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Numerical: received power with cable and mismatch loss

Short answer

Use Pr = Pt - cable loss + Gt - free space loss + Gr - mismatch loss. Cable loss is 0.6 dB and free space loss is 108.36 dB, giving Pr = 40 - 0.6 + 9 - 108.36 + 4 - 0.2 = -56.16 dBm. The delivered power is about 2.42 x 10^-9 W (2.42 nW).

A BS transmitter has a power output of 10 watts operating at a frequency of 250 MHz. The transmitter is connected by 20 m of an RF coaxial cable, which has a loss of 3-dB/100 m specification, to an antenna that has a gain of 9 dBi. The receiving antenna is 25 km away and has a gain of 4 dBi. There is negligible loss in the receiver feeder line, but the receiver is mismatched; the receiving antenna and feeder cable are designed for 50 ohm impedance. The receiver impedance loss due to mismatch is of about 0.2 dB. Calculate the power delivered to the receiver, assuming free-space propagation.

Using the link equation in dB with free space loss:

Pr = Pt − Lcable + Gt − LFS + Gr − Lmismatch
LFS = 20 log(4πd/λ) = 32.44 + 20 log f(MHz) + 20 log d(km)

Given: Pt = 10 W, f = 250 MHz, cable 20 m at 3 dB/100 m, Gt = 9 dBi, d = 25 km, Gr = 4 dBi, receiver feeder loss negligible, mismatch loss 0.2 dB.

Step 1: Transmit power in dBm

Pt = 10 log(10 W / 1 mW) = 40 dBm

Step 2: Transmitter cable loss

Lcable = 20 m × (3 dB / 100 m) = 0.6 dB

Step 3: Free space path loss

λ   = 3×10⁸ / 250×10⁶ = 1.2 m
LFS = 20 log(4π × 25 000 / 1.2)
    = 108.36 dB
(check: 32.44 + 47.96 + 27.96 = 108.36 dB)

Step 4: Power at the receiver

Pr = 40 − 0.6 + 9 − 108.36 + 4 − 0.2
   = −56.16 dBm
Pr = 10^(−56.16/10) mW = 2.42 × 10⁻⁶ mW = 2.42 nW
ItemdB / dBm
Tx power+40.00
Cable loss−0.60
Tx antenna gain+9.00
Free space loss−108.36
Rx antenna gain+4.00
Mismatch loss−0.20
Received power−56.16 dBm

Answer: Power delivered to the receiver ≈ −56.2 dBm ≈ 2.42 × 10⁻⁹ W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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