Sound Wave Review - 2023-07-27 Meeting Notes PDF

New channel meeting-20230727_085933-Meeting Recording July 27, 2023, 12:59PM 4h 0m 0s Richard Zhao started transcription Richard Zhao 0:23 Good morning. Shanti Dhakal Poudel 0:27 Good morning, doctor. Sayed Aimal Hashimi 0:27 Good morning. Bishnu Shrestha 0:28 Good morning. Lamia Elkalifaabdal...

New channel meeting-20230727_085933-Meeting Recording July 27, 2023, 12:59PM 4h 0m 0s Richard Zhao started transcription Richard Zhao 0:23 Good morning. Shanti Dhakal Poudel 0:27 Good morning, doctor. Sayed Aimal Hashimi 0:27 Good morning. Bishnu Shrestha 0:28 Good morning. Lamia Elkalifaabdalla 0:29 Good morning. Richard Zhao 0:30 OK so. Uh, today, you know, I I got some feedback from some of you. So probably we were proceeding too fast and some did. Students would prefer, you know, we slow down. Sayed Maqsood Kamyar joined the meeting Richard Zhao 0:49 Uh, today we're going to review the contents. Tamana Asey joined the meeting Richard Zhao 0:56 We we we covered in the last two classes and then we if we have time we will proceed and we're also going to do some questions regarding the contents we've already cover. OK. Enas Adil Abdelhai Salih 1:16 OK. Thank you. Richard Zhao 1:17 And no problem. So I see. Just turn the camera off. So in chapter 2. We talked about the the uh basics of waves post wave. Uh. Possible option. Umm. At the beginning, we showed you the the metric table that you need to know how to. Uh, so they they units, you know, what do this? Seyedehmaryam Mashalchi joined the meeting Arash Jamshidi joined the meeting Richard Zhao 2:19 Mean. So how? How can you switch this units? Arezou Azizi joined the meeting Richard Zhao 2:24 And you also need to know the uh reciprocal relationship in this table. OK, for example, uh, mega and Michael are reciprocals. This is very convenient if you do some calculations like sometimes they give you the frequency and they ask you they. Period. So if you know the reciprocal relationship relationship is very convenient for you to do the calculation. Huina Cui joined the meeting Richard Zhao 3:06 So for the wave, we know that wave is actually a traveling variation of some quantity or quantities. Uh, so we know that the sound sound wave is a longitudinal mechanical wave. Sayed Aimal Hashimi joined the meeting Richard Zhao 3:27 So first we have two cans of waves, mechanical and electromagnetic wave. So sound, uh. It's one of the one type of the mechanical waves. Uh. Among mechanical waves, we have longitudinal and transverse and sound is a longitudinal wave. So we give you some examples for uh, the uh electoral magnetic waves, including light, heat. Actually. Uh, you know other forms of electromagnetic waves? Uh, we said that for a mechanical wave to propagate in the media, it has to a foreign mechanical wave to propagate. There has to be a medium, so the the wave has to, uh, interact with the media. But for uh electromagnetic wave to propagate in the the the there could be a medium, but if there's only a if there's no medium, it can also propagate. Ying Cui left the meeting Bana Otaki joined the meeting Ying Cui joined the meeting Richard Zhao 4:48 So the medium is not necessary or electromagnetic view to propagate. So we introduced. Some acoustic variables for sound wave, so including four of them, pressure, density, particle, motion, temperature. Though all these are considered as acoustic variables. So. Umm. And we talked about the difference between transverse wave and the longitudinal wave. So the the motion and the direction of the particle movement is different if the the direction of the particle movement movement is perpendicular to the direction of the view propagation, we call this transverse view. Omid Ahmad Mohammadi joined the meeting Richard Zhao 5:53 If they particular movement is parallel to the direction of the provocation, we call this longitudinal wave. So in this longitudinal waves, we have areas of verification and the compression. Umm compression means that you know here the density of the medium is more dense and the and the pressure is also high and the very affection in that you know the particles here are more sparse and the the uh. Density is low in this area. The pressure is also low. Umm, you addition to those four acoustic variables? We also introduce some descriptive terms that you can describe waves. So for continuous wave, uh and also for possible we have frequency period, wavelength propagation speed, amplitude intensity. This actually parameters to describe a view. What is frequency? Uh, frequency really means that how many cycles appear in one second. So a cycle is a complete variation. Uh in either of the for acoustic acoustic variables? Uh, what is? Uh, we know that Percy can be represented. Uh, you know, in this units, so for for in AutoZone, we use megahertz to describe the fundamental frequency of the transducer or the waves you transmitted. But use kilohertz to to describe Doppler ships or GIF pulse repetition frequency. Dayani Jayapathy joined the meeting Richard Zhao 8:23 So this is showing you you know how a high frequency uh, wait, look, look, look. Looks like OK, so you have uh short period. And also, uh, if it's uh propagating in the same media with high frequency waves you have, you also have short wavelength. So, uh, we said that the ranges, the different ranges for some, you know the audible sound would be between 20 and 20 kilohertz. Uh. Infrasound. Uh, you know any frequency less than 20 Hertz we consider as infrasound and any frequencies above 20 kilohertz, we call it ultrasound. And the range of a a time, not that diagnostic ultrasound is between 2 and 20 megahertz. Of course we we use frequencies below this and above this about 20 megahertz uh. In AutoZone 2, but usually we we, we we consider this range is the range for diagnostic ultrasound. So language introduce the formula to calculate the frequency. So actually the frequency is calculated by divide propagation speed of the sound. Uh by the wavelength in this medium? So please add that the frequency of the sound and the penetration of the sound. Uh. You look goes uh to different direction, meaning that if the frequency is high, the penetration would be low because of higher attenuation. So from here we know that if you're looking at his structure, that which is deep, you need to lower your frequency, otherwise the penetration would not be good enough. And then we talk about period. The period is just, you know, the reciprocal, our frequency. It means that how much time it takes for one cycle to occur. So in ultrasound, we usually use microsecond to to be the unit of the pier. And we use. Millisecond to to to be, uh, the unit of PRP pulse repetition period. Because you know that pulse repetition frequency. UH-15 kilohertz, they reciprocal kilohertz, would be millisecond. You know the the the formula for period is just the reciprocal of frequency. As we said, you can see here because we said, you know we use megahertz as the unit for frequency. So the reciprocal of megahertz would be microsecond. So to calculate, uh, you know, these kind of questions. If they give you paper, they say they ask you to calculate the period you you just get the reciprocals of both the number and the unit. OK, so for example this this one, what is the period of a 5 megahertz transducer? So first we need to know the reciprocal of file is thereupon 2. Right and, uh, the reciprocal of uh uh megahertz would be microsecond. So the the, the, the, the period of this transducer would would would be 0.2 microsecond. So you you, you get the reciprocal of the number and then you get the reciprocal of the unit. So this is telling you that the frequency and the period are are, you know grouped together. You can consider them as actually one parameter. So. Frequency and the period are determined by the source of the sun, so they they doesn't have any relationship with the medium, no matter in what kind of medium the sound is propagating. Uh, the frequency and the period, you know, doesn't change. So then we talk about the wavelength. Wavelength is actually, you know, the length of space one cycle uh. Takes up in space, so we usually we use millimeter uh as a as a as a unit for uh, wavelength. So the formula for wavelength is you you divide propagation speed by uh, frequency. So far. Safia Addou joined the meeting Richard Zhao 14:27 Stop tissue. We consider we, we, we uh, suppose the propagation speed in the soft tissue is uh 1.54 millimeter per microsecond. So sometimes you you see questions asking you to, uh, calculate the wavelength. Uh, you know, they they give you a frequency? Umm and uh, suppose the sound is traveling in soft tissue and then they ask you to calculate the actual wavelength of this sum. Uh, they they they easiest way to do this is to just remember, you know, uh, if they frequency of the transducer is one is 1 megahertz, the wavelengths would be insultation would be 1.54 millimeter. So anything larger than one megahertz, I mean the frequency is larger than megahertz, you use the number of the megahertz divide the 1.5 millimeter, OK, the parts ample, if they ask you what's the wavelength of three megahertz transducer, no song. Omid Ahmad Mohammadi joined the meeting Richard Zhao 15:51 If the branch itself sound is 3 megahertz, what's a wavelength? The wavelength will be you divide 1.54 millimeter by three. So uh, the propagation speed of sound? Uh, but usually? Uh, use the we we we actually use 2 units for it. So you need to remember both of them. So is 1.5 kilometer or, you know, one uh Southend, 540 meters per second or 1.54 millimeter per microsecond? So we start that the propagation speed of sound is determined. Uh by many two factors. One, it's stiffness and the other one would be density. Uh, the higher the stiffness, they they. Faster the the the sound propagation. So the higher the density, uh, the lower the propagation speed. So in this formula you can consider this part as a stateless of the media. But actually, uh, this is called bulk modulus. This ones telling you how much pressure you need to cause a certain extent of volume change. So to cause, let's say this much, this amount of volume change, how much pressure you need. So the larger the number is, you know the the steeper the medium is. So this is showing you you know what is bulk modulus. So you divide stress or the pressure change by, you know the the the percentage of volume change. Umm, you don't have to remember. You know the how to calculate the bulk modulus. This is just a help you understand the stiffness and what is stiffness. Xuewen Jiang joined the meeting Richard Zhao 18:26 So and then we introduce what is Thursday, so it's it means that how much mass in a unit volume. So without that high bulk modulus means you know incompressible means it's in elastic means that the the medium is stiff low, uh bulk modulus mean that it's compressible, the medium is elastic or not stable. So in AutoZone, you can consider you know bulk modulus and stiffness as the same thing. So we said that you know the appropriate speed. Ohh it's different in you know gases, liquids and solids. Umm, his highest in solids because solids have the highest stiffness. Our party is lowest in gases because the stiffness uh in gases is the lowest. Umm so the the relationship between stiffness and density and the you know the propagation speed is shown here. So if you increase uh, stiffness supposed density doesn't change. Uh, you will increase the propagation speed, so if you increase density. Umm, while the stiffness keeps constant, you will decrease the speed propagation speed. But in reality, you know, if you talk about the tissues in human body. See. Umm what? Uh. Usually happens is that if you increase the density of the teacher like like if you choose a higher density tissue, usually the stiffness of that tissue would change would increase more significantly. Uh, then the density does so in that case, uh, it higher density tissue usually has a higher propagation speed of sound, which is counter intuitively to, you know, if you look at that formula. Umm you would think that if you increase the density, the public speed would decrease. But that is not true in uh human body, because this one will not keep constant if you increase density. This one the stiffness will change more significantly. And that's about the parameters for, uh, continuous