 So, in the last lecture what we had covered was a little bit about microphones we had gone through a range of microphones classified them we also went into details of two specific types of microphones pressure mics and pressure gradient mics and explain how each of these mics works and under what circumstances each of these two different categories of mics are appropriate for usage. We will continue on that journey today what essentially we will cover today is a explore couple of other instruments which are similar to mics, but they help us measure vibrations or accelerations even though they operate on a very similar principle principle as that for microphones. Following that what we will do is we will do some numerical calculations and manipulations on a term called microphone sensitivity and also on the decibel scales how do you add up decibels subtract decibels and so on and so forth. So, that is what we will cover today essentially what we are doing is we started with physical acoustics now what we are doing is trying to figure out the electro acoustics part we covered loudspeakers now we are covering microphones which help us measure sound levels and then maybe in later part of the course what we will go for is controlling sound, dampening it or reducing the sound levels and that relates to not only physical acoustics, but also a little bit of outside was to acoustics. So, what I am going to start with is an instrument called seismometer and like a microphone seismometer. So, like a very much like a microphone what it has is it has a moving coil we do not call voice coil in this case because it does not sense sound, but it senses vibrations and this coil is embedded in a magnet structure. So, as the coil moves up and down it current gets induced and we measure voltage across the coil and we sense vibrations. So, the way it is is that let us consider that this is a structure and it is rigidly connected to ground and let us say the vibrations in the ground are y, the construction of seismometer is something like this I have a motor structure and within this motor structure I have a voice coil not voice coil just a electrical coil this coil is connected to a mass let us call it m and the mass is connected to the outside frame by a spring of stiffness s and we could also have some dampening and let us call this damping coefficient is r resistance. So, what you have here is that in through this motor structure magnetic flux cuts I mean this is how magnetic flux moves and it cuts the conductor which is bound around the bobbin called the so that is the coil and that induces motion in the coil and because of vibrations the coil moves relative to this magnet and that induces voltage and we connected to it external voltmeter and sense the vibrations. The mass can have its own motion and let us call it x and x may not be necessarily same as y because of these stiffness elements which may not be infinite in number. So, if s is infinity then x equals y otherwise s x may not be same as y let us define a term beta equals x minus y it could be of this form e j omega t times a constant beta o the question is that if I can measure this term through voltage that is what the voltage will get translated I mean it is a reflection of this x minus y how can I figure out the value of y I am interested in how much the structure is moving so that is what we will do from equation of motion we know that mx double dot mass times acceleration equals minus s times the relative displacement or the compression or tension in the spring minus the viscous force and from one we can write x equals eta plus y. So, if I put that back into two what I get is m theta double dot plus m y double dot equals minus s theta minus r theta dot I can rearrange this entire equation and write it as m y double dot equals mass times theta double dot plus the damping force plus the force due to the spring and if I have y of the form y equals y naught e j omega t then what I get is basically my left side becomes m omega square y naught e j omega t equals this entire term right now in reality y does not have to have this exponential form it can be any complex signal which could have any shape but we know from Fourier transformation principles that we can break up any complicated signal into a Fourier series. So, if I have a solution for this I can use that in understanding to decompose any incoming excitation into a sum of different waveforms this is my equation I will label it three and if I do a little bit of math on this I am skipping a few specific mathematical steps operations what I get is relative velocity meter naught equals m omega square y naught e j omega t minus alpha over m omega minus s over plus r square and this entire thing is under the square root sign and what I have done is here is I have introduced a new term alpha and basically alpha is tan inverse the magnitude which is this term is this expression meter naught equals m omega y naught times m omega minus omega square plus r square all this four. So, now I introduce another term and