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It was assumed that the pressureat the lips is zero and the volume velocity source is ideal  no energy loss at the input and output. For radiation impedance:

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Presentation on theme: "It was assumed that the pressureat the lips is zero and the volume velocity source is ideal  no energy loss at the input and output. For radiation impedance:"— Presentation transcript:

1 It was assumed that the pressureat the lips is zero and the volume velocity source is ideal  no energy loss at the input and output. For radiation impedance: Morse and Ingard (1986), Beranek (1954), Flanagan (1972). Lip radiation is modelled by a piston in an infinite wall. The acoustic impedance is a resistance R r and an inductor L r in parallel. For the infinite wall R r = 1289 / 9  2 and L r =8a / 3  c Boundary Effects

2 The impedance equation can be converted to a differential equation via Laplace transform. Portnoff (1973) numerically simulated the above equation coupled to the wave equation.  Broader bandwidths and lowering of the resonances are observed. Higher frequencies are affected most. (This can be seen by considering Z r  0 when   0 (small)  p(l,t) = 0 but for large ,  L r >>R r  Z r = R r ) Boundary Effects

3 Glottal Source and Impedance Flanagan-Isızaka (1978) proposed a linear model for the impedance. This is a differential equation in the time-domain. The solution of this equation together with the wave equation yields broadening of the bandwidths at low frequencies. This can be seen from Z g = R g + j  L g becomes Z g  R g for small  ; (purely resistive). Overall Frequency Response introduces a highpass filter effect

4 Summarizing the results 1)Resonances are due to vocal tract. Resonant frequenciesshift as a result various losses. 2)Bandwidths of lower resonances are controlledby wall vibration and glottal impedance loss. 3)Bandwidths of higher resonances are controlled by radiation, viscous and thermal losses.

5 (Energy loss is assumed to be only at the lips.) For the k th tube (*) Boundary conditions are Model of Concatenated Tubes

6 The boundary conditions and (*) yield : Let  k = l k /c Using in (2) Model of Concatenated Tubes

7 (4) – (3)  Let  Both equations contain a component due to reflection and one component due to transmission. Model of Concatenated Tubes

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10 Lips: Suppose radiation impedance is real. At the N th tube Since There is no backward going wave from free space. Boundary Relations

11 Therefore Outgoing wave from the lips: Let the outer space be represented by an infinite length tube of cross section A N+1, and in particular, if (Also, if Z r (  ) = 0  r L =1, A N+1    no radiation from the lips.

12 Glottis: Suppose the glottal impedance is purely real, Z g =R g In particular, if glottal impedance is modeled by a tube of cross section A 0 and A 0 is chosen such that By chosig A k s properly it is possible to approximate formant bandwidths. Boundary Relations

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14 Consider a vocal-tract model consisting of N lossless concatenated tubes with total length l. length of a tube =  x = l/N propagation time in a tube =  =  x/c Let v a (t) be the impulse samples of the impulse response with 2  intervals. At the end of the tube v a (t) can be written as: Since the samples up to time N  will be zero. The Laplace transform and the frequency response are The frequency response is periodic with 2  /2  ; V a (  + 2  /2  ) = V a (  ) Discrete Time Realization

15 If the impulse response of the system is not band limited then aliasing will occur. The discrete-time frequency response can be obtained by  =  /2  Discrete Time Realization

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17 The effect of aliasing can be reduced by decreasing the individual tube lengths and hence the sampling period. It can be shown that the transfer function of the discrete-time model has the form Discrete Time Realization When A k > 0 all poles are inside the unit circle.

18 Ex: Let l=17.5 cm, c=350 m/sec. Find N to cover a bandwidth of 5000 Hz. (Assume that vocal tract impulse response and excitation are bandlimited to 5000 Hz.) Soln:  /2  is the cutoff bandwidth. 5000 Hz  10000  rad/sec  /2  = 10000    = 1/20000 N = l/c  = 10 Since N is the order of the system, there are 5 complex conjugate poles  There can be 5 resonances in the given bandwidth. Discrete Time Realization

19 Comparison of numerical simulation of differential equations and concatenated tube model (N=10) Discrete Time Realization Reflection coefficientsTube cross sections

20 Simulation Concatenated tubes


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