Noise bandwidth (equivalent noise bandwidth)

The noise bandwidth, or equivalent noise bandwidth (ENBW), of a filter or measurement system is the width of an ideal rectangular passband that would pass the same total white-noise power as the real system, with the rectangle's height set to the system's gain at its reference frequency. It is the bandwidth $B$ that belongs in the standard noise formulas: $4kTRB$ for thermal noise, $2qIB$ for shot noise, and the conversion from a noise-equivalent power in W/√Hz to a noise power in watts. Real filters roll off gradually and pass some noise beyond their 3 dB frequency, so the noise bandwidth is larger than the 3 dB bandwidth: by a factor of 1.571 for a single-pole low-pass, 1.220 for two identical cascaded poles, and 1.064 for a Gaussian response.

Definition

For a low-pass system with power transfer function $|H(f)|^2$,

$$B_n = \frac{1}{|H(0)|^2}\int_0^\infty |H(f)|^2\,df .$$

For a band-pass filter the reference is the gain at the center frequency. White noise with a one-sided power spectral density $S$ then delivers a mean-square output of $S\,|H(0)|^2 B_n$, exactly as if the filter were rectangular with width $B_n$.

Common filters

Single-pole RC. With $|H(f)|^2 = 1/[1 + (f/f_c)^2]$ and $f_c = 1/(2\pi\tau)$ the 3 dB frequency, the integral gives

$$B_n = \frac{\pi}{2} f_c = \frac{1}{4\tau} .$$

A photoreceiver whose response is a single pole at 10 GHz therefore collects noise over 15.7 GHz, 1.96 dB more noise power than the 3 dB bandwidth would suggest.

Two cascaded poles. Two identical RC sections with time constant $\tau$ give $B_n = 1/(8\tau)$. Their combined 3 dB frequency is $0.644/(2\pi\tau)$, so $B_n$ is 1.220 times the 3 dB bandwidth. Steeper filters approach a rectangle, and their noise bandwidth approaches the 3 dB bandwidth.

Boxcar integration. Averaging a signal with equal weight for a time $T$ has a sinc-shaped response with

$$B_n = \frac{1}{2T} ,$$

so a 10 ms integration time passes 50 Hz of white noise.

Lock-in amplifier. The output filter of a lock-in amplifier with time constant $\tau$ has $B_n = 1/(4\tau)$ for a single pole (6 dB/octave) and $1/(8\tau)$ for two poles (12 dB/octave): 2.5 Hz and 1.25 Hz at $\tau$ = 100 ms. A single-pole lock-in filter has the same noise bandwidth as a boxcar of length $2\tau$.

Spectrum analyzer filters. The resolution bandwidth filters of spectrum analyzers are close to Gaussian; for a Gaussian the noise bandwidth is 1.064 times the 3 dB width, and practical analyzer filters lie roughly in the range 1.06–1.13 depending on their design. Noise-marker functions apply the instrument's own factor when they normalize a reading to 1 Hz.

Worked numbers

The open-circuit thermal noise voltage of a 50 Ω resistor at 300 K in a noise bandwidth of 10 GHz is

$$v_\text{rms} = \sqrt{4kTRB_n} = 91\ \mu\text{V},$$

from a density of 0.91 nV/√Hz. The equivalent noise current, $\sqrt{4kTB_n/R}$, is 1.82 µA rms. A photocurrent of 1 mA in the same bandwidth carries a shot-noise current of $\sqrt{2qIB_n}$ = 1.79 µA rms, nearly equal; the two are equal at $I = 2kT/(qR)$ = 1.03 mA, the crossover given in the thermal noise entry. With a single-pole receiver of 10 GHz 3 dB bandwidth, both values should be computed with 15.7 GHz, which raises each rms current by a factor of 1.25.

At the other extreme, a detector with an NEP of 10 pW/√Hz read through a lock-in with a single-pole 100 ms time constant has a minimum detectable power, at a signal-to-noise ratio of 1, of $10 \times \sqrt{2.5}$ = 15.8 pW.

Where it matters

Any signal-to-noise ratio computed from a noise density needs a bandwidth, and using the 3 dB value for a gently rolling-off system underestimates the noise. The error is 1.96 dB in noise power for a single pole and 0.86 dB for two poles, and it is common in receiver sensitivity estimates, in noise floors computed from datasheet densities, and in comparisons of lock-in readings taken with different filter slopes. In optical measurements the same idea applies to optical filters: the amplified spontaneous emission passed by a filter in front of a receiver is set by the filter's noise bandwidth, its integral over the passband divided by its peak transmission.

Common questions

Why is noise bandwidth larger than the 3 dB bandwidth?

A real filter does not cut off sharply; it passes a decreasing fraction of the noise above its 3 dB frequency, and that tail adds to the total. Only an ideal rectangular filter has equal noise and 3 dB bandwidths.

What is the noise bandwidth of an RC filter?

$\pi/2$ times the 3 dB frequency, or $1/(4RC)$. For $RC$ = 1 ms the 3 dB frequency is 159 Hz and the noise bandwidth is 250 Hz.

Which bandwidth goes into the shot-noise formula?

The noise bandwidth of the whole receive chain, detector, amplifier and any filtering, measured from the reference gain. For a detector read by integrating for a time $T$, it is $1/(2T)$.

References: P. Horowitz and W. Hill, The Art of Electronics, 3rd ed. (Cambridge University Press, 2015); B. E. A. Saleh and M. C. Teich, Fundamentals of Photonics, 3rd ed. (Wiley, 2019); C. D. Motchenbacher and J. A. Connelly, Low-Noise Electronic System Design (Wiley, 1993); J. H. Scofield, "Frequency-domain description of a lock-in amplifier," American Journal of Physics 62, 129 (1994).