Ultrasonic modellingUltrasonic modelling

Using the Huygens – Fermat Principle

Philippe Rubbers SCM

Huygens Principle Huygens Principle In 1678 the great Dutch physicist Christian Huygens (1629-

1695) wrote a treatise called Traite de la Lumiere on the wave theory of light:

He stated that the wave front of a propagating wave of light at any instant conforms to the envelope of spherical wavelets emanating from every point on the wave front at the prior instant.

Diffraction patternsDiffraction patterns

Single slit: using Huygen’s principle

Equations not trivial

Near field: Fresnel diffraction

Far field: Fraunhoferdiffraction

Ultrasonic equivalentUltrasonic equivalent

Probe on a block of steel

Ultrasonic equivalentUltrasonic equivalent

Set of point sources User defined:

amplitudelocation (x,y,z)

E.g. 4 sources

Ultrasonic equivalentUltrasonic equivalent

Amplitude: probe area / no of user points

Position:random / equi-spaced

Defined by probe being simulated and user requirements

Building a diffraction patternBuilding a diffraction pattern

For each display location:define location = pixel

For each pixel:calculate distance to eachsource (e.g. 4 per pixel)

DistanceDistance

Calculate distance: trivial But require 3D Source : S(x,y,z) Pixel: P(x,y,z)

Distance: d(p-s)=Pi-Sj

AmplitudeAmplitude

Amplitude effect of each source

dAmplitude

dIntensity

1

12

∝

∝

Complex amplitudeComplex amplitude

Each source has an absolute maximum amplitude acting at the pixel of

Pixel also has a phase relationship to each source

Combining into a complex value

)(),(

sp

ssp d

AA−

=

λπθ 2)(),( spsp d −=

( ))sin()cos( ),(),(),(),( spspspsp AA θιθ +=

For each pixelFor each pixel

For each display pixel: Summation of effect of each point

source

∑=

=n

sspp AA

1),()(

Diffraction patternDiffraction pattern

For an image with x,y co-ordinates for each pixel: pixel

=

===

===

===

ynxnpynpynp

xnppp

xnppp

yx

AAA

AAAAAA

A

,,2,1

2,2,22,1

1,1,21,1

,

Amplitude Amplitude ⇒⇒ magnitude magnitude

A plot of |Amplitude| vs. position x,y for a single wavelength

Argument Argument

Similarly a plot of phase vs. position x,y for a single wavelength

yx,θ

Complex amplitudeComplex amplitudeCombining |Amplitude| and phase we obtain a single complex number for each location

Ultrasonic = broadbandUltrasonic = broadband

Probes excited using an impulse or other waveform.

Fourier transform: obtain Amplitude and phase of component frequencies

Complex number

For λ1

For λ2

.

.

.

.

For λn

Array of diffraction patternsArray of diffraction patterns

Fourier transformFourier transform

λcf =

wavelengthvelocitycfrequencyf

===

λ

We can use wavelength and frequency interchangeably:

1D array of complex values

Fourier transformFourier transform

Multiply diffraction patterns with pulse

UT pulseUT pulse

Addition of corrected diffraction patterns

SummarySummary

Find distances from each point source to each display point. 

Find Amplitude and phase for each point for each wavelength (frequency) of interest.

Multiply 3D array with excitation waveform. Sum the various frequency components and

get the resultant wave front pattern

½ way: any questions?½ way: any questions?

Fermat's principleFermat's principle

Principle of least time: the path taken between two points by a wave is the path that can be traversed in the least time.

Can be used to describe the properties of waves: Reflection and refraction through different media. It can be deduced from Huygens’ principle. It can be used to derive Snell’s law of refraction and reflection

Diffraction: complex geometryDiffraction: complex geometry

We now have a structured approach to imaging ultrasonic signals, however, what happens when we have multiple surfaces:

ReflectionRefraction

Shortest path (2D)Shortest path (2D)Known: c1, c2, y1, y2, xtUnknowns: x1, x2, θ1, θ2 Equations trigonometry:

21 xxxt +=

21

21

11sin

yxx

+=θ

22

22

22sin

yxx

+=θ

2

2

1

1

sinsin θθcc =

Shortest path (2D)Shortest path (2D)

Highly non-linear. 4th order polynomial Iterative solution

( )

+=

+−

−21

21

122

22

1

11

yxxc

yxx

xxct

t

DistanceDistance

Solve for each display pixel to each source !

Distance in wedge:

Distance in steel:

21

211 yxr +=

22

222 yxr +=

AmplitudeAmplitude

As before:

Where d(p-s) = r1+r2

Attenuation / reflection+transmission coefficients

)(),(

sp

ssp d

AA−

=

Amplitude: effect of interfaceAmplitude: effect of interface

The Ultrasonic testing of materials by Krautkramer and Krautkramer

Transmission curvesTransmission curvesFor example: solid / solid interface with liquid couplant

D2l is the relative amplitude of the diffracted longitudinal wave in the second material.

D2t is the relative amplitude of the diffracted transverse wave in the second material.

The equations assume that the incident wave in material 1 is a longitudinal wave. i.e. the probe crystal generates L waves only

lt

tt

t

tl

t

tt c

cN 21

41

22

224

11

422

11

41

2

1 tansin22coscot2tan

sin22coscot2 α

αραρα

ρρα

ααα +++=

tt

ttl N

D 21

421

1

22 tansin

2cos2cos αα

ααρρ=

t

t

t

tt Nc

cD1

21

211

222

2 sin2cos2

αα

ρρ=

Corrected amplitudeCorrected amplitude

So we have:

Where d(p-s) = r1+r2Dll = transmission coefficient

)(),(

sp

sllsp d

ADA

−

=

PhasePhase

As before:

But where:

λπθ 2)(),( spsp d −=

211

2)( rrc

cd sp +=−

Typical applicationsTypical applicationsIn nearly all UT cases, a wedge is used, so there is a minimum of one refraction.However for most cases we want to see the effect of a flaw or change in geometry, or even multiple changes in geometry with a flaw

DemonstrationDemonstration

Questions for the audienceQuestions for the audience

Mathematical complexityMathematical complexity

1 planar refraction: 4th order equation 2 planar refractions: 6th order equation

How do you simplify the mathematics?

Non planar reflectorsNon planar reflectors

What is the shortest path?What is the shortest path?

Side drilled holeSide drilled hole

Work to date: L wave around SDH (T waves not shown)

Possible error here

Side drilled holeSide drilled holeWork to date: T wave around SDH (L waves not shown)

When a Transverse wave hits a side-drilled hole at a tangent, what proportion of energy is transferred to the Rayleigh wave? When a Rayleigh wave looses energy, what proportion of energy is lost into a head wave at any instant? I assume that this is a function of radius.As the Rayleigh wave travels along a surface, at what depth below the surface does it propagate? My tests have indicated that this is a function of wavelength (e.g. 0.55λ or so, which may also be an indication of surface roughness)

AcknowledgementsAcknowledgementsThank you to Eskom Resources and Strategy for support funding.

Thank you to Manfred Johannes, Konrad Hartmann, Heintz J. Hilger, Hugh Neeson, Ed Ginzel, Arthur Every, Graham Wilson and Willem Nel for many discussions and for their insight, excellent suggestions and great support