 Good afternoon, welcome to this course on acoustics. In the last lecture, we had touched upon the area of board plots, pole 0 plots and phase and magnitude plots for transfer functions. In the context of providing the students or people who are interested in learning about acoustics, some grounding in some of the basic concepts which will be quite often used later in the course of acoustics. So, with that intention in the last class, I had covered in somewhat detail on the concept of board plots. And specifically, I had developed board plots for magnitude and board plots for phase, for complex 0 functions and for transfer functions which include simple zeros or for transfer functions which include just simple poles. So, today we will use all that information which we talked about in the last class and we will go a step further. We will start with a transfer function which will have which will be essentially a combination of poles and zeros and for this transfer function, we will construct a board magnitude plot and a board phase plot. So, in this context, let us say that the transfer function for which we are interested in developing a board plot for magnitude and phase, it looks like this s plus 2 over s square plus 6 s plus 25. So, if I find the roots of the denominator essentially what I get is in the numerator I still have s plus 2 and in the denominator I have s minus 3 plus 4j times s minus 3 minus 4j. Now, remember that these board plots are for the case that s equals j omega. So, in this transfer function I replace every s by j omega. So, what I get is H s, excuse me H of j omega is 2 plus j omega in the numerator and in the denominator I have minus 3 plus 4 plus excuse me 4 plus omega times j. This is one linear factor of the denominator and the other linear factor is minus 3 plus omega minus 4j. So, this transfer function is essentially a product of three individual functions transfer functions which is H 1 j omega times H 2 j omega times H 3 of j omega, where H 1 j omega is 2 plus j omega, H 2 of j omega is 1 over minus 3 plus 4 plus omega times j and H 3 j omega is minus 3 plus omega minus 4 times j. So, H 1 is 1 over minus 3 plus 4 plus omega times j and this is the expression for H 3. Now, as we are trying to develop a board plot for magnitude and a board plot for this transfer function what we will do is first. So, let us start by just thinking about the magnitude part. So, we will construct a board plot for magnitude for H 1 we will construct same thing for H 2 and same thing for H 3 and then because these board plots are essentially especially the magnitude is a logarithmic function because on the vertical axis we are plotting decibels all we have to do is add up these three board plots and we will get the final board plot for magnitude for the original function which is H. So, that is what we are going to do now. So, now what we are going to do is we are going to construct for a board magnitude plot for H 1. So, once again that is my original function as omega is approaching 0 when omega becomes extremely small H 1 j omega approaches to H 1 j omega so in decibels in decibels this approaches 20 log of to the of in base 10 of the number 2 which is essentially 6.02 decibels. Similarly, as omega is approaching infinity H 1 of j omega approaches j omega. So, if I am going to calculate the strength of H 1 in terms of decibels that approaches 20 logarithm in base 10 of omega because the magnitude of j omega is omega. So, this is these are the two asymptotes once again this is the low frequency the first one is low frequency asymptote and the second one is the high frequency asymptote. So, for H 1 the board plot board magnitude plot is going to look something like this board magnitude plot for H 1. So, on the horizontal axis I am plotting logarithm of omega and again it is logarithm on base 10 and on vertical axis I am plotting decibels. I know that my low frequency asymptote is a horizontal line it is a horizontal line which cuts the y axis at 6.02 decibels also I know the crossover point is such that when omega becomes 2 then the high frequency then after that the high frequency asymptote becomes more important. So, my crossover point this is my crossover point and that is essentially log in base 10 of 2 and my high frequency asymptote is a positively sloped line the slope being 20 decibels per decade 20 decibels per decade. So, I have this is my high frequency asymptote this is my high frequency asymptote and this is my low frequency asymptote. So, this is my board magnitude plot for H 1 of j omega likewise now I am going to construct a board magnitude plot for H 2 and H 3 functions. We will rewrite the relation for H 2 and that is 1 over minus 3 plus 4 plus omega times j that is H 2 as omega approaches 0. So, now again we are trying to construct a low frequency asymptote and also a high frequency asymptote. So, as omega approaches 0 H 1 approaches 1 over minus 3 plus 4 j are in decibels this is essentially 20 log to base 10 1 over 3 square plus 4 square the whole thing under square root. So, once again the low frequency asymptote comes out as a constant and when I