Noise, signal-to-noise ratio, and filtering

Declare what is signal and what is noise

Start with the quantity or pattern you want to recover. Call its contribution the wanted signal. In this lesson, noise means an unwanted contribution modeled as random; a repeatable unwanted effect may instead require a deterministic correction or distortion model. A changing signal is not automatically noise, and noise is not defined merely by being fast.

For an original voltage-record model, write x_k = s_k + n_k. The integer k labels a record value, x_k is the recorded voltage, s_k is the stipulated wanted voltage, and n_k is the random noise contribution, all in V relative to the same electrical reference. Addition is a supplied model, not a universal law for every measurement chain. Without further information, one recorded sum alone does not identify its separate contributions.

We will specify the wanted pattern and the noise probabilities before calculating. The preceding sampling lesson's deterministic quantizer error is not automatically assigned this noise law. Neither a supplied noise standard deviation nor a smoother record is a complete uncertainty assessment of a real measurand.

Measure size by a declared mean square

Signed averages can conceal fluctuations: −1 V and +1 V have an average of 0 V, although neither value is zero. To measure their size, square first. For a deterministic pattern that repeats every m record values, define M_s = (s_0² + ⋯ + s_(m−1)²)/m, using one full period. A constant can be treated as a one-value period. M_s has unit V², and its root-mean-square value is A_s = √M_s, in V.

For a random noise value, E denotes its probability-weighted average: multiply each possible value by its assigned probability and add. Define M_n = E[n_k²] and A_n = √M_n. In our examples the noise has the same distribution at every k, so M_n does not depend on k. These expectations describe the stipulated model; a short realized record need not have these exact averages.

Let μ = E[n_k]. The variance is E[(n_k − μ)²] = M_n − μ². Thus M_n equals the variance only when μ = 0 V. The distinction matters if an unwanted random contribution also has a nonzero mean. Finite mean squares and a stated averaging convention are needed for the ratios used below; no Gaussian distribution or universal noise spectrum is assumed.

Separate power ratio from amplitude ratio

For this lesson's comparisons, define power-equivalent quantities P_s = M_s/R and P_n = M_n/R using the same exact positive reference resistance R = 100 Ω. V²/Ω has unit W. For an ideal resistor subjected to a voltage, this is the corresponding mean-power relation; for our numerical records it is a declared comparison convention, not a claim that a software filter delivers power to hardware.

When P_n > 0 W, the linear power signal-to-noise ratio is SNR = P_s/P_n = M_s/M_n. The watt units cancel, so SNR is dimensionless. The noise term is the denominator and must be named. An amplitude comparison is A_s/A_n; under this same-resistance convention its square equals the power ratio. Calling both quantities simply a ratio without saying which one can change the numerical answer.

For example, a constant wanted voltage of 2 V has M_s = 4 V². Supply noise n_k = −1 V or +1 V, each with probability 1/2. Then μ = 0 V, M_n = 1 V² and A_n = 1 V. Hence P_s = 0.04 W, P_n = 0.01 W, the power SNR is 4, and the RMS amplitude ratio is 2.

Express positive ratios in decibels

The decibel, dB, expresses a specified logarithmic ratio; it is not a watt or a voltage. The current SI Brochure requires the quantity and any reference value to be specified. For P_s > 0 W and P_n > 0 W, define SNR_dB = 10 log10(P_s/P_n) dB. Here log10 r is the real number z for which 10^z = r; its argument r must be positive and dimensionless.

Because our power ratio is (A_s/A_n)², the same value is SNR_dB = 20 log10(A_s/A_n) dB. The factor 20 is justified by the square relation; it is not a general instruction to use 20 for any voltage ratio across different impedances or incompatible averaging conventions.

For the preceding power ratio 4, SNR_dB = 10 log10 4 dB ≈ 6.0206 dB. A power ratio 1 gives 0 dB, and a positive ratio below 1 gives a negative value: for 1/4 it is approximately −6.0206 dB. Zero wanted power gives SNR = 0 when noise power is positive, but log10 0 is not defined as a finite real number. Zero noise power makes the displayed finite-ratio formula inapplicable. Neither limiting case proves perfect knowledge of a real measurement.

Apply an explicit causal average

Specify a numerical filter at times t_k = kΔt, with Δt = 0.25 s: y_k = (x_k + x_(k−1))/2. It is causal because its output uses the current and previous values, not a future one. The weights are dimensionless, so the output remains voltage-valued. A first output needs a previous input or an explicitly chosen initialization; the periodic cases below supply values for every integer k.

