Transcript for:
Ultrasound Attenuation Principles

hi learners it's em from sano nerds this video is on unit six a attenuation i decided to split unit six up into two units unit six a and six b they're both gonna have the overarching theme of what happens to sound as it travels through a medium 6a is going to focus more on attenuation or sound that weakens as it travels in the body understanding attenuation is going to help you as a sonographer to understand some of the physical limitations of creating a diagnostic image 6b is going to focus on the creation of echoes the whole reason diagnostic medical sonography exists is because sound in the medium which is our soft tissue body will interact in a way that returns echoes to the machine for processing and display understanding how echoes are created are also going to help us to better understand some of the physical limitations on creating a diagnostic image remember we are not here learning ultrasound physics to build a machine or repair one or make the next super fantastic super duper transducer we're learning the physics of ultrasound because it's going to impact the decisions we make every day as a sonographer every image we take every exam we start from transducer selection to image optimization physics is always going to be there and eventually i promise it will become second nature to you section 68.1 strength parameters again when we were discussing the seven parameters of a sound wave there was a subset of parameters that described the strength or bigness of the wave those included amplitude power and intensity we used amplitude to describe the maximum variation of an acoustic variable remember those were density pressure and movement when amplitude is high the wave is really strong it causes larger fluctuations in the variable from its normal resting place in all applications of physics power is described as the rate at which work can be performed or the rate that energy is transmitted in the ultrasound setting the machine uses a voltage to initiate the sound pulse this voltage is the driving force behind the power of the wave and will translate into sound power or energy intensity then describes how the power is distributed over an area if we have a set power used over a large area the intensity is weak however when that same power is used in a small area the intensity will increase all three of these parameters are determined by the machine and can be adjusted by the sonographer if they change the output power but the important part for this lecture is to remember all three of these parameters weaken with propagation section 6a point 2 attenuation attenuation causes a decrease in the amplitude and intensity of a sound beam as the beam travels through a medium all sound will attenuate the rate that which it attenuates is going to depend on the initial intensity followed by the frequency of the wave and the medium that it travels through let's discuss first how the loudness of a sound wave can relate to attenuation imagine you are talking to a friend and remember that your friend can hear you because the sound you are creating with your voice travels through the air which is a medium and into their ears now when we whisper the amplitude of our voice is very small and unless your friend is very close the sound would attenuate before they could hear you the amplitude of our whispers are small they will attenuate very quickly hopefully your friend is nice and close so they can hear the secrets that you have to tell now compare that to shouting to your friend when you shout to your friend there's going to be a much higher amplitude there's much more loudness bigness to your voice to the sound you are creating but it's going to take a little bit longer for it to do so because it started the higher amplitude your friend could be further away from you the sound will still attenuate but they will still be able to hear you the sound created by an ultrasound machine will attenuate in the body as well the amount of attenuation is going to depend on a few factors like the power amplitude intensity of the initial beam but it also has to do again with the frequency of the beam and the medium that it's traveling through attenuation on the sound beam is highly applicable to the stenographer and the decisions that they make about every exam and every image as we work through this course we're going to start to see how the knowledge of physics is going to help us to create better images two of the big things that you'll need to know as a sonographer is one attenuation is going to limit the depth to which an image can be created by a sound beam there's only so much energy within the sound beam once it's attenuated or weakens to a certain point it is no longer able to return meaningful echoes for example in this image here on the top picture we have a 12 megahertz transducer it is only able to image to about 8 centimeters into the body compare that to the two megahertz transducer underneath which is imaging very clearly down to 17 centimeters these images will make an appearance later as we talk a little bit more about how high frequencies attenuate quicker but it's a great example to show that as a sonographer you're going to need to choose the right equipment to perform your exams another thing we need to know about attenuation is that it is going to be compensated for by our machines we are going to learn in this unit that when sound leaves the transducer it very quickly weakens in a lot of cases after only a couple centimeters of propagation it's already half the strength of when it started and it still has to go to multiple depths and then return back to the transducer so the echoes that the transducer is getting are very very weak signals and because of that the machine is going to need to manipulate the signals to compensate for the attenuation so they can be processed and displayed another thing that we need to know is that we do have tools on our machine as a sonographer we can choose how we want to compensate for some attenuation artifacts for example in this top picture we have a picture of a liver and in this liver is a cyst and the gallbladder so this is gallbladder and this is a cyst these are both weak attenuating structures meaning that they are not going to take up a whole lot of sound energy as the sound passes through them what we see behind them are very strong reflectors because there's still a lot of sound energy imaging this portion of the anatomy compare that to where there was regular parenchyma traveling through and attenuating we see that we have very dark weak echoes returning from our far field contrary to that then we have items that can absorb a lot of sound energy and leave none for behind it so when sound is coming into these structures here these are gallstones they're absorbing all that sound energy and there's nothing left to image behind this is where we get shadowing from so these two images show very nicely how certain structures certain media within the body can really affect how the sound propagates through the tissue so as a sonographer you will be able to compensate for these or maybe you decide to leave them because they add to the diagnosis these four pictures show very nicely some of the key concepts behind attenuation choosing the right transducer and the media in which the sound will travel through as a stenographer you will need to choose the correct tools and you will need to know when to compensate and how to compensate for attenuated sound section 6a 0.3 decibels when we refer to attenuation and as we'll learn in a subsequent unit amplification we are referring to the change in amplitude power or intensity this relative change is typically measured in decibels decibels are used to define a change in two values by using decibels there is an understanding that there is a comparison of an initial value and a final value and decibels are logarithmic as i mentioned previously we were talking about a change in amplitude a change in intensity change in power we would take the initial intensity compared to the final intensity and be able to describe that change in intensity in decibels but decibels do not follow a linear scale they follow what we call a logarithmic scale a logarithmic scale is a non-linear scale that is typically used to show a range of large quantities instead of increasing by equal increments the scale increases by a factor of the base of the logarithm decibels is going to use a base of 10. so for example i have a linear scale imaged here we can see that each tick mark on this number line increases by the same value the value of 1. we could replace these numbers with 5 10 15 20. it is still a linear scale because each space would then represent a change of five in a linear scale every increase every increment is equal in the logarithmic scale though the values are going to increase by a factor of 10 which means to multiply by 10. so if we start at our first tick mark and increase one increment we have increased by a value of 10. moving to the second tick mark means that we have increased the value by a hundred moving to the third is a thousand so you can see very quickly that we are generating very large numbers with numbers on this fourth tick mark not simply being four greater than one rather one thousand times bigger decibels use logarithmic scales so if we have a value of a wave and the intensity increases by 10 decibels that new intensity is now 10 times greater than the initial let's say we add another 10 decibels now the new intensity isn't 20 units bigger but rather has a value of 100 times bigger add another 10 decibels to the initial value the new signal isn't 30 times bigger but a thousand times bigger so for every 10 decibels that we add we jump a factor of 10 in bigness to further show this concept let's