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.

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 $\left( {\frac{{5.6 \times {{10}^6}}}{{28 \times {{10}^{ - 3}}}} = } \right){\text{ }}2.0 \times {10^{ - 8}}{\text{ (m}}{{\text{s}}^{ - 1}}{\text{)}}$;

$n = \left( {\frac{{3.0 \times {{10}^8}}}{{2.0 \times {{10}^8}}} = } \right){\text{ }}1.5$;