Contents
1.
1.1.
2.
Preliminaries
1
Method of Teaching
1
What is ultrasound imaging?
2
2.1.
What do we mean by ultrasound?
2
2.2.
Uses of ultrasound imaging
3
3.
A-Mode (F51–68, C123–135, W343–347)
3
3.1.
Uses
3
3.2.
What causes the reflections? (F 21, C 26–36, W 325)
4
3.3.
What is the significance of all the reflection coefficients?
5
3.4.
What other aspects of wave propagation are important?
7
3.4.1. Scattering (F 22–25, C 189, W 324)
3.4.2. Absorption (F 23–25, C 214, W 323–324)
3.4.3. Diffraction (F 28–39, C 89–104, SW 328–333)
3.4.4. Worked Examples
3.5.
3.6.
4.
7
8
9
10
How is it all achieved in practice?
11
3.5.1. Master Clock (PRF Generator)
3.5.2. Transmitter/Transducer/Receiver
3.5.3. Time Gain Compensation (TGC)
3.5.4. Demodulator
12
13
13
14
What is the right ultrasound frequency to use?
15
3.6.1. Transducer Resonance
3.6.2. Transducer response characteristic in the frequency domain
3.6.3. Transducer response characteristic and the time domain signal
3.6.4. Q-values and pulse lengths
3.6.5. Resolution and Frequency (F184–185, C82–89, 104–112)
3.6.6. Attenuation and frequency
15
17
17
18
19
20
B-Mode (F69–96, C135–158, W348–351)
20
4.1.
Introduction
20
4.2.
What do we need to add to an A-scanner to turn it into a B-scanner?
21
4.3.
The Co-ordinate Generator
21
4.4.
Compression and Amplifier
22
The Beam-Steering Device
23
4.5.
4.5.1. Static B-Scanners
4.5.2. Real-Time Mechanical Scanners (F83–84, C144–146)
4.5.3. Electronic Steering — Transducer Arrays (F89–94, C146–158, W340–343)
i
23
24
25
5.
The Theory of Diffraction
25
5.1.
Making Huygens’ Wavelets Quantitative
25
5.2.
The plane piston transducer
28
Two very small transducers  Young’s Slits
29
5.4.
30
The linear array transducer
Electronic beam steering re-visited
31
5.6.
Lateral Resolution, Diffraction and Focussing
33
Artifacts … or … Why do ultrasound images look so awful? (F179–189)
37
5.7.
6.
5.7.1. Misregistration
5.7.3. Speckle
5.7.5. If ultrasound images are so bad, why use them?
37
39
40
Doppler Ultrasound (F117–164, C186–212, W351–358)
40
6.1.
Introduction
40
6.2.
The Doppler Equations
40
How do we put all this into practice?
42
Second transducer for reception
6.3.2. Doppler demodulator
6.3.3. Frequency meter / spectrum analyser
6.3.4. Real-time display
Problems with CW Doppler
43
44
44
45
45
6.3.
6.4.
Duplex Scanners
45
6.5.
Limitations of Doppler Ultrasound
46
7.
Bio-effects of Ultrasound (F201–212, C213–218, W375–377)
46
7.1.
Introduction
46
7.2.
Heating
46
7.3.
Radiation Pressure
48
7.4.
Acoustic Streaming
49
7.5.
Cavitation
49
7.6.
Therapeutic Ultrasound
51
7.6.1. Physiotherapy
7.6.2. Surgical Uses
7.7.
51
51
And finally ... is it safe?
51
ii
1. Preliminaries
1.1. Method of Teaching
 This is the fourth year of this course. Things in this handout have been improved
continually, as a result of the response of the previous year’s students, so I will be
very happy to hear any feedback you might have.
 Not everything you need to know is in the handouts. There are gaps to fill in with
extra notes which I give in lectures. For the exam you need to know these as well.
 There is an extra Appendix handout, which contains derivations and other
background material. None of the derivations in the Appendix will be examined,
but you will be expected to understand the definitions of various acoustical
quantitites such as acoustic pressure and characteristic acoustic impedance. The
Appendix will be useful for you to look at it so you can understand where the
results come from.
 PwA students are also taking the course Ultrasonics and will have some
background in wave propagation which the PMP students do not. The exam is
designed so that this should not matter.
Recommended books:
 Diagnostic Medical Ultrasound, Peter Fish, John Wiley and Sons 1990, ISBN 0-47192651-5 (about £20) — abbreviation in notes F followed by a page number.
 The Physics of Medical Imaging, Ed. S. Webb, Adam Hilger 1990,
361-1 (about £20) — abbreviation W.
ISBN
0-85274-
 Ultrasonic Bioinstrumentation, Douglas Christensen, John Wiley and Sons, 1988,
ISBN 0-471-60496-8 (about £80, try library) — abbreviation C. Introduction
1
2. What is ultrasound imaging?
 Ultrasound imaging operates on the same principle as radar and sonar and is similar
to the echo-location method bats use to navigate.
 An emitter sends out pulses of sound. These bounce of objects and the returned
echoes give us information about the object.
Reflector
Emitted pulse
Transducer
0
50
100
150
c
200
0
c
5 0
1 0 0
1 5 0
2 0 0
Lower amplitude
reflected pulse
d
Figure 1: Basic principle of ultrasound imaging
The fundamental equation of ultrasound is
d
ct
2
[1]
d = distance of the reflecting object from the source/detector of ultrasound;
c = speed of the ultrasound;
t = round-trip time of the pulse, from emission to reception.
2.1. What do we mean by ultrasound?
 Acoustic waves with frequencies above those which can be detected by the human
ear. In practice, 20 kHz < f < 200 MHz.
 An acoustic wave is a
propagating disturbance in a
medium, caused by local
pressure changes at a transducer.
Wavefronts
Rarefaction
Compression
c
 The molecules of the medium
oscillate about their resting
(equilibrium) positions, giving
rise to a longitudinal waves.

Figure 2: Longitudinal waves in gas
2
 c  1540 m/s  6.5 ms/cm in most body tissues.
 = c / f = 1.5 mm at 1 MHzGases and liquids support only longitudinal waves;
solids support transverse waves as well, but these are rapidly attenuated for non-rigid,
“soft” solids.
2.2. Uses of ultrasound imaging
 Most widespread use is medical
imagingNon-invasive,
low
riskObstetrics,
abdominal
problems, measurement of blood
flow and detection of constrictions
in arteries and veins.Also used in
non-destructive testing in industry:
e.g., cracks in structures.
 S
o
n
a
r
,
Figure 3: A typical obstetric ultrasound scan
underwater imaging (e.g., in
submarine echo-location devices).
3. A-Mode (F51–68, C123–135, W343–347)
 Simplest form of ultrasound instrumentPulses of ultrasound in a thin beam are
emitted from a transducer into the body and encounter interfaces between different
organs.Some of the sound energy is reflected at each stage and some continues
through to be reflected in turn by deeper organs.The returning pulses are detected
by the transducer and the amplitude of the signal is displayed on an oscilloscope. If
the time-base of the scope is constant, then the distance across the screen
corresponds to the depth of the object producing the echo, according to Eq. [1].
3.1. Uses
 Gives information very quicklyMinimum of sophisticated apparatusWeakness is
that this information is one-dimensional — i.e., along the line of the beam
propagation.Nowadays, this mode has been largely superseded by B-mode (see
later).
3
 A-mode still finds uses in ophthalmology, where the simple structure of the eye
makes it relatively easy to interpret the echoes and where what is required are
straightforward but accurate measurements of, for example, distance from the lens
to the retina.
Even this very primitive instrument is not as straightforward as it might seem. To
understand why, we need to look at a number of principles of physics, engineering and
signal processing.
3.2. What causes the reflections? (F 21, C 26–36, W 325)
 Reflections occur when the incident wave encounters a boundary between two
materials with different acoustic impedances.Acoustic impedance Z is the material
property which relates pressure changes p (in excess of atmospheric) to the
vibrational velocity u of the particles in the medium.
p  Zu
[2]
If we are looking at a single plane wave through a substance with density and
speed of sound c, then Z = c.See the Appendix for a more lengthy description.
 When an incident plane wave, travelling through a medium with acoustic
impedance Z1 hits a boundary with a second material of impedance Z2 at an angle i
to the normal, there is a reflected wave at an angle r, and a transmitted wave,
refracted at an angle t . The following important laws apply:
Law 1
The direction vectors of the incident wave, the reflected wave and the transmitted
wave are all in the same plane.
Law 2
The angle of incidence is equal to the angle of reflection. i = r
Law 3
The angle of incidence and the angle of transmission (refraction) are related by Snell's
Law:
sin i c1
Z /

