 Welcome to the 35th session in the first module of the course Signals and Systems where we continue our discussion on convolution. In the previous lecture we had looked at discrete convolution and we had used a very simple and yet effective analogy called the train platform analogy. So we thought of convolution as a point by point or step by step interaction between the passengers sitting in a train and the passengers standing on the platform where corresponding passengers shake hands and the combined effect of the handshake is what we record as the value of the convolution at every step. And then you had this train move by one step and there was a new effective handshake and there you are. Now we go on to continuous independent variable convolution and we would like to build the same idea and to evolve a methodology for calculating the convolution of two continuous time functions. In fact, here let us begin with an example again. And we will begin with a very simple example of convolving two rectangular pulses and for variety we will take those rectangular pulses to be of different lengths or different sizes so to speak. Let us put down the question that we want to address. So what we are trying to do is to create a continuous independent variable convolution. We have two continuous independent variable functions, let us call them x1 t and x2 t. And we wish to calculate the convolution which we will of course represent as usual by a star and that is calculated to be x1 lambda x2 t minus lambda d lambda integrated overall lambda. This is done for each t and now as I said we will take the example of two rectangular pulse. x1 t is a rectangular pulse, let us be relatively general now instead of trying to restrict it to 0 or anything of that kind. Let this rectangular pulse go from t1 to t1 plus capital T1 on the axis of t and of course let the rectangular pulse have a height of A. Let x2 t be the rectangular pulse beginning at t2 and going up to t2 plus capital T2 with a height of B. Now without loss of generality let t2 be greater than t1. So you know here there is a choice to be made. We can convolve x1 with x2 or we can notionally convolve x2 with x1. However, we must now review the idea that we had established earlier. We had asked this question before specifically when we talked about how we would process an input going through a linear shift invariant system, we had talked about the commutativity of convolution. Let us just review that idea once again. So I had asked you to prove at that time I had left as an exercise for you to prove that convolution is commutative. I am sure many of you must have attempted that exercise by now. But let me now carry out the proof because it is a very important idea. So let me prove that convolution is commutative. What I mean by that is it does not matter whether you are talking about convolving x1 with x2 or x2 with x1 they are the same thing. We prove this indeed x1 convolved with x2 evaluated at t is integral x1 lambda x2 t minus lambda d lambda integrated over all minus to plus infinity. Put t minus lambda is another variable alpha for every specific t. Now notice that when you do this let us identify what happens to the limits. Of course lambda is t minus alpha and lambda equal to minus infinity gives alpha equal to plus infinity lambda equal to plus infinity gives alpha equal to minus infinity moreover d lambda is minus d alpha so we have all these little changes to be made. So if we make these replacements you notice that you can replace d lambda by minus d alpha t minus lambda is just alpha so I will write that down t minus lambda is alpha lambda is t minus alpha and then you have a minus sign and these limits also reverse. So the minus sign together with the reversal of limits cancels one another out. So this is cancelled by reversal of limits so what I am saying is that we could rewrite this as minus infinity to plus infinity x1 t minus alpha x2 alpha d alpha and that is the same as x2 convolved with x1 evaluated at t. And since this is true for every t it is quite clear that convolution is commutative. So you know for every t we have established for every individual let us write that down we have established this for every individual hence established x1 convolved with x2 is the same as x2 convolved with x1 for all proved. Now a very similar proof can be given for discrete time convolution for the sake of completeness let us complete that proof also. So what we would have there x1n and x2n being convolved. So x1 convolved with x2 evaluated at n is just summation on all k x1k x2n minus now again for a fixed n put n minus k equal to l which means k is n minus l and of course summation k going from minus to plus infinity is equivalent to summation l going from minus to plus infinity for a fixed n and for every fixed n therefore what we have is x1 convolved with x2 evaluated at n is also summation l going from minus to plus infinity x1n minus l x2 l and that is the same thing as x2 convolved with x1 evaluated at n this is true for every n hence proved. So we have completed the proof for the discrete case as well. Now we go back to the continuous example that we had begun with I would like to calculate that convolution. So you know your you notice I said without loss of generality you know. So let me draw the two pulses again and then it will be clear what I am trying to say. I have these two pulses which I want to convolve a pulse of length t1 and height a to be convolved with the pulse of length capital t2 and height b and I said it is without loss of generality that we said that one of them is greater than the other. We could take as we said b to be greater than a as we did there is no problem with that or I mean the longer pulse is the second b to be greater than a and t2 to be greater than t1 actually well that is not without loss of generality. I mean which pulse is higher or lower in height is not without loss of generality I must correct myself there. In fact it does not matter what the height is convolution is actually a consequence of essentially linear operations being done on the system. So when I scale up one of the inputs the output is scaled by the same amount. So in fact I can say a stronger thing without loss of generality. If I take pulses of unit height albeit of different widths so capital t1 and capital t2 need to be different for the general case and I need of course to evaluate it for the general case where t2 is different from t1 but I do not need to evaluate differently for different a and b. I could without loss of generality take a to be equal to b equal to 1 and then if I want the answer for any other a or b I could simply multiply the output by a b. So let me without loss of generality take a equal to 1 b equal to 1 and t2 greater than t1 as I did let me write that down without loss of generality we will take a equal to b equal to 1 and t2 greater than t1 here. This t2 greater than t1 is without loss of generality because of the commutativity of convolution it does not matter which one is longer the convolution would not change if we interchange that is why I am saying t2 greater than t1 without loss of generality. I am saying a equal to b equal to 1 without loss of generality because if you have any other non unity a or b you simply multiply by that a and then by that b alright anyway let us now calculate this convolution so we will set up the process we will have to have a train and platform here too the only change is that the train and platform are now very dense in passengers occupancy by passengers passengers are as close together as they can get tending to touch one another tending to become a continuum. So you know unlike the discrete case where you could identify this passenger and the next passenger the notion of next is there in the discrete case the notion of next is absent in the continuous case. Take any two passengers no matter how close there is another passenger which is between them and distinct from them these passengers are indexed by the continuous independent variable. So are now the question since we know the convolution process is commutative let me choose the more convenient of the two x1 convolve with x2 or x2 convolve with x1 one of them is likely to be more convenient and naturally if we have assumed capital T2 is greater than t1 it will be convenient to fix the one which is longer onto the platform and then move the train which is shorter. So we will move x1 we will make x1 the train and we will put x2 onto the platform and we shall work this out in the following session we shall work out the convolution entirely in the session that will soon come. Thank you.