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  • The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I L oe minus ex where I0 is the intensity at x = 0 an alpha is the attenuation constant I equal L oe -ex

The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I=I_oe^{-ex} , where I0 is the intensity at x = 0 \: and\: \alpha is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of ( In \: 4/\alpha ).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 \log_{_{10}} (I/I_0).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?

Answers (1)

The given components in the question,

I = Ioe^{-\alpha x }

\log _e. I/I_o = -\alpha x \: or \: \log_e . I_o/I = \alpha x

Now, to calculate the percentage decrease in intensity, the formula used would be
% decrease in intensity 

=\left ( \frac{I_0 -I}{I_0} \right )\times 100 =\left ( 1-\frac{I}{I_0} \right )\times 100 =\left ( 1-\frac{1}{4} \right )\times 100=75\%

(ii) Let \alpha be the attenuation in dB/km. If x is the distance travelled by signal, the

10 \log_{10}.\frac{I}{I_0}=-\alpha x

Given

\frac{I}{I_0}= \frac{1}{2},x=50\: km \therefore 10 \log_{10}\left ( \frac{1}{2} \right )=- \alpha \times 50 \: or\: 10 \log_{10} \; 2 =50 \alpha\\or \log _{10}\:2=5 \alpha \; or \\\: \alpha= \frac{1}{5} \log_{10} \: 2=\frac{0.3010}{5}=0.0602 \:dB/km

 

So, the final answers for both parts are,

i) The decrease in intensity is reduced by 75%.

ii) The attenuation of the given optical fibre = 0.0602 dB/km

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infoexpert21

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