wave. Then we introduced some more uh parameters used to describe possibly, uh. So what is possibly possibly really means that you know the the transducer is not, uh transmitting waves continuously. So after sending out a pause, there would be a pause so that would be that would be the listening time. During that time, the transducer is actually is actually listening to the echoes from the body, so it's not sending out pauses. Farkhunda Ateel Siddiqi joined the meeting Richard Zhao 22:07 So this kind of operation is called Pulse wave operation. So for possible operation we have pause, rapid repetition frequency and pulse repetition period and also post duration duty factor and the special parts length and also bandwidth to describe uh the post wave. So you can see their time related parameters and also. Space related parameters. So this is showing you the difference between continuous continuous wave and the boss wave. Anahid Modaresi joined the meeting Richard Zhao 22:54 So for pulse wave you have on time and off time. During this time the the the system on the transducer is not sending out waves, is listening to the echoes from the body. Arash Jamshidi 23:07 Excuse me, doctor Richard. Richard Zhao 23:08 Yes. Arash Jamshidi 23:09 Uh, actually, I have a question in definition of the pulse repetition frequency. Richard Zhao 23:17 Mm-hmm. Arash Jamshidi 23:18 Is this the number of the cycles in one second or is this number of the pulses in one second cause both of them is totally different? Richard Zhao 23:24 Uh. It's a pauses. Yeah, it's not. Yeah. Yes, yes. So impossible. The pause reputation means how how, how many pauses happening in one second. It's not cycles, so the frequency of the pulse wave and the frequency of a continuous wave is the same. Umm, so if I OK, let let me Draw Something for you. So what's the frequency of a positive? Actually, it's a frequency of this part. You know, for this paper, for the pulse duration. So harmony cycles in one second. So if you consider the IT is the same as they continuously. If it's the same as it, if it's continuously sending out the wave in the same frequency. So if you count the cycles in this pulse, this is considered the frequency of this positive, but will if you are talking about they. PRP post reputation. PRF. Pulse repetition frequency. It means that how many pauses this is 1 pause. This is another pause. How many pauses in one second? If you don't count the cycles you know in, in in all the poses, you just count the number of the pauses. Arash Jamshidi 25:32 Thank you. Richard Zhao 25:33 OK. So please look at the waveform of the the, the the path here. So it's very regular. It's like a umm it's a regular Soundwave, but in reality let's say if the amplitude of this wave is low low enough the the waveform is very likely real Soundwave. But if the amplitude is high, the high pressure would cause some deformity. Uh the no it will change the shape of the the the waves. So in reality, in reality, the way from a pulse more likely be like this. So it's not a some way you can see some part is rounded some part very uh acute, uh. So this is because of the uh propagation in the tissue. So the shape of the waveform would be distorted. Uh, we said that the past repetition period. Uh, frequency is just, you know, the number of pauses occurring in one second. So usually use kilohertz to describe the PRF. Uh, of course. You know the the ERP and PRF are reciprocals just as frequency and period. Then we introduce the concept of pulse repetition period. As we said, you know, it's a reciprocal of the pulse repetition frequency. It means you know how much time uh, that the the time between, you know, the beginning of 1 falls to the beginning of next pass or the ending of 1 pause to the ending of the other parts. So usually. And then we we introduce the concept of post duration. Post duration means that the time it takes for one pulse to occur. So here you can see this is 1 pause. How much time it takes for this post to occur? Uh, like the period we also use microsecond to describe a pulse duration. How do you calculate the position? They need to know how many, how many cycles in one false and the period for the wave. Tamana Asey 28:48 These two. Richard Zhao 28:53 Hello. OK. And then we introduce the concept of duty factor. So. So what is duty factor? It means that you know the percentage of time the system is actually on to transmit transmit waves. Umm, so we calculate the duty factor by divide the post duration by the post repetition period. So it shows you the percentage of time, uh. The system is actually on to sending out pauses. So far for AutoZone for 2D imaging, you know this is the typical range of duty factor for popular. The duty factor would be higher because we use longer pauses in doctor. Meaning the post duration would be longer in Doctor. That's why the duty factor in doctor you, you doctor in Doctor would be higher than the duty factor in 2D imaging. So this is formula to calculate the duty factor. We have another formula to calculate it. We'll talk about just in a moment. And then we introduce the concept of spatial path length length. So it is just the length of space that a pulse takes up, just like you know the wavelength. Uh, it's only now you're not talking about one cycle, you're talking about. Uh, several cycles. So the space it takes for. Sarah Cycles to to to occur in space. The formula is that you multiply wavelength by the number of cycles in one house. And then we talked about the bandwidth, you have different bandwidth, the bandwidth of transducer, the bandwidth of your pause or something. Uh. A bandwidth is actually the range of frequencies can content in the pause. Uh, so we said that there is a reciprocal relationship between pulse duration and the the bandwidth. Uh, the shorter the pause, the broader the bandwidth, OK. The bandwidth is actually a range of frequencies that can be used. There's different kinds of bandwidth, so actually how do you can sometimes they will ask you to calculate the bandwidth. Uh, you know, it's actually calculated by, uh, you, you know. Uh you. It's the difference between the maximum frequency and the minimum frequency actually. And then we introduced another concept of fractional bandwidth. So if we divide bandwidth by the operating or central or center frequency, that would be the fractional bandwidth. We said that this number, if this number is larger than 80%, we consider it's a broadbands design for a transducer we prefer. We prefer I broader bandwidth for for transducers for some reasons. OK, if you get the reciprocal of this number of the fractional bandwidth, that would be considered quality factor. Uh, of course. If they qualify, quality factor is high. It means that your bandwidth is narrow. So for outbound transducers, usually we prefer the quality factor to be low. Uh, so why? Uh create uh. Would like to have a broader bandwidth transducers. It's because you know it has a couple of advantages. And so so the first one would be, yes. Enas Adil Abdelhai Salih 33:44 Accuse me Doctor Richard in the. The big example of Canada sonography? Richard Zhao 33:53 Umm. Enas Adil Abdelhai Salih 33:53 Do we have a question? They ask us to to solve the bandwidth and how to bring the. Richard Zhao 34:01 Umm. Enas Adil Abdelhai Salih 34:04 Fractional bandwidth and these things or no? Richard Zhao 34:08 Uh, there, there could be questions like that. Enas Adil Abdelhai Salih 34:13 Umm. Richard Zhao 34:16 Could be, so I'll show you some questions later. Enas Adil Abdelhai Salih 34:16 OK. Richard Zhao 34:18 So even in the chapter questions of our textbook, you know, there are some questions regarding this. Enas Adil Abdelhai Salih 34:25 OK. Thank you. Richard Zhao 34:31 Uh, so with the broader bandwidth of the transducer, you have the flexibility of doing multi Hertz operation. You know you can like the transducer we use at our college. Uh, you can choose different, uh, frequencies. No car, different structures. If you're looking at a deeper structure, you may want a lower frequency. You you you can change that. So for this purpose the the bandwidth of the transducer has to be high. That's to be white. Another thing is that you can do dynamic frequency tuning, so this is a technique used to, you know, get better image quality and you know for the superficial part of the image you want to use the the higher frequency signals or the deeper part of the image you want to use. The uh lower frequency signals, so the system will do this by dynamically figuring out the the frequencies you don't want, but for doing this you you have to again you know have a a wide bandwidth transducer. And also for harmonic imaging. So we know that harmonic imaging are using the. Harmonic frequency signals, which which are twice the fundamental frequency of the but the the, the the franchise sales are up. Harmonic signals are twice of the frequency of the, you know, fundamental frequency. So for that uh function to the upper ball, you you you have to have a wide bandwidth of the transducer. Uh. Another thing, as with higher with wider bandwidth you can do frequency building or frequency compounding. So which means that you, after sending out a wide bandwidth, pause. Uh, you know, in the echoes you receive from the tissue. Uh, you have signals in different frequencies. You use uh? Let's say uh signals at lower frequencies to form one image, and then you use signals at the mid range frequency to form another image and then you use signals at higher frequency to form another image. Then you average these three images to form one average image, so that's called frequency compounding. We do this to to reduce noise, to reduce speckles, OK. Uh, we we we know that we said. A band? A wide bandwidth designs, not always the preferable, so sometimes we prefer uh. He narrow bandwidth design because it has higher sensitivity for that specific currency. So like when we do continuous wave Doppler so we don't need a broadband width because we only use one frequency which is you know the driving the the frequency of the driving voltage. So we want the transducer to be very sensitive at this specific frequency. So we at in, in this case we prefer sensitivity over the bandwidth. It said that there there is a reciprocal relationship between. Pasteurization and. And uh, bandwidth, OK. With longer pauses, you have narrower bandwidth. Umm, this is telling you the same thing. You can see that this pause is much longer than this one, so the bandwidth of this pause is narrower than this one. OK. Uh, we know that the pause, the pause is in 2D mode are B mode. Uh, it's shortest. That's why you know the bandwidth in a 2D pass is the widest. Yeah, and for continuous wave, uh, the bandwidth actually only contains one frequency. OK. And again, that it is telling you the formula for fractional bandwidth and amity. What is amplitude? Amplitude is actually the maximum variation. Uh of any of the acoustic variables we talked about? So what is amplitude is the distance from the maximum values uh to to the mean actually. So you have two ways to calculate the amplitude, so the first one would be if you use maximum value minus the minimum value. So you divide the distance between the maximum and the minimum. Uh by two you divide the this number by two. That would be the amplitude. It shows you the distance between the maximum value to the. But they mean the other way is that you calculate the mean first and then you use the maximum value minus the mean. OK, you have two ways to to to to do it. And we talk about some, uh, we reviewed some knowledge of physics. What is energy? What is work energy is the ability to to do work. Arash Jamshidi 41:28 Excuse me, Doctor Richards. Richard Zhao 41:30 Yes. Arash Jamshidi 41:31 Uh, actually, I have a question about the relationship of bandwidth and the positive duration. Richard Zhao 41:38 Yes. Arash Jamshidi 41:38 I wonder how is it the reciprocal of the pulse duration? I sorry, I I don't understand it. Well, that the bandwidth is the reciprocal of the pulse duration. Richard Zhao 41:55 Yes, this is a actually you don't have to calculate anything but this. This is just, you know, you use this to help you understand the relationship. It's a reciprocal relationship, but usually you don't need to calculate the exact there. There won't be questions asking you if they tell you that the