MEDT 8007 Linear field analysis

(Acoustic field from an ultrasound transducer)

Ingvild Kinn Ekroll

Dept. of Circulation and Medical Imaging

NTNU

2

Linear field analysis?

• Applications of methods introduced in Cobbold’s chapter 2

• Chapter 3: How to calculate pressure fields from transducers (acoustic sources)!

3

Lecture overview

• Brief sum up of chapter 2.1

• Reminder: The Rayleigh integral and diffraction

Linear field analysis

• Integral methods

• Impulse response methods

• Angular spectrum method

• Approximate methods

Field from a focused transducer

• Properties of the focused ultrasound beam

• Focusing the beam (short)

4

Sum-up of chapter 2

• Derivation of Rayleigh-Sommerfeld eqs. and Rayleigh integral based on wave equation for velocity potential

• Green’s functions used to obtain solutions for the velocity potential given certain boundary conditions

– (surface velocity or pressure)

• The velocity potential formulation is mathematically convenient

– It is a scalar field for which the gradient gives the velocity vector field, whereas the temporal derivative gives the pressure

Velocity (vector) Pressure Pressure (monochromatic)

𝑣 𝑟, 𝑡 = −𝛻𝜑(𝑟, 𝑡) p(r, t) = ρ𝜕φ(r, t)

𝜕t 𝑃 𝑟, 𝜔 = 𝑖𝜔𝜌F(𝑟, 𝜔)

5

The Rayleigh integral

𝜑 𝑟, 𝑡 = 1

2𝜋 𝑣𝑛 𝑡 − 𝑅/𝑐

𝑅𝑑𝑆

vn = Normal velocity component of the transducer surface

Velocity (vector) Pressure Pressure (monochromatic)

P 𝑟, 𝜔 =𝑖𝜌𝜔𝑣02𝜋 𝑒−𝑖𝜔

𝑅𝑐

𝑅𝑑𝑆

𝑣 𝑟, 𝑡 = −𝛻𝜑(𝑟, 𝑡) 𝑝(𝑟, 𝑡) = 𝜌𝜕𝜑(𝑟, 𝑡)

𝜕𝑡 𝑃 𝑟, 𝜔 = 𝑖𝜔𝜌F(𝑟, 𝜔)

Assumed single frequency excitation, vn = v0eiwt

S

S

6

Diffraction

• Is what happens when a wave encounters a slit or obstacle

• Adding contributions at given locations from many point sources gives an interference pattern

Diffraction pattern!

LINEAR FIELD ANALYSIS

Evaluating the Rayleigh equation

8

Integral methods

• Direct numerical evaluation of Rayleigh integral

– Ex: Ultrasim (University of Oslo)

• More economical approaches?

– Assumptions: Plane transducer in rigid baffle + single frequency excitation

𝑃 𝑟, 𝑧, 𝜔 = 𝑖𝜔𝜌𝑣02𝜋 𝑒−𝑖𝑘𝑅

𝑅𝑑𝑆

Can be reduced to a line integral over the transducer periphery:

𝑃 𝑟, 𝑧, 𝜔 = 𝜌𝑐𝑣0𝑒−𝑖𝑘𝑧 −

𝜌𝑐𝑣02𝜋 𝑒−𝑖𝑘𝑅1(𝜃)2𝜋

0

𝑑𝜃

Plane wave part (in geometric shadow)

Edge/diffraction part everywhere

9

What is a baffle?

• The baffle is the material surrounding the transducer surface

• Provides boundary conditions for the field response

• It is typically classified as either hard (stiff material, i), soft (ii), or pressure-release (iii, no pressure on bounding surface, e.g. air)

The material given around the transducer also play a part in determining the response at a given point in the field

S

Baffle

10

Impulse response method

• Chapters 2.2 and 3.3 in Cobbold

• Reformulation of Rayleigh integral using the impulse response h(r,t) leads to:

𝜑 𝑟, 𝑡 = 𝑣𝑛 𝑡 ∗ ℎ(𝑟, 𝑡) 𝑝 𝑟, 𝑡 = 𝜌ℎ(𝑟, 𝑡) ∗𝜕𝑣𝑛(𝑡)

𝜕𝑡

• Impulse response? Velocity potential in observation point given d-exitation at all points of transducer surface

• Problem: We need to find h(r,t) in every spatial point

• Will be covered in upcoming lecture on Field II!

ℎ 𝑟, 𝑡 = 𝛿(𝑡 − 𝑅/𝑐)

2𝜋𝑅𝑑𝑆 𝜑 𝑟, 𝑡 =

1

2𝜋 𝑣𝑛 𝑡 − 𝑅/𝑐

𝑅𝑑𝑆

11

Angular spectrum method

• Chapter 2.3 and 3.1 in Cobbold

• Frequency domain method

• Interesting from computational perspective due to effective implementations of 2D FFT

• The determined velocity field can be shown to equal the Rayleigh integral in the frequency domain

• Ex: FOCUS (University of Michigan)

• ASM will be covered in separate lecture!

