| Date | May 2014 | Marks available | 1 | Reference code | 14M.3.SL.TZ2.17 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 2 |
| Command term | State | Question number | 17 | Adapted from | N/A |
Question
This question is about digital transmission and optical fibres.
A digital signal is to be transmitted along an optic fibre. The signal to noise ratio \(\left( {{\text{that is 10l g}}\frac{{{P_{{\text{signal}}}}}}{{{P_{{\text{noise}}}}}}} \right)\) in the fibre must not fall below 35 dB.
The following data are available.
Attenuation per unit length of the optic fibre \( = {\text{2.6 dB}}\,{\text{k}}{{\text{m}}^{ - 1}}\)
Power of the input signal is \({P_{{\text{signal}}}}\) \( = {\text{88 mW}}\)
Noise power in the fibre is constant at \({P_{{\text{noise}}}}\) \( = {\text{52 pW}}\)
State what is meant by attenuation.
(i) Determine, using the data, the greatest distance the signal can travel before it must be amplified.
(ii) The optic fibre has a total length of 5600 km. The total transmission time along the length of the fibre is 28 ms. Estimate the refractive index of the core of the fibre.
Markscheme
the loss of power in the transmission of a signal / OWTTE;
(i) power when signal to noise ratio is 35 dB is \(\left( {{\text{10lg}}\frac{{{P_{{\text{signal}}}}}}{{{\text{52}} \times {\text{1}}{{\text{0}}^{ - 12}}}} = {\text{35}} \Rightarrow {P_{{\text{signal}}}} = {\text{52}} \times {\text{1}}{{\text{0}}^{ - 12}} \times {\text{1}}{{\text{0}}^{{\text{3.5}}}} = } \right){\text{ 1.6}} \times {\text{1}}{{\text{0}}^{ - 7}}{\text{ (W)}}\);
attenuation is \(\left( {{\text{10lg}}\frac{{{\text{88}} \times {\text{1}}{{\text{0}}^{ - 3}}}}{{{\text{1.6}} \times {\text{1}}{{\text{0}}^{ - 7}}}} = } \right){\text{ 57.4 (dB)}}\);
distance \( = \left( {\frac{{{\text{57.4}}}}{{{\text{2.6}}}} = } \right){\text{ 22 (km)}}\);
Award [3] for a bald correct answer.
Accept alternative approaches eg:
\({\text{10lg}}\left( {\frac{{{P_{\text{s}}}}}{{{P_{\text{n}}}}}} \right){\text{ = 92.3 (dB)}}\);
\(92.3 - 35 = 57{\text{ (dB)}}\);
\(\frac{{57}}{{2.6}} = 22{\text{ (km)}}\);
Award [3] for a bald correct answer.
(ii) speed of light in core of fibre is \(\left( {\frac{{5.6 \times {{10}^6}}}{{28 \times {{10}^{ - 3}}}} = } \right){\text{ }}2.0 \times {10^{ - 8}}{\text{ (m}}{{\text{s}}^{ - 1}}{\text{)}}\);
\(n = \left( {\frac{{3.0 \times {{10}^8}}}{{2.0 \times {{10}^8}}} = } \right){\text{ }}1.5\);
