 So, in the last class, we had talked about several acoustic elements. We had developed relationships which were analogous for resistors, inductors and capacitors. Something similar to these electrical elements, we had developed acoustic elements, acoustic mass, acoustic inductor or compliance and then an acoustic resistance. Where we left in the last class was that we still have not developed some elements which bridge the gap between electrical and mechanical. So, in mechanical we have elements which behave very similar to electrical elements and then an acoustic beam we have elements which behave very similar to mechanical elements, but how do you convert from electrical, jump from electrical side to mechanical side and mechanical side to acoustic side. That piece is missing. Those specific elements we call coupling elements. So, we will talk about those today, coupling elements and these could exist between electrical and mechanical reams, electrical to mechanical or they can exist from mechanical to acoustic. Similarly, they can exist between mechanical and piezo and so on and so forth. So, in today's lecture, we will cover basically these two coupling elements and these are essentially transformers. So, they transform one form of energy into another form of energy or transducers. So, we will start with electrical to mechanical coupling. So, let us consider a magnet and this is its north pole and then I have the south pole. This is my south pole and there is a gap between the two poles and what you have is a conductor passing through it. And let us assume that the current it is having which is flowing through this is I and there is a potential difference across the conductor. And because there is a north pole and a south pole you have a magnet, there is this magnetic field which we label as B and this is my axis system x, y, z. So, this B has several names. So, some people call it B field, others specially electrical engineers they call it magnetic field density, electrical engineers call it. I think a B field is typically used by physics people. There are still others who call it magnetic induction, magnetic induction and then there is one more name in prevalence it is called magnetic field. So, notice its magnetic field density, but some people also call it magnetic and the units for this are Tesla in SI system and Gauss CGS system which is some nomenclature. So, we know from laws of electrical engineering that force. So, when you have a conductor through which current is flowing and that current is cutting across a B field then this conductor experiences a force from laws of electromagnetics we know that. And that force which is called Lorentz force equals B times L where L is the length of the conductor times current and the potential difference across this conductor would be voltage equals B times L times velocity. So, this is all standard electromagnetic and your V is your back emf and U is velocity, B is magnetic field density L is the length of the conductor. So, from these two relations if I have an electrical circuit and if it is coupled to a mechanical circuit through this what you can call it a motor which converts electrical energy into motion then I can make a transformer something like this. So, I have voltage being applied across this thing there is a current going through this and this voltage gets converted into velocity and my current is getting converted into force. And the turn ratio voltage equals B L times U to my turn ratio is B L. So, I can on this side put all sorts of inductors and capacitors and resistors and I can have as complex as an electrical circuit I can have. And here I can have in this on this side whatever number of mechanical elements which are there in real system and if I have to compute force if I introduce this element this particular coupling element then I am able to convert voltage into force and I into well I am sorry voltage into velocity and current into force for mobility analogs. If I have impedance analogy then I have the same equivalence, but my turn ratio will become 1 is to B L you can do the math also in that situation my voltage will be the through variable and my current will be the across variable and similarly on the mechanical side force will be the across variable and velocity will be the through variable. So, this establishes an equivalence between electrical and mechanical rings this is like a coupling element. So, current goes in you also excited with the voltage it gets converted into motion through this transformation and then that motion later gets converted into sound. So, now we will develop a coupling element which joins acoustic and mechanical networks. So, again I have a transformer and in more if I am using the mobility system of equivalence then my across variable is velocity or actually yeah and my through variable is force on the mechanical side and on the acoustic side my across variable is volume velocity and my through variable is pressure. So, from here you can see that v v we know that v v volume velocity is essentially velocity times area of what whatever membrane or you know surface which is pulsating is the area of that surface and