 So what we will cover today is couple of concepts, one is instantaneous power flow and the context of utilizing this idea is as we had seen in the last class, we had talked about two concepts kinetic energy density and potential energy density. And so the other concept is instantaneous power flow that at every given point of time, if you have an acoustic circuit or an electrical circuit, what is the instantaneous power at that point of time. So we know that volt is real j omega t and I can generalize it as half real then v e j omega t plus v star which is its complex conjugate e minus j omega t. Similarly, current and the reason we are talking about voltage and current is that we will draw these analysis into the acoustic domain later. So this comes in handy is real component of I e j omega t equals half real I e j omega t plus complex conjugate oh I miss yeah complex conjugate e minus j omega t. So we do not need to write real part because it will already be real because we are writing conjugate. So as you guys rightly pointed out, we do not have to explicitly write real because when I add these two v e and v star, it does give me the real component. So my power equals and then I use this capital letter upper case letter p t. So it is a function of time is basically a multiple of instantaneous current and instantaneous voltage. So remember, if you are doing about we are talking about instantaneous power and that is basically a product of instantaneous voltage and instantaneous current. If I have to do average power, I cannot multiply voltage and current, but when I am talking about instantaneous power, it is a valid thing to multiply instantaneous voltage and current. So you have real v e j omega t times half of real I e j omega t. So if I expand that relation, I get this long relation. So it is one fourth v I e 2 j omega t plus v I star plus v star I plus then I have another component which has both stars e minus 2 j omega t. So if I expand on v I star and v star I and I add them up, essentially what we will find is that the real portion of v I star is same as real portion of v star I and the imaginary components are negative of each other. So they cancel out. So we know that real of v I star equals real of v star I and imaginary portions of v I star equals minus imaginary portion of v star I. So these two guys, I can club them up and I can just sum them up and say that this is v I star plus v star I is twice of real of v I star something like that. So I get power is a function of time one fourth v I e 2 j omega t plus I get v star I star e minus 2 j omega t plus 1 over 2 real v I star. So now what I will try to do is, I will try to simplify this term. As simplified this term, this has been taken care of, I have to simplify this term. So again if I expand and if I expand v star I star oh I am sorry e minus 2 j omega t then I will again see that when I add up the real and imaginary components together, imaginary components you can do this math in your home. The imaginary components cancel out and only the real portion survive and they are same. So basically what I get is the final expression is power which is a function of time because it is instantaneous power is half of real portion of v I e 2 j omega t plus half real portion of v I star. This portion its average value is not 0 unless the resistance in the circuit is 0 and there will be any power dissipated. So this portion the average value is not 0 average value is not equal to this portion it is a cyclic quantity. So when I average it over a cycle complete cycle which is 2 omega t over a cycle of 2 omega t then it adds up to 0 when I average it over a period. So we will now see an example what we will do is we will take 2 examples one with a circuit which has pure resistance in it and another which has a circuit another circuit which has a resistive load and also an inductor and we will see how instantaneous power changes and also what it tells us a little bit more and how we can relate it to acoustics and also transmission lines in general. So let us look at this circuit I have a voltage source and it has a external resistance and I am closing the circuit the current is I and my voltage which is a function of time is let us assume 100 cosine omega t and omega let us say is 60 hertz or 60 times 2 pi 60 is the frequency and the value of this resistance let us assume it to be 100 ohms actually I will change my mind and I will make this only 10 volts or 10 so it is 10 cosine omega t. So my power instantaneous power is half real VI star so my V that is the magnitude of this whole thing is 10 so that is 10 times I star so I is V star over Z star it just happens because this is purely a resistive circuit so V star and V are same and Z star and Z is same as R which is 100 ohms so 10 over 100 that is I star plus and then I have a cyclic component real VI e 2 j omega t so V is 10 times I is 10 over 100 e 2 j omega t so if I do all this math I get 0.5 plus 0.5 cosine of 240 pi t so again this is a constant dissipation of power half watts that is being dissipated constantly and this power is a cyclic thing so in part of the cycle energy gets stored by the system and another part of the system energy gets dumped back into it does not get dissipated gets dumped back into the system so it stores and releases the energy going back and forth. The other thing you will notice that the frequency of this is twice the frequency of voltage or current or in acoustic domain whatever is the source signals frequency it will be twice that so if I plot this whole thing my plot will look something like this is my time axis this is the total instantaneous power which is PT so my axis when I plot it 0.51 so it will vary from 1 to 0 to 1 to 0 and so on and so forth and a complete cycle will be twice you know 2 times t which is t is the period of the voltage cycle or the current cycle. We will do one more example so here we have the same voltage source I have an external resistance which is connected also to also an external inductor this is resistance this is inductance this is voltage source volts equals let us assume 100 cosine omega t R equals 100 ohms oh I