 So, in today's class we will be capturing a few more review concepts and with this capture of review ideas once this is done then going forward we will move into formal acoustics area and we will start with one dimensional wave equation, but at least in today's class and may be one more class we will be essentially doing a review of some of the concepts which we will be using frequently in the area of acoustics. So, the first concept I would like to talk about is having a very brief overview of what is an over damped system, what is a critically damped system and what is a under damped system. And I will explain that in context of a spring mass damper system and hopefully with that illustration these terms over damped, critically damped and under damped will become clear. So, let us consider a dash pot and then I also have a spring. So, the dash pot viscosity is C, spring constant for the stiffness is member is K and both these elements they are connected to a rigid body mass which can move on a frictionless surface. So, now let us assume that I am trying to perturb this by a small displacement x. So, from our previous classes we know that in such a situation the system is governed by a differential equation of this type m x double dot which is mass times acceleration plus viscous coefficient times velocity plus stiffness times displacement equals 0. And to solve this equation if I assume that x is some constant times exponential of E s t where s could be a complex number and if I plug this in my differential equation what I get is m s square plus C s plus K times x naught E s t equals 0. And if this equation has to hold true at all times for all frequencies then the only natural conclusion we can draw is for this equation to hold true this term m s square plus C s plus K should be 0. What that means is I will write that m s square plus C s plus K equals 0. Now, we know what is m it is a physical quantity mass of the object we know C it is again a physical quantity viscous coefficient of the damper and we know K which is the stiffness of the spring. So, this is a quadratic equation it is a quadratic equation in s. So, I solve for s. So, s equals minus 2 C I am sorry it is minus C plus minus and then I have C square minus 4 K m and then I divide the whole thing by twice of mass which is minus C over 2 m plus minus and then I move mass under the square root sign. So, what I get is C over 2 m square minus K over m K over m. So, that is my s and s we know we call s is complex frequency. So, now we will come up with three scenarios. So, first is case 1 and in this case this term is the term under the square root sign is larger than 0. In other words C over 2 m square is greater than K over m and when that happens what it essentially means is that s is a purely real negative number. So, there is no imaginary component to s and which in turn means that for a system where you have C over 2 m square is greater than K over m there are no oscillations in the system and such a system where C over 2 m whole square is larger than K over m such system is called over damped system. Then there is another case. So, case 2 would be C over 2 m whole square equals K over m in that case my s will be minus C over 2 m again C is a real number m is a real number. So, s is a real number and it is also negative. So, once again in this case there are no oscillations happening in the system and the system is called critically damped system. A good example of a critically damped system is sometimes in the doors which have dampers. So, that the door closes silently without creating a lot of noise. Sometimes those dampers are tuned in such a way that the whole system is critically damped. Some other examples where people use critical damping is these military guns where the barrel coils back and the system is being fired. So, once you have a charge coming out of the barrel the barrel recoils and that recoil phenomena is critically damped. So, in critically damped system the response as you may have studied in your previous courses would be something of this nature a plus b t times exponent of minus omega naught t where omega naught equals C over 2 m and then we have a third case. So, it is case 3 where C over 2 m whole square is less than k over m and in this case s is a complex entity. So, it is a complex number and once you have a complex number what that also means is that the system is going to oscillate because a part of a component of the complex frequency is imaginary in nature and that will that essentially implies that the system will oscillate. So, that is there. So, let us plot some of these performances. So, if I have a system and I am trying to plot its displacement if I perturb a spring mass tamper system single degree of freedom and if I perturb it and along time axis the horizontal axis is for time and the vertical axis is for displacement then critically damped system would be the fastest to decay. So, this is critically damped then an over damped system will also decay fast, but not as fast as a critically damped system. So, it will be fast, but it will be a little slower and note that in both these situations the oscillations are not happening. This is an over damped system and then finally, you have an under damped system where they will be oscillations