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Numerical: reverse link distance by link budget

Short answer

Reverse link path loss limit is 34 - (-80) - 10 = 104 dB. Okumura model gives free-space loss 74.44 dB, and at 900 MHz the corresponding distance is about 140 m. Maintain roughly 140 m between BTS and mobile.

Estimate the appropriate distance that should be maintained for reverse link between one BTS and mobile with appropriate link budget diagram. i) Mobile is connected to antenna with 20 dBi gain, with a transmitting power of 15 dBm and a receive sensitivity of -75 dBm. ii) BTS is connected to antenna with 5 dBi gain, with a transmitting power of 20 dBm and a receive sensitivity of -80 dBm. iii) Cables in both systems are short, with a loss of 3 dB at each side at 900 MHz frequency of operation. Use Okumura's model for mean path loss where G(Area) = 9 dB, Amu = 43 dB, hre = 10 m and hte = 100 m. (Link margin (reverse link) = 10 dB).

On the reverse link the mobile transmits and the BTS receives. The largest allowed path loss is found from the link budget, and the distance is then found by inverting Okumura's model.

Link budget diagram

 Mobile Tx 15 dBm
   | −3 dB cable
   v +20 dBi antenna      EIRP = 32 dBm
   ~~~~~~~~~~~~~~ path loss L ~~~~~~~~~~~~~>
                     BTS antenna +5 dBi
                          | −3 dB cable
                          v
                  BTS Rx input = 34 − L dBm
   must be ≥ sensitivity (−80) + margin (10)

Step 1: Maximum allowed path loss

Tx power (mobile)          +15 dBm
Mobile cable loss           −3 dB
Mobile antenna gain        +20 dBi
BTS antenna gain            +5 dBi
BTS cable loss              −3 dB
-------------------------------------
Signal before path loss    +34 dBm
BTS sensitivity            −80 dBm
Link margin                 10 dB

Lmax = 34 − (−80) − 10 = 104 dB

Step 2: Okumura model

L50 = LF + Amu − G(hte) − G(hre) − GAREA
G(hte) = 20 log(100/200) = −6.02 dB
G(hre) = 20 log(10/3)    = 10.46 dB

104 = LF + 43 + 6.02 − 10.46 − 9
LF  = 104 − 29.56 = 74.44 dB

Step 3: Distance from free space loss (900 MHz)

λ = 3×10⁸ / 900×10⁶ = 0.3333 m
LF = 20 log(4πd/λ)

4πd/λ = 10^(74.44/20) = 5270.5
d = 5270.5 × 0.3333 / (4π) = 139.8 m

Answer: The reverse link supports a maximum BTS–mobile distance of about 140 m.

Note: Amu = 43 dB and GAREA = 9 dB are curve values normally read for a 50 km, 900 MHz link; they are used as given. This makes the predicted range short. With the curve values for short distances, the real range would be larger.

✓ Answer checked by Claude Opus 5.5 (high effort)

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