 1 1
sin t c2 Z2 /  2 ,
[3]
where the Z1 and Z2 are the characteristic acoustic impedances of materials 1 and 2
(see Appendix), c1 and c2 are the speeds of sound and 1 and 2 are the (equilibrium)
densities.
4
Law 4
The pressure amplitudes pr and pt of the reflected and transmitted waves are related to
that of the incident wave pi by
pr 
Z2 cos i  Z1cos t
pi ;
Z2 cos i  Z1cos t
pt 
2 Z2 cos i
pi .
Z2 cos i  Z1cos t
[4]
Note that pr and pt are linked by pi+pr=pt. At normal incidence, this becomes
pr 
Z2  Z1
pi ;
Z2  Z1
pt 
2 Z2
pi
Z2  Z1
[5]
For those interested, the derivation of these laws is given in the Appendix. You do not
need to know the derivation for the exam, but you do need to know Laws 1–4.
3.3. What is the significance of all the reflection coefficients?
(i) Refraction effects are important.
If the beam can be turned through an angle, then we can not guarantee the path of
the return signal. We don't know exactly where the echo has come from.
It might theoretically be possible to calculate all these effects, but the
phenomenon certainly complicates things. In practice, simple scanners ignore
these effects and errors are possible.
(ii) Too little reflection is bad. pr / pi  0
Useful images occur only where there is a difference in acoustic impedance.
Tissues with strikingly different properties in other respects may have similar
acoustic impedances.
From Fig. 5, you can see that there is virtually no reflection at a transition from
liver to spleen and so the two tissues will not be delineated one from the other.
Transducer
Z1
Normal
i
Z2
Image
t
Interface
Figure 4: Displacement of image due to refraction. (Here Z2 < Z1.)
5
Reflector
Tendon/Fat
Water/Muscle
0
-10
-20
-30
Liver/Spleen
-40
-50 dB
(pr/ pi)2
1
0.1
0.01
Bone/Soft Tissue
Air/Solid
Air/Liquid
Figure 5:
10-3
10-4
10-5
Muscle/Liver
Lens/Vitreous or
Aqueous Humour
Reflection coefficients at various tissue boundaries. Note that these are
power reflection coefficients (see later).
(iii) Too much reflection is bad. pr / pi  1
If the difference in acoustic impedance is too high, then virtually all the incident
ultrasound will be reflected. This means that the boundary is opaque to
ultrasound.
 The organ in question will show up very brightly, but we won't be able to see
through it to find out what is underneath.No ultrasound images of brain in
vivo; skull reflects ultrasound
 Images of the heart have to be taken “round” the ribs, which are also opaque.
 Finding the right “window” into the body is important.
(iv) The ultrasound transducer must be “coupled” to the body using a special gel.
Before an ultrasound scan, a thin layer of gel is smeared onto the skin. Why?
Answer: (F p.45–47)
 The material from which transducers are made has a very different acoustic
impedance Ztransducer to that of the body Ztissue and more importantly that of air
Zair.
 These large “mis-matches” between Ztransducer and Ztissue and between Ztransducer
and Zair mean that the refelection coefficients at these interfaces are close to
1. Little of the signal gets through gets through at a transducer-tissue
boundary (pr/pi0.86) and virtually none at a transducer-air boundary (pt/pi
0.9997).
 By applying the coupling gel, we exclude all air from the region between
probe and body. This means that the worst case scenario of reflection from a
6
transducer-air boundary is avoided. The reflection coefficient is still high
(0.86), but imaging is possible.
 Some manufacturers use a trick called impedance matching to increase the
amount of transmitted radiation through a transducer-tissue interface. Sec. A9
explains mathematically how this works. Inside the probe, there is a matching
layer of thickness /4 between the transducer and the tissue. The acoustic
impedance of the matching material is approximately:
Zmatch 
Ztransducer Ztissue
[6]
 The technique has analogues in optics (“blooming” of lenses), electronics
(coaxial transmission lines) and quantum mechanics (scattering of particles by
potential wells).
 Note that this technique is not suitable in all cases and, in particular, a /4
layer will match completely only a single frequency of ultrasound.
3.4. What other aspects of wave propagation are important?
3.4.1. Scattering (F 22–25, C 189, W 324)
 Formulae in Sec. 3.2 strictly valid only for an infinite plane reflecting surfaceIn the
body, there are many structures which are much smaller than this (e.g., lung tissue
(a)
(b)
Transducer
Soft Tissue
“Matching” gel
Transducer
pi
Soft Tissue
pi
pt
pt
pr
pr

|pt| << |pr|
|pt| >> |pr|
Figure 6: (a) A large degree of reflection occurs at the interface between the ultrasound
transducer and soft tissue. (b) If the correct thickness of an appropriate material is built
into the probe, much improved transmission can be obtained. Note that there is still a
thin gel layer (not shown) between the “matching” layer inside the probe and the
tissue. This has approximately the same acoustic impedance as the soft tissue and is
used to exclude air.
7
is a fine network of air-filled tubes).These give rise to a whole series of interfaces,
at random orientations, and the reflections from these scatter the incident wave.A
typical 1 MHz ultrasound beam does not penetrate lung tissue.
At a smaller scale where d<<(e.g., red blood cells), Rayleigh Scattering occurs:
 The degree of scattering varies as f4 
This means that low frequency ultrasound penetrates tissue better.
 The distribution of scattered radiation is approx. uniform (isotropic re-radiation).
 The exact shape of the particle does not matter very much.
3.4.2. Absorption (F 23–25, C 214, W 323–324)
 A phenomenon by which organised vibrations of molecules (i.e., ultrasound) are
transformed into disorganised, random motion.
Acoustic energy  Heat
 The mechanisms for this transfer include fluid viscosity, molecular excitations and
chemical changes.This is still largely uninvestigated for most tissues. Since
biological tissues are so complex, this is not surprising.
 It is difficult to measure the proportion of energy loss which occurs by scattering
and the proportion lost by absorption.
 Assuming that the tissue is uniform, we find that the reduction in amplitude of the
ultrasound over a small distance x, is a constant fraction of the amplitude at the
point x, wherever that is along the path.
p0  (p0 )x .
[7]
In the limit x  0, this gives the differential equation
dp0
dx
 p0
,
[8]
.
[9]
which integrates to
p0  p0 ( 0 ) e x
We can do the same thing for the scattering effect, so that overall
8
p0 ( x)  p0 (0) exp[( scatter   absorb )x] .
[10]
This also applies to the peak oscillation velocity u0 and the amplitude of
displacement a0 of the particles.
Notes:
(i) Attenuation is approximately proportional to frequency, so that the depth of
penetration goes down as f rises.
(ii) Instead of using amplitude, attenuation is often measured in terms of a reduction
in the power density transported by the wave. Consider the units of pu, where u is
the particle vibration velocity:
Pressure  Velocity = N/m2  m/s = (Nm)s-1/ m2 = W/m2 = Power/unit area
I.e., pu represents the power being transported by the ultrasound through a unit
area of the tissue normal to the direction of propagation. It is often also called the
intensity of the ultrasound and is represented by the symbol I.
If we look at the power (intensity) attenuation, we see that
I  pu  Re[ p0 e i (t  kx ) ] . Re[u0 e i (t  kx ) ]
 p0 u0 cos 2 (t  kx )
 I0 cos 2 (t  kx ) .
[11]
Now p0(x) = p0(0) e-x and similarly for u0. Hence
I 0 ( x )  I 0 (0) e 2x .
[12]
The power density transported decays twice as quickly as the vibration amplitude.
(iii) Attenuation is often measured on a decibel scale, where
 I( x)
Attenuation in dB  10 log10 