and the magnitude of all these oscillations will slowly decrease will decrease with time in an exponential way. So, this is my under damped system. So, that covers the review of different types of damping over damped systems under damped systems and critically damped system. So, now, we will move on to the next concept in the class and this is what we call a pole zero plot. So, what we are going to talk about pole zero plots. So, any transfer function and we had talked about transfer function in some of the earlier classes. So, if I have a system, there is an input signal going into the system and then I am observing and I consider that observation as an output of the system and if I take the ratio of output and input and if a system is linear, then that ratio is called a transfer function and this transfer function depends on can depend on the natural frequency not natural frequency of the in going excitation and it will also depend on the element properties of the system. So, any transfer function can be it is a ratio and because it is a ratio, I can decompose it into roots of the numerator and roots of the denominator. So, it could be s plus z 1 s plus z 2 s plus z 3 and so on and so forth depending on how complex the transfer function is and I can also decompose the denominator into its in terms of its roots. So, s plus p 1 s plus p 2 s plus p 3 and so on and so forth. So, here z 1, z 2, z 3 these are roots of numerator and they are called zeros. They are called zeros because at s equals z 1, z 2, z 3 the transfer function goes to goes to an identically zero value. Also the roots of denominator p 1, p 2, p 1, p 2, p 3 they are called poles and at these values of s that is which is complex frequencies. So, h s goes to zero at zeros and h s goes to infinity at poles. Actually I made a small error here. Actually these should be s minus z 1, s minus z 2, minus z 1, s minus z 2, s minus z 3. Similarly, here I should be having s minus p 1, s minus p 2 and s minus p 3. So, when I am so let us do an example. So, we will consider transfer function h which is function of natural which is function of complex frequency s and it is expressed as s plus 3 over s square plus 7 s plus 10. So, this is s plus 3 and then I can make I can factorize the denominator. So, it becomes s plus 2 times s plus 5. So, my zeros are s equals minus 3 which is yeah which comes from the numerator and the roots of the denominator will give us the poles. So, my poles are s equals minus 2 and minus 5. So, if I plot this on a complex plane that is my real axis this is my imaginary axis. So, 1, 2, 3, 4, 5 and I will designate a zero as a circle. So, at minus 3 I have a zero and then I can designate a pole as an x. So, at s equals minus 2 I have an x which is a pole and at s equals minus 5 I have another pole. So, I know that s equals sigma plus j omega. So, on the real axis I am plotting sigma and on the imaginary axis I am plotting omega. We will do another example and in this case the transfer function is h s equals s square plus 4 over s square minus 6 s plus 25. So, I can decompose the numerator into its roots. So, that is s plus 2 j times s minus 2 j and then the denominator is s minus 3 plus 4 j times s minus 3 minus 4 j. So, my poles which are designated as x s are roots of the denominator. So, it is 3 plus 4 j and the other pole is 3 minus 4 j and my zeros designated as o's are minus 2 j and 2 j. So, if I have to plot them on a complex plane that is my real component this is my imaginary component. So, my zeros are located along the imaginary axis plus and minus 2 j my poles are located in the complex plane as I am going to just plot just now. So, it is 3 plus and minus 4 j. So, let us say this is 4. So, I have a pole here I have another pole at 3 minus 4 j minus 4 and then I have plus 4 on the positive side. So, this is the representation of my poles and zeros for this transfer function h s. So, now that we have understood how to identify poles and zeros of transfer functions we want to go to the next step and which is plotting these transfer functions and I can plot the transfer function and it will have a phase component and the other component will be a magnitude component. So, that is what we are going to do the purpose of making these plots is that I have a graphical representation of how h s is varying as omega is changing. So, typically so let us say I have h s. So, a phase plot or a magnitude plot will be a plot of h when s equals j omega. So, I do not put a real part in s and I just plot h as a function of j omega. So, my horizontal axis is going to be omega and my vertical axis is going to be the value of h the magnitude of h or the phase of h. So, let us say h s equals s plus 3 over s square plus 5 s plus 10. So, we had broken this earlier as this s plus 3 over s plus 2 times s plus 5. So, now if I have to do a phase plot or a magnitude plot for h I set s equals j omega. So, then my h j omega becomes 3 plus omega j over 2 plus omega j times 5 plus omega j. Now, what we see here is that h is a function of is a product of three individual small transfer functions. Why is that? So, I will just write down h 1 times h 2 times h 3 where h 1 is 3 plus omega j h 2 is 1 over 2 plus omega j and h 3 is 5 1 over 5 plus omega j. So, and that is h j omega and we know from