 Over last several lectures, we have been discussing the notion of directivity. In that context, we have explored the directivity pattern associated with a simple source. We have also explored the directivity pattern associated with a sound source, which is having two individual simple sources separated by some distance. Let us say that distance is d, when with the condition that the value of this d is extremely small compared to the wavelength and also the radius or the distance between the point of observation and the source. So, the ratio of this distance and d is extremely large. And once we did that, then we moved further and then we explored the directivity pattern associated with more than two sources. We started with four sources, six sources and then generalized it for a situation where we have n sources. So, we explored that if we have a very large number of sources and all these sources are lined up along a certain one single straight line. And also if these sources are evenly spaced and the volume velocities of these sources are identical and the phase difference between all these sources is exactly zero. Then we can develop sound projection mechanism, where sound prefers to travel along the zero degree direction and the side lobes for the directivity pattern, their magnitude tends to get very significantly suppressed. So, today we will continue this discussion on directivity further. And today what we are going to explore is the directivity pattern associated with a doublet. Now, a doublet is essentially combination of two simple sound sources. It is a combination of two simple sound sources, which are separated by very small distance and also the phase difference between these two sound sources is pi radians or 180 degrees. So, that is what a doublet is. So, we will see what kind of directivity pattern is associated with doublets. Good example of a doublet could be a simple speaker, which is not mounted in a baffle. So, on the front side this particular speaker is producing a sound wave and let us say at an instant T near the diaphragm, the pressure is positive in magnitude with respect to reference pressure, which is atmospheric pressure. Then on the back side of the diaphragm, because this loudspeaker is unbaffled. So, the back side of the loudspeaker on the back side of the diaphragm, this pressure is negative. So, there is a phase difference between the front side and the negative back side of exactly pi radians. So, that is a good example of a doublet. So, this is the schematic. So, let us say we have two sound sources and let us say this is S 1 and this is S 2 and I have a far off point and the distance between S 1. So, let us say this distance. So, this is the point of observation and the pressure at point of observation is P, which depends on R theta and T and let us say this is the midpoint. So, I construct a line connecting the midpoint with the point of observation and let us say that distance is R, the distance of S 2 with respect to P is R 2 and the distance of S 1 with respect to P is R 1 and let us say this angle is theta. So, once again with this understanding we can write the relationship between R 1 and R 2 in terms of R such that R 1 equals R minus d over 2 cosine of theta and R 2 equals R plus d over 2 cosine of theta. This is true only if R is very large compared to d. So, then this is the relationship between R 1 and R and R 2 and R these are the relationships. Further because we are considering a doublet source. So, we know that volume velocities V V 1 is equal to same as volume velocity V V 2 and that is V V and the phase difference between that of V V 1 excuse me V V 2 minus phase of V V 1 is equal to pi radians. So, these are my basic assumptions and on the basis of these assumptions now I develop the pressure relationship at point P. So, P is a function of R theta and T equals V V over 4 pi R j omega rho naught E j omega t times E j phi over 2 where phi is the phase difference minus R 2 over c omega plus E j minus phi over 2 minus R 1 over c omega and that means this is equal to V V over 4 pi R j omega t minus R over c j omega rho naught. So, I have taken j omega t and I have also substituted in one single step R 2 and R 1 by these respective relations. So, in the parenthesis what I get is j phi over 2 minus omega over 2 c cosine theta plus E j phi over 2 this should be negative minus d omega over 2 c cosine theta and excuse me there has to be a positive sign here. So, the term in brackets this can be simplified as it can be simplified further because e to the power of j times some angle alpha is a cosine component and an imaginary component which has a sin term in it. So, what I can I can write the expression for pressure complex pressure as V V over 4 pi R E j omega t minus R over c times j omega rho naught and then in the parenthesis if I resolve these two terms in terms of their imaginary and real components and do the mathematics what I get is 2 cosine of phi over 2 minus d omega over 2 c cosine theta. Now, we