 So, in the last class we had just begun analyzing the electro-dynamic lumped parameter model of an acoustic transducer, where we left in the last class was that we had identified some break points or resonance points, specifically there was an electrical resonance between R e and L e and that was happening at around 1820 hertz that is what we calculated. Then there was a resonance as a consequence of C m and M m and that was happening at 89 hertz and then an acoustic resonance between because of M a and R a that was occurring at 717 hertz or something like that. The other thing we had done was that we had decomposed this transfer function v naught prime over v naught in terms of u coil over v times v naught prime over u coil. So, what we will do today is we will develop a board plot for the first ratio u coil over v and see how it changes as frequency changes and then we will also develop a board plot for the second transfer function v naught prime over u coil and then once we have these two board plots together then we will add them up and see how the overall transfer function of the overall transducer looks like. As we develop this relation you may want to remember that an ideal transducer should have a very flat transfer function over a range of frequencies. So, the wider that flat area is from an ideal standpoint the better it is from a listening purposes. The other feature of a good transducer would be that not only it should be flat, but it should be as high as possible because what that means is that if I send in a little amount of electrical energy it gets converted into more and more of sound. So, actually the efficiency of very good acoustic transducer does not exceed one or at the most if you are really smart 2 percent. So, that should give you some overall perspective the overall efficiency of sound system or a transducer hardly exceeds 1 or 2 percent. So, we will start developing a relation for u coil over v v naught prime over u coil. So, we will do this in two steps step one is we are going to do v naught prime over u coil. Now, when I look at this circuit my v naught prime is the voltage across this acoustic resistor r a over ac square and u coil is basically just the voltage across the transformer after the electrical circuit and the current in the mechanical circuit is essentially force and the current in electrical circuit is obviously current. So, when I look at this circuit what I infer is that to get this ratio all I have to consider is basically the element m a times ac square and r a over ac square that is the only thing I have to worry about. So, the equivalent circuit is something like this. So, this is m a times ac square this r a over ac square my input voltage will be in this case u coil and what I will be measuring and what I am interested in is v naught prime. So, this just for sake of simplicity is something like this I just simplified the notation right and if I compute impedance if I compute the impedance and I try to find the ratio the relationship which I get is v naught prime over u coil comes to is complex frequency times capacitance which I have noted as m times r over 1 plus s m r. So, now I construct a board plot. So, on my horizontal axis I have logarithm of omega and in my vertical axis I have decibels omega tends to 0 how will the board plot look like as omega or that is s which is j omega as omega goes towards 0 the low frequency asymptote how will it look like. It will go through the origin and it will be a straight line with a positive slope of 20 decibels per decade and once it crosses a certain threshold then as omega goes towards infinity what happens this ratio essentially converges to 1 right at infinity. So, log of infinity log of 1 is 0. So, the high frequency asymptote is something like this and this what is this value 0. 0 decibels what is this break point this point 1 over m r right what was the 717 hertz we had calculated ok. So, this is my board plot for v naught prime over u coil let us call this figure a. So, we have finished developing a board plot for the second portion which is this now we will do for this one. In the second step step 2 what we are going to compute is board plot for u coil over v in the last class we had found that there was a resonance happening at 89 hertz right. So, below 89 hertz u coil over v is this is my value v and this is u coil below 89 hertz what is going to happen is that most of the current is going to pass through C m the mechanical inductance the impedance due to m m r m and all these things will be extremely high for f less than 89 hertz my equivalent circuit is going to look like C m and that is my u coil why did I neglect L e in the circuit I dropped out L e in the circuit on the electrical side I have not included L e why have I done that L e becomes important as we had calculated after 1800 and 20 hertz below 1820 R e is going to dominate right. So, I can neglect the impact of L e for f less than 89 hertz the board plot in this case is going to look like how will it look like it will be a straight line slanting positively and the slope will be 20 dB per decade and this is 89 hertz. So, now