 I (0) 
.
[13]
3.4.3. Diffraction (F 28–39, C 89–104, SW 328–333)
 Huygens’ Principle states that each point on a wavefront can be regarded as acting
like a secondary source, emitting spherical wavelets. The new overall wave is
found by summing the contributions from all the individual wavelets.
 Interference between these wavelets gives rise to diffraction effects.
9
Incident wavefronts
Line of constructive
interference
Secondary
sources
Wavelet
Figure 7: Diffraction effects
 Diffraction becomes significant when the apparatus dimensions and objects
examined become comparable with the radiation wavelength. Thus acoustic
diffraction (0.1mm) is a much more significant effect than optical diffraction
(500nm) for biological tissues.
3.4.4. Worked Examples
A 1 MHz US wave with initial intensity 100 mW/cm2 (RMS) is travelling through fat. (We
shall assume initially that we have a semi-infinite slab so that there is no reflected
component.) Calculate: (a) the initial peak pressure, (b) the initial maximum velocity of
oscillation of the particles, (c) the initial maximum displacement. Now suppose that, the
beam hits a barrier with muscle 3cm from the starting point. Calculate: (d) the intensity of the
reflected beam. The tissue properties are: fat = 940 kg/m3, c = 1480 m/s, (amplitude) =
0.07 cm-1; muscle = 1070 kg/m3, c = 1566 m/s, (amplitude) = 0.15 cm-1.
(a) Peak Pressure
We have
I = p u = p2/Z = p02 cos2t / Z
We are given the RMS intensity, i.e., IRMS = p02 <cos2t> / Z = p02 / 2Z
For a plane wave, Z =c  p0 = (2cIRMS)
= (2  940 kgm-3  1480 ms-1  1000 Wm-2)1/2
= (2.78  109 kg2m-2s-4)1/2
= 5.27  104 kgms-2m-2
= 5.27  104 N/m2 or about 1/2 atmosphere.
Remember, this is the excess pressure above atmospheric. This calculation gives the
peak pressure. The RMS pressure is p0/2.
(b) Peak Velocity
p0  Zu0  u0 
5.27  10 4 kg m -1s-2
 0.038 m / s  3.8 cm / s .
1.39  10 6 kg m -3 . m s-1
10
(c) Peak Displacement
q
,
u  u0 e i (t  kx ) 
t
where q is the displacement of the particle from its mean position. Integrating this
expression gives
u
q  0 e i (t  kx ) ,
i
where the i in the pre-multiplying factor indicates that q is 90 out of phase with u (1/i
can be re-expressed as e-i2 and so the whole phase angle is ei(t - kx - ).
 q0 = u0 /  and for 1 MHz ultrasound, = 2f = 2  106 rad s–1.
 q0 =
3.8  10 2
 6.0  10 9  6.0 nm .
2  106
(d) Intensity of Reflected Beam
By the time it reaches the interface, the incident beam has peak intensity
I (3 cm)  I (0) e 2x
 100 mW/cm 2  e  2 0.07cm
-1
 3 cm
 65.7 mW/cm 2 .
This is equivalent to an attenuation of 10 log10 0.657  182
. dB .
The reflection coefficient for amplitude of pressure fluctuations is
R
pr Z2  Z1 1.68  1.39


 0.0945  10.2 dB ,
pi
Z2  Z1 1.68  1.39
where Z1 = c for fat and Z2 = c for muscle.
Since both pressure and velocity are reduced by this factor, the reflected beam has its
intensity reduced by (0.0945)2  20.4 dB.
 the intensity of the reflected wave at the interface is
65.7 mW/cm2  (0.0945)2 = 0.59 mW/cm2 to 2 s.f.
So, 1/3 of the incident power is dissipated by attenuation processes in the fat before it
reaches the interface. Of the remaining 2/3, most of the power is transmitted, because the
acoustic impedances of the two tissues are quite similar.
3.5. How is it all achieved in practice?
Practical design of an A-mode scanner
(F51–68, C124–135, W343–348)
Fig. 8 shows the block diagram of a practical scanner. The new additions as
compared with the simple diagram are concerned with the practical problems one
discovers when one tries to use the reflected ultrasound signal.
11
V
t
PRF generator
Reflecting objects
Pulse generator
Transducer
0
50
100
150
200
0
5 0
1 0 0
1 5 0
2 0 0
Protection circuit
70-80 dB
Variable gain
amplifier
(TGC)
TGC generator
40-50 dB
V(dB)
t
Demodulator
y
Display
’scope timebase
x
V
t
Figure 8:
Block diagram of a practical A-scanner. N.B. Not all A-mode scanners include a demodulator. The
dynamic range values at each stage are approximate and refer to the power range in the signal.
Take the square root (i.e., halve the dB value) for the corresponding amplitude ranges.
 How can the same probe both transmit pulses and receive the echoes?
 How do we cope with the signal attenuation by tissue?How do we display the
signal?Master Clock (PRF Generator)
 This synchronises the various parts of the scanner (e.g., transmitter, receiver,
oscilloscope timebase) so that each is triggered to act at the correct time.
 PRF stands for pulse repetition frequency, the frequency at which clock pulses
occur and at which ultrasound pulses are sent out into the sample.
12
3.5.2. Transmitter/Transducer/Receiver
 On the leading edge of each clock pulse, either a momentary voltage step, or a short
sinusoidal burst of voltage is applied to the transducer.
 The transmitter which performs this must have a short rise time, i.e., it must be
able to go from zero to its maximum voltage (100–200 V) very quickly (typically
<25ns), in order to produce ultrasound pulses with very high frequency
components.
3.5.3. Time Gain Compensation (TGC)
Problem:
Ultrasound is attenuated as it passes through tissue.
So, even for the same type of reflector, the signal is less for deeper objects.
This effect is very significant. The worked example (Sec. 3.4.4) showed a typical 
value of 0.15 cm-1, so on a typical return trip of 10cm, the signal is reduced by e-0.1510
= 0.22 compared with reflections coming straight from the skin surface.
Solution:
 Amplify the later-arriving signals (i.e., the ones from deeper in the tissue) more.
I.e., change the receiver gain
with time to compensate for the echo
attenuation.Achieved by making the gain of the amplifier dependent on a control
voltage. This input voltage is changed by the TGC unit.Because of the logarithmic
nature of the decrease in signal, the TGC should increase the gain a certain number
of dB each ms.
Worked Example
An ultrasound beam propagates in uniform liver tissue with  = 0.15 cm-1. If the speed of
sound in the tissue is c = 1540 m/s, what should be the rate of gain increase by the TGC?
 = 0.15 cm-1 is the attenuation coefficient for amplitude; the power attenuation is double
this. Amplifiers are specified in terms of power, so 2 = 0.3 cm-1 is what we want. In terms
of dB, we have 10 log10 ( e-0.3 ) = 1.3 dB/cm to 2 s.f.
So we want a gain increase of 1.3 dB for each cm of travel to cancel this out. Now
1540 m  1 s  1 cm 
1.3 dB
1
ms . This means that the require TGC rate is
154
1
ms  200 dB / ms. Clearly, a specialised amplifier is needed.
154
In practice, tissue type varies with depth and the situation is more complicated. The
user is given a range of controls to vary the TGC. The rate of increase of gain (i.e.,
d2G/dt2 ) varies with time and hence depth. This is not an exact science! Notice, too,
that by “tweaking” the time-gain controls to get a better images, we lose the
information provided by the attenuation coefficient. By using this compensation we
are “ignoring” the physics of the situation. The fact that one might not be able to see a
particular boundary tells us something about the properties of that boundary.
13
3.5.4. Demodulator
 At the output of the compression amplifier, the echo signal mirrors that of the
pulse, i.e., it oscillates at the ultrasound frequency of several MHz.The display is
much easier to understand if this high frequency modulation is removed.Another
way of describing demodulation is to say we want to change a signal oscillating at a
high frequency
Let us multiply
an original signal S = cos hight by cos 0t. A useful trigonometry theory states that:
cos  high t cos  0 t 