know that omega equals 2 pi f therefore, d omega over 2 c equals 2 d pi f over 2 c is equal to d times pi over lambda. So, I can replace d omega over 2 c with this. So, what I get for my expression for pressure in a doublet is equal to V V over 4 pi R E j omega t minus R over c times j omega rho naught 2 cosine phi over 2 minus pi d over lambda times cosine of theta. Now, we know that in case of a doublet the value of phi is pi radians. So, this term this equals pi over 2 and if this equals pi over 2 then this cosine of anything cosine of pi over 2 minus some angle is essentially sin of that angle. So, I can finalize this relationship as P R theta t equals V V over 4 pi R times j omega rho naught E j omega t minus R over c times sin of pi d over lambda cosine of theta. Now, this relation is valid if R is large compared to d. Now, as we saw in earlier cases this entire expression for complex pressure has a term which depends on theta which is circled in red and then there is another term which is circled in purple which does not depend on theta, but it changes with R. So, my directivity is going to be influenced by this term which depends on theta. Now, before we start analyzing this picture carefully this equation carefully we have to just revisit this picture where we had located S 1, S 2 and P. So, please bear in mind that S 1 and S 2 they are located along the x axis while and with respect to that x axis the angle theta is measured for point t. So, with this understanding let us look at how the directivity of this system changes once we start changing theta. So, we will rewrite P r theta t equals P r theta t equals E j omega t minus R over c divided by 2 pi r j omega rho naught and then sin of pi d over lambda cosine theta and we had said that this term which is circled in red is the component which influences the directivity of the doublet source very strongly. So, for theta equals pi over 2 or 3 pi over 2 this term sin of pi d over lambda times cosine theta is 0. So, cosine theta is 0. So, this becomes 0. So, what that means is that if I have two sources which are very close to each other and these sources are S 1 and S 2 then the overall strength of sound at this location because in this direction it is 0 here and also in excuse me the overall strength of sound at this point P 1 and also at P 2 it will be 0 with theta equals pi over 2 and at pi over 2 or 3 pi over 2 the contributions of these two sources they tend to cancel each other. Now for theta equals pi or 0 degrees sin of pi d over lambda equals cosine of either 0 degrees or it is pi degrees of 0 degree or pi radians. So, in that case this is equal to if it is 0 degree then cosine of 0 is 1 then it is sin pi d over lambda or if it is pi radians then it is minus sin pi d over lambda. So, it is plus or minus of sin pi d over lambda. So, what this means is that if I have two sources S 1 and S 2 then the strength of the signal along the in the 0 degree direction in this. So, this is point P this is one possible location then the strength of this signal at point P will be directly proportional to sin of pi d over lambda. So, pressure will be directly proportional to sin of pi d over lambda. So, this is how doublets behave that along the vertical axis the pressure field is 0 along the horizontal axis it is directly proportional to sin of pi d over lambda. So, it all depends on the ratio of d over lambda. Now let us look at more another more interesting case for doublets itself that what happens when d is extremely small compared to lambda it is extremely small compared to lambda. So, when d is very small compared to lambda how does a doublet behave what kind of a direct duty pattern do we see for such doublets. So, that is what we are going to explore in over next 15-20 minutes. So, to understand this we have to revisit our pressure relation and we will rewrite the equation and the equation is for pressure field is going to be contribution of first source plus contribution of second source. So, contribution of first source is V v over 4 pi j omega e j omega t times e j phi over 2 minus r 2 omega over c divided by r 2. Now earlier please bear in mind that in the denominator we had assumed r because we assumed that for large values of r, r 2 and r are almost equal. So, in that case we had approximated r 1 is approximately equal to r and r 2 is also approximately equal to r, but we are not going to make that assumption here and we are going to explicitly write term r 2. So, this is what I am write what I have written here is the contribution of to pressure from sound source s 2. Similarly, there is a contribution from the second source from the first source s 1 and that is V v over 4 pi e j omega t times e j times minus phi over 2 minus r 1 omega over c and then again in the denominator I do not put r rather I put explicitly r 1. I am not here going to assume that r 1 is same as r and r 2 is same as r. Now we know that if d if d is small distance between these two sources is extremely small