 So, starting today, and in next couple of lectures, what we will be touching upon is how does sound propagate in rooms, in real rooms. So far, we have tried to understand propagation of sound in narrow long tubes, but what happens when sound is in an actual room and closed space, what happens, what kind of modes it has, so that is something we will try to understand, and that will also lead us into the next segment of this course, architectural acoustics. So, what we will be doing starting from today is a little bit exposure to architectural acoustics, how do you design spaces where it sounds good, we will develop some equations and also criteria in terms of what means a good sound, what does not mean a good sound in an architectural space. So, what we will do is today, explore effect of walls on sound, and we will start with a 1D room. A 1D room is a very long room, the length is significantly larger compared to the other two dimensions, and also the wave propagating in it is a plane wave. So, we will start with 1D room, then we will go to 2D room, and then if we have time we will try to capture 3D rooms today, otherwise we will do it later, and all we will be doing in next couple of lectures is effect of walls, we will not be considering effect of medium, because as sound travels through the medium, it also in reality gets dampened, because air can also has some dampening, but we will be ignoring that effect. So, this is effect of walls, and the first thing is part A is 1D room. So, we can consider a long tube, one dimensional tube which we call a 1D room, and at x equals 0, it sees perfect reflection, the length of the room is L, and what I am doing is I am exciting it through a piston at x equals minus L. So, we know from our earlier coursework that pressure and velocity can be expressed as p plus p minus p plus over z naught, p minus over z naught is an egg design here times an exponential form which captures the forward travelling wave e minus j omega x over c, and then the reflected wave would be e j omega x over c, and notice that in this I am ignoring the time portion of it. So, all this thing in reality gets multiplied by e j omega t. So, we know that at x equals 0, the boundary condition is that there is no velocity u is 0. So, at x equals 0, u equals 0, and what that gives is e plus equals e minus. So, plugging this back into this relation, I get e and u can be expressed as 1 1 1 over z naught minus 1 over z naught, I have a p plus outside the matrix, and e minus j omega x over c e j omega x over c, and this if I expand and add up the terms what I get is 2 p plus times o sin omega x over c, and sin omega x over c minus this is j here p plus where in mind it can be a complex entity, we have not assumed that p plus is a real or a complex entity. So, if we have to figure out p plus, we have imposed the first boundary condition that at x equals 0, u is 0, and at x equals minus l we know what is the velocity. So, once we impose that boundary condition on this particular equation, I can get the value of p plus. So, I can figure out what is the value of p plus, because this is the velocity which I know at x equals minus l, I am inducing that velocity. Now, consider case that after I have excited this piston for a while a few seconds, then I bring the piston to a dead stop all of a sudden, what happens in that case, in conceptually what will happen is that the sound which is there in that one day room, it will travel back and forth and it will get reflected from the walls on both hands, and what is the behavior of that phenomena, that is what we are going to explore now. So, once I have brought this piston to a dead stop, so the velocity at x equals minus l becomes 0. So, then I can say that sin omega l over c equals to 0, why do I say that, because here velocity is 2 p plus times j sin omega l over c. So, at x equals minus l, this entire term has to become 0, which means sin omega l over c has to be 0, which means omega l over c has to be n times pi, which means l equals n pi times r l equals n pi times, I know c over omega I can transform it to lambda over 2 pi, and if I simplify it I get n lambda over 2. So, one relation is l equals n lambda over 2, and equivalently I can also show that frequency equals n c over 2 l. So, f equals c n over 2 l and l equals n lambda over 2. So, let us consider n equals 1, then my frequency equals c over 2 l and l equals lambda over 2. So, if my tube or this 1 d room is a long room like this, the velocity profile what that means is going to be something like this, velocity is going to be 0, and the waveform is going to be such that half of a wave will cover the length of the room for n equals to f equals 2 c over 2 l, l equals lambda and in this case if I draw the room mode it will be something like this. So, what you are drawing here is the room mode associated with the shape, and what you are computing here are natural frequencies or room normal frequencies of the room. So, f is normal frequency of that one dimensional room, and associated with that is