 Good afternoon, what we are going to talk about today, primarily about generalized elements and in the previous lecture I had laid out some context as to why we would need these generalized elements. Since I have not talked and provided much detail about this whole idea of generalized elements you will not know much about it, but as we go further in the lecture you will see what the generalized element is and why it is needed. So, primarily the logic behind developing this theory or these equivalences is that for instance in a sound device such as a loudspeaker you have electrical energy coming in it gets converted into motion or mechanical energy. So, you have first transformation of energy then you have another transformation where that motion gets converted into sound and to bridge the domain of electrical engineering, the domain of mechanical engineering where you are dealing with motion stresses and strains and the domain of acoustics you need some mechanism or theory. And if we use lumped parameter approaches which work fairly well for devices such as loudspeakers, microphones and a bunch of other acoustic devices then this generalized element approach comes in very handy. So, we talked about loudspeakers now let us look at microphones. So, in general a microphone is a device where you have sound which is being sensed as pressure and then it gets converted into electrical energy, 5, 6 major categories of microphones. So, the first one is a condenser microphone and what you have there is that you have a membrane which senses pressure and then as that membrane moves back and forth it causes the motion of a plate which is also moving with it. In some cases the membrane itself could be a metallic plate. So, you have one membrane moving back and forth and then the other plate is fixed. So, this membrane is moving back and forth. So, the capacitance between the two plates changes and it responds to changes in pressure fluctuations caused by sound disturbances. So, you sense that the change in capacitance and then you convert it into acoustic pressure through again some equivalences. Then another type of a microphone is dynamic microphone and there it is essentially an inverse of a regular loudspeaker. So, what you have in a dynamic microphone is a membrane which senses pressure because of that pressure fluctuation it moves back and forth which in turn gets converted into electrical energy either voltage or current and then that voltage or current is measured and then it gets translated back by some equivalence relations into SPL sound pressure level. Then there are piezoelectric microphones where you have the conversion processes sound sound causes piezoelectric crystal to experience some pressure and when that pressure grows up or it goes down that generates voltage in the piezoelectric crystal. So, again you have a transformation of energy sound to pressure to electricity. There is another device a microphone device fiber optic microphone. So, what you have in fiber optic microphone is again you have a membrane which is moving back and forth. One side of that membrane could be polished and it could be a reflective surface. So, light from some other place comes and it gets reflected and as this membrane is moving back and forth the intensity of that light gets modulated. So, then there is another sensing device which senses the change in light intensity and that change in light intensity is then interpreted back into sound pressure level. So, in all these measuring equipment different types of measuring equipment you have transduction happening. So, there is a need to have some sort of theory on an approach to which you can bridge the gap between electrical mechanical sound or electrical optics sound or electrical piezoelectric mechanics sound and so on and so forth. So, what we will cover today is essentially we will develop few types of generalized elements and then we will see how they map into electrical and mechanical domains. So, that will be the focus. So, I can have a generalized element and I can symbolize it something like this. It could be positive here and negative here and it could have three variables. So, the first variable let us call it A, the second variable let us call it B and the third variable let us call it C and what we says A is an across variable for instance in a resistor what is across the voltage. So, voltage could be an across variable B is a through variable it is flowing through the element and C we call it the element variable and it is essentially a property of that generalized element. So, again if we have a resistor through variable could be current as current is passing through it and across variable could be voltage and the element variable could be the resistances it is a property of that device right and this is all for linear theory these elements as we have developed we will be using in context of linear systems. So, what we will develop is we will develop four types of elements the first type will be two terminal two terminal element. The second element which we will talk about is a four terminal element the third element we will talk about will be source and this will be let us say type A and the fourth element we will talk about is again a source and it will be you can call it type one or type two. So, we will start with two terminal elements. So, the first two terminal element is as a relationship where A which is the across variable is through two variable times element variable and element which has these three variables linked in such a way is a two terminal element of this particular type. So, for instance you have a resistance there is a voltage across it and this current going through it. So, in this case my A is voltage B is current and C is resistance we can