From Resonance to Vowels

1. From Resonance to Vowels March 10, 2011

2. Fun Stuff (= tracheotomy) Peter Ladefoged: “To record the pressure of the air associated with stressed as opposed to unstressed syllables we need to record the pressure below the vocal folds. A true recording of the subglottal pressure can be made only by making a tracheal puncture.This is a procedure that must be performed by a physician. A local anesthetic is applied both externally and inside the trachea by means of a fine needle. A larger needle with an internal diameter of 2 mm can then be inserted between the rings of the trachea as shown in figure 3.3”

3. Figure 3.3 “As you can see from my face it is not at all painful. But it is not a procedure that can be carried out in fieldwork situations.”

4. Stopping by woods on a snowy evening Whose woods these are I think I know. His house is in the village though. He will not see me stopping here To watch his woods fill up with snow. My little horse must think it queer To stop without a farm house near Between the woods and frozen lake The darkest evening of the year. Robert Frost (1874-1963)

5. Stopping by woods on a snowy evening He gives his harness bells a shake To ask if there is some mistake. The only other sound’s the sweep Of easy wind and downy flake. The woods are lovely, dark and deep. But I have promises to keep, And miles to go before I sleep-- And miles to go before I sleep. Robert Frost (1874-1963)

6. Standing Waves • The initial pressure peak will be reinforced • The whole pattern will repeat itself • Alternation between high and low pressure will continue • ...as long as we keep sending in pulses at the right time • This creates what is known as a standing wave.

7. Standing Wave Terminology node node: position of zero pressure change in a standing wave

8. Standing Wave Terminology anti-nodes anti-node: position of maximum pressure change in a standing wave

9. Resonant Frequencies • Remember: a standing wave can only be set up in the tube if pressure pulses are emitted from the loudspeaker at the appropriate frequency • Q: What frequency might that be? • It depends on: • how fast the sound wave travels through the tube • how long the tube is • How fast does sound travel? • ≈ 350 meters / second = 35,000 cm/sec • ≈ 1260 kilometers per hour (780 mph)

10. Calculating Resonance • A new pressure pulse should be emitted right when: • the first pressure peak has traveled all the way down the length of the tube • and come back to the loudspeaker.

11. Calculating Resonance • Let’s say our tube is 175 meters long. • Going twice the length of the tube is 350 meters. • It will take a sound wave 1 second to do this • Resonant Frequency: 1 Hz 175 meters

12. Wavelength • New concept: a standing wave has a wavelength • The wavelength is the distance (in space) encompassing one complete “cycle” of the standing wave: • For a waveform representation of a standing wave, the x-axis represents distance, not time.

13. First Resonance • The resonant frequencies of a tube are determined by how the length of the tube relates to wavelength (). • First resonance (of a closed tube): • sound must travel down and back again in the tube • wavelength = 2 * length of the tube (L) •  = 2 * L L

14. Calculating Resonance • distance = rate * time • wavelength = (speed of sound) * (period of wave) • wavelength = (speed of sound) / (resonant frequency) •  = c / f • f  = c • f = c /  • for the first resonance, • f = c / 2L • f = 350 / (2 * 175) = 350 / 350 = 1 Hz

15. Higher Resonances • It is possible to set up resonances with higher frequencies, and shorter wavelengths, in a tube. •  = L

16. Higher Resonances • It is possible to set up resonances with higher frequencies, and shorter wavelengths, in a tube. •  = L •  = 2L / 3 • Q: What will the relationship between  and L be for the next highest resonance?

17. First Resonance Time 1: initial impulse is sent down the tube Time 2: initial impulse bounces at end of tube Time 3: impulse returns to other end and is reinforced by a new impulse Time 4: reinforced impulse travels back to far end • Resonant period = Time 3 - Time 1

18. Second Resonance Time 1: initial impulse is sent down the tube Time 2: initial impulse bounces at end of tube + second impulse is sent down tube Time 3: initial impulse returns and is reinforced; second impulse bounces Time 4: initial impulse re-bounces; second impulse returns and is reinforced Resonant period = Time 2 - Time 1

19. Doing the Math • It is possible to set up resonances with higher frequencies, and shorter wavelengths, in a tube. •  = L f = c /  f = c / L f = 350 / 175 = 2 Hz

20. Doing the Math • It is possible to set up resonances with higher frequencies, and shorter wavelengths, in a tube. •  = 2L / 3 f = c /  f = c / (2L/3) f = 3c / 2L f = 3*350 / 2*175 = 3 Hz

21. Patterns • Note the pattern with resonant frequencies in a closed tube: • First resonance: c / 2L (1 Hz) • Second resonance: c / L (2 Hz) • Third resonance: 3c / 2L (3 Hz) • ............ • General Formula: • Resonance n: nc / 2L

22. Different Patterns • This is all fine and dandy, but speech doesn’t really involve closed tubes • Think of the articulatory tract as a tube with: • one open end • a sound pulse source at the closed end • (the vibrating glottis) • At what frequencies will this tube resonate?

23. Anti-reflections • A weird fact about nature: • When a sound pressure peak hits the open end of a tube, it doesn’t get reflected back • Instead, there is an “anti-reflection” • The pressure disperses into the open air, and... • A sound rarefaction gets sucked back into the tube.

24. Open Tubes, part 1

25. Open Tubes, part 2

26. The Upshot • In open tubes, there’s always a pressure node at the open end of the tube • Standing waves in open tubes will always have a pressure anti-node at the glottis • First resonance in the articulatory tract glottis lips (open)

27. Open Tube Resonances • Standing waves in an open tube will look like this: •  = 4L •  = 4L / 3 •  = 4L / 5 L

28. Open Tube Resonances • General pattern: • wavelength of resonance n = 4L / (2n - 1) • Remember: f = c /  • fn = c • 4L / (2n - 1) • fn = (2n - 1) * c • 4L

29. Deriving Schwa • Let’s say that the articulatory tract is an open tube of length 17.5 cm (about 7 inches) • What is the first resonant frequency? • fn = (2n - 1) * c • 4L • f1 = (2*1 - 1) * 350 = 1 * 350 = 500 • (4 * .175) .70 • The first resonant frequency will be 500 Hz

30. Deriving Schwa, part 2 • What about the second resonant frequency? • fn = (2n - 1) * c • 4L • f2 = (2*2 - 1) * 350 = 3 * 350 = 1500 • (4 * .175) .70 • The second resonant frequency will be 1500 Hz • The remaining resonances will be odd-numbered multiples of the lowest resonance: • 2500 Hz, 3500 Hz, 4500 Hz, etc. • Want proof?

31. The Big Picture • The fundamental frequency of a speech sound is a complex periodic wave. • In speech, a series of harmonics, with frequencies at integer multiples of the fundamental frequency, pour into the vocal tract from the glottis. • Those harmonics which match the resonant frequencies of the vocal tract will be amplified. • Those harmonics which do not will be damped. • The resonant frequencies of a particular articulatory configuration are called formants. • Different patterns of formant frequencies = • different vowels