bandwidth is this. What would be the positive duration be? So you consider this as a just relationship? Not exactly how how you calculate the the the bandwidth, so they if the bandwidth is longer they I mean if the post duration is longer, the bandwidth would be narrower. So this this is just to help you understand this. Bishnu Shrestha 42:48 Also, Doctor Richard, so if there is a lock. Richard Zhao 42:50 Yes. Arash Jamshidi 42:51 It doesn't. Excuse me. Excuse me? It doesn't make sense cause for example if the pulse duration is lower, so in that case we can have also different types of the frequencies, so the bandwidth can be even either the same or lower or higher. Richard Zhao 42:59 Yes. Arash Jamshidi 43:17 It depends on the machine actually. As far as I understand, it's there's no but automatic relationship between those ones. Richard Zhao 43:31 Uh. So let's say if we consider, you know, other things doesn't change, you know the uh. So if I say for the same user right? So if we are sending out a longer pause or a short pause, a longer pause means there are more cycles in my pause. So in that case, you know for the longer pause there would be a narrower bandwidth for the shorter process that would be a broader bandwidth. This is talking about the bandwidth for the pause. Umm, but for a transducer you know the the bandwidth for transducer as long as the the the transducer is produced you know the the the the bandwidth doesn't change. So that's the bandwidth for the transducer, but for the pause, if you if you change you know the the the pause duration. Of the bandwidth would would would change. So actually this is related related to related to how you. Analyze the they they, they, uh signals. How? How how can you reconstruct a waveform like this? We actually we use a wider range of frequencies to top, you know, sound waves to reconstruct this kind of pause. So far, umm, for a very long pause it's very easy. You you only need a small small range of different frequencies to reconstruct the the the the the view form of a longer pause, but for a shorter pause you have to use a wider range of frequencies to reconstruct the the the waveform. Shanti Dhakal Poudel left the meeting Richard Zhao 45:25 This is very complicated, so and this is a back that you only need to remember, you don't have to dig too deep into this and it takes a lot of time to understand them then actually know how. How doesn't mean? Arash Jamshidi 45:44 Thank you. Richard Zhao 45:45 No problem. Enas Adil Abdelhai Salih 45:50 Ohh. Excuse me, Doctor Richard, can you return back just for to the amplitude again? Richard Zhao 45:57 Amplitude. Enas Adil Abdelhai Salih 45:58 I'm fit to you, dear. Richard Zhao 45:59 Yes. How many teeth? Ohh, OK. I was here so. Enas Adil Abdelhai Salih 46:38 Yeah. Umm. Richard Zhao 46:41 Uh, you want me to explain the calculation or? Enas Adil Abdelhai Salih 46:44 The yeah, the calculation please. Richard Zhao 46:46 Yeah, calculation actually, you know. Uh. The amplitude is actually. Shanti Dhakal Poudel joined the meeting Richard Zhao 46:52 It really means you know the variation. Uh, the maximum variation of this acoustic variable. So if you consider the distance between the maximum values, either the maximum and the minimum, so the distance between these two points and the the mean. This is the mean. The point between these two extreme values would be the mean, so this would be the amplitude and this would also be the amplitude. So the distance between the extreme values and the mean OK. Enas Adil Abdelhai Salih 47:26 And is it always equal to each other or no? Richard Zhao 47:33 Uh, it's equal to each other. You you mean the positive and the negative ones would be yes, yes. Enas Adil Abdelhai Salih 47:40 Umm. Richard Zhao 47:41 You you ultrasound, we consider you know it's a equal. Enas Adil Abdelhai Salih 47:46 OK. OK. Richard Zhao 47:50 So because if you, if you know the maximum value you know the minimum value, you use the maximum minus minimum and then you divide the value by two and you have another way of you know the mean or you can calculate the mean. You can. Uh, you know, you know, divide the sum of the maximum and the minimum by two. That would be your mean and you you say maximum value. Uh to minus the mean, that would be your amplitude too, OK. Enas Adil Abdelhai Salih 48:28 OK. OK. Yes, thank you. Richard Zhao 48:31 No problem. Uh. So this is you don't have to know what is work. This is just for you to. Understand. So power is the rate at which you know energy is transferred from one part of the system to another. Umm. If we divide power by the area, uh, the power spread in space, that would be the intensity. So in this state is power over area. Are you understand intensity? So let's say for the same amount of uh power, let's say when the area is large like, you know, at the surface of the this this length, uh, the intensity is not high, so that won't cause any. Uh, he showed that won't ignite the the wood, but if you make the area very small using this lens, so the intensity at that point the focal point would be very high. That's why you can ignite the food. So in the human body, you know that relates to uh, the focal point of your transducer. So where the intensity would be the highest and at the same time you have the highest risk of the thermal bioeffects at that point. So if the intensity is high enough, that could cause some some some injury to to the tissue. Uh, we know that power and intensity are directly proportional to amplitude square, so. Let's say if uh, the uh amplitude is increased by factor of two. Uh, the power or intensity would be increased by factor of 4. They also said that. You can see that the intensity and area. Uh goes through the opposite ways. You if the intensity. Uh. If the area increases, the intensity would be decreased. Umm. Then we introduced a uh, you know, several different combination of intensities we we we have uh space related intensives and we have time related intensities and you you have different combinations of that. Uh, different interfaces. Umm, So what is special intensity is relating to they location. Up you know the the the the location in the beam where you measure the intensity. OK, usually the peak, uh special intensity would be at the center at the focal point at the center of the focal point. Umm. At the surface of the transducer, because you know the beam hasn't been focused yet, you can consider the intensity here as the special average actually. So and then we will introduce some ohm temporal interstates. Uh, so this is you can see this as a pause. We know that in a pause. Uh. The amplitude of the cycles in the polls. This is not the same. It's they're not equal. You're ready at the two ends of the past. The amplitude are and the the the amplitude is lower than the amplitude at the. You know, let's say the center. So at the time where you know. Uh, the peak? The peak amplitude passes when uh. Location. Umm at this time. You you can have the temporal peak temporal peak intensity. You know that you know, the pulse has to go through the specific location where you measure the intensity. If at the time you know the peak of this pass passes through this ohh location. Uh, at that time you have the peak temporal resolution, temporal intensity, OK. And then you have the pause average intensity, which means that if you average the intent intensities among these pause, you know you get a average one. That's your pause average. And if you divide the intensities uh by the. Uh, by the pause reputation period. The whole time you get the temporal average. Intensity. So this one would be the lowest among the temporal intensities, OK, and the temporal peak would be the highest among the temper interstates. So this is showing you the where you know you have the. Special peak intensity is usually, you know, at the focal point and the in the center of the deep, OK. And you also have special average. So that's when you average all the intensities at different locations in the in in the beam you have the special average. If you ever reach the intensity intensives you know along the whole. False reputation. Period. You get the temporal temporal average. So we said that we use the S PTA meaning special peak temporal average as it indicator for thermal bioeffects and we use special peak pause average. Use this one to uh to be a indicator for mechanical biofact. Arash Jamshidi 56:04 Excuse me, now that you're, can you go back to the last slide for a moment? Richard Zhao 56:09 His son. Arash Jamshidi 56:11 Thank you. Richard Zhao 56:12 OK. You mean this one? Arash Jamshidi 56:21 Yes. Thank you. Thank you. I got it. Richard Zhao 56:23 OK, OK. No. Let's see what's the time now. OK, so now time. So then we introduce another way to pack. Enas Adil Abdelhai Salih 56:37 Excuse me again please. Richard Zhao 56:39 Yeah, yeah. Enas Adil Abdelhai Salih 56:39 The the previous one just you said the special post there and temporal average we use the thermal bioeffects. Yeah. Richard Zhao 56:51 This is not pause spatial peak special peak temporal average? Yeah. Enas Adil Abdelhai Salih 56:51 And it's special, especial peak or. OK, peak. OK. Just just a minute again, just a minute. Special Big Bulls average we use in mechanical bioeffect. Richard Zhao 57:03 Yes. Mechanical will use special Peak cause average OK. Enas Adil Abdelhai Salih 57:15 OK. Richard Zhao 57:18 Yeah, no problem. Enas Adil Abdelhai Salih 57:18 Thank you. Richard Zhao 57:22 So we introduce another way to calculate duty factor. So that's when you divide temporal average by pulse average. Uh to help you understand this, remember this you you need to know that. You know, temper average is the lowest, uh, intensity temporary intensity. So this one has to be smaller than this one. Then pause average so we know that duty factor is always less than one. So to calculate this you need to put you know the temporal average, the smaller one. Uh, over the bar, you know, put the larger one below the bar to calculate. That's how you get a number smaller than one. OK, that's about, you know the first class. So in the second class we introduced the concept of decibels. The decibel is actually OK let me do this. Is a ratio in the logarithm form OK? You transform a ratio into a logarithm form. That would be decibel. So like we we we we know the but we use decibels to describe noise, right? So we can see their uh, the uh strength level that our ears can barely hear so that that will you can see there as zero decibel. So anything about that stronger than that, you know it would be the number of decibels. So if you if you compare the intensities. Of the noises. Between you know the intensity, you can barely hear. So and you take a logarithm form of that. That would be the noise described in decibels. So for the decibels, you need to remember some numbers. Uh, black. 