 Hello again, in today's lecture we will be using some of the concepts which we have developed in previous lectures, specifically the transmission line theory and use this approach to figure out what happens when sound as it propagates through medium, it hits a wall. How does the sound get transmitted across a wall and to solve this problem we will use the transmission line theory. So, what we will analyze is first transmission of sound through a wall when the incident angle of the sound as it hits a wall is 90 degrees that is the first problem we will solve and then subsequently what happens if it is hitting at all sorts of random angles and then using this approach we will figure out how sound gets transmitted across a wall. So, that is what we are going to do today. So, I will make a very simple picture let us say I have a wall and the sound is hitting the wall at an incident angle which is 90 degrees. So, it is a normal incidence. So, this is my incident wave and that is my reflected wave. So, the complex magnitude of incident wave is P plus of reflected wave is P minus and then this is the wall in question and a part of this sound it gets transmitted to the other side and let us say the complex amplitude of this wave is P t. Let us also assume that m is equal to mass per unit area of wall. So, 1 square meter of this area would weigh m kilograms and this wall could also have some compliance. So, because when something hits a wall these walls may bend they are not perfectly rigid. So, they may have some compliance and let us say c is the compliance per again its per unit area and this thickness is t and we assume that the thickness of the wall is fairly small compared to the wavelength or wavelength of the sound which is hitting the wall. So, t is extremely small compared to lambda. So, this is the problem formulation and what we are interested in knowing is. So, question if the question is that if I know P plus then what is the value of P t because P plus is the incident energy going into the system and P t is something which is coming out beyond the wall. So, I want to figure out the value of this transfer function P t over P plus and this is what we are interested in finding out. So, once again we will start with the transmission line equations. So, for the transmission line equations we will write two sets of transmission line equation. So, this is the left side of the wall and this is the right side of the wall. So, I will have one set of transmission line equations for the left side and the set of transmission line equations for the right side. Also, what I am going to do is I am going to establish my coordinate system. So, this is 0. So, wall stands at the location x equals 0. So, with this now I am going to write my transmission line equations. So, for left side pressure and velocity and they are functions of space and time and this is equal to P plus P minus P plus over z naught minus P minus over z naught. So, this matrix is going to be multiplied by this vector E minus S x over C E plus S x over C times E S t and we know that S equals j of omega j times omega. So, this is my first equation and then I have likewise equations for the right side. So, on the right side let us say I call pressure as P 2 and velocity as U 2. So, P 2 now when we see this problem there is no reflected wave because I am assuming that there is no reflecting surface at the right side of the wall. So, there is no reflected wave on right side of the wall. So, my P plus will be same as P t and since there is no reflected component this element of the matrix is 0. So, again I have a 0 this is my second equation. So, I have now two sets of equations one for the left side other for the right side. Now, at this point of time I start imposing the boundary conditions. So, my first boundary condition is that whatever is the velocity on this side at x equals 0 that velocity is same on just the right side of the wall again at location x is equal to 0. So, boundary condition is that U 1 at the 0 t equals U 2 at 0 t and what that gives me is. So, if I substitute in equations 1 and 2 the relations for velocity and I put x equals 0. So, from first set of equations I get U 1 0 t equals P plus minus P minus over z naught times E s t. And, this is equal to U 2 of 0 to E and that is equal to P t over z naught e to the power of s t because s is identically 0. So, these terms E s x over c and E minus s x over c they go with the vanish and they become 1. So, now what I do is I equate these two terms and what I get finally is P plus minus P minus is equal to P t. So, this is let us call a consequence of this boundary condition that at x is equal to 0 the velocity on right side of the wall is same as velocity on the left side of the wall. The other condition at this x equals 0 location is that the wall is experiencing force. It is experiencing a force and this external force is essentially the difference of pressures on left side and right side. So, this is the overall external force which the wall is experiencing and this force will be equal to the acceleration of the wall times mass plus stiffness of the system times the displacement. So, now I am going to develop a relation for force balance. So, this is my wall and on the left side my total pressure is P plus plus P minus times E s t of course, at x equal to 0 and how I get