also say that B equals 1 over C dot A. So, these are the two forms of a two terminal element and there are different categories of two terminal element. So, this is first type let us look at another two terminal element. So, in this case the across variable A is equal to time derivative of B times C R B equals A over C if I integrate it. So, again if I map it to electrical domain this looks like an inductor. So, again my through variable is voltage no I am sorry current through variable is current across variable is voltage and the element variable is L or the inductance. So, this is another form a two terminal element and it is called a two terminal element because it has two physical terminals. Similarly, an inductor has two physical terminals to the third type of this two terminal element and as you would expect it could look something similar to a capacitor. So, here the through variable which is B equals 1 over C times time derivative of A. So, you have a capacitance capacitor there is current going through it and there is voltage and voltage and current are related in such a way as defined by this relation and I can also reformulate this relation such that A equals B times C d t when I integrate it. So, this is the third category of a two terminal element. We have talked about two terminal element now let us look at a four terminal element which is another thing we talked about. So, what we are going to four terminal element. So, can you guys guess A four terminal physical device which has four nodes transform. So, what you have is these are two terminals and then the other two terminals are here this is one across variable another across variable is G then there is a through variable B another through variable is H and the element variable is another number we call T. So, T in this case is turn ratio and what we have drawn is essentially an ideal transformer. So, in this case the relationship between A and G is A equals T times G B equals 1 over T times H and then we also know that A B equals G H. So, we are right now talking only about electrical side of things, but we will see how these ideas also translate into the mechanical world. So, this is a four terminal element because it has one two three and four terminals. So, the third category of terminals was source type A and here what it is is essentially what we are saying is that the element variable C is same as A and this is the case for an ideal voltage source as an example that is its internal resistance is virtually 0. Now, I think I wrote it incorrectly here because this is an arrow here. So, this should be B and this should be an ideal current source and the fourth one is source type B it should be B. So, what you have here is a source with an element value of C and through variable is B across variable is A then in this case you have C equals A and an example would be ideal voltage source. In the first case it should be B equals C which is what is there. So, again we will we talked about different generalized elements and now we will just formally even though we know in electrical engineering which is what we will just formally document. So, in electrical engineering if I have a resistance current and then this voltage across it then A equals B dot C maps to B equals I dot R. If I have an inductor with an inductance L current is I voltage across the inductor is V then A equals D B over D T times C this relation maps to voltage equals L times rate of change of current. If I have a capacitor seeing a current I again voltage across it being B then A equals integral of B dot C dot D T this relation maps to voltage equals I times 1 over capacitance times D T then I integrate it and my four point element my input current is I 1 voltage across is V 1 and then output voltage is V 2 current is I 2 and turn ratio is T. Then the relationships are V 1 I 1 equals V 2 I 2 V 1 equals the turn ratio times V 2 and I 1 equals 1 over T times I 2 and then finally I will formally list the two sources. So, I have a current source current internal current which is the element variable is I and the current going through the circuit is lower case I voltage across the source is V then here I equals I and similarly if I have a voltage source then my relationship will be. So, till so far it is all standard, but just to provide a context in I have a generalized element and I am mapping that generalized element into electrical engine. So, as the next step what I will do is map it into mechanical area. So, what we will find is that if I use a specific type of analogy then we will see that the true variable turns out to be force and we will see it how it comes out and the cross variable in mechanical word comes out as velocity. So, force is analogous to current it is a true variable and velocity is analogous to voltage. So, we will see. So, the first element was A equals B dot C which is a resistance. So, what is a resistance in mechanical language? It is a dashpot it is dissipating energy. So, let us look at it you have a dashpot the cross variable is U the true variable is force. So, U equals force times mechanical resistance or force equals 1 over R n times U. So, in laws of mechanics you have seen that F equals C x dot this is your x dot U and C is 1 over R n. So, we have seen this equation force is felt throughout the line flow line. So, this equivalence is preserved in spring and in a mass you will see. So, it just happens that our traditional damping coefficient which we use as C we will not use C as damping coefficient in context of acoustics as it is being taught in this particular course. So, if C is damping coefficient such that force equals C x dot equals C U then damping coefficient is essentially 1 over R m and I call this R m as mechanical resistance. What is