take a look at this chart now our threshold of hearing is zero decibels so the quietest thing that we could possibly hear is considered zero decibels leaves wrestling outside in the wind comes in at 20 decibels that 20 decibels doesn't mean that leaves wrestling is 20 times higher or 20 more higher than our zero decibel threshold but rather it means that it is 100 times louder than the quietest thing that we can hear and that again is because decibels are a comparison between two different values and it's expressed in logarithmic scale meaning that we are looking at factors of 10 expressed in these values when we compare rainfall to the threshold of our hearing it is not a direct 40 value more than our threshold of hearing but rather at 40 decibels rainfall is actually 10 000 times louder than the very quietest thing that we could hear compare that then to a rock concert at 120 decibels that means that the rock concert is actually one trillion times louder than the quietest thing that we can hear so we can see that decibels compare numbers increase by factors of 10 and by using logarithmic scales we can express really really large numbers there's no linear scale that would be able to express 1 trillion and 100 in a relative fashion like we see here on this logarithmic scale now that we know that decibels use logarithmic scales and report a relative change let's go ahead and look at some examples of how we can report that change in power and intensity in a wave let's first take a look at positive decibels when decibels increase the power and intensity of the wave also increase normal restaurant sound is around 60 decibels where the rock concert was about 120 decibels the rock concert sound is louder it's more powerful and more intense and so when we see positive decibels we can say that the sound is amplified there are two rules when increasing or using positive decibels that you need to know for ultrasound physics the first one states that when decibels increase by three decibels the intensity increases by a factor of two that means that the intensity is twice as strong the second rule is that when decibels increase by 10 decibels the intensity is going to increase by a factor of 10 so it is 10 times as strong in this first example we are starting with a decimal value at 56 decibels we then increase it to 59 decibels now remember increasing from 56 to 59 doesn't simply mean that we've increased the intensity by 3 or increased it even by a factor of 3. an increase of 3 decibels means that the intensity is doubled or increased by a factor of two so when we increase by three decibels we are increasing the intensity by a factor of two so if you are sitting at your ultrasound machine and you have your gain set to 56 and you turn the gain knob to make everything brighter you'll notice that your gain decibels will increase and let's say you increase them to 59 decibels and by increasing the gain you have increased the intensity at which those signals are processed at so by changing your gain from 56 decibels to 59 decibels you're telling the machine to make the signals twice as strong as they were so anytime we describe a waves change as positive three decibels we are doubling the intensity doubling the power of the wave so in this example we started with a 56 decibel wave we increased the decibel to 59 which means we doubled the intensity of the wave to go over the second rule then let's again say that we start at 56 decibels but this time we've increased to 66 decibels that means that there was a positive change of 10 decibels now the way that 66 decibels is going to be 10 times more intense or 10 times stronger than the wave at 56 decibels and that is because of our rule when we add 10 decibels the intensity is going to increase by a factor of 10. so again we start with our 56 decibel wave we've added 10 decibels to describe the change and by adding those 10 decibels what we've actually done is increase the intensity by 10 times now using the rules that whenever we add 3 decibels we are increasing by a factor of 2 and whenever we add 10 decibels we are increasing by a factor of 10 we can actually kind of combine those two things together to understand different decibel changes so in this example we started at 20 decibels and we went to 26 decibels that's a positive change of 6 decibels so in this example we started at 20 we added three decibels to get to 23 and then we added another three to get to 26. each of those three decibel increases mean that the wave has gotten stronger by a factor of two so 20 to 23 doubled the strength of the wave and 23 to 26 doubled the strength of the wave again so in this example when we added six decibels we have made the new wave four times stronger because we are taking two multiplied by two to equal four let's look at another example now we've increased the decibels to 29 so again we start at 20 the first three decibel increase means that wave doubled once the second three decibel increase means a second doubling and the third three decibel increase means a third doubling so 2 times 2 times 2 equals 8 which means that by increasing from 20 decibels to 29 decibels we have made the new intensity eight times stronger so remember it's not adding these together it's not just nine times stronger it's not six times stronger by adding all of these you have to multiply each one because each increase of three decibels means a doubling so two times two times two equals eight this wave is now eight times stronger than it originally was we're going to see the same thing using the 10 decibel rule for every 10 decibels that we increase we are going to increase by a factor of 10. so in this example we started at 30 and we've increased to a value of 60. so 30 decibels to 60 decibels so that's going to be 10 plus 10 plus 10 gives us our 30 decibel change for each of those 10 decibels that we increase we are multiplying by 10. so the first 10 decibel increase is a 10 times stronger wave the second 10 decibel increase makes it a 100 times stronger wave and the third 10 decibel increase means that we have increased the intensity by a thousand times because ten times ten times ten equals one thousand in this example then the thirty decibel change resulted in a wave that was 1 000 times stronger than it originally started we can also use our tens of threes to help us figure out the intensity changes so in this example we started at 42 and we increased the decibels to 65. 42 to 65 is a 23 decimal change so for the first 10 decibels going from 42 to 52 we've increased 10 decibels that's a 10 times increase we have another jump of 10 52 to 62 causing a second 10 times increase and then we have 62 to 65 which is only a three decibel increase which means it's two times to calculate the sun we would take the ten times multiplied by ten times multiplied by two times so ten times ten times 2 equals 200. by adding 23 decibels to the original wave we have made it 200 times stronger let's take a look then at negative decibels attenuation itself is going to imply a decrease in decibels or a subtracting or a negative move and decimal value when decibels decrease the power and intensity of the wave are also going to decrease so if we're talking in a normal voice at about 60 decibels but then we whisper which is right about 30 decibels the whisper is quieter meaning it has less power and less intensity than the normal talking voice the 60 decibel of my normal talking voice compared to the 30 decibel of my whispering voice is a change of negative 30 decibels which means that my whispering voice is 1000 times less than tens than my speaking voice and here is why there are again two rules when decreasing decibels that you're going to need to know for ultrasound physics the first one is that when you decrease by three decibels the intensity is going to decrease and it's going to decrease by either a factor of two or you can say it's half as strong as the original wave or you can say that it is two times weaker than the original wave i want to point out that all three of these mean the same thing it means that the new wave is weaker and it is by a factor of two now it might be helpful to know that dividing by 2 is the same thing as multiplying by half same thing dividing by 10 is the same thing as multiplying by 1 10. just remember anytime you multiply a number that has a value less than 1 it's going to make that number smaller that's why we see a decrease in the intensity when we use negative decibels so you can probably guess what our second rule is when decibels decrease by negative 10 decibels we're going to see the intensity decrease by a factor of 10 or we can say that it is 1 10 as strong as the original wave or we can phrase it that it is 10 times weaker than the original wave again all three of these phrases mean the same thing that the intensity has decreased by that relative change being at either by a factor of 10 or being 1 10 as strong again let's take a look at a couple examples in this example we are starting at 78 decibels and we have decreased it by three decibels making the new wave a 75 decibel wave so that's a change of negative three decibels that means that by decreasing from 78 to 75 causing that negative 3 decibel change this new wave is half as strong as the original wave or we can say that the intensity has decreased by a factor of 2 because remember decreasing by a factor of 2 means to divide we'll see the same thing happened then with the negative 10 change in this example we start at 89 decibels and we decrease the decibels to 79 so that is the change of negative 10 decibels the 79 decibel wave is going to be 1 10 as strong as the 89 decibel wave the intensity of the 79 decibel wave is going to be 10 times weaker or 1 10 as strong or as decrease by a factor of 10 when compared to that 89 decibel