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**ICMC 95, Banff (Canada) 1995**

Copyright © Ircam - Centre Georges-Pompidou 1995

Let us call x the vertical position of the mass, A the area of the orifice, r the air density and P

Then the acoustic pressure at the inlet (p

,Let d be the mass thickness and l the orifice length. The pressure force when the orifice is not closed is [Flanagan 68]

But the measurements of [van den Berg 57] are contested [Pelorson & al. 95]. Let us now consider a different hypothesis and assume a jet flow out of the orifice. Then, with sgn(P

If the tube is open to the free atmosphere after the mass, P=0, there is no vertical force on the mass! The jet flow is a simplification which leads to a non interesting case for our simplest one-mass model uncoupled to any tube.

Using [Flanagan 68], we now consider the particular case where the tube opens into the "free atmosphere" after the orifice between the mass m and the opposite wall of the tube. This constitutes a one-mass model not coupled to any resonator. Let us show that the volume velocity U passing through the orifice depends only on the pressure P

Because we are interested only in rather low frequency behavior, let us neglect Lg . Then

Therefore we can write U as a function of P

(1)

where k>0 is the spring constant and r>0 is the viscous damping. The corresponding state equation is:

It is easy to show that, since the force depends only on x, for r > 0 this system cannot oscillate in the sense that it does not have a periodic trajectory [Rodet & Steinecke 94]. Different possibilities exist to make it oscillate:

- Have a force term which depends on the velocity . The violin - i.e. bow-string interaction - is an example. It can easily be proved
that a one-mass model with a bow-string like force function has at least one periodic trajectory.

- Add one dimension to the system by adding a second mass leading to a two-mass model
where the two masses are linearly coupled (by a spring) and non linearly coupled
by the flow. This is made simple with the assumption of a jet flow through the minimum
opening [Steinecke 93]. Such two-mass model can oscillate

- Add a with memory
system (this adds some dimension) to the one mass. For instance a flow-pressure coupling
by a linear system, such as an acoustic tube, which can also be considered as a delay
term. We now focus on this case. It is simple since it consists of a one dimensional oscillator coupled to a delayed feedback loop [Rodet 93a]. It seems to model
some important behavior of trumpet like instruments.

In [Fletcher 91]'s notation we have

Let us now consider which pressure effectively acts on the mass. There are different ways to derive the expressions for the force. We want to discuss and compare the force given by Fletcher (a general equation to describe wind driven instruments), the force which Flanagan uses for the one-mass model of vocal cords and the equations which we developed to implement the model. All these models even if they do not represent a classical musical instrument, are nonlinear dynamical systems which may have interesting musical properties.

associated with a complicated "flow geometry". The resulting force is the sum of a flow independent term which displaces the mass even if there is no flow and a flow dependent term (a modified Bernoulli pressure). The "static", flow-independent, force is very approximately proportional to the pressure difference P

To calculate the volume flow velocity Fletcher uses the empirical equation

The parameters used for the trumpet correspond to a kinetic-only resistance in the orifice. With this parameters the equation for the force results into:

For calculating U, Flanagan considers the pressure drop over a viscous resistance and a kinetic resistance. Here we neglect the inductance which describes the acceleration of the flow.

For large opening the term R

For the resulting force we get:

where g6 and g7 correspond to Fletcher's notation. From our point of view g6 = ld as in Flanagan's model. For an idealized Bernoulli flow and for P

To include other losses we can use a parameter g

(2)The resulting force is

where are constants of the corresponding models.

The resulting forces are equal when:

Setting in the implemented model leads to Flanagan's model. Fletcher's model is really different, because F = CFle(P

(3)When the lips are closed there is no flow and the flow dependent term vanishes leaving only the force which results from the coupling to the pressure difference. During the impact we add an additional restoring force which is a linear function of x. The corresponding spring constant is c = 3k. The equation of the mechanical oscillator is therefore:

(4)

andwhere y is the direction of the planar wave, the pressure p is the sum of the outgoing pressure po and the incoming pressure pi, c is the sound velocity and the volume flow U at y=0, i.e. at the lips position, is the sum of the flow of the outgoing and of the incoming waves. The volume flow velocity and the pressure are coupled by the acoustical impedance in the tube Z = rc/Acyl where r is the air density and Acyl is the cross sectional area of the tube. The pressure pi is calculated as a convolution of the outgoing pressure po with a reflection function r(t):

The reflection function r(t) can be formulated so as to make explicit the delay T = 2L/c corresponding to one round trip when L is the effective length of the tube. To get a simple model we often use only a reflection coefficient designated by reflect:

Finally, the pressure p is:

The acoustic tube is then coupled to the mechanical oscillator by the given equations for the pressure P and the volume flow velocity U. The relationship for the Bernoulli flow and the relationship for the wave in the tube are to be taken in consideration. Following [McIntyre et al. 83], values for P and U are obtained by solving the following system of three equations: of variables P, U and ph. Then the system is solved by iteration. We take a given x and ph as the incoming pressure wave which is supposed to be not influenced by the outgoing wave. Knowing ph we get U and P, these values are used to calculate the acting force on the mass and solve the difference equation for x. Note that U and P depend on x which is taken as a parameter to solve the equation and calculate the force. In the implementation we have:

For we get

upper sign for PFor we get:_{s}>ph, lower for P_{s}<ph,

The graphical solution of U and P is shown in Fig. 4. For the curve 1 reaches the axis and the curve 2 results into P = ph. We can interpret the system as a circuit which consists of two loops. One loop describes the dynamics of the mechanical oscillator. For a known pressure P we can solve the differential equation for x which we use to calculate P which also depends on the evaluation of the acoustic wave in the tube.

This is represented by the second loop which gives the pressure of the incoming wave ph=2pi. Both loops are coupled by a nonlinear function to calculate P(x,Ph) and U(x, Ph) respectively. Finally, this can be simplified in P alone (Fig. 5).We think that this system captures an important part of the behavior of trumpet-like instruments in the interaction of the modes of the tube loop with the own resonance of the mass subsystem [Rodet 92].

Length of the lip: l= 0.01, thickness of the lip: d = 0.005. Values used for the 2-mass model of the vocal cords are smaller, the vibrating part of the vocal cords has a thickness of d = 0.003.

Mass of the lip: m

m = m

By keeping r constant, the Q-factor does not change. In the program we can use f

For the reflection at the open end we use:

According to Fletcher the upper limit of the mouth pressure is around 10 kPa.

Empirical parameters

As mentioned above we assume an idealized flow and use the parameter

Then from equation 3 in section V the force term becomes:

,Finally, to avoid a discontinuity in the force term, we use the function , (see equation 2 of section IV) in the right term also, as already written in section V:

This different relationship between F and P leads to a different dynamic behavior. If F increases with P, the mass follows the pressure, because the maximum of P corresponds to the maximal force leading to a maximum of the displacement. For

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