eta equals r over r critical r critical is my critical damping frequency and this is essentially twice of under school mass time system. So, if I plug this into four what I get is eta naught equals y naught times omega over omega naught square divided by omega over omega naught square minus 1 plus 2 eta times omega over omega naught whole square. So, I understand that I have skipped a few mathematical operations. So, if you remember in one of the very early lectures we had said that there is a critical damping that is associated with this and then omega naught is essentially stiffness over mass whole thing under square root. So, what I have here is an expression of eta and y naught in terms of omega and this factor which relates to damping. So, what I will do is I will now plot this relation. So, what this is telling me is how the magnitude of eta and y naught are related. So, what I will do is I will construct some plots on my x axis I have this parameter omega over omega naught and on my y axis I have eta naught over y naught and this is a linear scale. So, all the curves for and what I will do is I will construct different curves for different values of eta this term. So, for all these all the curves start from origin because when y naught is 0 then eta naught is 0 identically. So, let us say this is 1. So, these are very standard curves and what the construct the curves look like is this is eta equals 0 and then it again comes down and has some sort of an asymptotic relationship as omega goes to infinity. So, this is still 0. So, it blows up at omega over omega naught equals 1. I will draw two more curves this is eta equals 1 and I will draw third curve it goes up and then starts to move flat something like this and again this is eta equals 0.5. So, for value of eta equals 0.78 it will rely somewhere between green and blue curves. So, what we see here is that eta if it is in this range 0.7 to 0.5 essentially we get a fairly flat frequency response once my omega divided by omega naught is above somewhere 2 and omega over omega naught I get a flat frequency response. So, this seismometer will have essentially have a flat frequency response for if I set the damping coefficient this value of r in such a way that eta equals 0.7 in that range and if my omega over omega naught exceeds 1 or in the actually 2 then my response curve will be fairly flat. So, that is the magnitude part what else are we missing there could be a phase also there is a phase relationship and ideally we would like that whatever excitation is coming in which is why now why it is in phase with whatever we are measuring which is eta otherwise for each particular frequency we will have to correct we will have to fix a specific phase correction factor. So, we want ideally we would like to have constant I mean 0 phase difference for all frequencies or if we do not have 0 phase difference then as a second best solution it should be constant right. So, what we will now do is we will plot the phase value let us say tan beta equals minus 1 over tan alpha, but even before that this relation what we see is the phase difference between and eta dot and y is alpha right. So, the phase difference so between eta dot and y I have phase difference of alpha and if I integrate eta dot I get eta and then between eta and y my phase difference is alpha plus pi over 2 right given that we define this term beta equals phase difference between y and eta. So, beta equals tan inverse r over s divided by omega minus m excuse me times omega and this is tan inverse 2 eta omega over omega naught divided by 1 minus omega over omega naught entire thing squared. So, now what I will do is I will plot beta as I vary eta for different values of omega. So, on my horizontal axis I have normalized frequency and on my vertical axis I have phase difference between the motion of the rigid mass the mass and the motion of the structure pi over 2 0 1 2 3 4. So, when this thing is 0 eta then my beta is basically pi and then for other values I have an inflection point or so this is eta equals 0.5 and this is eta equals 1 and somewhere in the middle equals 0.7. So, again I mean by looking at this particular curve for magnitude we have said that what this curve shows is that I can reliably measure that the relationship between eta and y is pretty much a fixed constant for eta equals 0.5 0.6 0.7 in that range and also for omegas which exceed omega naught by factor of 2. What this fact curve shows is that let us say if I pick up eta anywhere between 0.5 and 0.7 I have a fixed phase difference which is a little less than pi and if I keep my omega divided by omega naught above a ratio of 2 then that fixed constant is same for all frequency. Based on this I know how to calculate the phase difference between source excitation and whatever the instrument is measuring and also I know how to translate that voltage into the actual displacement. One thing in this picture what I have not shown is how is eta connected to voltage that relationship has not been explicitly defined but in our earlier