 last class we were exploring the notion of directivity of different sound sources or different combinations of simple sound sources. And till so far what we have covered is how two sources interact with each other and the consequential directivity patterns which are which emerge because of the interaction of these two sound sources. As a follow up to that understanding we want to explore how more than two sound sources interact with each other and what kind of directivity patterns emerge when more than two sound sources interact with each other. In context of that we had developed the relation for directivity associated with two sound sources and one relation which we had developed was this. So, if there are two sound sources S 1 and S 2 and they are separated by distance D and the volume velocity of first sound source is V 1, volume velocity of second sound source is V 2 and the phase difference of two sound sources is suppose 0. Then for that kind of a situation we had found that the pressure at a point which is far away the pressure at a point which is far away and which is r distance away from the midpoint of these two sound sources. The expression for that pressure complex pressure can be written as 2 V V over 4 pi r j omega rho naught E j omega t minus r over c times cosine of pi d over lambda times sin theta and minus phi by 2. But, where this is phi is the phase difference between the volume velocity of these two sound sources, but because we have assumed that this phi is 0. So, I am going to drop this out and close the bracket. So, this is one relation which we had developed and this relation has been developed several lectures back in time. In last lecture we developed an equivalent relation and the relation for that is some term which is dependent on radius and omega and time times sin of 2 alpha divided by sin of alpha, where alpha equals pi d over lambda times sin of theta. So, this is another expression for the same pressure. Both these expressions give us same exactly the same answer, but they are in different forms. But, the reason we have developed the second form is because we will see later that the second form becomes helpful. The approach to you are developing the second form becomes helpful when we have a very large number of sources and especially if the number of sources becomes infinite. In this, in both these terms there is a term which depends on the angle and this term is highlighted in red and there is another term which depends on radius, volume velocity of source, omega density of air and time and I am going to highlight that term in per circle that term in purple. So, the term which has been circled by purple color that depends on radius, omega, time, density and volume velocity. While the term which is in red it depends on theta, lambda and the distance between two sound sources. So, our focus as we talk more about this will be on the term which has been circled in red because we are trying to understand the polar pattern or the directivity pattern of a combination of several sound sources. Now, these two expressions are for two sound sources which are simple in nature. So, these are simple sound sources and also the assumption has been that the radius or the point of observation is very far compared to the distance between these two sound sources S 1 and S 2. Now, we will go to the next step and we will develop a similar relation when we have four sound sources. So, suppose I have four sound sources and these are S 1, S 2, S 3 and S 4 and the distance between these sources is D, D and D. So, using a similar approach as I did earlier I am going to construct a midline and then there is a midpoint and from this midpoint and then my point of observation which is far away from this sound source is this and I am interested in finding again P R theta and T. So, as I did earlier in case of two sound sources I am going to construct lines which connect these two sources from S 1 to P, S 2 to P, S 3 to P and S 4 to P and I call these lines or vectors as R 1, R 2, R 3 and R 4. Just as I did earlier I am going to drop a normal from each of these sources and what I can say is and suppose my midline the midpoint when I connect to the point of observation is and that is distance R away then I can develop relations between R 1, R 2, R 3 and R 4 with respect to R. So, these relations are so if this is the angle theta then these relations are R 1 equals R minus 3 over 2 d sin theta, R 2 equals R minus d over 2 sin of theta. So, I am sorry this theta is this angle theta is this angle R 3 equals R plus d over 2 sin theta and R 4 equals R plus 3 d over 2 times sin of theta. We again assume that the volume velocity of all these sound sources is same. So, V V 1 magnitude equals V V 2 equals V V 3 equals V V 4 and we also further assume that the phase angle between each of these volume velocity sources is exactly 0. So, the phase is 0. So, with these assumptions I can develop an expression for pressure at point P and that depends on radius theta and T and that equals V V over 4 pi r j omega rho naught and then because I have four sources I have to add contribution of each source. So, the contribution of first source is E j omega T minus R 1 over C contribution of second source is E j omega T minus R 2 over C plus E j omega T minus R 3 over C plus E j omega T minus R 4 over C. So, there are four sources and these are the contributions of each of these sources individually. Now, I can plug in these relations for R 1 and R 2, R 3 and R 4 in this equation and because of that what I get is pressure equals V V over 4 pi r j omega rho naught and I can take E j omega T out. So, what I get is and in the parenthesis what