the mathematics of complex variables that if I have a relation for h in such a way then magnitude of h j omega is basically magnitude of h 1 times magnitude of h 2 times magnitude of h 3 and also the phase of h j omega. So, the phase of h j omega is basically phase of h 1 plus phase of h 2 plus phase of h 3. So, in this case the phase of, so again so magnitude of h is basically magnitude of h 1 and h 1 is 3 plus omega j. So, magnitude of h 1 is 3 square plus omega square times magnitude of h 2. So, I will write down that times 1 over 2 square plus omega square the whole thing under square root and then similarly I have magnitude I have to write down magnitude of h 3. So, it is 1 over square root of 5 square plus omega square. So, now we have a relation and I can plot the magnitude component of h in terms of omega. Similarly, the phase component I can plot and I know from complex variable theory that the phase of h is basically a summation of phases of h 1, h 2 and h 3. So, that is what I will do. So, phase of h 1 is tan inverse omega over 3 minus tan inverse omega over 2 minus tan inverse omega over 5. So, I have now two relations I can use these two relations to develop a phase plot and a magnitude plot for h. So, in a quantitative sense this is how the magnitude plot for this particular transfer function looks like. So, what I am plotting is magnitude of h on the horizontal axis I have omega and this function it peaks it peaks at 0 and the value of h at 0 is about 0.3 and then it asymptotically goes to 0 as omega tends to goes to infinity and the phase plot again on my horizontal axis I am plotting omega and my vertical axis I am plotting phase angle and the phase plot looks something like this these are asymptotes at pi over 2 and minus pi over 2 and my phase plot looks something like this. So, it is an anti symmetric curve anti symmetric along the vertical axis the magnitude plot is a symmetric curve symmetry being along the vertical axis. So, this is how we can construct phase and magnitude plots for transfer functions by step 1 decomposing the whole transfer functions into its poles and roots and factors associated with those poles and roots and b if it is we are trying to plot magnitude then we find individual you know factors themselves and multiplying them and if we are trying to find the phase then we find phase relations for individual factors and then just adding them up. We will do another concept which is called the board plot. So, right now what we did was we plotted h in on a linear axis plotted the magnitude of h on a linear axis the horizontal axis was linear and also the vertical axis where magnitude is being plotted that was also linear and same was true for phase. Now, what we have another concept and it is called a board plot and board plots are essentially graphical depictions of transfer functions when s equals j omega. So, they are graphical depictions of transfer functions at s being equal to j omega and more specifically they are asymptotic behaviors. Further when we are plotting board graphs then we essentially use logarithmic scales on both horizontal and also the vertical axis when we are plotting the magnitude and when we are plotting phase the vertical axis remains a linear scale but the horizontal axis still is a log axis plots. So, they can be two types of board plots one is magnitude plot and this is basically a plot of 20 logarithm on base 10 magnitude of h j omega versus on the horizontal axis we have log of omega again. So, log 10 and the second board plot is a phase plot and here we are plotting the phase of h when s equals j omega and on the horizontal axis we have log 10 omega and these plots are basically plotted in such a way that we have asymptotic representations. So, essentially we are trying to plot the asymptotic behavior of magnitude and asymptotic behavior of phase of h as omega is changing in case of the magnitude plot we are plotting asymptotic behavior of h on a log scale in the vertical axis and also on a log scale on the horizontal axis or phase the vertical axis is still linear but the horizontal axis is log scale. The point the central point when we are plotting these board plots is that we are focusing on asymptotic behavior because they are these plots very quickly help us a generate a curve which we can look and get a feeling of the approximate behavior of the system and once we have an approximate we have feeling for the approximate behavior of the system then we can use that information to validate our understanding of the model which we are trying to analyze. The other point I wanted to make is that some thought as to why we are using 20 times logarithm of h j omega especially when we are plotting the magnitude plot. So, this is because it is not directly a representation of the magnitude itself but rather it is a representation of the energy contained in the system. So, what do I mean by that? So, let us try to understand it a little bit more at a detail level. So, typically when we plot h in physical systems I have a signal going in and I have a signal coming out. So, let us say I have a acoustic system it is getting excited by some signal let us say it is an electrical signal. So, I try to monitor that electrical signal in terms of so my input parameter is voltage and my output