How this instrument works
Signal-to-noise ratio compares how large the wanted signal is against the unwanted noise riding along with it, expressed in decibels as SNR = 20·log₁₀(Vs ⁄ Vn). Vs and Vn are both voltage amplitudes measured the same way — both peak, or both RMS, never mixed — because the formula takes their plain ratio before it ever reaches the logarithm. An amplitude ratio of 1,000:1, a signal a thousand times larger than the noise floor beneath it, works out to exactly 60 dB: log₁₀(1000) is 3, and the formula's leading 20 multiplies that to 60.
The factor of 20 rather than 10 is not a stylistic choice; it falls straight out of the relationship between voltage and power. Ordinary decibels compare powers, dB = 10·log₁₀(P₂⁄P₁), and power delivered into a fixed resistance scales with the square of voltage, P = V²⁄R. Substitute a squared voltage ratio into that power formula and the exponent drops out of the logarithm as a multiplier: 10·log₁₀((Vs⁄Vn)²) = 20·log₁₀(Vs⁄Vn). Anyone who plugs a voltage ratio straight into the 10·log₁₀ form meant for power is quietly halving every decibel figure they report — the single most common slip in signal-to-noise arithmetic.
An audio engineer setting up a mixing console, a technician chasing hum on a sensor cable, or an amateur radio operator judging whether a weak station is readable all lean on this same ratio, because it can be read straight off an oscilloscope or a voltmeter without a calibrated power measurement. The formula does have a hard edge: it needs a noise amplitude strictly greater than zero, since log₁₀(0) has no finite value. A perfectly silent, noiseless channel would return infinite SNR in theory, and in practice a reading that looks like zero noise just means the noise floor is too small for the instrument to register, not that it is truly absent.
- Enter the size of the wanted signal — its peak or RMS voltage — in Signal amplitude.
- Enter the size of the unwanted noise, measured the same way as the signal, in Noise amplitude.
- Keep both fields in a consistent voltage unit, switching V and mV together if you use the unit menus.
- Read Signal-to-noise ratio, dB: a large positive number means the signal comfortably dominates the noise.
Worked example — a 2 V signal over 2 mV of noise
A bench test feeds a preamp a 2 V signal, then removes the signal and reads 2 mV of residual noise on the same output — a 1,000:1 amplitude ratio. The formula gives SNR = 20·log₁₀(2 ⁄ 0.002) = 20·log₁₀(1000) = 20·3 = 60.0 dB, since 1000 is exactly 10³ and the base-ten logarithm of a power of ten is just that power.
Sixty decibels is a genuinely good figure — roughly what a well-built analog audio path delivers, the kind of noise floor that keeps hiss inaudible at normal listening levels. Push the same 2 V signal against ten times more noise, 20 mV instead of 2 mV, and the ratio falls to 100:1, dropping the reading by exactly 20 dB to 40.0 dB — proof that every tenfold change in the underlying amplitude ratio is worth exactly 20 dB on this scale, never more and never less.
Questions
Why does the formula use 20·log₁₀ instead of the 10·log₁₀ used for power?
Because power scales with the square of voltage, P = V²⁄R. Squaring a ratio inside a logarithm brings the exponent out as a multiplier, so 10·log₁₀((Vs⁄Vn)²) simplifies to 20·log₁₀(Vs⁄Vn). Using the power form's 10·log₁₀ on a voltage ratio by mistake understates every result by exactly half its correct decibel value.
What happens if the noise amplitude is entered as zero?
The ratio Vs⁄Vn becomes infinite and log₁₀ of infinity has no finite value, so the instrument requires Noise amplitude to be greater than zero and will flag a zero entry rather than return a result. Physically a true zero-noise channel is unmeasurable anyway; every real voltmeter reading carries at least a trace of thermal or instrument noise.
Do Signal amplitude and Noise amplitude have to be RMS voltages?
No, but they must be measured the same way. Enter both as RMS, or both as peak, or both as peak-to-peak — the ratio Vs⁄Vn cancels whichever convention you pick as long as it stays consistent. Mixing a peak signal reading with an RMS noise reading introduces a fixed 3.01 dB error for a sine-wave signal, since RMS equals peak divided by the square root of two.
What SNR figure counts as good in practice?
It depends on the application, but rough benchmarks help: roughly 96 dB is the oft-quoted ceiling for 16-bit digital audio, a clean analog audio path often lands near 90 dB, a decent cassette tape sits around 60 dB, and speech grows hard to follow once a channel drops below about 20 dB. There is no universal threshold, only how audible or costly the noise is for the job at hand.
How do I invert a dB figure back to a plain amplitude ratio?
Raise 10 to the power of the decibel value divided by 20: Vs⁄Vn = 10^(dB⁄20). A 60 dB reading inverts to 10³, or 1,000, matching a 2 V signal sitting over 2 mV of noise. A 40 dB reading inverts to 10², or 100, and every additional 20 dB multiplies the underlying amplitude ratio by another factor of ten.
Can this same formula be used with current instead of voltage?
Yes. The 20·log₁₀ form applies to any amplitude-type quantity — current, sound pressure, or field strength — because each one is proportional to the square root of the power it carries. Only power-like quantities themselves, such as watts or acoustic intensity, call for the plain 10·log₁₀ form instead.