SOLVETUTORMATH SOLVER

Instrument MI-03-035 · Physics

Attenuation Calculator

Two power readings, one logarithm. Enter what goes in and what comes out, and read loss on a scale built so that cascaded losses simply add.

Instrument MI-03-035
Sheet 1 OF 1
Rev A
Verified
Type 03 — Waves SER. 2026-03035

Attenuation (dB)

3.010300

A = 10·log₁₀(P_in ⁄ P_out)

The working Every figure verified twice
  1. dB = 10·log10(1 ⁄ 0.5) = 3.010300
Worksheet log
  1. No entries yet — change an input to log a scenario.

How this instrument works

Attenuation records how much power a signal surrenders while crossing something — a run of coaxial cable, one kilometre of optical fibre, the brick wall between router and laptop, the slab of tissue beneath an ultrasound probe. Because it compares two powers rather than reporting one, any decibel figure is dimensionless: pure proportion, never watts.

Bell System engineers built this scale out of frustration. Until 1923 telephone circuits were rated in 'miles of standard cable', the unit welded to one specific 19-gauge wire measured at 800 Hz — useless once carrier systems and loading coils arrived. Bell Telephone Laboratories answered in 1924 with the logarithmic transmission unit, and by 1928 that unit carried its lasting name: decibel, one tenth of a bel, honouring Alexander Graham Bell. The whole bel proved far too coarse for bench work, while one tenth lands near what an attentive listener can just detect.

Logarithms earn their place here because losses multiply but decibels add. A budget of 3 dB at one connector, 12 dB down the fibre span and 1 dB across the splice totals 16 dB by arithmetic, sparing you 0.50 × 0.063 × 0.79. Useful magnitudes to carry: single-mode fibre near 1550 nm gives up roughly 0.2 dB per kilometre, RG-58 coax around 1 GHz sheds about 0.5 dB per metre, and soft tissue absorbs ultrasound at some 0.5 dB per centimetre per megahertz. Two limits deserve respect. This ratio stays silent about where missing power went — reflection, absorption, scattering and stray radiation all look identical from outside. And it holds for one frequency only, since nearly every real medium bites harder as frequency climbs, and since high drive levels push materials out of linear behaviour entirely.

A=10log10 ⁣(PinPout)A = 10\,\log_{10}\!\left(\frac{P_{in}}{P_{out}}\right)Pout=Pin10A/10P_{out} = P_{in}\,10^{-A/10}A=20log10 ⁣(VinVout)A = 20\,\log_{10}\!\left(\frac{V_{in}}{V_{out}}\right)
A — attenuation, decibels (dB), dimensionless ratio · P_in — input power, watts (W) · P_out — output power, watts (W) · V — voltage, volts (V). Both powers must share one unit, since only their ratio survives.
  • Enter Input power — whatever arrives at your cable, wall, filter or splice. Switch units between W, mW and kW as your meter reads.
  • Enter Output power, measured at the far end. Mixed units are fine; both figures convert to watts before division.
  • Read Attenuation (dB). Positive means loss. Negative figures mean your stage delivers gain instead.
  • Chaining several stages? Evaluate each section on its own, then add decibel figures — logarithms turn products into sums.

Worked example — finding a filter's 3 dB point

A low-pass filter sits on your bench, driven at 1 W. You sweep frequency upward and watch the power meter on its far side; at one particular frequency that meter settles on 0.5 W. Enter Input power 1 W, Output power 0.5 W, and Attenuation (dB) returns 10 × log₁₀(1 ⁄ 0.5) = 10 × log₁₀ 2 = 3.01029995664 dB, displayed as 3.010300 at six-figure precision.

That frequency is your cutoff. Filter bandwidth is defined at exactly this condition — half power delivered, 3.01 dB down — which is why datasheets quote a '3 dB bandwidth' rather than some rounder-sounding threshold. Writing 3 dB instead of 3.0103 costs 0.7% in power, and nobody on the bench minds. Half power also means amplitude has fallen to 1⁄√2 ≈ 0.707 of its original value, which is where that second famous number comes from.

Questions

Why does halving power give 3 dB rather than 5?

Because this scale counts logarithms, not fractions. Ten times log₁₀ 2 equals 3.0103, so every halving costs 3.01 dB no matter where you start. Handy companions: tenfold drop is exactly 10 dB, hundredfold drop exactly 20 dB, and four times down is two halvings, so 6.02 dB.

Should I use 10·log or 20·log?

Use 10·log for power quantities — watts, intensity, optical flux. Use 20·log only for amplitude quantities such as voltage, current or sound pressure, and only across matched impedance. Power grows as amplitude squared, and squaring inside a logarithm merely doubles it, which is all that factor of two means. Feeding watts into a 20·log expression doubles your answer, and this instrument works in watts.

Is a decibel an SI unit?

No. It is a dimensionless logarithmic ratio that SI accepts alongside its own units. NIST SP 811 defines the decibel as one tenth of a bel and insists that any absolute usage state its reference explicitly — otherwise '−30 dB' identifies a proportion without saying of what.

What separates dB from dBm?

A dB figure compares two powers you supply. A dBm figure fixes one of them at 1 mW, making it absolute: −30 dBm equals 1 µW. Subtract two dBm readings and you recover a plain dB loss, which is precisely what happens here when you enter watts at both ends of your link.

Can attenuation come out negative?

Yes, and that is gain wearing a minus sign. Whenever Output power exceeds Input power — an amplifier, an active repeater, a resonant rise — this formula returns a negative result. Same arithmetic throughout; only sign convention flips. Amplifier datasheets usually quote gain as positive dB, so watch which direction the spec sheet has chosen.

Why does cable loss change with frequency?

Two mechanisms stack up in copper. Skin effect confines current to a thinning outer shell, pushing resistive loss up roughly as √f, while dielectric loss climbs in proportion to f, so coax gets worse as you go higher. Glass behaves differently: Rayleigh scattering falls off as λ⁻⁴ while infrared absorption rises, and those opposing trends leave the loss minimum near 1550 nm — which is why long-haul optics live at that wavelength.

References