this? Essentially what I am doing is I am in equation 1 putting x is equal to 0 and I am finding the value of pressure. So, on the left side the pressure is this and then on the right side pressure is e to the power of s t times P t. So, my force balance equation is P plus plus P minus e to the power of s t this is the force on left side minus P t e to the power of s t. So, this is the overall external force and this external force equals and then this external force we know if we draw for free body diagram of this system is equal to mass times acceleration. So, what is the mass of the overall mass? So, excuse me this is a difference of pressures and I have to multiply this by the area of the wall which is a. So, this is the external force and this is equal to the mass times acceleration of the wall plus stiffness times displacement of the wall. So, mass is m times area this is mass times acceleration and if I know the velocity of the wall which is let us say v u wall then if I take a differential of this then that is the acceleration of the wall. So, this is mass times acceleration plus stiffness and stiffness of the wall is essentially 1 over compliance times area. So, c is specific compliance and then I am going to multiply this by area because it is compliance per unit area and this times again if I know the velocity of the wall and if I integrate that velocity then that is what I get as displacement. So, I am going to integrate d over dt u wall. So, my overall equation becomes and I see that area is common on left side and also right side. So, I can eliminate it. So, my overall equation becomes. So, I have to make a correction here this should have been integral of u wall with respect to time. So, I will make that correction here as well. Now, in this equation I am also going to put a negative sign a negative sign and the reason I am going to put this negative sign is because if I have a positive pressure then pressure is actually moving in the negative direction it always acts inwards a positive pressure acts inwards. So, to account for that reality I have to put a negative sign here. So, this is my equation of motion. Now, what we are going to do is on the right side of this equation I have u wall. So, I will find a relationship for u wall plug into this and then do some more mathematical manipulation. So, we know that u wall is equal to u 2 at x is equal to 0 and if I see this relation for u 2 then I know that u 2 is nothing but u 2 equals p t over z naught times e to the power of minus s x over c times e s t and if I put x equals 0 then I get u 2 s and because s is j omega what I get is this relation because s equals j omega. So, what I am going to do is replace this term and this term by this term because that is the value of u wall. So, what I get finally is plus the compliance term. So, in this relation p t and z naught they are constants p t can be a complex entity but it is still a constant. So, then I can very easily differentiate this term and I can very easily integrate this term and that is what I do in the next step. So, after integrating and differentiating these terms what I ultimately end up getting is so I omitted one thing here that I had an e s t here. So, I should have e s t here and I will replace this e s t in this relation by e j omega. So, my left side is this p plus plus p minus minus p t e j omega t and on the right side I get m p t over z naught times j omega e j omega t plus 1 over c j omega times p t over z naught times e j omega t again I have e j omega t common and so I can eliminate these terms and thus my equation simplifies to this term and this term is same. So, I take it outside the brackets. So, I call this equation 4 and what this equation tells us is the condition for force balance that because the forces have to balance out. So, on the left side of the term you have the extra force p plus plus p minus times minus p t. This is the summation of different pressure elements and on the right side you have inertial element and stiffness element in the equation and this equation is equation 4 it has been directly derived from Newton's second law of motion. So, now I see equation 4 and then I go back and I see equation 3. So, between equation 3 and equation 4 I have 3 unknowns one is p plus the other one is p minus and the third one is p t these are constants, but they are unknowns. I know what is m I know what is omega and I know what is c which is the specific compliance. So, then using these two equations this equation and this equation I will like to solve I get my transfer function which is this and please bear in mind that this transfer function will depend on omega. So, with that intention I what I do is I modulate this equation I reframe this equation in such a way that I can rewrite this equation as p minus equals. So, I take p minus right side is same as p plus minus p t and I call this equation 3 a. Now, I put this equation 3 a in equation 4. So, where so I replace here p minus by p plus minus p t and let us see what we get. So, p plus p plus minus p t this is the equation this is what p minus equals minus p t, this is the equation this is what p minus equals minus p t equals minus p t over z naught j m omega plus j omega p and let us call this function f or actually let us call this function capital D. So, with this what is this is equal