3 decibel uh, meaning that if you decrease uh, they intensity by three days ago, meaning that your decreasing the intensity uh by a factor of two. If you're increasing the increasing by three decibels, it's means that you are, uh. Increasingly, they they intensity by a factor of 2K. If your, let's say, if you're, uh, increasing the intensity bifactor by 10 decibels, it means that your, uh, increasing the intensity by 10 times. So this form this is a attenuation form actually, so this decibels are actually minus, you know, minus decibels. So this ratio is showing you the final you the the. The final number percentage you can get after attenuation. So, uh, let's say if you if the victim has been attenuated by three decibels, so you use the original intensity, uh to multiply this ratio, that would would be your final intensity. After attenuation OK. So if you know if you know the beginning in his day and you know you know how many decibels uh that's being attenuated, then you look for the ratio for that decibel and you multiply it by this ratio. You get the final intensity. But that's because we need to know the two. Enas Adil Abdelhai Salih 1:02:26 So if there is a, excuse me if there is a there an equation or or a in the exam they will give us this table. Richard Zhao 1:02:35 Umm. Uh, sometimes they they will. If they don't give you this table, usually you know that's the numbers. So easy to remember, like UH-3 decibels. So for this one you need to remember 3 decibels. Enas Adil Abdelhai Salih 1:02:49 Umm 50%. Richard Zhao 1:02:52 Uh, it's two times. You know, if it's decreased by three decibels it meaning that the final intensity would be 1/2 of the original. Enas Adil Abdelhai Salih 1:03:05 Umm. OK. Richard Zhao 1:03:08 And 10 decibels. You know, it means 10 times of the power or intensity ratio. Enas Adil Abdelhai Salih 1:03:15 Umm. Richard Zhao 1:03:15 Uh, for this number say they won't give you a table. Enas Adil Abdelhai Salih 1:03:21 OK. Thank you. Richard Zhao 1:03:23 Cool. Uh, for decibels, we need to uh, we need to know the two formulas. So if if if they tell you a power ratio, uh, you use this one the the decibel would be 10 times log of the power ratio. So for the power ratio, we need to know that the dynometer would be the initial or original power and the numerator would be the final uh power or intensity. So this is initial, this is the final. If they if they tell you the amplitude ratio, not the power issue, you need to use UH- 20 times this lot. Lot of this ratio. And again, you know here would be the initial amplitude here would be the final amplitude. This is the basic uh logarithm calculation rules that you need to know. And. Was a log of these numbers. It's very straightforward, OK. Uh, some special numbers. The log of two is 0.3 the log of 1 / 2 one half would be minus 4.3 K. So these are some examples for you to calculate this. And you know, if you want actually calculate the power ratios of a certain decibels, uh, you can use this, OK. You can remember, uh, certain numbers like 1 decibel, the power ratio of 1 decibel is 1.26 uh. The power issue of three decibel would be 2. The power issue of 10 decibels is 10. If you remember this, you can actually calculate calculate the power issue of any decibels. So let's say if in this case, uh, since that's all you can consider, since that's wells, as you know, uh, 3 + 3, uh. If we want to calculate the power issue of UH-6 decibels, uh, you can use the power issue of three decibels, which is 2 time times another parish. Of the three decibels, which is another 2K, you get 4. Uh and uh, for example. Two decibels. Uh, you can consider either. Uh, Yes, 3 -, 1 or 1 + 1. Uh, in this case you can divide the power issue of the three decibels by the power issue of 1 decibel. No. Uh, you you get the result of 1.6 and if you consider two as 1 + 1, you can use the power issue of 1 decibel uh to multiply the term ratio of another one. That's well, you, you you get the same result. And then we introduce the concept of attenuation. What is attenuation is the weakening of sound as it propagates along the tissue. Uh, we describe attenuation in decibels. We said that there are three uh contributing factors for attenuation, including absorption, reflection, scattering. Uh, among which absorption is the dominant factor. Uh installed tissue. So absorption is the conversion of sound to heat. So it depends on uh, frequency that's and. A characteristic of the medium, which is the viscosity of the medium. Uh, not. And then we introduced a concept of attenuation coefficient. It means uh, for a given medium. So how much attenuation occur? Uh. When the sound travel. UH-1 centimeter, OK. So in soft tissue, uh, for each megahertz, uh, the attenuation coefficient would be 0.5. Uh, decibel if you can. If you consider you know, you combine the frequency and distance together, you can actually define the attenuation coefficient in this way. So it's how much does it Bill? How many decibels they attenuation would be? Per centimeter per megahertz. So this is a easier way to to understand it. You don't. You don't have to calculate, you know. The attenuation coefficient for different frequencies, you just use this as the constantly they they definition of attenuation coefficient. So they attenuation per centimeter per megahertz. Farkhunda Ateel Siddiqi 1:10:16 Excuse me, doctor Richard. Richard Zhao 1:10:19 Yes. Farkhunda Ateel Siddiqi 1:10:19 I'm for this equations. Is it possible if we actually solve problems like with actual numbers, not just see it like it's? Right now we're studying the formulas like we're studying theory, like memorizing it. Richard Zhao 1:10:29 Here. Yeah, yeah. Farkhunda Ateel Siddiqi 1:10:34 But if we actually solve these formulas with numbers and examples, I think it would be easier for us. Is it possible if we could see, I don't know if you could record umm, if you record yourself solving few videos or you we can see some videos from YouTube and then we can even have like homework. Like if you could give us homework like three or four people each day, we can volunteer and we will try to solve them and try to understand them. If it if it's a good idea. Richard Zhao 1:11:08 OK, so actually this is just the the review of the contents we covered you know in the in the previous classes we we had real examples like this. So but I I'll send you some questions after class and also we'll solve some questions after you know we review this previous contents. OK, uh, let let let let's look at this. Farkhunda Ateel Siddiqi 1:11:31 OK, OK. Richard Zhao 1:11:34 Listen for example. To calculate the approximate attenuation. If a format the first ultrasound. Uh, with this intensity, it's applied to a soft tissue surface. What is the intensity? 1.5 centimeter into the the tissue. So now we'll actually do do this, OK. Farkhunda Ateel Siddiqi 1:12:02 Yes. Yeah. Richard Zhao 1:12:21 So anyone wants to to try? Farkhunda Ateel Siddiqi 1:12:29 Uh, Doctor Richard, if you could try this one yourself and then we we would go after you. Richard Zhao 1:12:30 Yes. Farkhunda Ateel Siddiqi 1:12:33 Maybe another for another example. Richard Zhao 1:12:36 OK. Umm. So how how? How how do we do this? We just, you know, uh. The the question asks you what's the what's the intensity? 1.5 centimeter into the tissue, the the first thing you need to calculate is the actual attenuation. Uh, what's the attenuation? And then you can calculate the the final intensity right? So first we will calculate the attenuation, so we use the formula shown here. So that would be 0.5. What's that? What's up? Dear. 010.5 Ah, I'm frequency, which is 4 megahertz. Then you time the distance which is 1.5, so that would would be the attenuation in decibel. So the number would be 3. You can see here 3. That's a bells. So after 1.5 centimeters of traveling into the tissue, the attenuation would be 3 decibels. We know that. Three, that's bells. Meaning they in intensity would be decreased by a factor of two, meaning that the the remaining intensity would be 1/2 of the original intensity, which is, you know, this one is 10. So the the, the the remaining intensity would be 5. There was per square centimeter. In that, OK. Farkhunda Ateel Siddiqi 1:15:04 We didn't use that 10 uh, that the the number 10, then 10 city and the equation. Enas Adil Abdelhai Salih 1:15:05 Uh it? Richard Zhao 1:15:14 So this this is the original the original intensity. Farkhunda Ateel Siddiqi 1:15:19 OK. Richard Zhao 1:15:19 So the question asks you the final intensity, so the intensity after the attenuation, so we know the attenuation is 3 decibels, meaning that the intensity would be decreased by a factor of two after they attenuation. OK. Farkhunda Ateel Siddiqi 1:15:30 Umm. Richard Zhao 1:15:42 Remember, what is the decibel? Arash Jamshidi 1:15:43 And excuse me, the final the final intensity would be 5. Enas Adil Abdelhai Salih 1:15:44 So the final result will be. Five, OK. Richard Zhao 1:15:49 Yeah, five. So because you know 3. Enas Adil Abdelhai Salih 1:15:52 So we we the because it's decreased 2 times, yeah, you mean like that? Richard Zhao 1:15:58 But no, no, no, it's not decreased to time. This is one way because the question ask you the intensity actually. After uh, the sound traveling 1.5 centimeter into the tissue. So at this location, what is the intensity so the the the the attenuation only happens one way. In this case, they're not not asking you if you're doing the imaging of a structure with the depth of 1.5 centimeters, so they're asking you what is the intensity 1.5 centimeter into the tissue at this location? Uh, what? We what will be the intensity there? Enas Adil Abdelhai Salih 1:16:49 I mean that we have the attenuation and the attenuation. Richard Zhao 1:16:54 Umm. Enas Adil Abdelhai Salih 1:16:55 It was three disabled and the three disable. Richard Zhao 1:16:58 Yes. Enas Adil Abdelhai Salih 1:17:00 The intensity will reduce by factor 2. Richard Zhao 1:17:04 Yes. Enas Adil Abdelhai Salih 1:17:06 And then so I will uh, divide 10 by 2. Richard Zhao 1:17:11 Yes, that would be the the the final intensity. Enas Adil Abdelhai Salih 1:17:20 OK. Richard Zhao 1:17:20 You know for three decibel, if you increase the intensity by three decibel it means that you know you increase the initial intensity by a factor of two. If you decrease by three decibel, it means that you decrease the original intensity by a factor of two, which means that you divide uh. They are really, really low intensity by two. In this case it would be 10 / 2 it which is 5. Enas Adil Abdelhai Salih 1:17:54 While funds. Umm. Arash Jamshidi 1:17:59 Excuse me, Doctor Richard, in the form. Richard Zhao 1:18:00 Yes. Arash Jamshidi 1:18:03 You the the the content of parenthesis next to a half one second uh. Decibel per centimeter minus megahertz. It's it's this parenthesis is a little bit confusing. Richard Zhao 1:18:16 Yes, yes. Arash Jamshidi 1:18:20 What does it mean inside the parenthesis? Richard Zhao 1:18:22 Uh, actually, it's means so this one. You, you, you. You don't have to know. This is actually per centimeter per megahertz. Uh, so this is the uh, you understand this as the attenuation coefficient. You only need to know you know you multiply this, which is 0.5 by the number of megahertz. Say if you have a four megahertz transducer, you multiply by 4 here and then the distance that the uh, that's the sound travels. You know the actual distance of the attenuation. Arash Jamshidi 1:19:01 So is, is it meaning that the one second or half or 0.5 is the constant number in the formula or it is going to change for each example? Richard Zhao 1:19:14 So so this this is this is constant constant for stop tissue, you know in soft tissue they the number is constant. Uh, it's per centimeter per megahertz. So now you hear on this part you you you need to have the number of the megahertz and the number of distance. So for example, for higher frequencies, let's say 6 megahertz, you use six to multiply by this, because for each megahertz, uh. And each centimeter you have a number of 0.5. So you use this number multiply to multiply the number of megahertz and the the distance. Arash Jamshidi 1:20:06 OK, it means that, uh, in other questions, they will give us this number for any, any specific tissue. Richard Zhao 1:20:16 Yeah, they have to tell you. You know, they have to tell you, uh, the frequency of the of the sound and also the distance. The sound travels. Mario Tool calculate the the attenuation. OK. So this this one is for each megahertz, you know if you don't use this formula you you, you, you, you if you want calculate the umm attenuation coefficient for each frequency first that's that's also OK you for example for a 4 megahertz transducer you first you calculate the. Attenuation coefficient for this frequency, that would be uh 0.5 * 4. So you get a 2, you get 2 as the attenuation coefficient for if frequency of four. So then you multiply by this coefficient, uh, multiplied by by the distance. That's another way to do this. So that's the definition of attenuation coefficient in the crime core book. Uh, but the in the Frank Miller book you just you you can see the the, I'm 10 years coefficient. As you know, 0.5 per centimeter per megahertz. So you don't have to calculate the different attenuation coefficients for each frequency first, you just use this formula to calculate. Lamia Elkalifaabdalla 1:22:15 Excuse me, I have one question. Richard Zhao 1:22:16 Yes. Lamia Elkalifaabdalla 1:22:18 If the answer of three decibel, if we got another number rather than one and three and 10. Like for one, it won't 11 point 2.5 and for three you said two and for 10 is 10. If we got three like not three or not one or not 10, what should we do? Richard Zhao 1:22:39 Yeah, with. So you usually in that case they would give you a table so you can look look into table as as we said you know previously umm, but you can actually if you want you can calculate by yourself. Lamia Elkalifaabdalla 1:22:53 OK. Richard Zhao 1:22:57 For example, if you get. Who does it bells, right? Lamia Elkalifaabdalla 1:23:04 Yeah. Richard Zhao 1:23:05 If the attenuation is too decibels, you calculate the actual power ratio of two decibels. I told you. You know you can consider two as 1 + 1. You know the parish of 1 Decibel would be 1. Would be one ohh what happened? 