Find. The received signal power level at the far end of the three-section chain.
Approach. Convert the absolute amplifier gain to decibels so every stage is expressed the same way, sum the dB gains/losses along the chain (decibels turn a cascade of multiplications into a single addition), then convert the net dB figure back to a power level.
Convert the amplifier's absolute gain to dB.
$$G_{1,\text{dB}} = 10\log_{10}(15) = \boxed{11.76\text{ dB}}$$
Sum the dB gains and losses along the chain. A loss is a negative gain in the dB domain:
$$G_{\text{net}} = G_{1,\text{dB}} - L_{2,\text{dB}} - L_{3,\text{dB}} = 11.76 - 12 - 3 = \boxed{-3.24\text{ dB}}$$
The chain is a net attenuator overall, even though its first section amplifies — the two loss sections together (15 dB) outweigh the 11.76 dB of gain.
Express the input power in dBm and add the net chain gain.
$$P_{\text{in,dBm}} = 10\log_{10}(100) = 20\text{ dBm}$$
$$P_{\text{out,dBm}} = P_{\text{in,dBm}} + G_{\text{net}} = 20 - 3.24 = \boxed{16.76\text{ dBm}}$$
Convert back to a linear power level.
$$P_{\text{out}} = 10^{16.76/10} = \boxed{47.4\text{ mW}}$$
Cross-check (linear domain, no dB conversions): $100\text{ mW}\times15 = 1500\text{ mW}$ after the amplifier; $\times 10^{-12/10} = 94.6\text{ mW}$ after Section 2; $\times 10^{-3/10} = 47.4\text{ mW}$ after Section 3 — the two independent routes agree.
Fig. Q2 — the three-section chain: one amplifying section (net gain) followed by two attenuating sections, with the running power level shown at each end.
