 So, in the last class we had talked in detail about how to develop complex models and what we will be doing today in the this specific class is essentially a follow up on that and we will be actually talking more about specifics constructing actual models and also solving those. So, in one of the earlier classes we had developed the lump mass model lump parameter model for a suspension system for a car. So, we will revisit that because we never solved it in that class. So, to recap let us say you have a mass this could be the mass of the car and since their car has four wheels I will only take one four one fourth of the structure. So, well I will only consider one wheel and the load shared by it. So, the mass of the car or the one fourth mass of the car is MMC and this is my wheel which is moving on a road which could be uneven. So, as this wheel moves on the road it moves up and down and what I am interested in this particular problem is as the wheel moves up and down the person sitting on this body what kind of motion does he or she experience that is what I am interested in. So, what I am interested is that if the velocity here is u road what is the velocity of the person. So, I can call it u car now between the car body and the wheel typically we have springs I will put a compliance member and also we have some times dash parts which damp the vibrations. I am calling this as CMS meaning suspension. So, it is compliance of the suspensions is mechanical compliance of suspension and this is RMS that is mechanical resistance of the suspensions. Now, the wheel itself has mass as it moves up and down it has some mass. So, I assign it a value moving mass of wheel which is W and there is air inside the wheel which acts as a stiffness member. So, it also has a stiffness C M W. So, again what I am interested in is if there is an input motion because of the uneven of the unevenness of the road which is u road then what is the motion which is going to be experienced by the person sitting inside the car that is what I am interested. So, let us construct a lumped parameter model of this and I am going to use the mobility equivalent to construct it. So, that is my input source u road and I have two masses and in mobility equivalence we always ground the mass. So, my first mass is this it is M M W and this is my second mass which is mass of the car or one fourth of the car mass. So, it is M M C and I am going to ground both of these also I have this stiffness or compliance element C M W and then I have suspension stiffness and suspension damping both are in parallel. So, I put them in parallel. So, here it is 1 over r m s and this C m s. Let us put some numbers on these terms. So, C m s is 4 times 10 to the power of minus 6 meters per Newton M M W which is the mass of the wheel around 30 kilograms. 1 over r m s is 7 times 10 to the power of minus 4 meter per Newton second. Compliance of the suspension stiffness is 1.7 times 10 to the power of minus 5 meters per Newton and mass of the car or one fourth of its mass. Let us say it is 455 kilograms, we have done this example. Yes, we have done only part of the example. We never solved it, we did only the break points. So, we will very quickly revisit those break points and then actually start solving it. So, the first break, so what we are interested in is this ratio u car over u road. This is my transfer function, that is what I am interested in. How does this ratio change as my excitation frequencies change? Now, we had in the earlier class talked about break points. The first break point is associated with the fact that for some set of frequencies current will be more or less passing through the resistance and as frequencies go up in number, then that the current going through the inductor will become less and less and most of the current will go through the resistor. I have incorrectly labeled it. This is 1 over rms and this is cms. The other break point is associated with this loop versus current going through the other loop. Very quickly we call this loop 1, we call this loop 2 and also we assign another variable here, which is the voltage difference across these two points and we call it u wheel. Remember velocity of the wheel, which is essentially velocity of this point is not necessarily identical to the velocity here. So, we consider loop 2 and in loop 2, in loop 2 I have this parallel circuit. This is 1 over rms, cms. So, as we talked earlier at low frequencies most of the current will go through the inductor. As frequencies go up, current will start going less current will flow through the inductor and more and more of it will go through the resistance. So, the crossover point will be such that 1 over rms equals 2 pi f times cms. So, this gives me the first break point f 1 is 1 over 2 pi 1 over rms times cms and that is 6.6 hertz. Again, below 6.6 hertz more current will go through the inductor, above 6.6 hertz more current will go through the resistor. So, this is first break point. Second break point is associated with the fact. So, above 6.6 hertz the current going through this circuit will also have some sort of resonance. This is MMC. So, let us compute the break point of MMC and the compliance and let