Sound 2: frequency analysis Tues. March 27, 2018 1 Speed of Sound - - PowerPoint PPT Presentation

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Sound 2: frequency analysis Tues. March 27, 2018 1 Speed of Sound - - PowerPoint PPT Presentation

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COMP 546 Lecture 19 Sound 2: frequency analysis Tues. March 27, 2018 1 Speed of Sound Sound travels at about 340 m/s, or 34 cm/ ms. (This depends on temperature and other factors) 2 Wave equation =

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COMP 546

Lecture 19

Sound 2: frequency analysis

  • Tues. March 27, 2018
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Speed of Sound

Sound travels at about 340 m/s, or 34 cm/ ms. (This depends on temperature and other factors)

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Wave equation

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𝑄𝑠𝑓𝑑𝑑𝑣𝑠𝑓 = 𝐽𝑏𝑒𝑛 + 𝐽(π‘Œ, 𝑍, π‘Ž, 𝑒) 𝐽(π‘Œ, 𝑍, π‘Ž, 𝑒) is not an arbitrary function. Rather:

πœ–2 πœ–π‘Œ2 + πœ–2 πœ–π‘2 + πœ–2 πœ–π‘Ž2 𝐽 π‘Œ, 𝑍, π‘Ž, 𝑒 = 1 𝑀2 πœ–2 πœ–π‘’2 𝐽 π‘Œ, 𝑍, π‘Ž, 𝑒 𝑀 = 340 m/s

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The wave equation + boundary conditions give complicated shadow and reflection effects. What happens when sound enters the ear ?

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plane wave + single slit sea waves + islands

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Musical sounds

(brief introduction)

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Modes are sin(

𝜌 𝑀 π‘˜π‘¦) where 𝑀 is the length of the string, π‘˜ is an integer.

Write one string displacement at t = 0 as sum of sines. Example: guitar

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Physics says: where constant 𝑑 depends on physical properties of string (mass density, tension)

πœ• = 𝑑 𝑀

𝑀

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Modes of a vibrating string each have fixed points which reduce the effective length.

πœ• = 𝑑 𝑀 2𝑑 𝑀 3𝑑 𝑀 4𝑑 𝑀

𝑀 2 𝑀 3 𝑀 4

𝑀 Physics says:

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πœ• = 𝑑 𝑀 2𝑑 𝑀 3𝑑 𝑀 4𝑑 𝑀

πœ•0

β€œfundamental” β€œovertones” (1st harmonic) The temporal frequency 𝑛 πœ•0 is called the 𝑛-th harmonic.

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For stringed instruments, most of the sound is produced by vibrations

  • f the instrument body (neck, front and back plates).

http://www.acs.psu.edu/drussell/guitars/hummingbird.html

The lines in the sketches below are the nodal points. They don't move.

These are vibration modes, not harmonics. The guitar sound is a sum of these modes.

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Difference of two frequencies πœ•1 and πœ•2 : π‘šπ‘π‘•2

πœ•2 πœ•1

  • ctaves.

e.g. 1 octave is a doubling of frequency.

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(Western) Musical Notes

Each β€œoctave” ABCDEFGA is divided into 12 β€œsemitones”, separated into 1/12 octave. C-D, D-E, F-G, G-A, A-B are two semitones each E-F, B-C are one semitone each.

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Q: How many semi-tones are there from πœ•0 to πœ• ?

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Q: How many semi-tones are there from πœ•0 to πœ• ? A: 12 π‘šπ‘π‘•2

πœ• πœ•0

πœ•0

Fundamental frequency of note

πœ•

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88 fundamental frequencies (Hz) on a keyboard The fundamental frequencies of successive notes define a geometric progression. This is different from the harmonics of a vibrating string which define an arithmetic progression.

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Speech Sounds

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What determines speech sounds?

  • voiced vs. unvoiced

β€˜zzzz’ vs. β€˜ssss’, β€˜vvvv’ vs. β€˜ffff’

  • articulators (jaw, tongue, lips)

β€˜aaaa’, β€˜eeee’, β€˜oooo’, …

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Voiced sounds are produced by β€œglottal pulses”.

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

𝑕 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

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Exercise 16 Q7.

𝑕 𝑒 βˆ’ 𝑒0 = 𝑕 𝑒 βˆ— πœ€(𝑒 βˆ’ 𝑒0)

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Voiced sounds are produced by β€œglottal pulses”.

