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Modeling of an acoustic transducer and calculation of force

  1. Aug 8, 2010 #1
    Dear all,

    This problem is on the modeling of an unfocused receiving acoustic transducer and compute the electric signal due to the pressure field. I have come up with this simple model in 2D. Basically, the transducer has a flat surface S and the surface is assumed moving in the direction normal to its surface, i.e. direction of the force F.

    To the best of my knowledge, the force F is proportionally to the induced electric signal. So the problem becomes computing the F on the transducer surface from the pressure field. The finite difference time-domain method is used to calculate the pressure P and the particle velocity Uy and Uz as shown in the figure.

    I have found two possible ways to calculate the force and need some feedback.

    a) Linearly interpolate the pressure field on S, and

    [tex] F = \int p dl [/tex]

    where dl is some small distance on the surface and p denotes the pressure.

    b) Linearly interpolate Uy and Uz on S. Since the transducer surface only moves in the direction of F, I could calculate the net velocity

    [tex]
    U = U_y\sin\theta-U_z\cos\theta
    [/tex]

    Then, calculate the force through

    [tex]
    F = -\rho \frac{\partial } {\partial t} U \Delta l
    [/tex]


    where the next velocity U at the two adjacent time steps are needed for the time derivative.

    Since I have not worked with transducer before, firstly, I am not sure if this modeling is correct (the surface is about 3~2 cm in diameter for ultrasound imaging application). Secondly, I am not sure which one (a) or (b) is the proper way to compute the electric signal. My feeling is (b) as it takes into account of the direction. Any discussion or any reference on this kind of modeling is welcome.

    Thank you.


    Elgen
     

    Attached Files:

  2. jcsd
  3. Aug 9, 2010 #2
    If it's an electro-magnetic pickup I think it will respond in proportion to the _change_ in pressure/velocity, so option 2 might be the only one that works. A piezo might be proportional to the pressure itself.

    Just guesses on my part though...
     
  4. Aug 10, 2010 #3
    Since it is a transducer, your model needs to included coupled differential equations.
    One for the mechanical system and one for the electrical system.
     
  5. Aug 11, 2010 #4
    I am not sure why this is private, the idea of PF is for all to see.

    Schip is correct that for most transducers it is change of pressure that creates a voltage, not sustained pressure. In the case of electrodynamic types it is rate of change.

    So you need to consider the transducer type.

    The actual electrical output will depend upon the circuit conditions so yes, you need to consider these.

    I am not sure what you mean by aperture? This implies it is a function of area. Do you mean the angle the receiving surface makes with your grid?

    Why can you not align your grid to the receiving surface? Pressure is always normal to this.
     
  6. Aug 11, 2010 #5
    That is clear. Thank you. I am concerned with the common transducers.

    By aperture, I refer to the surface area of the unfocused acoustic transducer, e.g. circular, rectangular. I want to draw some quantitative conclusion on the ultrasound image resolution with respect to this area.

    The design of the system, in 2-D is to have several transducers placed a few cm away from the circular object. These transducers are all receiving and their surfaces are not necessarily aligned with the grid.

    On a second thought, the pressure (here pressure refers to the difference between the instantaneous absolute pressure and the standard pressure) is a scalar quantity equal to

    force / area

    which assumes that the direction of the force is perpendicular to the area of the surface. If the linear acoustic wave equation gives pressure at a point and at a time, it does not have any indication on how to measure this pressure.

    The measured pressure actually depends on how the surface is placed. If the surface is perpendicular to the gradient of the pressure, the measured pressure is maximized. If the surface is parallel with the gradient, the measured pressure is 0.

    I guess, the pressure calculated from the wave equation really gives the maxim pressure that would be measured if a small surface is placed perfectly perpendicular to the gradient.

    Is this the correct interpretation?
     
  7. Aug 11, 2010 #6
    Since you have still not identified the transducer type I will offer a sample for electrodynamic type.

    The mechanical equation is

    [tex]m\frac{{{d^2}\xi }}{{d{\xi ^2}}} + r\frac{{d\xi }}{{dt}} + s\xi = DI[/tex]

    and the electrical equation is

    [tex]L\frac{{dI}}{{dt}} + RI + D\xi = E[/tex]

    Where
    E is the generated EMF,
    L is the circuit inductance
    R is the circuit resistance
    D is common to both equations and is dependent on the magnetic flux
    I is the current
    [tex]\xi [/tex] is the displacement of the surface
    m is the effective mass in oscillation
    r is the mechanical resistance to displacement
    s is the stiffness
     
  8. Aug 11, 2010 #7
    The transducer that I try to model is

    Panametrics V389-SU

    [PLAIN]http://www.olympus-ims.com/en/panametrics-ndt-ultrasonic/pdf/ [Broken] [/URL]

    From your words and some reading (Kinsler's book), I think it is a pressure transducer, immersion type. However, I am not sure if it is electrodynamic type or electrostatic type.
     
    Last edited by a moderator: May 4, 2017
  9. Aug 11, 2010 #8
    These are piezo tranducers. So long as the circuit resistance is very high, the output voltage is proportional to the displacement. The mechanical system should be stiffness controlled. This results in a considerable simplification of the equations.
     
  10. Aug 12, 2010 #9
    Thank you for these pointers on acoustic transducers. They have oriented my perspective.
     
  11. Nov 18, 2010 #10
    I believe there's an error in this equation.
    [tex]Emf=vLB[/tex]

    therefore:

    [tex]L\frac{{dI}}{{dt}} + RI + D\frac{d\xi}{dt} = E[/tex]
     
  12. Nov 18, 2010 #11
    Yes you are absolutely correct I missed a dot over Xi.

    Silly me.
     
  13. Jan 6, 2012 #12
    What exactly is this D? and shouldn't there be a term taking into consideration of capacitance?
     
  14. Jan 6, 2012 #13
    These equations refer to an electrodynamic voice coil of a loudspeaker so capacitance is not significant in the electrical system. Yes it probably will be in your piezo film.

    You have not said whether you are using your piezo film as an electrical to mechanical or mechanical to electrical transducer?

    D is the key since it connects both equations, along with the current and ([itex]\xi[/itex]) dot.

    D=Bl where B is the flux density in the coil gap, and l the coil length. It is therefore the force per unit current generated in the mechanical equation and hence when multiplied by the current is the forcing function on the right hand side of the mechanical equation.

    Since we are talking about a coil moving in a magnetic field this generates a back EMF. This EMF is proportional to the velocity of the coil ([itex]\xi[/itex]) dot.

    You will need to write similar equations and find quantities that appear in both for your analysis. The equations are likely to have complex roots.
     
    Last edited: Jan 6, 2012
  15. Jan 6, 2012 #14
    my piezo film is electrical to mechanical.

    The driving force is the applied alternating electric field which causes the vibration, so I don't think I can use D as the driving force. I will have to think of something else.

    By the way is this covered in standard text-books or the literature? Can't seem to find anything there which could act as a guide.
     
  16. Jan 6, 2012 #15
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