Medical Physics Tutorial 5 - Non-ionising Imaging 1 (Ultrasound Imaging)

Contents

Initial Diagnosis
Ultrasound Imaging

Generation and Detection

Ultrasound in the Body

Imaging in the Body

Scans
Advantages and Disadvantages

Acoustic Impedance

Refraction in Ultrasound

Reflection of Ultrasound

Blood Flow

 

 

Initial Diagnosis

For most people the doctor can diagnose an illness by:

The doctor may need to carry out further tests by taking samples of urine, or blood, which are analysed in the hospital laboratory.

 

For some patients more specialised diagnostic techniques are needed, which are what we are going to look at here.  They are usually carried out in a hospital, and require specialist personnel to operate the machines and to interpret the results.  Before these techniques were invented, the only way to diagnose problems was to feel the body.  An exploratory operation, a laparoscopy, could be carried out, but this invasive procedure carried serious risks.  Such operations can still be carried out as a last resort.

 

These techniques we will look at are non-invasive.  They enable the doctor to see under the skin without having to open up the body.  An invasive procedure requires cutting the skin.  In this tutorial and Tutorial 6 we will look at diagnostic techniques that do not require ionising radiation.  In Tutorials 7 and 8 we will look at those techniques that use ionising radiation such as X-rays and gamma rays.

 

 

Ultrasound Imaging

Ultrasound is any sound that is higher than the upper limit of human hearing. 

 

Question 1

What is the upper limit of human hearing?

Answer

 

Bats and dolphins use frequencies in the range of 30 - 100 kHz for echo location and in the Second World War experiments in ultrasound were tried out to detect submarines.  Ultrasound imaging is a proven method of investigating objects internally without causing damage.  It is widely used in medicine because it is non-invasive.

 

 

Generation and Detection of Ultrasound

The ultrasound probe or transducer is used to generate and detect the ultrasound waves.  The most common method is by use of a piezoelectric transducer.  If you squeeze or stretch a crystal of quartz (easier than might be thought) a voltage is induced.  It can be high enough for a spark to jump.  Gas lighters use the effect.  Conversely, if you apply a voltage to a piezoelectric material, you can make it change shape.  If the voltage is alternating, the crystal will vibrate.  Maximum energy transfer occurs when the crystal is in resonance.

 

Question 2

What are the conditions needed for resonance?

Answer

 

The material, usually lead zirconate titanate (PZT), an artificial ceramic, has a thickness of half a wavelength of the ultrasound wave.

 

The lens protects the PZT slice and acts to converge the beam slightly.  The beam consists of short pulses of frequencies of several megahertz.  The vibrations are damped by the backing block which is made of epoxy resin.  The whole is contained in a metal case which protects the probe mechanically and electrically.

 

Question 3

A sound wave has a frequency of 5.0 MHz and travels at 1500 m s-1.  What is the wavelength? 

Answer

 

The ultrasound probe sends out pulses, rather than a continuous beam.  This allows the reflected waves to be detected.  The same crystal detects the reflected waves.  The frequency of the pulses used in medical ultrasound scanners are between 2 to 15 MHz.  The probe emits trains of waves (pulses) that last about 15 ms.  Then no pulses are sent out for about 0.5 ms, so that the machine can detect the echoes.  The idea is shown below.

 

 

Worked Example

The frequency of an ultrasound wave is 2.5 MHz.  Each pulse duration is 15 ms.  How many waves are in each train of waves?

Answer

Each wave lasts 1 2.5 106 Hz = 4.0 10-7 s

In 15 ms, there are 15 10-6 s 4.0 10-7 s = 37.5 waves

 

The Pulse Repetition Period is the sum of the pulse and the "listening time".  In this case it is:

 

PRP = 15 10-6 s + 0.50 10-3 s = 5.15 10-4 s = 515 ms

 

Therefore the probe would send out 1941 such pulse repetitions every second.

 

The ratio of the pulse duration to the pulse repetition period is often called the duty factor.  It given as a percentage.  In this case:

 

DF = (15 10-6 s 5.15 10-4 s) 100 % = 2.9 % 

 

 

Ultrasound in the Body

When ultrasound passes into the body:

The extent to which the ultrasound is absorbed or reflected gives information about the structures below.

