 So, we have talked about radiation impedance for a sphere. Let us look at very quickly radiation impedance for an open tube. So, ideally p is exactly equal to 0. In reality, you can say that it is approximately equal to 0, because that change to 0, it happens over some finite distance. So, this is an open tube. Now, my impedance I know is equal to p over u. So, if p is very is approximately equal to 0, I can say that z is approximately equal to 0. So, what does that physically mean? What it means is that when sound is travelling here, then in case of sphere, we saw that for a wide range of frequencies, it is difficult to dissipate power into free air from a spherical source, because of that psi factor and psi becomes significantly large in that case. Do you remember that discussion? Because of the power factor, the value of power factor is significant at high frequencies and also at low frequency in case of a pulsating sphere. So, at low frequencies, you need very high energy to pump small amounts of energy into free air and at also high frequencies, you need large amounts of input energy to pump. But in case of an open tube, given that z is approximately equal to 0, it is relatively easier to pump energy at all frequencies into free air. So, if I have a vibrating piston here, it is relatively less difficult or easier to pump energy into free air. So, that is it. So, essentially when we construct systems which have to produce sound, then you want to ensure that the radiating impedance of that system is as low as possible and also the overall power factor of that system. The closer it is to you know psi being equal to 0, the closer is it equal to that value, the better it is, the more efficient that system. If you want to have stronger attenuation of sound, which is the inverse problem, then you will like to construct systems which are in the other direction. So, that you will like to increase the radiation impedance because that will essentially mean that less energy will get dumped into the environment and will be heard by individuals. So, there are several topologies to which sound gets propagated into the air, a very common topology is that you have an infinite baffle or a wall, you can call and then here I have diaphragm or a flat surface which is moving back and forth. Especially in some of the older sound systems, you would see very big boxes which will be having one speaker which is moving back and forth. Essentially, it acts like an infinite baffle because the stiffness of the box is very low. So, this is one topology. Another topology which we have talked about is u-s sphere which is pulsating, growing and contracting. That also emits energy into air. This is another third topology and fourth topology could be a plain circular piston without any baffle. So, this is one. Then, how does it emit sound? So, they are different approaches. So, what we will talk is a little bit more about this particular topology. What we call plain circular piston in infinite baffle? So, this is my infinite baffle and this is my plain circular piston. The radius of this piston could be r naught. So, again what we are trying to get an understanding is that as this piston is moving back and forth, it has a certain volume velocity, it generates some pressure. How does all that energy gets dumped into air and that is quantified through this number called radiation increase. So, this problem you can develop some finite element models through some numerical experiments and after good amount of computational modeling. People have come up with fairly standard values. So, for a plain circular piston in infinite baffle, the radiation impedance looks something like this. It is a bunch of one capacitor, two resistances series and and and this is for impedance models. For mobility model, the circuit looks now that you understand the idea of a dual. It looks something like this. This is and the values of these elements have been computed and so here the value of r a 1 is 0.1404 times rho naught c over r naught square r a 2 is my stylus. So, r a 1 is 0.141 rho c over r naught square r a 2 is this particular expression m a is 0.27 rho naught over r naught and c a another constant is 5.94 r naught cube over rho naught c square and because now we are talking about specific constants 0.14 you know. So, these relations are good to the extent we are in SI units. So, r naught has to be in meters rho naught c has to be rho naught is density. So, it is kg over cubic meters c which is velocity of air has to be meters per second. Otherwise, these relations have to be adjusted and r a 1 lower case r a 1 and lower case r a 2 are basically just inverses of upper case values. So, if you have a pulsating membrane fixed in a big on a big surface and air from back side is disconnected with air from the others from your room then if you use this model then it will fairly accurately capture the radiation of that membrane. So, what we will do is we will look at it and so this is mobility model. Now, what we will do is we will see what are the characteristics of this particular model at low frequencies and at high frequencies what does this mean physically. So, if I am in very high frequency line when omega is large then let us see what is going to happen. So, when omega is large m a becomes extremely large. So, very little current goes through here. So, I can drop this out. So, the current has choice between going through r a 1 or through this capacitive element at very high frequencies. The impedance offered by capacitor will become very low. So, current will prefer to go from c a and then it goes through r a 2. So, my equivalent circuit at very high frequencies becomes so my this is my potential difference p and the current is volume velocity and this is r a 2. So, what this shows is that when I have high frequencies of piston mounted on an infinite level behaves like a