 So, in the last class we had started talking about this idea called radiation impedance. So, any acoustic element when it radiates sound into free air or atmosphere, there is an impedance offered by the larger atmosphere onto that acoustic element and we had developed analytically a relation for a pulsating sphere, what its radiation impedance could be. We have not developed any analytical relationship for other acoustic elements. Let us say you have a membrane moving back and forth which is mounted on a flat surface. What is the impedance which that membrane sees as it radiates into free air? We have not developed relations for those types of elements. So, what we will talk today about will be, we will talk a little bit more about radiation impedance of pulsating sphere. After that we will take a detour and we will talk about a concept called duals which is relevant in context of radiation impedance particularly and in general about these electrical mechanical networks. And then we will go back to radiation impedance and see for other elements, other acoustic elements. So, that is what we plan to cover today. So, in the last class we had developed a relation for radiation impedance of a sphere and acoustic impedance looked something like z r and that was z naught times s r naught over s r naught plus c. This is for a pulsating sphere extending and contracting back and forth uniformly in all directions. And you remember impedance in general is essentially pressure over velocity. In context of radiation impedance we had also talked about the bar square area and stuff like that. So, the question we had posed in the last class was that if I have to, if I have this impedance in a circuit in my acoustic circuit and if I have to break it up into small resistors, capacitors, mass elements, inductors, how will I do that? In other words, if I have a pressure applied here and I have a black box which can constitute a group of different acoustic resistive inductive and capacitive elements, then how will this network look like so that the impedance offered by this network is equal to this. That is what we are trying to figure out because if we know that then we can put those resistors, inductors, capacitors in the whole circuit and we can solve. So, what you see here is z r equals z naught. Z naught is a real constant number rho naught times c. So, z r equals z naught times s r 0 over s r naught plus c. And as, so let us see how this z r behaves. As s goes to 0, as remember it is j times omega. So, as s goes to 0, z r approximates to z naught times s r naught over c. So, at low frequencies the radiation impedance of a pulsating sphere looks like that offered by a, of an inductor, right, because there is an s term and everything else is constant. So, this mimics like an inductor. Inductor in the context that z equals p over u, right. So, I am in the impedance analogy line. Also as s goes to infinity that is as my frequencies are very large, z r approximates to z naught which is independent of natural frequency. So, at higher frequencies the radiation impedance looks more like a resistance. We know two things that if I have to break this into electrical elements it is composed of an inductor and a resistor. So, the only question is whether it is in a parallel or in a series, right. So, if it was in a series the form would look something different. So, clearly it is in parallel. So, I construct a circuit and I put a value r here, put a value l here and this is my v v. So, if I use the impedance analogy the value of r would be z naught and value of l would be z naught times r naught over c. What is r naught? r naught is the radius of the pulsating sphere. It is a pulsating sphere we are talking about. Remember here r naught is radius of the pulsating sphere. So, this is the electrical circuit which mimics the behavior of radiating impedance when a pulsating sphere is pushing sound into larger atmosphere. And just you can double check to cross check. If you put these and you try to compute the overall z you will find that that number comes same as what you have developed here. So, this is for impedance analogy. So, if I move to mobility analogy then we had talked that my inductors become capacitors r becomes 1 over r and so on and so forth. So, my mobility analogy will be something like this. So, this is r equals 1 over r this is capacitance my across variable is v v through variable is pressure. So, you remember you see that. So, here see the value of c is same as value of r which we had said z naught times r naught over c and the value of r is 1 over r. I think I made a small correction is to do. So, here it is actually velocity. So, I have to adjust if I have to go to yeah. So, if I use volume velocity and the pressure then in impedance analogy my L which is the inductive part is basically z naught r naught over what is the transformation ratio a and my resistance is z naught. We see two things here that when we move from mobility to impedance and impedance to mobility capacitors we have seen this earlier on capacitors become inductors inductors become capacitors r becomes 1 over r and so on and so forth. But one thing which we did not very explicitly talk about is that if there are some elements which are in parallel then when we transform it to the other analogy they become come in series. So, that is what we are going to talk about and that is the context in which we introduce this term called dual. So, we will talk about duals. So, let us consider a simple circuit I have a voltage source V s which is connected to a bunch of elements electrical elements. So, I have a