Perturbation Theory: Why do /i/ and /a/ have the formants they have? Tuesday, March 3, 15

Transcription

1 Perturbation Theory: Why do /i/ and /a/ have the formants they have?

2 Vibration: Harmonic motion time A x 0 + A + A Block resting on air hockey table, with a spring attaching it to the side of the table. Such systems are effectively frictionless. Block will oscillate about neutral position Changes in pressure over time in a part of the vocal tract also oscillates (about neutral pressure) and obeys a similar equation of motion. What is equation of motion for the block, and how do physical laws give rise to it?

3 Relevant physical laws Physical system that opposes acceleration Newton s 2nd Law: F = m a Physical system that opposes displacement Hooke s Law: F = k(x x 0 ) k(stiffness) x Inertial Force is proportional to displacement from x0

4 Multiple Masses I One mass attached to two springs to walls I will vibrate at a single frequency, depending on mass and sti ness. I Two masses, each attached to the wall and to each other I will oscillate at two di erent frequencies, depending on initial conditions. why?

5 Two-mass system I In a vibratory system with 2 m and 3 k, there will be 2 modes of vibration: I In-phase mode: the middle spring just rides up and down with the masses. I Out-of-phase mode: the middle spring stretches and compresses. Therefore: OP Mode has more e ective sti ness (3 springs vs. 2) and therefore has higher frequency.

6 Three Masses

7 Multiple Masses I A vibratory system with more than one mass can have more than one mode of vibration. I For every mass that is added, an additional mode is found, and the additional mode will be higher in frequency.

8 Infinite modes of String A string can be divided into an infinite number of masses, so can oscillate at an infinite number of frequencies

9 Infinite modes of String

10 Perturbation Theory If you perturb a physical property of the system essential for vibration (m or k), how does that affect the frequency of the signal. how does the frequency change if the mass increases? How does the frequency change if the stiffness of the spring increases?

11 Principle 1: Mass Spring perturba4on If you increase the mass, you lower the frequency, because a heavy system accelerates less. If you increase the s4ffness, you increase the frequency, because a s4ff system moves back with greater force. = k m

12 Perturbation of mass and stiffness I Imagine you could increase the mass of a one-mass system over time, what would happen to its frequency? I Imagine you could increase the sti ness of a one-mass system over time, what would happen to its frequency? Spectrogram+of+one+mass+system

13 Perturbation of mass in 2-mass system I Two-mass system has two modes: I I Low: High: I Increase the masses in the low frequency mode. What happens to frequency? I Increase the masses in the high frequency mode. What happens to frequency? I Increasing mass decreases the frequency of both modes.

14 Perturbation of stiffness in 2-mass system I Increase the sti ness of either of the end springs in the low frequency mode, what happens to frequency? I Increase the sti ness of either of the end springs in the high frequency mode, what happens to frequency? I Increasing the sti ness of the end springs increases the frequency of both modes.

15 Perturbation of stiffness in 2-mass system I Now increase the sti ness of the middle spring (k2) in the low frequency mode, what happens to frequency? I Increase the sti ness of the middle spring (k2) in the high frequency mode, what happens to frequency? I Effect of increasing k2 Frequency (Hz) Principle 2: Effect of mass or s4ffness perturba4on depends on the posi4on of the perturba4on Time

16 Principle 2: effect of perturba4on depends on the posi4on Applies to formant movements in speech. synchronous asynchronous

17 I Perturbation of 3-mass system I Increase the mass of the middle mass of in the lowest frequency mode. What happens to frequency? I Increase tmass of the middle mass of in the next higher frequency mode. What happens to frequency?

18 Perturbing string with ends attached to walls I Walls fix motion to 0. I An increase of mass at a wall will have no e ect on frequency, since the string is not moving there. I Increasing mass will have a maximal lowering e ect on frequency where the string is moving most I I An increase in sti ness near a wall has maximal increase e ect on the frequency, since stretchiness is maximal there. Increasing sti ness will have a maximal raising e ect on frequency where the string is moving least I Distance from a wall will determines the degree of e ect of an m or k change on frequency. I Close to wall, mass ine ective. Close to no-wall, sti ness ine ective.

19 Perturbing string with ends attached to walls Effect of Mass Change on frequency, by location Ends: No e ect on frequencies, since there is no motion, and increased e ect of mass is through motion. 2. Middle: Odd Modes will decrease and even ones will not be a ected, since mass is fixed there. 3. One third: Modes 12 down, Mode 3 No change, Mode 4 small change. Mode 5 slightly higher change, Mode 6 no change?

20 Perturbing string with ends attached to walls Effect of k Change on frequency, by location Ends: Maximal increase, since there is max e ective k. 2. Middle: No e ect on odd modes. Maximal increase on even modes, since they maximally stretch at the center. 3. One third: Modes 12 small increase, Mode 3 max increase, Mode 4 less increase, Mode 5 almost 0, Mode 6 max?

