InterviewSolution
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InterviewSolution
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(i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation `I = I_0 e^(-alphax)`, where `I_0` is the intensity at ` x = 0 ` and `alpha` is the attenuation constant. Show that the intensity reduces by 75 percent after a distance of `(ln 4)/(alpha)` (ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB `= 10log_10 (I//I_0).` What is the attenuation in `dB//km` for an optical fibre in which the intensity falls by 50 percent over a distance of 50 km? |
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Answer» (a) Given, the intensity of a light pulse `I=I_(0)e^(-alphax)`, where, `I_(0)` is the intensity at x = 0 and `alpha` is constant. According to the question, I = 25% of `I_(0)=(25)/(100).I_(0)=(I_(0))/(4)` Using the formula mentioned in the question, `I=I_(0)e^(-alphax)` `(I_(0))/(4)=I_(0)e^(-alphax)` or `(1)/(4) = e^(-alphax)` Taking log on both sides, we get `ln1-ln4=-alphaxlne" "(becauselne=1)` `-ln4=-alphax` `x=(ln4)/(alpha)` Therefore, at distance `x=(ln4)/(alpha)`, the intensity is reduced to 75% of initial intensity. (b) Let `alpha` be the attenuation in dB/km. If x is the distance travelled by signal, then `10log_(10)((I)/(I_(0)))=-alphax " ....(i)"` where, `I_(0)` is the intensity initially. According to the question, I = 50 % of `I_(0)=(I_(0))/(2)` and x = 50 km Putting the value of x in Eq. (i), we get `10log_(10).(I_(0))/(2I_(0))=-alphaxx50` `10[log1-log2]=-50 alpha` `(10xx0.3010)/(50)=alpha` `therefore` The attenuation for an optical fibre `alpha = 0.0602` dB/km |
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