us see where it falls. So, for this circuit which is essentially as current is going to this route, let us see if the resonance is below 6.6 hertz or not. For this one my frequency is f 2 is 1 over 2 pi 1 over MMC times cms and when I plug in the numbers I get 1.81 hertz. So, the point what I am trying to make is that below 6.6 hertz most of the current in this circuit will flow through the inductor and within that range from 0 to 6.6 there it will cross a certain break point again which corresponds to 1.8 hertz. If this number had been more than 6.6 hertz then this analysis would have been meaningless because current would have not gone through this would not have gone through cms anyway. For third break point what we are interested in knowing is what happens when current goes through CMW and MMW that is another inflection point. So, for that one f 3 equals 1 over 2 pi CMW times MMW and that gives me 14.5 hertz. So, these are my three frequencies 1.8 hertz, 6.6 hertz, 14.5 hertz. So, I have to find in from 0 to 1.6 how is the circuit behaving from 1.8 to 6.6 how is the circuit behaving and from 6.6 to 14.5 how is it behaving and what happens above 14.5. So, that is what we will do today and our aim is to find this transfer function. Yeah. Sir why are we interested in taking CMW and MMW? We should have considered because current has to all pass through CMW. So, we should have considered the MMW branch or the rest of the branch. So, instead of considering CMW and resonant between CMW and MMW we should have considered MM between MMW and rest of the branch. So, we will see that how it becomes just hold on. So, you will see that. CMW will be in the circuit regardless. Yes sir. So, we have to consider this and this. Yes. So, just hold on to your thoughts. So, you will see the relevance. So, what we are interested in is this ratio. So, u car over u road equals u car over u wheel times u wheel over u road. So, what we will do is in general our approach will be we will find how this transfer function is changing with respect to frequency and how this transfer function is changing with respect to frequency. We will construct both plots for both of these two transfer functions and then we will add them up and then we will get the total transfer. And in this context the value you will see the importance of 14. So, what we will start with is u car over u wheel this ratio. So, again u car. So, to make it clear this is my u car. So, u car over u wheel in that context what we are interested in is essentially loop 2. Whatever happens on loop 1 does not have a bearing on how the ratio u wheel over u car or u car over u wheel will get impacted. Now, the way the reason why we are doing all this is that if you do not have a computer and you do not have to solve very complicated equation. Then you can get reasonably good both plots using this approach you break the circuit in small pieces and analyze them piece by piece. Given that we have a lot of computers we have software tools which you can make all these circuits and the software tools and the software will solve for complex quantities very quickly. But to get a physical insight as to what is happening that is why we are doing. So, to get my u car over u wheel this is my sub assembly and what I have is m m c this is c suspension stiffness and that is my damping term. So, switching screens. So, this is my u car and that is my u wheel. So, let us see what happens above a if my frequency exceeds 6.6 hertz. If my frequency is above 6.6 hertz if it is less than 6.6 hertz then in this circuit I can ignore the impact of c m s. So, all I have is 1 over r m s and m m c. So, then my distilled down circuit becomes 1 over r m s and m m c c m s this is u wheel and that is my u car. The true variable going through it is force. So, the current which is analogous will be something like u wheel times divided by the impedance of this entire circuit z which is u wheel over l write the impedance here 1 over s which is my complex frequency j omega times c m s plus 1 over s m m c. So, I get that as u wheel over s square c m s m m c plus 1 into s m m c and my voltage difference across the moving mass of the car is basically this current times m m c times s the impedance offered by 1 over whatever that number. So, u car is this whole term u wheel over the whole denominator times s m m c to 1 over s m m c. So, my u car over u wheel equals 1 over 1 minus and once I put s equals j omega I get omega square c m s m m c. So, this is for frequencies less than 6.6. So, now I make a plot of it here I will construct a board plot and a board plot has two parts low frequency asymptote and high frequency asymptote. When the frequency goes low then this value becomes 1. So, my 20 log of this value becomes 0. When this frequency becomes very high I get a negative slope of how many dB's 40 dB's because it is an omega square. If it was just omega times some number it would have been a 20 decibel negative slope, but I will get a minus 40 dB slope and the threshold the crossover point. So, my board plot will look like if I plot it here itself. So, this is my decibels log omega my frequency response will look like something like this and the slope is minus 40 dB