HW5 Solution

1) a) a = 25 mu m, n_1 = 1.485, n_2=1.465, and multimode
D_{modal} = {Delta n}/ c = 0.02 / {3 times 10^8} (s/m) = 6.667 times 10 ^{-11} (s/m) = 6.667 times 10^-8 (s/{km}) = 66.67 ({ns}/{km})
Delta tau = D_{modal} ({ns}/{km}) L (km) = 66.67 L (ns)
b) R_b times L = 1/ {4 Delta tau} times L = 1 /{ 4 times 66.67} ({km}/{ns}) = 3.75 ({Mb}/s - km)
c) a = 4 mu m, n_1 = 1.467, n_2 = 1.464, lambda = 1.3 mu m, lambda_0 = 1360 nm, S_0 = 0.09 {ps}/{km-nm^2}, Delta lambda = 2nm. This fiber operates at single mode.
D_{"intra"} = S_0 (lambda / 4 ) [ 1 - ({lambda_0} / {lambda})^4] =0.09 ({1300} / 4) [ 1 - ({1360} / {1300})^4] = -5.79 ({ps}/ {km-nm})
Notice that we can drop the sign since it only means the long wavelength has less delay and does not affect the overall delay within the linewidth of 2nm.
Delta tau = D_{"intra"} ({ps}/ {km -nm}) Delta lambda (nm) L (km) = 11.57 L (ps)
d) R_b times L = 1/ {4 Delta tau} times L = 1/ { 4 times 11.57} ({km}/{ps}) = 0.0216 ({Tb}/s - km)= 21.6 ({Gb}/s - km)

2) a)

b)

3) P_{"in"} = 300 mu W, alpha_{dB}=0.5dB/km at 1330nm and P_{"in"} = 150 mu W, alpha_{dB}=0.2dB/km at 1550nm.
(P_{"in"})_{dB} = 10 log_{10} ( {0.3 mW}/{1 mW}) = -5.229 dBm and (P_{"in"})_{dB} = 10 log_{10} ( {0.15 mW}/{1 mW}) = -8.239 dBm
a) At 1km,
For 1330nm: (P_{out})_{dB} = -5.229 (dBm) - 0.5 ({dB}/{km})times 1(km) = -5.729(dBm) P_{out} = 10^{-{5.729}/{10}) = 0.2674 (mW)
For 1550nm: (P_{out})_{dB} = -8.239 (dBm) - 0.2 ({dB}/{km})times 1(km) = -8.439(dBm) P_{out} = 10^{-{8.439}/{10}) = 0.1433 (mW)
b) 5km,
For 1330nm: (P_{out})_{dB} = -5.229 (dBm) - 0.5 ({dB}/{km})times 5(km) = -7.729(dBm) P_{out} = 10^{-{7.729}/{10}) = 0.1687 (mW)
For 1550nm: (P_{out})_{dB} = -8.239 (dBm) - 0.2 ({dB}/{km})times 5(km) = -9.239(dBm) P_{out} = 10^{-{9.239}/{10}) = 0.1192 (mW)

4) P_{tx} = 5mW,  P_{rx} = 0.5 mW
a) alpha_{dB} = 0.6{dB}/{km}
(P_{tx})_{dB} = 10 log_{10} ({5mW}/{ 1mW} ) and  (P_{rx})_{dB} = 10 log_{10} ({0.5 mW} / {1 mW}) (P_{rx})_{dB} = (P_{tx})_{dB} - | losses |_{dB} and | losses |_{dB} = alpha_{dB} L => L = [ (P_{tx})_{dB} - (P_{rx})_{dB} ] / alpha_{dB} = 10 (dB) / 0.6 ({dB}/{km}) = 16.67 (km)
b) L = 40 km alpha_{dB} = [ (P_{tx})_{dB} - (P_{rx})_{dB} ] / L = {10 (dB)} / {40 (km)} = 0.25 ({dB}/{km})

5)  P_{rx} = 0.4 mu W, L = 55 km, alpha_{dB} = 0.55 ({dB}/{km}), 2 connectors with 1dB loss / connector, 0.2 dB loss / splice and splice every 5 km.
number of splices = {55}/5 - 1 = 10
Total losses = 2 times 1 + 0.2 times 10 + 0.55 times 55 = 34.25 (dB) and (P_{rx})_{dB} = 10 log_{10} ( {0.0004 mW} /{1mW}) = - 33.98 (dBm)
(P_{rx})_{dB} = (P_{tx})_{dB} - | losses |_{dB} => (P_{tx})_{dB} =(P_{rx})_{dB} + | losses |_{dB} => (P_{tx})_{dB} = -33.98 + 34.25 = 0.27 (dBm) => P_{tx} = 10^{0.027} =1.064 (mW)

6) There is a typo on the original homework which refers to the fiber in problem 2 HW4 and later on is corrected to problem 4 HW 4. I will consider both here.
a) Problem 2 HW 4, L = {R_b times L} /{ R_b} = {3.75} / {10 times 10^3} = 3.75 (m)
Problem 4 HW 4, L = {R_b times L} /{ R_b} = {21.6} / {10 } = 2.16 (km)
b) We need to consider only changing transmitter and receiver.
Problem 2 HW 4, we need to change to a longer operation wavelength, e.g. 16 mu m, so that the fiber operates at single mode where V=2.385 < 2.405. Now the maxium transmission distance will be the same as the result for problem 4 HW 4 in a), exceeding the requirement.
Problem 4 HW 4, we can reduce linewidth of the laser by 50%, i.e. Delta lambda = 1 nm and {R_b times L} = 43.2 ({Gb}/s - km)
c) We need to double the power budget [ (P_{tx})_{dB} - (P_{rx})_{dB} ] of the transmitter and receiver from 10dB to 20dB. For example, transmitter power is increased to 50mW or receiver required power is decreased to 50 mu W.