did the mistake this is again 10 also the value of inductance is 1 over 1.2 pi Henry's I am assuming that number so my impedance of the whole circuit is Z is basically 1 over R plus 1 over SL take the inverse of this and I know that S equals j omega equals 2 pi times 60 times j yeah so my Z becomes if I do the math basically I get 1 over 100 plus 1 over 100 j and then I have to take inverse of this equals 1 over 100 I take it out 1 plus 1 over j so then I have to do the inversion of this entire thing so my impedance is 100 over 1 minus j now if I look at the denominator it is basically there is a constant a real portion to it and an imaginary portion and I know that if I use the complex principles I can express 1 minus j as square root of 2 times e minus pi over 4 j so you have your real component which is 1 then you have the imaginary component minus j that is your so the magnitude is root 2 and this is the angle so e minus j pi over 4 so my Z is 100 times root 2 e minus j pi over 4 everything is in the denominator or it is 100 over root 2 e j pi over 4 I bring it up in the numerator so my Z star is complex conjugate of Z which is 100 over root 2 e minus j pi over 4 so now I can use Z star to evaluate what is I star so I star equals P star over Z star root 2 over 10 e j pi over 4 that is my I star so moving on I go back and start calculating power dissipated so instantaneous power equals half of real component of V I star plus half of real component of V I e 2 j omega t so this I get as half real component of voltages V is 10 times root 2 over 10 e j pi over 4 plus half real component of 10 times root 2 over 10 e minus j pi over 4 e 2 j omega t where omega is 125 so I get if I take the real component of thing I get 1 over root 2 times real component of exponent of j pi over 4 plus again I get 1 over square root of 2 real component of exponent of j omega t minus pi over 4 going back e j pi over 4 is cosine pi over 4 plus j sin pi over 4 the real component is cosine of pi over 4 which is 1 over root 2 so that becomes half and then the oscillating component of the power is again 1 over square root of 2 cosine omega t minus pi over 4 yes 2 omega t 2 omega t minus pi over 4 so again this is my average power and this power this component of power is fluctuating and there is a phase difference of pi the pi over 4 here so now if you plot this this entire expression you will see that power instantaneous power not the average power averaged over a whole cycle instantaneous power can become positive and also it can become negative but the fact that it is becoming negative is not that it is getting absorbed permanently somewhere it is getting saved into system and then the it gets done back into the system back and forth so if I plot it my graph will look something like this so this is my time axis and this is my vertical axis is let us say watts so power and so it will so the maximum value will be half and then cosine 2 omega t minus pi over 4 at the maximum it will become 1 so half plus 1 over square root of 2 that is the maximum value so that is the maximum value which is half plus 1 over square root of 2 now at t equals 0 the value will be half plus 1 over square root of 2 cosine pi over 4 right so it will be half plus half that is 1 so at time t equals 0 this is there and then my power cycle will look something like this and so on and so forth so in certain parts of the cycle energy is getting stored into the system in some other parts energy gets done back into the system but the average power dissipated does not change from a purely resistive circuit which is half watts even though peak power does shift and it does exceed 1 watts first of this number so this is something we have to understand also because we will be mapping some of these concepts into the acoustic domain as we move on which you will see either in the next class or the class after that one so again average power remains same if the resistive load does not does not change the other thing is that peak value can shift because of presence of inductors and capacitors or because of presence of elements in the circuit similar to inductors and capacitors and the third thing is and it has a strong bearing especially in the electrical domain is that this term v over z so v is voltage z is the total impedance that basically influences the total current right now z in a purely resistive circuit we saw was 100 ohms but in case of the other circuit where we had an inductor the magnitude of z was 100 over square root of 2 so it was not 100 ohms it went less than 100 ohms by a factor of 1.414 what that means is that in general the circuit is drawing more current into the system now in transmission lines and this does not have a strong bearing on acoustics itself but it is worthwhile to know in transmission lines you have a power generator unit and that current gets transported over let us say 1000 kilometers and it comes to your home if your home has a purely resistive circuit then it will consume some amount of watts if your home has resistors and also let us say inductors it will still heat the same amount of heat but it will draw more current what that means is that the heat dissipated along the transmission line goes up significantly and most of the loads in your houses what do they have your refrigerator has a motor it has means it has an inductor vacuum cleaners they have inductors mixy it has an inductor most of the circuits in household appliances television it is an electronic device but it also has inductors most of the circuits which have motors they have inductive loads so what that means is that it draws more current so that and what that means is that the heat dissipated in the transmission lines goes up so basically the system overall efficiency the system goes down so what people do is they try to rectify by putting some capacitive load also in parallel to the inductor and that kills the influence of inductive load so that the current comes back to the original number so that we will very quickly capture and see what happens so if I have a resistor an inductor and a capacitor and let us say my R is 100 ohms L again is 1 over 1.25 