dB/m
How Attenuation Per Meter Sets the Cable Budget
The reason engineers normalize loss to a single meter is that transmission-line attenuation is a distributed property: a uniform line loses the same fraction of power in each successive segment, so total loss grows in direct proportion to physical length. Quoting dB/m lets a designer multiply by run length and immediately know the path loss before any hardware is built. A 75 m antenna feed in 0.10 dB/m cable contributes 7.5 dB on its own, which can be the difference between closing and missing a link budget.
Two physical mechanisms set the value. Conductor loss arises because RF current crowds into a thin surface layer (the skin effect), and that layer gets thinner as frequency rises, so conductor loss scales roughly with the square root of frequency. Dielectric loss arises from the insulating material between conductors and scales linearly with frequency through its loss tangent. The sum of the two is why a single cable carries a whole curve of dB/m values across its usable band rather than one number, and why datasheets always print loss versus frequency.
dB/m is closely related to the attenuation constant from transmission-line theory. The fundamental constant is expressed in nepers per meter (Np/m); multiplying by 8.686 converts it to dB/m. This lets the same physical quantity move freely between the field equations engineers solve and the practical loss numbers printed on a spec sheet.
Converting and Scaling dB/m
Ltotal (dB) = α (dB/m) × ℓ (m)
dB/m to dB per 100 ft:
L100ft = α (dB/m) × 30.48
Nepers to decibels per meter:
αdB/m = 8.686 × αNp/m
Frequency scaling (approximate):
α(f) ≈ k1√f + k2f
Where ℓ = run length, k1 models conductor (skin-effect) loss and k2 models dielectric loss. Example: 0.15 dB/m over a 12 m run = 1.8 dB; the same cable reads ≈ 4.6 dB per 100 ft (0.15 × 30.48).
Representative Loss Per Meter by Media
| Transmission Media | Loss at 1 GHz | Loss at 10 GHz | dB per 100 ft (1 GHz) | Typical Use |
|---|---|---|---|---|
| RG-58 flexible coax | 0.52 dB/m | 1.9 dB/m | 15.8 dB | Short bench jumpers |
| RG-213/U coax | 0.21 dB/m | 0.72 dB/m | 6.4 dB | HF to UHF feeds |
| 1/2 in. corrugated hardline | 0.07 dB/m | 0.25 dB/m | 2.1 dB | Tower runs |
| Semi-rigid 0.141 in. | 0.30 dB/m | 1.0 dB/m | 9.1 dB | Module interconnect |
| WR-90 waveguide (X-band) | n/a | 0.03 dB/m | n/a | Low-loss radar runs |
Frequently Asked Questions
How do I convert dB/m to dB per 100 feet?
Multiply by 30.48, since 100 ft = 30.48 m. A 0.20 dB/m cable loses 0.20 × 30.48 = 6.1 dB per 100 ft; divide a dB/100ft value by 30.48 to recover dB/m. North American coax datasheets often use dB/100ft while European sources use dB/m, so this conversion comes up constantly. Always confirm the test frequency, because attenuation per meter is only meaningful at a stated frequency.
Why does dB/m increase with frequency?
Conductor loss rises with the square root of frequency from the skin effect, and dielectric loss rises linearly with frequency through the loss tangent. The two add, so a cable showing 0.05 dB/m at 1 GHz can show 0.18 dB/m at 10 GHz. That is why long high-frequency runs use larger or low-loss cable, and why millimeter-wave systems favor short runs or waveguide.
How do I turn dB/m into total cable loss for a run?
Multiply attenuation per meter by run length: a 0.15 dB/m cable over 12 m contributes 0.15 × 12 = 1.8 dB. Add connector and adapter losses (0.1 to 0.3 dB each) for total path loss, which is subtracted from transmit power in the link budget. The relationship is linear in length, so halving the run halves the cable loss.