I have to go from 89 to what is the next break point 700 and 17 hertz right. So, for f 89 hertz is larger than is smaller than f and that is 700 and 17 hertz. So, my electrical circuit is going to look like. So, now I will have the capacitor m m both these capacitors are going to be because very low current is going to pass through this inductor because I am above 89 hertz and the capacitors will become active. Which one in an asymptotic sense. So, as s becomes infinite as s becomes infinite the numerator is becoming more and more as s tends to infinity numerator and infinite denominator they become more and more equal. So, the ratio goes to 1 regardless of the values of m and r. So, this is my reduced circuit for this frequency range 89 hertz to 717 hertz again this is v and this is u y. So, my board plot in this case and as frequency goes higher and higher the impedance offered by these. So, I get my this was my original line going up to 89 hertz then I have a negative slope and the negative slope is 20 dB per decade and this is going up to 717 hertz. So, again on the horizontal axis I am plotting logarithm of omega and on the vertical axis I have decibels and this is still u coil over v. So, we know response below 89 hertz response above 89 hertz up to 717 hertz. You would ask ourselves a question that what happens at 89 hertz specifically. So, that is what we will once if we know the answer to that we will know this value how many dB is at 89 hertz. So, that is what we will find. So, what happens when f equals 89 hertz. So, at 89 hertz on the other side above the beyond the transformer the circuit will look like there will be a capacitor. There is an inductor which is mm capacitance is yeah. So, I have capacitance of mm inductance of C m and resistance of 1 over R m in a general sense I am just calling them L C r and when L and C resonate and they are in parallel. Their overall impedance does it go to 0 or does it go to infinite when they are parallel. If it goes to infinite we will see that if we just find the impedance of this block Z block is 1 over S L plus S C inverse equals S L 1 plus S square C L. So, at resonance the impedance offered by this entire block is infinite because 1 plus S square C L becomes 0. So, no current is going at resonance when an inductor and capacitance capacitor they are in parallel and they resonate no current flows through them and all the current goes through the resistor. Our original question was that what is happening at 89 hertz what that means is all current is going through the resistance. So, just at 89 hertz my equivalent circuit is u coil for this one if I actually compute the value I get u coil over v equals 1 over 9 times 162 over 162 plus 8. I am just basically plugging in the values which we had recorded earlier. So, this comes to 0.1059 in decibels it becomes if this is w 1 then w 1 becomes 20 dB 0.1059 log minus 20 dB about. So, I go back and modify this graph and I make this peak value minus 20 dB. One note of correction in all these cases the value here is a resistor below 1820 hertz it is going to be a resistor because that is larger than the inductor and both are in series. So, this also needs correction. So, I apologize. So, now we have figured out the board plot from 0 hertz at 89 hertz this value becomes minus 20 decibels beyond 89 hertz again it will start rolling off at a slope of minus 20 dBs per decade. So, our next one is for the range 717 hertz to 1820 hertz. So, again looking at making judgments which components to include which components should not be included my equivalent circuit looks like the only thing I have is m m this is my u coil is everyone clear why this is the case the board basically what this picture tells me is that the board plot will not change its characteristics beyond 717 hertz also the slope will continue to fall at the same rate the value. So, my modified board plot becomes 89 hertz and it keeps on going till 1820 hertz that is my omega this is in decibel and what I am plotting is u coil over v this is up to 1820 hertz. Finally, for f greater than 1820 hertz I get. So, here I have a resistance and a capacitor above 1820 hertz what is the change going to happen inductor becomes active the electrical inductor becomes active and the slope of this line will be what it will have a slope of minus 40 decibels because now I have to earlier we had seen that if I put a capacitor and inductor in series which is the same case here the slope becomes minus 40. So, I will go back and modify it will make a new one clearly this figure is not to scale and this value is minus 20 decibels. So, again this is u coil over v we started with the fairly not a very complex circuit, but the fairly complex circuit, but just by making identifying the break points decomposing our transfer function into small manageable ratios. We have been able to plot the board plots for all the important all the components which are going to play a role. So, now what I am going to do is I am going to find the board plot for the entire function which is essentially a sum of those two board plots. My final board plot will look like