Question

(i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I_oe^–αx, where I_o is the intensity at x = 0 and α is the attenuation constant. Show that the intensity reduces by 75 per cent after a distance of `((In 4)/α)`

(ii) 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 per cent over a distance of 50 km?

Answer 1

(i) Given, `I = I_0e^(-αx)`

or `I/I_0 = e^(-αx), log_e * I/I_0 = - ax`

or `log_e * I_0/I = αx`

 When `x = 1/α log_e 4`

Then `log_e * I_0/I = α xx 1/α log_e 4`

or `I+0/I` = 4

or `I/I_0  = 1/4`

% decrease in intensity = `((I_0 - I)/I_0) xx 100`

= `(1 - I/I_0) xx 100`

= `(1 - 1/4) xx 100`

= 75%

(ii) Let α be the attenuation in dB/km. If x is the distance travelled by signal. the `10 log_10 * I/I_0 = - αx`

Given `I/I_0 = 1/2, x = 50 km`

∴ `10log_10 (1/2) = - α xx 50`

or `10log_10 2 = 50α`

or `log_10 2 = 5α`

or `α = 1/5 log_10 2 = 0.3010/5` = 0.0602 dB/km.