1
cos( high t   0 t )  cos( high t   0 t ) .
2
[14]
Variety of simple pre-programmed
shapes for increasing the gain
Gain / dB
60
40
Vclock
20
0
Vin
0
2
4
6
8
10
Depth / cm
Vcontrol
1/PRF
t
Vcontrol
Set of levels all adjustable by
operator for more flexibility
Variable gain
amplifier
(TGC)
t
Gain / dB
60
40
Vout
20
0
0
2
4
6
8
Vout = G( Vcontrol ) . Vin
10
Depth / cm
Figure 9: Principle of operation of time gain compensation (TGC)
14
Original Signal
Reference Signal
Low-pass filter
Result at difference frequency
Figure 10: Principle of demodulation by multiplication by a reference frequency
Now if we use a low pass filter, , we can cut out the first term leaving a signal of
the form
cos( high t   0 t )  cos( t ) .
[15]
 Any echo can be made up of a sum of such terms. The multiplying and filtering
process simply shifts all the terms down in (angular) frequency by 0..
 This gives us the envelope of the signal, as in Fig. 10.
3.6. What is the right ultrasound frequency to use?
 No universal “right” frequency
 Ultrasound frequency chosen is determined by a number of factors: scan time for
cos  t
images, resolution, the length of the Vultrasound
pulses and the transducer design.To
~
understand how these interact, we start with a simple explanation of how
transducers work.
0
Transducer
Desired ultrasound
insonating sample
Ultrasound coming from
back of transducer
3.6.1. Transducer Resonance
 Transducers exploit the piezo-electric effect.When a voltage is applied across
certain types of crystal, the crystalline structure of the crystal is distorted and a
Interfering US
in transducer
~ /2 for fundamental
15
Figure 11: Principle of operation of a transducer
change in size occurs. If the crystal is made such that the change in size happens
perpendicular to the faces.
 Apply voltage perpendicular to faces oscillating at the ultrasound frequency. These
move in and out generating pressure waves.Ultrasound waves both outside and
inside the crystal
 Waves inside the crystal are reflected off the opposite face and return to interfere
with the original vibration.Waves may come back either in phase with the original
vibration, out of phase, or somewhere in between. Only if the waves reinforce each
Frequency Domain
Time Domain
Observed
Vibration
Spectral
Response
A0
Exponential envelope
R0
Q = 0 / 2
R0/2
A0 / e
2
0
0
50

100
150
200
00

0
50
50
00 / Q
10
100
150
150
200
200
T=2/0
Fig. 12: Ideal transducer characteristics in the frequency and time domain
other will we get an ultrasound signal.
 Relative phases of the original vibration and the reflected wave depend on the
wavelength of the ultrasound in the crystal. So the amplitude of the signal
generated by the transducer depends on the frequency of the oscillating voltage.
 Transducers exhibit resonance properties.Because of the relative Z values for the
transducer and the external medium, we require the faces of the transducer to be
nodes (almost). This might at first seem counter intuitive.For simple cases, the
width of the transducer must be an odd multiple of
Fundamental frequency and
.
 In more realistic situations, we want all the radiation to exit from the front of the
transducer. This means that a backing material is placed at the rear face to absorb
backward travelling ultrasound. This changes the boundary conditions for the wave
equation and hence the width of the transducer at the fundamental frequency.
16
t
3.6.2. Transducer response characteristic in the frequency domain
 Around each of its resonant frequencies, the transducer will have a response curve
like that shown in Fig. 12.
Typically, the curve will have a Lorentzian form such as
R 
R0
1  (   0 ) 2 / (  ) 2
,
[16]
 is the (half) width of the response function at half height.
 We define a quantity called the quality factor or “Q-value” of the transducer by
Q
0
Resonant frequency

Linewidth
2
.
[17]
 A high-Q circuit has a very sharply peaked resonance, while a low Q corresponds
to a broad line. The size of Q depends on the losses from the resonant circuit
containing the transducer. Since transducers are generally constructed out of
materials having low internal loss, it follows that the largest contributor to losses is
radiation of power into the surrounding material, i.e., tissue in a medical
instrument.
 In cases where we want very low losses, such as in highly stable quartz oscillators,
the transducer is bounded by air or a vacuum and the Q can be up to 106. For
pulsed ultrasound, where the aim is to radiate energy into the tissue, Q  5–15.
3.6.3. Transducer response characteristic and the time domain signal
 How does the above response characteristic relate to the actual time variation of the
vibration of the crystal when we create a pulse of ultrasound?
 It can be shown (Fourier Transform theory) that if the crystal is “excited” by a very
short, intense pulse of voltage (mathematically, a -function), then the observed
signal will be an exponentially-decaying cosine wave.
Excitation =  (t )  Signal =
+
 R( )e
-
it
d
,
[18]
where R() is the frequency response function. If R has the Lorentzian form above,
then this Fourier integral leads to
17
Signal  e  t cos  0 t .
[19]
3.6.4. Q-values and pulse lengths
 A second way of looking at the signal is in terms of an alternative definition for Q:
Q  2
Energy stored
Energy lost per cycle
.
[20]
 N.B. The denominator is actually shorthand for instantaneous rate of energy loss,
expressed in units of energy per cycle
Energy loss per second  dE / dt
2 dE


No. of cycles per second
f
 0 dt
[21]
Hence, Q becomes
dE

  0 E  E  E0e  ( 0 / Q ) t
dt
Q
.
[22]
 The amount of energy stored goes down exponentially with time constant 0/Q.
 In the ideal case, all the energy loss of the transducer goes into the ultrasound
beam. Hence we have
I  I 0 e  ( 0 / Q ) t  e i  0 t .
[23]
 Finally, as we have seen, I = P2 / Z, so the pressure variations in the pulse of
ultrasound have the form
p  p0e  ( 0 / 2Q ) t  ei 0 t .
[24]
This explains the way in which ultrasound pulses are drawn in the illustrations.
 It is convenient to define a “nominal” pulse length, the time taken for “almost all”
of the power to have gone from the pulse. This choice is arbitrary, but it is often
convenient to choose a length of time t = Q / 2 periods of the waveform. This leads
to
I  I 0e( 0 / Q )( 2 /  0 ) / 2  I 0e
18
,
[25]
i.e., the intensity decays to around 4% of the original.
3.6.5. Resolution and Frequency (F184–185, C82–89, 104–112)
 It is the distance apart which two ultrasound objects must be to give distinct signals
at the receiver.
 Axial (often called “longitudinal”) resolution is the quantity that is important for Amode scans and refers to resolution along the beam direction;
 Consider two reflectors a distance d apart, as in Fig. 14.
 When the pulse hits Interface 1 (a), some is reflected and some transmitted (b).
 The pulse then arrives at Interface 2, where, again, some is reflected and some
transmitted (c).
Interface 1
Interface 2
(a)
 Eventually, the two echoes arrive back at the transmitter, separated by twice the
distance between the reflectors.
0
50
100
150
200
Incident Pulse
 When d is reduced to the point such that the echoes
d are just resolvable, then
(b)
AR
200
150
100

50
Q
4
00
50
.
[26]
100
150
200
(c)
200
(d)
200
150
100
50
Echo 1
150
100
50
200
0
150
100
50
0
0
50
10 0
15 0
20 0
Echo 2
0
200
150
100
50
0
0
5
0
1
0
0
1
5
0
2
0
0
0
50
10 0
15 0
20 0
2d
19
Figure 13: Reflection of a pulse of ultrasound from two closely separated interfaces
 The important points to take away are:
Axial Resolution  Wavelength
Signal
200
150
100
200
50
+
150
0
Two peaks
just resolved
Demodulation
100
50
0
t
AR
Minimum resolvable
distance
 Using a smaller wavelength (higher frequency) gives a better resolution.
Figure 14: Criterion for resolution of two ultrasound reflectors
3.6.6. Attenuation and frequency
 In Sections 3.4.1 and 3.4.2, we noted that both scattering and absorption rise with
frequency. This is bad news because the penetration depth of the ultrasound goes
down.
 When choosing the frequency to use, we thus have to make a compromise between
resolution and signal attenuation.
4. B-Mode (F69–96, C135–158, W348–351)
4.1. Introduction
 The commonest form of ultrasound imagingThe ultrasound equivalent of the radar
pictures you see in those old war movies.A thin beam of ultrasound is scanned
across the object and the strength of the returned echoes is displayed on the
monitor.Notice that whilst in radar, full 360 coverage is required, in medical
ultrasound, where only the body in front of the transducer is of interest, we look at
a limited “pie-shaped” sector.
 A B-scan is “simply” an A-scan in which the ultrasound beam is moved and the
results are displayed differently.
20
 “B” means “brightness”: the ultrasound signal changes the brightness of a spot on
an oscilloscope screen instead of the amplitude of the trace in A-mode.
4.2. What do we need to add to an A-scanner to turn it into a Bscanner?
 As soon as we try to turn the idea into a working system, we find a number of
problems lurking!
 How do we display the data received?How do we make the beam sweep across the
sample?What do our data mean?Fig. 16 is a block diagram of a generic B-scanner.
Only three new items have been added: the co-ordinate generator, the video
amplifier and the beam-steering device.
4.3. The Co-ordinate Generator
 This device is often also called the scan converter.
 It takes information about the instantaneous orientation of the beam and turns it
into the co-ordinates of a line on the display monitor.
0
50
100
150
200
Reflecting
Surfaces
0
50
100
150
200
Ultrasound
beam
Boundaries giving
rise to echoes
Image Formed
Other orientations
of ultrasound beam
Figure 15: Principle of B-scanning
21
V
t
PRF generator
New
Pulse generator
Beam steering
device
Protection circuit
Probe
Variable gain
amplifier
(TGC)
TGC generator
V(dB)
t
New
Demodulator
Compression and
Video Amplifier
Brightness
Co-ordinate
Generator
(x,y)
Display
Figure 16: Block diagram of a B-mode scanner
 On simple systems, the CRT electron beam is physically scanned up and down the
desired line (i.e., the co-ordinate generator acts as a variable voltage source to the
’scope x- and y- plates).
 On more modern systems, the co-ordinate generator gives the memory location in
which signal information is stored. The data is then displayed on a monitor by an
computer program.
4.4. Compression and Amplifier
 Even after passing through the TGC, the range of signals in the data is still large.
22
 This is due to the range of reflector strengths in the body — see Fig. 5.
 The compression amplifier transforms the data by some rule Vout = f(Vin), which
reduces the dynamic range of the data (i.e., compresses the scale).
 Typically, a 40–50 dB dynamic range for Iin (i.e., the ratio Iin max/Iin min  104–105) is
transformed to an output dynamic range of 10–20 dB (10–100). Remember: take
the square root of these values to get the corresponding voltage amplitude ranges.
 This allows low intensity echoes to be seen on the same display as high intensity
ones, i.e., strongly reflecting organ boundaries and weakly reflecting internal
structure can be seen on the same image. A video monitor can display only about
256 values simultaneously and the human eye can distinguish only about 32
different shades of grey.
 This means that:
a huge amount of information is lost as in the case of the TGC;one should not
normally interpret B-mode image intensities quantitatively.
4.5. The Beam-Steering Device
 This is what distinguishes the different types of scanner.
 There are various levels of distinction. The most basic is between static and realtime scanners.
4.5.1. Static B-Scanners
 The transducer is moved manually by the
operator.
Hinge



