 A warm welcome to the 36th section in the first module of the core signal and systems where we now carry out the convolution of two rectangular pulses upon which we are embarked in the previous session. We agreed on a few things. We agreed that we will make the longer pulse put onto the platform and we will make the train out of the shorter pulse and we will now start moving the train across the platform. There we are. We have this x2 here fixed. So, you know we are going to write the convolution integral as x2 lambda x1 t minus lambda d lambda for every individual t. So, lambda is the indexing on the platform and with respect to lambda I have already drawn x2. We have agreed to make the height 1. Now let us draw x1 t minus lambda. You see x1 t minus lambda is equal to x1 of t1 when t minus lambda is equal to t1. And of course, similarly x1 t minus lambda is equal to x1 t1 plus capital t1 where lambda is equal to t minus t1 plus capital t1. And of course, t1 plus capital t1 is more than t1. So, this point will occur before. So, this is the situation that we have t minus t1 and t minus t1 plus capital t1 and we put the pulse in between. We have agreed the pulse has a height of 1. We do not want to complicate too many things at once. Now we put them together, train and platform together. Train here, platform there. Now visualize the picture. You need to visualize when they will begin a handshake. They can only begin a handshake when the front post point on the train has reached the back post point on the platform passengers trip. In other words, let us show it on this graph. When these match, that is when t minus t1 is equal to t2 or when t is equal to t1 plus t2. And where does the handshake end? Let us identify that too. The handshake ends when this matches this. In other words, when t minus t1 plus capital t1 is equal to t2 plus capital t2 or t is t1 plus capital t1 plus t2 plus capital t2. So, we now know the extent over which the handshake lasts. It lasts from well, small t1 plus small t2 to capital t1 plus small t1 plus capital t2 plus small t2. That is right. And there too, we need to take difference to prove. You see now you remember the people on the platform form a longer strip and you are moving the shortest strip in the form of the train. So, there will be 3 regions. The train just entering the platform strip. The train completely within the platform strip and the train leaving the platform strip. Let us write down where those 3 regions will be. So, train entering the platform strip. The situation is this. This is x2. This is x1. And clearly, this has not reached this. In other words, t minus t1 plus capital t1 is less than t2. In fact, you could put less than equal to it. In other words, t is less than equal to t2 plus t1 plus capital t1. And we take this region first. Now in this region, what is the situation? We need to do the handshakes of corresponding passengers. Where are the corresponding passengers? Let us mark. So, this is the region of corresponding passengers. Only the green belt. And when you multiply the functions only in the green belt and integrate over that green belt, how much do you get? When you multiply them, you get more. You get essentially 1 into 1. And you integrate them, you are integrating them essentially from t2 to t minus t. So, what do we have here? The integral, the convolution x2 convolved with x1 evaluated at t is t minus t1 minus t2 for all t less than equal to t2 plus t1 plus capital t1 and starting from t1 plus t2. So, in this region, the first region where the train enters the strip. Now the next region is where the train is within the strip. So, the frontmost part of the train has not reached the end of the strip, but the backmost part of the train has entered. That is because that is possible because the train is shorter than the platform strip. So, how do we depict that situation graphically? Let us draw it here. Second region, train within the platform strip. So, what would the situation be? t minus t1 there, t minus t1 plus capital t1 there. This is essentially x2 lambda, x1 t minus lambda. And this has not reached the end of the train. So, this has reached this, but this has gone after this. That is the situation. So, how much is the total effect of the handshake? The entire train is in the platform strip. So, you have complete handshakes from the train, but not all over the platform. So, let us look at that. Let us mark that in green. This is the portion of the handshake, all extending the region of handshake. And how much does that contribute? The contribution is all of t1, 1 into 1 integrated over a length of t1. So, contribution is very simple. So, there we go. You can write down the convolution. So, you know, in what region are we talking about? We are talking about the region t minus t1 is less than equal to t2 plus capital T2. So, t is less than equal to t1 plus t2 plus capital T2. And of course, t is greater than the previous case. Of course, t, as I said, is greater than t2 plus t1 plus capital T1. So, we have the region t between t1 plus t2 plus capital T1 and t1 plus t2 plus capital t2. Naturally, since t2 is greater than t1, this inequality makes sense. And in this region, the convolution is simply equal to capital T1. And finally, what is the region? Finally, we have the region where the train is now leaving the platform. Let us draw that region. Region 3 train leaves the platform. Here is the situation. So, this has now gone beyond this. In other words, t minus t1 is greater than t2 plus capital T2 or t is greater than t1 plus t2 plus capital T2. But on the other hand, this is before this. So, the last part of the train should not have left the platform. So, let us write down the inequality completely. And well, on the same graph, before we write down the region of time, let us identify how much of convolution is contributed there. So, let us mark that region. This is the region contributed by the convolution. How much is this contribution? It goes from t2 plus capital T2 to t minus t1 plus capital T1. So, it is essentially t2 plus 1 into 1 integrated over that length. So, t2 plus capital T2 minus t minus t1 plus capital T1, which is of course, equal to t2 plus t2 plus t1 plus t1 minus t. Now, we have an expression. We need to write down the region over which it is valid. Let us do that. You know, you need t minus t1. So, t minus t1 has gone beyond t2 plus capital T2. But or and t minus t1 plus capital T1 is still less than equal to t2 plus capital T2. So, this says t is greater than t1 plus t2 plus capital T2, but t is less than equal to t1 plus t2 plus capital T1 plus capital T2. So, this is the region. The region specified algebraically is described as follows. Now, I have identified three regions for you. The exercise that I am leaving to you is to sketch the entire function. I encourage you to do that, but we will do it at the beginning of the next session just to help you on. But I would encourage you to do it and then come to the next session. Thank you.