1.26 Another you can another uh coverage of one that's got. Which would? Which would be another 1.26. So this would be the power ratio of two decibels. And then now you have this number. You divide because the the initial intensity would be decreased by a factor of this. So if you know the initial intensity, let's say is this one, you divide it by by this number you will get the final intensity. OK, but usually they want they they won't ask you to to do this kind of calculation because it takes time. Lamia Elkalifaabdalla 1:24:16 OK. Thank you. Yeah, and. Richard Zhao 1:24:24 But yeah, so you're really it's very, you know the number would be easy to calculate like 3 something 10. Lamia Elkalifaabdalla 1:24:32 OK. Thank you. Richard Zhao 1:24:33 No problem. OK, so I think we can have a 15 minutes break, OK. Omid Ahmad Mohammadi 1:24:46 OK. Anahid Modaresi 1:24:49 OK. Thanks. Kateera Mustafa 1:24:51 OK. Thank you. Ying Cui 1:24:53 And killed. Richard Zhao 1:41:07 OK, uh, getting back. Anahid Modaresi 1:41:17 Welcome back. Richard Zhao 1:41:19 Yes. Not see. And now here it shows you the steps that you can you can you can take to through these kind of questions. OK, you can uh, do it ording to the steps. You know, this is the criminal Bob Book UM showed you this way of doing. So let's do another we will spend more time on this. OK, let's do another question. OK. In soft tissue, what is the attenuation of a bio megahertz? Trans ocean beam at depth of 10 centimeter. Try to do it. Uh, don't look at the you know how they they they did this. So this one uh? Uh, quite straightforward, right? So you have this number 0 point file times file which is uh, frequency 5 megahertz, then uh, multiplied by 10, which is the distance. Farkhunda Ateel Siddiqi 1:42:52 Yeah. Richard Zhao 1:43:05 You get 25 decibel the generation. Bana Otaki 1:43:12 Doctor Richard, do we have to memorize like a the attenuation for like a soft tissues or like stiff surface or do we have to memorize this like so we will know how to solve such problems? Richard Zhao 1:43:29 You mean this number? Bana Otaki 1:43:31 No, I mean like how? That it will sound beam like how it will. How is it in soft tissues or on stiff surface? Or like the formula we saw before, do we have to memorize it so to help us to solve this or or what? Richard Zhao 1:43:53 Uh. Farkhunda Ateel Siddiqi 1:43:54 I think so. Yeah. Yeah, that's the 0.5 we have to memorize it. Richard Zhao 1:43:58 Yeah. So for this specific number you you need to memorize. Farkhunda Ateel Siddiqi 1:43:58 That's for soft tissue. Richard Zhao 1:44:01 You need to know this is for soft tissue. The attenuation coefficient in soft tissue would be this one. Another number. You you you better know. Is that the attenuation coefficient muscles which is 11 decibel per centimeter per megahertz, that one you better also memorize? Umm other than these two, you you you don't have to know. So the. Bana Otaki 1:44:31 Because because there are a lot of formulas and a lot of information and the during like they listen sometimes we understand that we don't have to memorize this. We don't have to memorize that, just to know if you just could you please like point out to us which formulas are like the one we will always use and which one we have to memorize and which ones are secondary, just like general understanding. Richard Zhao 1:44:41 Yes, that's. Bana Otaki 1:44:59 So that will help us if you should highlight to us the formulas we will always be used and like the table, just like this attenuation. Richard Zhao 1:44:59 Uh, yeah. OK, so. Umm. Umm. Bana Otaki 1:45:11 0.5 disable whatever so we will know what to memorize. Richard Zhao 1:45:11 OK. Umm. Umm OK so. Bana Otaki 1:45:16 Thank you. Richard Zhao 1:45:20 OK. We'll we'll do this later. I'll tell you, you know. UM, the formulas that that you need to actually memorize. OK. Yeah, but in my slides, you know, a lot of information you is for you to understand this. Bana Otaki 1:45:28 Yes, thank you. Richard Zhao 1:45:35 You don't have to actually memorize those so so it's formulas that so I'll tell you. Bana Otaki 1:45:39 But we don't. We don't know that we get over when it's too much and it's like a just a doesn't make sense, at least for me. Richard Zhao 1:45:44 Umm. Bana Otaki 1:45:49 So if you just tell us what your memorized, we will concentrate on the formulas that will help us to solve the exercises. Richard Zhao 1:45:49 Umm. Mm-hmm. Bana Otaki 1:45:57 Thank you. Richard Zhao 1:45:57 Hmm. OK, OK. Ohh I'll do this after class. OK. I'll umm specify you know what's the permanence that you need to memorize? You know in in the previous classes, so I'll do this after class. Let's look at the. And also for this kind of questions you need to pay attention to the wording, so if they're asking you, they when they attenuation or UH-2 way attenuation for example for. For this question, for this example. Uh. They ask you in soft tissue what is the attenuation of megahertz ultrasound beam imaging to a depth of 10 centimeters. So as long as you see the word imaging, it means that you know the sun has to, uh, reach the the structure 1st and then the echoes need need to come back to the transducer for the system to actually imagine the the to image the the the structures. So this is always mean means that you need to do the two way attenuation so that that let's do this one OK. Hamid Reza Sharif mousavi joined the meeting Richard Zhao 1:47:59 OK, you know you you see attenuation coefficient. Uh hi and say frequency, which is π. And then in this case, uh, you use two times this steps because now is 2 way attenuation. So the the the sun needs to reach the depths here and then the echoes need to come from this steps back to the transducer. That's two way attenuation. You use 20 instead of 10 here. Hey. So we know that there are. 3 contributing factors for attenuation? Uh, so one is reflection so. Uh, we talked about what's reflection, you know, and the type of reflection that occurs at an interface depends on the wavelength relative to the surface, the diameter of the surface. Uh, if the surface is large, smooth and flat, we consider the reflection as a specular reflection. Uh, if the service is rap, we consider it as scattering and the if the reflecting structures are small. Umm, like the red blood cells which is very small relative to the wavelengths. We consider this as relay scattering. So this is showing you in this case. This is a large and smooth flat surface. You know, this is considered as specular reflection. Uh, this surface is rough. Uh, you can see that the reflection can be, you know, the beam can be directed to different directions. It's not towards One Direction, so you consider this as scattering. Uh, so scattering is less angle dependent than the specular reflection. Or this is the case of uh, release scattering. You know the the, the, the, the, the, the the particles vary. The reflectors are very small relative to the wavelength, so actually the uh reflection would be, you know, directed to all of the directions. Anahid Modaresi 1:51:08 I have a question. Doctor returned Umm in in reality in human body most of the surfaces are rough year and we have umm and the second is scattering right second reflection. Richard Zhao 1:51:10 Yes. Yeah. So actually, you know, in the solid organs, uh, you can consider there are many small rough surfaces. So the reflection happening in the pragma of the organs we consider as the scattering, it's only at the very, you know, significant. Uh. Interfaces between organs like the diaphragm diet diagram. We consider that as a specular reflection and also you know the for example the reflection happening at the interface of the tissue and bone and are, uh, let's say the the, the uh large vessel walls. Sometimes we consider that as specular reflection, most of the reflections in human bodies would be scattering. Anahid Modaresi 1:52:24 OK. Thank you. Richard Zhao 1:52:30 And we have different services, specular scattering surface, which is rough. Umm for the specular reflection? Umm, we we talk it we we talk about this in different scenarios. One is when they. Instead, when when they beam is not perpendicular to the surface, meaning it's a oblique incidence. In that case, there could be a reflection. There could also be reflection if you know the propagation speeds in the two media is different. There would be refraction if the beam is perpendicular to the to the interface. There won't be refraction. Refraction means the the the, umm, the parent of the sound beam. So if the beam is perpendicular to the surface or or the interface, there won't be refraction. Uh, there would only be transmission and reflection. How much of the energy would be transmitted and how much would be, uh, reflected. Uh, just depending on, you know, the impedance mismatch of these two media. So this is showing you the case of perpendicular instance. Enas Adil Abdelhai Salih 1:54:00 So ohh excuse me? In oblique and they are oblique. One, there is no transmission. Richard Zhao 1:54:11 And there is transmission. So you can see there are regression is a type of transmission. It's only the direction of the beam is change. Enas Adil Abdelhai Salih 1:54:21 OK, OK. Richard Zhao 1:54:22 But it is. This is also a type of transmission, OK. Enas Adil Abdelhai Salih 1:54:27 OK. OK. OK. Richard Zhao 1:54:32 So this showing you the the case of perpendicular instance. You can see there's reflection. Uh, and there's no refraction. There's transmission, but no refraction. Umm, scattering happens in two situations, ones that you know the drop surface. Another case is when they reflectors are very small. So this is the same thing. What? What's you know, perpendicular instance is when. The emails perpendicular through the interface. In that case, the pause is partially reflected, with the remainder continuing into the second medium. Then you know how much of the energy would be reflected and how much uh there's transmitted is depending on the impedance mismatch of these two media. So in this case, the transmitted sound does not change direction as it moves into the second meeting because it's perpendicular instance. Uh. Again, the strengths strengths of the reflected and transmitted policies are determined by repentance of the two media at the boundary. So then we introduce the concept of impedance. Uh, we said that impedance is a property of a medium. So the mismatch of the impedance in the two media determines how how much of the incident sound would be reflected and how much it's translated into the second medium. Uh, you can see here. You know the unit for impedance is this one rails. Actually this is from the name really. You know the really scattering. This is the same person. Uh, impedance, we use the formula. Uh to calculate the impedance by uh multiply density by propagation speed. Umm actually this this this way of defining the impedance is called characteristic impedance acoustic characteristics, acoustic impedance. OK. You guys see that? Uh. The higher the density, the higher the impedance, the higher the speed, the higher the impedance. And the introduce another important concept, which is, uh, the intensity reflection coefficient. Uh, so it it, it means you know how much of their energy would be reflected. So this is the percentage, OK, uh. So we know that for perpendicular instance, if the impedances of the two media are the same, they're they're there, won't be any reflection. All of the energy would be transmitted into the second medium if there is a very large difference. Uh. Meaning impedance mismatch between the two media. There would be nearly total reflection in that there barely any energy would be transmitted into the uh, second medium. Almost all of them would be reflected at the interface. So that's why. We we we need to use coupling medium like like the gel we use when we scan the patient. So the purpose of the gel is to reduce the the the error between the transducer and the skin because you know the impedance mismatch