I get is exponent j times omega times 1.5 d sin theta over C plus E j omega d sin theta over 2 C plus E minus j omega over 2 C d sin theta plus E minus j omega times 1.5 d sin theta over C. Now, at this stage I put alpha equals pi d sin theta over lambda and I also put omega equals 2 pi f. So, once I do that what I get is P which is the function of R theta and T equals V V over 4 pi r j omega rho naught E j omega T and in the parenthesis I get E j 3 alpha plus E j alpha plus E j minus alpha plus E minus j times 3 alpha where alpha equals pi d over sin theta pi d sin theta divided by lambda. I call this term. So, once again this term it does not depend on theta. So, I call it A and essentially what I get is this expression E j 3 alpha plus E j 3 alpha minus plus E j alpha plus E minus j alpha and then I can subtract and divide this entire expression by a common term and this common term is same as that which I used in the expression when I was deriving the expression for the case when we had two sources. So, I can subtract multiply and divided by E j alpha minus E minus j alpha divided by E j alpha minus E minus j alpha. Now, if I do the mathematics what I get is outside of parenthesis I have A and then I multiply E j 3 alpha with E j alpha to get E j 4 alpha I multiply E j minus 3 alpha times minus E minus j alpha. So, I get E minus j 4 alpha. So, that is one set of terms then I multiply E j 3 alpha with E minus j alpha to get minus E j 2 alpha and then I multiply E j E times E exponent of minus j 3 alpha with exponent of j alpha to get E minus j 2 alpha and then I multiply this term and the multiplicative term to get E j 2 alpha minus E minus j 2 alpha all other terms we will find we find the cancel out. So, this term and in the numerator I still get E j alpha minus minus j alpha. So, in the numerator this term and this term they cancel out. So, my expression for P which depends on R theta and T that comes down to A times E j 4 alpha minus E minus j 4 alpha divided by E j alpha minus E minus j alpha and this becomes A and in the numerator I get 2 j sin 4 alpha in the denominator I get 2 j sin alpha. So, 2 j and 2 j they cancel out and essentially I am left with A times sin 4 alpha divided by sin alpha. So, this is for the case when I have 4 simple sources which are evenly spaced they lie in a particular straight line and these 4 sources have same volume velocity and the phase difference between the volume velocities of each of these sources is 0. So, to compare this with earlier case when we had 2 sources let us see the theta dependent term was essentially a ratio of sin 2 alpha divided by sin alpha when we had 2 sources the theta dependent term when we have 4 sources is the ratio of sin 4 alpha divided by sin alpha. So, expectedly we should get when we have 6 sources we should get something like sin 6 alpha divided by sin alpha. So, let us explore that. So, now we will do case for 6 simple sources simple sources. So, we have source 1, source 2, source 3, source 4, source 5 and source 6 this is my midpoint the distance between each of these sources adjacent sources is again d I have a far flung point such that the distance between the midpoint and this far off point is r such that r is very large compared to d and then now with this understanding I develop vectors from each of these source points and let us say these vectors are r 1, r 2, r 3, r 4 excuse me this is not r 4. So, that is r 3 the midpoint is r distance away. So, midpoint is r and then I have 4th source. So, this is r that is r 4, r 5, r 6 and these are 6 sources s 1, s 2, s 3, s 4, s 5, s 6. So, if we use the same logic and if this angle is theta azimuthal angle is theta then P of r theta t equals using the same logic we can say we can write that this is equal to some variable which does not depend on theta but it only depends on omega r and rho. So, that is the term A times a theta dependent term which is essentially a sum of j times 5 alpha plus exponent j times 3 alpha plus exponent j times alpha plus exponent minus j times alpha plus exponent minus j times 3 alpha plus exponent minus j times 5 alpha. I can rewrite this expression as P which depends on r theta and t equals a times. So, now I am going to organize these terms according to their powers. So, e to the power of j times 5 alpha plus e to the power of minus j times 5 alpha. So, that is 1 plus e to the power of j times 3 alpha minus e to the power of j times 3 alpha. There should be a plus sign here plus e to the power of j alpha plus e to the power of minus j alpha that is again and then I am going to multiply and divide it by the same term as I did in earlier expressions and this term is e to the power of j alpha minus e to the power of minus j alpha. So, I am going to multiply it and divide it by the same term and now when I do mathematical operations on this long equation what I get is P r theta t equals a times e to the power of 6 j alpha minus e to the power of minus 6 j alpha and all other terms cancel out and then in the denominator I have e to the power of j alpha minus e to the power of minus j alpha and this can be written as a times 2 j sin 6 alpha and in the denominator it is 2 j times sin alpha which equals a to the power of sin 6 alpha divided by sin alpha. So, this is the equation for pressure when I have 6 sources in a straight line. So, let us tabulate some of our results. So, when I had 2 sources then I had 4 sources and then I had 6 sources we have done 6 cases. In case of 2 sources the overall pressure was expression for pressure overall expression for pressure was a times sin 2 alpha divided by sin alpha. When I had 4 sources the expression for pressure was a times sin 4 