wave and just like positive decibels we can combine negative decibels to figure out different variations of the 3 and the 10 decibel changes so for example in this one we have started at an 18 decibel and we have decreased all the way to 6 decibels this is a total change of 12 decibels remember for every 3 decibels the wave gets to be half as strong so the first three decibels causes the intensity to be half as strong the next three decibels cause another half decrease the next three decibels again decrease it by a factor of two and then the last three decibels to get to that 12 decimal change will finally also decrease the strength by half so that whole negative 12 decimal change dropping from 18 to 6 decibels causes the intensity of the new successible wave to be 1 16 as strong as the original wave and we get to 1 16 strong by taking 1 divided by 2 we get 0.5 divide that by 2 we get 0.25 decreased by another factor of 2 we get 0.125 decreased by the last factor of 2 and we get 0.625 this is the same thing as one sixteenth we can see it in the other way if we are decreasing by half for each three decibels we do half times half times half times half and we will end up with one sixteenth as well let's look at an example using our tens this wave started at a 75 decibel and decreased to a 55 decibel that was a negative 10 jump creating a wave that is 10 times weaker and then another negative 10 jump causing the wave to become yet another 10 times weaker so we can say that the intensity has decreased by a factor of 10 once and then a factor of 10 again or we can say it's one-tenth as strong and then another one-tenth is strong so we can take one and divide it by 10 that gives us 0.1 divide by 10 again giving us 0.01 which is the same as 1 100 we can also see if we decide to use the fraction version of this 1 10 multiplied by 1 10 will also give us 1 100 because we just multiply across the bottom 10 times 10 is 100. so by decreasing the decibels by 20 we have made the intensity of the new wave 100 times weaker than the original and then we can also combine our tens and threes to calculate new values as well in this example again we're starting at 38 and we are decreasing all the way to 25 38 minus 25 indicates a negative 13 decibel change we need to figure out how we can fit a 10 in there how we can fit threes in there maybe it'll be more than one three more than 110. it really just kind of depends on how you can fit a 10 and a 3 into your decimal change in this case we can do 110 so one negative 10 decimal change means one tenth is strong or decreasing by a factor of 10 and then that last negative three decibels is going to cause the wave to become half as strong so we first see it ten times weaker and then we see a two times weaker we can look at that as being one tenth as strong followed by half as strong so to calculate that then we could take one divided by 2 for decreasing by a factor of 2 take that and decrease it by a factor of 10 and we will see that our new wave is 0.05 or 1 20th as strong as the original if we want to take our fractions and multiply we can do half times 1 10 and that will also give us 1 20th so the new wave at 13 decibels negative change is 1 20th the strength of the original wave all of those examples took us knowing the initial decibel value and seeing what the decibels change to to understand the relative change of the intensity however we can also take the actual change of the intensity and figure out what the corresponding decimal change would be for example in example one we have an intensity change of 30 milliwatts per centimeter squared to an intensity of 60 milliwatts per centimeter squared we can look at those values and see that the intensity has doubled so we would need to know our rules about decibels and we would need to figure out what decimal change means a doubling in intensity and because we are on our physics game we know that an increase of three decibels would cause the intensity to double or increase by a factor of two so we can say that the wave has undergone a change of three decibels now if you read through the rest of the examples and you're comfortable with the math please feel free to skip the next section but we are going to head to the board to show the math through these four examples and then we'll come back to review decibels example one tells us that intensity one is 30 milliwatts per centimeter squared that is going to increase to an intensity of 60 milliwatts per centimeter squared now we need to calculate what the factor of change was we can take 60 divide it by 30 and that tells us that intensity one has increased by a factor of 2 to become 60 milliwatts per centimeter squared next we're going to go back and take a look at our rules because we are increasing we are working in positive decibels so you just want to write down our positive decibel rules when we increase by 3 decibels we are seeing an increase by a factor of 2 and when we increase by 10 decibel we see an increase by a factor of 10. since intensity increased by a factor of 2 and 3 decibels also means an increase by a factor of two we can say that this wave has increased by three decibels and that three decibel increase resulted in the intensity increasing by a factor of two let's take a look now at example two we start with a five milliwatt centimeters per squared intensity and that is going to increase into a 100 milliwatt centimeters per squared intensity first we need to calculate the factor by which intensity one increased so we take one hundred and divide it by five that tells us that the intensity one has increased by a factor of twenty again we are working in increases so we know we are working in positive decibels three decibels is going to be an increase in a factor of two and ten decibels equals an increase by a factor of ten just like we did with the previous example we now need to figure out what decibels are going to cause a 20 factor increase our first example worked out really well that three decibels just happened to match up with the factor increase but we aren't seeing that for this example so now we have to use some multiplication to get to 20. whenever you are considering the factor increases with the decibels remember that we are multiplying them we're not adding them we are multiplying them we can see that an increase in a factor of 2 and an increase in a factor of 10 will result in an increase in a factor of 20 because 2 times 10 is 20. so with that knowledge then we know that this wave has gone under a 3 decibel change and a 10 decibel change 3 decibels plus 10 decibels equals 13 decibels so the wave in this example has undergone an increase of 13 decibels to affect the intensity making it 20 times stronger example 3 starts with an intensity of 30 milliwatts per centimeter squared and is going to decrease to 15 milliwatts per centimeter squared now we are interested in by what factor intensity one has decreased by and we know that decreasing in intensity means a decrease in the decibels let's first figure out the factor by which intensity one decreased by again we will take intensity 2 and divide it by 30. 15 divided by 30 is going to be a decrease by a factor of 2 or half as strong next then we need to remind ourselves of our rules since we are dealing with a decrease we are going to be dealing with negative decibels negative 3 decibels results in a half as strong or a decrease by a factor of 2 and negative 10 decibels will equal 1 10 as strong or decrease by a factor of 10. like our first example this one works out very nicely we see a reduction by half and we can see that a reduction by half matches up with our negative three decibels therefore in this example we see that intensity one dropping to intensity two must have experienced a weakening by a factor of two so that is going to match up with a negative 3 decibel change example 4 tells us that we are starting with a 10 milliwatts per centimeter squared and we are reducing to a 0.01 milliwatts per centimeter squared intensity we then need to calculate the change in the intensities so we'll take intensity 2 and divide it by 10 and we see that we end up with a beam that is 1 1000 or one thousand times weaker than the intensity one so we are working with negative numbers so we are going to remind ourselves again of our decibel rules minus three decibels equals half as strong and negative ten decibels is going to equal one-tenth as strong now any time you are seeing a factor of 10 you should immediately think that you're probably going to be using your negative 10 decibel rule now we see that the negative 10 decibel rule doesn't automatically apply because it's only a tenth as strong not one 1 000 as strong so we need to figure out how many 10 decibel changes does it take to make the beam 1000 times weaker and i've got a little hint for you notice that when we drop by one tenth that is one negative ten decibel change if we drop by another ten decibels we have now decreased the wave by 100 times notice 1 0 1 10 decibel two zeros two 10 decibel drops so going with our pattern if we have made the wave 1000 times weaker we have three zeros we need three 10 decibel changes and that is because 1 10 multiplied by 1 10 multiplied by 1 10 equals 1 1000 by using this method we see that this wave must have decreased by 30 decibels to cause the wave to become 1000 times weaker or 1 1000th of its original intensity i have a few things that i want you to focus on first remember that decibels are going to compare two values and we saw that when we were taking the intensity changes and figuring out the decimal change those two intensity changes were expressed by one decibel value know that increasing decibels means an amplification of the intensity or of the sound increase in the power that is going to be a positive decibel change where