calculate this value essentially what I get is minus 13.98 decibels and as omega approaches a very large number or infinity H 1 approaches 1 over omega j. So, in decibels essentially I get 20 log of 1 over omega which is essentially minus 20 log of omega. So, once again we see that the low frequency asymptote is essentially will be represented by constant horizontal line which is 13.98 decibels away negative 13.98 decibels away from the horizontal axis and the high frequency asymptote will be a straight line, but it will have a negative slope of minus 20 decibels per decade. Let us do the same thing for H 3 j omega. So the relationship is minus 3 plus omega minus 4 times j. So, as omega approaches 0 so excuse me earlier I had written H 1 this should be H 2. So, as omega approaches 0 H 3 approaches 1 over minus 3 minus 4 j or in decibels again it is 20 log 10 1 over 3 square plus 4 square whole thing under square root and that is again minus 13.98 decibels and as omega approaches infinity H 3 approaches 1 over omega. So, in terms of decibels I get 20 log to the base 10 1 over omega which is minus 20 log rhythm of omega. So, essentially what we are seeing here is that the magnitude part for H 2 and H 3 they are same. The low frequency asymptote for the magnitude of H 2 is minus 13.98 decibels constant line and same thing for H 3 and the high frequency asymptote for H 2 is essentially a negative slope line with a slope of minus 20 decibels per decade and it is the same thing for H 3. So, if I have to construct a board magnitude plot for H 2 on the horizontal axis I am going to plot log of omega on the vertical axis I am going to plot in decibels my horizontal so my low frequency asymptote. So, let us say I am plotting for I am doing a board magnitude plot for H 2 and H 3 because they are the same. So, my low frequency asymptote is again once again a horizontal straight line my high frequency asymptote is a negatively slope line and the slope of this is minus 20 decibels per decade and the crossover point is corresponds to this point and this number is such that when omega equals 3 the crossover point is 0.2. So, when omega equals 0.2 then low frequency asymptote and the high frequency asymptote after the crossover point the high frequency asymptote becomes more important. So, this is log of 0.2. So, this plot is for H 2 as well as H 3 now what I am going to do is I am going to develop a board plot magnitude plot H 2. So, it is not H 2 and H 3, but rather I will develop a plot for H 2 plus H 3. So, essentially I will add these two lines. So, my low frequency asymptote will be once again a constant line it will be a constant line such that the y intercept will be negative 13.98 times 2 decibels. So, this is my low frequency asymptote and that is minus 27.96 decibels on the horizontal axis I am plotting log of omega on the vertical axis I am plotting decibels low frequency asymptote once again is minus 13.98 times 2 which is negative 27.96. The high frequency asymptote is steeply sloped line and the slope is negative 20 plus negative 20 decibels per decade. So, the overall slope is minus 40 decibels per decade and once again the crossover point remains as is and it does not change when I am just adding up the board plots for H 2 and H 3. So, when I am summing up the board plots for H 2 and H 3 as I mentioned just now the crossover point still remains omega being equal to 0.2 which means on the horizontal axis it gets plotted as log of 10 of 0.2. So, this is this plot is a combination is a board magnitude plot for a combination of two functions or a product of two functions which are H 2 and H 3. So, now we will construct the final board magnitude plot which will be essentially a sum of three individual magnitude plots for H 1 H 2 and H 3 once I sum all these three up I get the board magnitude plot for a function H 1 times H 2 times H 3 which is same as the original H the transfer function which we had developed. So, for purposes of clarity I will just redraw the magnitude plot for H 2 times H 3. So, the low frequency asymptote is a horizontal line with a y intercept at minus 27.96 dB and this low frequency asymptote dominates the response of the system represented by H 2 times H 3 up to specific crossover point which is which corresponds to this point when omega is larger than 0.2 then the high frequency asymptote starts dominating the response. So, that is my crossover point and after that I have my high frequency asymptote and here the slope is minus 40 decibels per decade. This is mag plot actually I will be more specific board magnitude plot for H 2 plus H 3 and then we had constructed a board magnitude plot for H 1 and that was the low frequency asymptote was again a horizontal line with y intercept of 6.02 decibels. The crossover point was omega equals 2 and after the crossover point high frequency response starts dominating and that is represented by a straight positively slope line the slope being plus 20 decibels per decade. So, now I am going to construct a board magnitude plot. So, just to be mathematically consistent this is a board magnitude plot for a function which is