Implementación del Sistema Experto en la Etapa

It has been seen how the frequency dependence of the directivity of the array can be established by finding an expression that encapsulates the generated force distribu- tion, and then Fourier transforming it into the reciprocal k−space domain. Hence, in order to calculate the directivity of a PPM EMAT, such an expression must be found. As the length of the PPM array elements is much larger than either the width,a, or the depth, b, the problem can be reduced to two dimensions. Hence, by modelling the load distribution as a linear array ofN sources of alternating polarity, as shown in figure 3.4, the expression is given by:

f(x, ω) = f0(ω)p(x1, x2)~ N−1 X n=0 (−1)nδ2 x1− nd 2 , x2 ! , (3.8) where f0(ω) represents the frequency content of the drive signal, d is the peri-

odic distance of the array (equal to twice the pitch), δ2

x1− nd₂ , x2

is the two dimensional Dirac delta function representing the centre of the nth _{array element,}

p(x1, x2) is the spatial impulse response that describes the shearing force generated

the elements of the array, and _~ represents the convolution operation. The recip- rocal lattice can be calculated by applying a two-dimensional Fourier transform to convert the co-ordinates x into wavenumberk−space, leading to:

F (k, ω) = f0(ω)P (k1, k2)

N−1

X

n=0

(−1)neink21d, (3.9)

where P (k1, k2) is the array element impulse response in the Fourier domain.

The PPM array elements can be described as a rectangular function, as shown in figure 3.4, meaning their impulse response, in the far-field, can be calculated.

P (k1, k2) = ∞ ˆ −∞ ∞ ˆ −∞

Rect (a)e−ik1x1 _{Rect (b)e}−ik2x2_dx

1dx2 = 1 ab b 2 ˆ −b 2 a 2 ˆ −a₂ e−ik1x1_e−ik2x2_dx 1dx2 = sinc k1a 2 ! sinc k2b 2 ! . (3.10) However, the thickness of the force distribution, along the x2−direction, will be

limited by the electromagnetic skin depth [17]. Consequently, b is very small and hence sinck2b

2

≈1, meaning that this term can be neglected in most cases. Hence, the force load distribution is given by [101]:

F (k, ω) =f0(ω) sinc k1a 2 !_N−1 X n=0 (−1)neink21d. (3.11)

This expression can be used to investigate the reciprocal lattice of a PPM EMAT, which is plotted in figure 3.5. The array modelled in figure 3.5 is a six element array consisting of 4 mm wide elements, with a pitch of 5 mm. Unlike the atomic structure for crystals, which are periodic in three dimensions, the array only repeats along the

x1−axis. Consequently, the reciprocal lattice shows lines of constant intensity along

the k2−axis, with the lattice only varying alongk1−axis. There are two diffraction

‘rods’, which can be seen at ±628 m−1_{, corresponding to the fundamental spatial}

frequency of the 5 mm pitch array. In addition, due to fact that the reciprocal lattice is symmetric about k1 = 0, it can be seen that two equal magnitude beams

Figure 3.5: The force load distribution of a six element, 5 mm pitch PPM EMAT array, shown in the frequency domain. The circles represent the dispersion rela- tions of two bulk shear waves propagating in aluminium (vs= 3111 ms−1). The

smaller radius circle represents a 400 kHz shear wave, whilst the larger radius circle represents the dispersion relation for a 600 kHz shear wave.

about its centre as it contains an even number of elements. Hence, by symmetry, the generated wavefronts will also be anti-symmetric and the beams, whilst having equal magnitudes, will have a 180◦ phase shift relative to each other. By considering this phase shift and using a separate transducer to detect the wavefront, it is possible to determine whether the wave was launched in the +θ or the −θ direction.