the unit of R m? It will be inverse of unit of damping coefficient. So, that unit is meters per second Newton's right. So, as you are doing your calculations using this approach you should make sure that you do not equate C with R m it is there is an inverse relation that is a mistake which happens very often. So, we talked about mechanical resistance or a dashpot. So, the second two terminal element is a spring just happens that the symbol is very similar. So, my true variable is F my cross variable is U and we know that force equals stiffness times displacement. So, if I differentiate it I get this relation K times U. So, if I rearrange I get U equals 1 over K times time derivative of force or I can say if I have to find force is basically. So, we will go that go to that later, but 1 over K we term it as C m or compliance. So, again this relation looks very similar if I write this U equals C m times T f over D t. This relation looks very similar to voltage equals D i over D t times L. So, I am seeing again the same equivalence is being preserved. It is being preserved in the case of a mechanical resistance here i you have a derivative of i you have a derivative of f you know voltage you have velocity. So, that relationship is being preserved. So, this is called compliance or it is also called mechanical inductance. Most of people use the term compliance, but you should know if someone uses mechanical inductance you know C m. So, resistance we saw that in case of a resistance the relationships are preserved U being equal to voltage equivalent to voltage current being mapped into force and the element variables. So, now let us look at mass that is the third variable. So, we know that force for a mass which is accelerating is m m times acceleration which is a time derivative of velocity and I can rearrange it. So, I get U equals 1 over m m times f D t if I integrate it. So, again this is again looking similar to the relation as you have in a capacitor. So, the equivalence is a preserved amongst spring is similar to an inductor mass is similar to a capacitor and a dash spot is similar to a resistor. And how do I depict it? So, this is my mass I have a through variable f and the cross variable is U is the velocity this is the inertial frame this thing is your inertial frame. So, this is one difference whenever we measure acceleration we measure it with respect to an inertial frame. So, I have drawn an inertial frame here. So, whenever you have a mass in a system when you try to convert it into a you know capacitive something similar to a capacitor you have to make sure that you refer to you always have a provide with a reference frame in inertial frame this is important. So, we did what? So, is that clear how the analogy between force velocity and current and voltage work and also how mass gets translated into a capacitor a spring gets translated into an inductor and dash spot it is translated into a resistance. What could be a 4 node or a 4 terminal element in mechanical area? Think about it you have a transformer in electrical what could be a 4 node element here is one very nice here is one gears or levers you know you have a lever you have f 1 b 1 you know force and velocity on one side it gets translated into force and velocity on the other side and it is essentially same ratio of length the two lengths. So, that is what we will talk about and you can just map the same thing for gears also. So, a 4 node 4 terminal element and I have a lever and I can do the same thing for gear I have a fulcrum I have two input variables f 1 u 1 f 2 u 2 then the relationships are u 1 over l 1 equals u 2 over l 2 f 1 l 1 equals f 2 l 2 and this is analogous to a transformer. So, here it will be u 1 plus minus this is the through variable this is the second through variable u 2 and this is my turn ratio is the ratio of the length and similarly we have a current source and a voltage source in electrical land we can similarly have a force source and a velocity source in mechanical area what is a force source if you have there are some in you know machines tensile testing machines you can either have a force controlled experiment or you can have a displacement or a velocity controlled experiment. So, when you are having a velocity controlled experiment essentially you are having a 4 velocity source in the system if you have a force controlled experiment then you have a force source in the system. So, these equivalences are preserved transducer most of the transducers they control the velocity and the force is a as an output of external conditions and the overall impedance outside, but the transducer knows how much to move back and forth. So, if it knows how much to move back and forth then it is a velocity controlling. So, you have a velocity source and there are you know other types of transducers also. So, you have to know which type of sources you are placing in the system. So, we have seen different types of equivalences between mechanical elements and electrical elements. So, we should also see one more special thing about transformer which could be a gear or an electrical transformer or a lever that it is possible if you have drawn a circuit to eliminate the transformer at all by appropriately changing the values of some inductances. So, that is what we will see. So, you have a transformer that is its turn ratios t is to 1 again these are my 1 1 2 and this is connected to an external impedance which is z. So, we know that z equals v 2 over i 2. So, z equals v 1 over t over i 1 times t right equals v 1 over i 1 and bear in mind when I am writing v and i actually I should have been a little bit more careful these should be in upper place because these are vectors because there may be a phase component in z. So, v 1 over i 1 times 