it is actually normal to the direction of motion because volume velocity is a dot product of area and velocity also we know that pressure equals force over area. So, again you have this. So, my turn ratio will be related to area and so your turn ratio is 1 is to be in this case. So, this is my coupling element for acoustic and mechanical circuits and this particular ratio is valid for mobility equivalence. If my equivalence is of impedance type then again these relations are still holding, but my this ratio will become a is to 1 rather than 1 is to a. So, this is the last piece in the whole puzzle. So, now we know how to construct an electrical circuit. So, I can have an electrical impedance and then I can have a transformer is that current is getting yes. The turn ratio will be the same just we are changing the through variable and cross variable in the changing. Yes. Turn ratio will be then also come out to be. The turn ratio becomes inverted. So, when you do at the actual map you will see that it becomes 1 instead of 1 is to a. So, why f f will always be times now whether f comes as a through variable then also it is also. If you so explain in electrical transformers v times i. So, the ratio of a cross variable is turn ratio and ratio of through variable is 3. Yes, because v times is preserved voltage times i on one side of the transformer is same as voltage times i on the other side of the transformer. So, if you switch those then your turn ratio becomes. Like here if you switch f and u. Sir f will come as a cross variable. F will become the cross variable. Yes. You will come will become the through variable. Yes. And the same side also. Yes. We will become the through variable and we will become the cross variable. Yes. So, when you have a step up transformer where voltage is going up your n 1 and n 2 have n 2 has to be more than n 1 and then a vice versa. So, if you use. So, we will explain it a little later may be after the class we can go over it in little bit more detail, but the believe me this ratio gets inverted and we will show it very clearly mathematically. I have wanted to cover couple of other things, but we will capture explain it in more detail. So, what we have talked about is the coupling elements and that essentially in a very general sense complete the picture. So, you now have equivalences between electrical, mechanical and acoustic parts of the entire circuit. You can use similar philosophy if you have a piezoelectric transducer and develop similar equivalences. And so you can develop similar coupling elements between piezo and mechanical or piezo and electrical rings. So, the idea of showing all this exercise at this detail level is one to help you understand how these interdisciplinary circuits are constructed. And B if you run into some other transduction processes then you have at least a third process which you can use to develop similar equivalent variables in other beam of engineering. So, now we will talk about something called the Helmholtz resonator. So, in mechanical circuit you have a spring and there is a mass, there is a spring and if it moves on a friction layer surface and if I perturb it by small distance x then it oscillates at a natural frequency omega equals k over m or the frequency will be 1 over 2 pi times k over m. You can have similar acoustic mass, acoustic compliance systems in one degree of freedom in area of acoustics that is called a Helmholtz resonator that particular device. Physically what it looks like is in a very general sense you will have a volume and you will have a short tube and in mobility analogy we saw not mobility this volume we know it acts as a spring physically. And we know that the air in this tube it acts as a rigid mass or an acoustic mass we talked about it at length. So, again you have a mass and you have a spring here and this has a tendency to resonate by itself. So, if you have you know one of the soft drink bottles and they have a little bit like this and you blow over on top of it you will hear some sound coming out that is essentially the natural resonance frequency of this bottle may be in one of the exercise we will ask you to calculate what is that and essentially what is happening is. So, again very similar to a mass and spring system if I push the mass spring gets compressed and let us say that is my steady state condition then I release the force on it. So, then the mass moves out but then it because of inertia it goes above and beyond the neutral position. So, then the string goes into tension and at a point when mass stops moving the strings pulls it back then the mass comes back and it again it exceeds the neutral position it compresses the string and it goes back and forth. Similarly, if you press a bottle or a device something like this if you press the air in it and after time you release that pressure all of a sudden. The air the mass in this acoustic mass in this part will move back and forth and in theory it will move back and forth for infinite amount of time at a natural frequency which we will calculate because you have a string here and you have mass here. So, come some other shapes another shape of a Helmholtz resonator