MEDT8007 2015 Exercise 1

Exercise 1 - Suggested solution

Summary

1)We look at the CW diffraction pattern from a piston transducer using Huygens principle (the Rayleigh integral). Further, we compare it to the far-field (Fraunhofer) approximation.

2)By looking at a single observation point and repeating our calculations in 1) for all frequencies (same amplitude and zero phase), we are effectively calculating the response (the velocity potential) from a delta excitation on the transducer surface. As mentioned in the lectures, this is actually how we define the impulse response, which we can use to find the pressure at a certain observation point given a certain excitation of the transducer surface.

3)In 1) and 2) we had a numerical approach to the problem of finding the field response, as we modeled the transducer as a collection of point sources. For a piston transducer, an analytic expression exists, and in this final exercise we compare the numerical approach (Huygens) to the analytic expression.

Further details

  1. Huygen’s principle – single frequency diffraction pattern
  2. Parameters needed are the distance from all points of the transducer surface to the field points + the wave number
  3. Filling in the expression for the velocity potential in the frequency domain will do the trick for finding the diffraction pattern, assuming single frequency excitation (and neglecting any constants).

Time domain:

Frequency domain:

  1. Plot the normalized magnitude of the calculated diffraction pattern for the given observation points.
  2. Compare the beam-profile from 1 b) with the far-field directivity function given in eq. 3.33.
  3. We only care about the amplitude, so we take the absolute value of the directivity function. As mentioned in the lectures a rectangular aperture will result in a sinc shaped beam profile in the far field. For a circular aperture the far field beam profile is a Bessel function

In Matlab:

abs(2*besselj(1, k*a*sin(Theta))./(k*a*sin(Theta)))

  1. Can you explain the differences in the beam profiles? What happens if you change the resolution on transducer surface or the axial distance between the observation point and the surface?
  2. The far-field approximation is of course most accurate in the far field of our transducer, so the functions will be more similar when the axial distance is increased (beyond 3 cm). By increasing the resolution (number of point sources) on the transducer surface the accuracy of Huygens method will also be increasing.
  1. Pressurefield
  1. Calculate the full frequency response using the quadruple for-loop to iterate over a set of frequencies. Make sure that these are matched to the given excitation signal, s(t).
  2. We are going from a CW to a PW signal and need to do our calculations using a vector of frequencies, i.e.

f = linspace(-0.5,0.5-1/N_FFT,N_FFT)*f_Sample;

  1. Given the excitation signal, s(t), find the transient signal p(r,t) in a single observation point r1. Are there several ways to obtain the same result?
  2. Here we should be utilizing the relations between the impulse response h(r,t) and the momentum potential/the pressure, using s(t) as the velocity excitation of the transducer. The two solutions are given by: 1) Convolution in the time domain and 2) Multiplication in the frequency domain. NB: To find the pressure signal, the temporal derivative of the momentum potential needs to be found, either in the frequency domain, or in the time-domain.

Remember (using x to indicate convolution):

Look at the review script Review_ex1_Ingvild.m for Matlab details.

  1. Spatial impulse response
  1. Compare the matlab function piston_sir(a,x0,y0,z0,fs,c0) to the expressions in section 3.3.1 and use this function to recalculate p(r,t).
  2. Here we have a set of analytical expressions giving us the impulse response of a piston transducer.
  3. Compared to the results in part 2, what can you say about the accuracy of Huygen’s method?
  4. Pretty good, especially if the number of point sources is sufficiently large. However, it is a bit slow….
  5. What happens to the transient signal when you move the observation point towardsand beyondthe edge of the transducer’s geometric shadow?
  6. When we move out of the geometric shadow, the signals from the periphery of the transducer no longer overlap in time, so the transient pressure signal is split in two.