equal to H 2 times H 3 not H 2 times not H 2 plus H 3, but rather H 2 times H 3. So, now we will construct a board magnitude plot for H S which is our original transfer function when S equals j omega and that is nothing but product of three individual transfer functions H 1 j omega times H 2 j omega times H 3 j omega. In the way we are going to do it is we are going to add these two plots we add these two plots and we get our final answer. So, that is how I am going to construct it and this is how the final plot looks like. Once again my horizontal axis is a logarithmic axis where I am plotting omega on the vertical axis I am plotting decibels. So, what we see here is that up to omega equals 0.2 the response curves are essentially straight lines. So, let us say this is log of 0.2 this is log of 2 and up to this point I just add two horizontal lines one line is plus 6.02 decibels away from the x axis the other line is minus 27.96 decibels away from the x axis. So, my first segment of the line will be something like this and this is minus 21.94. Now how did I get this 21.94 essentially it is minus 27.96 which is this number plus 6.02 equals minus 21.94 decibels. So, the response curve for a system which is represented by this transfer function H is essentially a constant horizontal line which is minus 21.94 decibels away from the horizontal from the horizontal axis up to a frequency of 0.2 angular frequency of 0.2. Now after that I have another segment of frequency and the band of the frequency which is from 0.2 to 2 and for this region we see that this particular transfer function it is having a slope of negative 40 decibels per decade while this transfer function is still in the constant range. So, because this is let us say log of 10 of 0.2. So, the overall transfer function will be negatively slope line it will be negatively slope line and the slope will be minus 40 decibels per decade up to angular frequency of 2. Once I have exceeded the angular frequency of 2 then this part of the curve for H 1 also starts to kick in. So, my overall slope becomes minus 40 plus 20 decibels per decade. So, my slope becomes lesser and here the slope is minus 20 decibels per decade. So, this is my overall board magnitude plot for a transfer function H of j omega where H of j omega is essentially a product of three individual transfer functions H 1 times H 2 times H 3. So, we will use a very similar approach for also developing about phase plot for this complex transfer function H and that is what we are going to do in next several minutes. So, now our objective is to develop a board phase plot a board phase plot for H of j omega. So, once again my H where H of j omega is a product of H 1 j omega times H 2 j omega times H 3 of j omega and once again for purposes of clarity I am going to rewrite the functions for H 1 it is 2 plus j omega for H 2 it is 1 over minus 3 plus 4 plus omega times j and for H 3 it is 1 over minus 3 plus omega minus 4 j. So, the phase of all these specific individual functions will be something like this for H 1 phase will be tan inverse of omega over 2 for H 2 the phase will be negative because it is in the denominator of tan inverse of omega plus 4 over 3 actually in the denominator I will have negative 3 and for H 3 it will be negative of tan inverse omega minus 4 over minus 3. So, these are the three different relations for phase for each particular sub transfer function. Now we will start constructing board phase plots for each of these transfer functions and then finally we will just add them up we will get our final board phase plot. So, for H 1 when omega goes to 0 phase of H 1 goes to 0 when omega goes to infinity phase of H 1 goes to pi over 2. So, my board plot is going to look like for low frequency it is a constant line at 0. So, once again let me just plot the label the axis horizontal axis is log of omega vertical axis is phase and this is in radiance. So, this is my low frequency asymptote and my high frequency asymptote is going to look like this and the crossover point where the intercept is going to be pi over 2 this is 0 and my crossover point is log of 2. So, now I am going to do same thing for H 1 H 2 as omega is approaching 0 phase of H 2 is approaching negative of tan inverse of 4 over minus 3 and that is 0.93 here little more precise 9 to 8 7 radiance and as omega is approaching infinity phase of H 2 is approaching negative of tan inverse omega over minus 3 and when omega becomes extremely large then essentially it becomes plus minus pi over 2. So, once excuse me when omega has a positive value and it is extremely large then it becomes pi over 2 I will again. So, the board plot for this H 2 the phase plot for this will look like something like this. So, again I am having phase of H 2 being plotted here on y axis on the horizontal axis I am plotting log of omega and base tan and this is my low frequency asymptote the crossover point is log of 0.2 and my high frequency asymptote is once again a constant line and the value y intercept is pi over 2 for H 3 as omega approaches 0 phase of H 3 approaches minus tan inverse minus 4 over minus minus 3 and that