1 over t square now what this means is that I can replace this entire box by another box which will be something like this is that clear. So, whenever you have a transformer or whenever there is a transformation happening this is important to understand whenever there is a transformation happening from electrical to mechanical and if you can figure out what is that ratio for instance in a coil which moves back and forth force equals b i l right from laws of electromechanics. So, the turn ratio there is b l plus force equals b i l. So, then you can have a mechanical circuit you can also have an electrical circuit and they are initially broken because of the transformer but you can remove that transformer and combine the electrical and mechanical circuit into one single circuit and you can in change the impedance on the mechanical side by using this factor right. So, we will do some couple of examples and how these transformations happen. So, in case of a resistor and we will do more as we draw complete circuits of loudspeakers and microphones. So, you have v 1 and I have a resistance r and my turn ratio is t is to 1 i 1 i 2. So, this is equivalent to t square r. So, I can eliminate the transformer and I can just replace it by an increased resistive load or changed resistive load changed by a factor of t square. So, the other one is an inductor and in this case I have l as the impedance. So, this transforms to l times t square and the last one is of a capacitor. So, I have a capacitance c this transforms to c over t square because your impedance is inverse of c. So, again some of these concepts are drawn from electrical engineering, but they will find them extremely useful as we try to solve acoustic problems and so mechanical problems. We will do one more example and see how it changes. So, if my circuit is like this and the turn ratio is t is to 1 let us say it is a R L C circuit R L C then this is changing to my resistance value changes to R t square l goes to l t square and capacitance changes to c over t square. So, it does not matter you can solve this circuit or you can solve this circuit the answers will be same. So, depending on the situation you can figure out do I want to solve this circuit or do I want to solve this circuit whichever appears easier you can use that. So, we have talked about how resistances, inductors, capacitors transform into mechanical equivalence and back and forth. How transformers can be changed back and forth and also how sources in electrical area map into mechanical area and vice versa. As we are doing this analysis in circuit analysis for electrical engineering if it is a complex circuit then we use two laws to solve for different currents and voltage. One is the Kirchhoff's current law and the other one is Kirchhoff's voltage law. So, you can again we will need similar laws or mechanical area mechanics area to solve electromechanical problems. So, we have a Kirchhoff's current law where through a node if I have I 3 I 1 I 2 then the Kirchhoff's current law says that through a node I equals 0 that is I 1 plus I 2 plus I 3 equals 0. And this maps if I have a node a current node for in the mechanical area what this also translates to is f 1 plus f 2 plus f 3 equals and this is essentially your Newton's. Kirchhoff's law is similar to Newton's law they are not same, but they are similar. Similarly, you have Kirchhoff's voltage law that if you have three elements and a closed circuit you know and the voltage across third element is v 3 minus plus here you have v 1 plus minus and here you have v 2 plus minus then K v l says v 1 plus v 2 plus v 3 equals 0. And this comes from law of conservation of energy there is an assumption here that there is no external flux, but we do not have those conditions then this holds good. In mechanical land this translates to u 1 plus u 2 plus u 3 equals 0 and that is basically your law of conservation of mass. So, these are important rules to remember as you are trying to do an electromechanical circuit analysis. So, we will do one or two very quick examples and see how we can use some of these techniques. So, let us say I have a very simple spring mass system. So, the mass is m m the spring has a stiffness of K which is or a compliance of C m and it is being excited by some source velocity. So, let us say if I am interested to find what is velocity of the mass you want to know what is the velocity of mass. In other words I want to find if I map it to electrical word what is the potential difference across the mass right because velocity is mapping to voltage. So, I make a circuit out of it. So, u s is like a voltage source what is going through the circuit course is flowing through and then I have a spring or an inductor of inductance C m and then I have a mass which is grounded. So, I can now convert it to do formally electrical circuit. So, I have a u s my i equals force u s is velocity source velocity I know this and I can put a capacitance here and again because capacitance is grounded that condition is being satisfied and this value of capacitance is m m. So, my total impedance in the circuit z equals i equals m m d v over d t you should say it will be m m. So, z equals s l s is j omega plus 1 over s m m equals s square l m m plus 1 over s s. So, I put s as j omega. So, what I get is 1 minus omega square l m m over j omega m m is that right. So, now I want to find the velocity of this mass or the voltage difference across this capacitor. So, I can find the current I multiply by the impedance offered by m m. So, the force which is same as current when I map it is u s over z and that comes to be j omega m m over 1 minus omega square l m m times u s. So, my velocity across the mass u m is this is current force right this is equivalent to current. So, is