Reading a spectrum: scaling and units

One raw spectrum, turned into volts and then into power in three steps. Watch the tone’s reading go from a bare count to its true amplitude.

An FFT’s numbers have no unit

The DFT (13.2) gives $X[k]$, and nothing about it says volts. Its size depends on how many samples you took and on which window you used. Read it as it comes, and the same 1 V tone shows a different number every time you change $N$ or the window.

This page turns $\lvert X[k]\rvert$ into three quantities that do have units. The amplitude spectrum is in volts (V), and reads a tone’s amplitude. The power spectrum is in V², and reads a tone’s power. The power spectral density, or PSD, is in V² per hertz, and reads how much noise power sits in each hertz. Each reads one kind of signal true, and I will show you which and why.

I use one test signal for the whole page: a 1 V tone at 1000 Hz plus white noise of RMS 0.1 V, sampled at $f_s=8$ kHz and cut by a Hann window (Window functions compared, 15.2). Its truths are known in advance. The tone’s amplitude is 1 V. Its power is $A^2/2=0.5$ V², which is $-3.01$ dB re 1 V². The noise power is $P_\text{noise}=0.1^2=0.01$ V².

White noise spreads its power evenly from $-f_s/2$ to $f_s/2$ (Oversampling and noise shaping, 11.3). Folded onto positive frequencies only, which I call the one-sided form, that is a density of $2P_\text{noise}/f_s$. Here it is $2(0.1)^2/8000=2.5\times10^{-6}$ V²/Hz, or $-56.02$ dB re 1 V²/Hz.

Two facts from earlier pages carry the rest. A tone of amplitude $A$ sitting on bin $k$ reads $\lvert X[k]\rvert=A\sum w[n]/2$ (Window functions compared, 15.2), where the sum runs over the window’s $N$ samples. Half of the tone sits in bin $k$ and half in its mirror bin $N-k$ (The DFT, 13.2).

From raw FFT to volts

Three plain steps turn $\lvert X[k]\rvert$ into the tone’s amplitude. Divide by $\sum w$, which removes the window’s gain and the length $N$ together. Fold the negative half onto the positive half by multiplying by 2. Then the peak reads volts. Squaring and halving that amplitude gives power. It is like a kitchen scale: a raw count, minus the bowl, times the calibration, is grams.

The picture at the top of this page follows these steps for the page’s test signal, with $N=1024$. The raw $\lvert X[k]\rvert$ of the tone’s bin reads 256, or 48.2 dB: a count with no unit. For $N=1024$ the Hann window sums to 512, so the tone reads $1\times512/2=256$, which is $20\log_{10}256=48.2$ dB.

Dividing by $\sum w$ slides the whole curve down 54.2 dB, which is $20\log_{10}512$. The tone now reads 0.5, $-6.0$ dB re 1 V. The other half of the tone sits in the mirror bin at $-1000$ Hz.

Folding the mirror half on raises the curve 6.0 dB, the factor 2. The tone reads 1.00 V, 0.0 dB, its true amplitude: this is an amplitude spectrum. Bins 0 and $N/2$ have no mirror partner, so they are not doubled.

Squaring and halving drops the curve 3.0 dB, and the floor moves with it. The tone reads 0.500 V², $-3.0$ dB re 1 V², its power, which is its RMS squared. A power spectrum reads tones true. The noise floor, near $-45$ dB here, is another matter: it depends on $N$.

Written out, the amplitude and power spectra of a one-sided bin $k$ are

The second is the power spectrum. Its unit is V², and for a tone it reads the power, $A^2/2$. The noise floor is a separate matter, and the next section turns to it.

Why the floor depends on N

A white noise sample of power $P_\text{noise}$ is multiplied by the window and added up, so a bin of pure noise has mean square $P_\text{noise}\sum w^2$. Put that in the power-spectrum formula and you get a floor of

because $\mathrm{ENBW}=N\sum w^2\big/\big(\sum w\big)^2$ (Window functions compared, 15.2). The tone’s reading did not change with $N$, but the floor falls as $N$ grows. Another way to say it: each bin is $f_s/N$ wide, so it collects the noise of $\mathrm{ENBW}\cdot f_s/N$ hertz. A longer record has narrower bins and each one catches less.

For the Hann window at $N=1024$ the floor is $2(0.01)(1.5)/1024=2.93\times10^{-5}$ V², which is $-45.33$ dB re 1 V². Four times the length lowers it by $10\log_{10}4=6.02$ dB.

Longer records: the tone stays, the floor moves

The cure is to divide by the width of a bin in hertz. A bin’s noise width is $\mathrm{ENBW}\cdot f_s/N$, so dividing the power spectrum by it gives

This is the power spectral density, in V²/Hz, written $S_x^{\text{1s}}(f)$, where the superscript “1s” means one-sided. Put the noise floor into it, and the factors of $N$ cancel: $2P_\text{noise}\sum w^2/(f_s\sum w^2)=2P_\text{noise}/f_s$, the truth from the start. Now I will show the trade at three lengths and then in reverse.

Watch the dotted floor: in the power spectrum it falls as $N$ grows, and in the density it sits on the truth line at every $N$.

As a power spectrum, the tone reads $-3.0$ dB re 1 V², its true power, at every length. The floor sits near $-39$ dB at $N=256$, $-45$ dB at 1024 and $-51$ dB at 4096, 6 dB lower each time $N$ grows four times. The noise did not change; only the bins did. A power spectrum misreads noise by an amount that depends on $N$.