compared to r and that is a valid assumption then we know that r 2 equal r plus d cosine theta over 2 and r 1 equals r minus d cosine theta over 2. So, we are going to put these two expressions for r 2 and r 1 in my expression for p. So, what I am going to do is I am going to erase this and I am going to replace r 2 with r plus d cosine theta over 2 and I am going to replace r 1 as r minus d cosine theta over 2. So, I am going to now rework this equation and I am going to organize some terms. So, v v over 4 pi j omega e j omega t and from the denominator I am going to take out r as the common number. So, I am going to place r here. So, what I am left with in the parenthesis is e j phi over 2 minus r 2 omega over c divided by 1 plus d cosine theta over 2 r plus contribution from the source s 1 and that is e j minus phi over 2 minus r 1 omega over c divided by 1 minus d cosine theta divided by 2 r. Now, we know that d over 2 r is very small compared to 1. We know that because d is extremely small compared to r and because of that and cosine theta never exceeds 1. So, d cosine theta over 2 r is extremely small compared to 1. So, if that is the case then 1 over 1 plus d over 2 r cosine of theta is approximately equal to 1 minus d over 2 r cosine of theta and similarly 1 over 1 minus d of cosine theta divided by 2 r is approximately equal to 1 plus d over 2 r cosine of theta. So, we put these relationships in this equation. These relationships in this equation in the denominator and we what we get out is p r theta t equals v v j omega rho naught e j omega t divided by 4 pi r and in the parenthesis what I have is 1 minus d cosine theta divided by 2 r times e j phi over 2 minus r 2 omega over c plus 1 plus d cosine theta over 2 r e j phi over 2 minus r 1 omega over c. At this stage I in the exponent terms I replace r 1 and r 2 by these relations r 1 equals r minus d cosine theta over 2 and r 2 equals r plus d cosine theta over 2. So, what I get is so I have this v v over times j omega rho naught e j omega t over 4 pi r times 1 minus d cosine theta over 2 r times I have to be careful here e j minus phi over 2 and then I have r 2 is r plus d cosine theta over 2. So, minus omega over c times r plus d cosine theta over 2 this is my first term and this is because of the source s 2 and then I add contribution of s 1. So, 1 plus d cosine theta over 2 r e j here I have excuse me I should not have had phi should be positive with a positive sign. So, here I have minus phi over 2 minus omega over c and then parenthesis r minus d cosine theta over 2. So, this is my overall relation now we take taking e j omega r over c out of this parenthesis because it is common to both these terms what we get is p r theta t equals v v j omega rho naught e j omega t minus r over c divided by 4 pi r. So, I have taken out e j omega r and I have combined it with the time term. So, I get omega times t minus r over c and then parenthesis I am left with 1 minus d cosine theta over 2 r times exponent of j phi minus omega d over 2 c cosine theta plus 1 plus d cosine theta over 2 r exponent of j phi oh they should be phi over 2 and then this is minus phi over 2 and then I have a positive sign omega d over 2 c cosine theta. So, this is my relation now I am going to simplify this relation further and by decomposing this term and this term and then again clubbing terms which look similar together. So, what I get is p of r theta t equals v v j omega rho naught e j omega t minus r over c divided by 4 pi r times. So, what I am going to do is club terms which have coefficient of 1 together and coefficient d cosine theta over 2 r again together. So, what I get is e j phi over 2 minus omega d cosine theta over 2 r plus e minus j minus phi over 2 plus omega d over 2 r cosine theta. So, this is 1 set of terms plus d cosine theta over 2 r. So, this is 1 set of terms and then another set of terms is having a coefficient of d cosine theta over 2 r and actually there should be a negative sign here and then in parenthesis I get e j phi over 2 minus omega d cosine theta over 2 r excuse me should be 2 c omega d 2 over 2 c. So, it should be 2 c here as well and minus exponent j minus phi over 2 plus omega d over 2 c cosine of theta. Now, if I add these two terms after decomposing them into their real and imaginary parts, these two terms if I after decomposing them into their real and imaginary parts then what I get is cosine term and in this case I am left with only the sign term multiplied by j. So, what I get from such complex algebra manipulations is v v j omega rho naught exponent j omega t minus r over c divided by 4 pi r and in the parenthesis I am left with 2 cosine phi over 2 minus omega d over 2 c cosine theta plus actually it should be a negative sign here minus d cosine theta over 2 r times 2 j sin of phi over 2 minus omega d over 2 c cosine theta. Now, we know that this is a doublet and for a doublet phi equals pi. So, phi over 2 equals pi over 2 radians. So, if that is the case then p r theta t