The source–filter model of speech

A voice sets its note and its vowel separately. Watch a synthetic voice below sing three vowels on one note: its harmonics stay 125 Hz apart while the peaks of the curve over them move.

A buzz shaped by a tube

Sing “ah” and then “ee” on the same note. The note stays and the vowel changes. So one part of your voice sets the note, and another part sets the vowel. This page pulls the two apart.

Air from the lungs passes the glottis, the gap between the vocal folds in the larynx. During a vowel the folds open and close many times a second, letting the air through in puffs. That stream of puffs is the source, a buzz.

The buzz then passes through the vocal tract, the tube of throat, mouth and lips from the larynx to the open air. The tract is a filter with a few resonances. They are called formants, numbered from the lowest up: $F_1$, $F_2$, $F_3$, in hertz.

Moving the tongue and lips changes the tube’s shape, and that moves the formants. A vowel is, to a first approximation, a set of formant positions.

So speech is a source through a filter: the source–filter model, set out by Gunnar Fant in 1960. The source sets the pitch and the filter sets the vowel. They are separate parts, so either can change while the other stays.

Every sound on this page is synthetic, built from the formulas below; nothing is recorded. The formant values come from Peterson and Barney’s 1952 study of American English vowels.

The source: a pulse train with soft puffs

I work at $f_s=8$ kHz, the telephone rate. Start with one unit impulse every 64 samples. That is a period of $64/8000$ s, which is 8 ms, so the buzz repeats 125 times a second.

In Signals as sums of sinusoids (7.1), “A second arrow, three times as fast” showed that a periodic signal is a sum of harmonics, at whole multiples of its fundamental. Here the fundamental is $f_0=125$ Hz, and it is the voice’s pitch. So the train’s spectrum is a comb of lines every 125 Hz.

A pitch of 125 Hz lies within the range of men’s voices: Peterson and Barney’s men averaged 124 to 141 Hz across their vowels. Finding the pitch from a recording is the job of Pitch detection (29.1).

Real vocal folds do not let out sharp clicks: each puff swells and fades. So I shape every impulse with two real poles at 0.95, the filter $1/(1-0.95z^{-1})^2$. Its impulse response is $(n+1)\,0.95^n$. It rises for 18 samples, about 2.3 ms, and then dies away.

The glottal source $x_\text{glot}[n]$ is the impulse train through this filter. A smooth puff has little fast detail, and the filter’s gain falls steadily with frequency. So the comb’s lines get shorter as they go up.

One more step happens at the lips. The pressure that reaches a listener follows how fast the airflow leaving the lips changes, not the flow itself. So the model adds lip radiation: the first difference $1-z^{-1}$.

Oversampling and noise shaping (11.3), in “What a first difference does to slow and fast wiggles”, showed that its gain rises with frequency. It tilts the source back up a little. With both filters, the source is strongest near 65 Hz and about 26 dB weaker at 4 kHz.

The tract: three resonators

Each formant is a resonance, so I build it from the two-pole resonator of Transfer functions, poles & zeros (16.3), from “Where the pair sits is how h[n] moves”. Formant $i$, for $i=1,2,3$, gets

The pole angle $\theta_i$ sets where it rings. 16.3’s rule $f=\theta f_s/2\pi$ turns a formant frequency into an angle: $\theta_i=2\pi F_i/f_s$.

The radius $r_i$ sets the peak’s width. Each formant has a bandwidth $\mathrm{BW}_i$, its 3 dB width in hertz. I take

Why this radius? In Resonators, notches and combs (17.4), “Poles behind the zeros narrow the notch” gave the width rule $\Delta f\approx(1-r)f_s/\pi$. When $r_i$ is near 1, $e^{-\pi\mathrm{BW}_i/f_s}\approx1-\pi\,\mathrm{BW}_i/f_s$. So $1-r_i\approx\pi\,\mathrm{BW}_i/f_s$, and the rule gives a width of $\mathrm{BW}_i$.

For /a/’s first formant, 730 Hz with a bandwidth of 60 Hz, that gives $r_1=0.9767$ and $\theta_1=0.5733$ rad.

The tract is the three resonators one after the other, so their bottoms multiply:

Multiplied out, $A(z)$ is one polynomial of degree 6. So the tract is an all-pole filter, the $1/A(z)$ of Parametric models and linear prediction (25.3). That will matter in the second half of the page.

Here are the three vowels, with Peterson and Barney’s averages for their 33 men. Each vowel is named by the word their speakers read. The columns F1, F2 and F3 give $F_1$, $F_2$ and $F_3$ in hertz.