Probe
 The probe slides backwards and forwards
over the patient, changing its angle.The
image is built up line by line. Each time,
the co-ordinates 1, 2 and 3 tell the
display where on the screen to show the
results. See Fig. 17
Patient
Fig. 17: Co-ordinate generator for static Bscanner
 The advantage of the system is that the
operator can choose which bits of the
picture to update most often and to tailor
23
Oil bath
Oscillating mirror
Ultrasound
beam
Rotating
transducers
Window
Oscillating transducer
Fixed transducer
Figure 18: Various types of beam steering device for real-time scanners
the scanning motion to view the feature of interest from several different directions
 It is also very cheap.
 However ...
The scans take several seconds to build up and form a complete picture. This is a
problem if the object in question moves in the meantime. Static B-scanners are not
suitable for imaging, for example, a beating heart.
4.5.2. Real-Time Mechanical Scanners (F83–84, C144–146)
 “Real time” scanners acquire anything from a few frames (images) per second up to
several hundred. They are ideal for imaging motion.
 In a motorised scanner, the transducer is moved mechanically by a motor.
 Because of the difficulties of maintaining contact between the skin and a moving
transducer, a larger probe is used, which contains the transducer “suspended” in a
bath of oil, with a window to allow the pulses to leave.
 There are several different designs, as shown in Fig. 19. In all cases, the final
device will depend on obvious mechanical engineering questions like:
How do you make a probe rock backwards and forwards very fast? Can you make
it do so uniformly? How do you get leads to three transducers on a ring without
everything getting tangled up when they rotate?
24
 The major disadvantage of this type of device is that mechanical systems have an
inherent speed limit.
 The advantage is that there is no complicated (and expensive) electronics.
4.5.3. Electronic Steering — Transducer Arrays (F89–94, C146–158, W340–343)
 We shall not go into any detail until later, but the basic principle is that a number of
very small transducer are placed into a line and are then fired separately.
 By “firing” (i.e., sending out a pulse from) the transducers at different times, one
can make composite wavefronts (Huygens Principle again!) which mimic that
given out from one of the moving transducers above.
 Electronic beam steering is potentially much faster than mechanical steering and
also has the advantage that the order of sampling of the different lines is much
more flexible.
5. The Theory of Diffraction
5.1. Making Huygens’ Wavelets Quantitative
 We are all used to the simple concept of Huygen’s wavelets, as shown pictorially in
Section 3.4.3. But how can we use this mathematically and without having to draw
all those fiddly diagrams?
 Fresnel proposed that we could represent all the points on the wavefront by
individual sources of spherical waves.In principle, each spherical wave would give
rise to a contribution to pressure
p

e i (t  kr )
r
where r is the distance from the point on the wavefront (i.e., the origin of the
secondary wavelet) to the point where we are making a measurement.
25
[27]
y
x
O
X
R
Y
r
Q
P
Figure 20: Notation for diffraction theory
 In practice, there are a number of complications and proportionality factors that we
will not consider. Consider instead , which we may regard as being the same as p,
except for a few constant factors and 2, which we will take to be the same as I.
 Consider the situation shown in Fig. 20. Waves are incident from the left
 ( x, y )  ei (t  kx )
[28]
 These hit an obstacle at x=0 and we look at the Huygen’s wavelet at a typical point
Q with height y compared with the origin O.
 We are interested in evaluating the contribution of this wavelet at point P.
 We need to describe the obstacle. This is done by an aperture function, a(y). In the
simplest case, a(y) is 1 where the wave can get through and 0 where it is blocked.
In more complex examples, we can vary the amplitude of a(y) to represent partial
transmission or introduce an arbitrary phase.
d (at P)  d ( X , Y )  a( y )dy 
ei (t  kr )
r
,
[29]
where r = (X2 + (Yy)2)1/2 is the distance from the point Q to the point P.
 Overall value of is the result of summing all the individual contributions d, i.e.,
doing an integration.
26
Y-y
 Difficult to go further without approximations … Replace r as follows:
r

X


R 1 

2
 Y
2
 2 yY
2 yY 

R2

1 /
2
where R is the distance from O to P, i.e., (X2 + Y2)1/2. Now use a Taylor expansion
 yY

r  R 1  2  
 R

 R
yY
R
 R  y sin 
where  is the angle between the straight-ahead direction and the point P, as
measured w.r.t O.
 Hence Eq. [29] becomes
d ( X , Y )  a ( y ) dy 
eik ( R  y sin  )
R  y sin 
 We then argue that, whilst the extra phase given by the ysin term is important, the
extra amplitude is not (for y<<R) Finally, the expression for the total signal at P is
given by
ei (t  kR )
 ( X ,Y ) 
R

 a( y ) e
iky sin 

dy
[30]
Fraunhofer integral
 This important equation tells us that the waveform depends on the angle rather
than on the particular values of X and Y. So we make a further notational change
and rewrite  as
27

 (q) 

a ( y ) eiqy dy

[31]
 Here, I have been very cavalier and “got rid of” the term before the integral,
because it’s not very interesting. The key bit is the integral. Introducing q = k sin 
is a minor but important change, the purpose of which is to make the equation look
like a Fourier transform.
 This is the form of the Fraunhofer integral that we will use in practice to tell us
about our diffraction pattern.
5.2. The plane piston transducer
 Simplest type of transducer — a rectangular block.
 We will look at the problem in 1-D but extension to 2- and 3-D is straightforward.
1 - b/2  y  b / 2
Aperture function a ( y )  
 0 otherwise
Notice that this is just the same as for a single slit of width b.
 The Fraunhofer integral becomes
 (q) 
b/2

1  eiqy dy
b / 2
b/2
 eiqy 
 
 iq   b / 2
 b sinc b2 q
 Notice that what we end up with is a function of q = k sin This can be displayed
in one of several ways — see Fig. 21.
28
(a)
1
2
1
(c)
0.8
0.6
0.6
0.4
0.4
0.2
sin 
0.2
0
-0.2
0
-1
-0.5
0
0.5
sin 
0.8
0
0.2
0.4
0.6
0.8
1
cos 
1
-0.4
(b)
-0.6
-0.8
-1
Y = R sin 
Figure 21: Possible methods of display for the function . (a) is a plot of intensity ( v. sin . (b)
is the equivalent pattern of fringes as might be observed in an optics experiment. (c) is a typical
transducer beam profile. (Note that in order to visualise the sidelobes (which are very small, I have
plotted not 2. These diagrams were simulated for the case of b =
5.3. Two very small transducers  Young’s Slits
1 - b  y  a and a  y  b
Aperture function a ( y )  
0 otherwise