to minus p t over z naught times d where d is this whole term in brackets. So, now I simplify this. So, I get on the left side 2 p plus minus 2 p t is equal to minus p t over z naught times d or if I move this on this side I get 2 p plus equals 2 p t minus p t over z naught and that is same as p t 2 minus. So, I missed the d here d over z naught. So, with this I can write p t over p plus is nothing, but 2 over 2 minus d over p plus is nothing, but 2 over 2 minus d over z naught. This is my relation for this question where d equals j m omega plus 1 over j omega c, where c is a specific compliance. So, let us call this equation 5. So, now I am going to pan this further. So, I get p t over p plus 2 minus and I am going to replace d by this whole relation. So, what I am going to get is this is equation 6. So, this is my transfer function and this transfer function it depends on omega its value changes with omega or the angular frequency of the sound wave which is striking the wall. So, using a relation 6 I can calculate what will be the intensity of what will be the value of p t if an incident wave of a strength p plus is hitting it. So, now what I am going to do is now I am going to do some further processing on this relation and based on weird frequencies we are talking about. So, we see from this relation is that this term m j omega plus 1 over j omega c j omega as omega is. So, this I can approximate as m j omega for all omega or just to make it simpler for omega which is large compared to 1 over c m. So, if omega is extremely large and by large I mean if it is extremely large compared to the natural frequency of the system this term approximates to m omega. This range is called mass controlled region. Similarly, m j omega plus 1 over c j omega it approximates to m j omega plus 1 over c approximates to 1 over j omega c for omega which is extremely small compared to natural frequency of the system that is 1 over square root of c m and this range of frequencies is called stiffness controlled region. So, we will try to understand how this function works this function works this equation 6 in stiffness controlled region and also in mass controlled region. Now, once again to recap if frequency is extremely small compared to the natural frequency over this of the system which is 1 over square root of c m then that range of frequencies where it is extremely small to the natural frequency it is called stiffness controlled region. If the natural frequency is if the frequency we are considering is extremely large compared to natural frequency of the system then that frequency that range of frequencies is called you know it is supposed to lie in mass controlled region because the mass term dominates the response of the system. So, we will look at stiffness controlled. So, again I will write down the relation p t over p plus equals 2 over 2 minus and we know that in the stiffness controlled region I can omit m j omega and I have to just include 1 over j omega c. So, what I get is 1 over 2 z naught c j omega. So, I can simplify this. So, there is j in the denominator of this term. So, I can simplify this by moving j upwards, but also eliminating the negative sign what I am doing is that in this term I am multiplying this and also dividing this term by j. So, what I am getting is 2 z naught c omega. And now I am further simplifying this I am rationalizing this by taking the j in the numerator. So, what I am going to do is I am going to multiply this whole function and also divide this whole function by 2 the complex conjugate of the denominator. So, I multiply numerator by 2 minus j over 2 z naught c omega and I divide the numerator by the same term. So, what I get is this 2 should not be there and here what I will do is I will again divide the numerator and denominator by 2. So, I get this plus j over 2 z naught c omega and now I am going to rationalize it. So, what I end up getting is this relation. Now we know that if I have a wall if this is my incident wave this is reflected wave and this is transmitted wave then the incident energy is directly proportional to magnitude of p plus whole square and transmitted energy is directly proportional to magnitude of p 2 So, the attenuation that is the loss of sound as sound travels across the wall is equal to intensity of the transmitted wave divided by intensity of incident wave and that is nothing but this ratio. So, this is my attenuation and I can also call this as this. So, now I have this equation let us call this equation A from equation A I am going to find what is the level of attenuation. Attenuation I am writing it is at A t n and 1 minus j over 2 z naught c omega divided by 1 plus 1 over 2 z naught c omega. The whole thing squared and what I do is I take the magnitude of numerator which is 1 plus 1 over 2 z naught c omega this is the magnitude of numerator magnitude of denominator is same as the denominator because it is a real number and then I square the whole. So, my attenuation comes out to be 1 over 1 plus 1 over 2 z naught c omega whole square. So, now that I have developed a relation for attenuation of sound at is as it gets transmitted across a wall now what I will try to find is the transmission loss in decibels. So, I call it t l and that is defined as 10 log of 10 over 1 over attenuation and that is essentially 10 log 10 of this whole