I give every vowel the same bandwidths: 60, 90 and 120 Hz for the first, second and third formant. Real bandwidths differ from vowel to vowel; holding them fixed lets only the formant frequencies change.

The whole chain is the train through the glottal filter, the lips and the tract. In z-transforms the factors multiply:

On the unit circle the gains multiply too. The source gives the comb of harmonics; the rest gives a smooth envelope over it. So the spectrum of a vowel is harmonics under an envelope, and the envelope’s peaks are the formants.

Same pitch, different vowels

The picture at the top of the page sings “ah”, “ee” and “oo” on one note. Watch the harmonics’ spacing and the envelope’s peaks as the vowel changes.

Watch the harmonics first. They do not move: 125 Hz apart for all three vowels, because the source never changed. Only the envelope moves.

Now the formants. /i/ puts its first two far apart, with envelope peaks at 268.6 Hz and 2292.1 Hz. /u/ puts them close together, at 299.5 Hz and 864.9 Hz.

Those peaks are a few hertz from the set values, 270 and 2290, 300 and 870. In the chain, each resonance’s sloping sides tilt its neighbours’ peaks and pull them slightly. The largest pull is on /a/’s second formant: 1090 Hz shows up at 1083.8 Hz.

A formant need not sit on a harmonic. /a/’s first formant, 730 Hz, falls between the harmonics at 625 and 750 Hz. The comb only shows the envelope where a harmonic is. Here the harmonic at 750 Hz, the nearest to 730 Hz, is the tallest of all.

The waveform tells the same story in time. Each 8 ms period starts with a kick from a puff. Then the resonances ring and fade until the next puff. A higher pitch would bring the kicks closer together; a different vowel changes how each period rings.

The tract, recovered from the sound

Now turn the problem round. Suppose I am handed only the sound. Can I find the formants?

The tract is an all-pole filter, $1/A(z)$, driven by the source. That is the model 25.3 fits.

“One order at a time” there found $A(z)$ from a signal’s autocorrelation by the Levinson–Durbin recursion. Used on speech, this is called LPC, for linear predictive coding. The plan has four steps.

Four steps from a frame to formants

1. Pre-emphasis. The source’s falling tilt would take up some of LPC’s poles. So first I lift the high frequencies of the whole vowel with $1-0.97z^{-1}$, a first difference that leaks a little. This is pre-emphasis.

Here it nearly undoes the source’s shape. The glottal filter, the lips and the pre-emphasis together are flat to within 1.7 dB from 125 Hz to 4 kHz. Without pre-emphasis, the source with the lips spans 24 dB over that range.

2. One frame. From the pre-emphasised vowel I take 256 samples, which is 32 ms, from sample 1800, 225 ms into the vowel. I multiply them by a Hamming window, from “Five windows, one trade” in 15.2. Thirty-two milliseconds hold four periods of the buzz.

Why 32 ms? Spectrograms & the STFT (15.5), “Short window or long window”, showed the trade. A longer frame sees finer frequency detail; a shorter one follows faster changes.

A real voice moves its tongue and lips within tens of milliseconds, so speech is analysed in frames of about 20 to 30 ms. My synthetic vowels never change, so here the frame length only has to hold a few periods.

3. Fit $A(z)$. I estimate $\hat R_x[0]$ to $\hat R_x[10]$ from the windowed frame, as in 25.3, and run Levinson–Durbin to order $p=10$. That gives $A(z)=1+a_1z^{-1}+\dots+a_{10}z^{-10}$.

Why 10? A man’s vocal tract has about one formant per kilohertz. Up to $f_s/2=4$ kHz that makes about four, and each takes a pole pair: eight poles. Two more cover what is left of the source’s tilt.

The usual rule is $p\approx f_s/1\,\text{kHz}+2$, which is 10 at 8 kHz.

My vowels have only three formants, so LPC has poles to spare. They land well inside the circle, as we will see.

4. Read the roots. Each root $p_k$ of $A(z)$ is a pole, $\lvert p_k\rvert e^{j\theta}$. Its angle gives a frequency by 16.3’s rule, $f=\theta f_s/2\pi$. Turning $r=e^{-\pi\,\mathrm{BW}/f_s}$ round gives its bandwidth:

A formant’s poles sit close to the circle, so I keep the three roots with a positive angle that lie nearest to it. Those are the formant estimates. On all three vowels the choice is clear: those three lie at radius 0.947 or more, and every other root at 0.495 or less.