decreasing decibels refers to attenuation which is why we're talking about all of these decibel things in this unit and when we have attenuation we're going to see a negative decibel change and lastly remember the four rules regarding the positive and negative decibels we can see that when we have a positive three decibel change we are going to double the intensity and when we have a 10 decibel increase the new intensity and power will be 10 times stronger opposite of that then when we have a negative three decibel change we will see the new intensity be half as strong and a negative 10 decibel change will result in a new intensity one tenth as strong as the original so mostly focus on the threes and the tens those are more likely what are going to be on your tests and on your boards but it's also a good idea to understand how a positive or negative successful change might change the intensity and power and how the positive or negative 20 decibel change will affect intensity and power like most of our practices you are going to be filling in a chart using the new knowledge that you just gained the chart contains columns for an original value a new value and a spot to describe how the wave has changed so you will be filling in those new values and then describing how the signal has changed based on the other values that are already filled in go ahead and pause this video once you have finished your answers unpause and let's go check out the answers here are the answers to your decibel practice feel free to pause if you need to go through and double check your numbers remember you have filled everything in that is highlighted in light blue now if you feel comfortable with these values understand how to add and subtract decibels and then describe those decimal changes based on how the intensity is affected feel free to jump forward to section 68.4 otherwise if you want to see some of these practice problems worked out we are going to head over to the board to look at a few of them we're going to start with the first half the chart these numbers are working with changing values in a three decibel factor so all of our original signals have started with 100 decibels we have one part that is going to require us to calculate the new signal and then we have another part that's going to require us to describe the new signal for the first three rows notice again that we are starting all with 100 decibel original signals and in our changes we're seeing a two times stronger a four times stronger and an eight times stronger the fact that the wave is getting stronger tells us that we are working in the positive decibels remembering our rules then we know that an increase in three decibels equates to a two times increase or increasing by a factor of two making it twice as big that actually makes our first one very easy we know that we just need to add three decibels to make this 100 decibel wave twice as strong so 100 plus three is 103 decibels as the new signal jumping down to the four times stronger remember that the change in strength is telling us the factor by which the intensity is going to change by so when we say something is four times stronger it has increased by a factor of four or multiplied by four and because we're working in multiplications we should always be multiplying then our decibel factors so we are not looking for two factors that add to 4x we are looking to find factors that are going to multiply to 4x using our three decibels and knowing that that creates a factor of two we can easily take two times two to figure out that we will see an increase in four times when we take two multiplied by two each of these doublings means that there has been a three decibel increase so one doubling three decibel increase the second doubling another three decibel increase we then will add our decimals together to show that a positive six decibel change will result in a wave that is four times stronger so we will add six decibels to the 100 to get the new signal lastly then we are looking at an eight times stronger signal again we are going to use our three decibels knowing that each three decibel results in a doubling you may end up using some trial and error with this but if you understand powers and understand how to multiply things you'll see very quickly that 2 multiplied by 2 multiplied by 2 again does indeed equal 8. 2 times 2 is 4 four times two is eight and just like we did with the four times stronger each of these doublings means a change in three decibels so accounting for each of those increases we take three decibels plus three decibels plus three decibels and we see that to create a wave that is eight times stronger we will need to increase it by nine decibels add that nine decibels to our original signal and we can calculate 109 as our new signal looking at the bottom three rows we are seeing that we are now decreasing our original signal we are decreasing from 100 to 91 194 and 100 to 97 these decreases in signals means that we are working with negative decibels so remember that negative three decibels results in a wave that is half as strong so let's figure out what our decimal changes were for our original to new signals 100 minus 91 is a negative 9 decibel change 100 minus 94 is a negative 6 decimal change and 100 minus 97 is a negative 3 decimal change we can very quickly see that we are using negative three decibels negative three decibels we will see a wave that is half as strong and that is what we've written in this column going up to negative six decibels we need to figure out how negative three decibels can fit into that we know that negative three decibels plus another negative three decibels would equal that negative six and for each negative three decibel change we are going to see a reduction in half of the strength of the wave so we have a half for that three decibels we have a half for that three decibels and now we need to multiply these because these are factors by which these waves will change one-half times one-half equals one-fourth so the new wave will be one-fourth as strong lastly then we have a change in negative nine decibels again we want to figure out how can we add our decibels together in a way that makes sense that we can then use our rules to figure out the factors well negative three decibels plus another negative three decibels plus another negative three decibels equals nine and each one of these negative three decibel changes will result in a halving of the strength of the beam and because we're talking about factors here we are going to multiply these together to understand their true effect on the intensity so one-half multiplied by one-half multiplied by one-half is going to equal 1 8. and that is what we get when we decrease by negative nine decibels we've created a wave that is 1 8 as strong or eight times weaker working with the bottom half of this chart then we are going to use some of the other rules that we've been learning about the negative and positive 10 decibels and look at how we can describe and combine these rules the first row tells us that we are starting with a three decibel wave and increasing to a six decibel wave that is a change of three decibels knowing our rules a positive three decibel change results in a wave that is two times stronger that goes right along with our rules pretty easy the next one down tells us that we started with a 20 decibel wave and that it became 10 times stronger and we have to calculate what the new signal is knowing our rules that a positive 3 decibel equals an increase by a factor of 2 and a positive 10 decibel equals an increase by a factor of 10 or 10 times stronger we should be able to follow those rules to understand that we simply need to add 10 decibels to calculate the new signal for a wave that is 10 times stronger so 20 plus 10 equals 30 decibels the next row down tells us that we started with an original wave of 50 and have decreased to 47 that is a change of negative 3 decibels let's finish writing the rest of our rules here negative 3 decibels is going to equal half as strong and negative 10 decibels is going to equal 1 10 as strong negative 3 decibels lines up well with our rules so we can see then that by decreasing by negative three decibels we end up with a wave that is half as strong now the next row flips things around on us just a little bit we have been given the new signal and we are told that it is one tenth as strong so looking at our rules we know that if the new signal is one tenth as strong as the original signal then it has to be a negative 10 change so we were trying to figure out then what was our original value so that when we subtract 10 it equals our new value well 70 minus 10 equals 60. so that's how we calculate the original value to be 70. the next row down tells us that we have started with a 25 decibel wave and that it is 100 times stronger so we have to calculate how many decibels this wave increase to create a wave that is 100 times stronger so we are going to work on figuring out what factors can get us to a 100 factor remember we need to multiply items together to understand how factors are going to affect one another and remember that rule that i gave us earlier that when we are working in tens hundreds and thousands ten thousands we can look at the number of zeros in that final factor to tell us how many times we had to change by a factor of ten so there are two zeros so we know we are going to increase by a factor of 10 once and then increase by another factor of 10. each of these increases by a factor of 10 requires a 10 decibel increase so 10 times 10 equals 100 so that checks out with our making it 100 times stronger we know that each of those increases required a 10 decibel increase so 10 plus 10 equals a 20 decibel increase so 25 plus the 20 decibel increase is going to