No Slide Title

[Audio] Hi there! My name is Dr Louis Lee, I'm one of your lecturers in the course of ' Physics of Medical Imaging". I'm responsible for three areas namely, the Ultrasound Imaging, Computed Tomography ( CT) and X-ray Imaging. The first part will be on the Ultrasound Imaging. Due to the disruptions brought by the COVID- 19, this year we'll take a different approach on lecturing. Approach 1: If you already have some basic knowledge in Ultrasound, you can do the self reading first and use my notes as a revision. Approach 2: If you have difficulty in going through the chapter, you might first read my notes to grasp some basic ideas and then go back to the chapter, you would find the chapter much easier to understand. Either way will work depending on the type you prefer. Make sure you understand the basic concepts pertaining to the syllabus listed on page 4. The book chapter contains more than enough you need for this course, you can skip those not included in the syllabus..

Dr Louis Lee, PhD, CSci. Physics of Medical Imaging Ultrasound Imaging.

[Audio] This is the slice which is the most important slice in all my lectures. This powerpoint is for personal use only and not to be disseminated in any public domain. I'm not going to read this out, so please have a look yourself..

[Audio] The syllabus is divided into three major sections. The first one is some basic knowledge of ultrasound and I'll talk about the ultrasound physics, interactions of ultrasound with tissue and some basic ultrasound instrumentaion. The we'll move on to some practical ultrasound imaging, to learn about the pulse-echo ultrasound machine and the ultrasound image artefacts. Finally, we'll go to study the Doppler ultrasound to learn about the Doppler physics and the instrumentation..

[Audio] Before we start, I'd like you to look at these 2 images and remember their patterns. On the left, it is a liver image, you'll see the liver tissue, the diaphragm and the haptic veins. On the right, this is a foetal image, this is the baby's head, the thorax, the spine, the humerus and the femur. And this is the placenta. After the lecture, you might be able to tell how the images of these structures are formed. Please also note the ultrasound images are in a sector form and I hope you can tell why after the lectures..

[Audio] Basic ultrasound physics.. Basic Ultrasound Physics.

[Audio] What is sound? In fact, we learnt this at high school, right? Let's do a revision or recall. Sound is a mechanical disturbance that propagates as a wave in a medium, so we need a medium for the sound to travel. You can image a medium is made up of molecules in a lattice, and if I put it in a one dimensional sense, I can represent this idea by this drawing; the molecules are joined by a spring. In a very loose sense, the molecule have a mass and you need some kind of force to move it, that is, it is reluctant to move, we can say it has inertia. Also in a very loose sense, the spring arises from the intramolecular force that gives the medium kind of compression and recoil properties. Now in this way, I can describe a medium as having inertia and compressibility. Inertia is associated with mass per unit volume, i.e. density. Compressibility is elasticity if you like. So when a sound travels through a medium, the molecules in the medium do not move from the source to the detector, only the disturbance travels. The molecules are oscillating about its mean position in a concerted way..

[Audio] OK sound frequency is represented by Hertz. Hertz is the no. of oscillations per second. The sound frequency that we can hear is from 20 Hz to 20 thousand Hz. Any frequency larger than 20 kHz is classified as ultrasound ( ultra means outside the range of frequency that we can hear, I guess) In diagnostic ultrasound, the useful range is from 2 to 25 MHz. kilo-Hz is 10 to the power 3 Hz and mega-Hz is 10 to the power 6 Hz. By the way, throughout the powerpoint, you'll notice that there are some club symbols indicating some important key points to remember..

[Audio] Now we come to the propagation of ultrasound in a medium, if we can take a snapshot before and after a sound wave propagates, you'll see something like that. Snapshot 1: the molecules are at rest. Snapshot 2: a wave or a disturbance is introduced, you'll notice that some of the molecules move the the right , some move to the left and some are in their original (mean) position. In the region where the molecules are crowded together, it's called a compression and where the molecules are far apart, it's called the rarefaction. In fact, during the wave propagation, each molecule moves back and forth about its mean position along the direction of the wave travel and the adjacent molecules move in a ' harmony' way so the wave can propagate out. This is a longitudinal wave. So if you image again to see a movie of the movements of the molecules during a sound propagation, you'll see the rarefaction and compression moves along the line in a orderly way. Again, please remember, it is the wave that propagates, the molecules are oscillating about their mean positions, not travelling along the line. How about water wave? When you flow a stone in the water, water crests will be travelling out but actually the motions of water molecules are moving up and down, perpendicular to the direction of the wave. They are called transverse wave..