equals. So, once I put pi over 2 in place of phi over 2, this cosine becomes sin term and this phi over 2 term goes away. Similarly, this sin term becomes cosine terms and phi over 2 term goes away. So, what I get is v v j omega rho naught e j omega t minus r over c divided by 4 pi r and then in the parenthesis what I have is 2 sin omega d over 2 c cosine of theta minus d cosine theta over 2 r times actually I will write this in the next line minus d cosine theta over 2 r times 2 j and then I have a cosine term omega d over 2 c times cosine of theta. Now, yeah so I will rewrite this equation p r of theta equals v v j omega rho naught e j omega t minus r over c divided by 4 pi r times 2 sin omega d over 2 c cosine of theta minus d cosine theta over 2 r times 2 j and sin of omega d cosine of theta over 2 c. Now, at this stage I replace omega by 2 pi f. So, omega is equal to 2 pi f therefore, omega d over 2 c equals 2 pi f times d over 2 c equals pi d over c over f equals pi d over lambda. So, I rewrite this relation as p r theta t equals this whole term I am not going to rewrite it times 2 sin and omega d over 2 c is nothing but pi d over lambda. So, pi d over lambda cosine theta minus d cosine theta over 2 r times 2 j times. So, I am going to rewrite this relation using this approximation omega d over 2 c equals pi d over lambda and what I get from there is p r theta t equals v v j omega rho naught e j omega t minus r over c divided by 4 pi r and in parenthesis I have 2 sin omega d over 2 c cosine theta minus d cosine theta over 2 r times 2 j and then I have to put the parenthesis here cosine of. So, here omega d over 2 c I will replace it by pi d over lambda and here it is cosine of pi d over lambda cosine of theta. Now, we know that if d is very small compared to lambda if d is very small compared to lambda then we can also say that pi d over lambda is very small compared to 1 and because cosine theta is at the max only 1. So, pi d over lambda times cosine theta which appears here as well as here. So, this is also very small compared to 1 if that is the case then this term cosine of pi d over lambda times cosine theta this term this becomes approximately equal to 1 because the term in parenthesis pi d over lambda times cosine theta is very small and approximately equal to 0 and also. So, I write this approximation so if that is the case then cosine pi d over lambda times cosine theta is approximately equal to 1 and sin of pi d over lambda cosine of theta is approximately equal to whatever value is there within the parenthesis is approximately equal to pi d over lambda cosine of theta. So, now I put these approximations in this equation and what I get is P r theta t equals V V j omega rho naught E j omega t minus r over c divided by 4 pi r times 2 and then sin of this entire term is nothing but same as the term. So, in parenthesis I have 2 times pi d over lambda cosine theta minus d cosine theta over 2 r times 2 j because cosine theta of pi d over lambda times cosine theta is 1. So, I have simplified this even further. So, I will rewrite this equation. So, P r theta t equals V V j omega rho naught E j omega t minus r over c divided by 4 pi r times 2 pi d over lambda cosine of theta minus d cosine of theta over 2 r times 2 j. Now, I can take d cosine theta I can take this term out of the parenthesis with this is common. So, what I get is I replace this parenthesis with the following relation. So, I am taking d cosine theta out what I am left with in the parenthesis 2 pi over lambda minus j over r and 2 pi over lambda is k which is the wave number. So, I rewrite this equation further and I write it here as k and of course, I have also taken this 2 term out. So, this 2 term cancels with 4 and I have I am left with no excuse me and if I now process this equation further and replace omega by frequency then what I get is I replace omega by frequency. So, f omega equals 2 pi f and once omega equals 2 pi f if I put this relation here then pi in the denominator goes away. So, my denominator becomes 2 r. So, this is my overall relationship this is the overall relationship. Now, for a simple source for a simple source the directionality pattern is symmetric around origin. Now, we will look at some special cases as to how this directionality pattern looks like for different values of r for different values of r. So, magnitude of p r theta t equals magnitude of this portion. So, it is v v f rho naught e excuse me. So, magnitude of this portion is 1. So, I put 1 here and then divided by 2 r magnitude of d cosine theta is d cosine theta in vertical brackets times magnitude of k minus j over r and that is nothing but k square plus 1 over r square root or if I process this relation further what I can get is k square r square plus 1 and then I take r out. So, I get 2 r square in the denominator. Now, if k square r square is very large compared to 1 which means k square is very large compared to 1 over r square which means now k is 2 pi over lambda. So, it is 4 pi square