equal 45 decibels for the new wave the next row down kind of combines a few things that we've talked about now we are seeing that the wave is 100 000 times stronger and the new signal is 80 decibels so what we need to calculate is the original let's first work with our 100 000 increase in strength using the trick i told you if you count the number of zeros that's going to tell us how many factors of 10 we need to get to our total change so we are going to see five increases by a factor of 10 to get to a hundred thousand for each of these increases by the factor of 10 we are adding 10 decibels 10 times 10 times 10 times 10 times 10 equals 100 so we see that that works out that there's been five increases by a factor of ten to make the new wave 100 times stronger to do that change then we will see an increase of 10 plus 10 is 20 plus 10 is 30 plus 10 is 40 plus 10 is 50. so we're going to see a total increase of 50 decibels to create a wave that is 100 000 times stronger now the problem gave us the new value of the signal so we have to calculate what number plus 50 is going to equal 80. and we can see then that 30 plus 50 equals 80 making 30 decibels the original value the next row says that we started with a value of 100 decibels and that the news wave is going to be one-tenth as strong that matches up with one of our rules so we know that we just need to subtract 10 decibels from this value to get a new value of 90 decibels the last row gives us an original value of 60 decibels that has increased to 80 decibels so that is a plus 20 decibel change when we are working on the decibel side of things we are going to try to figure out what decimal numbers will add up to that 20 decibels using the rules that we know we know that 1 10 decimal change plus another 10 decibel change would equal that 20 decibels for each 10 decibel change then we are multiplying by a factor of 10 so 10 times 10 equals 100 when we increase the wave by 20 decibels we will see that it becomes 100 times stronger section 68.4 causes of attenuation when a pulse is emitted from the transducer it's going to travel through the acoustic gel and then into the body as sound propagates through a medium it will attenuate there are three physical phenomena that cause attenuation they include absorption scattering and reflection there are also two factors that determine the amount of attenuation and that is going to be frequency and propagation distance now all sound waves will encounter absorption scattering and reflection but it'll depend on the sound waves frequency and how far it has to travel to determine the total amount of attenuation that will occur when a sound wave comes to an interface of two media sound can either be absorbed reflected or scattered which typically cause attenuation we can see sound being refracted which redirects the sound into a different direction or it can be transmitted which means that sound continues through the structure a lot of times a lot of interfaces are going to do multiple of these some is absorbed some is reflected and some is transmitted the first attenuation cause that we're going to talk about is absorption absorption is the main cause of attenuation absorption happens when a reflector takes in a lot of that sound energy and transfers it into another form of energy this is typically going to be the form of energy called heat and it's that transfer of energy into heat that is going to cause tissues that are highly absorbing of sound energy to heat up and this is where we get concerned about thermal bio effects if we have a really strong sound beam with a lot of energy in it and it interacts with a structure that can absorb a lot of energy that is going to cause that structure to increase in temperature and with that increase in temperature we're worried that there's going to be some cellular damage now there are multiple structures around the body that are more likely to absorb sound we call these highly attenuating structures because they take up so much of that energy and we typically see that bone and lung and air are going to be some of our greatest ultrasound energy absorbers so when a sound wave interacts with a highly attenuating reflector the transducer sends in that sound wave and it's going to propagate to this reflector once it hits the reflector is typically going to send back a very strong echo our highly attenuating reflectors tend to have a very strong reflection on its anterior surface but then what we typically see behind that very strong reflection is absolutely nothing and that's because the reflector has absorbed the rest of the sound energy the reflection that went back was only a little bit of the sound energy that came in the rest of it got absorbed by the reflector and when that happens it leaves no sound energy to be transmitted and show what's behind it this is why if you are imaging over a rib you will see just dark shadow coming from behind it that rib has absorbed most of your sound energy there's no way to get meaningful reflections from behind that rib the second cause of attenuation that we're going to discuss is scattering and scattering is the whole reason that we can see the parenchyma of the organs now parenchyma is just a fancy word meaning tissue so it's why we can see the liver tissue why we can see the spleen why we can see the organs within the abdomen cavity same reason that we can see the muscles of the heart muscles and the limbs any sort of solid structure of an organ is viewed by ultrasound due to scattering now scattering is a little bit more helpful in the sense of actually creating images by ultrasound but it does account for a little bit of attenuation of the sound beam as well so when sound interacts with these interfaces that are found in these structures these interfaces are really really small they're less than one wavelength in size and as that sound interacts with these small interfaces it's going to cause the sound to go in all sorts of directions so as the sound interacts with these small interfaces and it gets scattered around a little bit of sound energy is going to be taken from that main beam to create those scattered echoes now scattering doesn't cause a whole lot of attenuation but it does add to the process for most tissue the scattering of the sound beam is what allows us to see the actual organ tissue and that is when it is very useful to us however if we have very dense organs or if we have a combination of air and very tiny interfaces like the lung we will then see scattering and absorption weaken the beam very quickly so on the setting of scattering the transducer again will send an ultrasound beam into the body it propagates through and as it interacts with very small reflectors the sound is going to bounce off of them and get sent in multiple directions so a little bit of energy gets pulled away from that main beam now some of those reflections are going to end up back up at the transducer and that's how we're going to get pictures of our organs some of it's going to head off into directions that aren't going to matter whatsoever and some of it is going to continue forward so it's slightly attenuated but we still get forward transmission of the beam and this forward transmitted beam is going to go until it interacts with another reflector and then some of that sound will be scattered or sent back or absorbed and it just keeps going until this has been completely attenuated and the third cause of attenuation that we discussed is reflection now reflection is what allows us to see the borders of our organs and that is because reflections are going to occur at large interfaces typically those that are more than one wavelength in size the reason that reflections cause better border visualization is because as the sound interacts with these larger interfaces it is going to cause the sound to be sent back the way it came from so scattering scent sound all over the place reflection will typically send sound back towards the transducer so we get more echo information back at the transducer which results in crisper stronger echoes also just like scattering reflection is going to take some of that sound energy out of the beam which will slightly attenuate it but if we didn't have reflection then we wouldn't have ultrasound pictures there are two types of reflections that occur in the body the first type is specular reflection which we can think of as a mirror like reflection and these are going to be very strong reflectors because they send sound back in one direction very organized back the direction from which it came we see specular reflections from surfaces that are smooth so think like the diaphragm vessel walls those are very smooth structures and the second type of reflection is diffuse reflection diffuse reflection kind of combines the concept of scattering and specular reflection together it is going to send sound back towards the transducer but it does so in multiple directions and this is typically because there is an irregular or rough surface at this reflector diffuse reflectors are the more common reflectors that we see within the body very few are truly a smooth surface both scattering and reflection do divert some of that sound energy causing attenuation but both concepts are imperative to being able to create echoes that return to the transducer so we can create our ultrasound images so we're actually going to cover both scattering and reflection again in much more detail in unit 6b in the case of specular reflection the ultrasound transducer is going to emit the beam that will propagate into the body and then it's going to interact with a very large smooth reflector now the particular thing about specular reflectors is that however the sound beam hits it is how the echo is going to leave it so if a specular reflector were to come in at 90 