purely resistive circuit. The first thing is that the inefficiencies associated with the power factor they become minimized at high frequencies. So, I do not have any inductive or capacitive element which alter the phase that is one thing. The other thing is that in the near field again what this shows is that it the transmission of sound happens in kind of like a beam because this impedance is something very similar in nature qualitative nature to plane waves as they are moving through tubes open tubes. If you remember the impedance in a open tube is purely resistive thing and it does not attenuate in strength as you move forward. The strength of sound as it moves in an infinitely long tube it does not decay with traveling. What this shows is the it is the same thing happening here also. So, what that means is that at high frequencies the behavior of sound is like a beam travels like a ray of light at high frequencies this is what it means physically. Now of course, when you go away from the source when you are in far field then it again starts to radiate, but for a fairly good amount of distance sound moves like beam at high frequency this is the physical condition. So, that is case one. Case two happens what if omega is low omega is small. So, what happens? Let us look at this circuit again when omega is small I can fairly easily drop this term out C a. So, I what I have left what I am left with is m a and r a 1 and r r a 2 at extremely low frequencies between r a 1 and r a 2 and this mass it will still prefer the inductive part. So, in a little more general sense I can say that my circuit will be something like this. So, for extremely low frequencies current will still like to go through m a which means that for very low frequencies the dissipation of power through an infinite baffle will be very less. Energy will go into the system back into the system it will not get dissipated. So, it is very difficult to dissipate for very low frequency sound into 3 a in infinite baffle. All what you are seeing is that different approaches of dissipating low frequency sound into medium is a not an easy thing this efficiently doing the same thing efficiently. Similarly, attenuating low frequency noise is not easy. So, what we will do is we will yeah in the first case. So, you understand why I dropped out m a right at high frequencies. Now, at high frequencies capacitance the impedance offered by capacitance will be virtually 0 because it is 1 over s times c a. It will be there, but at high frequencies the impact of that negative magnitude with respect to r a 2. Yes with respect to r a 2 what I mean I can keep on increasing the frequency and I will hit a number where it will be small enough compared to. So, what that crossover point is will depend on the ratio of r a 2 and c a right. So, you have to figure that out, but at high frequencies c a you can ignore it and replace it by a short. Sir, can you repeat the point that you mentioned about attenuation of low frequency sounds and the other equivalent in making the sound at high frequencies. Yeah what we have seen is that when we try to dissipate sound through some of these sources into free air what we are seeing again here is that at very low frequencies m a dominates all the current goes through m a which means that the dissipation of power is less. So, all what happens is energy goes in goes back and forth into the system. We had seen earlier that the average power dissipated in general is essentially dependent on the r term if very low energy is going through r the energy dissipated into the sound into the into free air will be very less that is what. So, what we will also do is here we will compare the efficiency of this system for low frequency with a spherical source is it is a spherical source easy to dissipate power into free air or is it easier to do it using an infinite that is what we will do. So, this is for infinite battle. Now, for a spherical source we had developed a relation something like this the impedance was this is the radiation impedance for spherical source. So, if the size of the source both the sources are same that is r o is same then look at let us look at the numbers. So, we will construct a small matrix spherical source and the other column could be infinite battle and what I will do is I will list r and l. Again the circuits looks very similar you have a resistor and in the inductor here in parallel same thing in case of a spherical source all we will be doing is comparing the magnitudes of these elements and see how they look like. So, in case of spherical source the resistive value was z naught over a that equals over 4 pi r naught square is 0.084 rho c over r square that is my number 0.084. In case of infinite battle it is 0.458 rho naught c over r square the inductive value will be z naught r naught over 4 pi r naught square and if I do the math it comes approximately to 0.084 rho naught over r naught I just assume pi to be 3 it comes close to 0.084 may be 0.083 in this case it comes to 0.27 rho naught over r naught. So, what do you see when you look at these numbers that the impedance offered by a spherical source is substantially less for the same size of the object substantially less than that offered by source of sound mounted on a baffle. So, spherical source by itself is not a very efficient way to disseminate sound into air at low frequency and infinite baffle is even poorer approach of disseminating sound at low frequency. This is the because your r is less your l is less for a spherical source your baffles numbers are significantly high five times or more. So, we have talked about spherical sources we have talked about radiation impedance of a tube at the end of it. So, now we will very quickly consider radiation impedance for a plane circular disc in free air. So, for low frequencies I will very quickly give the impedance circuit looks something