[32]
 Since the Fourier transform is linear, we can regard this as the sum of two terms:
 (q) 
b

a
1 e
iqy
dy 
b

1  e iqy dy
a
 b sinc b2 q  a sinc a2 q
[33]
 Note that when the slits are very thin, we can formulate the problem in a different
way:
a ( y )   ( y  b)   ( y  b)
[34]
which leads to the Fraunhofer integral
 (q) 


 ( y  b) e iqy dy 


  ( y  b) e
iqy
dy

Remembering that the integral of the delta function has the effect of sampling the
accompanying function at the location of the delta function, we get
29
 (q)  e iqb  e iqy  2 cos qb
[35]
 This result is very famous and describes the pattern of fringes that is seen in the
classic optics experiment Young’s slits.
5.4. The linear array transducer
 A key question arises: Can we generalise the result above for more than two
transducers?
 This is important for work using linear arrays.
 Consider the array in the Fig. 22. Note first that this is a 2-D problem. Diffraction
both in the X- and Y-directions.
 It turns out that the result is simply the product of one function for Y, which we
have already calculated as h sinc qh/2, and a term corresponding to the ax(x)
aperture function.
 In class we will show mathematically how we can use the Convolution Theorem to
find a general result that will apply to linear arrays with any number of elements.
 The result is the beam pattern shown in Fig. 23. It possesses an important feature
— the existence of major beam sidelobes. These will be considered in more detail
later.
h
L
Y
d
X
b
Figure 22: Schematic diagram of a linear array transducer
30
1.2
(a)

(b)
sin 
1
1
0.8
0.5
0.6
0.4
0
0.2
0
0.2
0.4
0.6
0.8
0
-1
-0.5
0
-0.2
0.5
1
sin 
-0.5
-0.4
-1
Figure 23: Beam pattern of the linear array transducer presented in two different forms. Here
the results have been simulated for the case where b = and d = 3.
5.5. Electronic beam steering re-visited
 The quantitative theory of diffraction developed above can explain how to steer an
ultrasound beam. The diffraction pattern above is what we get if we fire the
elements all at the same time; beam steering involves firing the elements at
different times.
 If the elements are made to fire at intervals t, such that
t 
d sin 
c
,
[36]
then the wavecrests will form a line at angle . This corresponds to a plane
wavefront moving in the direction , i.e., we have steered the beam electronically.
 Fig. shows this pictorially, but we can also calculate the results using the
Fraunhofer diffraction theory.
 Each time delay t corresponds to a difference in the phase at which the Huygens
wavelets are emitted.
  2
j t
 (ck t ) j
T
,
where j is the number of the transducer.
31
[37]
1
cos 
Phase shifter introduces
time delay t in signal
Transducer
t = 0
For this transducer,
t = (6d sin /c
d
t = (d sin /c
t = (2d sin /c
t
0
0
0
50
10 0
50
15 0
0
50
10 0
15 0
0
50
10 0
10 0
15 0
20 0
50
15 0
Effective plane
wavefront
6d
10 0
15 0
20 0
20 0
20 0

20 0
Figure 24: The principle of electronic beam steering
 The expression for ax(x) now becomes
 
a x ( x)    ei
 j  
j ( ck t )




 ( x  jd )   hb ( x)  hL ( x)
[38]
 Showing how this expression leads to beam steering is one of the problems on the
examples sheet.
 Electronic beam steering has the following advantages:
(i) The beam motion can be faster than by mechanical steering.
(ii) The order of sampling does not have to be a single sweep backwards and
forwards. Hence the screen refresh rate can be more uniform, as opposed to a
sort of “windscreen wiper” effect. In addition, we avoid confusion between
outgoing and incoming signals from adjacent lines.
(iii) It is easy to arrange to update some lines more frequently than others, to
concentrate on, for example, a beating heart.
 Set against these, there are two main disadvantages:
32
Motion of transducer
Image
Profile across image
Small transducer
Signal strength
Ideal, thin
US beam
Object
a
a
a
x
Blurred image
Signal strength
Real US beam
width b
b
a
a+b
x
a+b
Figure 25: Blurring of an image due to a wide ultrasound beam
(i) The processing electronics are more complicated. This is becoming less of a
problem as electronics prices fall.
(ii) The side-lobes associated with the linear array lead to problems. There is
ultrasound intensity where we don’t want it. If this bounces of a strong
reflector and gets back to the receiver, we get “ghost” images, as described in
the next section.
5.6. Lateral Resolution, Diffraction and Focussing
 Lateral resolution is the smallest distance that we can distinguish perpendicular to
the beam and in the image plane.
 Obtaining the best possibile resolution depends on getting the smallest possible
beam diameter at the point of reflection.
 Diffraction effects mean that we can’t get the beam as sharp as we would like. If
we start off with a narrow beam, it will spread out.
33
Far-field (“Fraunhofer”) region
Transition zone
d2/4
d
Diverging beam(sin  = 1.2/d),
but smooth variation of intensity
within the beam.

Near-field (“Fresnel”) region
Tightly collimated beam, but
internally, a complicated structure
containing lots of maxima and
minima of signal intensity - not
much good for imaging
Figure 26: The three regions in the beam pattern of a circular transducer
 A mathematical treatment shows us what the beam from a simple transducer will
look like. We can divide it into three regions: near-field, transition zone and farfield, as shown in Fig.
 Only the far-field region is suitable for ultrasound imaging.
 Consider the intensity variation of the beam along its axis. A mathematical
treatment of the diffraction (see 2US) shows that at a given time t,
I( x)  I0[ cos(t  kx)  cos(t  k a2  x 2 ) ]2
.
[39]
The important feature is the “envelope” or time-average of this function. You do not
need to know this formula, nor the exact details of the pattern shown in Fig. 27, but
you should be aware of the consequences:
(i) In the near field, the rapid oscillations of the signal intensity make it useless
for imaging.In the far field, the intensity varies smoothly.(iii)
The closest
point for imaging is approximately
34
Fresnel
Transition
Fraunhofer
Monotonically decreasing
ultrasound intensity
Complicated
intensity patterns
d2/4
0
Figure 27: Simulation of signal intensity along beam axis. This is a “snapshot” at time t of Eq.
[39]. In the Fraunhofer zone, we can regard the function as a cos kx function with a slowly
decreasing envelope.
dmin  a2 /   D2 / 4 .
[40]
(iv) Since we use the Fraunhofer zone for imaging, with a diverging beam, the
lateral resolution gets worse with increasing depth.
 A limited improvement can be achieved by using ultrasound lenses to focus the
beam.
 However, the mathematics of diffusion say that you can only go so far:
(i) No matter how good the lens is, there is a minimum beam diameter.
In the focal plane of the lens, the beam size for a circular transducer is
d
2.4 flens 
,
D
[41]
where flens is the focal length of the lens and D the diameter of the transducer.
This is typically of the order 0.2mm.
(ii) The aim of focusing is to get a focal spot with size d<D. Hence,
35
2.4 flens
D
D

flens
D2
.