 

The table below gives some ultrasound properties of some body tissues:

 

Material

Density (kg m-3)

Velocity (m s-1)

Acoustic Impedance (Z = rc)  (kg m-2 s-1)

Air

1.3

330

429

Water

1000

1500

1.50 106

Blood

1060

1570

1.59 106

Brain

1025

1540

1.58 106

Fat

925

1450

1.38 106

Muscle

1075

1590

1.70 106

Bones (varies)

1908

4080

7.78 106

PZT Transducer

7650

3791

29.0 106

Quartz Transducer

2650

5736

15.2 106

 

 

The acoustic impedance is the product of the density and speed of sound in the material.

 

Z - acoustic impedance (kg m-2 s-1); r - density (kg m-3); c - speed of sound (m s-1)

 

Use the table above for information to use in the questions below.

 

Question 4

What is the thickness of a slice of PZT material if its fundamental resonant frequency is 1.5 MHz? 

Answer

Question 5

The time delay for a pulse going through fat is 0.133 ms.  How deep is the fat?

Answer

 

Remember that the waves emitted from the ultrasound probe are reflected and picked up by the probe.  Therefore the time taken for the waves to get through a layer is half the time measured by the probe.

 

 

If the probe is placed straight on to the skin, almost all the energy will be reflected.  So the probe has to have a coupling medium between it and the skin.  This is a gel or an oil.  If there is gas anywhere, it can cause big problems for imaging.

 

When the waves reach a boundary, a small amount, about 1 % gets reflected.  However if the difference in acoustic impedance between two tissues is large, a high proportion of the ultrasound waves are reflected.  For example the boundary between lung tissue and the air in the lungs leads to a 99.9% reflection, making it impossible to view structure behind the lungs.

 

 

Imaging in the Body

The ultrasound passes through the tissue and is reflected at various boundaries as shown:

 

 

There will also be refraction, as the wave speed changes as it passes across boundaries.  For example, the speed of sound through water is 1500 m s-1 compared with 1450 m s-1 through fat. 

 

The resolution of the beam means the smallest distance that can be discriminated in the image.  The higher the frequency, the better the resolution.  However as the beam passes through, the sound waves get scattered and absorbed by the molecules.  This attenuation is more marked with higher frequency.  Therefore a compromise has to be made.  The optimum frequency for imaging the brain and abdomen is roughly 1 - 3 MHz.  

 

When the return pulses are received, the transducer turns them into electrical signals to be stored:

 

Scans

There are a number of ways in which ultrasound can be used.  We will consider:

The two are compared in the diagram below:

 

 

A-scans are used where the anatomy of a section is well known and a precise depth measurement is needed.  One example is where the position of the midline of the brain is needed.  Any delay could indicate the presence of a tumour or a fluid filled space.

 

The B-scan is the basis of two-dimensional scanning.  The transducer is moved about to view the body from a variety of angles.  The probe can be moved in a line (linear scan), or rotated from a particular position (sector scan).

 

 

 

The two movements can be combined to give a compound scan.  It requires considerable skill and a good knowledge of anatomy for the sonographer to get a decent image and to interpret it.  As you can see, it takes some interpreting.  The head is on the bottom right, while the legs are to the top left.

 

Image from Wikimedia Commons (Melimama)

 

However it is a widely used technique for assessing the growth of the prenatal foetus.  It can give early indications of any problems that may arise.  

 

Ultrasound is used in other investigations such as detections of cysts, abscesses, and tumours.

 

Real time B-scans  use a linear array of up to 100 transducers to get a cross section of the body.  Moving images are possible.

 

Ultrasound can be used with the Doppler effect to watch the movement of blood through the blood vessels.

 

 

Advantages and Disadvantages of Ultrasound Scanning

Ultrasound is generally a very safe diagnostic technique:

 

However:

 

High energy ultrasound waves can be used for therapeutic purposes.  Low intensity ultrasound can be used in healing wounds and relieving discomfort and pain in conditions like arthritis.  High intensity beams can shatter kidney stones.  Ultrasound treatment has to be done with care because:

While this would not cause many problems for an adult, it must be avoided where there is a growing foetus.

 

Question 6

Write down two advantages of using ultrasound as a diagnostic tool and two disadvantages.

Answer

 

Photo by Joseph Caballero, Wikimedia Commons

 

Acoustic Impedance

We have seen before how acoustic impedance is defined as:

the product of the density and speed of sound in the material.

 

Z - acoustic impedance (kg m-2 s-1); r - density (kg m-3); c - speed of sound (m s-1)

 

The unit for acoustic impedance is sometimes called the Rayl (named after John William Strutt (1842 - 1919), 3rd Baron Rayleigh). 