Then the scaling changes to density. The floor now reads $-56$ dB re 1 V²/Hz, the true noise density. At $N=4096$ each bin’s noise width is $\mathrm{ENBW}\cdot f_s/N=1.5\times8000/4096=2.93$ Hz, and the slide is $10\log_{10}\big(N/(f_s\,\mathrm{ENBW})\big)=-4.67$ dB.

Going back down in length, the floor stays at $-56$ dB, and the tone’s peak drops, to $-13.7$ dB at $N=1024$ and $-19.7$ dB at 256. A density reads noise true at every $N$; a power spectrum reads tones true. Choose the scaling for what you measure.

Why does the tone’s peak move in the density? The tone’s power is fixed at 0.5 V², and it all lands in one bin whose noise width is $\mathrm{ENBW}\cdot f_s/N$ hertz. Its density there is the power divided by that width, which changes with $N$, so a density cannot read a tone. Its peak is $0.5/2.93=0.171$ V²/Hz at $N=4096$, which is $-7.68$ dB.

When the clip has finished, drag across the plot to set $N$ yourself, from 256 to 8192. In the density view the dotted floor sits on the truth line at every $N$ you set. The tone’s peak is what changes.

One spectrum, four units

Here are all the scalings in one place, for one-sided bins $1\le k\le N/2-1$.

Fig. Bins 0 and N/2 are not doubled. SciPy’s periodogram(…, scaling=‘spectrum’) returns the power spectrum and scaling=‘density’ the PSD, both one-sided for real input.

For a digital signal with full scale $x_\text{FS}=1$, the amplitude in dB becomes dBFS (Signal size, 1.3), and the PSD becomes dBFS/Hz. A full-scale sine reads $-3.01$ dBFS by RMS and 0 dBFS by peak, and this site uses the peak form unless it says otherwise. Amplitudes and RMS use $20\log_{10}$. Powers and densities use $10\log_{10}$. Always write the reference after the dB.

Window sums connect to the numbers of 15.2: $\sum w=N\,\mathrm{CG}$ and $\sum w^2=N\,\mathrm{CG}^2\,\mathrm{ENBW}$, where $\mathrm{CG}$ is the coherent gain. NumPy’s np.hanning is the symmetric Hann, so its sums differ slightly from the periodic window used here.

Getting a tone’s power from a PSD

A PSD does not read a tone, but you can still get the tone’s power out of it. Add the density over the tone’s lobe and multiply by the bin spacing $f_s/N$. For this page’s tone the sum is $\sum S_x^{\text{1s}}\cdot f_s/N=0.500$ V², the true power.

Do the same with the power spectrum and you overcount. The lobe of a power spectrum adds to $0.750$ V², which is $0.5\times\mathrm{ENBW}=0.5\times1.5$, because every bin of the lobe is already a power, not a density. With no noise, SciPy’s periodogram(x, 8000, window='hann', scaling='spectrum') reads 0.500 V² at the tone’s bin. With the noise of the instrument’s signal, that bin reads 0.5003 V² for one draw: 0.5 plus the noise in that bin.

Noise density in everyday terms: $\sqrt{2.5\times10^{-6}}=1.58$ mV per root hertz. That is how data sheets quote it, and squaring it gives back the PSD.

dBFS/Hz

A converter’s noise floor is usually quoted in dBFS/Hz. Take a 16-bit converter at 48 kHz. Its quantisation noise has power $\Delta^2/12=7.76\times10^{-11}$ re full scale squared, which is $-101.1$ dB (Quantization & noise, 11.1). Spread over 0 to 24 kHz, the one-sided density is $-144.9$ dBFS/Hz. A full-scale sine is $-3.01$ dBFS, so the signal-to-noise ratio is $98.09$ dB, which is the $6.02B+1.76$ of 11.1 with $B=16$ before rounding.

Worked example

1. Chain, $N=1024$, Hann ($\sum w=512$, $\sum w^2=384$, ENBW 1.5). The tone is 256 (48.16 dB), then 0.5 ($-6.02$ dB re 1 V), then 1.000 V (0.00 dB), then 0.500 V² ($-3.01$ dB re 1 V²). The expected noise floor at those four steps is 5.84, $-48.34$, $-42.32$ and $-45.33$ dB.
2. Floors and peaks. For $N$ of 256, 1024, 4096 and 8192 the power-spectrum floor is $-39.31$, $-45.33$, $-51.35$ and $-54.36$ dB re 1 V². The density tone peak is $-19.72$, $-13.70$, $-7.68$ and $-4.67$ dB re 1 V²/Hz. The density floor is $-56.02$ dB at every $N$. The bin noise widths are 46.88, 11.72, 2.93 and 1.46 Hz.
3. Truth. $S_x^{\text{1s}}=2P_\text{noise}/f_s=2(0.1)^2/8000=2.5\times10^{-6}$ V²/Hz, which is $-56.02$ dB re 1 V²/Hz, or 1.58 mV/√Hz.
4. Tone power from the PSD. Lobe sum times $f_s/N$ is 0.5000 V². The power-spectrum lobe sum is 0.7500 V², which is $0.5\times\mathrm{ENBW}$.
5. dBFS/Hz. A 16-bit converter at 48 kHz gives $-144.9$ dBFS/Hz, and at 44.1 kHz $-144.5$. A 24-bit converter at 48 kHz gives $-193.1$ dBFS/Hz. The SNR of a full-scale sine is 98.09 dB at 16 bits and 146.26 dB at 24 bits.

Where you’ll meet this

Spectrograms & the STFT (15.5) draws these levels in dB for every frame. The periodogram (25.1) and Averaged periodograms: Bartlett and Welch (25.2) estimate a PSD and ask how much it scatters. Power spectral density (24.4) defines the PSD of a random process.

Reference card