Question 9.3: The power reduces by 1.5dB every 100[m] on a transmission li......

(a) From Eq. (9-32) we obtain

dB  power  loss = -20 \log _{10} \left[e^{- \alpha z}\right] = 8.69\alpha z                          (9-32)

\alpha = \frac{dB  power  loss}{8.69 z} = \frac{1.5 }{8.69 \times 100} = 0.0017 [Np / m] .

(b) From Eq. (9-30) we obtain

\boxed{ \left\langle\mathscr{P}(z)\right\rangle = \left\langle\mathscr{P}(0)\right\rangle = e^{-2 \alpha z} }          [W]                 (9-30)