over lambda square is very large compared to 1 over r square and this. So, in that case or I can say or if r square is extremely small compared to lambda square over 4 pi square which is approximately equal to lambda square over 36. So, for this condition which means k square r square is extremely large compared to 1 for this kind of a condition I can omit 1 in the magnitude and in that case this imaginary component in the parenthesis it goes away. So, for this case p r theta t equals v v j f rho naught E j omega t minus r over c divided by 2 r times d cosine theta times k where k is equal to 2 pi over lambda. So, what we see here is yeah. So, this is one relation and I can rework and play with lambda f and c and what I can get is the following relation v v over 4 pi r c times omega square rho naught j E j omega t minus r over c times d cosine theta. So, this is one relation and this is good if r square is extremely small compared to lambda square over 36. So, for very small values of r such that r square is extremely small to lambda square over 36 my pressure for a doublet is given by this relation and what we have to note here is in this case the dependence on radius is 1 over r. Now, for the case for the case r square is extremely large compared to lambda square over 36 if that is the case then in that case the other term goes away. So, I have to make a small correction here that we had seen that if k square is extremely large compared to 1 over r square that means that it is 4 pi square divided by lambda square is very large compared to r square which means that r square should be extremely large and not extremely small it should be extremely large compared to lambda square over 5 4 pi square. So, this relation which has been boxed in red is good for the condition when square of radius is extremely large compared to lambda square over 36. What this means is that if I am far away from the doublet source then pressure decays in a 1 over r fashion because I have r in the denominator. So, that is first condition the second condition is that what happens if r square is extremely small compared to lambda square over 4. So, in that case so the second condition is if r square is extremely small compared to lambda square over 36. So, in that case using similar approach if r square is extremely small then this term dominates versus this term. So, I can drop the k term here and essentially what I get is that P r theta t equals V V f rho naught j over 4 pi r square excuse me 2 r square times d cosine theta times E j t minus r over c minus pi by this is the second relation. So, I will rewrite on a clean slide these two relations r square very large compared to lambda square over 36 P r theta t equals V V omega square rho naught j over 4 pi r c times d cosine theta times E j omega t minus r over c and if for so this is one set of relations and then if r square is extremely small compared to lambda square over 36 then P r theta t equals V V f rho naught j over 2 r square t cosine theta E j omega t minus r over c minus pi over 2. So, we make couple of observations here first observation is first observation that if lambda square over 36 is very large compared to r square then pressure of the doublet which depends on r theta t decays in a 1 over r square fashion as r grows. Then if lambda square over 36 is extremely small compared to r square then P d pressure in doublet which again depends on r theta t this also depends on r theta t is directly proportional to 1 over r. So, this is like a simple source. So, if I am far away from the doublet then the decay of pressure follows the magnitude of that decay follows something similar to that of a spherical source. And third thing is P d is directly proportional to d cosine of theta. So, that puts some polarity in the system and the fourth and the last point which I like to make here is that the phase difference between the pressure when I am near to the system near to the doublet and far away from the doublet that equals pi over 2. So, what that means is that there is a phase difference of pi over 2 as we move from near field to far field. And that phase difference comes from this term this is the source of phase difference. So, this concludes our discussion on directivity. And what we have seen over last several lectures is several aspects of directivity as they pertain to a simple points of the source which has spherical directivity pattern which is symmetric about the origin in all the directions. And also we have explored what kind of directivity patterns are depicted by two sources which are close to each other multiple sources. And finally, we have also covered a doublet source. So, this covers directivity in a fairly comprehensive way especially in context of simple and in context of simpler systems of sound sources. And we can extend this understanding if we have more complex arrays of sound sources we can use this information and this knowledge to predict their directivity patterns as well. Thank you very much.