could be like this. So, this is again you have an acoustic mass element and here you have a string. So, if I want to understand this and solve for its natural frequency I can if I use mobility analogy then this is my inductance which acts like a string and the value of this is rho naught c square and this is my capacitance which is behave similar to a mass and the value of this as we developed in the last class is rho naught l over a where a is the cross section of this tube cross sectional area of this tube this is my L and cross section is rho a. So, total impedance between these two points is j omega times v naught over rho naught c square plus 1 over j omega rho naught l over a. Now, for resonance z equals 0 in this case for resonance conditions. So, j omega v naught over rho c square equals 1 over I can bring my a up j omega rho naught l minus and then I can eliminate the negative sign by bringing j up in the numerator and now if I solve for omega essentially what I get is omega equals c times cross sectional area of the tube divided by length of the tube times v naught. I can use this Helmholtz resonator in a lot of ways it has its applications we will talk couple of applications about it the first application where it is used a lot is in sound systems. So, we will draw two boxes. So, I can have a box a excuse me and it has speaker mounted here and it is a seal box a lot of traditional boxes you have a speaker and just seal you may have seen it in your homes and you try to measure the pressure at this location over a frequency range and then another system you can have is again a seal box with the Helmholtz resonator and we will see what it does. So, I still have exactly the same transducer, but here I have a port also or a tube when I plot the frequency response for system a and this is my system b typically when you do the actual measurement and in later parts of this course you will actually develop relations which will show. So, if I am plotting pressure at this location s p l and I am plotting it against let us say frequency it will be something like this and just to put some numbers let us say this is 85 and this is all decibels and this could be 100 hertz and note that this is a decibel scale. So, a drop of 6 dB is a big change 6 dB means like twice the energy content is 2 times less in the whole system. So, if this drop is 6 dB essentially what it says is that at 100 hertz at may be a higher frequency it is producing 85 dB, but at 100 hertz it is producing 6 dB or even less than 6 dB from that 85 dB threshold which is not a good thing which means that low frequency base frequency which people like to hear more about more of they are not being produced at an equal level. So, to solve this problem what people do is they introduce a Helmholtz resonator and it is a very inexpensive solution they just introduce a tube. So, you have a volume here and you have a tube a port. So, you have a spring mass system and what it does is that because of the and they tune the length of the port and volume in such a way that the resonance is below some number. Let us say if I want to improve performance at 100 hertz or 80 hertz then I tune this in such a way that my resonance frequency is at around 80 hertz if I want to improve the performance at 80 hertz. So, when I do that my this transducer is producing SPL something similar to that, but my resonator is producing a heightened output at 80 hertz wherever I am tuning it to be. So, for on 80 hertz wherever whatever is the tuning frequency and in the neighborhood of that I get a little bit of extra push. So, I get improve my base performance. So, once you have so this is a and once I put a Helmholtz resonator this was something like this once I have a resonator it could be something like this. So, I have increased my frequency content the magnitude frequency in this range and on the negative side you have a steeper slope the on the low frequency end. So, you start killing lower frequencies the center then certain threshold much more steeply, but you are able to get some extra base in this area. So, lot of sound product companies they use these pipes. So, you have a sealed volume they put the pipe sometime they put two pipes to solve to add extra content for specific frequency bands. So, this is first application. Second application is in musical instrument. So, we will show see two pictures this is one of the original Helmholtz resonators constructed by Mr. Helmholtz in around 1850s. So, you have volume and you have a little bit of a tube here. Now, look at so this was 1850s this is the picture of Wiena. So, you have two pictures of Wiena and these round big spheres are essentially some hollow either pumpkins or locky you know that gold. So, in traditionally they would hollow it out and they will have a small tube going up and this they have been using it over at least last 2000 years, because people in ancient drawings also people have found pictures. So, people have been using Helmholtz resonators prior to birth of Mr. Helmholtz this is a very long time. Other thing is this is another Wiena called Kinnari Wiena and it has three resonators. So, again these resonators increase