a mode shape which is called as room mode. For n equals 3, f equals 3 c over 2 l and l equals 3 lambda over 2. So, initially the frequency excitation is dependent on us, and suddenly as we stop it, it suddenly assumes these frequencies. We will talk about that, I am going to come to that point, let me just make this picture that is exactly the point what I am trying to get you. Because your excitation frequency could be anything, these frequencies have discrete values, it could be let us say 100 hertz, 200 hertz, 300 hertz, 4 hertz, but I can excite it at 40 hertz or 2 hertz or 101 hertz. So, we will talk about that. So, all what we have shown here is that a long one dimensional room has certain specific natural frequencies associated with it. This is what these pictures and this analysis is showing, and these natural frequencies are dependent on the length of the room, c is constant 345 meters per second. So, they are only dependent on the geometry of the room which is the length, and associated with the length is a particular natural frequency, and they are infinite in number, because n theoretically you can run up to infinity. So, you have a infinite number of, but discrete frequencies, and each frequency has a specific mode shape. Another thing is, so this is the mode, and this is actually your velocity envelope. So, velocity at any given point at this particular value of x, it will not exceed the envelope. But as Mayank was saying that my excitation frequency could be anything, let us say that we choose a particular value of l such that f 1 is 100 hertz, then f 2 will be 200, f 3 could be 300, and so it will be 100, 200, 300, 400 and so on and so forth. But I could excite the room, my piston could be initially getting excited at 45 hertz, which is not an integral multiple of this. So, the question is what happens when the piston is exciting at specific frequencies, which are not same as this. So, we have two situations, if the piston is generating frequency or was in this case, because now it has come to a dead stop. If it was generating frequencies, which are coincident with the room modes, then you will have these types of standing modes in the room, and in theory they will go on till infinite time, because there is nothing which is dampening the vibrations in the room. If the frequencies of piston, of the frequency of the piston was not equal to the exciting, no the natural frequency of the room, then you can develop a Fourier solution, which where the, these will be your natural frequencies, and in terms of these natural frequencies, you can develop a Fourier solution, which will tell you the overall response of the room. Does that answer your question? So, in this analysis we did not include any damping, we did not include damping in the air, and we also assumed that there was perfect reflection at both the walls. So, our next step will be that 1 d cube or room with damping in walls, again so even here we are not considering damping in the air, it still assuming to be 0. So, I will draw another long room, it has an impedance at the end of the room Zl, the excitation is coming through a piston, and here also the impedance is Zl, x is 0 at the fixed end, x equals minus l, that is my axis system. So, total length of this tube is l, and what I am doing is, I am putting a pressure microphone here, at the distance l over 2, I am putting a pressure microphone, and I am measuring pressure, what happens. So, my pressure and velocity relationships are something like 1 1 1 over Z naught minus 1 over Z naught e plus e minus j omega x over c, e minus e j omega x over c, till so far I have not imposed any boundary condition. Now, we know that at x equals 0, my impedance total impedance is, what is the impedance at x equals 0 Zl, so p over u x equals 0 equals Zl. So, I use this figure out the relationship between p plus and p minus, so essentially what that tells me is that Zl equals, I put x equals 0 in this relation, so I get p plus plus p minus over p plus minus p minus times Z naught. So, Zl equals p plus plus p minus in this entire relation, and if I introduce that term which we had earlier defined, reflection coefficient gamma is essentially p minus over p plus, so let us number this equation 1 and this is 2. So, if I introduce gamma in this expression for Zl, essentially what I get is 1 plus gamma Z naught equals 1 minus gamma Zl and p plus gets cancelled out. So, therefore, and then I manipulate this and I get gamma equals Zl minus Z naught over Zl plus Z naught, question gamma minus 1 into Zl, if I take p plus out this will be 1 and then minus. So, this is my expression, this is essentially the same expression which we have shown earlier for perfect reflection p plus equals p minus, what that means is and that can happen p plus equals p minus which means gamma is 1 and that can happen only when Zl equals infinity. So, for perfectly rigid surface Zl is infinity for imperfect reflection gamma is less than 1 and for walls which are very stiff gamma is still less than 1, but almost equal to 1 