something like this. So, this final board plot will be a plot of v naught prime over v what is the slope of the first segment 40 decibels and this is 20 decibels and this guy is minus 40 decibels per degree looking at this picture what conclusions do you draw? That is the first thing that from 89 to 717 hertz you can use it and whatever is going in it will be faithfully reproduced at the same level output response will be same will have a flat response. Now, in this entire analysis I mean we have plotted it up to infinite hertz this range could be as high, but as we were developing the lumped parameter model of the entire system we had made certain assumption and the most central assumption was my size of the transducer or radiating area shall not exceed lambda over 2 power. So, I have not done the calculation, but the validity of this board plot is only good to the extent that whatever let us say lambda over 2 pi comes to 1200 hertz I have not calculated but you can go and calculate what that number is. So, this board plot can be relied upon only up to that particular frequency that is the second thing. The first one is that I have flat frequency response from 89 to 717 hertz if everything else is good then I can rely I can use this in this way. The second thing is that I may have to restrict the usage of this board plot at a cutoff frequency which corresponds to lambda over 2 pi to be hyperbolic. The third thing is lumped parameter model is especially from mechanical standpoint is that I am assuming that the masses are rigid I have a diaphragm and it just moves like this. Now, in reality when you have any membrane or any structure it has its own modes it could be that the diaphragm starts breaking up meaning that it starts exhibiting its own modes at 300 hertz 400 hertz I do not know if that is the case then that is another word of caution that this particular board plot has its validity as long as rigid masses remain rigid. We have assumed that the diaphragm is infinitely stiff and it does not bend and twist as it moves back and forth. So, someone while this analysis is done at a more detailed and a refined level someone has to go into a modal analysis of the diaphragm using some finite element method and figure out what are the modes at which the diaphragm starts exhibiting its own it is no longer rigid. So, that is the third thing. In what sense the lumped parameter model? No lumped parameter model it gives you a guideline if because it is a relatively straight forward way just doing the diaphragm finding modes of diaphragm all it will tell you is that the diaphragm is going to have. So, when you do this is your diaphragm right and then here you have a voids file I am exaggerating can everyone see this or can you see. So, and then there is a suspension here so it is like a spring. So, when you do and then there is also a suspension here. So, when you do modal analysis of this using finite element method you model you may create elements on the spider you create elements on the surround on the phone everywhere you have elements and then you do modal analysis naturally the first mode which is the finite element analysis shows it will fit out will show will be what it will be equivalent to your mechanical resonance k over m right. So, there is a k associated with these springy elements and then there is the overall mass. So, the first mode will be let us say m 1 will be I do not know what we calculate k over m 75 right. So, it will calculate it is 75. So, what that means is that this entire structure and when you draw the mode shifts you will see that finite element will predict that this entire thing moves up and down in the modal analysis that is a pistonic or a rigid body motion that is fine and may be some other mode m 2 what it will do is that it starts instead of moving like that the cone also has a propensity the structure to do this right. I mean it can move like this it can also do this because it is hinged here, but at a higher frequency. So, this was so that so this can do this. So, this is called rocking mode this is going too deep into transducer design. So, we are not developing some systems in this class since you are, but it is still higher what starts happening is may be in the next class I will show you some pictures that this cone it no longer remains rigid. So, it starts doing suppose this is a cone surface it starts doing something like this and then you have to this 3D when you do a 3D picture of this 3D if I look at it from the top side you will see something like this. So, some part is coming up some part is going down and so on and so forth what that means is that the cone is no longer rigid. So, at frequencies above that particular number the value or the faith which I will have in the predictions of my lump parameter model will be limited. They may still be valid in a quantitative, but they will still qualitative, but they will have a limited value. So, but again we start typically some lump parameter model because it gives you an overview gives you an overview how the system is going to behave and it is very easy to tweak specific parameters and adjust this