One change from the first picture: here the dashed curve is the tract alone, $\lvert H_\text{tract}\rvert$ from the set formants, labelled “tract”. Pre-emphasis took the source’s tilt out before the fit, and an all-pole curve cannot follow the lips’ zero at 0 Hz. So the tract is what LPC can recover, and the tract is what it is compared with. The harmonics still carry the source’s shape, so at low frequencies they stand above the dashed curve.

Look at the /a/ frame. The thick LPC curve has its three peaks beside the dashed tract’s, and three crosses sit almost on the three rings. LPC never saw the formant values; it found them from 256 samples of sound.

The two curves stay close. The largest gap is 3.9 dB for /a/ and 3.4 dB for /i/. For /u/ it is 7.9 dB, at 4 kHz, where the tract is 67 dB down and drawn on the floor; on screen it is 5.6 dB. Within one bandwidth of each formant the gap is at most 5.6 dB.

Across all three vowels, the largest miss is 2.5 %, /i/’s first formant: 263 Hz for a set 270 Hz. Its neighbouring harmonics are at 250 and 375 Hz, and the estimate lies between the set value and the harmonic at 250 Hz.

The misses are small but not zero. One reason is that the model is only nearly right. 25.3 fitted $1/A(z)$ to white noise through the filter; here the input is a buzz, and the frame shows the envelope only at harmonics, every 125 Hz.

Now the spare poles. For /i/ they are two pairs, at 1139 and 3058 Hz. For /a/ and /u/ they are one pair and two real roots. None of them is near the circle, and none looks like a formant.

What is left after LPC? Filter the pre-emphasised vowel by $A(z)$ and you get the prediction error of 25.3.

In this frame its four largest samples fall at the four puffs, 64 samples apart, for every vowel. Those four samples hold over 90 % of the error’s energy. What the predictor could not guess is the source’s pulse train.

That split is worth money. A coder can send $A(z)$, the pitch and a loudness for each frame instead of the samples, and the receiver rebuilds a buzz through $1/A(z)$. This is an LPC vocoder.

CELP coders, used in phones, send $A(z)$ plus a coded version of the prediction error. Both need only a few kilobits per second, against 64 kbit/s for plain 8-bit samples at 8 kHz.

Worked example

1. A formant’s poles. Take /a/’s first formant, $F_1=730$ Hz with $\mathrm{BW}_1=60$ Hz, at $f_s=8$ kHz. The radius is $r_1=e^{-\pi\cdot60/8000}=e^{-0.02356}=0.9767$. The angle is $\theta_1=2\pi\cdot730/8000=0.5733$ rad, which is 32.85°.

So the resonator’s bottom is $1-2r_1\cos\theta_1\,z^{-1}+r_1^2z^{-2}$, which is $1-1.6411z^{-1}+0.9540z^{-2}$. Going back, $0.5733\cdot8000/2\pi$ gives 730 Hz, and $-\ln(0.9767)\cdot8000/\pi$ gives 60 Hz.

2. A root read back. For /a/, LPC returns a root $p_k=0.8198+0.5413j$. Its size is 0.9824 and its angle 0.5836 rad. So the formant estimate is $0.5836\cdot8000/2\pi=743$ Hz, with a bandwidth of $-\ln(0.9824)\cdot8000/\pi=45$ Hz.

The frequency misses the set 730 Hz by 1.8 %. The bandwidth, 45 Hz against a set 60 Hz, is further off. On these vowels LPC places a peak more reliably than it measures its width.

3. The order. At $f_s=8$ kHz the rule gives $p=8000/1000+2=10$: four formants below 4 kHz at two poles each, plus two for the source. At 16 kHz the same rule gives 18.

Where you’ll meet this

Mobile phones code speech with LPC. GSM’s full-rate coder fits order 8, and AMR and G.729, both CELP coders, fit order 10 to frames of 10 or 20 ms.

Phoneticians track formants through recorded speech with LPC frame by frame. The free program Praat, much used in phonetics, fits an LPC model to each frame to find them.

Formant synthesisers run this page backwards: a source through resonators, with the formants moved over time. Dennis Klatt’s 1980 synthesiser chains resonators like the ones above.

I left out much of speech. Nasal sounds add zeros as well as poles, and consonants such as “s” use noise as the source instead of a buzz. Detailed models of the glottal flow, such as Rosenberg’s and the LF model, and tube models of the tract, are left for the books.

Speech features (30.2) turns frames of speech into the features used to recognise words and speakers. For more, see L. R. Rabiner and R. W. Schafer, Theory and Applications of Digital Speech Processing (2011), chapters 5 and 9; G. Fant, Acoustic Theory of Speech Production (1960); and J. Makhoul, “Linear prediction: a tutorial review” (Proc. IEEE, 1975).

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