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}=0.5`dB/km at 1330nm and `P_{"in"} = 150 mu W`, `alpha_{dB}=0.2`dB/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`.