Indoor attenuation using FSPL and partition losses

Indoor attenuation using FSPL and partition losses#

Data from the exercise:

C = 300e6  # Speed of light in m/s
F = 2e9  # Frequency in Hz

According to the article How to Measure Wall Attenuation For Spotless Wi-Fi Network Designs:

LOSS_GLASS_PARTITION_DB = 2  # dB (1-3)
LOSS_CONCRETE_DB = 12  # dB
WALL_PIECE_VERTICAL = 3  # m
WALL_PIECE_HORIZONTAL = 5  # m

The longest paths are from the top left and right to the base station. On the path we have:

(partition_loss_db := 2 * LOSS_GLASS_PARTITION_DB + LOSS_CONCRETE_DB)

16

The total loss based on:

  1. free-space path loss \(L_\mathrm{fs}\)

  2. indoor partition losses

Define \(L_\mathrm{fs}\)

from sympy import Eq, pi, solve, symbols

l, d, f, c = symbols(r"L_\mathrm{fs} d f c", positive=True)
(eq := Eq(l, (4 * pi * d * f / c) ** 2, evaluate=False))
\[\displaystyle L_\mathrm{fs} = \frac{16 \pi^{2} d^{2} f^{2}}{c^{2}}\]

Based on the triangle hypotenuse the distance is

from math import sqrt
a = 2 * WALL_PIECE_VERTICAL
b = WALL_PIECE_HORIZONTAL * 1.5
(distance := sqrt(a**2 + b**2))

9.604686356149273

Calculate \(L_\mathrm{fs}\)

(free_space_loss := eq.rhs.evalf(subs={c: C, f: F, d: distance}))
\[\displaystyle 647446.048711462\]

in dB

from sympy import log
(free_space_loss_db := (10 * log(free_space_loss)/log(10)).evalf())
\[\displaystyle 58.1120358476393\]

Total loss:

(total_loss_db := partition_loss_db + free_space_loss_db)
\[\displaystyle 74.1120358476393\]

We used the linear formula above to calculate the loss. We can also use the dB version:

l_db = symbols(r"L_\mathrm{fs[dB]}")
(eq_db := Eq(l_db, (20 * log(d, 10) + 20 * log(f, 10) - 147.55), evaluate=False))
\[\displaystyle L_\mathrm{fs[dB]} = \frac{20 \log{\left(d \right)}}{\log{\left(10 \right)}} + \frac{20 \log{\left(f \right)}}{\log{\left(10 \right)}} - 147.55\]
(free_space_loss_db2 := eq_db.rhs.evalf(subs={f: F, d: distance}))
\[\displaystyle 58.1202636615906\]

Is there an equivalent free-space distance of a partition loss?#

This does not make sense because, for different distances \(a\) and \(b\), we cannot break the free-space losses down like this \(L(a+b) \neq L(a) + L(b)\). This only applies to partition losses.

If you still try to calculate an equivalent distance as follows, the result corresponds to the distance from the transmitter.

(loss_concrete := 10**(LOSS_CONCRETE_DB/20))

3.9810717055349722

Solve for the distance:

(solution := solve(eq, d)[0])
\[\displaystyle \frac{\sqrt{L_\mathrm{fs}} c}{4 \pi f}\]

Evaluate:

distance = solution.evalf(subs={c: C, f: F, l: loss_concrete})
f"{distance:.3f} m"

'0.024 m'

This result only means that the attenuation in the first 24 mm would correspond to a concrete wall. However, this is even a simplification, because radio waves may behave differently in the very proximity of a transmitter.

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