Substituting x_k = s_k + n_k gives y_k = y_s,k + y_n,k, where y_s,k = (s_k + s_(k−1))/2 and y_n,k = (n_k + n_(k−1))/2. This separation follows from the supplied linear rule. To assess the filter, calculate what happens to both contributions. Moving-average filters can change a sinusoid's amplitude and phase rather than merely remove noise.

Now add the condition that the original ±1 V noise values are independent at different k. Each pair (n_k, n_(k−1)) then has probability 1/4. The filtered noise is −1 V, 0 V or +1 V with probabilities 1/4, 1/2 and 1/4. Its mean square is 0.5 V². A constant 2 V wanted signal passes unchanged, so its power SNR rises from 4 to 8, or from approximately 6.0206 dB to 9.0309 dB. The gain is conditional on preserving this wanted signal and on the stated noise law.

Check independence and shared effects

Zero mean does not establish independence. As a contrasting original model, draw one N that equals −1 V or +1 V with equal probabilities, then let n_k = N for every k. Each input still has zero mean and mean square 1 V², but the filter gives (N + N)/2 = N. Its noise mean square stays 1 V² and the constant signal's SNR stays 4.

The general square expansion is E[y_n,k²] = (E[n_k²] + E[n_(k−1)²] + 2E[n_k n_(k−1)])/4. The cross-product term has unit V². For our independent zero-mean inputs it is zero; for the perfectly shared inputs it is 1 V². This directly explains why the two models give different results. Independence is a substantive assumption, not something inferred from a neat-looking average.

Moreover, in our independent-input case neighboring filtered outputs share an input and have covariance 0.25 V². The input independence must not be silently reused for a later calculation involving those outputs.

Test the wanted frequency pattern

Use the same Δt = 0.25 s filter and independent ±1 V input noise, but change the wanted signal. Supply s_k = (2 V) cos(πk/2), defined for all integer k. Its phase is in radians; the values over k = 0, 1, 2, 3 are 2, 0, −2, 0 V, with s_(−1) = 0 V. These are the values of the original 1 Hz cosine s(t) = (2 V) cos(2πft) at t_k, because fΔt = 1/4. This statement supplies a signal; it makes no reconstruction claim.

The filtered wanted values are 1, 1, −1, −1 V over the same four-value period. Its mean square falls from 2 V² to 1 V², while the noise mean square falls from 1 V² to 0.5 V². Thus the power SNR stays 2. Lower output noise has not increased this SNR, and the wanted pattern's values have changed.

For another original wanted signal, s_k = (2 V)(−1)^k, every adjacent pair cancels. This is the supplied 2 Hz cosine's alternating record on the same grid. The filter removes the wanted contribution completely while retaining noise mean square 0.5 V²: output SNR = 0. The table compares these exact periodic models. It is the measurement purpose that decides whether a changing component is wanted; this filter does not know that purpose.

Retain time labels and the purpose

A causal average can also affect the interpretation of a changing quantity. For a separate original noise-free ramp, supply s(t) = 1 V + (4 V/s)t only on −0.25 s ≤ t ≤ 1 s. At the declared grid times, y_s,k = (s(t_k) + s(t_k − Δt))/2 = s(t_k − Δt/2), whenever both inputs lie in that domain. The equality follows from this linear ramp; it does not mean that the filter is a pure time delay for every signal.

At the output tagged t_k = 0.50 s, the inputs are 3 V and 2 V, so y_s,k = 2.5 V. That is the ramp value at 0.375 s, 0.125 s earlier than the output tag, and it differs from the current wanted 3 V by −0.5 V. This determined filtering effect is not random noise. We have specified numerical record values, not a physical instrument's continuous-time settling behavior or processing latency.

Choose a filter by declaring the wanted behavior, noise information, units, input history and output time convention, then checking the changed signal as well as the changed noise. Multiplying both contributions by the same nonzero gain leaves their power SNR unchanged. A favorable SNR in one stipulated case is neither universal filter improvement nor complete measurement uncertainty. The next subject, Controls, blinding, and experimental design, considers how the measurement comparison itself is arranged.