alpha divided by sin alpha and when I had 6 sources it was a times sin 6 alpha divided by sin alpha. So, by process of inductive reasoning we can actually mathematically prove that if there are n sources in a in a in an actual case then if there are n sources then for n sources I can actually extend this and I can have a very generic relation my relation for pressure is going to be p r theta t equals a sin n alpha divided by sin alpha. So, this is the relation when I have n sources all in a straight line all separated by equal distance which is d and all sources are working in such a way that the phase difference between all of them is exactly 0 and their volume velocities are identical. In such a case the overall pressure at a far at point which is far away from the this array of sources is a times sin of n alpha divided by sin of alpha. Now, let us look at the case when we have infinite number of such sources. So, we can consider an array and the overall width of this array is b and let us say it has n sources and if these are n sources then the distance and if all these sources are equally spaced then the distance and if the distance between two adjacent sources is d. So, in this case if n is very large then the relationship between b and d is nothing but b equals n minus 1 times d. Now, if n is very large then b equals n times d for large value of n if n is extremely large then b equals n times d. So, in such a case we can write that pressure equals a times sin n alpha divided by sin alpha. This is the relation which we had developed earlier and now I am going to write the expression for alpha. So, that is a times sin of n pi d over lambda times sin theta divided by sin pi d over lambda times sin theta. Now, we know that n d approximates b especially if n is very large. So, then my expression for p r theta becomes a times sin and n times d n times d equals b. So, I get b pi over lambda times sin theta divided by sin pi d over lambda times sin of theta. This I can further rewrite this as excuse me there is no sin of pi d over lambda times sin of theta. Now, if n is extremely large then this term if n is extremely large and if b is finite then d will be extremely small and when d is extremely small then pi times d divided by lambda will also be very small and that very small term is being multiplied by sin of theta which never exceeds 1. So, in that case again when n is very large then this term is very small and if that is the case then sin of this extremely small term is essentially same as the small term itself. So, I can again further simplify this relation as sin b pi over lambda times sin theta divided by pi d over lambda times sin theta moving on. So, I will just rewrite this equation p of r theta t is equal to a sin a times sin b pi over lambda times sin theta in the numerator and in the denominator I have pi d over lambda sin of times sin of theta. Now, I can multiply this and divide this by n and what I get is a n times sin b pi over lambda times sin theta divided by. So, n this n comes in the numerator and in the denominator I am left with pi d n divided by lambda times sin of theta. So, this is equal to a times n times sin b pi over lambda times sin of theta divided by pi and we know that d n equals or approximately equals b when n is large. So, this is essentially pi b over lambda times sin of theta. So, this is the expression for the pressure field at a point which is very far away from an array of sound sources which are all lying in a straight line and separated by very small distance and such that the overall length of the array is b then the expression for p is this thing. Now, a times n does not change with respect to theta and that essentially defines that the overall maximum possible pressure in the field will be determined by this term. So, I can rewrite this equation as that p r theta t equals some constant or some radius dependent number p naught times sin b pi over lambda times sin of theta divided by pi b over lambda times sin of theta. So, this is my expression and here I have assumed that r is extremely large compared to b. So, this is my expression. Now, again in this expression this term it depends on theta, this term depends on b angular frequency omega n t and r. So, it depends on the size of the array depends on angular frequency time and radius how far are we away, how far away we are from the array it depends on this. So, if I have to know what is the overall polar pattern or the direction, directivity diagram of the array then I have to essentially rely on this term which has been circled in real color and if I am able to plot it that will give me a very good idea of the directivity of the overall array. So, what we will do is that now that we have developed this expression for n sources all lined up in a straight line, we will look at the directivity patterns of such sources and compare them with some of the sources which we had studied earlier. So, that is what we are going to look at. So, this first slide is for is relates to the directivity pattern for two sources, both sources acting in phase, both sources having same volume velocity and these two sources are separated by a distance d which equals the wavelength lambda. So, what we see here is that we have a principal lobe and we have a principal lobe and we have maximization of pressure at theta equals 0 degrees and at theta equals 90 degrees and the width of the lobe in 0 degrees is very sharp, the width of the lobe at 90 degree direction is fairly wide. Now, let us look at the directivity pattern of a source where d equals where we where of a source or a set of sources where we have a very large number of transducers emitting