[Audio] I want to introduce a term called Acoustic Velocity and it is usually represented by a small letter c. This is the speed at which a wave propagates through a medium, it is NOT the particle velocity. It also represents the rate of wave energy transmitting through a medium. c depends on the density and compressibility. The density is related to the inertia and the compressibility is something related to the 'spring' I mentioned before. So if I put it in a mathematical way, c is equal to the square root of kappa over rho where kappa is the bulk modulus of elasticity, a measure of compressibility. And rho is the density. Now you know c is related to the properties of the medium, you should now expect the acoustic velocity will be medium dependent..

[Audio] Let's have a look on the speed (acoustic velocity) in some media. Water is 1480 m/s, soft tissue is 1540 m/s, lung is 650 m/s and bone is 4080 m/s. By the way, why do we need to specify the water temperature for the speed above? It is because the density changes with temperate and hence the speed. PZT is the crystal we use to generate ultrasound, the speed is 3791 m/s. In the diagnostic ultrasound dealing with human tissues, it just happens that c is proportional to the density. A higher density medium will have a higher c. Soft tissue is 1540 m/s, lung is less dense, so it is 650 m/s, bone is of course denser, so it has a c of 4080 m/s. Again, in diagnostic ultrasound on human tissue, c has NO dependency on the body temperature, c has NO dependency on the frequency use. Therefore, if you're using 3 MHz, the speed of ultrasound in soft tissue is 1540 m/s. If you change to use 5 MHz, what would be the ultrasound speed in soft tissue? It is 1540 m/s..

[Audio] What are the relations amongst the speed, frequency and wavelength in ultrasound? The seed of ultrasound is constant for a particular medium. c is equal to f lambda. Can you then work out the wavelength of diagnostic ultrasound if the frequency you're using is from 3 to 15 MHz? If you cannot get the right answers for that, ask me during the tutorials. One important thing to remember, the frequency remains the same when a ultrasound wave passes from a medium to another medium with a different speed. Also you need to work out how long does it take for the ultrasound to travel one cm in soft tissue. It is 6.5 microsecond. So ultrasound is kind of transmit-and-receive matter, you send out an ultrasound pulse and you listen for an echo to come back, if the echo comes back 13 microsecond later, it means the surface reflecting the echo is 1 cm from you because it takes 6.5 microsecond to reach the surface and it takes another 6.5 microsecond for the echo to travel back making a total of 13 microsecond. Remember 13 microsecond is a magic number for 1 cm in ultrasound in soft tissue. If you need wait 26 microsecond to hear the echo, how far is the reflecting surface from you? It is 2 cm, so how about 39 microsecond? It is the 3 cm..

[Audio] Another parameter of ultrasound is intensity which is a physical parameter that describes the amount of energy flowing through a cross-section area per second. Frequency, wavelength and the speed are not affected by a change of intensity in an ultrasound beam. As the instantaneous intensity in an ultrasound is oscillating between high and low values, we use time-averaged intensity to describe the intensity of an ultrasound beam. Power = Intensity x Area But in the ultrasound community, power and intensity are used interchangeably, acoustic power means acoustic intensity or vice versa. There is power control knob on most ultrasound machine, it adjusts the intensity of ultrasound beam that is transmitted into the patient..

[Audio] The absolute values of the intensity or power of an ultrasound beam are difficult to measure, particularly true as our ultrasound beam is operated at pulsed mode. So we use a relative measurement that compares the value at a point with a reference value. A good way is to use decibel which is 10 times log of the Intensity at a point of interest over the original or reference intensity. By using dB, we can handle a wide range of levels . We have a 10 here because we don't want to handle decimal point. And dB is 'additive' as well..

[Audio] The ratio of 10,000 to 1 is 40 dB. The ratio of 1 to 10,000 is -40 dB. And this is the level we encounter in ultrasound. The intensity of the returned echo is in the order of 0.0001 of the original pulse sent out. i.e. -40 dB..