degrees it would bounce right back at 90 degrees and we would get an amazing image from it however most of the time the sound beam hits at an oblique angle which means that the strong reflection gets sent off at an oblique angle very organized singular very strong still heading back in the direction towards the transducer but most likely is not actually going to reach the transducer because of that oblique angle and with that reflection some of the sound energy is taken away from the beam but a lot of it will still continue through the tissue to go and interact with a new reflector diffuse reflection then works very similarly again the sound beam is going to travel through the body until it hits the diffuse reflector when the sound beam interacts with the diffuse deflector it is very large it's going to have rough irregular edges to it as that sound beam interacts with those rough and irregular edges it's going to cause the sound beam to go in multiple directions they're mostly directed back towards the transducer but only a few of them are actually going to come in contact with the transducer to be processed these reflections are typically very weak so they don't pull a lot of sound energy from the beam but a little bit does which causes the beam to slightly attenuate however that slightly attenuated beam is going to transmit through until it hits another reflector so we talked about absorption scattering and reflection being the causes of attenuation absorption being the greatest cause of attenuation and we can see that because if you look at diffuse reflection specular reflection and scattering you'll see that a lot of sound is still being transmitted forward if we didn't have this transmission of sound through a reflector then ultrasound wouldn't exist we would lose it the second the sound left the transducer phase diffuse reflection specular reflection and scattering are the whole reason that we have ultrasound images but because they redirect some of that sound energy either back to the transducer or other places into the body a little bit of sound is pulled from the main beam causing a little bit of attenuation but not enough to completely attenuate the beam now that story is a little bit different when we look at absorption and this is why absorption is the biggest cause of attenuation remember when that sound beam comes into the body and interacts with a highly attenuating or high absorber of sound energy it's going to take in most of that instead of allowing it to go through without that transmitted sound there is no sound energy left to get more echoes or to create more of an image and this is why when we are imaging we want to avoid those high absorbing structures whenever possible so absorption scattering and reflecting are all interactions that sound can have within a medium that would cause it to weaken or attenuate however the amount that the sound weakens is going to be more dependent on the frequency of the wave and how far the sound beam has to travel and because frequency and distance are the main determining factors of attenuation there are four things that i want to cover about them the first one is that higher frequencies are going to attenuate more so this makes attenuation and frequency directly related if frequency increases attenuation will increase if frequency decreases attenuation decreases and most of us have actually probably experienced this if you've ever lived in an apartment building or a dorm or other close quarters if a neighbor starts playing really loud music what can you hear typically just the bass notes and that's because those higher frequencies of our melody are going to attenuate more quickly compared to the low frequency of the bass notes that are going to continue transmitting going through the walls secondly higher frequencies are going to be absorbed more high frequencies will cause the particles in a medium to move more if you think about it a high frequency wave causes the particles to oscillate more that means they're moving more and with more motion more energy is going to be transferred into heat so if more of that sound is absorbed then less sound can be transmitted this is yet another reason why high frequencies tend to attenuate quicker third high frequencies are going to cause more scattering when a high frequency transducer interacts with those very small reflectors the scattering echoes are more intense that means that it draws more energy out of the beam if we scatter more energy that means that there's less energy that can be transmitted and lastly longer distances are going to cause more attenuation the further that a beam needs to travel the more it will attenuate and this makes sense if you think about it the more distance that the beam travels the more reflectors it's going to come in contact with which means there's more opportunity to be absorbed scattered or reflected so this also makes distance and attenuation directly related now we touched on these pictures briefly at the start of the lecture but i wanted to bring them back because they nicely show us how frequency and distance are going to highly affect attenuation on the top image we have a 12 megahertz transducer creating an image in the abdomen now i want to use this as an opportunity as well but if you don't recognize the structures that are in these pictures that's okay physics boards cannot ask you anatomy questions i'm just using these as examples to show you how the sound can attenuate i will typically try to describe any anatomy that we see in our ultrasound images but again remember you do not need to know anatomy to take your physics boards anybody who wants to be registered in msk breast cardiac peds admin and obgyn all have to take the spi so they can't ask you anatomical questions all right so that aside this first picture is of the 12 megahertz transducer on the abdomen kind of in the area of the long right kidney so in this picture here we have the skin line then we have a layer of fat and then some connective tissue abdominis rectus muscle here we are getting into the capsule of the liver this is liver parenchyma and that's really about it and that is because the sound has attenuated there's just not any more information really beyond this point and this is again a 12 megahertz transducer now compare that to the 2 megahertz transducer behind it we can see the same structures we've got the skin line the layer of fat the rectus abdominis muscle getting into the capsule of the liver liver tissue but the sound hasn't attenuated yet at this point and we can actually see more so now we're seeing more of the liver we're seeing the right kidney we're seeing bowels surrounding the kidney we're actually imaging all the way down to the diaphragm here and then lung behind it so again on the top here we are seeing that we're losing a lot of signal by the time we get to eight centimeters and i drew a picture here this eight centimeters corresponds with right about here these eight centimeters on the two megahertz transducer the two megahertz transducer just provides so much more penetration i also want to point out that we talked about lungs being a high attenuator you can see that here the diaphragm is kind of the last high reflector and then we have really nothing behind it and that's because once the sound interacts with the lung tissue it gets scattered and absorbed very quickly which leaves very little sound energy left to create any meaningful echoes we can then kind of think about these interfaces that the sound beam is interacting with first we've got maybe some scattering and some reflectors because we've got parenchyma and we've got some really bright reflectors when we get to the liver capsule we're getting a very high reflection we're getting that specular reflector looking at the tissue of the liver we're getting that scattering that's why we can see the solid organ when we look through blood vessels we get very little interaction and so that's why they're anechoic they're not really returning any echoes back to the transducer and we're all the way down to the diaphragm which actually is another specular reflector so we get that very bright reflection from it as the sound interacts with each of these interfaces it's going to attenuate a little bit more a little bit more a little bit more until it's completely gone what's cool about the 12 megahertz transducer though is that if you compare these first few centimeters this is really where high frequency transducers shine they can give us a ton of detail so the highest frequency that we can use is the frequency that we want as long as we can see the organs that we need as well in this image here we have kept the two megahertz transducer in the same picture on the bottom here but now we've switched to a little bit lower frequency for the top picture now this one gives us a little bit more information we are imaging to about the same depth we are 16 on the top 17 on the bottom but this is that attenuation in play here we are just not getting meaningful echoes any point past this and that's because that nine megahertz has become weak enough from interacting with tissue that we're just not getting strong enough echoes returning from this area compare that to our two megahertz transducer just powers through it all and sends reflections back from very deep into the body to further exemplify why we don't just use low frequencies on everybody i'm switching now to the neck we are in the neck is a relatively small structure in the body therefore we don't need as much penetration so we have a 12 megahertz image on top and using the large curved linear transducer we're probably using about a three megahertz wave on the bottom these are both of the same organ here we have the thyroid so a little bit of the thyroid coming here that's a little bit