per decade. My actual curve because at a particular value of when this denominator becomes 0 my actual value will look something like this because this resonance happens. But what we will be constructing is essentially the dark line which is asymptotic force B. When frequency exceeds 6.6 hertz what will happen? I can drop the compliance term and include the resistive term. So, my equivalent circuit will be this will be my equivalent. Can you infer what will be the shape of the board plot for this one? For this circuit without going through the analysis I mean we can construct you know we can develop again as we did here find the overall impedance and find the voltage difference across the moving mass. But just based on whatever we did here if I have an inductor and capacitor in series I get a minus 40 dB slope. If I have a resistor and a capacitor in series what kind of a slope will I get? 20 minus 20 or plus 20. So, the slope across if it was purely a resistor what will be the slope across it? It will be 0 because there is no omega term. If there was no capacitor or inductor the slope of the response curve will be 0 because there is no omega term. You have omega square. So, that generates a 40 dB. So, it will be only 20 dB. Also what is this number? What is this frequency? 1 over C m s n m s. And what is that number? Is that 6.6? We calculated 3's frequencies right? 2 pi into 1.81 2 pi into 1.81. Yeah, yeah, yeah. We calculated 3 frequencies. One was where is it? 6.6 hertz, 1.8 and 14.5. Which one is this? 2.8 2 pi into 1.8. 1.8. Yeah, I am just calling the frequency. Yes, if you want to make it omega it will be 2 pi. So, it is 1.8 hertz. So, this is 1.8 hertz. It keeps on going with a negative slope of minus 40 dB. Above 6.6 hertz as you guys mentioned. So, I will construct this again. Logs omega. This is decibels. This is 0 dB. 1.8 hertz. I have a slope of minus 40 dB up to 6.6 hertz. And above 6.6 hertz I get minus 20 dB slope. Here the slope is minus 40 dB. I am plotting a resistor and a capacitor in series. Yes. It should be constant after a while. Because it would be 1 upon s 3 plus r. What I am plotting is a ratio of the voltage difference across this element and the input which is u v. So, there will be an omega dependency always because u r is dependent on omega. So, this is my overall plot up to 6.6 hertz. Now, what we will do is. So, we have plotted this one u r over u wheel. Now, we will work on u wheel over u road. So, u wheel over u road and the effective circuit for that is essentially this. So, this is u road that is mass of the car, mass of the wheel, M M W. And this is compliance of the wheel. What will the boat plot for this circuit look like? Minus 40 dB. Minus 40 dB. High frequency or low frequency? High frequency. And low frequency? Zero. It will be a zero dB. Same thing which we did earlier. What will be this number? 14.5 hertz. This is my decibel scale. Glotting log of omega on this one. This is zero dB slope is minus 40 dB per decade. The actual curve may look something like this because at resonance the denominator goes to zero. So, this is u wheel over u road. So, I have a plot for u wheel over u road and I have a plot for u car over u w u wheel. So, all I have to do is now just add these two up. So, my final plot looks something like this. I have three break points 1.8, 6.6, 14.5. I have initially a flat line at zero dB. Then I get a negative slope of minus 40 hertz, minus 40 dB decibels. Then my slope is still negative, but it reduces. This is minus 40 dB per decade. This is minus 20 dB per decade. And then I get a very acute minus 60 dB per decade. So, what this picture is showing is that given the parameters which we had listed earlier in the mechanical circuit, first thing is that you have virtually no attenuation at very low frequency. It is all zero dB. Whatever is going in is being experienced by the messenger. And as frequencies go up, you start seeing attenuation. And that attenuation starts very rapidly increasing after this 14.5. So, you have to pick the right amount of damping, the right amount of stiffness as you are designing. Whether it is a mechanical, it is a suspension circuit or an acoustic system. What this is telling you is that you have to make the proper choices of some of these parameters, so that you get the right amount of damping. It also tells you is that if the frequency of input signals, which is essentially coming from the road, is in this region. Let us say your road is such that most of the frequencies are coming from 0 to 1.8 hertz, then the suspension is doing not much. So, depending on the road conditions, you have to construct, you have to develop or identify the right parameters of your suspension stiffness and suspension damping. That is what you do. So, whatever you have to understand the topography and based on that you construct the right damping. Another thing you should think about is that and this will help you understand this. Suppose you are driving, suppose this road, this is perfectly flat road, but in real conditions it is doing this. When your car is going like this, the car is seeing the same amount of vibrations, whether it is going at slow speeds or at high speed, but the