the performance at specific frequencies. So, in this Kinnari Wiena it just happens that this is a picture of a Wiena which is sitting in a museum in France. It is an instrument which was made something like in 18, 19 century and some French sound engineers and scientists what they found was that they found the tuning frequency of this little pumpkin and also the larger pumpkin and they found that the ratio is 2. So, they are off by 1 octave. So, even in earlier times people had a very clear understanding of octave. This is a proof it is sitting there in the museum you can go and measure it. People had a clear understanding of octave and also had figured out how to tune. So, that the natural resonance of one resonator is twice or half of the other resonator. So, this is. So, again in a lot of musical instruments in guitar there is a hollow box below the string. Now, you will not see an explicit tube in a guitar, but we will talk about it later that hollow box also kind of acts as a resonator. So, that is another application of the same resonator. Third application is in air boxes. So, in engines earlier air used to go to a carburetor and then there it used to get mixed with petrol fuel and then it used to get. Nowadays people are using less of carburetors and more of air boxes and there also the aim is that if they can figure out somehow as a way to smartly suck air more efficiently into the system then it requires less energy. So, again they use concepts rooted in the idea of a resonator and they suck air. Another example is and we will do some math about it is architectural acoustics. So, you have a big hall and someone is playing music or speaking and if the walls in the hall are parallel then you can have standing waves because they are parallel and the wave length of those standing waves will depend on the distance between the two walls. So, we saw earlier if you have two fixed structures they could be a standing wave. So, people try to eliminate those standing waves because those standing waves are not pleasing to hear a by using resonators. So, in the first three examples in case of a loudspeaker in musical instruments and air suction we used resonator to enhance the frequency performance for certain pitches for certain frequency. In case of architectural acoustics they use the same tool to kill specific frequency and we will do a little bit of math on around it and see how it that they accomplish it. So, let us see that we have a long tube and the cross section area here is a 1 here it is a 2 and here the cross sectional area is a 3. Also this length is l 1 this length is l 2 and this length is l 3 and then I have a big volume here not necessarily big volume here v naught and I am sending signals through this side let us say the volume velocity is v v and then there is a guy sitting here observer and he is measuring v v o. So, this combination of volume and this small tube on top of this circle is your resonator. So, the question is fine v v o over v v. So, we draw an acoustic circuit for this. So, in this case because v v happens to be through variable we will use impedance analogy. So, my pressure is analogous to voltage and v v is analogous to current this is impedance approach impedance analogy. So, now I construct a circuit. So, essentially I have two tubes short tubes here another short tube here a volume here my volume in impedance analogy will be modeled as a capacitor and my tubes will be modeled as inductors. So, let us say this inductance is l 1 l 2 l 3 and this capacitance is c. So, my l 1 equals rho naught times length of the first element over a 1 l 2 is rho naught l 2 over a 2 l 3 is rho naught l 3 over a 3 and capacitance is volume over rho naught c square. So, through variable as we talk this v v and here I am measuring v v naught. So, the question is what is v v naught over v v magnitude of this. So, at a very subjective level when the frequencies are very low how will v v get split along these two paths v v will split at this junction some of it will go through l 2 some of it will go through l 3, but when frequencies are very low what do you think will be the magnitude of current going through this circuit the l 3 c it will be virtually 0 because impedance offered by c will be very high. So, for very low omega v v naught is approximately equal to v v. So, my ratio is 1 for high values of omega there will be some. So, at very high there will be virtually 0 current, but what we are interested is what will be the ratio of v v naught over v v. So, whatever little current is going how will it get split in the two directions that is the question c will be acting as a short circuit. So, it will split based on the ratio of l 2 and n 3. So, c kind of acts as short. So, whatever v v is going in it will split like this the influence of capacitance will be virtually negligible I am just ignoring that and if l 2 is fairly large l 2 is very large compared to l 3. Remember l 2 and l 3 are inductances which are these numbers they are not they are not lengths by lengths. So, if l 2 is very large compared then v v naught over v v over v v naught is essentially l 3 over l 3 and finally, what will happen at