is essentially minus 0.9 to 8 7 and as omega approaches infinity from the positive side then H 3 phase approaches negative of tan inverse omega over negative 3 and that is pi over 2. So, just for purposes of clarity let me just label this is phase plot for H 1 this is phase plot for H 2 and now I am going to construct a phase plot for H 3. So, my low frequency asymptote is again a horizontal line with the y intercept of minus 0.9 to 8 7 once again I am plotting here phase and here I am plotting log of tan of omega. So, my low frequency asymptote for phase is a horizontal line with a y intercept of minus 0.9 to 8 7 then after the crossover point which corresponds to omega being equal to 2 I have a high frequency asymptote and the high frequency asymptote also has a y intercept of pi over 2 and once again this crossover point is log tan of 0.2. So, this is H 3 I am plotting phase plots. So, I have plotted phase plots for H 1 H 1 H 2 and H 3 on log scale. So, my next logical step is just to add these three up because we know that if there is a function which is a product of three individual functions then the phase for this complex function is nothing but a sum of phases of each individual sub function. So, once I do this exercise my final curve it looks like this on the horizontal axis I am plotting log of omega on the vertical axis I am plotting phase of the original transfer function and I know that for the range 0 to 0.2 the contribution of H 1 and H 2 cancel each other and the contribution from H 1 is identically 0. So, for the range 0 to 0.2 my curve for the phase is a horizontal line which is exactly cutting the vertical axis at 0. Then for the range 0.2 to 2 the contribution from H 1 is 0.2 to 2 the contribution from H 1 is pi over 2 no excuse me the contribution from. So, for the range of frequencies or angular frequencies between 0 and 0.2 we have seen that the phase plot is essentially coincident with the horizontal axis and the x and the y intercept is 0 radians. So, this is what it corresponds to this dark blue curve for the angular frequency range for from 0.2 to 2 the contribution from H 2 and H 3 is pi over 2 radians we see it from here contribution from H 2 and H 3 is pi over 2 radians while contribution from H 1 still remains 0 because the low frequency asymptote for H 1 is 0 up to 2 radians per second. So, what that means is that for this range for this range my phase curve will once again be a constant line, but it will be a constant line. However, the x a y intercept will be at pi radians right then for the range. So, this is excuse me then for the range where angular frequency exceeds to which is for this range for this range let us look at the contributions from H 1 from H 1 the contribution is pi over 2 for H 2 the contribution is pi over 3 again a constant and for H 3 from H 3 the contribution is once again pi over 2. So, pi over 2 times pi over 2 the pi over 2 plus pi over 2 plus pi over 2 is 3 pi over 2. So, for this range the phase response is 3 pi over 2. So, once again I have a horizontal line and it extends to infinity and it extends to infinity and that is my phase response curve my phase curve board phase curve for a transfer function which is again a product of 3 individual transfer functions H 1, H 2 and H 3. Now, bear in mind that this is the asymptotic response curve. However, in real systems the transitions from this range to this range to this range they are more or less fairly smooth. So, the actual response curve will be bounded by this asymptotic response curve, but however this asymptotic response curve it gives us some qualitative understanding how the system is going to behave. The actual curve in this case may very well look like something like this and that is what I am going to plot in light blue. So, at extremely low frequencies it is bounded by this dark blue line curve which is essentially a horizontal line cutting the vertical axis at 0 at extremely high frequencies it is bound by another horizontal line which has a y intercept of 3 pi over 2. So, this is how we can construct board magnitude plots and board phase plots for transfer functions which can be broken up into individual poles and zeros and once we have an individual curve for every single pole and every single 0 then all we have to do is we have to add these up and we get a board magnitude plot and we also get a board phase plot for the original transfer function which could be a combination of several individual pole and zero related transfer functions. So, that is pretty much what I wanted to cover in terms of board plots and I think this introduction will help you develop board plots for magnitude and phase for fairly complex transfer functions and that will come in really handy once we start talking about acoustics in detail. So, we have covered a slew of topics complex algebra complex numbers board and board magnitude plots board phase plots whole zero plots and so on and so forth and all these concepts will come in fairly handy once we start talking about acoustics in detail. So, having said that now I will move to the wave equation which is the first concept we are