parameter in case it is a microphone or some other displacement sensor or velocity sensor. At the end of the whole measure of a transduction process I end up measuring some more voltages. So, my input is a voltage and my output in a lot of cases is voltage as well or in other cases it could be current and current or combination of voltage and current. V square which is square of voltage is essentially proportional to energy. For instance across a resistance if I have voltage V then the energy dissipated is V square over R. Similarly, I square is again proportional to energy in a general sense. Again if I have the example of a resistance the current going through it is I and the value of resistance is R then energy dissipated or power dissipated is I square R. So, when I am taking ratios so let us call this a decibel and this is defined as 10 times logarithm of ratio of energies. So, as energy is directly proportional to V square and current is directly proportional to energy is also directly proportional to I square. So, I can so at this point of time now that we have seen that energy is proportional to the square of voltage or energy is proportional to the square of current we introduce a term called decibels and decibel is essentially defined as 10 times log of ratio of energies and the log is on log 10 scale. So, what that means is that it is essentially log 10 and ratio of energies could be V 2 square over V 1 square and I have to take the magnitude of it as this is log and I can similarly have a relation for current also now note that V 2 and V 1 are functions of j omega they can be functions of j omega. So, when I have a square relationship what this essentially means is that it is 20 log 10 of magnitudes of V 2 over V 1. So, therefore similarly so when we are taking and we are plotting transfer functions, but they are also ratios of whatever is going in and whatever is coming out. So, I have if I am converting transfer functions on a decibel scale then what I get is decibels equals 20 log 10 of h which can be a function of j omega. So, as we plot 20 log 10 j omega h j omega on a y space on vertical axis we are plotting a law on a log scale and on the horizontal axis why we are plotting on a log scale is because typically the range of frequencies for acoustics problems they can start from 20 hertz or maybe even sometimes less than 20 hertz and then they can go up to tens of thousands of hertz. So, it is a wide range so it is much more convenient to use a log scale on the horizontal axis. So, we were talking about both plots and we had said that these plots could be of two types magnitude plots and phase plots. So, now we will see how to plot how to construct these plots. So, as we had discussed earlier any transfer function could be decomposed into its factors of the numerator and factors of the denominator. So, let us do that. So, it is this is a general transfer function and then on the denominator I have poles. So, now what we will do is we will as step 1 construct a board plot for a transfer function which is made up of only a factor of the numerator and in another case it is only a factor of the denominator and see how it behaves and then we can try to develop an approach to which we can combine plots of different factors. So, let us consider a transfer function s plus z, z implying it is on the numerator and it relates to a 0 and I will construct a magnitude plot for this. So, you have to construct a magnitude plot I have to convert this into decibels which is as we had explained earlier is 20 log on scale 10 magnitude of h s where h equals j omega which is 20 log 10 z plus j omega and as we mentioned earlier we have to develop a plot which reflects the asymptotic response. There will be two asymptotes for this relation. So, asymptote as omega tends to infinity omega tends to 0 is 20 log and asymptote as omega goes to infinity is 20 log omega. So, I will plot these. So, my vertical axis as I had said is on a decibel scale my horizontal axis is on a log scale. So, I am plotting log omega on the horizontal scale on the vertical scale I am plotting decibels and I am plotting decibels for the transfer function h s as s equals j omega equals j omega plus z. We had seen that as omega goes to 0 the value of d b tends to 20 log and as omega goes to infinity the value of d b goes to 20 log omega. So, these are the two asymptotes this is the low frequency asymptote and this is the high frequency asymptote. So, the low frequency asymptote is converging to a straight horizontal line 20 log z. So, let us say that value is 20 log z and the low frequency asymptote is a horizontal line and that is my low frequency asymptote and my high frequency asymptote is 20 log omega and it and it cuts that the high frequency asymptote cuts the low frequency asymptote at omega such that log omega equals log z. So, at this line my high frequency asymptote cuts the thing also this is my high frequency curve also the slope is 20 d b per decade. We had explained earlier a decade is the space between two frequencies which are off by a factor of 10. So, when omega goes increases and let us say we are starting from omega equals omega 1 to omega equals omega 2 where omega 2 is 10 times omega 1 then this number goes up by such in such a way that the slope of this line is 