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A $1 \mathrm{~kW}$ signal is transmitted using a communication channel which provides attenuation at the rate of $-2 \mathrm{~dB}$ per $\mathrm{km}$. If the communication channel has a total length of $5 \mathrm{~km}$, the power of the signals received is [gain in $\mathrm{dB}=10$ $\left.\log \left(\frac{\mathrm{P}_0}{\mathrm{P}_{\mathrm{i}}}\right)\right]$
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Verified Answer
The correct answer is:
$100 \mathrm{~W}$
$100 \mathrm{~W}$
As given, that power of signal transmitted is given $P_i$ $=1 \mathrm{~kW}=1000 \mathrm{~W}$
Rate of attenuation of signal $=-2 \mathrm{~dB} / \mathrm{km}$
Length of total path $=5 \mathrm{~km}$
So, gain in $\mathrm{dB}=5 \times(-2) \mathrm{dB}=-10 \mathrm{~dB}$
Also, gain in $\mathrm{dB}=10 \log \left(\frac{\mathrm{P}_0}{\mathrm{P}_{\mathrm{i}}}\right)$
where $\mathrm{P}_0$ is the power of the received signal Putting the $\mathrm{dB}$ values in Eq. (i),
$$
\begin{aligned}
&-10=10 \log \left(\frac{P_0}{P_i}\right) \Rightarrow 10 \log \left(\frac{P_i}{P_0}\right)=10 \\
&\log \frac{P_i}{P_0}=1 \Rightarrow \log _e\left(\frac{P_i}{P_0}\right)=\log _e 10 \quad\left(\because \log _e 10=1\right)
\end{aligned}
$$
taking antilog
$$
\begin{aligned}
\text { So, } & \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_0}=10 \\
& \frac{1000 \mathrm{~W}}{\mathrm{P}_0}=10 \quad\left(\because \mathrm{P}_{\mathrm{i}}=1000 \mathrm{~W}\right) \\
\Rightarrow & \mathrm{P}_0=100 \mathrm{~W}
\end{aligned}
$$
Rate of attenuation of signal $=-2 \mathrm{~dB} / \mathrm{km}$
Length of total path $=5 \mathrm{~km}$
So, gain in $\mathrm{dB}=5 \times(-2) \mathrm{dB}=-10 \mathrm{~dB}$
Also, gain in $\mathrm{dB}=10 \log \left(\frac{\mathrm{P}_0}{\mathrm{P}_{\mathrm{i}}}\right)$
where $\mathrm{P}_0$ is the power of the received signal Putting the $\mathrm{dB}$ values in Eq. (i),
$$
\begin{aligned}
&-10=10 \log \left(\frac{P_0}{P_i}\right) \Rightarrow 10 \log \left(\frac{P_i}{P_0}\right)=10 \\
&\log \frac{P_i}{P_0}=1 \Rightarrow \log _e\left(\frac{P_i}{P_0}\right)=\log _e 10 \quad\left(\because \log _e 10=1\right)
\end{aligned}
$$
taking antilog
$$
\begin{aligned}
\text { So, } & \frac{\mathrm{P}_{\mathrm{i}}}{\mathrm{P}_0}=10 \\
& \frac{1000 \mathrm{~W}}{\mathrm{P}_0}=10 \quad\left(\because \mathrm{P}_{\mathrm{i}}=1000 \mathrm{~W}\right) \\
\Rightarrow & \mathrm{P}_0=100 \mathrm{~W}
\end{aligned}
$$
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