sound excuse me the overall length of this array is b and that b equals lambda just as here b equals lambda in this case the width of the overall array is b and that equals lambda and the number of such sources is extremely large theoretically infinite and we have developed the directivity pattern for this using the relation which we just developed. So, this is how it looks like. So, what you have here is there are two curves the curve in blue represents the case when we have just two sources these two sources are off by a distance lambda the curve in red represents a very large number of sources theoretically infinite and the overall width of the array is lambda and we observe some several interesting points. One is that the red curve has only two lobes and both these two lobes are in the principal direction while the curve in blue has four lobes two in principal direction two in 90 degree direction or off axis direction. So, because of this distributed array when and because of the presence of a very large number of sources what we find is that the intensity of sound which is being propagated in 90 degree direction and also in minus 2 and also in 270 degree direction it has gone down to virtually zero as deflected by the red curve and this is what is known as side lobe suppression. So, now let us look at another curve and this is the curve this is the directivity pattern for two sources emitting sound both sources in phase same volume velocity for each source and the distance between these two sources is 1.5 times lambda or 3 lambda divided by 2 and what we find is that we have six overall lobes there is a lot of sound getting emitted in 0 degree direction and also in 180 degree direction, but there is also sound getting emitted in 45 degree, 135 degree and so on and so forth. Now, let us compare this pattern with the case when we replace these two sources by distributed source whose overall width is 3 lambda over 2. So, let us look at the comparison of this particular directivity pattern with a pattern which is once again 1.5 times wavelength and let us see how that compares. So, again that is the that is shown in this slide. So, what you see in this slide is basically two curves the first curve is in blue color or light blue color and this corresponds to emission of a directivity pattern associated with two sources separated by a distance 1.5 lambda and both sources are in phase and say same volume velocity. The curve in red the directivity pattern depicted in red corresponds to a distributed source where the number of sources is extremely large theoretically infinite. The overall length of the array is 1.5 times lambda same as the overall length of earlier array, but what you find here is that instead of six lobes as we saw when we had just two sources we have only four lobes here. And the second important thing to note is that the overall intensity of sound which is being emitted in 90 degree direction is extremely small compared to the overall intensity of sound when we had only two sources. So, from when we had only two sources. So, once again we find that once we have if we have the larger if we have a very large number of sources in all standing all arranged in a straight line. And if they are all in a phase then side lobes tend to get suppressed and the on axis sound propagation does not get affected it does not get affected. Now, we will close this lecture, but before we will close we will just do a very quick calculation of what is the extent of this noise suppression. So, to do that let us just observe on this curve that the blue curve and 45 degree direction something has an overall pressure level of something like 0.7 while the red curve in the off axis pressure is something like 0.2. So, while in the 0 degree direction the overall pressure for blue curve as well as for red curve is identical and it stands at 1. So, we will based on this we will do some quick calculation and let us see what is the amount of attenuation we get in off axis direction because of this effect. So, our aim is to find extent of side lobes suppression when d equals 3 lambda over 2 and phase equals 0 and volume velocities are identical. So, in 0 degree direction pressure magnitude is equal to 1 we have normalized it and we call it 1 in off axis direction when we have 2 lobes when we have 2 sources only then pressure equals 0.7 and if I have to calculate it in decibels then it is d B off axis equals 20 log of 10 0.7 divided by reference pressure which is 1 and that works out to be minus 3.1 decibels. Next we look at the off axis SPL when we have this whole array and because of this we have a very large number of sources. So, in off axis direction that is when we have n sources and n is large and that equals pressure we saw we measured it was something like 0 point approximately 0.2. So, in decibels off axis that is equal to 20 log of 10 0.7. So, 2 over 1 and that works out to be minus 14 d B. So, side lobes suppression that works out to be minus 14 plus 3.1 is equal to something like 10.9 decibels. So, just by virtue of having a distributed source which implies that we have n sources very large number sources over a distance of 1.5 lambda my side lobes suppression went down to was as high as 10.9 decibels and what essentially this tells me is that I can use this kind of an approach to develop acoustic transducer arrays which can throw sound in a particular direction and this sound moves like a beam moves like a beam because the amount of sound which gets radiated in off axis direction is very little. So, with this we close today's lecture and we will further continue our discussion and directivity in the next class. Thank you very much.