21 Perturbing string with one end free I Maximum motion at open end I No motion at wall I The vocal tract is like a tube filled with air that is closed at one end (larynx) open at the other end (lips). I Modes of vibration of air in a tube are like those of the vibration of a string that is fixed to a wall at one end. Modes are what we call formants!

22 Perturbing string with one end free m! k! m!k! )#Open# End All#F i #" ###### All#F i #" 2)#Closed# End #### All # F i #! All#F i #! 3)#1/3 F 1 :#Sm#" F 2 :#Max#" F 1: :#Lg#! F 2 :### F 1: :Mid##! F 2 :#Max#" 4)#2/3 F 1 :#Lg#" F 2 :#### F 1 :#Sm#! F 2 :#Max#! F 1 :#Mid#" F 2 :#Max#! 5)#Mid F 1 :#Sm" F 1 :#Sm! F 1 :#### F 2 :#Sm" F 2 :#Sm! F 2 :####

23 Effect of constriction in VT: m k I Portions of air have mass and springiness. I Constricting a portion of air by constricting a tube: I Raises the mass, since packed molecules are harder to move, i.e. a constriction raises density. I Raises the sti ness (as in a tire), i.e., a constriction raises pressure. I So a constriction in a tube amounts to raising both mass and sti ness at the location of the constriction.

24 Effect of constrictions for /i/ and /a/ m! k! m!k! iy 1)#Open# End All#F i #" ###### All#F i #" aa )#Closed# End 3)#1/3 F 1 :#Sm#" #### All # F i #! All#F i #! F 2 :#Max#" F 1: :#Lg#! F 2 :### F 1: :Mid##! F 2 :#Max#" 4)#2/3 F 1 :#Lg#" F 2 :#### F 1 :#Sm#! F 2 :#Max#! F 1 :#Mid#" F 2 :#Max#! 5)#Mid F 1 :#Sm" F 1 :#Sm! F 1 :#### F 2 :#Sm" F 2 :#Sm! F 2 :####

25 String and air in vocal tract Closed-Open vocal tract is analogous to fixedfree string. Cons. at the lips /u/ = reduce all formants Cons. at the glottis = raise all formants Cons. at 1/3 VT /a/: F1 up and F2 down Cons. at 2/3 VT /i/: F1 down and F2 up

26 Adding a little Math Now we want to actually calculate how much each formant frequency changes as a constriction is introduced. The simplest perturbations are perturbations of a neutral schwa-like tube. What do we need for the math? 1. Mode Frequencies for schwa 2. Function representing the perturbation 3. Function representing how sensitive each mode frequency is to a perturbation at each location in tube

27 1. Formant frequencies of unconstricted tube I What are formant frequencies of vocal tract with no constrictions? I Mode frequencies are evenly spaced, depend on length of tube and speed of sound (c = 33, 400cm / s). I F n = nc 4L n =1, 2, 3, 4... I For a 17 cm vocal tract F=500, 1500, 2500,

28 2. Perturbation Function The perturbation itself can be expressed as a list of numbers (vector), with 0 s through it all, except for the point in space at which a perturbation exists, e.g.: Simple /a/: [ ] Simple /i/: [ ] Simple /u/: [ ]

29 3. Sensitivity Functions 1 0 F standing wave mass effect stiffness effect 1 0 Velocity Standing Wave (V) Pressure Standing Wave (P) is 90 degrees out of phase Effect of M on ω α - (abs(v)) 2 Effect of K on ω α (abs(p)) 2 SF = Keffect + Meffect sensitivity: stiffness + mass effect glottis lips 29

30 Sensitivity Functions standing wave stiffness effect mass effect glottis F2 sensitivity: stiffness + mass effect 1 0 lips glottis F standing wave mass effect stiffness effect sensitivity: stiffness + mass effect lips 30

31 % sens.m % Louis Goldstein % 3 Oct 2015 % compute sensitivity function % input arguments % i = formant number % L = length of vocal tract % step_size = size of steps along vt in cm step = [0:step_size:L]; lambda = 4*L/(2*i-1); spatial_freq = 1/lambda; rad_per_cm = spatial_freq*2*pi; wave1 = sin(rad_per_cm*step); wave2 = -wave1; stiff=(cos(rad_per_cm*step).^2); mass = (sin(rad_per_cm*step).^2); S=stiff-mass; %wavelength of mode %cycles/cm %standing wave %stiffness effect %mass effect 31