 

When ultrasonic sound waves are applied to any system consisting of any material, the materials will present some kind of opposition to the passing of the sound waves.  Therefore the wave energy is lost.  The intensity of the reflected waves received at the probe will always be less than the intensity of those that are transmitted.  The extent to which the energy is lost in a material is called the acoustic impedance.

 

Question 7

Show that the acoustic impedance of air is about 400 kg m-2 s-1, assuming the density of air is 1.2 kg m-3  and the speed of sound in air is about 340 m s-1.

Answer

 

When wave energy strikes the interface between two different materials, a certain amount of the energy is reflected.  This is why you can see your reflection in the glass of a window when it's dark outside.  We saw that when we did refraction with light.  There was always a weak reflected ray, which we tended to ignore.  The same is true with sound waves, which is why you hear echoes from walls.  (You may remember doing an experiment to measure the speed of sound by standing 30 metres from the wall of the gym, banging two bits of wood together, and banging them again when you heard the echo.  You timed 10 bangs, etc.  I hated doing that experiment, as I could never get the rhythm...)

 

Refraction in Ultrasound

When ultrasound is emitted by the probe, the probe acts as a point source, with the waves propagating as a cone.  The idea is shown below:

 

As the waves pass across the boundary between the abdominal cavity and the organ, they pass from water of acoustic impedance 1.50 106 kg m-2 s-1 to a tissue of a different acoustic impedance.  For the liver, Z = 1.69 106 kg m-2 s-1

 

Question 8

Calculate the speed of sound in the liver.  The density of the tissue for liver is 1.06 103 kg m-3.  Give your answer to an appropriate number of significant figures.

Answer

 

From Waves Tutorial 6 we know that:

 

 

Do NOT use the refractive indices that you might know for optical refraction.  We are using sound waves, not light.  Most tissues in the body are opaque to light.

 

Question 9

An ultrasound beam is incident on the surface of a liver at an angle of 2.0 o as shown:

 

 

The speed of sound in water is 1500 m s-1.  Use your answer to Question 8 to calculate the angle of refraction.

What is the path of the beam after the refraction?  Explain your answer.

Answer

 

It is quite possible that you might get asked a question like this in the exam.  Refraction will spread the waves out as they pass boundaries (if the speed of the waves is higher in the second material).  This will have the effect of reducing further the intensity of the waves.  It could also affect the results if there are several probes being used to pick up transmitted waves.

 

Reflection of Ultrasound

We have discussed how the waves can be scattered by reflection and refraction.  However, for a single probe ultrasound scan, that isn't really an issue.  The sonographer is only interested in the waves that are transmitted and received by the probe.  The received waves are those that are reflected by the tissue or organ under study.  The acoustic impedance, Z, is the important quantity in this case.

 

The fraction of the energy reflected is sometimes called the reflection fraction, or the reflection coefficient.  It is expressed as a fraction or a percentage.  It is always less than 1, because some energy is lost by reflection, refraction, and absorption.  If you get an answer greater than 1, you have done something wrong.  The reflection fraction is sometimes given the code G (Gamma, a Greek upper-case letter 'G').  It is related to the intensity in the equation:

 

 

G - reflection fraction; Ir - Intensity of the reflected wave (W m-2); Ii - Intensity of incident wave (W m-2)

 

We can extend this further by relating the reflection fraction to the acoustic impedance with this relationship:

 

 

Z1 - acoustic impedance of material 1 (kg m-2 s-1); Z2 - acoustic impedance of material 2 (kg m-2 s-1)

The way to remember this is write:

Reflection fraction = (difference in acoustic impedances sum of the acoustic impedances)2

 

It doesn't matter if the difference yields a negative number, because the whole thing is squared.

 

Worked example

An ultrasound probe has a power of 3.0 W.  It forms a circular beam of diameter 2.5 cm, and can be modelled as having parallel sides

(a) Work out the intensity of the of the beam.

(b) The density of aluminium is 2700 kg m-3 and sound travels through aluminium at a speed of 6320 m s-1.  Calculate the acoustic impedance of aluminium.

(c) An ultrasound probe is placed near a block of aluminium  as shown:

 

 

     Calculate the reflection fraction from the aluminium.

 

 (d) Calculate the intensity of the reflected wave, assuming that the transmitted wave has not lost any of its energy in the water.