flat range. If I want my flat range from let us say 40 hertz to 1000 hertz suppose the requirement is that I want my transducer particular transducers to start working from 40 hertz to 300 400 hertz then I can use this kind of approach to figure out what are the optimum parameters how much is my moving mass going to be once I know that I can figure out the size and the thickness of my cone in different part and I can design them. I know from this approach what is going to be the stiffness of my system. So, then accordingly once I have that target number I can design the shape of my spider shape of this around and figure out how to get that particular number. It also tells me what is the value of BL going to be which helps me design the magnet. The same model also tells me what is going to be the optimum value of RE which helps me design the voice well. It also tells me what is going to be the optimum value of AC radiating area and that essentially helps me understand how big my transducer has to be. So, there is a very strong value in using this approach in figuring out the overall design at a coarse level and once I have figured out the coarse level important parameters of the design then I go part by part and design them at a more refined level. So, that is the value of this system overall at a system level modeling the entire system. So, what we will do again in next 10, 15 at the most 20 minutes is we will change play with a couple of parameters in this circuit and see how it impacts the overall response of the system. So, that will again give you a close layer the value of this particular way of modeling transducer. What we had assumed in this particular model was that the transducer is mounted in how is it mounted? Infinite. Infinite backfill. So, you have an infinitely large wall it mounted there the wall is rigid it does not move back and forth and the speaker is radiating sound outside in the positive half plane. Now, in a real application you do not have infinite backfill. So, in a lot of conventional systems you may have seen big boxes right where the mount is the speaker. So, what we will try to find out is what does that do to the performance? Transducer in a box or sealed enclosure. So, I have a box this diaphragm is moving in and out and let us say the volume of this box is V naught. Physically we have talked about this earlier what does a volume do? It acts as a spring. So, I have this entire circuit and the stiffness of this spring is V naught over rho c square and if I have to remove that transformer I also put an ac square in the denominator. So, this is acting like a inductor. This inductor is going to be parallel will it be in parallel to the mechanical inductor or in series to the mechanical inductor which is the spring? It will be in parallel if you add two springs in series the overall stiffness does it go up or does it go down? It goes down right if you have the two springs in parallel the overall stiffness goes up. So, what this box is doing is it is going to increase the overall stiffness. Physically I am trying to explain all I have to do here is I have to add one more element. I just erase some things for so that I can put in more stuff I will put one more compliance number and this will be C B compliance of the box. And the compliance of the box is this value to make my life simpler for analysis purpose. I can assume that C B equals C M it makes things simpler. What that does is that my F M mechanical resonance it goes up by how much? Basically if C B equals C M then I can add these two up together right C M and C B I can just make that make the overall thing as two C M. My mechanical resonance will go up by a factor of my overall system response will be this was my original curve 89 hertz. Basically what I will get is I will get another parallel line this is 89 times root 2 89 times root 2 that is this by omega log omega. So, the dark black line solid black line is the original line this dotted line and it corresponds to the performance of the transducer in a box whose stiffness is same as the stiffness of original transducer. So, if that is the case then this gap A A prime is off by half of A and off cave is factor of 2 right half of root 2 times root 2 this drop of 6 decibels. So, basically what you have done is once you put system in a box have reduced the operating back of the transducer. So, that helps you understand why in conventional systems will come in boxes which are very large. So, that the additional stiffness exerted by the box onto the whole system if my system becomes large and large this A prime moves towards point B A if I move towards B. So, if I have to bring it closer to A then I have to keep on increasing the size of the box. So, that is why in a lot you have seen in auditoriums I mean sometime they have boxes as a book as a room it is something I mean height of the room something comparable to the height of the box. So, that is the motivation will do one or two another one is what they call a ported box. We have talked about this ported box earlier in one of the earlier lecture again in context of boxes we have developed. So, I