that comes over across the trachea that's called the isthmus of the thyroid this is the carotid artery here and it is surrounded by skin and fat and muscles and other connective tissue in the neck the trachea is in the middle here this is full of air which is why it's causing a lot of attenuation so our 12 megahertz transducer is giving us a beautiful amount of detail so much information can be seen in these first three centimeters this is why we use the correct transducer for the correct job compare that to this three megahertz transducer if i wanted to image this patient's neck and the bed and maybe the floor behind them then great give me the three megahertz transducer but this will do nothing for a diagnostic picture again we have the thyroid here the isthmus this is the trachea this is the carotid artery here we're still seeing all the connective tissue that's in there but we just don't see it in the same amount of detail and then we can actually see all the way back to the vertebrae in the neck here as well so the answer is not always use the lowest frequency to avoid attenuation because we need to balance out with using the highest frequency to get more detailed images section 68.5 reporting total attenuation attenuation is reported in decibels and implies a negative decimal change and we know that the sound beam is going to get weaker which means that the power intensity are also decreasing we also know that there are a lot of factors that go into how the beam will attenuate did it cross any high attenuators that absorb the sound what's the frequency of the transducer how far did it have to travel what other reflectors did it come in contact with and how many so there just isn't an easy way to say in conditions a b and c we're going to get this attenuation in conditions a b and d we're going to get this attenuation so instead science helped us out and we have a term and an equation that's going to help us to define how much of beam attenuates based on the frequency how far the frequency travels and the type of medium the term is attenuation coefficient and the attenuation coefficient is the decrease in decibels a beam will experience for every one centimeter of travel we can then take the attenuation coefficient and calculate the total attenuation because the total attenuation will be determined by not only the attenuation coefficient but the distance that the beam travels so the attenuation coefficient is going to tell us the number of decibels sound attenuates for every centimeter of travel in a particular medium what's neat about this is that the value once it's calculated is a consistent value at every depth so it's really helpful for calculating that total attenuation however every medium is going to have its own attenuation coefficient so of course in ultrasound we focus on the average soft tissue so the two formulas that i'm going to present to you are the two formulas that are going to help us calculate attenuation coefficient of the ultrasound beam in soft tissue we can say that the attenuation coefficient with the units decibels per centimeter is equal to the frequency in megahertz divided by two or we can say that the attenuation coefficient is going to be a half a decibel per centimeter per megahertz looking at these two formulas we can see then that frequency is directly related to the attenuation coefficient if we have a higher frequency we're going to have a higher attenuation coefficient which means more attenuation occurs which we have already said but now are proving with math it's awesome when it all works out like that and these equations are super easy to use as well so for example if we have a 10 megahertz transducer we divide that by 2 and that tells us that for every centimeter the 10 megahertz wave travels into the body it is going to attenuate 5 decibels because 10 divided by 2 is 5. so once we've calculated the attenuation coefficient then we can look at the distance traveled by a beam and figure out the total attenuation so we have the formula for total attenuation being equal to the attenuation coefficient multiplied by the distance a beam has traveled from this formula we can see that an increase in distance is going to directly be related to the total attenuation again proving that distance and frequency are the biggest factors when determining how much attenuation will occur so using our attenuation coefficient and knowing a distance traveled we can figure out how much attenuation has occurred at any given distance remember that the formula for attenuation coefficient only applies to soft tissue therefore calculating the total attenuation from this formula also only applies to soft tissue so let's take a look back at our 12 megahertz picture that we saw earlier we have a 12 megahertz frequency to calculate the attenuation coefficient we take 12 megahertz and divide it by 2. 12 divided by 2 is 6. so the attenuation coefficient for a 12 megahertz transducer is 6 decibels of attenuation per centimeter that the beam travels we can then plug that six centimeters and take a depth and calculate the total attenuation at any point in the beam's path so we can take six multiplied by one and at one centimeter we'll see that the total attenuation is six decibels at two centimeters it's twelve two times six is twelve at three eighteen decibels of attenuation at four centimeters there's 24 decibels of attenuation 5 there's 30 and so on all the way down to 8 we can see that there's 48 decibels of attenuation by the time this 12 megahertz beam reaches 8 centimeters to take this a step further then these total attenuation decibels are the decibel changes that we talked about when we were looking at that logarithmic scale and figuring out how the intensity is affected so looking at one centimeter we have six decibels of attenuation one centimeter into the body a 12 megahertz transducer is already reduced to a quarter of the original intensity if we look all the way down to eight centimeters then looking at 48 decibels of attenuation by the time the wave gets down there it's 1 64 000th of the original intensity and that's just going into the body it's got to get back to the transducer as well so i think this is really interesting how we can see the attenuation in our picture and then we can use the math to figure out how much attenuation has actually occurred and that verifies that this high frequency is going to attenuate very quickly being just a fraction of its original intensity at just eight centimeters well let's compare that then to our two megahertz transducer so our two megahertz transducer if we want to figure out the attenuation coefficient remember it's frequency in megahertz so two divided by two so two divided by two is one which tells us that for every centimeter that this sound beam travels it will attenuate one decibel which is very different from the six decibels so now at one centimeter into the body you only have one decibel of attenuation at three centimeters we're going to have three decibels of attenuation at five we see five decibels of attenuation at 10 there's 10 decibels because we're taking the attenuation coefficient multiplying it by the depth in centimeters when we get down to 15 centimeters we're at 15 decibels and to the max imaging depth we're at 17 centimeters with only 17 decibels of attenuation so we actually have a lot of sound energy that could go past 17 centimeters with the 12 megahertz transducer we saw that it was a quarter as strong just one centimeter into the body with our two megahertz transducer we can get three centimeters in and only be half as strong and then we can get all the way to ten centimeters and so only be a tenth of as strong so this is the math behind why we can see further into the body with a lower frequency there's less attenuation meaning there's more sound energy that can keep propagating and keep sending echoes back now there's one more term that describes the rate of attenuation and that's called the half value layer thickness you're literally never going to think about this term again after you take your boards or unless you come back and teach ultrasound physics but of course we got to cover it just in case you have a question about it on your boards because it does relate back to our attenuation so half value layer thickness actually has a lot of synonyms or other names for it we can call it the half intensity depth the penetration depth depth of penetration or the half boundary layer now the key word in this is the half part of it half value half intensity half boundary and what the half value layer thickness is going to determine is where in the body does the intensity become half the strength of the original beam or another way to say the exact same thing is at what depth has a change of negative three decibels occurred because remember when we are half the original strength that aligns with a negative three decibel change or three decibel attenuation since the half value layer thickness is a distance it comes with a unit of centimeters we typically see in ultrasound the clinical average is between 0.25 and 1 centimeter so for most of our ultrasound equipment by one centimeter the intensity of the beam has been decreased by half if we're looking at half value layer thickness in general terms in any setting of physics we can say that the half value layer thickness is three decibels divided by the attenuation coefficient remember our attenuation coefficient is in regards to soft tissue knowing that we actually have a special half value layer thickness that we can use and we can simply take six and divide it by the frequency to also calculate the half value layer thickness so as i was saying the half value layer thickness will change based on the medium so far we've been discussing mostly soft tissue up until this point we're