resistance here R 1 another resistance R 2 inductor L fester C the current going here is I 1 the current going in this loop is I 2. So, the current in this loop will be I 1 minus I 2 in a vectorial sense. So, I have one loop here this is one loop and this is another loop. So, what we do here is because I have an external voltage source I will try to solve this problem using the curve voltage conservation laws. So, I will apply that law for loop 1 and I will apply that law for loop 2. So, I will get two different two equations I have two variables I 1 and I 2. So, I can solve for 2. So, for loop 1 what do I see? So, first I move to frequency space by mapping this into their impedances. So, C becomes gets mapped as 1 over C s L gets mapped as L times s. So, once I have moved there then for loop 1 my equation is going to be 1 over C times s times I 1. So, that and then in this part I have R 1 times I 1 minus I 2 and this entire thing equals and then for loop 2 I have R 2 times I 2 plus s L times I 2 plus. So, here I take I 2 minus I 1 times R 1 equals 0 and now I rearrange this and just construct a simple matrix. So, my unknown variables go into a vector the unknown variables are I 1 and I 2 on the other side of the equation I have my known variables V s and 0 and here I have 1 over C s plus R 1 minus R 1 R 1 and then here you have R 1 plus R 2 plus s times L. Look at this matrix equation carefully. So, you have this part what is this physically it is 1 over C s plus R 1. So, it is essentially the overall impedance of first loop this same thing here also R 1 plus R 2 plus s L is the overall impedance of the second loop and these two terms this is like a finite element equation not strictly like a finite element equation. These are coupling coefficients. So, minus R 1 in the first equation and minus R 1 in the second equation is the common impedance which is common to loop 1 and loop 2. So, remember this because then when we go to the to another example you will see very good similarity. So, you have constricted a set of equations and then you can now invert this matrix and solve for I 1 and I 2. So, now we consider another here instead of a voltage source I have a current source I s and the circuit looks something like this. So, the value of this inductance is L this is R 1 this is R 2 these are resistors and then I have a capacitor. In this I have a current source. So, instead of using Kirchhoff's voltage conversion potential conservation law I will use the other one current conservation. So, I have two nodes here E 1. So, let us say the potential at node 1 is E 1 and this is another node E 2. The potential at this node is E 1 potential at the second node is E 2. So, using conservation of charge I develop relations again I will have I have two unknowns E 1 and E 2 if I know E 1 and E 2 then I can compute everything. So, I have to figure out what is E 1 and E 2 I know what is I s L R 1 R 2 and C. So, the current going. So, the current at going coming into node at this first node will be same as current leaving this node that is essentially what I am trying to do. So, my first equation becomes E 1 over S L which is essentially I 1 plus E 1 minus E 2 which is the potential difference across the resistive element R 1. So, the current going through that will be that whole thing divided by R 1 that equals what will come on the right side I s. Similarly, the second equation becomes something like this E 2 minus E 1 over R 1 plus E 2 over R 2 plus E 2 times C s equals 0. So, now I rearrange some of these terms I construct and put them in a matrix form. So, my unknowns will go in a vector E 1 and E 2 on the right side I put my known quantities I s and 0 and my first term in the matrix is 1 over L s E 1 plus 1 over R 1 then I have a 1 over R 2 negative again I get negative 1 over R 2 then I get here 1 over R 1 plus 1 over R 2 plus C times s. So, again we have a matrix of different equations two equations. So, what is the first term in this matrix it is 1 over L s plus 1 over R 1 physically what that means is it is the total admittance at this node admittance is like 1 over yeah 1 over R 1. So, this term let us call this term a is the admittance of elements at node 1 admittance is your 1 over impedance. Similarly, this term which corresponds to node 2 is admittance of all the electrical elements which are coming in at node 2 and this term minus 1 over R 1 is the admittance of the term which is common which connects physically connects node 1 and node 2 that is one thing other thing is now let us look at the values of these terms. So, what you are seeing here is just let us compare. So, you have 1 over C s plus R 1 right that becomes 1 over L s plus 1 over R 1. Similarly, you have R 1 plus R 2 plus s times L here what you are getting is 1 over R 1 plus 1 over R 2 plus C times s. So, what is what essentially you are seeing is that a the circuits look similar, but they are not identical the matrix equations look similar they are not identical, but if you compare these two circuits the first circuit where we had a voltage source and the other one where we had a current source then V transforms to I and back and forth C transforms to L and back and forth R transforms to 1 over R and back and forth series elements which are in series they go in parallel and vice versa elements which are in series they become go in parallel elements which are in parallel they go into series like R 1 and R 2 and then you see here. So, R 1 and N goes in parallel to R 2 what else yeah a current source transforms to voltage source and back and forth and one thing which remains constant in this is power