classes we are shown that this BL term connects the velocity with the voltage. So, we can use similar transformation relationships and figure out develop the entire electromechanical model what here we have talked about is just the mechanical motion. So, that was a seismometer what we will talk about is now how the same device the same device could be also used as an accelerometer and it helps us measure accelerations. Neater naught equals y naught second derivative and I will define what that term means divided by omega square over omega over omega square this entire thing is squared plus 2 eta omega over omega naught and this entire thing is also squared what y naught is double dot is acceleration amplitude. So, it is the amplitude but of acceleration. So, I can rearrange this equation in such a way that I get a term omega naught square divided by y naught double dot and this I call q and lot of people call this q as mobility and the you know the radical sign the terms are same. So, now if I plot this equation q on the vertical axis versus normalized frequency and this is what I get. So, what I am plotting here is q on the horizontal axis I have normalized frequency omega over omega naught. So, that is my q equals 1 0.5 1.5 and my normalized angular frequency is here 0.5 1 1.5 and so on and so forth. So, in this case all the curves start from when omega is 0 q is identically 1. So, all curves start from here. Yeah, you can do that but what you will measure is the neta naught right to the volt meter and that you have to convert it into acceleration. So, what we are trying to do here is essentially see again where is the response curve flag that is all we are trying. So, this is one curve then for eta equals 1 something like this and then it becomes flatter 0.7 and then it becomes even more flat at and then if I reduce my eta further then it becomes something like this. So, what this shows is essentially that the operating range for the accelerometer. So, the operating range for the accelerometer is a very narrow band it is a narrow band. So, if you want to make your accelerometer measure large very high frequencies you have to increase the stiffness very significantly omega naught has to go up right. That is what a lot of accelerometers do they try to increase the stiffness as high as possible and they do it by using solids like piezoelectric crystals they act as stiffness members also. So, they have stiffness and they also generate voltage they do both the function they act as stiffness members and they also generate voltage across themselves. So, after 1.5 they also seem to be flatter. So, this is a better looking curve should be something like this something like that does not become very asymptotic, but it is still flatter here in this range. Now the problem with this. So, the challenge is that if I have to measure at 100,000 hertz the acceleration then I have to may be increase my omega naught up to may be 10 million or a million something like that because it has to be very small. I can do that by making by reducing the size of the structure you know if I reduce the size of the structure the stiffness starts increasing significantly. So, I can do that, but then challenge becomes that how do I make sure that the signals are large enough. So, that they can be measured here we are assuming that I can measure any amount of voltage it could be millivolts or micro volts or nano volts, but my sensitivity also starts decaying very fast. So, that is the challenge and a lot of times what they do is that within the accelerometer itself or very close to the accelerometer they have some amplifiers charge amplifier. So, some devices. So, whatever is the signal which is being generated it gets immediately amplified. So, that it can be measured reliably because otherwise if that device amplification device is far from the location of measuring it may get corrupted by the time it reaches them. So, we mentioned about sensitivity and so what we will do is we will define what is sensitivity. So, for a microphone we want a flat frequency response which essentially means that for a large range of frequencies the response curve is flat you know and we also want that it should be sensitive enough. If it has very low sensitivity and a very flat frequency response it may still have limited use because then it starts becoming difficult to measure the signals generated. So, if you go to a small microphones very tiny small pressure type of microphones they have very flat frequency response, but their sensitivity is very low. If you make the microphones large because of this lambda over 2 pi factor their frequency range becomes of a limited width, but their sensitivity goes up. So, it is an interplay my sensitivity let us say S is electrical output divided by mechanical input which is your pressure in this case right. There is a term called microphone sensitivity level L S microphone sensitivity level and that is defined by this term 10 logarithm in base 10 times not times I mean logarithm of e out over e divided by e ref