resonance you have the circuit what happens when the there is a resonance in this part of the circuit impedance goes to 0. So, what happens to v v naught. So, at resonance v v naught equals 0. So, v v naught over v v is 0 and the value of this resonance frequency will be 1 over 2 pi times 1 over l c that is 1 over 2 pi square root of l 3 times c. So, I at a very coarse level I plot this transfer functions magnitude I am having here frequency and here I am having let us plot it omega. So, here I have omega and here I have v v over v v that low frequencies this value is 1 at resonance which is 1 over l 3 c this value comes to 0 and for the condition this condition if l 2 is very large then large compared to l 3 this should have been v v naught over v this is an asymptotic value will be l 3 over l 2. So, my transfer function will look something like this. So, in all other examples of the resonator I was using the resonator to boost the performance at specific frequency here I am using the same device to kill specific frequencies in this case something in this band I am able to attenuate the magnitude of that. So, resonance can do all sorts of things. Sir. So, in the in the actual hall the it is the walls are designed that the architectural this thing is designed to kill this frequency. So. So, this is used to be used especially in earlier times because people did not have a very clear understanding of how standing waves are interacting what is the damping coefficient of different materials at different frequencies. Nowadays there is a more higher emphasis on the shape of the room also the damping materials being used how the speakers are placed in the auditorium where should people who are generating music or sound they should sit all that. The reliance is not that much on resonators, but the point what I am trying to make is you can use the same resonator in totally opposing you know with totally opposing or different purposes you can boost frequency response or you can kill frequency. Sir, I am just trying to say that the system in the hall sir. It should also be tuned to produce this frequency. So, that it it can be killed with this. No, the tuning of the room will depend as I said suppose there are two parallel walls and they are separated by some meter length. So, it is tuned to have a standing wave which is related to that dimension. So, it is in that sense it is tuned to a specific frequency. So, what you want is that those standing waves they should not be heard loudly by the people sitting in the auditorium or in that listening space. So, that is a consequence of the geometry of the room the natural mode. So, we will do one very quick one more example. This is not a Helmholtz resonator just for practice purposes. So, I have connected to a volume pardon my very poor drawing and this is connected to another sag and here you have an acoustic resistance. In mobility the resistance will be 1 over r a this is all this is right now impedance model. Here the volume is v naught this is v 1 and this I call l 1 and this length l 2. So, we will very quickly construct an acoustic circuit using impedance analogy. So, my across variable is pressure through variable is volume velocity it goes through the first tube which acts like an inductor such that l 1 equals rho naught l 1 over a 1 and then I have an acoustic capacitor here or a spring element which is v 1 over rho naught c square. Then I have another acoustic inductance of value l 2 such that that being equal to rho naught l 2 over a 2 and then I have finally one more. So, I should have put a resistance here and if I do a mobility model. So, this is my impedance model. So, in my mobility model I have volume velocity inductors get replaced by capacitors and vice versa the values do not change and the resistor is the same, but its value becomes inverted. So, this is rho naught l 1 over a 1 and this is rho naught l 2 over a 2 this is mobility model. So, in a lot of these circuits just wanted to recap couple of very simple, but important points you will see combinations of inductors and capacitors. So, you can have a capacitor of value l I am sorry an inductor of value l and a capacitor of value c and when you see you should start thinking how is this thing going to behave. So, again for low values of omega it will behave purely as a open circuit and the capacitive part will dominate for low frequency. The capacitive part will dominate for extremely low values it will become an open circuit it will go towards an open circuit, but capacitive part will dominate and for high values of omega inductive part dominates. So, if I plot the impedance and this is my omega the capacitive component will be infinite at 0 and it will fall like this. So, here z equals 1 over j omega c and the inductive component will grow with frequency and here z equals j omega l I am just plotting this magnitude. So, again I am just plotting the magnitude of c. So, this is the cross over point and the value at which these two become equal is 1 over square root of l c and the last point I would like to make here is all this approach where we are having lumped parameters does not matter whether it is electrical