driver experience changes. You can try it, even on a scooter or a bike, you could try it. You have a small bump when you hit and you try to cross it at a high speed or try to cross it at slow speed and how do you feel and why do you feel, you should think about it and figure out based on this understanding. At different speeds, even though the amount of bumps and the frequency of bumps is same, the perception of the person who is driving the car or bike or motorcycle changes and the amount of vibrations he or she experience also changes and it also depends on speed and what that has to do is it relates to the Doppler effect that as you are moving at higher speeds, the apparent frequency of these bumps which are coming from the road, it goes high or low, the frequency it goes high. So, that has a bearing and that you can relate that fact to this curve and figure out why you feel better at higher frequencies or at higher speeds or you will feel more uncomfortable at higher speeds. So, we have done one example and this one was overdue, we had constructed this. The next example we will today do is of a loud speaker, given that this is an acoustics course. What we will talk in the remaining part of the lecture is model of the speaker and we will also try to, if hopefully we will try to finish this solving it today itself, at least one model of the speaker. So, what you are seeing here is a magnet which is sandwiched between two metal plates. The plate on this side is called the rear plate or the back plate and the plate on the front side is called, it is on the front, so it is called front plate, because this is the front of the speaker. So, it is on the front side, so it is called front plate and between the front plate and the center you have this magenta or pink metallic piece. This is also made of a regular steel, which could be magnetized and there is a small gap between this blue front plate and the pink pole piece and in this gap you have this voice coil. So, just to zoom in, you have front plate, you have pole piece, here you have voice coil on which the windings are there and here this could be a magnet, I am just cutting a cross section. This is your back plate and back plate is connected with the pole piece something like this. So, the way magnetic circuit works is, this is magnet, it acts like a battery. So, you have magnetic flux line going here, then they can also go in air, but air is much more resistive, so they do not give there, they get bent, they come here, they cut the voice coil conductor, they come here like that, this is how the circuit comes. So, it analogous to like an electrical circuit. This gap, people try to minimize this gap as little as possible, because the more gap you have, the impedance to the magnetic lines will be higher and higher. So, typically this gap between the voice coil and this is about half a millimeter to account for tall winds as another things also. Also, you have gap on this side, you have gap on front of the voice coil and also this gap is there, this is gap too. So, people call it outside gap, inside gap that is also not typically more than half a millimeter. If you have big gap, which it means that you have to put bigger magnet in this, more stronger magnet and magnets are expensive. So, people try to minimize the gap, so just wanted to give you some flavor of that. There is an electrical part, current is coming in, it goes through this voice coil winding. What will winding have in it, electrically what elements will it have in it? It is a conductor, so it will have resistance, if you have to model it, you will model it as a resistor and it also it is a winding, it can also be an inductor. So, people of model it as typically resistor and a inductor in series, that is the electrical side. Then the electrical side, the electrical energy gets converted to motion, so the transformation factor that is what D L, we saw it in the earlier class. And on the mechanical side, what do you have? This voice coil is moving up and down, it has some mass, so you can model it as a mass. Also, you are seeing that this green element, it is a suspension element, it acts as a spring. Similarly, this curve here, this curve here, this also acts as a spring, so this is also, so these are two spring elements. So, you have mass because this voice coil is moving back and forth, it is physically moving back and forth, so it has a mass, there is an inertia effect. Also, this entire diaphragm is moving back and forth, that also has mass. So, you have mass because of all these things, you have stiffness because of this spider and what they call the surround. Also, spider itself has a little bit of mass, it is not like much, but it has a mass which is not negligible, so people account for that mass also. So, it has a mass, it has a spring element in it and then also anything, any mechanical thing moves, piece of paper, rubber, metal, does not once you strike it, it vibrates for a while and then it damps down because there is always internal damping. So, people put damping elements on side. So, on the electrical side, there is a resistor and inductor, on the mechanical side, there is, you have to think physically how I am going to model