The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I0e-ax, where I0is the intensity at x = 0 and α is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of (In4/α).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 log10(I/I0).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation I = I0e-ax, where I0is the intensity at x = 0 and α is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of (In4/α).
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB=10 log10(I/I0).What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
-
2 Answers
-
This is a Long Answer Type Questions as classified in NCERT Exemplar
as we know that I= I0
And I= 25%of I0=
I=I0/4
I0/4= I0
I0 cancel from both sides
¼=
Taking log on both sides log1 -log4= - loge
X= log4/
-
This is a Long Answer Type Questions as classified in NCERT Exemplar
as we know that I= I0
And I= 25%of I0=
I=I0/4
I0/4= I0
I0 cancel from both sides
¼=
Taking log on both sides log1 -log4= - loge
X= log4/
Similar Questions for you
According to question, we can write
Increment in height of tower = h2 – h1 = 500 – 125 = 375 m
Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through
µ = A? /A? = 0.5 {A? = 20 volt, A? = 40 volt}
m (t) = A? sin ω? t {ω? = 2π×10? }
c (t) = A? sin ω? t {ω? = 2π×10×10³}
C? (t) = (A? + A? sin ω? t) sin ω? t ⇒ A? {1+ µsin ω? t} sin ω? t
Kindly consider the following figure
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers