Question 4 of 7: 40 km Digital Fiber Link — Dispersion and Attenuation Limits, Repeaters, Modes

Question 4: 40 km Digital Fiber Link — Dispersion and Attenuation Limits, Repeaters, Modes (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.

Find. (a) dispersion-limited max repeater spacing, (b) attenuation-limited max repeater spacing, (c) number of repeaters and link length, (d) number of guided modes.

Approach. First convert the analog signal to a bit rate (Nyquist sampling $\times$ 9 bits/sample); the dispersion limit then compares the fiber's ray-theory intermodal spread against half a bit period, and the attenuation limit compares the full link power budget (laser power, minus coupling losses, margin and receiver sensitivity) against the fiber's dB/km rating. The smaller of the two spacings sets how many repeater sections the 40 km route needs.

1. Bit rate. Nyquist sampling of the 100 kHz baseband signal requires $f_s=2B_a=200$ k samples/s; at 9 bits/sample, $$B=f_s\times9=1.8\times10^6\ \text{bit/s}=1.8\ \text{Mb/s},\qquad T_b=\frac{1}{B}=555.6\ \text{ns}.$$
2. Part (a) — dispersion-limited span. The fiber's ray-theory intermodal dispersion coefficient is $$\frac{\Delta\tau}{L}=\frac{n_1\Delta}{c},\qquad\Delta=\frac{1.500-1.495}{1.500}=0.00333,$$ $$\frac{\Delta\tau}{L}=\frac{(1.500)(0.00333)}{2.998\times10^{8}}\times1000\ \text{m/km} =16.68\ \text{ns/km}.$$ With the maximum allowed spread capped at half a bit period ($T_b/2=277.8$ ns), the dispersion-limited span is $$L_{disp}=\frac{T_b/2}{16.68\ \text{ns/km}}=\boxed{16.7\ \text{km}}.$$
3. Part (b) — attenuation-limited span. The receiver sensitivity corresponding to 400 photons/bit at $\lambda=900$ nm and $B=1.8$ Mb/s is $$P_{min}=400\,h\nu\,B=400\left(\frac{hc}{\lambda}\right)B=1.59\times10^{-10}\ \text{W} =-68.0\ \text{dBm}.$$ The laser's 10 mW output is $+10$ dBm, so the total loss the link can tolerate (power minus coupling losses minus margin minus receiver floor) is $$\text{Loss}_{allowed}=10-(3+2)-5-(-68.0)=68.0\ \text{dB},$$ and dividing by the fiber's 3.5 dB/km rating, $$L_{atten}=\frac{68.0\ \text{dB}}{3.5\ \text{dB/km}}=\boxed{19.4\ \text{km}}.$$
4. Part (c) — repeaters and link length. The dispersion limit (16.7 km) is smaller than the attenuation limit (19.4 km), so the link is dispersion-limited and no single span may exceed 16.7 km. Covering 40 km therefore needs $$n_{links}=\left\lceil\frac{40}{16.7}\right\rceil=3\ \text{links}\ \Rightarrow\ \boxed{2\ \text{repeaters}},$$ each link being $40/3=\boxed{13.3\ \text{km}}$ long — comfortably under both the 16.7 km and 19.4 km limits.
5. Part (d) — number of modes. For this fiber, $\mathrm{NA}=\sqrt{1.500^2-1.495^2}=0.122$, and with core radius $a=50\ \mu\text{m}$ at $\lambda=900$ nm, $$V=\frac{2\pi a\,\mathrm{NA}}{\lambda}=\frac{2\pi(50\times10^{-6})(0.122)}{900\times10^{-9}}=42.7,$$ $$M\approx\frac{V^2}{2}=\frac{42.7^2}{2}\approx\boxed{912\ \text{modes}}.$$