it, there is literally this mass, this spring and there is damping. And then this whole speaker could be mounted in an infinite baffle or in a tube or whatever. So, once we go to the acoustic side, what do we put? Radiation impedance because that speaker is going to radiate energy out into free field and that generates radiation impedance. So, based on that physical understanding of the system, we will construct a lumped parameter system, model of a, we call it speaker. So, my input energy is voltage, I can measure it, 14 volts, 10 volts, whatever it goes through and this is the electrical element, electrical side of the story. So, I call it L e, electrical inductance and this is R e, electrical resistance. Then electricity gets converted into motion. So, what is my through variable here? On the mechanical side, if I use impedance analogy, voltage corresponds to velocity, right, u and the through variable is? Of course. Of course, turn ratio is B L is to 1, B is magnetic flux, L is what? Length. Length of what? It is conductor. It is conductor. It is conductor. Total length of the wire, the total length of all these coils, it could be in several meters. There is a springiness in it that I represent by mechanical compliance C m. There is mechanical mass M m and there is damping. I have question here. Does it make sense to put C m, M m and R m in parallel or should they be in series? Based on your physical understanding, this is the picture of the model. So, should I put the mechanical elements, string, mass and damping in series or in parallel or should I put them in whatever? Why should I put them? We are talking about the magnet. Magnet. M m is what they call mechanical mass. Mechanical mass. M m is the mechanical mass, which is a sum of this voice coil. It has some mass. Also part of this diaphragm, which is moving back and forth. All the element moves with same velocity. So, we have to keep it in parallel with that. So, because we are using mobility model, u across each of these variables is same. And then this mechanical circuit is coupled with an acoustic circuit. In the turn ratio here is 1 is to AC. What is AC? It is area of the diaphragm, area of the cone. Whatever area is radiating sound, it is the area of that. What is the across variable here in the acoustic lamp? Volume velocity. And the through variable here is pressure. Here I have that. And I am assuming that this speaker is mounted in an infinite baffle. What that means is that there is an infinite wall extremely rigid. There is a small hole in it and there is a speaker mounted in it and it is firing sound on one side. On the other side there is back. That is what I mean. So, for that from standard texts, the radiation impedance is I have to put an acoustic mass here and then I have to put an acoustic resistor. Whatever is the energy dissipated across the acoustic resistor is something I will be listening to. So, what we are interested in is v4, which is the voltage across this acoustic resistor. So, given a value of v, what is the value of v naught? That is what I am interested in. So, the transfer function which I am interested in is H s equals v naught over v. This is what I am trying to find. We will make this circuit a little simpler not by dropping any element because we have not done any element. What we will do is we will eliminate this transformation here. And we had seen earlier in an earlier class that if I want to eliminate this transformation, my r a will get converted into r a divided by this turn ratio square of turn ratio. So, r a over a c square, my capacitor will get multiplied by that and my voltage will get divided by naught square just by a c. So, what we will do is I will just construct only this part of the circuit. So, this is m a times a c square. This is r a over a c square and my voltage will become v naught prime, which is basically v naught over a c and this part of the circuit is same. So, if I can find instead of v naught over v, if I can find v naught prime over v, I am fine. I can very easily transform that into v naught. So, now what I will do? I will assign some values. So, my b, b is magnetic field density is 1 tesla. These are fairly close to some of the real numbers if you go and buy a speaker. So, what and I have drawn this from Barenek book. It is about 10 inch big speaker. Length is 9 meters. So, these are long words. Just gives you an idea. There is a lot of copper while boundary. R e is 8 ohms and L e is 0.17 millihenries. Moving mass or mechanical mass is about 11 grams. Compliance is 1.79 times 10 to the power of minus 4 meters per Newton. R m is 0.5 Newton second over meter. A is 0.13 meter. A is the diameter of the dipole. So, from that I can calculate a c. So, a c is pi a square over 4. Did I miss anything? R a over a c square equals 0.32. So, what will I do first when I start trying to analyze the circuit? What did you do in the case of this suspension system for a car? Whatever are interesting points you have to figure them out. So, the first one could be we will just look at the electrical circuit. On the electrical side you have L e and R e. Those are the two elements. So, there will be a certain threshold above which inductive part of the