dissipated in the circuit. So, power dissipated in circuit does not change also power dissipated in every network and across each element that also does not change. So, circuits like the two which we saw they are called one is called circuit A is called dual of circuit and vice versa. Every electrical circuit which is composed of these three types of elements R L and C you can have it is another thing to notice that if I have a matrix of equations only if I have just the equations for instance if I am just looking at this equation and I do not know the circuit then it is fairly straightforward for me to draw an actual circuit of the whole system. How will I do that when I look at the circuit I will see oh there is a current source here. So, I will put a current source then I will say it is a 2 by 2 matrix. So, there are two potential nodes in the system. So, I will put two dots then I will say how what is the impedance of the two nodes which are connected to each other. So, I will say that the admittance when I look at these blue boxes is one over R 1. So, I connect them by a resistor of value R 1. Then I say at node 1 I have two elements an inductor and a resistor and they are in parallel. So, using that kind of a thought process I can construct a circuit if I just know the equations of the overall system. If I know the system I can develop equations if I know the equations I can construct overall. So, we will very quickly do one example for illustration. Let us say it gives me 7 volts and this is a purely resistive circuit. So, this is 8 ohms the value of resistance here is 4 value of resistance here is 2. So, the equations so I have two loops loop 1 and this is loop 2. Let us say the current going in here is I 1 and the current going here is I 2. So, my equations become 8 plus 4 minus 4 minus 4 4 plus 2 times I 1 I 2 equals my external voltage is 7 and 0. So, this is a DC circuit. So, everyone can see the equations. So, solving this I get I 1 equals 3 4th of MPS I 2 is half an M. Given this can we construct a dual. So, my first step will be that I write the equation of dual. So, what will go in this vector V 1 and V 2 or E 1 and E 2 what will go in the on the right side it will be 7, but this is a current source what will be the first element of this matrix. It is remember its admittance. So, it is 1 the resistance value will be 1 over 8, but the admittance will be 8. So, it will still be 8 plus 4 you see this here these elements are impedance values admittance is 1 over 8. So, this is a resistance of 8 ohms which will transform to a resistance of 1 over 8 ohms in the dual, but the admittance of that 1 over 8 ohms will be 8. So, it will still remain 8. So, 8 plus 4 minus 4 minus 4 4 plus 2. So, now we construct a Rengen you got it. So, now we construct a circuit for this. So, my first thing is that there is a current source 7 M current source. How many nodes does this have a element have 2 nodes. So, first one is at a potential E 1, second guy is at a potential E 2. What is the element which is connecting E 1 and E 2 1 over 4 ohm resistance. The other thing is 4 and 2. So, 1 over 2 and then on the first node I have 8 and 4 2 elements each of impedance 1 over 8 and 1 over 4. So, that is my equivalent circuit. So, the current here in going through 8 ohm resistance is 3 4 M. What is the voltage of E 1 voltage at E 1. What is the potential at E 1? No, E 1 will be 3 by 4 based on right E 2 is half volts. What is the potential drop across this resistor 8 ohm 8 times 3 by 4 right how much is that 6. So, voltage drop across 6 8 ohm resistance is 6 volts. So, the question is how much is current going through 1 8 this resistance 6 M and so on and so on. So, let us very quickly find the power dissipated. So, in the first circuit power dissipated will be 8 times 3 over 4 8 square times 3 over 4 plus 2 square no half square times. Sir, from the voltage diagram say it should be 1 over R times 1 over R. Say it is 8 plus ohm. So, in the voltage when there is a voltage. And sir just now we have seen that if the resistance is R 1 it becomes 1 by R 1 M. Yes. So, what I will do is I will erase this now. So, that we can find the power dissipated on this page itself rather than going back and forth. So, power dissipated here in the first circuit is 3 over 4 square times 8 plus half square times 2. And what is the current going through the 4 ohm resistance, what is the value of the current going through 4 ohm 1 4. So, everyone sees that the current going through 4 ohm resistance is quarter ohm times 4. So, if I do all this correctly my total heat is 5 or 5.25 and in the second case. So, here my current is 6 ohms and here my current is 1 ohm. So, my total power in this case is 36 times 1 over 8 to the first resistor plus 1 square times 1 over 4 plus 1 square times 1 over 2 and that again comes to 5. So, these duals are good to know. Why because in your real complex circuit analysis you may be you may find that oh it is easier to move from mobility to impedance or impedance to mobility and so on and so forth. So, when you are transforming one model into the other models these some of these laws have to be preserved. It is just not that capacitors become inductors R becomes 1 over R, but also series goes into parallel and so on and so forth. So, it was for that reason we transformed this particular network when we move to mobility model the R and R became 1 over R L became C, but because they were in parallel they moved became they went into series. So, that is the context I wanted to introduce that for very sound engineering reasons you go from parallel to series and so on and so forth.