We have a surface of this continuity can be regarded as a. And we have two medium, okay. Now we have a limpor that has a mass per unit area M, that is mass per unit area. And I have medium Z0 over here and medium Z1 over here. I have instant way, PI that can be expressed as complex amplitude, PI. And exponent- j omega t- k 0 x, and I have reflected [INAUDIBLE] which I denote p r x t and that I can write captive pr. That is complex amplitude and exponential- j, sorry about this. Omega t- k 0x, and minus? No, that has to be plus because it is going to the negative x axis. And then there will be some transmitted wave that is also function of x and t. That is capital P. Kept the complex pressure exponential- j omega t- k x. This is okay because the transmitted word will propagate along the positive x axis. But, K in this different case has to be K1. Because K zero, and one has to be omega over C, zero, or one. And omega has to be the same in this two medium because we are handling linear acoustics. See, and linear acoustics, if I shake this one by certain frequency, then what you will hear or the response has to be same excitation frequency. All right, that's good. And as you note over there the boundary condition, [SOUND] Now let's look at, as the boundary condition, as the condition that give you the solution. Okay? Boundary determine The ratio between incident and reflected, the ratio between transmitted and incident, that concept holds for every discontinuity case. So that is why I am using this example. And rather deep science because this will provide solely the foundation to understand what's going to happen on every discontinuity case. Now first boundary condition What would be. I will use different chalk. Okay, first. First one, what would be? Force Force balance. Okay, force balance must be achieved between these two medians. So I can say, so pi plus pr at x equals zero Minus p t at x equals zero. That is the force acting upon this per unit area, okay? The reason why I put minus is because I am taking positive sign over here and a pressure by definition is acting normal to the surface there for I put plus sign over in here Even if the reflected wave is propelling toward this direction, that does not, necessarily, mean that the pressure of a reflected wave is acting in this direction. It has to act upon the surface in normal direction. Okay, that's very Fundamental concept of newton. I can see lot of mistake because of the misunderstanding of how the definitions of operation. Okay, not has to be equal to one or could newton’s said if you have imbalance force the mass will loft So mass times x [INAUDIBLE]. What is mass in this case? That is m per unit area, because we are handling per unit area force that is m times the acceleration that is y double And what is y double dot? Note that y double dot is y exponential minus j omega t double dot. So I got minus j omega t minus j omega t. Therefore I get minus omega squared and Y exponential minus j omega t. That's what I have. And all this term also has, for example, Pi exponential minus j omega t. Because I am looking at the pressure x equals zero. This move away, so I got the moving of the I got Pi complex. Plus pr complex minus pt complex that equal to minus m Y. That's the first equation I had. And you can see this equation in the text too. I want the students follow the derivation, because the reason why I am doing derivation is, because by doing derivation you can observe how the physics is following. How the physics is following into your derivation. Derivation is not just mathematical practice. Derivation is very effective demonstration of managing the physics, okay. So this is first equation, I denoted this as equation number one. And second equation that I can get, from what? So second equation, [SOUND]. Not only force balance is required, but also kinematic relation has to be imposed on this problem. What is it? Velocity continuity [SOUND]. [INAUDIBLE] On x equal zero plus as well as zero minus. That means I am assuming the thickness of is very, very small with respect to wavelength lambda. So this theory cannot be applied for the case of which the thickness of is very large compared with the wave lengths. So that is one limitation. And another limitation is the case when the wave is instant to the surface normal. To the flat surface of this continuity. Hey, but then the result that we will obtain from this spatial case is useless, is too spatial. No, I don't think so because in general case, for example, I have this kind of instant y. Then I can decompose this instant y into this two-component, and the transmission loss for example due to this instant y is Very small because we assume that the fluid we are handling has no viscosity. Therefore, no problem. What we obtained really, the major primary part that dominate the physics of this kind of flat surface. Okay lets go back to velocity continuity case. What is the velocity of the fluid particle over here. This is the fluid particle we have considered. Velocity due to instant wave would be Pi over Z0. Because we assume plane wave. And what about the velocity due to reflected wire. That is Pr / Z0 and minus. Because the velocity is a vector quantity. And the velocity due to reflected is approaching this way. And that has to be equal to the velocity of over here. And what is it? That is y dot, and y dot is y minus j omega exponation minus j omega t. Therefore magnitude of the velocity of is minus j omega y, y is the complex amplitude. And that is the velocity on x equals zero minus. Zero minus means, this is zero and this is zero minus, right before zero. And I want to apply the velocity continuity at zero plus, which I can say the velocity due to transmitted wave is Pt over now Z1. That has to be equal to minus j omega y. Okay? I