and capacitance is 1 over 12,005 then my total inductance total impedance will be 1 over 100 plus 12 pi over j times 120 pi so this is the inductive part plus 120 pi j over 12,000 pi and then this whole thing I have to invert and if I do the math correctly it again brings back brings the overall system impedance to a purely resistive number which is 100 ohms so by adding a capacitor you can change the phase back to 0 degrees similarly if you have a predominantly capacitive circuit and if you want to reduce the power consumption in an overall sense then you put some inductors appropriate value of inductors and you can bring it back so in this context there is another term and it is called power factor correction and this also used in acoustic it is good to know that it is power factor correction so we know that we saw that power is half real component of v times i star plus half of real component of v i e 2 j omega t so this is half real component of v v star over z star plus this oscillating component and I know that v v star is square of the magnitude of v so I can bring it out now let us say that z is magnitude of z times e j psi so if I plug this relation in here I get v square magnitude of v square over 2 and then I get real component of so I will still have and then 1 over e minus j psi and then the oscillating component right and so it becomes v square over 2 into 1 over magnitude of z times cosine psi plus other things this number cosine psi is power correction factor if psi is 1 then my transmission losses in the circuit are minimized is not psi cosine of psi is 1 then my transmission losses get minimized if it is more than 1 not more than if it is more than a psi is more than 0 degrees then I start seeing losses in the circuit especially in the transmission so their country is especially in Europe which have legal requirements by for all product manufacturers that you have to buy a refrigerator it has to have this correction factor built into its system so that transmission losses in the country do not get big do not become large in India we lose about 5 percent of total electricity just because of this power correction factor so we have this nuclear deal and all these nuclear plants will be coming up so total energy generated by these plants will be something equivalent to if we just fix one small power correction factor in our product that much energy can be conserved by one change so wanted to share this so we have seen electrical power at an instant of time so now we map the same idea into acoustic now in acoustic domain it happens that we use pressure and velocity so pressure times velocity is essentially not energy but energy per square not power but power per square area if you do the dimensional analysis of that so a lot of times people use the term acoustic power but what they are in reality implying is instantaneous power per unit area so you have to be careful that when you are doing the math that scaling happens accurately so acoustic power this common language again essentially basically power per unit instantaneous power per unit area so let us call that w acoustic power so is so it is basically a dot product of pressure at a given point in time times velocity so that is my instantaneous acoustic power per unit area and again the velocity and pressure variables they have to be normal to each other that so if the pressure is acting on this plane the velocity has to be normal so so I expand this actually so I know that t x t is real component of t x omega e j omega t similarly q x t equals real component of u x omega e j omega so I substitute these pressure and velocity functions back into the equation for w and what I get is w x t equals half real e dot u star plus remember I am using exactly the same analogy which I had in electrical engineering p dot u e to j omega t okay so now what we will do is we will evaluate what is the power dissipated in a closed tube and also in a in an open tube of infinite length and see how it works out so we will go to a closed tube so I have a tube closed at this end it is a rigid perfect closure so everything is getting reflected such that reflect reflective coefficient is 1 and the thing is getting excited by transducer here like this this is my x equals 0 here x equals minus l my input excitation is that at x equals minus l t the velocity is u s superior number cosine omega t so in complex notation I can write that as real component of u s e j omega t okay so I calculate instantaneous acoustic power for unit area so that equals half u z that is my pressure right times u star plus oh I missed real real component of u z which is the pressure function times velocity e to j omega t now we know that impedance in a closed tube we had seen in earlier classes at minus l omega equals minus j z naught cotangent of omega l over z so my power equals 0 plus u s square over 2 real component of a very big number very big term j z naught cotangent omega l over c times e to j omega t so my acoustic power is 0 plus u s square over 2 cotangent of omega l over c sin 2 omega t so in a closed tube the total acoustic power being dissipated if there are no dissipative losses in our wave equation we assume that there are no dissipating losses dissipative losses happening because of the medium is 0 average power being consumed by the system is 0 however energy transfer does happen between potential and kinetic states so my plot for the power would be something like this where this this is time so this will correspond this distance will correspond to twice the frequency of the excitation frequency so we will do one more example and that is of an infinite tube and again I am exciting it at this location to a velocity source x equals 0 at this location so also I know that at x equals 0 velocity equals u s cosine omega t and in complex notation I can write it as real component of u s e j omega t so again instantaneous power per unit area is half real u times z times u star plus half real component of u z times u e j omega t closing the bracket so again it will transfer between kinetic and potential energy but theoretically because you are assuming that damping