between air and the soft tissue is very significant and the question. Bana Otaki 1:58:48 Sound. I don't know. They had to come. Very good. Mr Cutler came goodbye. Anahid Modaresi 1:59:02 I didn't get there. Impedance of medium. Richard Zhao 1:59:06 And he does or what? Anahid Modaresi 1:59:07 I impedance in the medium what it is it exactly? Richard Zhao 1:59:14 Umm, OK, I'll explain later. So you have different ways to be banned impedance, but you know for like you said for for your exam purpose you only need to remember uh this some someone said to you you you don't want to be overwhelmed. You know the impedance. Actually you have different ways to define impedance. Anahid Modaresi 1:59:32 Umm. Richard Zhao 1:59:34 Uh, so this one is what you need to remember. This is called characteristic. Anahid Modaresi 1:59:39 Umm. Richard Zhao 1:59:42 Of course they can be dance. You also have a specific acoustic impedance, which is depend a different way. Umm, so that one helps you. You know how much pressure you need to have to cause uh, they particle to move at a certain, you know, speed. Anahid Modaresi 2:00:06 Umm. Richard Zhao 2:00:07 So if the pressure you need is high, it means that the impedance is high, so you need to you know that for the sound to propagate in the medium you need to, uh, you, you you need the particles in that medium to move. Anahid Modaresi 2:00:13 Umm. Yeah. Richard Zhao 2:00:26 If you cause the the movement of the particles to a certain speed, you have to apply how much pressure that defines the specific impedance. Anahid Modaresi 2:00:35 Umm. Richard Zhao 2:00:41 So uh, probably there. Anahid Modaresi 2:00:42 Umm. Richard Zhao 2:00:43 There's a slight later you can you can see that. Anahid Modaresi 2:00:47 Ah. Richard Zhao 2:00:56 OK, so this one you need to know this formula. You need to remember, OK. Anahid Modaresi 2:01:04 OK. Richard Zhao 2:01:06 And this one you also need to know that there will be some calculations using this one, uh. Anahid Modaresi 2:01:13 Umm. Richard Zhao 2:01:13 How do you calculate the? So. So they, they, they, they intensity reflection coefficient is calculated by another uh formula using impedance but this one usually is for you to calculate. Uh, they intensities like umm they give you they yesterday intensity and they gave you the two impedances of the two media. They ask you what would be they reflected intensity in that case. Uh, you need to calculate the intensity reflection coefficient first using the impedances and then you use this coefficient to calculate the reflected intensity. Because you are already know the incident intensity. Anahid Modaresi 2:02:08 Umm. Richard Zhao 2:02:12 Umm, so we have another concept which is transmission coefficient. You know, anything not reflected would be transmitted into the second media. So you just use 1 minus the reflection coefficient. That would be your transmit transmission coefficient. So this is a percentage, OK. So this is the actual equation you use to calculate the reflection coefficient. So you can see that. Uh, you. You you use the. Impedance in the second media minus the impedance of the first media and on the dynometer there's the sum of the two impedances. So this one you also need to remember. So because sometimes you you you have to use this one to calculate the actual. Reflection coefficient. So pay attention that this is the reflection coefficient for the intensity, so that's telling you how much of the intensity would be reflected at the interface. It's not the pressure we have if we don't square this one, that would be the. Pressure reflection coefficient. So this one is the intensity reflection coefficient. Are you getting me? Enas Adil Abdelhai Salih 2:03:53 Ohh actually no sorry. Richard Zhao 2:03:56 So this is telling you how much of the intensity would be reflected at the interface. So that's why it's called intensity reflection coefficient. If you don't square they this this number, this ratio, if there's no square, this part would be your pressure refreshing coefficient. It's telling you how much of the pressure is reflected, reflected at the interface. Enas Adil Abdelhai Salih 2:04:30 Pressure. Refraction. Richard Zhao 2:04:33 If pressure reflection coefficient, but that that's fine. Enas Adil Abdelhai Salih 2:04:36 Thanks. Richard Zhao 2:04:37 Again, you don't need to remember. The only thing important is the intensity reflection coefficient in the exam. OK. Tamana Asey 2:04:46 But Professor would surface. Are you talking about? Is it the second medium? Which medium? Enas Adil Abdelhai Salih 2:04:51 And. Richard Zhao 2:04:53 It's any any interface so you have you for example in the human body you have two different tissues, one for example is fatty tissue, the other potential, uh let's say is soft tissue. Tamana Asey 2:05:03 Superficial. Richard Zhao 2:05:09 So we know that they impedance uh in the two tissues. You know there's different. So when you have a interface, there would be reflection and there would be ohm transmission. So when the beam is perpendicular to the interface, you can use this kind of formula to to to to calculate how much of the energy would be reflected at that interface. But if the the beam is not perpendicular perpendicular to that interface, meaning that it's a oblique instance, you cannot use this one. There is another formula for that if there's an angle, you know it's not perpendicular to the interface. There's another formula to calculate that, but you don't have to remember that there won't be. You know, questions asking you to do that kind of calculation. Tamana Asey 2:06:03 Got it. Thank you. Richard Zhao 2:06:04 And no problem. Enas Adil Abdelhai Salih 2:06:06 Thank you. Richard Zhao 2:06:07 Uh, if you look at this formula like we said, if the impedance impedances in the two media you know are are the same are the same meaning that you know the the the difference is 0. In that case, there won't be any energy reflective at that interface, meaning that all of the energy would be transmitted into the second medium. So for reflection to happen, there has to be some difference between, you know the the impedances of the two media. Otherwise, all of the energy would be translated into the second medium. And also view like we said, if you look at the formula here, uh, if the difference between. Between the two. Uh, impedances is very large, so let's see this one impedance 2 is very big and this one is close to 0. So in that case you can neglect this one, so the result would be very close to to one, because this one is very small, so very close to one, meaning that almost all of the energy would be reflected. It so there's no barely any energy would be transmitted into the second media. So this is the case. Uh, so when you have, you know, large impedance mismatch happening at the interface of let's say air and tissue or tissue and bone? Uh, in that case, you know the the impedance mismatch is very big, so almost all of energy would be reflected at that interface. Bana Otaki 2:08:13 So talk to Richard. It's in the example like 1000 -. One is almost one 1000 right. Richard Zhao 2:08:21 Yeah, almost myself. And yes, if you divide, if you divide by another 1000, it's close to very close to to one. Bana Otaki 2:08:22 Yeah, that's what you are saying, yeah. Richard Zhao 2:08:42 Uh, let. Let's do some. Some uh actual calculation regarding this? OK, for the the first one, uh, for impedances of 40 and 60 rails determine the intensity reflection and the transmission coefficient. So you you pick up your pen and do this, OK? Very straightforward, right? So you use the formula. You know the. So the difference of the two impedances, the sum of the two impedances because there is a square, you know it doesn't matter if we use which number to Manus switch. So this is the result and for the transmission, remember that this formula is for reflection intensity, reflection coefficient. If you want to calculate the intensity transmission coefficient, you need to use one and minus, uh. The reflection coefficient that will be our transmission coefficient. So let's look at this one. So they ask you if they instantly intensity is 10 milliwatts per square centimeter. Calculate the reflected and transmitted. It's intensities. So for this kind of question, first you need to calculate the coefficients and then you use this number. The original intensity. Uh, multiply multiplied by the different coefficients to calculate the uh reflected or transmitted intensity. Locate for this one. Bill Barr? Samina Gul 2:11:02 Talk to Richard. Zhichao Wen 2:11:02 Yeah, yeah. Richard Zhao 2:11:03 Somehow, yes. Samina Gul 2:11:04 So in the second example, like as you said, then sit and intensity is 10 milliwatts. Richard Zhao 2:11:08 Yes. Samina Gul 2:11:12 So far this we have to solve, we have to know about the intensity transmission coefficient and reflection coefficient. So all these values will be given in the examples in the question like that. Richard Zhao 2:11:21 Yes. Samina Gul 2:11:26 This will this media has this intensity relation reflection coefficient and this has intensity transmission coefficient. Richard Zhao 2:11:30 Uh. Samina Gul 2:11:33 Now find out the incident intensity is at 10 milliwatt and calculate the reflected and transmitted intensities. Richard Zhao 2:11:41 Sometimes they they they, they they will tell you they will tell you what's the, let's say the transfer reflection or transmission coefficient. Sometimes you need to calculate by yourself like in this question. Uh, they tell you that they yesterday you said would be 10 and also, you know the impedances of the two media would be 40 and 60 and they they they don't tell you directly the coefficients, they tell you the impedance. Samina Gul 2:12:13 OK, so we have to solve first, right? Richard Zhao 2:12:15 Yes, you have to calculate the the coefficients by yourself and then you calculate the reflected and transmitted energy the intensity. Samina Gul 2:12:15 Right. But it, yeah. OK. Thank you. Richard Zhao 2:12:28 No problem. But you know, in most cases the number will be very easy for you to calculate the numbers. You know, like this is very easy to calculate, right? Bishnu Shrestha 2:12:40 Uh, Doctor Richard the impedance will be given or we have to remember. Richard Zhao 2:12:46 Sometimes you you're sometimes they. They ask you even to calculate the impedances they give you the density. They give you the propagation speeds. You calculate the impedance first and then you calculate the coefficient that. That would be very complicated. So usually they won't ask you to do that. It takes a lot of time. Bishnu Shrestha 2:13:07 So in this case. Richard Zhao 2:13:12 Most likely they will tell you the impedances and the initial the intensity and ask you to calculate the reflected and transmitted intensity. But you know, uh, you can actually calculate the impedance impedance by, you know using that formula. You know, if it is equals density times propagation speed, right? If they tap yes. Samina Gul 2:13:40 But Doctor Richard is there enough time in the exam so we can calculate all these things because these require the time. Richard Zhao 2:13:46 Uh, yes. Like I said, they're they're usually there, won't be very complicated calculations in the that like there's calculation usually is very straightforward. The calculation itself won't be that complicated. Umm, the the most likely questions would be, you know, asking you the relationship between different parameters. Umm, they're not a lot of calculations actually. So, like uh, they will ask you, for example, uh, if uh, the frequency increases, what happens to the other parameters like, you know, wavelengths, SPL, uh, like, uh pause duration something like that. So you, you, you, you, you, you in most cases you don't have to actually calculate. But there will be. Sometimes there will be some questions. Very you how to do some calculation. Like I said, there different ways of defining the acoustic impedance. You know the one, the formula that you need to remember. Umm, it's the one I showed you previously which is called the characteristic acoustic impedance. So this way of defending the impedance is called specific acoustic impedance. So this one you don't have to remember, but it's good for you to understand what is actually the acoustic impedance. So it's actually it's it, it measures the response of the particles of the medium