_r = 10^{-30.76/10}\ \text{mW} = 0.839\ \mu\text{W}, \qquad I_p = R\,P_r = (0.6)(0.839\times10^{-6}) = \boxed{\;0.504\ \mu\text{A}\;}$$
5. Part (b): the rise-time budget for RZ. A return-to-zero pulse occupies half a bit slot, so the whole system may rise in no more than 35 % of a bit period: $$t_{sys,\max} = \frac{0.35}{B} = \frac{0.35}{250\times10^{6}} = \boxed{\;1.40\ \text{ns}\;}$$
6. Evaluate the fibre contribution. Chromatic dispersion acting on the 5 nm source linewidth over 75 km gives $$t_{\text{mat}} = D\,\Delta\lambda\,L = \left(2.5\ \tfrac{\text{ps}}{\text{nm}\cdot\text{km}}\right) (5\ \text{nm})(75\ \text{km}) = 937.5\ \text{ps} = 0.9375\ \text{ns},$$ easily the largest single term. The fibre is single-mode, so there is no modal contribution.
7. Choose the receiver bandwidth. The budget leaves $$t_{rx} \le \sqrt{t_{sys,\max}^{2}-t_{tx}^{2}-t_{\text{mat}}^{2}} = \sqrt{1.40^{2}-0.50^{2}-0.9375^{2}} = 0.912\ \text{ns},$$ i.e. a receiver bandwidth of at least $0.35/0.912\ \text{ns} = 384$ MHz. Because an RZ pulse is only half a bit slot wide, its spectrum extends to about twice the line rate, and the standard choice is therefore $$\boxed{\;B_{rx} = 2B = 500\ \text{MHz}\;}$$ which clears the 384 MHz floor while keeping the noise bandwidth as small as the format allows. The corresponding receiver rise time is $t_{rx} = 0.35/500\ \text{MHz} = 0.700$ ns.
8. Show the requirement is met. The three independent contributions add in quadrature: $$t_{sys} = \sqrt{t_{tx}^{2}+t_{\text{mat}}^{2}+t_{rx}^{2}} = \sqrt{0.50^{2}+0.9375^{2}+0.700^{2}} = \boxed{\;1.27\ \text{ns}\;} \ \le\ 1.40\ \text{ns}. $$ The requirement is met with 0.13 ns to spare. Dispersion alone consumes two-thirds of the budget, so this link is dispersion-limited rather than loss-limited — a narrower source, not more power, is what would extend it.
9. Part (c): the same link in NRZ. A non-return-to-zero pulse fills the whole bit slot, so the allowance doubles: $$t_{sys,\max}^{\text{NRZ}} = \frac{0.7}{B} = 2.80\ \text{ns}.$$ NRZ also needs only about half the bandwidth, so a 250 MHz receiver ($t_{rx} = 1.40$ ns) suffices and $$t_{sys} = \sqrt{0.50^{2}+0.9375^{2}+1.40^{2}} = 1.76\ \text{ns} \ \le\ 2.80\ \text{ns}. $$ Yes, the design would work, and with considerably more room. Halving the receiver bandwidth also halves the noise power, improving sensitivity by about 3 dB and adding that much to the 5.24 dB margin. The price is a data-dependent spectrum with no energy at the line rate: long runs of identical bits carry no timing information, so NRZ requires scrambling or a line code (and the accompanying clock-recovery circuitry) that RZ does not. For this link, where dispersion is the binding constraint, NRZ is the better engineering choice.
10. Part (d): the direct-modulation limit. The rate at which a laser diode can be modulated by its drive current is set by the relaxation resonance between the carrier and photon populations: $$f_r = \frac{1}{2\pi}\sqrt{\frac{1}{\tau_{sp}\tau_p} \left(\frac{I}{I_{th}}-1\right)} = \frac{1}{2\pi}\sqrt{\frac{(10/2)-1} {(1\times10^{-9})(5\times10^{-12})}} = \boxed{\;4.5\ \text{GHz}\;}$$ Modulation is usable up to roughly this frequency, so the laser could in principle carry of the order of 4.5 Gb/s — eighteen times the 250 Mb/s the link requires. Direct modulation is therefore not the limitation here; dispersion is.

Problems that arise with direct modulation. Four matter in practice. Frequency chirp is the most damaging: modulating the drive current modulates the carrier density, which modulates the refractive index and hence the instantaneous lasing frequency, so each pulse sweeps in wavelength and its effective spectral width broadens well beyond the nominal 5 nm. On a dispersion-limited link such as this one, chirp translates directly into extra pulse spreading, and it is the usual reason high-rate 1550 nm systems move to an external modulator. Relaxation-oscillation ringing appears as overshoot and damped oscillation at the 4.5 GHz resonance whenever the drive changes abruptly, closing the eye at high rates. Turn-on delay occurs if the laser is biased below threshold: the carrier density must first be built up to threshold, which for this device takes $t_d = \tau_{sp}\ln\!\left[I/(I-I_{th})\right] = 0.22$ ns and, worse, varies with the recent bit pattern, producing data-dependent jitter. Biasing just above threshold removes it at the cost of a poorer extinction ratio. Finally, the quoted relative intensity noise of $-120$ dB/Hz sets a floor on the achievable signal-to-noise ratio that no increase in received power can beat, and in a multi-longitudinal-mode device mode partition noise adds a further dispersion-coupled penalty as power hops between modes from bit to bit.