APPROXIMATIONS

From exact methods to

13

Approximations – why?

• Quick predictions of radiation patterns (beyond near-field)

• Simpler expressions can provide clearer insight into important parameters that govern radiation patterns

• Provides «rule of thumb» also for focused beams and fields from arrays

14

How?

x0 x

y0 y

Transducer surface, S Field point, P(x,y,z) |r-r0|

𝑅 = 𝑟 − 𝑟0 = 𝑧2 + (𝑥 − 𝑥0)2+(𝑦 − 𝑦0)2

≈ 𝑧 1 +1

2𝑧2𝑥2 + 𝑦2 − 2 𝑥𝑥0 + 𝑦𝑦0 + 𝑥02 + 𝑦02 + …

1)

P 𝑟, 𝜔 =𝑖𝜌𝜔𝑣02𝜋 𝐴𝑝𝑜𝑑(𝑥0, 𝑦0)𝑒

−𝑖𝜔𝑅𝑐

𝑅𝑑𝑆

P 𝑟, 𝜔 =𝑖𝜌𝜔𝑣02𝜋𝑅 𝐴𝑝𝑜𝑑(𝑥0, 𝑦0)𝑒

−𝑖𝜔𝑅𝑐𝑑𝑆

S

15

• Mid- to far-field approximation

• vn(t) = Apod(x0,y0)v0eiwt (Surface velocity of transducer)

Fresnel approximation

𝑃(𝑥, 𝑦, 𝑧, 𝜔) = 𝑖𝜔𝜌𝑣0

2𝜋𝑅𝑒−𝑖𝑘 𝑧+

𝑥2+𝑦2

2𝑧 𝐴𝑝𝑜𝑑(𝑥0, 𝑦0)𝑒𝑖𝑘

𝑧𝑥0𝑥+𝑦0𝑦 −𝑥0

2−𝑦02𝑑𝑥0 𝑑𝑦0

• Given uniform excitation and piston transducer of radius a, the on-axis response is:

𝑃(0,0, 𝑧, 𝜔) = cρ𝑣0𝑒−𝑖𝑘𝑧(1 − 𝑒−𝑖𝑘𝑎

2/2𝑧)

S

16

Fraunhofer approximation

x0 x

y0 y

Transducer surface, S Field point, P(x,y,z) |r-r0|

𝑅 = 𝑟 − 𝑟0 = 𝑧2 + (𝑥 − 𝑥0)2+(𝑦 − 𝑦0)2

≈ 𝑧 1 +1

2𝑧2𝑥2 + 𝑦2 − 2 𝑥𝑥0 + 𝑦𝑦0 + 𝑥02 + 𝑦02 + …

1)

P 𝑟, 𝜔 =𝑖𝜌𝜔𝑣02𝜋𝑅 𝐴𝑝𝑜𝑑(𝑥0, 𝑦0)𝑒

−𝑖𝜔𝑅𝑐𝑑𝑆

17

• Far-field approximation (see ch 3.4 for more equations!)

• 𝑧 ≫𝜋(𝑥02+𝑦02)

l (or, less stringent 𝑧 ≫

𝐷2

2l)

• W is aperture function, enables full space integration

𝑃 𝑥, 𝑦, 𝑧, 𝜔 ≈𝑖𝜔𝜌𝑣02𝜋𝑅𝑒−𝑖𝑘 𝑧+

𝑥2+𝑦2

2𝑧 Á 𝐴𝑝𝑜𝑑W (𝑥, 𝑦)

Fraunhofer approximation

𝑃(𝑥, 𝑦, 𝑧, 𝜔) ≈ 𝑖𝜔𝜌𝑣0

2𝜋𝑅𝑒−𝑖𝑘 𝑧+

𝑥2+𝑦2

2𝑧 W(𝑥0, 𝑦0)𝐴𝑝𝑜𝑑(𝑥0, 𝑦0)𝑒𝑖𝑘

𝑧𝑥0𝑥+𝑦0𝑦 𝑑𝑥0 𝑑𝑦0

Resulting in the convenient formulation:

18

Fraunhofer approximation

In words:

In the far field, the diffraction pattern from an ultrasound transducer is proportional to the Fourier transform of the aperture (times the apodization) function

19

Accuracy of approximations

Far field

(Fraunhofer error <5%)

20

Example – rectangular aperture

Fresnel approximation – lateral beam profile

• Assuming uniform excitation (Apod = 1)

• Valid only for single frequencies, but give important insight in diffraction/radiation patterns from any rectangular aperture