20 decibels per decade. So, this is my again to reiterate the black dotted line is my low frequency asymptote the red dotted line is my high frequency asymptote the asymptotes cut at an important point which corresponds to omega such that omega is where omega equals z or log omega equals log z. So, I have a high frequency asymptote red line low frequency asymptote black line and the combined line I will draw is the green line. So, this is my total response curve. So, my response curve for transfer function h s where s equals j omega equals j omega plus z where z is the 0 is this green line what this line tells me is that as long as my omega is less than z. So, in this region the low frequency asymptote is going to dominate and when my omega exceeds z then the high frequency asymptote is going to dominate. So, my total response curve is this green line which is a combination of low and high frequency asymptotes. So, what I have drawn here is a response curve a board plot for magnitude of a very simple transfer function where transfer function is essentially z plus j omega. So, now we will draw another transfer function h j omega or actually we will start from s. So, h s equals 1 over s plus t. So, remember we can decompose a very complex transfer function into a bunch of simple factors s plus z 1 s plus z 2 s plus t 3 which all go in the numerator and in the denominator we can decompose it as s plus p 1 times s plus p 2 times s plus p 3 and so on and so forth. So, what here I am doing is I am going to plot the the board plot or a very simple transfer function h s which equals 1 over s plus p such that s equals j omega. So, this becomes h j omega equals 1 over p plus j omega. So, again I have to draw asymptotic curves because I have to develop an asymptotic response. So, asymptote as omega tends to 0 gives me d v becomes 20 log of this whole relation is 20 log 1 over t is basically minus 20 log e and asymptote as omega goes to infinity gives me in decibels if I have to do that is 20 log 1 over omega equals minus 20 log. So, now if I have a transfer function which has only a pole then this is how the asymptotes of the transfer function are and now I will plot it I have a horizontal axis on which I am plotting logarithm here in mind these are all log to the base 10 omega and my vertical axis I am plotting decibels and I am going down because both these are negative numbers. So, my low frequency asymptote I will draw a black line black dotted line this is my low frequency and the y intercept of this is minus 20 log e and the high frequency asymptote I will draw in red and this has a slope of minus 20 d v per decade again I will draw it as a red dotted line. So, this is high frequency asymptote the slope is minus 20 decibels per decade and the combined response before I talk about combined response the intersection point corresponds to a value of omega when omega equals t. So, this is log of e so this is my intersection point. So, my combined response I will plot in green is for low frequencies below omega equals p my low frequency asymptote is going to dominate and for high frequencies for high frequencies the high frequency asymptote is going to dominate. So, this is my combined curve for transfer function which is expressed as h s equals 1 over s plus p. So, once again I have a low frequency component which is this part and I have a high frequency asymptotic response which is this part. The slope is minus 20 d v per decade the break point corresponds to logarithm of p where p is this entity and the horizontal flat curve is basically a straight line in cutting the y axis at minus 20 log p. So, using this approach so we have seen that a complex transfer function can be expressed as a sum of as a you know as a bunch of products on the numerator and also a bunch of products in the denominator. Now, we have seen how to construct plots both magnitude plots for each of these numerator components and also we have seen how to construct magnitude plots for each of these denominator components. So, now the next step would be that you synthesize all these individual plots into one single plot and that will be the overall board magnitude plot for the whole transfer function. We will illustrate that through an actual example my transfer function let us say is h s equals s plus 2 times s square plus 10 s plus 109 which is equal to. So, now I factorize the denominator and what I get is s plus 2 over s plus 3 plus 10 j and s plus 3 minus 10 j. So, now what I do is s equals j omega. So, my h which is a function of omega becomes 2 plus j omega over 3 plus 10 j plus j omega times 3 minus 10 j plus j omega and this I can write it as h 1 omega times h 2 omega times h 3 omega where h 1 equals 2 plus j omega h 2 equals 1 over 3 plus 10 j plus j omega and h 3 is 1 over 3 minus 10 j plus j omega. So, I have expressed a complex transfer function such as h in terms of 3 simple transfer functions h 1 h 2 and h 3. My next step will be to construct board plots for h 1 a board plot in magnitude for h 2 and a board plot in magnitude for h 3 and then I will sum these board plots up because when I take a logarithm of h 1 is essentially a logarithm of h 1 if I take a logarithm of h it is same as log of h 1 plus log of h 2 plus log of h 3. So, it is a valid mathematical exercise to