It should also noted that, due to the alternating polarity of the array, the second order diffraction terms have been eliminated from the reciprocal lattice, although third order diffraction bands can be seen at ±1885 m−1_{. These third order diffrac-}

tion rods are much lower in intensity than the first order rods, due to the frequency dependence of the element impulse response. The directivity of an individual ele- ment, encapsulated in the P (k1, k2) = sinc (k1a/2) term, allows lower frequencies to

be transmitted at a greater efficiency than higher frequencies. Consequently, the higher order diffraction modes are suppressed, relative to the first order diffraction terms. There are also a number of ‘side lobes’ around the main diffraction rods, and these are a consequence of the limited number of elements in the array. As there are only six elements in the array, the interference could not build up to a sufficient level to eliminate them completely. An increased number of elements would act to reduce the side lobe levels, as well as sharpen up the main diffraction beams [39,101], although this would increase the length of the signal in the temporal domain, as a consequence of time-frequency duality.

In addition to the reciprocal lattice of the array, an expression for the dispersion relation of the ultrasonic signal used to excite the array must be found. This will depend upon the material into which the array is operating, and special consideration must be paid to the relevant dispersion curves. However, for bulk waves in an isotropic medium, the dispersion relation is given by:

˜ D(k, ω) =k2₁+k₂2− _ω c 2 = 0. (3.12) This means that, in 2Dk−space, the dispersion relation reduces to a circle, with a radius, kT: kT = q k2 1 +k22 = ω c. (3.13)

Whilst it is not strictly isotropic, aluminium can be considered to be macro- scopically isotropic with variations in shear speed of typically less than a few per- cent [166, 167]. Anisotropy can be easily handled by this model, provided that the dispersion relation can be calculated and parametrised [159]. The dispersion relation

shown in figure 3.5. From equation 3.7, the maximum displacement amplitudes are achieved when both the dispersion relation and the reciprocal force distribution are maximised. Graphically, it can be seen when this condition is met as it corresponds to the locations where the dispersion relation overlaps the bands of strong intensity of the reciprocal lattice. As this happens within confined regions ofk−space, the ul- trasonic beam is steered to a particular angle,θ = arctank1

k2

. The total directivity of the array being driven at a particular frequency can be calculated from equation 3.11. For a given drive frequency ω =kTc, then k1 =kT sinθ and so equation 3.11

gives the directivity functiond(θ, ω):

d(θ, ω) =f0(ω) sinc ωasinθ 2c !_N−1 X n=0 (−1)neinωdsinθ/2c . (3.14) However, a real signal will not contain a single frequency, ω, but a range of frequencies. If the frequency spectrum of the excitation signal is given by S(ω), then the directivity of the array driven by such a signal is:

d(θ) = ˆ ∞

0

S(ω) sinc ωasinθ 2c !_N−1 X n=0 (−1)neinωdsinθ/2c dω. (3.15) Equation 3.15 can be used to numerically calculate the directivity by truncating the integral to an appropriate, finite limit. Example directivities in figure 3.6 show how the beam is steered by simply changing the input frequency. A clear shift in the peak position of the main beam can be seen, with the steering angle changing by simply varying the frequency of the excitation signal. It is also clear that the beam is brought into tighter focus at the lower angles, which is achieved by activating the array with a higher frequency signal. This is due to the inverse proportionality between frequency and angle, that can be seen from the diffraction grating equation; the angular spread is much smaller, for a given bandwidth, at the higher frequencies than the lower frequency values.

The driving signal obviously also has an effect on the beam width obtained by frequency steering. If the array is driven with a narrow bandwidth signal, the ultrasonic energy will be steered over a very narrow range of angles as there are only few frequencies present, and so there is only a limited range of angle over which the signal can constructively interfere. However, a narrow bandwidth signal necessarily implies that it will have a large temporal range, hence leading to poor axial resolution. Conversely, a temporally well-defined pulse, whilst having good axial resolution, will contain a range of frequencies and therefore have poor beam steering capabilities. Therefore, there is an explicit compromise between angular and axial resolution caused by time-frequency duality; increased angular resolution directly causes a reduction in the axial resolution that could be achieved, and vice versa.