[Audio] Another term to know in ultrasound is Acoustic Impedance (a large capital Z). It is a measure of the resistance to ultrasound passing through a medium. It depends on density rho and speed c. Z is equal to rho times c. The unit is kg per meter squared per sec. Air is 0.0004 times 10 to the minus 6 kg per meter squared per sec and soft tissue is 1.63 times 10 to the minus 6 kg per meter squared per sec. Any acoustic mismatch at a interference will cause some portion of ultrasound to be reflected and that's exactly what allows us to see a boundary. If there is no reflection, there will be no echo and you'll see nothing..

[Audio] Let's look at this diagram. On one side, we have a medium of Z1 and on the other side Z2. The green big arrow represents the incident ultrasound beam, on hitting a interface between the media of different acoustic impedances, some of the ultrasound will transmit through (the blue arrow) and some will be reflected (the red arrow). The degree of which will be dependent how mismatch the two impedances is. Alpha R is the reflection coefficient which is equal to the square of the ratio of the difference of Z to the sum of Z. Since there is a square here, it doesn't matter which side has a larger Z, only the difference between the impedances matters. That is only the interface is of concern. It follows that alpha T ( transmission coefficient) is equal to 1 – alpha R. Air between the transducer and the patient will cause a total reflection as almost 100% of the ultrasound is reflected. Can you work this out? Put in the values of Zs for the air and soft tissue and work this out. That's why during ultrasound scanning, we apply coupling gel to eliminate air, to provide acoustic matching and also reduce some kind of friction. If you do not have coupling gel, what would you use for coupling? Water or paraffin oil will do the same job except water will dry up quickly..

[Audio] We finish the most boring part – the basic physics of ultrasound. Now we come to the interactions of ultrasound with tissue which is essential knowledge to the understanding of ultrasound image formation..

[Audio] Ultrasound is reflected, refracted and attenuated as it propagates through a medium. Very simple, isn't it? The same thing happens for the propagating and returning echo waves. These interactions might affect the direction and amplitude of both waves. And the degree of various types of interactions in the body depends on the acoustic properties of the tissues which are Acoustic Impedance, density of compressibility, shape of the interface, ' roughness' of the interface and size of organelles, cells & etc. For the last three properties, we'll come to them later in the lecture..

[Audio] Specular reflection and transmission. Specular means 'mirror-like'. It is similar to what we learnt at high school in light reflection. We have here two media with different impedances and speeds. The green arrows represents an ultrasound beam incident on the interface at an angle theta i, the ultrasound is partly reflected with an angle theta r which is equal to theta i. Part of the incident ultrasound will be refracted with an angle theta t which have a relation with theta i in the following way. Sine theta t over sine theta i is equal to c2 over c1. The specular reflection occurs when the spatial extent of the interface is much larger the wavelength of the ultrasound and the cross-section of the incident beam. Can any one recall the magnitude of the wavelength of diagnostic ultrasound. It is in the order of mm. Specular reflection is also frequency INDEPENDENT..

[Audio] Here is another type of reflection called non-specular reflection, that is non-mirror like reflection. We can call this scattering. When ultrasound encounters structures with dimensions small compared with the wavelength of the ultrasound, scattering will occur. Just like the following on the left, the spatial extent of a very rough surface is considered as small compared to the wavelength, the ultrasound is scattered. On the right, it is the organelles in an organ, they scatter the ultrasound for the same reason. Note that the scattering is equally well in all directions and it has strong frequency dependent, from f to the power 2 to f to the power 6, depending on the size of the scatterers..

[Audio] How do we describe a pattern like this below? We say it is a speckle pattern with white and black dots being dispersed in a random manner. Do you remember the ultrasound image of the liver I showed to you at the beginning, the tissue texture is similar to a speckle pattern. Now let's see how this speckle pattern is formed in ultrasound imaging. Our human tissue is made of intracellular organelles and cells; and these all act as scatterers. Whilst scattering occurs from all these scatterers at the same time, the scattered ultrasound itself will undergo multiple backscattering by many other scatterers, these backscattered waves might interfere constructively or destructively among themselves causing maxima or minima in the echo signal picked by the transducer. The maxima corresponds to the 'white dots' and minima to the 'black dots'. Please note that the placement of the echo is always along on the beam with a distance based on the ' 13 microsecond for one cm' rule. So the echo position is never correct because the signal is from multiple reflections occurring not exactly on the beam axis. That said, if the organ being examined is normal without any mass lesion, the ultrasound image will show a fine and uniform texture so as to speak..