Worked examples

Try it yourself

1. Under the same positive-resistance convention, a wanted RMS voltage is 3 V and noise RMS voltage is 1 V. Find the amplitude ratio, power SNR and logarithmic SNR. State why the factors 10 and 20 give the same decibel value here.

   Show the answer and reasoning

   The amplitude ratio is 3, the power SNR is 9, and SNR_dB = 10 log10 9 dB = 20 log10 3 dB ≈ 9.5424 dB.

   With the same resistance and compatible RMS definitions, the power ratio equals the square of the amplitude ratio. Both logarithms therefore have positive dimensionless arguments and describe the same comparison.

2. The wanted signal is constantly 2 V, and a single random N = ±1 V with equal probabilities is added to every record value. Does the two-value average halve noise mean square? Calculate its output SNR.

   Show the answer and reasoning

   No. Output noise is N, its mean square is 1 V², and the power SNR remains 4.

   The two noise inputs are identical, not independent. Averaging N with itself leaves it unchanged, although its mean over the stipulated probability law is zero.

3. For the constant 2 V wanted signal and independent zero-mean ±1 V noise, replace the filter by y_k = (3/4)x_k + (1/4)x_(k−1). Find output noise mean square and power SNR. Does this alone establish the better filter for every wanted signal?

   Show the answer and reasoning

   The output noise mean square is 0.625 V² and the power SNR is 6.4. These results alone do not choose a filter for every wanted signal.

   The weights sum to 1, so the constant wanted signal remains 2 V. Independence and zero means make the cross-product average zero; noise mean square is [(3/4)² + (1/4)²](1 V²) = (5/8)V². Dividing 4 V² by (5/8)V² gives 32/5 = 6.4. Changing-signal response must be checked separately.

4. A wanted period is now 4, 0, −4, 0 V, repeated for all integer k. Under the equal two-value average, find its output period, input and output wanted mean squares, and power SNRs with the same independent ±1 V noise.

   Show the answer and reasoning

   The output period is 2, 2, −2, −2 V. Wanted mean squares are 8 V² before and 4 V² after filtering; both power SNRs are 8.

   The preceding wanted value is 0 V. Squaring and averaging the four input values gives 32 V²/4 = 8 V², while the output gives 16 V²/4 = 4 V². Noise mean square changes from 1 V² to 0.5 V², so 8/1 = 4/0.5 = 8.

5. For the supplied noise-free ramp s(t) = 1 V + (4 V/s)t on −0.25 s ≤ t ≤ 1 s, use the two-value average with Δt = 0.25 s at the output tagged 0.50 s. Find its voltage, matching ramp time and signed difference from the current wanted value. Is that difference random noise?

   Show the answer and reasoning

   The filtered voltage is 2.5 V, matching the ramp at 0.375 s. Its difference from the current 3 V is −0.5 V. This is a determined filtering effect, not random noise.

   The inputs at 0.50 s and 0.25 s are 3 V and 2 V. Their average is the ramp's midpoint value, 0.125 s earlier than the output tag. Both input times lie in the declared domain; no behavior of a real instrument or pure-delay rule for other signals follows.

6. An ideal numerical stage multiplies both the constant 2 V wanted signal and the independent ±1 V noise by a dimensionless gain 3. Under the unchanged common-resistance convention, what happens to the two mean squares and power SNR? What if a different case has power SNR 1/4?

   Show the answer and reasoning

   The wanted mean square becomes 36 V² and noise mean square 9 V², so SNR remains 4. A positive power SNR 1/4 corresponds to approximately −6.0206 dB.

   Multiplication by 3 multiplies each mean square by 9, which cancels in their ratio. A negative decibel value means a positive power ratio below 1; it does not mean negative power. A larger voltage scale alone is not an SNR improvement.

What to carry forward

• Define the wanted contribution, unwanted effects and additive-model assumptions before naming or calculating noise.
• RMS is the square root of a specified mean square; noise mean square equals variance only for zero-mean noise.
• Power SNR and amplitude ratio are different numbers; their square relation requires compatible definitions and the stated common resistance.
• Use 10 log10 of a positive dimensionless power ratio, or 20 log10 of its corresponding positive amplitude ratio, when reporting this SNR in dB.
• The two-value average reduces the stipulated independent noise but leaves perfectly shared noise unchanged; preserving the wanted signal is a separate check.
• Retain input history and output time labels. Lower noise, larger gain or a smoother record alone does not establish universal SNR improvement or complete measurement uncertainty.