A signal of
Question:

A signal of $0.1 \mathrm{~kW}$ is transmitted in a cable. The attenuation of cable is $-5 \mathrm{~dB}$ per km and cable length is $20 \mathrm{~km}$. The power received at receiver is $10^{-x} \mathrm{~W}$. The value of $\mathrm{x}$ is _________

[Gain in $\mathrm{dB}=10 \log _{10}\left(\frac{\mathrm{P}_{0}}{\mathrm{P}_{i}}\right)$ ]

Solution:

Sound level decreases by $5 \mathrm{~dB}$ every $\mathrm{km}$ so sound level decreased in $20 \mathrm{~km}=100 \mathrm{~dB}$

$\beta_{2}-\beta_{1}=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}}$

$-100=10 \log _{10} \frac{\mathrm{I}_{2}}{\mathrm{I}_{1}} \Rightarrow \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=10^{10}$

$\mathrm{I}_{2}=10^{-10} \mathrm{I}_{1} \Rightarrow \mathrm{P}_{2}=10^{-10} \mathrm{P}_{1}=10^{-8} \mathrm{~W}$

$x=8$

Ans. 8

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