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
a.
i. Parameters needed are the distance from all points of the transducer
surface to the field points + the wave number
ii. 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 𝑣 = 𝑣0 𝑒 𝑖𝜔𝑡 (and neglecting any
constants).
Time domain: 𝜑(𝑟, 𝑡) = ∬
𝑣(𝑡−𝑅/𝑐)
2𝜋𝑅
Frequency domain: 𝜙(𝑟, 𝜔) ∝ ∬
𝑑𝑆 ∝ ∬
𝑒 −𝑖𝑘𝑅
𝑅
𝑒 𝑖𝜔(𝑡−𝑅/𝑐)
𝑅
𝑑𝑆
𝑑𝑆
b. Plot the normalized magnitude of the calculated diffraction pattern for the
given observation points.
c. Compare the beam-profile from 1 b) with the far-field directivity function
given in eq. 3.33.
i. 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)))
MEDT8007 2015 Exercise 1
d. 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?
i. 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.
2. Pressure field
a. 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).
i. 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;
b. 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?
i. 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):
𝜑(𝑟, 𝑡) = 𝑠(𝑡) × ℎ(𝑟, 𝑡)
𝜕𝜑(𝑟, 𝑡)
𝜕ℎ(𝑟, 𝑡)
= 𝑠(𝑡) ×
𝜕𝑡
𝜕𝑡
𝑃(𝑟, 𝜔) = 𝑆(𝜔)𝐻(𝜔)𝑖𝜔
Look at the review script Review_ex1_Ingvild.m for Matlab details.
3. Spatial impulse response
a. 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).
i. Here we have a set of analytical expressions giving us the impulse
response of a piston transducer.
b. Compared to the results in part 2, what can you say about the accuracy of
Huygen’s method?
MEDT8007 2015 Exercise 1
i. Pretty good, especially if the number of point sources is sufficiently
large. However, it is a bit slow….
c. What happens to the transient signal when you move the observation point
towards and beyond the edge of the transducer’s geometric shadow?
i. 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.