have my original transducer and then I have a port length is L its area is A T what this does is it add following elements to the box. So, it is V naught over rho naught C square A C square this is basically the stiffness of the box and then rho naught L over A T times A square this is the acoustic mass which we had talked about earlier port side act as acoustic masses. So, in this circuit what I am going to do is I am going to add this additional circuit this value the inductance is V naught over rho C square A C square and this is rho naught L over A P times A C square. This is square will be a numerator. It is an LC circuit how I am interested in two parameters one is whatever is the volume value in physical sense sound coming out of the port that is one thing I am interested in the other thing which I am interested in is whatever what is the spectrum of sound as it comes out of the speaker there are two sources which are radiating energy. So, this is the I call it V cap sound is coming out of the speaker that one surface sound is also coming out of the port that is another surface I am interested in both of them and in a very quality sense we will try to see how things change at resonance below and above resonance and so on and so forth. So, the first thing is that this additional circuit has its own there. So, it resonates which is there in blue because the capacitor and inductors are in parallel are in series their overall impedance goes down to 0 when they are parallel their impedance shoots to 0. So, when that happens almost all current goes through this loop. So, at resonance of this LC circuit you will have a peak in terms of d B s field, but it will be in a narrow band around the resonance it is in. So, one at LC resonance all current goes through port I am putting port in parallel this is because I am using mixing current with port. Second think about it physically what happens when frequencies are extremely low when you have very low frequencies meaning below mechanical resonance when the speaker is moving out there will be a vacuum which will get created in the or low pressure condition which will get created in the box and air will come in from the port and it will go in you have a suction condition happening when speaker is moving out. So, the transducer is trying to generate a positive pressure and port is trying to generate just the inverse of it and both these things will almost cancel out each other at low frequencies. So, the output at low frequencies will be negligible point about frequency is important because at higher frequencies the time it will be required the inertia effects will become important enough. So, that the time required for air to come in and compensate for the pressure will not be large enough that high frequencies this will not happen, but at low frequencies this will happen at low frequencies very low output. What this case happen with all all sorts of frequency. All sorts of frequency what is low frequency depends on the system parameters, but it will it is not specific to this particular transducer, but it will happen in any case. Third at mechanical resonance mechanical resonance implying C M and M M when they resonate the overall impedance we had seen was infinite because C M and M M are in parallel. What that means is physically that at that point the piston does not move much even though current is going through there is impedance and will not move much if the impedance is infinite. So, the piston does not move much and all the energy it gets channelized through the port all the energy gets channelized through the port. So, the volume velocity at mechanical resonance of the port is very high. So, you can design the system in such a way that the at resonance the volume at the contribution of port is high at specific frequencies where the transducer itself is not contributing and at other frequencies where port does not do much your transducer plays an important role. So, a lot of so what this system does is its performance is still not as great as that of an infinite baffle, but in reality you cannot have an infinite baffle and if you want to get closer and closer to infinite baffle you need very large boxes which is not practical. So, people have started using in a lot of cases ports even in small sound systems boom boxes which you may have you may see a small tube going right support there is a volume inside and there is support inside that is all physically. The third example will consider is increased mass by mass I physically mean mass of the phone mass of voice well part of the mass of the these springs what happens if I increase them without increasing or changing other engine ring parameters specifically resistance stiffness and all those numbers when you increase your mass the resonance is going to come down. So, that is my original curve when my mass goes up then in theory my curve looks like this what I get here is that at a reduced resonance I get some extra decibels this is about if I increase my mass by if I double it then I get an enhancement of about 3 decibels which is significant my overall broadband response goes down