like this. So, this is my impedance model. So, my cross variable is pressure here my true variable is pressure and the cross variable is volume velocity and r is 0.019 r square times density times omega over c cube. And m 1 m m equals 0.271 times rho naught over r naught and in this case r equals 0.261 omega square r naught 4 over rho naught c cube. You see that things start getting more and more complicated. So, you will find out more exhaustively some of different geometries and what each of these geometries have for the radiation impedance in a book like Baranet which is the press rate test for this course reference test and for high frequencies the circuit looks simply like this where r equals to rho naught c over pi r naught square and this is this is valid for which model impedance of mobility which one. So, if I have to use for high frequency something similar for mobility what will I do r will get replaced by its inverse. So, what we will do is one example and see how radiation impedance improves our prediction of some of the acoustic properties of systems. So, let us consider a bottle this is actually a physical bottle which I measured. So, it has two parts the badly bit bottle. So, there is a volume here this length is l 1 let us call it l 2 this length is l 1 its mouth has a diameter of 2 times r 1 and the volume has a diameter of 2 times r 2. So, what I would like to know is what is the natural frequency of this bottle this is the volume is like a stiffness member and the open tube on top of it is like a mass right. So, it is kind of a mass spring system and I am I like to see how good my model is in terms of predicting. So, we will put some numbers l 1 equals 6.7 C m r 1 equals 1.43 centimeters volume of the lower portion is 530 times 10 to the power of minus 6 cubic meters I mean I can also find l 2 is p naught over pi r 2 square. So, first thing I will check is that is my all these lengths significantly smaller than what number lambda over 2 pi right. So, l 1 has to be less than lambda over 2 pi which gives me a frequency number corresponding to 819 hertz is everyone sure how to jump from lambda to f. So, what this means is that if I am significantly below in frequency from 819 hertz then if I approximate my tube which is l 1 long as an acoustic mass I will be that is what it means. Similarly, l 2 has to be less than lambda over 2 pi which means frequency equals 365 hertz. So, if using this mass spring approach if my resonance exceeds 365 hertz then my model is not valid basically that is what it means. So, now we will find what is the frequency of this system. So, if I construct an impedance model pressure that is my mmc now I am sorry acoustic mass. So, ma we have seen from earlier lectures and discussions is rho naught l over pi r naught square and my capacitance value is or acoustic stiffness is v naught over rho naught c square. So, my z bottle is s times ma plus 1 over s times c a. So, if I plot its poles and 0s I have a pole at 0 and a pole at origin and I have 2 0s symmetrically along the imaginary axis and this value is 1 over m a c a. When I do the math and I plug in all the numbers my first estimate on frequency resonance f naught comes to be 210 hertz and when I actually did the measurement. So, f m it was 233 hertz I should have shown this. So, this is clearly it is off and in terms of estimation frequencies they should not be off by this much of an amount frequency estimates come fairly close to reality. So, next we will see how we can improve our model to get a better and closer estimate. So, what we will do is I went to this book Berenek and I said there is this tube I have just modeled it as a pure mass element, but there is some radiation this is a tube, but it may be seeing some radiation impedance here because it is dissipating sound into free air. So, what is that value? So, once I do that my refined circuit becomes something like this. So, I still have my this is m a which we have already calculated this is c a and then I put. So, the extra elements based on more regress computational modeling are these terms r 1 and m 1 where r 1 is 0.479 rho c over r naught square and m 1 is 0.195 rho naught over r naught. So, putting the values of r and everything in this I get r 1 equals 953 times 10 to the power of 3 Newton second over meter 5 and m 1 is 19 times 10 to the power of 3 something like this. So, for low frequencies clearly r 1 is very large compared to m 1. So, I have to make correction this is omega m 1 omega m 1 for I just chose a small number 190 hertz where I said 190 is fairly close to 230. So, I wanted to get an estimate that in the neighborhood of resonance which term is dominating r 1 or omega m 1. So, what I am seeing is that in the neighborhood of resonance this is very small this is very large these are in parallel. So, what that means is I can drop r 1 out at around 192 hertz. So, then my approximation of this circuit becomes m a m 1 c I still have pressure. So, I get these three elements is everyone comfortable with why I drop the r term. So, I know m 1 I know m a I know c. So, my second I try to get different estimate. So, my second frequency estimate natural frequency estimate came to be what was it 219. So, I was able to improve my estimate from 210 by another 10 hertz or so. So, that is how close I got to the actual number after that I did not bother. So, what this shows is that incorporation of radiation impedance in your actual model will help you get a better picture of how the circuit is going to be. Because the outside space does impede flow of sound as sound emanates from vibrating member or a tube or whatever. So, some of these elements have to be captured in your circuit analysis. So, that is all I wanted to talk about today. Starting next lecture we will start talking about more detailed analysis of the whole speakers, microphones and all. See what they show us.