in terms of their velocity. Uh. Who are given pressure, you know, to, to, to, to, to cause you know, this much of velocity of the particles. How much pressure you need to apply to the particles. So that's that. That's called, you know, specific acoustic impedance. If you look at this. Uh, let's say you know the larger ball. Larger balls mean me, me, me, me. Meaning that you know they have larger, uh, density, smaller ones, meaning that they have lower density. Uh, so the thinner? Uh. A sprain means that you know the stiffness is low, so the thicker spring means that you know the stiffness is yes, it's it's higher. So you you can you can figure out if you wanna cause this kind of particle to reach. To push this up particle to a certain velocity cause it, uh, move at a certain speed? Umm, so you you you need a smaller pressure in this case if the density is high. If the stiffness is high, you need more pressure. Uh to generate a similar velocity of this particle, right? But in that case the impedance of this media is high. The impedance of this kind of media is low. So we can actually define. The impedance as this the square root of the product of density and stiffness because you know the if it's if it's the depth is high. The thing is high it's harder to to move the particles. You need to apply more pressure to cause a certain velocity of the particles. If you can remember, when we define the speed of sound, we use this formula. It's the square root of steepness divided by density. If you combine these two uh, you can get the formula for impedance we we want to use. So this one you have to remember all the other ones is just help you to understand. You don't have to actually remember the formula, but this one you need to remember, OK. So this is showing you you know the process, how how how you understand, you know where is this one coming from? Tamana Asey 2:18:53 Professor, can you please explain what I see and what is PNC again, OK. Richard Zhao 2:18:59 This is impedance. OK, we use this to represent impedance. This one is density. This one is stiffness. He is thickness again, impedance, density propagation, speed of the sun. Tamana Asey 2:19:09 OK, good. Thank you. Richard Zhao 2:19:14 OK, let's go. Yes. Tamana Asey 2:19:16 OK. Thank you. Richard Zhao 2:19:17 So this one is the pressure you apply on the particles. This one is the velocity of the particle movement. This one against impedance. Bana Otaki 2:19:35 Sorry, Doctor Richard, what is the V? What is the V stands for? Richard Zhao 2:19:39 This one is the velocity of the particle you want to cause the protocol to move at a certain speed. Bana Otaki 2:19:47 I still. I still. I still don't. I still don't get the word. What is it velocity? What does it mean? Richard Zhao 2:19:55 It means speed. Bana Otaki 2:19:56 Yeah. Richard Zhao 2:19:57 The difference between velocity and the speed is that you know for velocity has a has a direction. If you talk about speed it, it doesn't have to have a direction but velocity, velocity. There, there has to be a a direction, OK. Arash Jamshidi 2:20:14 Excuse me, Doctor Richard, what is the difference between the you. Richard Zhao 2:20:16 Umm. Arash Jamshidi 2:20:20 You told that the K stands for the stiffness but. Richard Zhao 2:20:25 Yes. Arash Jamshidi 2:20:26 Uh, in the past you we had another formula that told that stiffness is the same as balk module and it's B. Richard Zhao 2:20:37 Yes, yes, this one is yes, this. Arash Jamshidi 2:20:39 What is the difference between B&K? Richard Zhao 2:20:42 Uh, so same, it's just how you you use what? What kind of uh, you know, uh character to represent that that that, that, that, you know, bulk modulus. Here's the key it's also block modulus. OK. Arash Jamshidi 2:20:59 OK. Richard Zhao 2:21:00 This is bulk modulus. You consider it as stiffness. You you you don't need to differentiate between. Pop modular says stiffness. You consider them as same. Same thing here. Tamana Asey 2:21:19 Got it. Thank you. Richard Zhao 2:21:20 OK, no problem. Then you know I don't want to overwhelm you. So, uh, so this is, you know, this is just showing you you know you have different ways to define uh. They of course they impedance. So if we defend the acoustic impedance using this way is called specific acoustic appearance. You know that if the pressure you need to apply on this particle to to to generate a certain speed of the particle, uh, if the number is large, you you you see that the acoustic impedance is is is higher. Umm. By knowing this you can you can. Uh. I know that what matters to the impedance is that you know the density of the medium and also the ohh steepness of the media. If the stiffness is high, uh or the density is high, you need more pressure. Uh to cause the particle to move at a certain speed. So that's why we get the formula. Here we define you know the impedance as the square root of the product of density and the stiffness, uh. If you combine these two formulas, this is for the. Uh propagation, speed of sound. The formula for the speed of sound, so we use this one. You can get this one OK. This one you need to remember. Density and propagation speed. Ohh, that's about the normal instance, meaning the beam is perpendicular to the interface and So what happens to oblique instance? Meaning that the beam is not perpendicular to the interface. Uh for a smooth flat reflector, meaning specular reflector? Uh, the reflected sound will not return to the transducer. It will be directly to another way and. The system will not receive echoes from that reflector. OK, in that case. Umm in in this case, if there's an angle between the beam and the normal line, meaning that it's not a perpendicular instance. Uh, there could be a refraction of the beam. Uh, meaning that the the the beam, the direction of the beam would be changed in the second medium. That happens when they propagation speeds in the two media are different. So if the propagation speeds of the two media are the same, uh, in the two media, there won't be refraction. So there's transmission, but the direction of the beam won't be changed. So it's only, uh, when the propagation speeds into media are different. Uh, that you know refraction will happen. Or the these two angles the instant instant angle and the reflection angle. These two angles are all always the same. They are. They are equal to each other, OK. And we talked about the, you know, you have two ways to defend the incident angle. One is at the angle between the instant the, the, the, the sound direction and the normal normal line is a line you draw perpendicular to the interface. Another way is that you can consider the incident angle as the angle between the wave front and the the interface. The the view front at is specific spot. You can consider the view front is perpendicular to the beam to the direction of the beam. So far, the transmission angle. How how can you calculate the transmission angle? So we need to know the two speeds in the two media. So we use a lot called snails log to calculate uh they transmission angle. If you already know the incident angle and you know the two speeds in the two media, you can calculate the transmission angle. OK, if the speed in the second medium is faster than the speed in the first media, the transmission angle will be larger than the incident angle. If the speed in the second medium it is lower than the speed in the first media, they transmission angle would be smarter than the incident angle. So this is a snail's log. Log umm, the actual formula is like this, so the ratio of the two propagation speeds in the two medium is actually they ratio. Umm. Of the sun's uh of instant angle and the sun of transmission angle. Umm, you know if the the angle is small enough you can use the actual number of the angle to to to do this kind of calculation you don't need to use the signs, OK. Where he said that. But yes. Enas Adil Abdelhai Salih 2:27:42 Oh, sorry. Sorry. Sorry, excuse me. You. You said we don't need to use the sign just to calculate. Richard Zhao 2:27:50 Yeah. Enas Adil Abdelhai Salih 2:27:51 Oh, good. Richard Zhao 2:27:51 So, because that's that would be the result would be pretty close if the if the angle is small. Enas Adil Abdelhai Salih 2:27:52 Thank you. Umm. Richard Zhao 2:28:08 So what's the refraction? As we said, is the change in direction, uh. And the refraction can cause a lot of problems. We do scanning, so for example. Uh, so the structure could be. Registered at a different location than where it should be. OK. Literally. Uh. Displaced. So far refraction to happen you have to how first oblique instance means that you're beam is not perpendicular to the interface. Another condition is that there has to be different propagations speed in in in the two uh media. Bana Otaki 2:28:59 Doctor Richard, where for example, in the body we will have refraction more like than in which which organs more than the others. Richard Zhao 2:29:09 Uh, actually, you know. Umm, there's there's a lot of births. Uh, refractions happening in a very small scale, you know. Actually in, in in the tissue, the small structures where the the appropriate speeds are are not the same. Umm. Refactoring actually happens everywhere in the body. But uh, if you consider the most significant refraction, uh, and they're just some cases. If you have, let's say if you have a lipoma. So for which you know the program speed is not the same as the program speed in soft tissue. Ah, or any fatty layer. So between the fatty layer and the soft tissue, the there would be refraction and also a very common case is the uh the rectus uh abdominal. You know the the, the, the, the. Uh, rectus abdominis muscle, you know. You scan, let's say they Yota. Ohh. In the midline of the abdomen. So there could be refraction. So in that case, there could be a artifactual aorta. You know, other than the you you could see two air tasks. So one is real structure, the other one would be a artifactual. Uh Erica by uh. Caused by reflection. So you this. So this is the rectus muscle here. So you can consider this as a lengths when you know the sound beam you you're you, you put your transducer here. Uh, there could be refraction. The being like would be reflected to another direction here. Bana Otaki 2:31:38 So this refraction is like, uh, we'll make it more difficult for us to do, like, uh, the right. Yeah, I'll try sound. I mean, is it something negative, right we should avoid it. Richard Zhao 2:31:49 Yes. So so refraction can cause a lot of problems, so it can degrade lateral resolution. It can cause a artifacts, which means that you you see a artifactual structure in your image. If you understand, you know the common situations that you you could see they artifactual structures by, you know, the mechanism of refraction when you see it you, you, you, you, you understand that this is not a real structure. This is probably caused by refraction, so. Bana Otaki 2:32:28 You know, you know Doctor Richard, this way. Even you know it's simple with this simple drawing you did. Now it makes me understand much better the concept. Richard Zhao 2:32:37 Umm. Bana Otaki 2:32:37 So even with you, yeah, if you draw it like us. Richard Zhao 2:32:37 Dark because. Yeah. So yes, yes. Umm. Bana Otaki 2:32:40 Like if you treat us like a a children and even with this simple drawings the understanding is like a really much more thank you. Richard Zhao 2:32:45 Umm. No problem. So we're still at the very beginning of this course. You know, we actually we'll talk about this much more in detail when we talk about artifacts. So we we have a lot of artifacts. Umm, I don't worry too much, you know, at the beginning. So we're just introducing the very basics and the the concepts, OK. Let's see. Again, you know how if you know the two? Uh probably speeds in the two media and they incident angle. You can calculate the transmission angle. And also we introduced a the concept of critical angle let's say. Uh. If the propagation speed in the second medium is faster than the propagation speed in the uh first media, when you increase the incident angle so you increase this angle at some point they are transmission angle would be uh, 90 degrees at that. At that point, you know there there there wouldn't be energy transmitted into the second medium. All of energy would be reflected. So if you the increase the yes angle just a little bit more, all of energy would