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core24 min studyLesson 953 in Physics

Noise spectral density and equivalent bandwidth

Integrate a noise spectrum through a declared filter and distinguish noise bandwidth from signal cutoff frequency.

Before you begin
On this page

The question for today

What you’ll understand

  • Convert consistently between one-sided and two-sided spectra.
  • Compute the equivalent noise bandwidth of a one-pole low-pass filter.
  • Distinguish two identical cascaded poles from an unspecified second-order filter.

A noise density is not an RMS noise until a frequency weighting has been supplied. Real filters usually admit some noise beyond their nominal cutoff and attenuate some below it. Equivalent noise bandwidth replaces that weighting with an ideal rectangular band that transmits the same variance for a white input, after normalizing the reference gain. It is a property of the full transfer function and its normalization. Calling both the signal cutoff and the noise bandwidth simply bandwidth hides a predictable error in a sensitive measurement.

Start with the variance convention

For a real stationary voltage fluctuation, define a two-sided PSD on positive and negative frequencies, with the integral over all frequencies equal to the variance. Away from special DC components, the one-sided PSD is twice the positive-frequency two-sided PSD. Their amplitude spectral densities differ by square root two. No extra factor of two belongs in the variance integral when a one-sided density is already being integrated over nonnegative frequency.

Use frequency f in hertz. An angular-frequency spectrum needs a Jacobian when changing the integration variable; it cannot simply be relabeled from hertz to radians per second. A deterministic DC offset or coherent spectral line also needs separate treatment from a smooth broadband density. For a linear time-invariant filter, a noise component at each frequency is weighted by the squared magnitude of the corresponding gain.

σy2=∫0∞Sx(1)(f)∣H(f)∣2 df=∫−∞∞Sx(2)(f)∣H(f)∣2 df,Sx(1)(f)=2Sx(2)(f) (f>0)\sigma_y^2=\int_0^\infty S_x^{(1)}(f)|H(f)|^2\,df=\int_{-\infty}^{\infty}S_x^{(2)}(f)|H(f)|^2\,df,\qquad S_x^{(1)}(f)=2S_x^{(2)}(f)\ (f>0)
The transfer H is dimensionless here because input and output are voltages. For a real physical filter, its magnitude is symmetric in frequency. These formulas describe the stationary random component and exclude any separately specified deterministic mean.↔ Scroll wide equations horizontally.

Normalize the gain before defining a noise bandwidth

For a low-pass response with nonzero DC gain G, divide the squared response by G squared before integrating. The resulting equivalent bandwidth has units hertz. White input density then gives output RMS equal to the input amplitude density times the magnitude of G times the square root of that bandwidth. This separates amplifier gain from frequency weighting and keeps a voltage gain from being counted twice.

For one RC pole, the normalized squared magnitude is one divided by one plus the squared frequency ratio. Substituting the frequency ratio as integration variable gives an arctangent integral. Its value from zero to infinity is pi over two, so the noise bandwidth exceeds the minus-three-decibel cutoff by that factor. Equivalently, expressing the cutoff through the time constant gives a noise bandwidth of one divided by four time constants.

BEN=∫0∞∣H(f)G∣2df,H(f)=G1+if/fc,BEN=π2fc=14τ,σy=∣G∣enBENB_{\mathrm{EN}}=\int_0^\infty\left|\frac{H(f)}{G}\right|^2df,\qquad H(f)=\frac{G}{1+i f/f_c},\qquad B_{\mathrm{EN}}=\frac\pi2f_c=\frac1{4\tau},\qquad \sigma_y=|G|e_n\sqrt{B_{\mathrm{EN}}}
en is a white one-sided input amplitude density in V/√Hz, G=H(0), and fc=1/(2πτ). The expression assumes an ideal one-pole response over the frequencies contributing appreciably to the integral; additional real high-frequency behavior can change the result.↔ Scroll wide equations horizontally.

Cascaded filters require their actual pole arrangement

Two buffered identical RC poles have squared response proportional to the inverse square of one plus the squared frequency ratio. Their equivalent bandwidth is pi times the individual pole frequency divided by four. The overall minus-three-decibel frequency is lower than that individual pole frequency. A two-pole Butterworth response has a different shape and therefore a different bandwidth factor; the phrase second order does not determine either result alone.

For colored input noise, equivalent bandwidth for white noise is not enough: integrate the actual spectrum with the actual transfer function. Noise introduced between stages is weighted by only the stages that follow it. A lock-in output filter likewise needs a stated response and normalization after demodulation. Reporting a noise number together with the filter configuration, spectral convention, and reference point makes it reproducible and allows a later measurement to be compared without guessing hidden square-root factors.

Work it through

Correct a one-pole white-noise estimate

A declared toy amplifier has DC voltage gain G=10, a single pole at fc=100 Hz, and white one-sided input density en=5 nV/√Hz. No other noise is present. Compute its equivalent bandwidth, input-referred integrated RMS noise, and output RMS noise. Compare with treating 100 Hz as a rectangular noise band.

  1. The correct equivalent bandwidth is BEN=(π/2)(100)=157.0796B_{\mathrm{EN}}=(\pi/2)(100)=157.0796 Hz, rather than the 100 Hz signal cutoff.

  2. The integrated input-referred noise is 5157.0796≃62.66575\sqrt{157.0796}\simeq62.6657 nV RMS. Gain ten gives 626.657 nV RMS at the output.

  3. A rectangular 100 Hz assumption gives 500 nV RMS at the output. The correct result is larger by π/2≃1.25331\sqrt{\pi/2}\simeq1.25331, because the real one-pole tail passes extra noise.

The result

The equivalent bandwidth is about 157.08 Hz and the output noise is about 626.66 nV RMS, 25.33% above the rectangular-cutoff estimate.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A noise variance weights amplitude gain quadratically

For each fresh comparison, Reset and set BOTH Coefficient a=1 and Position x. Compare x=0.50 and x=1.00, predicting y=0.25 and 1.00. Interpret x as a normalized amplitude transmission at one frequency and y as its variance weighting. This quadratic graph is not a frequency sweep, filter response, PSD integrator, or noise-bandwidth readout.

A noise variance weights amplitude gain quadraticallyThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAmplitude transmission and variance transmission are related by a square, not by the same numerical factor.

Pause & reason

Check your understanding

For a white input and one ideal RC pole, how does equivalent noise bandwidth compare with its signal cutoff?

Your turn

Try it yourself

Replace the one-pole response by two buffered identical poles, each at 100 Hz, keeping total DC gain ten and the same white input density. Use the integral from zero to infinity of (1+u²)⁻² equal to π/4. Compute the new noise bandwidth and output noise. Also find the overall minus-three-decibel frequency.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The two-pole magnitude is |H/G|=1/[1+(f/100 Hz)²]. Its minus-three-decibel point has magnitude 1/√2.

Compare with the solution

The noise bandwidth is 100π/4=78.5398100\pi/4=78.5398 Hz. Output noise is 10(5)78.5398≃443.11310(5)\sqrt{78.5398}\simeq443.113 nV RMS. Solving [1+(f/100)2]−1=1/2[1+(f/100)^2]^{-1}=1/\sqrt2 gives f=1002−1≃64.3594f=100\sqrt{\sqrt2-1}\simeq64.3594 Hz. The 100 Hz number labels each individual pole, not the cascade's overall cutoff. These results use the stated identical-pole topology; substituting an unrelated second-order response could change both factors.

Keep the idea

Take it with you

  • A PSD convention and a squared transfer function determine the variance integral.
  • Equivalent noise bandwidth and signal cutoff are different filter properties.
  • Colored noise and intermediate noise sources require their own weighting.

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