Distance Change Correction dB


by David Lutton, 2020/10/14

Purpose

Apply a correction for distance change

Presumptions

All distances are in far field for the frequiencies being considered

Where

Applied in CISPR 16-1-4 for sVSWR calibration

Product

Is dB loss/gain for distance change
Loss for increasing distance
Gain for reducing distance

Python

from numpy import log10

distance_measurement = 340  # cm
distance_reference = 300  # cm
x = 20 * log10(distance_measurement / distance_reference)
# dB loss/gain for distance change

Example

\begin{gather*} \begin{aligned} \mathrm{distance}_{measurement} &= 340 \; \\[10pt] \mathrm{distance}_{reference} &= 300 \; \\[10pt] x &= 20 \cdot \log_{10} \left( \frac{ \mathrm{distance}_{measurement} }{ \mathrm{distance}_{reference} } \right) \\&= 20 \cdot \log_{10} \left( \frac{ 340 }{ 300 } \right) \\&= 1.087 \; \;\textrm{(dB loss/gain for distance change)}\\ \end{aligned} \end{gather*}