be reflected at the surface at the interface. This is called total internal reflection, so this angle is called critical angle. So, uh, these kind of questions you look could occur in our exams. You you need to know what is critical angle and how can you calculate the critical angle. How how do you do that? Uh, so we know that we we use the Snell's law. So now you know, uh, the propagation speeds of the two media and the, you know, when the critical angle occurs, they are transmission angle has to be 90 degree. So you put the sign up 90 degrees. Here you can get the sun of the critical angle. So here the sign of incident angle you can consider the sign of critical angle because this is how you know the critical angle is defined. Let me draw. Let me draw Draw Something. So so this is the first media. This is the second meeting. So we know the propagation speed in the second media has to be faster than the propagation speed in the first media for the critical angle to to occur. So when the speed is higher in the second medium, you know the transmission angle has to be larger than the incident angle. So according to the Snells law, when you increase the incident angle, the transmission angle has to increase as well. So at certain point, you know when you increase the incident angle big enough, the transmission angle would be 90 degrees. So this is the angle which can cause the transmission angle to be 90 degrees. This angle is called critical angle. So when this angle is rich, all of the energy would be reflected. There won't be energy transmitted into the second media. This is called total internal reflection and this angle is called critical angle. So because carding tool the snells law. Bana Otaki 2:37:45 Doctor Richard, disregarding like the surface, if it's like a stiff or or soft. Richard Zhao 2:37:53 Uh, it's it's no matter. It's a stable soft so so this is related to uh the oblique instance and also you know the the propagation speed in the second medium is larger than the propagation speed in the first media. So usually why this is important? Sometimes you know for the vessel walls. Uh, if your like, but let's say if it's a vessel wall. OK. Vessel wall. This angle we call Doctor Angle. If the doctor angle is small enough, sometimes all of the beam or energy would be reflected. There would be ohh total internal reflection in that case. There won't be any. Energy transmitted into the vessel. Uh, you're not getting any Doppler signal in that case. So that's when you know the critical angle is rich. Anahid Modaresi 2:39:05 So we're seeing very bright. Richard Zhao 2:39:08 Umm. Anahid Modaresi 2:39:09 We see it very bright in our image. Richard Zhao 2:39:14 Uh, sorry. Anahid Modaresi 2:39:15 We see we see actually because all of the energy is back to the transducer, we see it very bright. There was a wall. Richard Zhao 2:39:24 Uh, so not not right. So when that happens, when when you do doctor, you know if the critical angle is rich, meaning your double angle is too small. Uh, you top language is not in this language, so this is incident angle. This is doctor angle. If the doctor angle is too small, meaning your E angle is too large, so it reached the creative angle. All of the insulating. Some things would be reflected in that case. You're not getting a doctor signal from the vessel, from the blood flow, because there's just no sound transmitted into this vessel. But in case you are not getting, uh Doppler signals, so there's no spectral signals showing on your image. Bana Otaki 2:40:14 Because it's 90 degree, right? Because it's reached 90 degree. Richard Zhao 2:40:20 What? Bana Otaki 2:40:20 What? Richard Zhao 2:40:20 Degree. Bana Otaki 2:40:21 What? Because of the 90 degree, because you said if it's 90. Richard Zhao 2:40:25 It's not 90 degree. It's not that difficult because it reached the critical angle, meaning that you know all of the energy would be reflected at the interface between soft tissue and the vessel vessel wall. Bana Otaki 2:40:31 The young. Richard Zhao 2:40:42 There won't be sound transmitted into the vessel. The lumen of the vessel. So you cannot detect the the the the information. The velocities in the vessel. You didn't get me glass. Tamana Asey 2:41:04 So how can we avoid it? Richard Zhao 2:41:07 So in that case you need to increase your doctor angle, meaning that you need to decrease the instant angle so that you know the angle is below the critical angle. Uh, in that case, you know you you can avoid the total internal reflection. Some of the energy would be transmitted into the lumen of the vessel. Now you you're getting the signals from from the blood flow. Tamana Asey 2:41:38 OK. Richard Zhao 2:41:38 You got it. Welcome. Tamana Asey 2:41:38 Thank you. Yeah, thank. Enas Adil Abdelhai Salih 2:41:40 And in this case it has to the, umm, the philosophy in medium to it has to be more than medium one. Richard Zhao 2:41:53 Yes, yes, if it's smaller. If let's say if the program speed in the second medium is smaller than the program speed in the first media there, there won't be greater than angle, so the total internal reflection won't happen in that case. Enas Adil Abdelhai Salih 2:42:07 Umm OK, thank you. Richard Zhao 2:42:11 No problem. Uh, OK, let's see. Sorry. You know, when we talk about Doctor will uh explain more in detail regarding this. So this is about the critical angle. And then we signed out there a lot of uh effects of refraction. You can have ACD shadowing. You can have displacement of the structure. Uh. Naturally, it will affect uh your lateral resolution. And you, you, you, you will lose some signal intensity because of dispersion of the bin. Again, you know when we talk about artifacts, we'll explain more on this, yes. Enas Adil Abdelhai Salih 2:43:14 Excuse me, doctor Richard. I notice just now in our book that uh, given by by the college, all this it's not including on it so. I don't know if I got the previous version of of this one or or for all this one it's it's the, yeah. Richard Zhao 2:43:43 I so the information I am teaching here is from various sources. You know, I I read a lot of books and I put anything I think important here. So the the book, you know, the card give you is the slides used by Presley used by Doctor Fariborz. So I I I added something into this. Enas Adil Abdelhai Salih 2:44:08 Yes, it's it's not in our book. Richard Zhao 2:44:12 The which book you mean that this is from a lot of several books, I would say. Enas Adil Abdelhai Salih 2:44:14 He. Yeah. So we need you. We need the slides off of of this lecture or or all your slide. Richard Zhao 2:44:26 Yeah, I will upload you know upload that everything. I I I teach here so you can find them in, in my classroom teams. Enas Adil Abdelhai Salih 2:44:32 OK. Richard Zhao 2:44:35 OK, no problem, no problem. Enas Adil Abdelhai Salih 2:44:35 Thank you. Thank you. Kateera Mustafa 2:44:37 Thank you. Richard Zhao 2:44:44 Uh, OK, you know, you need to know refraction artifact results in lack of displacement of structures, but again, you know we we we will explain this later. Uh, OK, so you see if there's a a strong reflector in the second medium and the the speed in the second media is faster than the speed in the first medium. So after refraction, the beam would be directed to this direction and the echoes generated from this strong reflector would the return back to the transducer along the same pathway. Uh, like this? But you know the system the other system would consider any echoes returned to the transducer as coming from a straight line. So although the echoes are coming from actually this reflector, uh, the system would consider they are coming from here. So along the street line. So that's why you know this structure would be displaced laterally to this location because of refraction. OK. Arash Jamshidi 2:46:02 Excuse me. Doctor Richard. Richard Zhao 2:46:04 Yes. Arash Jamshidi 2:46:05 It's a little bit confusing. What? So if I why the transducer cannot calculate the the the the the refraction angle changes between the median. So if this one happens, it means that nothing in the transducer shows the exact measurement of the exact location of any particle, any anything. Richard Zhao 2:46:35 Yeah. So that that's the the reality because the transducer don't know if there there's a, you know difference in the propagation speeds. Uh. The transducer the system doesn't know. Uh, what kind of tissue? Uh, the beam? You know is going through. So the the only thing you know. Umm, it can can do is that it assume the the tissue is homogeneous, meaning that the space the speed of the tissue is the same along the propagation pathway. So the system will consider, you know any echoes returning back to the transducer as coming from a straight line, and that's how the system works. And and that's also how you know why the different artifacts. Uh, uh. January is so actually that's the cost because there's some assumptions by the autonomic system. Uh, which are not the not the reality. Uh in real life? So that's why artifacts happen. Uh, you're you're you're right. Yeah, you're right. That's not exactly the reality, but in most cases they are pretty close to reality. Arash Jamshidi 2:48:03 OK. Richard Zhao 2:48:09 So, but in some cases the difference would be very significant, so that would be our artifacts that that you need to be aware of. Arash Jamshidi 2:48:20 OK. Richard Zhao 2:48:22 Our design is not perfect and there are a lot of. Artifacts which you know is not the real structure in the human body. What? What do you see in the image? Sometimes it's not real thing that you need to be aware of. Arash Jamshidi 2:48:43 OK. Thank you. Richard Zhao 2:48:44 No problem. So this showing you the edge shadowing. So it's caused by refraction. You don't have to know this. The the You're the only thing you need to know is at shadowing. Is caused by refraction. Umm, here again we are talking about scattering and the scattering is the redirection of sound in many directions. Uh by either rock surfaces or, uh, heterogeneous media? Or where is small reflectors. So like we said, you know, in the solid organs in the human body, you know, the reflections will be considered as scattering. Bishnu Shrestha left the meeting Richard Zhao 2:49:56 But at you know. Some very. Strong reflecting borders between organs like diet. Uh diagram. Uh. Like between soft tissue and bones, we consider as ohm specular reflection. Scattering is relatively independent of the direction of the instant sound. We know that for the specular reflection, uh, if the the uh instant beam is not perpendicular perpendicular to interface, they reflected beam cannot return back to the transducer. Umm, but for scattering, you know, if you're scanning uh, they solid organs of the human body, so no matter what, what? What they. The incident angle is, you know that there would always be, uh, reflection from scattering coming back to your transducer. So that's how you can you can actually to the scanning. Again, this is showing you, uh. You know there's rough. Uh. Surface. Umm, the reflected. Echoes could be directed to any a lot of different directions. So the strength of scattered echoes is depending on these factors, the frequency of your beam. Uh, the testing of the scatterers and the size of the scatterers. So the higher the frequency, the stronger the the, uh, scattered echoes. The higher the density of the scatterers, the higher the strength of the scattered echoes. They higher the larger the scatterers, the stronger the scattered echoes. And if the mismatch acoustic miss my meaning, the impedance mismatch? Uh. And you know, uh, between the scatterers and the surrounding tissues are large. This acoustic impedance mismatch is large. Uh. The strength of the scattered echoes would be higher. And then we talked about really scattering for really scattering. It happens. It happens when the reflectors are very small compared with the wavelength of the sun. Uh, we need to know that. Uh. Really, scattering is a very big reflection. Umm. And it's very frequency dependent. OK, this is formula you can see here. Uh, if you increase the frequency, the reflected, uh, being that the strength of the reflected beam would be, uh, much stronger? So this one is the diameter of the refactor. This one you cannot change. For example, if we talk about the release scattering happening, ohh you know on red blood cells the diameter o