by 6 dB, but my operating width becomes a little larger operating width becomes a little wider. So, this is a good story as well as a bad story good story because my operating width has become a little larger. So, what I can do is that to get to the same sound performance pressure level I can put in more power and I can get more energy more sound energy for a wider band the bad thing is that I have to put in more power this is what I wanted to cover in today's lecture I also wanted to quickly recap on an assignment problem which I had given in one of the earlier assignments. And the question was I think that merits a little bit explanation the question was that if I have a ball of radius r naught this was in assignment 5 and it is pulsating uniformly and it emits an energy such that at radius r 1 the total pressure not pressure the power output is something like this this corresponds to t time period what I am plotting here is power per unit area this value is 3 times r positive number m this value is negative m and one of the questions was that if I have this graph what is the phase difference between pressure output going pressure and velocity what is the phase difference phase angle of the output going to be. So, there were some questions on that particular this particular part of the problem I just want to capture that for starters what is the average power in this case m times 4 pi r square 4 pi r square because this is power per unit area some people miss that we know that power per unit area we had developed this relation is half of real e u star plus half of real e u e 2 j omega t. Now, before I go more into detail I just wanted to just by looking at this picture we can make some inferences that given that there is minus 1 this graph for power is not symmetric if it was perfectly symmetric then the average power will be 0 what that means is that it will not have any resistive element in it this is a combination of some resistive element and it is an inductive or capacitive element in the overall system. And we know that for a point source or a source of finite size as it emit radially into far field there is a real component in its impedance and there is an imaginary component this asymmetric function is a consequence of the presence of real and imaginary components. So, this picture should give us some idea how to figure out what is the phase relation if it was perfectly symmetric then the phase would have been either minus 90 or plus 90 if it was everything was 0 0 or positive then the phase would have been 0 because it is purely resistive. The expectation is to figure out what is the phase difference and finding it in actual number based on the information provided. So, this is I develop this further real p plus over r 1 e minus j omega r 1 over c that is my p times u star. So, that is p plus star over r 1 e j omega r 1 over c times 1 over z star plus half now I will expand the second component this one is p plus over r 1 e minus j omega r 1 over c times u. So, it is p plus e minus j omega r 1 over c over r 1 u is basically p over c. So, I 1 over z and then I have e to j omega t 1 over z star is 1 over what is that relation rho naught c minus 1 over j omega r 1 rho naught 1 over z is this entire thing there is no negative sign, but there is a positive sign. So, z star is just basically I put a m. So, I plug this up here I know that power over area is also the other thing is the 3 m 2 m 2 m this is also power over area right some this graph the average power is m which is and the transient component is 2 m cosine 2 pi 2 pi t over t. So, I put all this in this original relation and what I get is m plus 2 m cosine 2 pi t over t equals my first component which is real p u star half. So, that gives me p plus r 1 square 2 why do I do that because p plus times p plus star gives me modulus real 1 over rho naught c minus 1 over j omega rho naught r 1 plus. So, now I am writing the transient component 2 r 1 square now before I expand I just wanted to make a small comment I am assuming when I am developing this relation the transient proportion of I am assuming that z is modulus times e j phi I can assume that. So, with that assumption I get 1 over z e j 2 phi minus 2 omega r 1 over c plus 2 omega t. Say that again p plus times p plus second term does not contain star I am assuming yes. So, as I said I am assuming that p plus is p plus is what I get is. So, I have 2 sides in the equation and now what I will do is I will just equate the steady state part with the steady state part as the first step. So, let us call this equation 1. So, comparing steady state parts I get I am just talking about the first part yes you are right. So, comparing only the steady state parts on both sides I get times 1 over rho naught c. So, my p plus becomes 2 r 1 square rho naught c times m. So, now I use this equivalence put it in transient and then equate the two transient portions sides. So, what I get is using that exercise 2 m cosine 2 pi t over t equals m r 1 square n 2 they get cancelled out rho naught c real portion of this entire thing exponent j 2 p minus 2 omega r 1 over c plus 2 omega t over z.