construct individual log plots for h 1 h 2 h 3 and then sum them up and that is my total board plot for the entire system. So, we will construct the board plot for h 1. So, we know that h 1 omega equals 2 plus j omega 2 plus j omega. So, limit as omega goes to 0 and essentially developing an asymptotic response for h 1 j omega is 2. So, my decibels is 20 log 10 of 2 is about 6.02 and then my limit as omega goes to infinite for h 1 is basically omega. So, my decibels is 20 log 10 of omega and my break point. So, before I talk about break point this is going to give me my low frequency asymptote. This will give me my high frequency asymptote and the break point where these two asymptotes will cut each other will be such that omega equals p that is omega equals 2 or log omega equals 0.3 o 1. So, now I plot my low frequency asymptote my board plot for this function h 1. So, my low frequency asymptote I will do it in black the y intercept is 6.02 this is my log omega on the vertical scale I have decibels and my high frequency asymptote which has a slope of 20 decibels per decade is the red dotted line and this break point corresponds to log of 2 and my overall curve for h 1 will be this green line. So, this is the plot for h 1 now I will draw a plot for h 2. So, h 2 we have seen in terms of j omega is this whole term 1 over 3 plus 10 j plus j omega. So, it is 1 over 3 plus 10 j plus j omega. So, limit as omega goes to 0 for h 2 j omega is 1 over 3 plus 10 j. So, in decibels this term goes to 20 log magnitude of 1 over 3 plus 10 j is basically 10.44 basically I can what do is I can find 3 square plus 10 square that gives me 109 take its square root find the log of it minus of that and I get. So, this is this number is minus 20.37 and then similarly limit as omega goes to infinite and this is a high frequency will give us a high frequency response is h 2 j omega that will go to 1 over j omega. So, my decibels will be basically 20 log 1 over j omega is minus 20 log omega. So, my boat plot my horizontal axis is log omega vertical axis is in decibels my low frequency line is going to be again a straight line cutting the y axis that minus 20.37 and my high frequency asymptote is again straight, but slanted line of slope minus 20 dB per decade slope is minus 20 dB per decade and my overall response is going to be this green line. And the break point is going to be such that this is log of 10.44 where 10.44 is the magnitude as we had seen of 3 plus 10 j I can do the same exercise for h 3 s. So, my h 3 is essentially 1 over 3 minus 10 j plus j omega 1 over 3 minus 10 j plus j omega and my boat plot will be identical. So, this is my boat plot for h 2 j omega and when we do all the math we will find that the boat plot for magnitude for h 3 j omega will be same as this particular plot because the magnitude of the denominator does not change with the change in sign here. So, I have constructed the plot for h 2 and h 3 and I have constructed the plot for h 1 j omega. So, my next step is to construct the final plot. So, h of j omega which is essentially a sum of plots for h 1 h 2 h 3. So, if I have to construct a boat magnitude plot for h it is essentially a sum of boat magnitude plots for h 1 h 2 and h 3 and that curve looks like something like this. So, initially I have a horizontal line and I have a upward sloping segment and then I have a downward sloping segment. The break points my horizontal axis is log omega my vertical axis is on the decibel scale the break points are at log 2. The second break point as you have expected is going to be log 10.44 and the y intercepts are going to be minus 34.72 and the other peak this peak point is going to be minus 20.37 decibels. So, this is my asymptotic response which corresponds to transfer function h which is a product of three individual transfer functions h 1 h 2 and h 3. Now, again this is an asymptotic response for the magnitude the actual response is not this the same as this dark red line. In this case if we actually plot the actual magnitude response that is shown here in the purple line. So, asymptotically on this side the line converges to the red line this is the asymptotic response. The actual response is something like this and at infinity the red line and the purple line converge and meet. Another observation from this is that because the purple line which represents the actual system response is above the red line it happens to be an under damped system and we will explore this more in our later lectures. So, in this class what we have covered till so far is a little bit about what is an over damped under damped and critical system. Then we went into the area of phase and magnitude plots. You also understood what is the meaning of terms poles and zeros poles being the roots of the numerator factors I am sorry zeros being the roots of the numerator factors and poles being the 0 roots of denominators. And then we constructed phase and magnitude plots and then we moved to both plots and what we have covered today is how to construct a Bose board magnitude plot for a transfer function which is a multiple of several small simple transfer functions. So, what this is what we have covered today in the next class we will learn how to construct board plots for the phase of the same transfer function. Thank you.