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Numerical: 50 W in dBm/dBW and received power

Short answer

50 W equals 46.99 dBm and 16.99 dBW. With unity gains at 900 MHz, lambda=0.333 m, Friis gives 3.52 microwatts at 100 m, or -24.54 dBm. At 10 km free-space loss adds 40 dB, so received power is -64.54 dBm.

If a transmitter produces 50 watts of power, express the transmit power in units of (i) dBm, and (ii) dBW. If 50 watts is applied to a unity gain antenna with a 900 MHz carrier frequency, find the received power in dBm at a free space distance of 100 m from the antenna. What is Pr(10 km)? Assume unity gain for the receiver antenna.

(i) and (ii) Transmit power

Pt(dBm) = 10 log(50 W / 1 mW) = 10 log(50 000) = 46.99 dBm
Pt(dBW) = 10 log(50 W / 1 W)  = 10 log(50)     = 16.99 dBW

Received power at 100 m

Friis equation with Gt = Gr = 1, f = 900 MHz:

λ = c/f = 3×10⁸ / 900×10⁶ = 0.3333 m

Pr = Pt·Gt·Gr·λ² / [(4π)²d²]
   = 50 × 1 × 1 × (0.3333)² / [(4π)² × 100²]
   = 3.52 × 10⁻⁶ W = 3.52 µW

Pr(dBm) = 10 log(3.52×10⁻⁶ / 10⁻³) = −24.54 dBm

Received power at 10 km

In free space power falls as 1/d², i.e. 20 dB per decade. Using d0 = 100 m as reference:

Pr(10 km) = Pr(100 m) + 20 log(100/10 000)
          = −24.54 − 40
          = −64.54 dBm

Answer: Pt = 46.99 dBm = 16.99 dBW; Pr(100 m) = −24.5 dBm; Pr(10 km) = −64.5 dBm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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