and that is the case of a lot of structures where you have brick walls or concrete walls or stone walls that is the type of structures you have, where gamma is almost equal to 1, but it is still a little less than 1 and in that case Zl is not infinity, but Zl is extremely large compared to Z naught, once you introduce this relationship in this one you get gamma is almost equal to 1. Physically what Zl equals infinity means is Zl is essentially at x equals 0, Zl is the ratio of pressure and velocity. So, Zl being infinity means that pressure is non-zero, but velocity is 0 because the wall is perfectly rigid, so nothing is moving, so velocity is 0, so p over u becomes infinity that is what it means physically. So, now we will consider a scenario where Zl is very large compared to Z naught and this is a thought experiment think about it that instead of this piston which is getting excited continually, what is happening is that the piston at t equals some time t equals some not the initial value t naught. It generates an impulse pressure, what is an impulse pressure that the input signal is something like this t and then this is pressure. So, the input signal is 0 to begin with and at some critical time at some specific time it shoots up to a particular value it is p naught and then it drops again and it remains 0 forever. So, in this tube what this guy this piston is doing is that it generates an impulse pressure at a specific moment and then it falls silent it does not vibrate anymore and what we are interested in figuring out is that the wave generated because of that impulse pressure how does it change with time as sensed by this particular microphone this is the only instrument we have in the setup. So, what we are interested in is that once this impulse pressure has been generated at x equals l what is the pressure sensed by the microphone which is located at x equals minus l over 2 with time. So, this is how the pressure will look like. So, here what I am plotting time and here I am going to plot pressure. So, just for clarity purpose is my 0 is little above x axis this is p equals 0 this is l over 2 c l over 2 c is the time it takes for the pressure to start from here and reach this point right. The distance is l over 2 the velocity of wave propagation is c. So, the total time it takes to reach the microphone is l over 2 c right what will be the pressure sense by the microphone at time l over 2 c. So, this is p naught I am sorry. So, before that time the pressure microphone will sense 0 pressure at that time it will sense p naught and after that it will fall then the same pressure wave will traverse from here it will come to x equals 0 location. So, it will travel a distance of l over 2 it will get reflected and come back to the microphone right. So, my next time interval when the pressure will be sense by the microphone will be after a time interval l over c. So, this distance is l over c what will be the pressure sense by microphone at this time gamma times p naught because the reflection coefficient is. So, and gamma is a little less than 1. So, this is p naught times gamma then the microwave will continue to travel in negative direction it will get reflected by the piston which also has reflective coefficient of gamma because z l is same. So, at after another l over c seconds the microphone will again sense pressure and this time the pressure will be gamma square and then p naught gamma cube and so on and so forth. So, my wave will continually decay with time this is for an impulse function. So, we will do we will develop some mathematics around. So, we can say that p microphone is p t equals n delta t p is a function of t where t equals an integral number n times delta t will define delta t equals p naught gamma n right where delta t equals l over c and n equals 0 1 2 3 and so on and so forth. Also this equation is valid at equals 0 if you start measuring time from this point right. If I measure time from this then in this zone this relation is not valid because there is a l over 2 factor, but after t this point and so on and so forth this relation holds good. So, my t equals 0 is when p equals p naught at microphone at mic. So, my original position is when pressure equals p naught at microphone that is my t equals 0 position. So, I get p naught gamma t over delta t because n equals t over delta t I have which is p naught gamma t c over l because delta t equals l over c. In this relation what I have done is all of a sudden I have converted a discrete time function right to a continuous time function this is a big assumption I have made. Here it was discrete here for an impulse function this may not be this may be more accurate this may not be that much accurate, but we will talk about this later. So, now we say that gamma is approximately equal to 1 and this is based on reality most of the walls are fairly rigid and their gamma is fairly close to 1 I can say that is same as 1 minus 1 minus gamma this is an exact relationship, but then I can make a big approximation and we will explore