[Audio] We called the speckle pattern we see on an ultrasound image ' Speckle Noise' instead. And according to definition of artefact, speckle pattern ( noise) is considered as artefacts because there is no one-to-one correspondence between the image and the structure being examined. That is to say, if you have a structure in an object, then you do an ultrasound image, you see the structure being depicted and shown on the image, this is a faithful representation of the structure in the image. On the other hand, if you see something on an image but in fact there is no such thing in the object, this is called an artefact. This is exactly the case similar to speckle noise, no one-to-one correspondence between the white dots and the structures they represent. I think this is the most difficult part to understand in ultrasound imaging, hope you can got hold of this concept..

[Audio] About the absorption in ultrasound, we can simply consider absorption as a process of converting acoustic energy into thermal energy. The simplified idea is that it is due to internal friction, try to remember the lattice model of the medium I introduced to you at the beginning, image a wave is passing through the lattice, there will be internal friction amongst the lattice elements during the propagation. Absorption is related to frequency, viscosity and relaxation time of the medium which you can image something to do with the springs in the lattice. Remember there is an approximate linear relation here, the higher the frequency, the higher the absorption..

[Audio] Attenuation refers to removal of acoustic energy from the beam. Broadly divided into absorption and scattering, scattering means the acoustic energy is re-radiate off the beam axis, so it is also attenuated. Scattering accounts for only 2 to 10% of the total attenuation in soft tissue, so most attenuation is absorption, therefore people tend to use the terms absorption and attenuation interchangeably..

[Audio] Attenuation coefficient is a metric to measure the attenuating power of the medium. I null is the initial intensity, x is the thickness and alpha is the attenuation coefficient, I is the transmitted intensity. I is equal to I null to the power exponential minus alpha x where alpha has two components, one for absorption and one for scattering..

[Audio] Let's do an exercise, on the LHS curve, the intensity is plotted against the thickness x, and you can seen the intensity is decreased in an exponential way, can you figure out which curve corresponds to a medium with a high attenuating power? RHS curve. In an ultrasound pulse, there is a spectrum of frequencies with different intensities. The distribution is like the full line curve. If this pulse passes through a medium of certain thickness, what would be the transmitted pulse like? It would be like the dotted curve. The overall intensity is of course reduced with those intensities at a higher frequencies much more reduced, it is because absorption is higher at higher frequency making the dotted curve shrinks towards to the left..

[Audio] How do we quantify attenuation loss? Loss is equal to mu times frequency times x where mu is the intensity attenuation coefficient in dB/cm/MHz, f is frequency in MHz and x is the distance traversed in cm. Loss is denoted in dB. This is a crude approximation but good enough in clinical use. Typical soft tissue is 1 dB/cm/MHz. By looking at the formula, you'll notice that loss is higher at higher frequency, that is, penetration reduced at a higher frequency. And also a reflector at a greater depth will generate a weaker returning echo because it also suffers from the loss when it travels back. You would probably hear in an ultrasound department, the ultrasonographer might ask to change from a 5 MHz probe to a 3.5 MHz probe when scanning an obese patient..

[Audio] Now you almost grasp all the basic physics in ultrasound imaging, let's move on to more practical knowledge in ultrasound imaging. In ultrasound imaging, there are only two things that make up an image. Specular reflection that allows you to see the interface so that you visualize the boundary between the organs. Scattering allows you to see the ' dots' so that you can visualize the tissue parenchyma in an organ. This is that simple..

[Audio] Finally we come to the principle of Ultrasound Imaging. To determine the echo intensity, that is echo strength To determine the position of the interface, that is echo ranging Distance is equal to speed multiplied by time, if you know the time and speed, you can calculate the distance. Suppose this is the ultrasound probe that gives out an ultrasound pulse at time t equals to zero, the pulse travels a distance d, and hits an interface, part of the ultrasound is reflected, part is transmitted, the reflected ultrasound travels back a distance of d and detected by the probe at time t prime. So 2d equals to speed times the time t prime and you can easily work out d and the ultrasound will place interface at a distance d from the probe surface in the image. There is an assumption, the speed is constant and is 1540 m/s. We're actually using the golden rule of 13 microsecond one cm'. If we hear the echo in 26 microsecond's time, how far is the interface? It is 2 cm..