0 i /4 x xx

x L 2 x L 2p (x,z, ) e F F

2 z 2 z 2

-

w - l l

2z

i t /2

0

F(z) e dt-

21

Example – rectangular aperture

x 0 i /4 xx

L L xp (x,z, ) e sinc

zz

w ll

Fraunhofer approximation (lateral beam profile) Similar results are given for the y-direction (elevation direction)

Fourier transform of a rectangle, i.e. flat apodization and finite aperture

• Assuming uniform excitation (Apod = 1) • Valid only for single frequencies, but give important

insight in diffraction/radiation patterns from any rectangular aperture

22

Sound field characterization

Beam profile plots – 1-D / 2-D plot of axial and

lateral pressure

– Typically used to look at the beam width versus depth

Axial pressure plots – The pressure along the main

axis (x=0)

– Typically used to investigate depth penetration / uniformity

Fresnel

Fraunhofer

23

Quick note on spatial resolution

• The ability of the imaging system to resolve two nearby objects

• The axial resolution (along the beam) is determined by the pulse length

• The lateral resolution is determined by the beam width

• In ultrasound imaging, the axial resolution is typically better (2-4x) than the lateral resolution

Lateral resolution

Axial resolution

24

Quick note on contrast (resolution)

• The ability of an imaging system to discern (small) differences in scattering amplitude

• Mainly determined by:

– The beam side lobes

– Grating lobes (arrays)

– Acoustic beam distortion (beam aberration)

– Multiple reflections

Example: Liver tissue Carotid image courtesy of J. Rau

25

Apodization

• Apodization results in reduced side lobes, but also widenes the main lobe. Reduces lateral resolution, increases contrast resolution.

• In conventional (array) imaging, apodization is mostly applied during reception (easier to implement)

No apodization -13 dB to first side lobe Hamming apodization -40.6 dB to first side lobe

26

CW vs PW excitation

• Previously assumed: Continuous wave (CW)

• Pulsed wave (PW) excitation has finite length and a given bandwidth of frequency components

• How is the beam profile affected?

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5 6 7 8 9 10-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Power

Frequency

Power

Frequency

27

CW vs PW comparison

• CW excitation gives the ”expected” diffraction / interference pattern

• PW excitation has interference at different spatial locations for different frequency components, which smoothens the beam profile

CW beam profile PW beam profile

28

CW vs PW comparison

• Axial pressure profile comparison

29

Unfocused field calculation

Cardiac imaging – Aperture width D = 2 cm, transducer frequency f0 = 2.5 MHz,

sound speed c = 1540 m/s – Region of interest = 2-20 cm Far field, z_far = D^2/(2*lambda) = 32.4 cm (!) In other words: • Medical ultrasound imaging takes place in the near field of a

transducer • This would lead to a very wide (and complicated) beam in our

region of interest

How then do we get a well defined and narrow beam width in our region of interest?

THE FOCUSED BEAM

Controlling the interference pattern

Controlling the interference pattern

Two point sources: • Constructive and

destructive interference • Interference pattern

Multiple sources

• Constructive interference in one direction

Curved sources:

• Focused energy

• Constructive interference in a narrow area

One point source:

• Spherical waves

32

Focusing the beam

• Focusing can be achieved by:

– Curving the transducer itself (previous slide)

– Using a lens in front of the transducer surface

– Electronically delaying the emitted signal on smaller elements in an array (coming up if there is enough time)

• Bringing the far-field into the near-field

– Accellerating the diffraction pattern

• In the focal region far-field relations hold

The beam pattern in the focal point is equal to the Fourier transform of the transducer aperture (times the apodization) function

33

The focused transducer

Focal number (F-number/F#)

• The ratio of the focal depth to the aperture width

• Determines the lateral resolution (beam width)

– Lower F-numbers more narrow beam width

• Determines the depth of field (LF)

– Lower F-numbers shallow depth of field

34

Depth-of-field

• The range of depths over which the beam width is approximately constant

Depth [cm]

Azim

uth

[cm

]

Beam profile - oneway/transmit/azimuth/RMS

0 0.5 1 1.5 2 2.5 3 3.5 4-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

-60

-50

-40

-30

-20

-10

0

2

F, 3dBL 7.2 Fnum- l

LF

The focused transducer

35

Angular / lateral resolution

• Determined by the beam width

• In the focal point, we can use the Fourier transform of the aperture function to estimate the resolution

• Example: Rectangular aperture, flat apodization: – Fourier transform of rect sinc function

– First zero at F#*lambda = z/D*lambda

– This is also approximately the (full) -6dB beam width (power)

F#*lambda

~ -6dB beam width (power)

The focused transducer

Rectangular aperture

Conventional beamforming Static Tx, dynamic Rx

Static Tx, static Rx

-5 5

15

50 [mm]

Dep

th [

mm

]

-5 5 [mm]

Focus

Poin

t scatterers