going to talk about in context of this course on acoustics. So, before I start talking about the wave equation per se just wanted to give you a brief understanding of what this equation means and what is it that we are trying to do through the wave equation. So, as we had talked about earlier in acoustics we can break up this whole range of all issues and problems which we try to solve in the area of acoustics into three groups. The first group relates to situation where noise or sound is being produced. So, you have a source and this source produces sound and then I have a listener and this sound and this listener could be a human being an animal a microphone or whatever. So, I have a source, I have a listener and I have a medium and there are different branches of acoustics which deal with these three things. There is a branch which helps us understand how sound is generated and a lot of that information comes from ideas in electro acoustics. So, electro acoustics along with some related areas of acoustics they help us understand how sound gets generated. Then in terms of listening or recording sound or analyzing sound or sensing sound here we deal with two broad areas one is electro acoustics. Electro acoustics helps us design for instance microphones which help us sense sound and then that can be recorded and then another very big area is psycho acoustics. And what psycho acoustics helps us do is it helps us understand sound from the standpoint of the human individual. So, it is one thing to record data and see it on a computer screen, but how does brain understand sound and how does brain interpret sound signals that is what psycho acoustics helps us understand. So, this is about sound generation, this whole area is about sound sensing of sound and interpreting of sound. And as sound travels from the source to the listener through a medium the branch of acoustics with which deals with this propagation of sound is called physical acoustics. So, that is physical acoustics. So, sound gets generated and there we seek the help of electro acoustics to understand that phenomena it gets propagated from point A to point B and there we seek the help of physical acoustics to understand that propagation phenomena. And then finally, there is a listener who is listening to sound and there the understanding is derived through principles of once again electro acoustics and or psycho acoustics. So, wave equation is it helps us understand how sound gets transmitted through a medium. So, that comes in the ream of physical acoustics. So, that is a very broad context in which I wanted to place wave equation where does wave equation fit in the overall scheme of things. So, what we will be doing in this course is that first now starting from today onwards we will do several lectures on physical acoustics and then we will try to understand how sound gets propagated through a medium. Once we have done that then we will move on to electro acoustics and there we will try to understand how we can generate sound. So, we will address this part and then finally we will go to the listener and there we will try to understand how sound gets recorded interpreted and so on and so forth. So, that is the overall course landscape and that is the context where we can place wave equation. So, what we are going to talk about is wave equation and once again what this equation helps us do is helps us understand how does sound get propagated as it travels through a medium. Now, this medium could be a piece of solid as in a piece of steel or it could be volume of gas. So, most of times when we speak we are in sitting in air and sound gets propagated from our mouth to an individual's ear. So, that is again where wave equation comes in handy and there are different versions of wave equation for different types of medium or it could be the case of a fluid not to be more specific the case of a liquid where you have sound getting propagated in water through miles and miles or kilometers and kilometers of distances and how does get sound propagated there in that type of a medium. So, in each of these different media types it is the wave equation which helps us understand how sound gets propagated. What we are going to talk about today is specifically propagation of sound in elastic media and more specifically in air at atmospheric conditions. And even more specifically we will talk about the 1 D wave equation. So, typical example of 1 D wave equation could be I have a long tube and I am speaking in that tube and the tube may be bent, but sound gets it travels through that tube through the air inside the tube and it can be heard at the other end of the tube. So that is typical example of 1 D wave equation. Another example could be again rectangular pipe or a channel let us say you have HVAC duct at one end of the duct you have a blower which is blowing air into the duct and in that process it is also generating some noise and that noise moves through the duct and once sound gets it comes out from the duct into an air conditioned room that sound gets heard. So what is the phenomena and how is that propagation happening through