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

𝑕 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

= 𝑕 𝑒 βˆ—

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

πœ€ 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

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π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

𝑕 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

decrease π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š by increasing tension in vocal cords

increase frequency of pulses ≑

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𝐽 𝑒

= 𝑏 𝑒 βˆ— 𝑕 𝑒 βˆ—

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

πœ€ 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

Let 𝑏 𝑒 be the impulse response function of the articulators.

(jaw, tongue,lips)

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π‘œπ‘žπ‘£π‘šπ‘‘π‘“ βˆ’1 π‘œπ‘žπ‘£π‘šπ‘‘π‘“ βˆ’1

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π‘œπ‘žπ‘£π‘šπ‘‘π‘“ βˆ’1 π‘œπ‘žπ‘£π‘šπ‘‘π‘“ βˆ’1

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Oral and nasal cavity have resonant modes of vibration, like air cavity in guitar does.

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Time domain Temporal frequency domain Peaks are called β€œformants”

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π‘ˆ

𝑕 is the period of the glottal pulse train.

The pulse train has π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š pulses in π‘ˆ time steps, i.e. π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š = π‘ˆ.

Assume that the Fourier transform is taken over π‘ˆ samples.

𝐆

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

πœ€ 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

= ?

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π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

Assignment 3: Show

𝐆

π‘˜=0 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘šβˆ’1

πœ€ 𝑒 βˆ’ π‘˜ π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

= π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š

𝑛=0 π‘ˆπ‘•π‘šπ‘π‘’π‘’π‘π‘š βˆ’1

πœ€ πœ• βˆ’ 𝑛 π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š

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Units of temporal frequency πœ•

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π‘ˆ

π‘•π‘šπ‘π‘’π‘’π‘π‘š is the period of the glottal pulse train.

π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š pulses in π‘ˆ time samples. To convert π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š to β€˜pulses per second’, we divide π‘ˆ (to get pulses per sample) and then multiply by β€˜time samples per second’. High quality audio uses 44,100 samples per second.

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π‘œπ‘•π‘šπ‘π‘’π‘’π‘π‘š is the fundamental frequency of the voiced sound. It determines the "pitch". Adult males : 100-150 Adult females : 150-250 Hz Children: over 250 Hz

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πœ•0 = 100 𝐼𝑨 πœ•0 = 200 𝐼𝑨 glottal pulse spectrum formant spectrum sound spectrum glottal pulse spectrum β€œformants” sound spectrum

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Voiced vowel sounds

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Unvoiced sounds

noise instead of glottal pulses

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Unvoiced sounds

noise instead of glottal pulses

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Flat amplitude spectrum

  • n average ( β€˜white noise’)
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Restrict flow of air by moving tongue, lips into contact with the teeth & palate. Fricatives

  • voiced z, v, zh, th (the)
  • unvoiced ?

Stops

  • voiced b, d, g
  • unvoiced ?

Nasals (closed mouth)

  • m, n, ng

Consonants

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Restrict flow of air by moving tongue, lips into contact with the teeth & palate. Fricatives

  • voiced z, v, zh, th (the)
  • unvoiced s, f, sh, th (theta)

Stops

  • voiced b, d, g
  • unvoiced p, t, k

Nasals (closed mouth)

  • m, n, ng

Consonants

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I did not have time to cover the following slides properly. I will present them again in lecture 22.

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Spectrogram

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Partition a sound signal into 𝐢 blocks of π‘ˆ samples each (i.e. the sound has πΆπ‘ˆ samples in total). Take the Fourier transform of each block.

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Spectrogram

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Partition a sound signal into 𝐢 blocks of π‘ˆ samples each (i.e. the sound has πΆπ‘ˆ samples in total). Take the Fourier transform of each block. Let 𝑐 be the block number, and πœ• units be cycles per block.

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Cycles per second (Hz) Time (samples) πœ•0 =

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e.g. T = 512 samples (12 ms), πœ•0 = 86 Hz T = 2048 samples (48 ms) πœ•0 = 21 Hz

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e.g. T = 512 samples (12 ms), πœ•0 = 86 Hz T = 2048 samples (48 ms), πœ•0 = 21 Hz You cannot simultaneously localize the frequency and the time. This is a fundamental

  • tradeoff. We have seen it before (recall the Gaussian).
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Narrowband

(good frequency resolution, poor temporal resolution … ~50ms)

Wideband

(poor frequency resolution, good temporal resolution)

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Examples: Spectrograms of 10 vowel sounds