is not happening so even in pure in purely spring mass systems when you excite it at a resonance theoretically I am again giving me a parallel analogy when you excite a system which has a pure which has which is purely having a mass and a spring at resonance the displacement goes to infinity but that does not mean that the energy required to induce that kind of thing has to go to infinity because it gets compensated by between I mean it fluctuates between kinetic and potential so all is have what is happening is energy transfer from potential to kinetic so going back that is my acoustic power per unit area is half real and then I get u s square times z plus half real u s square times z e j omega t I missed 2 here so again I have u s square over 2 real component of z plus u s square over 2 real component of z e 2 j omega t now we know that for an infinite tube z equals z naught equals rho naught c we have seen this right and it stays constant throughout the length of the tube so if I make this replacement then instantaneous power equals u s square over 2 z naught plus u s square over 2 z naught cosine omega t so here average power is non-zero and this is the oscillating component of the power so in a close tube we saw that because all the energy gets reflected by the close end surface the consumption of power by the source point at the source point is 0 but here because you have infinite tube and if I am exciting a membrane back and forth that energy travels keeps on traveling and never comes back so my average power consumption is non-zero so physically also it makes sense even though in both the situations you do not have any dampening happening. So if you have an infinitely long tube you have a membrane in an infinitely long tube and it is exciting air so that membrane is initially exciting air at this location then after sometime it excites air at this location so it has to induce kinetic energy and potential energy in medium which is infinitely long so it will keep on consuming power that energy will not come back to it at all because the tube is infinitely long if you have a tube which is close at the end then that energy comes back it gets reflected because this t minus is non-zero so then the energy required to excite is not non-zero it is zero in that case in a close tube even though in both the systems you have no dampening. So this is what I wanted to capture today in terms of instantaneous power and also about how it relates to acoustics how it helps us understand total dissipated total power consumed in an open tube in a close tube the other thing I wanted to share with you is a brief introduction to how transducers acoustic transducers are made so that you become familiar with the structure of it the notion of transduction and then how do we start establishing equivalences between electrical and mechanical components. So we will just for a few minutes go over this then in the next class we will go deeper into it so what you have here is let us say you have a metallic structure what I call frame this is made of metal on which the whole transducer or the speaker loud speaker is mounted so this is a metallic structure and this is I have made a cross section so you can rotate it by 360 degrees and you will get a visualization. Then at the back side you have what we call a back plate or a rear plate this blue thing in some of these speakers there is also a hole because heat gets generated and this hole helps dissipate the heat so this is what they call wind then between back plate and there is another plate called front plate this magnet and then there is also a metallic piece which is this magenta in color and between the magenta piece and the front plate you have this thin film like thing and it is called a bobbin which we talked about in the last class and on bobbin is bound the voice well so as electricity comes into it AC current comes into it it goes through and spins around the bobbin and then gets out and because of electromagnetic induction the voice well moves back and forth as the voice moves back and forth its motion is opposed by spring force which is generated by this green thin film called a spider and it is also opposed by at this brown piece of material called a surround and then you have a rigid diaphragm called a cone and finally we have a blue membrane called a dust plate so what is happening is that you have electrical energy it comes into the system it gets transformed into mechanical energy and the transformation is called electromechanical transduction then the mechanical energy gets transformed into sound through mechanoacoustic transduction transducer is any device which transforms one type of energy to another type of energy so if these are electric transducers where you have electricity being transformed into motion or vice versa so this is a three dimensional view of the same object so you get a better understanding that is your spider this is your surround it is typically made of foam this is a diaphragm which moves in and out to cover close the vent hole you have this dust cap you have a voice well here magnet is here magnet is sandwiched between the front plate and the back plate so I just wanted to give you a brief understanding of how a transducer is constructed because in next class what we will try to develop is that as electric electricity is coming in and it is a signal it is not a DC current it has all sorts of frequencies in it it goes through that winding which has an inductance in it there may also be an external capacitor mounted on the speaker so how do you bridge the electrical element of the circuit to the mechanical element to the acoustic element how do you make that bridge between these three domains electrical mechanical and acoustic because at the end of the day what you want to know is that oh so much is current is coming into the system at such frequency what is the pressure I am going to hear that is what is needed that is what we want to know so how do we bridge that gap and what kind of a theory we should use to understand that fundamental question so that is what I wanted to share today and with that I think we will close.