Basic Ultrasound Instrumentation.

[Audio] In ultrasound, a very important property you need to know is piezoelectric effect. In a very simple sense, piezoelectric property allows certain kinds of materials to absorb mechanical energy and transform it into electric energy and vice versa. Some commonly known piezoelectric materials are quartz, PZT ceramics and PVDF plastic. Quartz is naturally occurring whilst the PZT ceramic and PVDF plastic are artificial. These materials are usually crystalline materials that have dipoles. If you look at the diagrams here, the dipoles are aligned in a regular way, if you apply an electric field in this way, the applied polarities will 'press' on the crystal and in the next instance, you apply an opposite polarity, the applied polarities will 'elongate' the crystal. If we keep changing the polarity in a rapid manner, the dimension of the crystal will be changing accordingly generating a mechanical wave out – that is the ultrasound wave. On the other hand, if a mechanical wave hits on the crystal, an electrical voltage will be detected. Therefore, the piezoelectric crystal can be used for generating ultrasound and at the same time it can behave like a detector for ultrasound. PZT and PVDF are used in ultrasound probes..

[Audio] Ultrasound is all about pulse and echo things, you send out a pulse and the you wait for an echo, if any, to return. What is a transducer? A transducer is a device that converts energy from one form to another. An ultrasound probe is a transducer as it changes electrical energy and mechanical energy and vice versa at different times. In ultrasound, we can further say that the ultrasound transducer is a transceiver – a transmitter and a receiver together. If you look at the diagram here, this is the ultrasound transducer or probe, this line represents the patient's skin, there is a switch here. At the beginning, the switch will turn to the excitation mode and the transducer sends out a short burst of ultrasound, that is the green thick arrow here. The transducer acts as a transmitter. After that, the switch will quickly turn to receiving mode, and listens for any echo returns to the transducer. The transducer now acts as a receiver. The so called receiving mode is actually a period of ' silence' for listening echoes. In a typical ultrasound unit, 99.9% of the time, the transducer is in the receiving mode or in the period of 'silence'..

[Audio] This is a typical single element ultrasound probe. Let's look at the function of each component in the next slices..

[Audio] The metal case is for structural support and electrical shielding. The acoustic insulator avoids transmission of ultrasound into the metal case. The crystal is made of piezoelectric material. The thickness of which must be an odd multiples of a half wavelength for complete construction interference. The usual thickness is half of the wavelength, so it is extremely thin and fragile. Therefore, you should handle the probe carefully and not to drop it..

[Audio] For ultrasound imaging, we are using a short burst of ultrasound with a short pulse length, so once after we excite the crystal, we want the crystal to die down and not in a ringing mode, therefore, we need to add backing material to damp down the crystal after excitation. But at the same time, if the transducer acts as an receiver, the damping will reduce the sensitivity. There is a balance between the two. Note that the rear surface is slanted to prevent reflection of sound energy back to the crystal. But the angle should not be at 45 degree. Can you figure out why? We need a matching layer in front of the crystal to reduce the acoustic impedance mismatch between the crystal and the tissue. It also have a requirement on the thickness of quarter wave-length to enhance the transmission of ultrasound into the body..

[Audio] About the pulse operation I mentioned before. This is an ultrasound pulse plotted as amplitude against the time. If we put this pulse through a mathematical operation called Fourier Transform, you'll see how the frequencies are distributed within a pulse together with the amplitude associated with each frequency. This bell shape is typical of an ultrasound pulse. We have here a nominal frequency f naught, and at the full width half maximum, we have f1 and f2; and the range of f2 minus f1 is called the bandwidth. The shorter the pulse, the wider the bandwidth; so in ultrasound, we want a short pulse, we like to have a pulse with wide bandwidth..