beam, mechanical beam or acoustic beam. They work to the extent that your frequencies are less than compared to lambda over 2 pi the lengths of the elements are less than lambda over 2 pi the length of the element is less than lambda over 2 pi. So, if in an electrical beam you are starting to violate some of this condition you will not be able to accurately model the electrical circuit using lumped elements because then they become continuous system. As we saw earlier in several lectures back in transmission line as we were developing the transmission line theory capacitors stop behaving as lumped capacitors when at very when the wavelengths violate this fundamental condition. The same thing is true for mechanical and same thing is true for acoustic beam also. So, as we are solving some of these equations and constructing these complex circuits we have to be cautious that in every single ream are we sticking to some of those guidelines or not. So, that is what I wanted to talk today in context of Helmholtz resonator and a little bit about overall electrical circuit. Now, I wanted to introduce this notion of acoustic radiation impedance radiation impedance. What is it? An electrical circuit let us say an amplifier and it is receiving electrical energy. You know how to convert that electrical energy and model that conversion process to mechanical energy we saw this whole transformation process. Then this mechanical energy gets converted into sound which is generated by a loud speed and we know how that conversion is happening and we have developed elements, lumped elements for short tubes, closed tube short volumes and acoustic resistors. Once the sound leaves that box or sound system whatever and it reaches my ear there is some impedance in that travel process also that is called radiation. So, far we have not captured that if I have to accurately model the entire circuit from source to the listening point or the observation point that is one last piece of puzzle which is left and that is radiation and the radiation impedance will change depending on if you have a sphere which is pulsating back and forth it will have one type of radiation impedance. If I have a sound source mounted on an infinite wall the radiation impedance will be different. If it is a sound source just moving back and forth may be for some frequencies low frequencies the person at the other end may not hear anything because as a piston is moving forward it is creating positive pressure on the front surface and on the back surface it is creating suction. So, those two waves add up and they reach your ear you will hear very little sound. So, all that is about radiation impedance and that depends on the distance between the listener and the source. How the sound source is mounted? What is the boundary condition at the source point? The geometry of the source and several other variables. What we will start today is radiation impedance of a pulsating sphere because we have developed some relations in past how sound propagates as it emanates from a simple spherical small source. We have developed some relations and then in subsequent lectures we will do some more examples of different types of radiation. So, we will start from the definition of instantaneous power. So, instantaneous power we had talked about it earlier is instantaneous voltage times instantaneous current and an acoustic beam will be instantaneous pressure times instantaneous velocity it is p t times u t. Then in some of the earlier lectures we had developed this relations such that this is equal to p times u star plus real component of e u e 2 j omega t and then the u star is essentially complex conjugate of pressure over complex conjugate of z that is z star plus half real e and then again u is p over z e 2 j omega t. This one I simplify so I get half times magnitude of pressure star real of 1 over z star plus half real e times p over z times e 2 j omega t. In one of the earlier lectures we had said that z which is the impedance could be a magnitude times exponent j psi where psi was related to this power factor notion. Also pressure will have a magnitude which is a real positive number times exponent of j theta. So, if I put these two relations in my above equation I get instantaneous power equal magnitude of pressure whole square times and real of 1 over z star would be times 1 over z times cosine psi plus times 1 over magnitude of z times cosine 2 omega t plus 2 theta minus psi e square over 2. Also so now this is what we had defined as average power and this is the fluctuating portion and there is a psi component to it also I mean this influence of psi also. So, when I plot this and this is to a certain extent a recap of some of the earlier work for psi not being equal to 0 this is my time axis and that is my power the total power will be something like this where my average power is going to be positive nonzero entity and this is my total power and this area below the axis is essentially the energy which is getting dumped back into the system. Also we had in the last class or one of the earlier lectures talked about that because this is not symmetric along the horizontal axis the time axis