and I can make this 20 log e out over e and I will explain that how I made that jump when the 2 became once I remove 2 this becomes 20, but I also eliminated from here e ref. So, e out is output voltage from the instrument e ref is reference voltage and typically it is 1 volt for if my incident pressure is 1 micro bar 1 bar is 10 to the power of 5 Pascal's which is 1 atmosphere 1 micro bar is 0.1 Pascal's. So, e ref is a reference voltage and it is about like 1 volt for incident pressure of 1 micro bar. So, because of that e ref goes away and then P is RMS pressure it is not P pressure RMS pressure on microphones in micro bars. So, it is not in Pascal's, but it is in micro bars from this relation I get e out equals pressure times 10 to the power of L s divided by 20. We also know that SPL sound pressure level is 20 logarithm of P divided by P ref. So, which gives me P equals P ref times 10 to the power of SPL divided by 20 and my P ref it is industry standard is 2 times 10 to the power of minus 5 Pascal's. So, from this what we get is P and I will make it explicit in micro bars equals 2 times 10 to the power of minus 4 times 10 to the power of SPL over 20 and if I plug this back into the original equation there is this 10 also yes, but then it also gets multiplied by 10. Because we are doing in micro bar, so because of there is a factor of 10. So, that because of that reason it becomes 2 to the power of minus 4. Let us call that equation 2 and this is my equation 1. So, from 1 and 2 what I get is e out equals 2 times 10 to the power of minus 4 times 10 SPL over 20 times 10 to the power of L s over 20 or e out equals 0.002 times 10 to the power of SPL plus L s over 20 volts and one thing I omitted earlier was that the unit of this is decibel voltage over micro bar. They are not of the same decibel voltage over micro bar. So, we will do an example. So, let us consider microphone. So, there is a microphone which has a sensitivity L s of minus 50 decibel voltage for each micro bar and the question is that what will be the output of this microphone when my SPL is 85 So, if SPL equals 85 decibels what is e out? So, I just essentially use this relation e out equals 0.002 times 10 to the power of SPL plus L s divided by 20 and what I get is 2 times 10 to the power of minus 4 times 10 to the power of 85 minus 50 divided by 20 and if I solve this I get 0.0112 volts 11.2 millivolts SPL and L are different units. So, how can we substitute? But essentially what this process tells you is that you will know if you have some idea that your sound pressure level is going to be 120 dBs at the peak level and the minimum sound pressure level is going to be let us say 40 dBs. And if you want to measure it you should be able to figure out what will be the output voltage from that and when will my instrument be able to actually measure that voltage or not that is what. Yes, but to his point is. So, if it is decibel then there is no problem that is true. So, the answer to that question is to your question is e over e ref becomes non dimensional and p is in micro bars which is again a number is in micro bar. So, it becomes a non dimensional one. So, a typical sensitivity for large microphones is something like half a micro volt for each micro bar or you can convert it into is 125 decibels for each micro bar. So, that is on the low end and then on the high end you it goes up to 3 millivolts for each micro bar. Other thing we will now talk about is waiting this relates to this stream of psycho acoustics. So, let us say I am playing a particular instrument at a very high frequency and I produce 100 dB SPL from that instrument. Now, we use another instrument let us say a bass guitar or a tabla and it produces sound at 100 dB SPL, but it is at a less frequency. My ear even though the SPL at different frequencies is the same. My ear will perceive the loudness of higher frequency is different from that which is coming from a lower frequency. This is how ear works and this is start yeah. So, 40 hertz 100 dB SPL will sound will be perceived it is not different, but we perceive to have different loudness compared to say a 1 kilo hertz compared to a 15 kilo hertz and so on and so forth. The other thing is that this difference in perception for different frequencies becomes more amplified if my over sound loudness level is low. So, if I am playing the whole thing at let us say different different frequencies then that difference in perception will be much of much larger magnitude compared to if I am playing the everything at higher dB. So, one perception is related frequency band and the other related difference in perception is related to overall loudness level. So, this is how ear works. So, in a lot of measurements which relate to human ear if you are trying to reduce the overall noise in a machine you know machine shop where you are generating all sorts of frequencies. What people try to do is that they wait different frequencies by different numbers. So, that the ear based on whatever is the difference which is being perceived by