Received power: 50 km, 10 GHz link

We use the Friis equation in dB form. The radiated power is taken as 150 W (the "150 dB" in the question is read as 150 W, the usual value in this problem).

Given

• d = 50 km = 5×10⁴ m, f = 10 GHz
• Gt = 25 dB, Gr = 20 dB, Pt = 150 W

Step 1: Wavelength

λ = c/f = 3×10⁸ / 10×10⁹ = 0.03 m

Step 2: Free space loss

4πd/λ   = 4π × 5×10⁴ / 0.03 = 2.094×10⁷
LFS(dB) = 20 log(2.094×10⁷) = 146.42 dB

(Check: 32.44 + 20 log 50 + 20 log 10000 = 32.44 + 33.98 + 80 = 146.42 dB.)

Step 3: Transmitted power in dB

Pt = 10 log 150 = 21.76 dBW

Step 4: Received power

Pr = Pt + Gt + Gr − LFS
   = 21.76 + 25 + 20 − 146.42
   = −79.66 dBW
   = −49.66 dBm
Pr = 10^(−7.966) W ≈ 1.08×10⁻⁸ W = 10.8 nW

Answer: Pr ≈ −79.66 dBW (−49.66 dBm), i.e. about 10.8 nW.

(If the transmitted power were literally 150 dBW, the result would simply be 150 + 45 − 146.42 = 48.58 dBW; such a power is unrealistic, so 150 W is used.)