Advanced Imaging 1024 Jan. 7, 2009 Ultrasound Lectures

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 Lecture 2:Interactions of ultrasound with tissue and image formation DG: Jan 14 • Absorption • Reflection • Scatter • Speed of sound • Image formation: • - signal modeling • signal processing • statistics

2. Lecture 3: Doppler Ultrasound I DG: Jan 21 • The Doppler Effect • Scattering from Blood • CW, Pulsed, Colour Doppler Lecture 4: Doppler US II DG: Jan 28 • Velocity Estimators • Hemodynamics • Clinical Applications 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 u+Δu u x z u y x t u Particle Displacement = Particle Velocity = Particle Acceleration =

7. Equation of Motion dV ¶ p + dxdydz ¶ x Net force = ma p = pressure = P – P0 density or (1)

8. Definition of Strain S = strain (2) bulk modulus B C 2 3 = + + + Also p A S S S ... 3 ! 1 (3) Nonlinear Terms Taking the derivative wrt time of (2) = (4)

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

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) g é ù ê ú r - r + P ê ú 0 1 P = P0 r ê ú 0 ê ú ë û Condensation = S Gamma=ratio of specific heats

11. P = P , Expand as a power series 1 1 ( ) P = P 2 + g + g g - + ' ' 3 P S 1 P S CP S ' 0 0 0 2 3 ! . . . . A B Depends solely on thermodynamic factors B/A =

12. Champagne (Bubbly liquid)

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

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

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

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

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

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 Hamilton and Blackstock Nonlinear Acoustics 1998

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: (13) Where is a complex phasor function satisfying the Helmholtz Equation (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 Field Point Conjugate field point n Aperture as the Greens function use = 0 in the aperture

25. Equation 15 can now be written: (16)

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

32. M: 54 3 2 1 0 Eq. 17

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 [ ] jk jkz e 2 2 + x y = 0 0 P ( x , y ) e 2 z 0 0 l j z ( ) K z

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

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

38. (21) Where FWHM

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 ie and Thus we have: (next page)

50. Planar Circular Aperture