the ear because the aim is that the ear should perceive that the over sound level has gone down. If you want to make it quiet means that the ear is sensing that everything has become quieter. If you reduce the frequencies by the same amount not the frequencies by the amplitude by the same amount the ear will not necessarily perceive that all frequencies have gone down by the same amount because some frequencies it will perceive are still louder compared to others. So, that is what is what relates to this idea of waiting. But these weights cannot be like these weights will keep on changing with less SPL level. And also loud SPL level yes. So, they cannot be standard. So, that is what these three curves tell you. So, you have this A waiting level is in blue then you have this yellow curve which is B weight level and then you have this C waiting level which is in red. So, A waiting curve basically is used for sound levels in general which are less than 55 decibels. Then your B is used for 55 to 85 degrees and C is used for 85 degrees in excess of 85 degrees. And there is also a D waiting level and Z waiting level. Z is not shown here. We will talk about that a little later. But it is important to know that in industry when you measure sound you will say what type of waiting level you used A, B, C. So, you should be able to know what is being talked about. Now the most prevalent waiting level in industry today is A. Most prevalent is A because as you see in case of C which is in red it is pretty much flat across a very wide spectrum from let us say 30, 40 hertz to 3000 hertz is pretty much flat. So, A is the most widely practiced waiting level and B also has fallen into issues. It is used in very specific cases for some aircraft estimation. And these curves have been developed based on how ear perceive sound at different frequencies and at different levels. And based on that on some statistical samples people have developed these curves and now they are the industry standard. And a lot of environmental agencies say in India or in United States or in Europe actually enforce that if you are trying to solve noise related problems you have to wait your measurements based on some of these curves. So, that is the purpose. You also see that there is this D waiting level which is in black. And what this is for is that if you have a bunch of not a bunch if you have a random noise where noise is at all frequencies then the loudness perception of that random noise is different than the noise which is around 6 kilo hertz. At 6 kilo hertz the loudness perception is at a higher level compared to random noise which is a broadband noise. So, that is why you have this peak at around 6 kilo hertz otherwise it is again more or less flat. I wanted you to get exposed to some of these curves and you can also get charts for you know these curves would be translated in chart the industry standard tables. You can use them and wait your sound pressure level measurements accordingly. Also in a lot of sound systems you may have seen this loudness control. There is a loudness now in some sound systems. It is not same as volume loudness there is a loudness and that also it does the same thing. If you are playing it at a higher level it uses some of these waiting curves to enhance certain frequencies and enhance not so much or not to the same extent some other frequency. And if you play it at a lower level then this loudness knob does something which relates to k-waiting level. Equalizers that is used. Yes something. We will come to the last portion. How do you add and subtract decibel levels? So, if I have an 85 dB sound level coming from one speaker and another source is generating 100 dB sound level. What will be the overall SPL? Will it be 195? Because it is logarithm you cannot just add them up. So, we know that dB equals 10 log P i over P ref. So, P i over P ref equals 10 to the power of dB SPL over 10. So, if I have a bunch of decibel sources then total SPL I can write is 10 log summation of P i over P ref equals 10 log. So, this is fairly straightforward. A question here could be here what I am doing is I am basically using this 10 log relationship. But what I could also do is I could remove this 2 and make this 20. Will that be an appropriate thing because I will get a different number. Will that be right or will that not be incorrect? Which one you will only have one correct answer? So, is this the right way to do it or is that where you eliminate this power of 2 and put that into this on the other side of the equation. Will that be the right approach and why would that be the case? That is a defined even in such a way. Mathematically you can do this. Mathematically you can do this. Decibel if you remember in our original definition is a ratio of energies. It is not ratio of pressures or ratio of velocities. It is ratio of energies or power and energy is directly proportional to V square or pressure square or displacement square of displacement and so on and so forth. So, I am basically when I am adding decibels I am adding up