2.4
[42]
I.e., focusing does not work at all for distances into tissue greater than a certain
value.
(iii) The tighter you focus the beam, the less long it stays focused for — see Fig.
29. We have to make a compromise between width of beam at focus and depth
of field.
Focal Plane
Ultrasound lens
Diverging
Beam
D
Transducer
flens
Focal Spot
(diameter, d)
Figure 28: Use of an ultrasound lens to focus the beam
36
(a)
(b)
Figure 29: (a) Strongly focused and (b) weakly focused ultrasound beam
5.7. Artifacts … or …
Why do ultrasound images look so awful? (F179–189)
 It takes an expert to “read” most ultrasound images. Why is this?
 There are a number of artifacts to which ultrasound images are prone.
5.7.1. Misregistration
 A bright spot in the image appears in the “wrong” place. I.e., it does not correspond
to the true position of the reflector.
 There are several reasons for this:
(i) Refraction effects — see Section 3.3
37
Transducer
Ultrasound Image
Main
beam
Weak, angled beam
(“side-lobe”)
Weakly reflecting
object
Strongly reflecting
object
Figure 30: “Ghost” image caused by transducer side-lobe
(ii) Velocity differences
Since depth in ultrasound images is given purely by the time between the
transmission of the pulse and the reception of the echo, then if any part of the
medium between the transducer and the reflector has a non-standard velocity,
then the wrong depth will be recorded. The bright spot will appear at the
wrong place in the image.
(iii) Side Lobes
In Sec. 5.4, we noted that the transducer beam had a central lobe, plus sidelobes of reduced intensity at other angles. If one of these beams reflects off
something, then an echo will be received from one angle when the detector
“thinks” it’s pointing at a different angle — see Fig. 30.
(iv) Multiple Reflections
It is possible for the ultrasound beam to bounce backwards and forwards
several times between the transducer and shallow reflectors. At each stage, the
transducer will register an echo. This gives rise to a number of evenly spaced
spots on the final image.
38
Figure 31: Typical abdominal
ultrasound image. The arrow is
pointing to an example of
acoustic shadowing, but this is
not obvious to a non-expert.
(Image
source:
Brigham
Women’s Hospital, Harvard)
5.7.2. Shadowing and Streaking
 This is a consequence of the TGC unit — see Sec. 3.5.3
 Suppose we have a region of very low attenuation. This lets through more
ultrasound than the TGC is expecting. When this reflects off the objects behind, we
get a streak with higher intensity than normal.
 Suppose we have a region of higher than normal attenuation. This lets through less
ultrasound than expected the signal from objects behind it will be less than
expected. This gives rise to a shadow.
 Both these artifacts can be useful diagnostically.
5.7.3. Speckle
 Possibly the worst artifact.Caused by having small reflectors (not the infinite planes
used to derive the equations of reflection), which are closely spaced.An
interference effect. The reflected pulses interfere with each other in a random way
— see Fig. 31.BlurringThis is caused by the limited resolution of the apparatus —
see Sec. 3.6.5.Separate blurring occurs in the axial, lateral and slice directions.
39
5.7.5. If ultrasound images are so bad, why use them?
(i) As an imaging method ultrasound is relatively cheap.(ii)
It’s
quick
—
continuous, real-time images of, say, a beating foetal heart, which is difficult by
any other technique.
(iii) It’s portable — you can take the machine right to the patient’s bedside.It’s very
patient friendly
(v) The human eye is very good at making sense of noisy data and recognising the
patterns which correspond to the features of interest. Trained radiologists can
obtain very large quantities of information.
(vi) A general rule of imaging: pretty pictures are to impress the punters.
If you can get all the information you want easily from the given data, then you
don’t need anything more sophisticated. E.g., obstetric ultrasound: the recognised
measurements are diameter of foetal skull and length of femur. These can be
made to within a few mm with cheap, standard equipment.
6. Doppler Ultrasound (F117–164, C186–212, W351–358)
6.1. Introduction
 Measurement of blood flow is the second big use (after obstetrics).This is done
using the Doppler effect.When ultrasound waves strike a moving target (e.g., a red
blood cell), the reflected wave is shifted in frequency.This frequency shift is
proportional to the velocity of the target. Hence we can measure the speed of blood
flow.
6.2. The Doppler Equations
 Fig. 32 shows the notation we will be using.
 Because the blood is moving w.r.t. the transducer, the effective velocity of sound
relative to the red blood cells (RBC) is
vrel,inc  c  component of v along direction of beam
 c  v cos  inc
So, wavecrests arrive at the RBC with a frequency
40
[43]
fRBC 
=
vrel, inc
(normal v  f with changed speed)
 inc
c  v cos  inc
 inc
[44]
 These wavecrests are scattered off the RBC and are re-radiated back to a second
transducer. Clearly, since the source is moving away from the receiver, these waves
are “stretched out and so scat is greater than inc.
 Since the frequency of re-radiation of wavecrests by the RBC is fRBC, then in a time
t, fRBC t wavecrests have been emitted.
 During this time t, the first wavecrest will have travelled a distance ct in the
laboratory frame of reference, but ct + v cos scatt relative to the RBC, which is
receding from it.
 Thus, there must be fRBC t waves contained in a distance ct + v cos scatt. The
reflected ultrasound beam has wavelength
 scat 
ct  v cos scat t
fRBCt
 (c  v cos scat )
41
c  v cos inc
 inc
[45]
Figure 32:
Statement of the problem in Doppler ultrasound. A moving boundary (RBC)
scatters an incident ultrasound beam and the reflected beam is shifted in
frequency.
 The final result can be written either in terms of the reflected wavelength, or, more
usefully, in terms of the frequency of the reflected wave in the laboratory frame of
reference.
 scat 
c  v cos scat
 inc
c  v cos inc
or
fscat 
c  v cos inc
finc
c  v cos scat
[46]
 We often make the simplifying assumption that v<<c, in which case, the frequency
difference between the two beams is
v
f  finc  fscat  (cos inc  cos scat ) finc
c
[47]
6.3. How do we put all this into practice?
 Fig. 33 shows the block diagram of a Doppler ultrasound scanner.
 This most primitive type of Doppler machine is called a continuous wave (CW)
scanner. It does not emit pulses like an A or B mode scanner.
42
Figure 33: Block diagram of a CW Doppler ultrasound machine
 What is reflected back is a continuous signal which contains some ultrasound
reflected by moving spins and some reflected back by stationary spins.
The extra components of the Doppler scanner are listed below.
6.3.1. Second transducer for reception
 The Doppler probe in the diagram above consists of two transducers.
 The first transducer is transmitting ultrasound continuously. There is no “spare”
time to receive the signal bounced back.
 The second transducer is normally very close to the first and offset at a very slight
angle. incref and so Eq. [47] becomes
f 
2v
cos  inc  finc
c
43
[48]
(a)
Spectral
Amplitude A
f0 = 2(v/c) cos inc
v
 f0
Demodulated
frequency f
A
(b)
fpeak = 2(vpeak/c) cos inc
vpeak
fpeak
(c)
Rapid flow
past obstruction
f
A
Stationary object
peak
Slow flow downstream
of obstruction
Stenosis
(blockage)
“Negative” 0
flow in eddy
Reverse flow
in eddies
f
Figure 34: Three different flow patterns with their corresponding Doppler frequency profiles, as
given by a spedtrum analyser: (a) plug flow, only a single velocity present; (b) Poiseuille
flow with quadratic velocity distribution — common in vessels with slow flow; (c)
complex flow pattern around an obstruction in the vessel. (a) and (b) represent
“idealised” patterns; in practice, one would not obtain such “clean” patterns.
6.3.2. Doppler demodulator
 This works in exactly the same way as described in Sec. 3.5.4.
 Note that the frequency which we subtract from the signal is finc, which is used as a
reference. Thus, in the block diagram, the output from the oscillator goes both to
the transducer and the demodulator.
6.3.3. Frequency meter / spectrum analyser
 After the demodulation process, we have an oscillating signal of form cos (2ft).
We still have to extract the value of f.
 On primitive machines, radiologists just listened to the signal, since f is in the
audio frequency range.
44
f v
Range of flow velocities
present at time t0
t0
t
Figure 35: “Real time” graph of flow velocities in a vessel.
 Slightly more sophisticated is a frequency meter, which counts the number of zerocrossings of the function, divides by 2 and displays the result as a frequency. This
leads to problems with multiple frequency components and noisy data.
 A real-time spectrum analyser is the ideal solution. It gives a graph of all the
different frequency components.
6.3.4. Real-time display
 Displays such as Fig. 34 show complete information for a “snapshot” view.
However, it is often more useful to show more limited information, but in real
time.
 Modern Doppler machines can acquire the information very rapidly, so we often
plot a graph with time along the x-axis and velocity range up the y-axis.
 By analysing such plots we can find out about the flow properties of blood vessels
and diagnose problems.
6.3.5. Problems with CW Doppler
(i) Signal comes from the whole sensitive region — see Fig. 33 — we have no way
of deciding where the flowing material is. What happens if there are two vessels
with different velocities?We have to estimate the value of
[48], so the
derived velocities are not certain.We can only find the component of velocity
along the beam direction. If the scatterer is moving perpendicular to the beam,
there is no Doppler shift.Duplex Scanners
 The ideal solution to problems (i) and (ii) above is to combine a Doppler ultrasound
scanner with a B-mode imager.
45
 To do this, we need to select where we acquire the Doppler information from.
 This process is called range gating and works as follows:
1.
A pulse of ultrasound is emitted at angle
2.
We wait for echoes to come back, but turn the receiver on only between times
The
frequency shift of the signal is analysed as before (although there are a number
The whole procedure is repeated for
region.The velocity information is often colour-coded and superimposed on a
normal greyscale B-scan.
6.5. Limitations of Doppler Ultrasound
 Although Doppler ultrasound has revolutionised many aspects of diagnosis, it is
important to bear in mind that there are a number of limitations:
(i) We can measure only the component of flow along the direction of the
ultrasound beam.We need to be sure of the angle between the beam and the
vessel of interest. Blood vessels are rarely straight and this can lead to
problems in assigning the velocity correctly.The blood cells from which the
ultrasound is scattered are not all moving at a uniform velocity, so a spread in
values will be obtained. Flow is often turbulent.In practice, although it is
possible to undertake much more sophisticated forms of analysis, clinicians use
Doppler ultrasound in a very crude way. For example, a typical decision might be:
flow reduction < 50%, don’t operate; flow reduction > 50%, think about operating.
7. Bio-effects of Ultrasound (F201–212, C213–218, W375–377)
7.1. Introduction
 Sufficiently strong intensities of ultrasound have the potential to affect living
tissue.In some cases, these effects are beneficial and can be used for
therapy.However, in diagnostic imaging, we wish to keep exposure as low as
possible.Heating
46
Tissue mass m
A
Beam
propagation
x+x
x
Figure 36: Notation for calculating temperature rise in a tissue caused by absorption of
ultrasound energy
 In Sec. 3.4.2, we saw that an ultrasound beam is attenuated as it passes through
tissue and that the beam’s acoustic energy is changed into heat.
 How does this change the temperature of the tissue?
We have IRMS(x) = IRMS(x) exp(2x), or, in differential form,
I RMS ( x)  2I RMS ( x) x .
[49]
 IRMS(x) is the power loss to the beam, so, by conservation of energy, it must be the
power gain by the tissue.
 Remember: is the attenuation coefficient for pressure, I0(x) is the peak power
flux through a unit area (measured in W/m2). It can be shown that the average
power is half this.
 The energy gained by the shaded tissue element in time t is
Q  2 AIRMS ( x) x t.
[50]
 Assume that heat conduction by the tissue is quite poor and that we look at the
temperature rise only for the first few seconds after the beam is switched on
[51]
where c is the specific heat capacity and m is the mass of the volume element, i.e.,
(Ax). This gives
47
T 
2 AIRMS ( x ) x t 2IRMS ( x )
I