what it means is that it will eat more than the norm quote unquote the normal current which would have been needed if cos psi was equal to 1 because there is lack of symmetry along the horizontal axis it will consume more power the case where cos psi would have been identically 1 that is if psi was equal to 0 this was a recap of some of the stuff which we have done. So, now remembering this we look at a spherical radiating. So, I have a small sphere we had developed its radiating impedance to be e r 0 s over u r 0 s what this means is that this is the radiating impedance of small spherical source of radiating of radius r naught on the surface of itself. So, this is what the body will see as its impedance on its surface if you move away from the surface this r naught will grow will become whatever r value is. So, my z r a which is acoustic impedance is essentially z r over area. So, that is e r naught s over u r naught s times 1 over area and remember s is equal to s equals j omega z r we had seen was 1 over rho naught c times c over s r naught plus 1 the whole thing inverse for a purely resistive circuit for purely resistive circuit the what would be the value of psi power factor psi would be 0 you see here that there is an s component here which means that the impedance offered by a pulsating sphere is not purely resistive. So, whenever you have a pulsating sphere it will by nature of its presence introduce a nonzero psi in the entire circuit which means that at the amplifier level it will draw more than optimal value of current and it will require more than optimal value of voltage. So, you are introducing just because by virtue of that it is a spherical surface radiating in free space you are introducing a nonzero value of psi and may be to fix that at the electrical level you can put some capacitors or you know to rectify that this is z naught if I inverted where z naught is rho naught times c times s r naught over c plus s r naught and if I replace s by j omega I get z naught times j omega r naught over c plus j omega r naught. What we do here is we try to rationalize this term. So, I get basically what I do is z naught times j omega r naught over c plus j omega r naught times I multiply and divide by c minus j omega r naught and I also divide it and what I get is if I do simplifications omega square r square over c square plus j omega r naught over c. So, I will now plot this and I will plot its real component and also its imaginary but before I do that I will rewrite it z r equals z naught omega square r square over c square did I I think j omega r naught over c. We are assuming that omega is we are assuming that is only then these relations will apply. Why because if we do the exact match there is a omega square r naught square term in the denominator. So, this is z naught and this is omega square r square r naught square over omega square r naught square plus c square plus j times omega r naught c over omega square r square plus c square yeah that was a simplification I had done it. If omega is low then this is my relation and if omega is large then I get z r equals z naught 1 plus j c over omega r naught. Regardless whether omega is small or omega is large you have complex imaginary components in the entire term except for very low frequencies what you see is except for very low frequencies it becomes purely resistive in nature and at extremely high frequencies again it becomes purely resistive in nature but in the intermediate range it has a reactive component and what that means is that most of the at most of the operating frequencies what these two approximations show is that psi will remain non-zero. So, power factor will be non-zero will be not equal to exactly equal to 1 for most of the frequencies we are talking about. So, I will just plot. So, I have here what I will plot is this is my real component of z r and my imaginary component will be something like it will be asymptotic. So, what this shows is again this is a plot of different components for z r real and imaginary this is my omega axis and what we have shown here is that regardless except in very extreme conditions extremely high frequencies or extremely low frequencies in most of the range my spherical sources are fairly reactive in nature which means that if I have to produce significant amount of d b s p l from these sources spherical sources it is a difficult thing to do because they eat a lot of power at the amplifier level that is the implication. So, if you are designing complex system and if you are thinking of using a spherical source then you have to think a little bit more carefully how you develop. So, that is all we want we will be covering today in the next lecture what we will talk about is. So, this is my z r I still have not approximated this into a set of resistors inductors and capacitors in the acoustic world. So, you have if I have electrical mechanical and acoustic I have all that conversion, but I still have till so far not converted this into some lumped systems which I can construct in the acoustic range. So, that is what we will do for a spherical source in the next class and then we will also consider couple of other.