energies. I am not adding up pressures. So, that is why eliminating power of thus power number and putting it on the other side of the equation will be invalid because that does not help me add energy levels. So, we will do an example. Let us say I have 3 sources 96 dB's, 87 dB and 90 dB even E 2, E 3. So, what is the total SPL? So, total SPL equals 10 log of 10 to the power of 96 over 10 plus 10 87 over 10 plus 10 to the power of 90 over 10 and that gives me 97 dB. So, it is just an increment of 1 dB. Similarly, if I have to do subtraction then difference of dB's 2 decibels is 10 times logarithm of 10 to the power of SPL 1 over 10 minus 10 to the power of SPL 2 over 10. So, suppose we will do another example. Let us say we have a machine shop, a large machine shop and there is some ambient noise and then there is some also noise coming when a particular milling machine is running. So, when we run the machine, when we run the machine then the total SPL level is let us say 93 dB's when machine runs and when the machine is not running let us say the total dB SPL is 85 dB's. So, a logical question could be what is the overall contribution of the machine to the overall noise level. So, SPL due to machine is 10 log 10 to the power of 93 over 10 minus 10 times 10 to the power of 85 over 10 and this gives me 92 dB's. So, these are varies I mean it does not become very apparent right away and a very I mean from transfer would be what 8 dB's, but the numbers are significant. Another example, so if I have this is the most common usage that you take a measurement in any experiment, its value is V1 then you take 4, 5 same type of measurement V1, V2, V3 you want to be sure that the measurement is right right and these measurements will be different. Each time you do an experiment the measurement will be different. So, what is the average value? Now in linear systems we just add V1 plus V2 plus V3 the entire thing divided by number of measurement, but here averaging requires a little bit different mathematical operations. So, average SPL equals 10 to the power of log 1 over n I equals 1 to n 10 to the power of dB SPL over 10 I's value. So, it is 10 to the power of 10 times logarithm of this entire expression. So, another example I will do is let say my 5 measurements are 96 dB's, 88 dB's, 94 dB's, 102 dB, 90 dB then my average SPL equals 10 to the power of log. Now 10 times logarithm of 1 over 5 10 to the power of 9.6 plus 10 to the power of 8.8 plus 10 to the power of 9.6 plus 10 to the power of 10.4 plus 10 to the power of 10.2 plus 10 to the power of 9 and that gives me 97 dB's. So, each of these operations are important. They have practical usage. They are cases where you will be expected to subtract contribution of source A from source B. There will be cases where you will have to take averages, there will be still some other cases we will have to add them up and you have to be aware that when you are manipulating decibels, then the rules of the game are little different. The last point I wanted to cover very briefly are two terms correlated sound and uncorrelated sound. So, a correlated sound is essentially when you have a bunch of sources and each source is generating sound. In fact, you precisely know the phase relationships for each frequencies coming from different all of these sources and also at a specific time, if source 1 is generating f 1, then source 2 is also generating f 1 frequency, source 3 is also generating f 1 frequency. Then the sound coming from all these different sources and the phase difference is also same and also the phase difference due to the variation of location is also not significant. Then you can add up the pressure. In practical applications, most of the sources of sounds which we have, they are not correlated. So, you run a machine here, a person is talking about disrupting, there is no correlation between these. So, you have a correlated sound, P total correlated is essentially P 1 times T plus P 2 times T plus P 3 times T and so on and so forth. A good example of this could be, I have one speaker, I have another speaker and both are having the same frequency response curves and they are being fed from the same amplifier, the same signal and also the distance between these two is not significantly large. In context of the wave lengths which these speakers are generating, then I can use this relation to add up the pressures. If I have uncorrelated sound, I could have one speaker or actually a human being. So, I could have three human beings, they are generating sound, they are talking and the person who is listening is situated a little away and in this case, P total uncorrelated equals P 1, it is a root square root of the squares of sums. So, here we do not have any information on the phase. So, again while we are doing measurements, we have to understand, are we measuring correlated sources or are we measuring uncorrelated sources and be aware, what is it that we are trying to do when we are summing up signals. That is all I wanted to cover for today and thank you very much.