force times impedance of the capacitor 1 over omega c j this gives me u s over 1 minus omega square l m m. So, instead of doing a lot of force equilibrium and solving differential equations I am relatively easy able to solve for the velocity of moving mass by using some concepts of circuit analysis right. So, that is the whole point let us do one more example. So, now you have a spring of compliance C m a dash pot and both these springs and the dash pot are connected to mass a moving mass m m the damping coefficient is C which is inverse of mechanical resistance we talked about. So, if you know the damping coefficient you should be able to figure and I am at this point I am exciting it with a known velocity. So, it is source velocity. So, then the question is find the velocity of this moving mass. So, what is this value that is the question. So, first I draw circuit and what I get is. So, again I have u s here which translates into a voltage source of a strength u s current is f and then I have an inductor I have a resistance and then I have a capacitance the value of capacitance is same as the mass of the object value of inductance is C m and this value is 1 over r m. So, my overall impedance across these two points is 1 over s times moving mass which is because of the capacitance and then for the inductance and the resistance I get r m plus 1 over s times inductance which is C m then I take inverse of that. So, I can I have to take inverse of it this resistance and this inductance is R m that is 1 by C. Yes, I am sorry this should be R m. So, here I should make it 1 over r m. So, through this I can figure out what is the value of z using that z I can figure out that f equals u s over z. So, I know the current flowing through the capacitor or the force seen by the mass and then from here I can evaluate what is the value of velocity. So, which is so u m equals 1 over s m m times relatively simpler or a straight forward process. So, in this whole exercise what we have done today is establish couple of equivalence not couple several equivalence between an equivalence between resistance and a dash spot a spring and an inductor and between mass and the capacitor and then force translates to or becomes equivalent to current and velocity transforms to voltage. This is one way of establishing the equivalences. Now, if you play with the variables in a slightly different way you can establish equivalences which are just the opposite. So, resistance will still map into a dash spot because it is dissipating energy, but a spring can map into here we saw that spring is mapping into an inductor, but you can you will be able to map a spring into a capacitor. So, either you use those set of principles or you use this set of principles which we spoke today you will get the same answers, but you have to be as we are doing this mapping exercise we have to be sure that we are consistent. So, there are people who will do the same sort of analysis, but in their calculations force will translate into voltage and velocity will translate into current and in that particular scenario spring maps into a capacitor and so on and so forth. So, they are two different very similar, but still different approaches of doing these conducting these transformations, but at the end of the day what we have shown today is that we can bridge the gap between electrical engineering and mechanical engineering and we can create one single large circuit where electrical energy is coming in and as an output what you are having is motion being exhibited by the mass. So, if you have a transducer or a speaker which is moving back and forth and if you want to know that if I am putting in 100 watts of energy at 20 hertz how much this diaphragm is going to move I should be able to calculate using this approach. Now, I still have not bridged the gap between motion and sound we have not talked about sound at all today, but in some of the subsequent lectures we will speak about how that motion gets transformed into sound based on some of the transmission line theories we studied earlier 1 d wave equation and so on and so forth, but through this way we have an approach how we can bridge the beam of electrical engineering mechanical and acoustics and in case of most of not most a lot of electro mechanical acoustic systems like speakers, microphones, sensors this approach works fairly well, because most of the elements are very close to lumped parameters. So, 1 d approaches and lumped parameter approaches work very well in a lot of applications. So, it is really worthwhile understanding the other thing is that once you have created an elaborate circuit you may find that it has 7 5 or 6 resistances 6 or 7 capacitances several inductors how do you solve that. So, one approach could be that you go through you know do it through the hard way and solve and take a lot of time, but there are standard software packages in electrical engineering one package is a piecewise. So, you take and it is it has graphic tools. So, you pull a resistance put it on the screen connected to a capacitor connected to you know a resistance connected to another capacitor or an inductor the way you have constructed the overall circuit excited through a voltage and you will get an output number. So, all that can be automated, but we have to know how to construct those circuits. So, what we are learning is how to construct those circuits and then at some point of time we will see how we can use some of these standard tools. We dump all that those circuits into those software tools and then solve for whatever variables you want to know. So, that is what I wanted to talk to you all today and in the next lecture we will further develop some of this stuff which we have been talking about. Thank you.