thing and because it is 1 over attenuation. So, this denominator comes in the numerator. So, I get 1 over and please bear in mind that this relation is good if the frequency of the incident sound is extremely small compared to 1 over 2 pi c m extremely small compared to the natural frequency of the system. The other thing we see from this is that if I am below the natural frequency of the system then as I keep on reducing my omega then my transmission loss keeps on increasing. So, as I get closer and closer to 0 hertz I get an improved transmission loss. So, if I plot it on a log log scale. So, this is frequency and this is in decibels I am plotting transmission loss and frequency I am plotting in hertz. So, and because this is a log scale my board plot will something will look something like this. It will be a straight line because it is an asymptotic response and the slope will be 6 decibels per octave because as omega goes down I get this 6 decibels slope because as omega is extremely small the term in the parenthesis this term is extremely large compared to 1 and when I take its log I get a 6 decibels per octave slope because there is a square term there. So, this is the response of the system for a stiffness controlled region. So, now we move on and now we start looking at mass controlled region and the definition of mass controlled region is that my frequency should be large compared to 1 over 2 pi times 1 over C m square root which is essentially the natural frequency of the system. So, once again my original relation for P t over P plus was this thing and in the mass controlled region I ignore this term because the product of m and omega is extremely large compared to the number 1 over C times omega because omega is extremely. So, if that is the case then I can simplify this as and the mathematics is very similar. So, I am not going to repeat that mathematics I can simplify this as 1 minus j m omega over 2 z naught this is the term and as we had calculated earlier attenuation in this case is once again the square of this term and this comes to be like this. And finally, we will calculate the relation for transmission loss T L and that is essentially 10 times log 10 of 1 over attenuation. So, what that is? So, here in this relation we see that unlike like in a stiffness controlled region if I reduce the stiffness my attenuation would go up and so would my transmission loss in case of a mass controlled region I have to increase the frequency to ensure higher transmission losses. So, if I again plot this in decibels and if I construct a board plot then it will be something like a straight line the asymptotic response as omega becomes very large will be straight line and this slope will be 6 decibels per decade. So, this is how the system is going to respond as omega goes up. So, what that means is that if I have a wall and if I am striking it normally with frequencies which are large compared to you know the natural resonance of the system natural frequency of the system then as I keep on increasing my frequencies less and less sound passes through that barrier because my transmission loss doubles transmission loss goes up by 6 decibels every octave every octave it goes up by factor of by factor of 2 that is 6 decibels. Now, the question is what happens at the resonance point? So, derivative question is so when I have the condition for resonance then this term it essentially m j omega plus 1 over c j omega it becomes 0. So, the influence of mass counter acts and it cancels the influence of stiffness exactly at the resonance frequency. So, when this term becomes 0 then my p t over p plus is essentially exactly equal to 1. So, p t over p plus equals 1 and thus my at resonance what mathematics tells us is that p t over p plus equals 1 attenuation equals 1 based on this relation and transmission loss is 0. So, I do not have any loss I do not have any transmission loss across the wall. So, if I am hitting a wall normally with a frequency which equals the resonance of the system of the wall then most of the sound will just go across the wall uninhibited you know without any damping out without any damping out. But, in reality what happens is that at resonance frequency the damping is starts playing damping of the system starts playing a role. So, the overall transmission loss curve the overall transmission loss curve it looks something like this. So, let us say this is my resonance point where f naught equals 1 over 2 pi then in the mass controlled region the asymptotic response the boat plot will look something like this and this is how the boat plot will look in the stiffness controlled region. So, this is my stiffness dominates and in this region mass dominates and this is my resonance point. Now, this is these straight lines they have a slope and same slope is on this side in the stiffness control region, but it is a negative slope. So, this is my ideal curve, but in reality we have damping and also we have this approximation at lower frequency is not exactly true. So, the real response the real transmission loss curve looks something like this. So, asymptotically it goes and merges to. So, this is my real response curve real TL curve and this height is it depends on depends on damping. Thank you very much for your patience and we will meet you once again in our next lecture. Thank you.