 t  0  t,
A x 


[52]
or, alternatively, the initial rate of temperature rise is
dT 2IRMS

.
dt

[53]
 In practice, selective heating of tissue is used as a method of treating cancer. The
ultrasound beam is focused so that where the tissue is not tumorous, IRMS(x) is
small and the heating rate is low, but at the tumour a large amount of power is
concentrated.
 In fact, the assumption that we made above is a gross over-simplification. The
conduction by tissue is not negligible and in practice, one needs to use a heat
diffusion equation. In addition, the body has its own thermo-regulatory system,
which tries to keep the body at its normal temperature. Thus, flowing blood can
carry away excess heat quite efficiently. The organs most susceptible to clinical
hyperthermia are those with a poor blood supply.
 With a less focused beam, the heating effect is lower, but more widespread. This is
often used to apply localised deep heat to damaged muscles in physiotherapy.
7.3. Radiation Pressure
 You will find in many branches of physics that a flow of energy via a wave motion
brings with it a flux of momentum. The normal relationship is
E  Mc
[54]
where M is the momentum “carried” and c is the speed of the radiation.
 For ultrasound, the energy carried per second is AIRMS(x). (Remember: A is the
cross-sectional area of the beam and I0(x) is the peak power flux per unit area —
the time-average is half this.) The momentum carried is
M
AIRMS ( x )
t.
c
48
[55]
 Newton’s 2nd Law tells us that Force = Rate of change of momentum and the
definition of pressure is Force/Unit area. Hence
Prad 
I RMS ( x)
.
c
[56]
 If, instead of being absorbed, the ultrasound is reflected, then the momentum
changes from M to -M (i.e., the total change is 2M) and so the pressure is doubled.
 This force is extremely small, but may affect delicate structures in the body.
7.4. Acoustic Streaming
 Putting together the arguments of the previous two sections, it is clear that each
element x absorbs a power aIRMS(x)x per unit area, which is equivalent to a
momentum of IRMS(x) x / c per second.
 Hence, there is a force F = IRMS(x)x / c acting on each element xof the sample.
 If that sample is a liquid, it
moves. If the liquid is in an
enclosed vessel, we get a
characteristic type of motion
called acoustic streaming.
Streamline
Fluid motion
Transducer
Figure 37: Acoustic streaming in an enclosed container
 A delicate structure in such
a stream may be affected
both by the direct force of
the ultrasound, or by
viscous drag from the
liquid, or even Bernoulli
forces.
7.5. Cavitation
 Suppose that there are tiny bubbles in the sample, with diameters <<l.
 When an ultrasound wavefront (pressure maximum) coincides with the position of
the bubble, the bubble is squashed.Half a cycle later, there will be a rarefaction
(pressure minimum) and the excess pressure in the bubble causes it to expand.So,
49
as the peaks and troughs pass by, the bubble oscillates in size. Overall, the bubble
may grow by a process known as rectified diffusion. During the accoustic pressure
cycle, gas alternately diffuses in and out of the bubble. In the rarefaction phase, the
surface area of the bubble is larger and more gas diffuses in during this part of the
cycle than diffuses out during the compression phase. Hence the volume of the
bubble increases.
 There are two types of cavitation, stable and unstable. Both types have the potential
to cause damage to tissue.
 In stable cavitation, the bubbles persist for a long time (i.e., a relatively large
number of acoustic cycles) and can grow quite large. They cause micro-movements
of the surrounding liquid and tissues. This may be harmful.
 In unstable cavitation, the bubbles burst violently and this can cause important
changes at a cellular level. The bursting is associated with high local temperatures,
emission of light, chemical changes and the formation of free radicals, which can
be very destructive.
(a)
Bubble
Crest of
ultrasound wave
c
Low external
pressure
(b)
High external
pressure
Figure 38:
Mechanism of cavitation. (a) In a low external pressure (rarefaction), the vapour
pressure inside the bubble causes it to expand. (b) When the external pressure is
high, the bubble contacts. As the ultrasound wave passes, the bubble oscillates in
size.
50
7.6. Therapeutic Ultrasound
7.6.1. Physiotherapy
Several of the effects described above are believed to play a role in the treatment of
muscle damage in physiotherapy. Although not all of the mechanisms have been fully
explained, it is known that ultrasound is beneficial for:
 increasing blood flow and nutrient supply to the damaged muscle (by a factor of up
to 2 or 3);breaking up unwanted/damaged/fibrosed parts of the muscle;aiding
transport of chemicals across cell walls;
 relieving pain — this is not well understood;increasing tendon flexibility;reducing
muscle “spasm” — cramp is an extreme form of muscle spasm.Average power
(IRMS) used in physiotherapy ranges from about 0.125 to 3 W/cm2.Surgical Uses
This makes use of two main properties:
 The heating effect is used in the treatment of cancer by controlled clinical
hyperthermia. Essentially, the tumour is “cooked” to kill it.
 The mechanical vibration and cavitation effects (also used in “sonicator”
cleaners/stirrers in chemistry) can dislodge unwanted tissue or particles. It is often
used to fragment kidney stones (lithogtripsy) and sometimes cataracts as well.
7.7. And finally ... is it safe?
One of the reasons why diagnostic ultrasound is used in pregnancy is that it is
claimed to be completely safe. However, as we have seen above, it is used at higher
powers to kill cancerous tissue deliberately. So what power is “safe”?
We can give the following rough guidelines:
IRMS << 0.125 W/cm2

0.125 < IRMS < 3 W/cm2 
IRMS >> 3 W/cm2

diagnostic — “completely safe”, “does not change
tissue”
physiotherapy — “safe”, “does not harm tissue”
surgery — damages/kills/shatters tissue in a
localised area
51
As with any other technique, it is impossible to prove that ultrasound (even
diagnostic ultrasound) is completely safe. We can only demonstrate the scarcity of
recorded cases demonstrating that it is dangerous.
52