Perturbation Theory: Why do /i/ and /a/ have the formants they have? Wednesday, March 9, 16

Transcription

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

2 Components of Oscillation 1. Springs resist displacement from rest position springs return to rest position ẋ = k(x x 0 ) Stiffness (k) determines how quickly they return to rest Money in Bank Time

3 Components of Oscillation 2. Masses resist changes in velocity v = F m Larger masses resist velocity change more strongly

4 Combining components What happens if you combine a mass and a spring? Pull the object at the end of the spring, and it will return to its rest position, but because the mass is in motion, it wants to stay in motion. (That is what masses do). Motion causes spring to compress, then the spring wants to return to its rest position again Result is oscillation around rest position.

5 Vibration: Harmonic motion time A x 0 + 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 = k m 5 + A Changes in pressure over time in a part of the vocal tract also oscillates (about neutral pressure) and obeys a similar equation of motion.

6 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?

7 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.

8 Three Masses

9 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.

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

11 Infinite modes of String

12 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?

13 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

14 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

15 Summary Principle 1: PT for 1- Mass 1- Spring is mé fê ké fé Principle 2: A vibratory system with more than one mass can have more than one mode of vibra4on, and the higher modes have higher frequencies of vibra4on.

16 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.

17 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.

18 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 3: Effect of mass or s4ffness perturba4on depends on the posi4on of the perturba4on Time

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

20 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?

21 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.

22 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?

23 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?

24 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 :####

25 String with one end free and vocal tract 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.

26 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.

27 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 :####

28 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

29 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

30 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,

31 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/: [ ]

32 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 - V 2 Effect of K on ω α P 2 SF = Keffect + Meffect sensitivity: stiffness + mass effect glottis lips 32

33 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 33

34 function S = sens_demo(i, L, step_size) % 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 34

35 figure (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' 35

36 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 :

37 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.

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

39 function F= pert_point (L, step_size, loc) % pert.m % Louis Goldstein % 16 Oct 2012 % calculate effects of a point perturbation (consgtriction) on the % resonances of a 17 cm neutral tube % % input arguments: % L length of vocal tract in cm % step_size distance between successive tube sections % loc location of perturbation in cm from glottis % % output arguments: % F new (perturbed) formants c= 33400; n = [1 3 5]; Fneut = (n.*c)./(4*l); step = [0:step_size:L]; % steps along tube from glottis to lips [val, pert_step] = min(abs((step)-loc)); % pert_step is where the perturbation is P = zeros(1,length(step)); P(pert_step) = 1; for i = 1:3 S = sens(i, L, step_size); S(pert_step) F(i) = Fneut(i) + ((P*S')/2)*Fneut(i); end 39

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43 Results /i/ Loc 11.5 Neutral Freq /a/ /u/ schwa P * SF New Freq

44 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) 44

45 function P = makep (L, step_size, CL, C_width, CD, lip_cd) % makep % Louis Goldstein % 9 March 2016 % 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) % CD degree of contriction (arb units) 0=none; 1=narrow % lip_cd separate value of CD for lips Pval=1; 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 = CD.*(P./sum(P)); P(end) = lip_cd; plot (step, P, 'o-','linewidth',3); xlim ([0 L]); xlabel ('Distance from Glottis (cm)', 'FontSize', 18, 'FontWeight', 'bold') ylabel ('P','FontSize', 18,'FontWeight', 'bold') shg 45

46 % make_sens_nomo % Louis Goldstein % Plot Nomogram based on sensitivity function % set parameters L = 17; step_size =.5; C_width = 4; CD =.75; lip_cd = 0; L = 17; step_size =.5; C_width = 4; CD =.75; lip_cd =.5;... 46