32 igure (1); clf; subplot (4,1,1) plot (step, wave1, 'k', step, wave2, 'k', 'LineWidth', 2); xlabel ('standing wave', 'FontSize', 18, 'FontName', 'Arial') subplot (4,1,3) plot (step,stiff, 'g', 'LineWidth', 2); hold on plot (step,zeros(length(step),1), 'k'); hold off xlabel ('stiffness effect','fontsize', 18, 'FontName', 'Arial') ylim ([-1 1]); subplot (4,1,2) plot (step,-mass,'r', 'LineWidth', 2); hold on plot (step,zeros(length(step),1), 'k'); hold off xlabel ('mass effect','fontsize', 18, 'FontName', 'Arial') ylim ([-1 1]); subplot (4,1,4) plot (step,s,'linewidth', 2); hold on plot (step,zeros(length(step),1), 'k'); hold off xlabel ('sensitivity: stiffness + mass effect','fontsize', 18, 'FontName', 'Arial'); 32

33 Sensi4vity Func4ons mé ké ) Open End All F i decrease ) Closed End All F i increase 3) 1/3 F 1: : Mid é F 2 : Max ê ) 2/3 F 1 : Mid ê F 2 : Max é 5) Mid F 1 : F 2 :

34 Weight Perturbation Function by SF Now we have two functions of x, the perturbation function (P) and the sensitivity function (SF). At each x, multiply the one by the other: P.* SF and sum over x (inner product): P * SF The points at which PF = 0, don t contribute to the inner product. x-points at which PF = 1 for which SF = 1, will contribute a 1 to the sum and should increase freq. x-points at which PF = 1 for which SF = -1, will contribute a -1 to the sum and should decrease freq.

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38 Ehrenfest s Theorem: effect of perturbation F i = NF i + (P'*SF) 2 NF i NF = neutral formant value

39 Results Neutral Freq P * SF New Freq /i/ /a/ /u/ schwa

40 P Nomogram value of formants when constric4ng at loca4ons along the tube P Distance from Glottis (cm) Distance from Glottis (cm) 2000 Hz Distance from Glottis (cm) 40

41 No lip constriction Added lip constriction Hz 1500 Hz Distance from Glottis (cm) Distance from Glottis (cm) 41

42 function P = makep (L, step_size, CL, C_width, Pval, lip_val) % makep % Louis Goldstein % 23 OCT 2012 % Calculate perturbation function from parametric specification % % input arguments: % L length of tube (cm) % step_size length of tube sections (cm) % CL ` location of constriction from glottis (cm) % C-width Width of constriction (length along tube) % Pval maximum value of P % lip_val separate value of P at lip end step = [0:step_size:L]; w = fix(c_width/(2*step_size)); [val, c] = min(abs(step-cl)); P = zeros (1, length(step)); P(c-w:c+w) = Pval; P = P./sum(P); P(end) = lip_val; plot (step, P, 'o-','linewidth',3); xlim ([0 17]); xlabel ('Distance from Glottis (cm)', 'FontWeight', 'bold') ylabel ('P','FontWeight', 'bold') shg 42

43 % make_sens_nomo % Louis Goldstein % 23 OCT 2012 % Plot Nomogram based on sensitivity function L = 17; step_size =.5; steps = L*step_size; C_width = 4; Pval = 1; lip_val = 1; Fall = []; first = floor(c_width/(2*step_size))*step_size; last = L-floor(C_width/(2*step_size))*step_size; CL=[first:step_size:last]; for CL=first:step_size:last; P=makeP(L, step_size, CL, C_width, Pval, lip_val)/1.5; F = apply_pert (L, step_size, P)'; Fall = [Fall F]; end CL=[first:step_size:last]; plot (CL, Fall, 'LineWidth',3); xlim ([0 17]); xlabel ('Distance from Glottis (cm)', 'FontWeight', 'bold') ylabel ('Hz','FontWeight', 'bold') 43

44 function F = apply_pert (L, step_size, P) % pert.m % Louis Goldstein % 16 Oct 2012 % calculate effects of a point perturbation (consgtriction) on the % resonances of a 17 cm neutral tube % % input argument: % length of vocal tract in cm % L length of vocal tract in cm % step_size distance between successive tube sections % P perturbation function % % output arguments: % F new (perturbed) formants Fneut = [ ]; step = [0:step_size:L]; for i = 1:3 lambda = 4*L/(2*i-1); spatial_freq = 1/lambda; rad_per_cm = spatial_freq*2*pi; % steps along tube from glottis to lips %wavelength of mode wave1 = sin(rad_per_cm*step); %standing wave wave2 = -wave1; stiff=(cos(rad_per_cm*step).^2); %stiffness effect mass = (sin(rad_per_cm*step).^2); %mass effect S=stiff-mass; % total sensitivity end F(i) = Fneut(i) + ((P*S')/2)*Fneut(i); 44