have, how many equations do I have? I have, this is equation two, sorry this is This is equation two, and this is equation three. Now I have one, two, three equation, and I would like to have for example, PT/ PI which is transmission. And what is this? Okay, how to get it? Well let's do it, let's do it because this is the case which is different with what we have, Z0 and Z1 we have. Okay, so we have to eliminate a Pr to get the relation between Pr and Pt. How to eliminate to the Pr? Okay what I can do is I will just, Divide the forced equation by Z0. Then I will have minus omega squared m y/Z0. Let's call, that is equation four, then I will add equation four to equation two. To eliminate Pr, what we will have. Can you give me some hint? Okay, What we will have is. Okay I have a Pi/Z0 and I have a Pi/Z0 two. So i have two Pi/Z0 and I have plus PR/ Z0. And I have minus PR/Z0. So those two terms Cancel out, and I have over here minus Pt/Z0. And I have now this one minus omega square mY/Z0. And I have what? - j omega, y. That's what I have. Okay? That's correct. Now, we have a relation Pi and Pt, but still there is another one which I want to eliminate. That is y. So I want to eliminate y and express y in terms of Pi and Pt. How can I do to it? There is another equation. Okay. So in y equation three, that expresses the relation between y and pt. Okay, j omega y has to be, according to equation 3, I will write down equation 3 over there, has to be Pt over 0. And then what about this one? This is a headache. What I can see is simply,- omega square my, is what? I mean, what we can do is, we will substitute this y over here, over there. I'm using equation 3 again. [SOUND] Y is Pt over Z1 and I have to divide this 1 over- J omega. Okay. That's what I obtained. Looks not bad because I can see that this term will change to, because J omega- omega squared divided by J omega gives me J omega, right? Where? Over here, right? So you got price, where is my wallet? So I will give you after the [LAUGH] lecture. So minus and minus go away and the 1 divided by J is- j omega, okay? And then I have m and then I have a Z0 and a Z1, and I have a Pt. Wow, this is nice. This is nice, and I put this one over here. Then I got following relation. So what I obtain is 2 pi over Z0- Pt over Z0 that is equal to the first term is this,- j omega m/Z0 Z1 Pt. And then I have a second part that is simply + Pt/Z1. So, what we obtain now is there is a Pi and there is a Pt, and a Pt and a Pt. So let's summarize what it means. Okay, I have Pt look very complicated and Pi. So let's write down what is Pt/Pi. Okay, there. And the Pi is simply 2/Z0 and a Pt, I have many interesting component, okay? Pt, this one go over there. Then I have 1/Z0 + 1/Z1, and I have- j omega m/Z0Z1, therefore, if I multiple Z0Z1, then I will have 2Z1 over there. And then I have -J omega m + Z1 + Z0, and this is very interesting. Okay, those three term is very interesting, as well as this term. So let's go back. What we have for the case of when we have just two mediums, Z0 and Z1, at that time the tau was Z1 + Z0, and I have over there 2Z1, right? Reason, let's look at this carefully. Try to relate this mathematical expression with physics, okay? This term expresses sort of Averaged impedance. And this expresses the impedance of a transmitted medium. So, keeping that in your mind, then this is the same, right? This is 2 times the impedance of a transmitted medium that is 2Z1 and this is Z1 + Z0 because we have two medium, Z1 + Z0. And also we have additional term that is proportional to- j omega m. That is the highlight of physics that happen when we have limp war. All right? Why this is -j omega m? What is this? What is the dimension of this? The margin of this has to be equivalent with the margin of that is impedance. So this is another form of impedance that represent dynamics of limp world. So that's very interesting,that's is tau. So, tau proportional to medium, related to the medium, as well as tau is strangely related with the mass and omega of lumpur. Alright? The reason why it has minus j over j. Why? One divide minus j is? J? So that means It has face difference between instant and transmitted wave. Okay? So if you look at the magnitude squared of this term, transmission power transmission coefficient. Then we can come up with what? How much power will be transmitted. That will be related with, anyway this one has to go to omega m square, plus magnitude the square of this one and this one. So if Z1 and Z0 for the case of air and water, that should be row 0, c0, and row 1, c1. Has on a real path, so I can argue that this one would be z1 scale press z0 scale and a full z1 scale. Okay, that's the power transmission coefficient. So if I look at the log of 10, 1 over tau magnitude of scale and tau, and call this transmission loss, what is it? This transmission loss and other representative measure that describes the measure, the performance of lumpur. The reason why I use inverse of 1 / tau square is because if I have Large tau, that means I have more large transmission, that often we do not want to have large transmission. We often have as noise control engineer, we often want to have a small tau, maybe we want to have a zero tau. So, if I measure like that, in dB scale, that is the transmission loss I obtain by using the the lumpur. So if I have 10 dB RTL, 10 dB RTL, that means my war reduce the noise 10 dB. So suppose here I have 100 dB, and I design the war that give me at certain frequency through the dB transmission laws, then I expect to have 70 dB over there, in other words I am successful to reduce necessary dB. So, transmission loss, sometimes this is called transmission losss that is very useful measure or index that normally used for most control engineer. Okay, so this is interesting.