[Audio] You might have heard the Q-factor of a transducer. It is defined as the energy stored per cycle divided by the energy lost per cycle. It is equal to f naught divided by the bandwidth. In diagnostic ultrasound, we generate a short pulse with wide bandwidth and hope most of the energy is lost during the first few vibration to generate a short pulse, therefore the Q-factor would be small for a diagnostic ultrasound probe. The Q-factor is usually 2 to 3..

[Audio] Remember that the crystal is excited to give a pulse and then turns into a listening mode. After that, the crystal is pulsed again and listens again, the cycle keeps on repeating to gain enough signals to create an ultrasound image. In this connection, we have defined a term called Pulse Repetition Frequency which is the number of times the crystal is pulsed in one second. Since the transducer acts as a transceiver, this frequency is limited by the maximum depth ( R) of the tissue to be sampled and the velocity of the ultrasound (c) in the medium. That is to say, if you want to detect an echo which is 10 cm deep, you need to wait for all the echoes that are generated shallower than 10 cm to return to the transducer before you give another excitation. At the time of excitation, any echoes deeper than 10 cm will not be heard. The total time taken for an pulse to reach a 10 cm deep interface and for the echo to return is 2 times R divided by c, let's call it t. So how many times I can pulse the crystal in one second is simply an reciprocal of t. Therefore, PRF max equals to c over 2R. It follows that the Pulse Repetition Period is simply a reciprocal of Pulse Repetition Frequency. The PRP is in fact the t mentioned above, the time required to transmit a pulse plus the time to listen for the echo..

[Audio] Now we come to a very important property of ultrasound – Spatial Pulse Length ( SPL) SPL is the length of the short-duration pulse and is equal to n lambda where n is the number of cycles in a pulse. So if we need a pulse of short SPL, we can reduce n and increase the frequency as well. Since c equals to f lambda, c is fixed, a high f will give a shorter lambda. A short SPL will increase the axial resolution. I'll explain this in the next slice..

[Audio] Axial resolution is a measure of the ultrasound machine's ability to separate the echoes from reflectors which are closely spaced along the beam axis. Or you can say it is the distance between the reflectors which are just resolvable. The order of axial resolution in ultrasound is smaller than 1 mm. The axial resolution is half of the SPL. That is to say any reflectors which are place closer than SPL/ 2 will not been seen as two entities, they are merged as a single echo. Any reflectors placed apart more than SPL/2 will be clearly distinguished as two separate entities in the image. Let me try to explain this to you by using the schematic diagrams below. The yellow pulse has a certain SPL. And we have two interfaces at a separation of SPL/2, one yellow and one green. Suppose the head of the pulse hits the yellow interface, part of the pulse will be transmitted like the green arrow and part will be reflected and travelling back, by the time the head of the yellow pulse has reversed and travel a length of SPL, the green arrow would have already travelled half of the SPL and bounced back and travelled another half of the SPL, so the head of the returning pulse will be just touching the tail of the yellow pulse. This is the closest of the green pulse that comes to the yellow pulse and still can be distinguished as two echoes. Image that if the two reflectors are placed closer than half of the SPL, the two echoes will overlap and the two interfaces are not resolvable. If the two reflectors are placed further apart than half of the SPL, the two echoes are definitely resolvable..

[Audio] The lateral resolution is the ability to resolve 2 objects that are placed adjacent to each other perpendicular to the ultrasound beam axis. It is determined by the beam width at the measurement depth. Sometimes, an intuitive way is used to describe this lateral resolution is the elongation of the point in an object being recorded as an image. That is to say if two points exist in the object, their separation must be cleared from their elongated distance in order to be seen as two entities, otherwise they will be merged as one entity. Let me explained why there is an elongation using the schematic diagrams, first of all, we need to appreciate an ultrasound beam is divergent in nature, it is wider as it travels further down. Suppose there is a small dot we want to image in the tissue, the blue circle, below is the screen showing the ultrasound image. When the probe is moved along the skin to do ultrasound imaging, you will find the dot becomes a sausage shape object instead of a dot anymore, the reason for that is the beam has a finite width and any echo detected within the width of the beam are displayed as it were on the beam axis. In the second diagram from the left, the beam edge barely hits the dot and the reflection back to the detector causes a registration of the echo placed at the beam axis location, which is in fact not the case. Similarly, on passing along the dot, echoes will be generated as far as the dot is within the beam width, and the placement of the echoes are always on the beam axis as the ultrasound machine has no idea a beam width actually exists. Note that the beam axis is actually displayed in the screen and moves along with the probe movement, the echoes placed at the central axis are spread out as a sausage shape. Therefore a dot is elongated into a sausage shape in this way. And you see immediately, the wider the beam, the longer the elongation, the worse the lateral resolution. Also the deeper the dot, the worse the lateral resolution as the beam is wider at a deeper depth..