the duct that is something 1 D wave equation will help us understand. So, what we will be talking about is wave equation in one dimension. Now initially we will start this discussion by having propagation of the wave in one dimension where my coordinate system is a rectangular Cartesian coordinate system, but later I will use mathematics to also explain the wave equation in a radial or a spherical coordinate system. So let us consider a 1 D piece of air, one dimensional piece of air where sound is getting propagated and let us say the initial pressure and initial pressure and initial pressure, initial density, initial volume and initial velocity of this air are p naught or p naught, rho naught, v naught and u naught and then because of the disturbance due to sound propagation these values change. So the initial values have a subscript of 0 and the final values have a subscript of p t. So my p t which is the final pressure and it can depend on x which is the position. So again it is not x, y and z, but it is just x because it is only in one dimension. So my final pressure which is p t is dependent on x and it is also dependent on time. So that is equal to p 0 which is my initial pressure which was again dependent on x and t, excuse me it was p 0 plus some disturbance which is dependent on x and t. So once again here I am saying essentially that p 0 does not change with time and x because it corresponds to standard atmospheric pressure which is approximately 10 to the power of 5 Newton per square meter. So that is for that is the relation for pressure likewise the density of the air under consideration, the final density could be rho subscript t and that is equal to my initial density which is rho naught plus incremental change in the density which is rho which is again a function of position and time. So that is my relation for density. Then my volume of air, let us say my final volume of air, let us say my final volume of air is V subscript t and that is equal to initial volume V 0 plus increment some change in volume which is tau again a function of position and time. So that is the relation for volume and finally I have a relation for velocity, velocity being u. So my final velocity is ut which is equal to an initial velocity u naught plus some change term which is represented by u which is again a function of position and time. Now usually u naught is essentially 0 because we assume that in the initial state of affairs there is no motion in air. So u naught is 0 so I can drop this term because u naught is 0 usually. Also it is important to provide some values at this point of time. So my p naught equals approximately 10 to the power of 5 Newton's per square meter my rho naught which is the density of air at MSL is 1.18 kg's per square meter and initial volume could be variable because I can consider as much volume as I want. So once again p naught, rho naught, v naught and u naught which is 0 in this case they are my initial pressure values of pressure density and volume and velocity while p, rho, tau and u are my incremental values for pressure density volume and velocity. And when I add these two up the initial value and the incremental value I get final values which are designated by a subscript t. So this is how I am going to frame my problem. So what I am interested in knowing is how are p, rho, tau, u all interrelated and as I am trying to develop these relationships I will make two very important assumptions. So I will make two important assumptions. The first assumption is that this is a one dimensional system and so my system is one dimensional. And what that means is that del over del y that is the partial derivative of any variable with respect to y is equal to 0 and del over del z is also equal to 0. Second assumption I am going to make is that I have constant mass particles. So I have a piece of air which has initial mass its mass is not changing as it moves through the system. So essentially what that means is rho t, v t is same as rho naught plus rho times v naught plus tau and that is same as rho naught v naught plus incremental terms and that is equal to rho naught v naught. So these are the two key assumptions and then we will also talk a little bit more about some other assumptions in the next class. But this is how I am going to frame my problem and in the next class what we will do is we will try to develop an equation of how pressure and density and velocity and all these variables are interrelated through three important equations. One is the Newton's equation which is an equation for force equilibrium. So if I have a piece of fluid it has to have equilibrium from using Newton's laws of motion. The second equation I am going to use a relationship I am going to use is that for conservation of mass and that is what I call continuity equation. And finally I will use a relationship which links pressure and volume through a material constitutive equation and through these three equations. Equation of momentum that is Newton's law, equation of mass conservation and material constitutive equation will develop wave equation in one dimension. Thank you very much.