impedance will become significantly larger compared to the resistive part. So, above that threshold I can only consider inductive in the electrical part. Below that threshold I can only consider resistive as a very general gross approximation. So, that what is the threshold? So, that threshold will be when R e equals 2 by F e L e. From here I get my F e equals 1820 hertz. So, above 1820 hertz in my circuit analysis I will be ok as a first level of approximation if I can drop the resistor. Below that I will have to consider the resistor and I can ignore inductive. Second one is acoustics. Now of course, if you have a software you can do it you do not have to do all these things. So, for the acoustic part what should I be doing? This is acoustic. You have a capacitor and a resistance. So, I will do the same thing there R a equals 1 over 2 pi F a times m a. This gives me F a equals 1 over 2 pi m a R a equals 770 hertz. So, I started with the easy part. There are very few elements here. There are very few elements here. Now we will look at the middle one. The middle elements are what are there in the mechanical? You have a compliance. You have stiffness that is the stiffness. You have mass and you have a dashboard. If my transducer is in vacuum again abstraction. The rule of R a and m a will be 0. The mechanical mass acoustic part I can ignore that. So, if my system is in vacuum they will the mechanical part will resonate in such a way that is it has two elements mass and spring. So, natural frequency in vacuum is mechanical. In vacuum basically my F vacuum will be 1 over 2 pi k over m k is 1 over compliance. So, it is m m times c m is 115 hertz. What this physically means is that if I place my speaker in a vacuum and I strike it, it will resonate at 115 hertz. If I place it in freer it may not resonate at the same time because there is a radiation impedance. Second one is in infinite baffle. When I place an infinite baffle I have m m and this m a times a c square in parallel. So, I can add these guys and I can find the total the final resonance when this thing is in an infinite baffle. Do you understand it? F infinite baffle is 1 over 2 pi times as the compliance changed the compliance is not the mass is changing. Does everyone understand why is a c square here? Because we eliminated the transform. So, this brings down my resonance to 89 hertz. It is a significant difference. The air in the system is acting as a mass and that brings down the resonance from 115 to about 90 hertz. That is a very significant change in frequency. So, you cannot ignore the impact you know role of radiation. Third one this is again. So, this is my actual condition. If I place my speaker in a hand or I place it on a table then it what frequency it is going to resonate? National frequency will it resonate at f vacuum, f infinite baffle or at some number. National frequency forget the table I mean I can put it in hand or I can hang it in here do not worry about table. See what you have here that right. So, this membrane is moving back and forth right. It has two surfaces one surface is pointing here let us say my observer is here and this is back and forth. This membrane is moving back and forth it has one surface facing this side this way and the surface facing this side right. So, this side you have the radiation impedance which we accounted for here by m a times a c square. On this side there is no radiation radiative impedance because there is vacuum. Now, if I place it in hand or I place it on a table both the surfaces of the same membrane they will see the same radiation impedance on back side and also on the front side. So, instead of m a times a c square what will be the new number the mass this it will be twice that in hand my f is 1 over 2 pi 1 over c m times m m plus 2 m a a c square that comes to 75 hertz. So, that is so these are so what we have identified is this is something we will care about as we develop the board plot for the whole system. We will also care about this one we will not care about this one because the actual speaker is mounted in a infinite battle. We will care about this one 89 hertz and again we will not care about this one because the actual mounting condition is in an infinite battle. So, my board plot will have it will be something I do not know what it is going to be like, but it will have several break points 89 hertz 717 hertz this is from acoustic circuit this is from mechanical circuit and this is from 1820 hertz and this is from electrical circuit at these three these are the these are the three important break points. We have short of time so we will not develop the actual board plot today, but think about it that if you want an ideal speaker loudspeaker what kind of board plot would you prefer should it be flat, should it be going up, should it be going down, should it be wiggly what kind of a board plot would you like to have an ideal perfect loudspeaker and why will it be good what does it mean physically same sound whatever you are getting in is being faithfully reproduced at equal levels for all frequency. So, bear in mind as we are developing that board plot and then we will see the type of response we get for this particular system.