[Audio] Spatial is related to space in context, therefore spatial resolution includes axial and lateral resolution as well. In a ultrasound phantom, you often see an array of pins spaced in ever-decreasing space, it is used to measure the axial and lateral resolution, in the axial direction, the two pins that are resolvable is reported as the axial resolution. Therefore, this one is 0.25 mm and this one is 0.5 mm. For the lateral resolution you just measure the length of the pin recorded as the lateral resolution. So for the bottom right images, which one has a better lateral resolution? Of course, it is the right side one as the measured length is shorter. Note that the pins are never aligned in a straight line in order to avoid the attenuation from the preceding pin above..

[Audio] This is typical tissue equivalent ultrasound phantom. This is an image of the image where you can measure a lot of the ultrasound image qualities. Note that the three clusters of pins at different depths on the right are the arrays of pins I mentioned in the previous slice..

[Audio] Scanning of eyeball on the left, say to study ocular structures such as the retina, since the structure sizes are small, you need to use a high frequency to give the resolution you need. Remember that the higher the frequency, the better the axial resolution. Therefore 15 to 20 MHz is used. How about the absorption at such a higher frequency? We're not too worried about the high absorption at this frequency range as the eyeball is small and we still have enough intensity to do the scanning. How about the scanning of the pelvis, can we use 10 to 20 MHz ultrasound to give us an excellent axial resolution? First of all, it is not necessary to have such a high resolution as we're just studying the pelvic organs and this frequency range cannot even penetrate the patient due to high absorption. So the frequency used in pelvic scanning is about 5 MHz, the resolution is good enough for pelvic organs. For thick and obese patients, you might even want to change to 3.5 MHz to penetrate down the thick tissue..

[Audio] Let's look at an ultrasound beam profile in a schematic way. Suppose this is the crystal of a diameter 2a. There exists a Near Field which is quite confined so as to speak, the length of this Near Field is D. D is equal to a square over lambda the wavelength of the ultrasound. Beyond D, the beam starts to diverge and we define an angle theta here to indicate the degree of divergence, and sin theta is equal to 0.6 lambda divide by a. The field from here is called the Far Field. From here, you will notice that the near and far field depend on the wavelength of the ultrasound and the crystal size. Sometimes, the Near Field is called Frensel zone and the Far Field is called the Fraunhofer zone. We can also conclude here the higher the frequency, the longer the D and the beam tends to remain confined without divergence; in this way, we can say the lateral resolution is better..

[Audio] This follows what we talked about from the previous slice. On the left hand side, the size of the crystal will affect the shape of the beam profile. A larger crystal, or if a is larger, it will give a longer near field and a less divergent far field. And on the right hand side, a higher frequency probe with a smaller lambda will give a longer near field and a less divergent far field..

[Audio] In diagnostic ultrasound, we are using Low Q transducer, the axial intensity profile will be like this in the graph below. Note that the intensity is not uniform along the beam axis. It rises to a maximum at the junction and decreases as we go further away..

[Audio] As we know the lateral resolution depends on the beam width, we might want to have a narrow width of beam despite we know there is divergence in a beam. We can create a focal zone with a narrow width to improve the lateral resolution. We can accomplish this by using an acoustic lens, curved crystal or acoustic mirror. The width can be worked out from this formula, it is equal to R lambda over a where R is the radius of curvature. We define the focal zone as the region where the intensity is plus and minus 1.5 dB of the maximum intensity along the axis. And of course, the large the focal zone, the better the latera resolution. However, focusing on the one hand narrows the width, but on the other hand it limits the near-field depth because the beam diverges rapidly beyond the focal zone..

[Audio] About the degree of focussing, you can tell right away the magnitude of focussing by looking at the curvature of the crystal..