A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $$-5$$ dB per km and cable length is 20 km. The power received at the receiver is $$10^{-x}$$ W. The value of $$x$$ is ______. [Gain in dB = $$10 \log_{10}\left(\frac{P_O}{P_i}\right)$$]

The transmitted power is $$P_i = 0.1$$ kW $$= 100$$ W. The attenuation of the cable is $$-5$$ dB per km, and the total cable length is 20 km. Therefore, the total attenuation is $$-5 \times 20 = -100$$ dB.

Using the gain formula in dB: $$\text{Gain (dB)} = 10 \log_{10}\left(\frac{P_O}{P_i}\right)$$, we substitute the total attenuation: $$-100 = 10 \log_{10}\left(\frac{P_O}{100}\right)$$.

Dividing both sides by 10: $$\log_{10}\left(\frac{P_O}{100}\right) = -10$$.

Taking the antilogarithm: $$\frac{P_O}{100} = 10^{-10}$$, which gives $$P_O = 100 \times 10^{-10} = 10^2 \times 10^{-10} = 10^{-8}$$ W.

Comparing with the given expression $$P_O = 10^{-x}$$ W, we get $$x = 8$$.

Therefore, the value of $$x$$ is $$8$$.