---
title: "Numerical: received power with cable and mismatch loss"
subject: "Wireless Communication"
chapter: 3
key: "semester_7/wireless_communication:received-power-coax-mismatch"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/received-power-coax-mismatch
site: BCT Engineering
---

# Numerical: received power with cable and mismatch loss

Wireless Communication · Chapter 3: Radio Wave Propagation in Mobile Networks

- Asked in: 2072 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/received-power-coax-mismatch
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-3

## Question

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.

## 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).

## Answer

Using the link equation in dB with free space loss:

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

#### Step 2: Transmitter cable loss
```text
Lcable = 20 m × (3 dB / 100 m) = 0.6 dB
```

#### Step 3: Free space path loss
```text
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
```text
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
```

| Item | dB / 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which expression gives the received power in dBm for this link?
   A. Pt - Lcable + Gt - LFS + Gr - Lmismatch
   B. Pt + Lcable + Gt - LFS + Gr + Lmismatch
   C. Pt - Lcable - Gt - LFS - Gr - Lmismatch
   D. Pt - Lcable + Gt + LFS + Gr - Lmismatch
   Answer: A. Gains are added and losses (cable, free space, mismatch) are subtracted from the transmit power in dB. (Key point: Link equation in dB)

2. What is the transmitter cable loss for 20 m at 3 dB/100 m?
   A. 6 dB
   B. 0.06 dB
   C. 3 dB
   D. 0.6 dB
   Answer: D. 20 m x (3 dB / 100 m) = 0.6 dB. (Key point: Cable loss)

3. What is the free-space loss at 250 MHz over 25 km?
   A. About 108.4 dB
   B. About 88.4 dB
   C. About 128.4 dB
   D. About 54.2 dB
   Answer: A. LFS = 32.44 + 20 log 250 + 20 log 25 = 32.44 + 47.96 + 27.96 = 108.36 dB. (Key point: Free space loss)

4. What is the power delivered to the receiver?
   A. -46.2 dBm
   B. -56.2 dBm
   C. -66.2 dBm
   D. +56.2 dBm
   Answer: B. 40 - 0.6 + 9 - 108.36 + 4 - 0.2 = -56.16 dBm. (Key point: Final result)

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