Question 20.14: The attenuation of a 6.0 mW signal is 30 dB. What is the fin...

Attenuation = 10  lg\left(\frac{P_{1}}{P_{2}}\right)

Substituting known values gives

30 = 10  lg\left(\frac{6.0  \times  10^{-3}}{P_{2}}\right) meaning 3 = lg\left(\frac{6.0  \times  10^{-3}}{P_{2}}\right)

Taking inverse logs, 10^{3} = \left(\frac{6.0  \times  10^{-3}}{P_{2}}\right)

Rearranging gives P_{2} = \left(\frac{6.0  \times  10^{-3}}{103}\right) = 6.0 \times 10^{-6}  W