Harmonic motion. time 1 - A. Wednesday, September 30, 2009

Transcription

1 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 What is equation of motion for the block, and how do physical laws give rise to it? Changes in pressure over time in a part of the vocal tract also oscillates (about neutral pressure) and obeys a similar equation of motion.

2 Relevant physical laws Newton s 2nd Law: F = m a Sum of Forces Applied to an Object = Object Mass x Object Acceleration Inertial Force Hooke s Law: F = k(x x 0 ) k(stiffness) x Inertial Force is proportional to displacement from x0

3 Deriving the Equation of Motion x=f(t) What is the function? F = m a F = k(x x 0 ) ma = k(x x 0 ) Differential Equation (DE): Second derivative of x is proportional to x with a constant = -kx/m a = k m (x x 0) ẍ = k m (x x 0) x = x x 0 To solve, find x=f(t) such that it satisfies the DE. ẍ = ( k m )x

4 Complex Exponential (phasor) as a solution to differential equation x = e jωt ẋ = jωe jωt d dt (act )=ca ct ẍ =(jω) 2 e jωt ẍ = j 2 ω 2 e jωt ẍ = ω 2 e jωt if ω 2 = k m ẍ = ( k m )x Relation is satisfied. Mass will oscillate with frequency k/m k increases, ω increases m increases, ω decreases

5 A simpler discrete approach... Position Velocity Acceleration

6 Discrete Velocity and Acceleration

7 Deriving discrete difference equation using Newton s and Hooke s Laws ma = k(x x 0 ) X = x x 0 a = k m X D(D(X n )= k m X n 1 X n 2X n 1 + X n 2 = k m X n 1 X n =2X n 1 k m X n 1 X n 2 Looks like feedback filter equation: Future points predictable from past ones. X n = (2 k m )X n 1 X n 2

8 Feedback filter can implement oscillator Feedback Filter equation: Y(k) = b(1)*x(k) + b(2)*x(k-1) b(m)*x(k-m+1) -a(1)*y(k-1) - a(2)*y(k-2) a(l)*y(k-l) Y(k) = -a(1)*y(k-1) - a(2)*y(k-2) X n = (2 k m )X n 1 X n 2 Y(k) = Xn ; Y(k-1) = Xn-1 ; Y(k-2) = Xn-2 ; a(1) = (2 k m ) a(2) = 1 Examine impulse response of this filter. (Give it some initial value and it will oscillate): Y (k) =X(k) + (2 k )Y (k 1) Y (k 2) m

9 function undamped_osc(k,m) % % Louis Goldstein % 29 Sept 2009 % % simulate oscillator with freq k and mass m % using feedback filter approximation % generate impulse x=[1 zeros(1,99)]; % coefficients of feedback filter a(1) = 1; a(2) = -(2-k/m); a(3) = 1; k/m =.1; ω=.316; period = 19.9 % filter impulse y = filter(1,a,x); figure (1); plot (y,'or-'); grid w=sqrt(k/m); period = 2*pi/w; title (['k/m = ' num2str(k/m)... '; omega should equal ' num2str(w)... ' radians/sample; period should be '... num2str(period)]); Approximation begins to be inaccurate for ω >1 k/m =.05; ω=.223; period = 28.09

10 Damped Oscillations Oscillations will lose energy to friction Friction Force + Spring Force = Inertial Force (resists motion)

11 Equation of motion of damped oscillations X n 2X n 1 + X n 2 = b(x n X n 1 ) k m X n 1 X n + bx n =2X n 1 X n 2 + bx n 1 k m X n 1

12 Equation of motion of damped oscillations X n 2X n 1 + X n 2 = b(x n X n 1 ) k m X n 1 X n + bx n =2X n 1 X n 2 + bx n 1 k m X n 1 (1 + b)x n = (2 + b k m )X n 1 X n 2 X n = 2+b k m 1+b X n b X n 2 feedback filter coefficients

13 function osc(km,b) % Louis Goldstein % simulate oscillator with k/m and damping b % using feedback filter approximation % km -- ratio k/m % b -- damping coefficient % generate impulse x=[1 zeros(1,99)]; % calculate coefficients of recursive filter using approximation a(1) = 1; a(2) = -(b+2-km)/(1+b); a(3) = 1/(1+b); % filter impulse y = filter(1,a,x); figure (1); plot (y,'ro-'); grid % plot tranfer function of filter figure(2); transfer(b,a); % compare predicted value of w (=sqrt(k)) with actual pole frequency of filter allroots = angle(roots(a)); normroots = allroots; normroots(allroots<0) = pi + normroots(allroots<0); pole = min(normroots)/pi; w=sqrt(km)/pi; title (['k/m= ' num2str(km) '; w = sqrt(k/m) = ' num2str(w) ' * pi rad/sample; pole = num2str(pole)]);

14 Damping and Bandwidth k/m=.1, b=.01 k/m=.1, b=.1

15 Formants as Damped Oscillators Formants can be modeled as damped oscillators, with a frequency and damping. Vowel-like sounds can be produced by filtering a buzz through a cascade of three resonators, corresponding to F1, F2, F3. The filter coefficients can be determined from the frequency, damping, using the equations derived above. buzz F1 F2 F3

16 function signal = vowel (F1,F2,F3) % create a crude vowel by filtering a buzz % through a cascade of 3 damped oscillators % one corresponding to each of the first three % formant frequencies % input arguments: F1-F3 in Hz % output argument: signal-- synthesized vowel signal b=.01; % resonator damping sr = 20000; % sampling rate buzz= make_buzz(sr,200); %f0=200 % F1 oscillator %(1) convert frequency from Hz(cycles/sec) to w (radians/sample) w=hz2w(f1,sr); %(2) calculate filter coefficients for damped oscillator of freq=w, damping=b a(1) = 1; a(2) = -(b+2-w^2)/(1+b); a(3) = 1/(1+b); %(3) perform filtering signal = filter(1,a,buzz); [your code here] soundsc(signal,sr)