actually going to discuss some other types of tissue in the body and how it affects attenuation in the next section but in general the half-value layer thickness will depend on two things frequency and the medium when the frequency increases we will see that the half value layer thickness is going to become thinner or have a lower numerical value when a medium's attenuation rate increases that is also going to cause the half value layer thickness to become thinner or decrease in numerical value so we can say that both the frequency and the attenuation rate of a medium are inversely related to the half value layer thickness the math is relatively straightforward for all three of these equations it is knowing the frequency of your transducer and knowing the distance that the beam has traveled with knowing those two factors you can then figure out the attenuation coefficient the total attenuation and the half value layer thickness by plugging in your numbers so we are coming up on a practice where we will have the opportunity to fill out the rest of this chart however i am going to jump to the board real quick show you one example and then we'll come back to the answers for our practice the equations for attenuation coefficient total attenuation and the half value layer thickness are relatively straightforward but i did want to show you just one quick example in soft tissue the attenuation coefficient is going to be equal to the frequency in megahertz divided by 2. in our example we have 15 megahertz divide that by two and we will see that for every centimeter that this way travels seven and a half decibels of attenuation will occur total attenuation then is going to be the attenuation coefficient multiplied by the distance in this example we have given you that we are traveling 6 centimeters 7.5 multiplied by 6 is going to give us a total attenuation of 45 decibels half value layer thickness had two ways that we could calculate we could simply take 6 and divide it by the frequency and that's because we are working in soft tissue or we can take three decibels and divide it by the attenuation coefficient regardless of how you do it it would either be 6 divided by 15 or 3 divided by 7.5 and that will equate to 0.4 centimeters remember that some of this math is a little bit more than what you would need to do on your boards or on your test calculating a decimal from 6 15 or dividing 3 by 7.5 might be a little bit tricky but remember we also learned how to convert fractions into decimals and if you are just dying to know how that happens i can actually show you here first we can look at 6 divided by 15 remember you'll take the 15 put it on the outside of your long division put the numerator on the inside 15 cannot go into six so we are going to place our decimal and a zero now we know that we are creating a value less than one we then will add a zero in our long division fifteen goes into sixty four times and that works out that makes our 0.4 centimeters if you're doing it the other way and you have 3 divided by 7.5 we can do the same thing 7.5 divided by 3. now you might be looking at that and thinking that looks a little bit tricky i don't really know what to do with that other decimal well remember if we multiply this by 10 and this by 10 we would get the same fraction 30 divided by 75 so now we can do the same thing make it 30 divided by 75. 75 cannot go into 30 so we're going to put our decimal in and our zero we need to add another zero 75 now can go into 300 and it happens to go in four times doing the math to get to the 0.4 centimeter half value layer thickness is probably just a little bit harder math than you would need to do on your boards but i just want to be able to show you during our practice times how we can still get to the same answers to better prepare us for those easier equations that you'll see in your tests if you have not already done so go ahead and fill in the rest of your chart pause the video because up next we do have the answers to our attenuation practice and here are the answers so remember that attenuation coefficient is the frequency divided by two two divided by two is one five divided by two is two point five twelve divided by 2 is 6. attenuation coefficient unit is decibels per centimeter using that attenuation coefficient we can then calculate the total attenuation at the depths given you will need to take the depth multiply it by the attenuation coefficient so 4 multiplied by 1 is 4 decibels 16 multiplied by 1 is 16 decibels and you should be able to figure out the rest of the column by completing those simple multiplications the half value layer thickness column then can be calculated one of two ways you can either take six and divide it by the frequency because we are working in soft tissue or you can take three decibels and divide it by the attenuation coefficient whichever way that you choose to calculate it you should get three centimeters as the half value layer thickness for a two megahertz transducer meaning the intensity is half as strong at three centimeters it's half it's strong at 1.2 centimeters with a 5 megahertz transducer and it's half as strong at half a centimeter in our 12 megahertz transducer the total attenuation calculations and the half value layer thickness calculations both align with the fact that increased frequencies attenuate more and the more distance the sound beam travels will also cause more attenuation and our last section of unit 6a is going to be section 6 8.6 attenuation in other tissues now the ultrasound machine assumes a soft tissue average to calculate the propagation speed of the sound beam and we also use the idea of soft tissue to calculate attenuation however we know that the sound beam is going to interact with other types of tissue some of the tissue in the body is going to cause sound to attenuate faster than the soft tissue average and we've talked about a few of these already for example air air is a very high attenuator it attenuates very quickly it's going to attenuate quickly due to absorption in fact frequencies over 1 megahertz can't even propagate through air they are immediately absorbed immediately attenuated and that's a big reason why we use gel if we had just a sliver of air in between our transducer and the skin we would attenuate very quickly and we would get very little information back from the actual body so we use a water-based gel to help move the sound beam from the transducer face and into the body now the lungs are also filled with air but they also have very small interfaces so the lungs are going to attenuate sound quickly due to absorption because of the air within them and those very tiny small spaces causing scattering of the ultrasound beam another strong attenuator in the body is bone bone is going to absorb a lot of the sound energy not allowing any of it to travel through causing dark shadows behind bony structures muscle will also cause sound to attenuate quicker than soft tissue muscle is kind of a funny one though if we are perpendicular to the muscle fibers the sound is going to attenuate faster versus being parallel to them the sound will attenuate a little bit slower so on this side of tissues that attenuate sound faster air is the absolute fastest with bone and lungs kind of tying for the next attenuation level muscle being a little bit more closer to soft tissue and then we have our soft tissue average of attenuation on the flip side of things then we have some tissues in the body that will barely cause any attenuation at all now the least attenuator in the body is going to be water water in fact causes basically zero attenuation in frequencies under 10 megahertz so water is an excellent transmitter of sound this again is why we like to use water-based gel to help us get our images moving towards the soft tissue average we find that fluids in the body like blood urine and bile do attenuate sound a little bit but not nearly as much compared to soft tissue as a sonographer it's very important that we remember fluid filled structures do not attenuate sound it can actually help us in making a diagnosis of cystic structures that we see around the body in the setting of gynecological ultrasound we have our patients fill their bladder their urinary bladder so when we go to look at their uterus it acts as a window to the uterus we use that very little attenuation through the urine to get a very nice look at the uterine muscle so fluids around the body cause very little attenuation and can be very helpful for not only diagnosing but serving as windows to other structures in the body and then lastly moving closer to soft tissue we have fad fat will attenuate sound a little less than soft tissue but more so than fluids or water as a sonographer knowing that some tissues can cause attenuation and some will not cause attenuation it really helps us to determine how we want to go about choosing an imaging window for example in the abdomen with all the bowel and air circulating around going directly over the intestines causes a lot of attenuation causes a lot of problems for seeing structures that sit behind the intestines in echocardiography if somebody has a lung that kind of flops over everything it makes it really difficult to see the heart because the lung is scattering so much of our ultrasound information so by changing windows angling around high attenuation structures or using structures like a full bladder to help us visualize ends up being very critical in our decision making process and how to obtain diagnostic images and that brings us to the end of the first half of unit six this is its own standalone workbook so in the workbook you will find activities related to attenuation as well as open-ended study questions in the nerd check that'll help you to test your recall on the information that we covered