the goodness of this approximation later. I can say 1 minus 1 minus gamma plus 1 minus gamma square over 2 factorial minus 1 minus gamma cube over 3 factorial plus 1 and so forth. If gamma is very small is almost close to 0 then the contribution of all these terms will be very small these extra terms and when you look at this entire series you can say that this is essentially same as exponent and we will see the goodness of this approximation later. So, my P equals P naught exponent minus 1 minus gamma C T over L again all this approximation is good to the extent C L is very large compared to C naught. So, what we have done is we have made a big jump we have approximated gamma as exponent of minus 1 minus gamma and we will see how good this approximation is. So, what I will do is I will just select different values of gamma and then in another table I will take minus 1 minus gamma and then I will compute the error. Error is basically this term minus this term divided by this term times 100. So, I am computing error in percent. So, when gamma is 1 this guy is 1 error is 0 gamma is 0.98 I am not going to list these values here error comes to 0.005 percent gamma is 0.9496 my error is 0.082 percent this is in percent is less than one tenth of a percent gamma is 0.90 my error is 0.54 percent gamma is 0.86 my error is 1.09 percent gamma is 0.8 my error is 2.3 percent. So, this approximation is not that bad that is what I am trying to show to the extent I am comparing the closeness of gamma with respect to E times minus 1 minus gamma, but in the actual equation it is not just E minus 1 minus gamma, but there is also time C T over L. So, this C T over L may amplify the error also. So, I will very quickly did another table. So, what I am going to do here is I will pick different values of gamma I will compute gamma to the power of C T over L I will compare this value with the exponential form and then I will compute the error in percent and let us see three values in this my C is 345 meters per second standard value actually for this one I assume 350 just to make my calculations simpler I assumed L to be 14 meters 14 meter long and I took T to be 1 second before I construct this table I wanted to you to understand what does T equals 1 second mean. So, my delta T what is going to be my delta T delta T is L over C right. So, what is it going to be 0.04 seconds 14 over 350 0.04 seconds. So, T equals 1 second means 25 reflections right T it will that is a very big number of reflections. So, let us see for 25 reflections. So, T equals 1 means 25 reflections it is not 1 or 2 reflections 25 reflections it is a lot of reflections. So, I will just list values of gamma 1 0.99 0.98 0.96 0.92 and I am just going to site the errors actually this is 1 0.78 0.6 0.36 0.124 and my error is going to be obviously 0 in this case in this case it is 0.125 and this is again in percent here it is 0.51 here it is 2.07 and finally it is 8.82 percent. So, for 25 reflections this particular approximation you know this one is not that bad even for a high value of or for a low value of gamma which is like 92 percent the error is still 8.8 percent which is not that bad. So, this approximation fairly decent approximation now of course, if gamma starts becoming very small and it is no longer close to 1 and or this C T over L parameters starts becoming very large then the approximation will not be valid. But in that case see this after 25 reflections for a little high value of damping this is only 1.124 what that means is that the 25th reflection will be only 12 percent as strong as the original reflection. So, once you go down this scale if you go down this scale further these numbers become so small they actually start going into 1 e minus 6 1 e minus 12 1 e minus something like so on and so forth that the comparison of those very small numbers with another set of extremely small numbers it becomes meaningless because essentially what the story tells you is everything has died down to this approximation is fairly good the plot of this expression. So, what I am going to plot here is time and here I am plotting p. So, it looks like something like this. So, what this is is p equals p not exponent 1 minus gamma C T over L and this is called envelope of decay envelope of decay. What this shows is that the decay will follow this curve but the actual pressures and this is T equals 0. The actual pressures will be like this for each reflection you will have a discrete pressure fluctuations in the mic, but the pressures will be bounded by this particular envelope of decay these are actual measurements in mic. One last thing this particular envelope of decay will shift if you change the location of the mic because we had just assumed that it was in center. So, wave going on the right side and wave on the left side they type equal amount of height. So, it will change its shape and so will these discrete values also change if I shift the position of mic. So, position of the instrument is very important at least in these type of measurements where