 

Answer

(a) I = P A

     Area = pD2/4 = (p (2.5 10-2 m)2) 4 = 4.91 10-4 m2

     I = 3.0 W 4.91 10-4 m2 = 6112 W m-2

 

(b) Z = rc = 2700 kg m-3 6320 m s-1 = 17.0 106 kg m-2 s-1.

 

(c)  Z for water = 1.50 106 kg m-2 s-1 Z for aluminium = 17.0 106 kg m-2 s-1.

 

      Formula:

     

      G = ((17.0 106 kg m-2 s-1 - 1.50 106 kg m-2 s-1) (17.0 106 kg m-2 s-1 + 1.50 106 kg m-2 s-1))2 = ((15.5 106 kg m-2 s-1) (18.5 106 kg m-2 s-1))2

     

      G = 0.8382 = 0.702 = 70.2 %  (Remember that you have to square it.)

 

(d)  Formula:

    Ir = GIi = 0.702 6112 W m-2 = 4290 W m-2 = 4300 W m-2 to 2 s.f.

 

Note that this example uses a non-biological material.  The principles of ultrasound are equally valid in engineering.  Ultrasound probes are widely used in the engineering industry to check for flaws in casting, etc. 

 

 

 

Remember to square the term [(Z2 - Z1) (Z2 + Z1)]

 

Question 10

In the human body, the kidneys are in close proximity to the muscles of the back.  An ultrasound probe sends waves to the kidney.  By the time the waves reach the junction between the kidney and the muscles, they have an intensity of 260 W m-2.

 

Using the data below, calculate:

(a) the acoustic impedance for the kidney tissue, and the muscle tissue.

(b) the intensity of the reflected wave.

 

Density of kidney tissue = 1050 kg m-3; Speed of sound in kidney tissue = 1570 m s-1

Density of muscle tissue = 1065 kg m-3; Speed of sound in muscle tissue = 1590 m s-1

Answer

 

From this answer, we can see where the acoustic impedances are similar, the fraction of energy reflected is small.  They will not show up well against the muscle.  Ultrasound is useless to look things going on in the lungs, because they are full of air.  The next question shows why.

 

Note that the frequency of the ultrasound is NOT affected by reflection, absorption, refraction, or attenuation.

 

Question 11

An ultrasound probe is being used to diagnose a problem with a patient's lung. 

Lung tissue has an acoustic impedance of 2.60 105 kg m-2 s-1.

Air has an acoustic impedance of 429 kg m-2 s-1.

 

(a)  Calculate the percentage of the intensity that is reflected.

(b)  Comment on whether an ultrasound scan would be able to pick up any problems within the lungs. 

Answer

 

Investigations of the lungs require different techniques for investigation, for example, endoscopy.

 

Blood Flow (OCR, Eduqas, and Welsh)

Ultrasound can be used to check the flow of blood into an organ.  If there is insufficient flow of blood, the organ might not work properly.  It could be seriously damaged which can cause permanent damage to the patient. 

 

The blood flow can be measured using a probe held at an angle as shown below:

 

 

 

The ultrasound waves are reflected from the red blood cells.  Since the red blood cells are moving from right to left (towards from the probe), the reflected frequency f ' is higher than the transmitted frequency f0, due to the Doppler Effect.  (If the probe were in the other direction, the reflected frequency would be lower than the transmitted frequency.)

 

The factor we are interested in is the difference in the frequency, given the physics code Df.  This is shown in the equation below:

 

The Doppler Effect equation tells us that the ratio of the difference in the frequency to the original frequency is the same as ratio of the speed of the moving object to the speed of the wave.  This is shown here:

[Df - change in frequency (Hz); f0 - original frequency (Hz); v - speed of blood flow (m s-1); c - speed of ultrasound (m s-1)]

 

Ideally we would place the probe along the axis of the blood flow.  However that is not possible, so we have to have the probe at an angle, q

 

To get the speed of the blood flow we use the equation:

 

The v term is doubled, because the waves are being transmitted from a fixed point, rather than a moving object.

 

Question 12

An ultrasound probe transmits at a frequency of 2.0000 MHz.  It is set up an angle of 20 o to a blood vessel to check the speed of flow.  The received signal at the probe is found to be 2.0055 MHz. 

(a) Calculate the speed of the blood, assuming the waves travel through the body tissue is about 1600 m s-1.

(b) What would be the effect if the probe were held at 90 o to the blood vessel.

Answer