Presentation is loading. Please wait.

Presentation is loading. Please wait.

Advanced Imaging 1024 Jan. 7, 2009 Ultrasound Lectures History and tour Wave equation Diffraction theory Rayleigh-Sommerfeld Impulse response Beams Lecture.

Similar presentations


Presentation on theme: "Advanced Imaging 1024 Jan. 7, 2009 Ultrasound Lectures History and tour Wave equation Diffraction theory Rayleigh-Sommerfeld Impulse response Beams Lecture."— Presentation transcript:

1 Advanced Imaging 1024 Jan. 7, 2009 Ultrasound Lectures History and tour Wave equation Diffraction theory Rayleigh-Sommerfeld Impulse response Beams Lecture 1:Fundamental acoustics DG: Jan 7 Absorption Reflection Scatter Speed of sound Image formation: - signal modeling - signal processing - statistics Lecture 2:Interactions of ultrasound with tissue and image formation DG: Jan 14

2 The Doppler Effect Scattering from Blood CW, Pulsed, Colour Doppler Lecture 3: Doppler Ultrasound I DG: Jan 21 Velocity Estimators Hemodynamics Clinical Applications Lecture 4: Doppler US II DG: Jan 28 Lecture 5: Special Topics Mystery guest: Feb 4 or 11

3 ULTRASOUND LECTURE 1 Physics of Ultrasound Waves: The Simple View

4 ULTRASOUND LECTURE 1 Physics of Ultrasound: Longitudinal and Shear Waves

5 ULTRASOUND LECTURE 1 Physics of Ultrasound Waves: Surface waves

6 ULTRASOUND LECTURE 1 Physics of Ultrasound Waves 1)The wave equation y z u t u x u+Δ u Particle Displacement = Particle Velocity = Particle Acceleration = u x

7 Equation of Motion dV dxdydz x p    Net force = ma p = pressure = P – P 0 density or (1)

8 Definition of Strain S = strain Also  TermsNonlinear S C S B SAp... !3 32  bulk modulus Taking the derivative wrt time of (2) = (4) (2) (3)

9 Substituting for from Eq (1) (in one dimension) Substituting from or ; wave velocity 3 dimensions (3) (5)

10 For the one dimensional case solutions are of the form ; (6) for the forward propagating wave. A closer look at the equation of state and non-linear propagation Assume adiabatic conditions (no heat transfer) P = P 0 P                  0 0 1 Condensation = S Gamma=ratio of specific heats

11 P = P, Expand as a power series P = P  3 0 ' 0 ' 0 ' !3 1 1 2 1 2 SCPSPSP B A .. B/A = Depends solely on thermodynamic factors

12 MaterialB/A Water 5 Soft Tissues 7.5 Fatty Tissues 11 Champagne (Bubbly liquid)

13 Additional phase term small for small x and increasingly significant as = shock distance (8) In terms of particle velocity, v, Fubini developed a non linear solution given by: Nonlinear Wave Equation (7)

14 Mach # (9) At high frequencies the plane wave shock distance can be small. So for example in water: Shock distance = 43 mm Shock Distance, l

15 Where (11) Thus the explicit solution is given by (12) We can now expand (Eq. 8) in a Fourier series (10)

16 Hamilton and Blackstock Nonlinear Acoustics 1998 Aging of an Ultrasound Wave

17 Hamilton and Blackstock Nonlinear Acoustics 1998 Relative Amplitude Harmonic Amplitude vs Distance (narrow band, plane wave)

18 Focused Circular Piston 2.25 MHz, f/4.2, Aperture = 3.8 cm, focus = 16 cm Hamilton and Blackstock Nonlinear Acoustics 1998

19 Propagation Through the Focus

20 Nonlinear Propagation: Consequences _______________________ Generation of shock fronts Generation of harmonics Transfer of energy out of fundamental

21 RADIATION OF ULTRASOUND FROM AN APERTURE We want to consider how the ultrasound propagates in the field of the transducer. This problem is similar to that of light (laser) in which the energy is coherent but has the added complexity of a short pulse duration i.e. a broad bandwidth. Start by considering CW diffraction theory based on the linear equation in 1 dimension

22 The acoustic pressure field of the harmonic radiator can be written as: Where is a complex phasor function satisfying the Helmholtz Equation (13) (14)

23 To solve this equation we make use of Green’s functions (15) s = surface area= normal derivative = pressure at the aperture

24 Rayleigh Sommerfeld Theory Assume a planar radiating surface in an infinite “soft” baffel Aperture Field Point Conjugate field point n use as the Greens function = 0 in the aperture

25 (16) Equation 15 can now be written:

26 Example Consider the distribution of pressure along the axis of a plane circular source: radius = a

27 From Equation (16) in r,, z coordinates The integrand here is an exact differential so that

28 The pressure amplitude is given by the magnitude of this expression (17)

29 To look at the form of (17) find an approximation for:

30 Maxima m = 1, 3, 5,...

31 Minima ; m = 0, 2, 4, 6

32 Eq. 17 M: 54 3 2 1 0

33 THE NEAR FIELD (off axis)

34 Fresnel approximation (Binomial Expansion) (19) From (16) we have

35 Note that the r’ in the denominator is slowly varying and is therefore ~ equal to z Grouping terms we have  )( 2 00 2 0 2 0 ),( zK yx z jk jkz e zj e yxP  

36 Where (20) We can eliminate the quadratic term by “focusing” the transducer Thus the diffraction limit of the beam is given by:

37 Consider a plane circular focused radiator in cylindrical coordinates radius a Circular Aperture

38 Where FWH M (21)

39 Square Aperture

40 Need to consider the wideband case. Returning to Eq. 16 we have:

41 This is a tedious integration over 3 variables even after significant approximations have been made There must be a better way!

42 Impulse Response Approach to Field Computations Begin by considering the equation of motion for an elemental fluid volume i.e. Eq. 1 (22) Now let us represent the particle velocity as the gradient of a scalar function. We can write

43 Where is defined as the velocity potential we are assuming here that the particle velocity is irrotational i.e. ~ no turbulence ~ no shear waves ~ no viscosity Rewrite as (22) (23)

44 The better way: Impulse response method

45 Impulse Response where (24) Thus (25)

46 Useful because is short! convolution easy Also is an analytic function No approximations! Can be used in calculations You will show that for the CW situation (26)

47 IMPULSE RESPONSE THEORY EXAMPLE Consider a plane circular radiator is the shortest path to the transducer near edge of radiating surface far edge of radiating surface

48 Also let So that

49 (27) * * a very powerful formula while the wavefront lies between and Thus we have: (next page) ie

50 Planar Circular Aperture

51 otherwise small large Consider the on axis case:

52 Recall Equ 26 so that the pressure wave form is given by small large

53 Off axis case

54 5 mm radius disk, z = 80 mm

55 Spherically focused aperture - relevant to real imaging devices

56 Spherically focused aperture impulse response

57 Spherically focused aperture impulse responses

58 Spherically focused aperture pressure distribution a.f/2 b.f/2.4 c.f/3 Frequency = 3.75 MHz


Download ppt "Advanced Imaging 1024 Jan. 7, 2009 Ultrasound Lectures History and tour Wave equation Diffraction theory Rayleigh-Sommerfeld Impulse response Beams Lecture."

Similar presentations


Ads by Google