A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10-xW. The value of x is…………………
[Given in dB = 10 log10
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is -5dB per km and cable length is 20km. The power received at receiver is 10-xW. The value of x is…………………
[Given in dB = 10 log10
-
1 Answer
-
Sound level is given in dB = 10 log
As sound level decreases 5 dB every km,
So, in 20 km sound level will decrease by 20 × 5 = 100 dB.
x = 8
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
- 686k Reviews
- 1800k Answers