you are measuring the impact of reflection. So, this gives you some flavor of 1 D e rooms. One thing is that this an exponential decay in the room if there is a wave standing wave in the thing also each long room is associated with a bunch of standing modes and in terms of those standing modes you can decompose any set of externally imposed frequencies through Fourier expansion. So, now what we will do is we will develop something similar for two dimensional we are still not talking about real room three dimensional 2 D rooms. We will talk about natural or normal modes in 2 D rooms. So, let us say I have a room length is l x and its other dimension is l y that is my x axis that is my y axis this is my origin and my incident wave is travelling like this at an angle theta with respect to the x axis let us say the frequency is f. So, as that wave travels and hits a wall it will get reflected obeying Snell's law angle of incidence equals angle of reflection same law which we use in optics. What will be the boundary condition along this wall? So, velocity normal to wall equals 0. So, along this wall when it hits the velocity which will be normal that is in this direction will die down the other thing. So, that is one boundary condition and the other boundary condition which is essentially a consequence of this particular first statement is that velocity at corners equals 0 because both the normal components have to be 0. So, we will draw another picture with this understanding. So, I draw my x direction this is my incident wave and I draw several lines. So, let us say this distance is lambda over 2 this is also a line parallel to it lambda over 2. So, just as we saw in case of 1 d room you have nodes at every interval of lambda over 2 you have a line of velocity nodes which are spaced by distance lambda over 2 distance. Now, this angle is theta. So, a b a b equals lambda over 2 cosine theta and similarly I can draw a vertical line c d c d equals lambda over 2 sin theta. So, using the analogy of 1 d room I am just extending that analogy this is not an exact proof extending that analogy condition for natural modes in room is such that l x equals a number n x times lambda over 2 cosine theta and l y equals another number n y and will physically interpret what n x and n y mean 2 sin theta. What this gives me is f cosine theta equals n x times c over 2 l x this gives me is f sin theta equals n y times c over 2 l y. So, if what this shows is that if l x which is a physical dimension of the room and l y they satisfy these conditions these conditions then f is the natural frequency of that room. So, I can also combine these two such that f equals c over 2 l x times n x then tan theta is n y over l y times n x over l x is that clear. So, in a two dimensional room a normal mode depends on three parameters one is the physical dimensions of the room actually there are two parameters length and width. So, those are the two parameters and then it also depends on the orientation a mode is associated with the orientation which is theta unlike in case of a 1 D room. The other thing is that n x and n y they are integers. So, a room can have infinite modes associated with every angle, but that does not mean that all frequencies are normal modes of a room. I mean it is an infinite set, but it is not a complete set. So, we will do some quick examples physical interpretation of this. Let us consider a room and it is a longish room. So, this is 690 meters and this is 345 meters. So, l x equals 345 l y equals 690. So, we will use this relation c over 2 l x and c over 2 l y. So, c over 2 l x equals 1. No, I am sorry c over l x is 1 and c over l y equals 2. So, f n x n y I should have put a subscript here associated with n x and n y. So, f n x n y equals n x over 2 where 2 n y over 2. No, I think c over l y is half and similarly tan theta equals n y over l y over n x over l x and that is n y over l y is 690 times 345 over l x is n y over n. So, I now start putting values. So, I pick up a specific value of n x is the book a specific value of n y and I construct a mode. So, f 1 0 is I put n x in this relation. So, basically this is half hertz. I put n x is 1 and y is 0 it gives me half hertz and my theta is 0. So, my room mode looks like this. It has 1 lambda over 2 in x direction and 0 lambda over 2 in y direction. What do you think about f 0 1? So, this is 0 n x is 0 n y is 1. So, 1 1 fourth of a hertz and theta equals 90 degrees and the mode looks like this another one f 2 0. So, this is 1 hertz. So, here the mode looks like it has 2 lambda over 2 in the length direction 0 lambda over 2 in the y direction. So, I will I will show you some other modes also. So, what you are seeing here in this picture is can you read? Maybe it is too small font. So, this is the normal mode this is how the picture of a normal mode looks like for a room which is 10 meters in the length direction x direction 15 meters in the height or the width direction and the value of theta is 57.265 degrees and n x is 3 and y is 7 and f 3 7 is about 96 hertz.