 Welcome to the 37th session of the first module of the core signal and systems. We will now complete the continuous time convolution that we are undertaken in the previous session. By complete I mean let us write down all the outputs of the convolution in the three regions that we are identified and then sketch them. You must get a complete understanding of that particular convolution. So, we are identified three regions. The first region was p greater than equal to t1 plus t2 and less than equal to t1 plus t2 plus capital T1. Region 2, t goes from t1 plus t2 plus capital T1, t1 plus t2 plus capital T. Of course, capital T2 is greater than capital T1 so valid. And finally, region 3 was p is greater than t1 plus t2 plus capital T2 but less than equal to t1 plus t2 plus the sum of capital T1 and capital T2. And the convolution value in region 1, 2 and t were as follows. I will just write them down in this horizontal fashion. Here it was t minus t1 plus t2. Here it was a constant at capital T1 and here it was essentially a dying or reducing line. So, for the last part we had written down the output of the convolution as essentially a difference. So, you know, let us by sketching this we will get an idea. First the convolution rises, reaches t1 and then goes to the end falling off as a dying line or a reducing line. Now, the convolution is a function of t. It begins at t1 plus t2. There is a transition of behavior at t1 plus t2 plus capital T1. One more transition of behavior at t1 plus t2 plus capital T2 and the final transition to 0 at this. In this region, the convolution takes the value capital T1, a rising line here and a falling line there, 0 beyond here and 0 before here. Region 1, region 2, region 3, all straight lines, 0 here and 0 here. Now, notice something interesting. Here our convolution was simplified because you were always multiplying two constant functions and finding the area under a constant product. You could, if you wish, think of it in a metaphorical sense as all passengers having the same strength, all the passengers inside the train and all the passengers outside the train have same unit strength. So, when they hand shake, they all contribute unity and then we integrate and find the combined effect. What if the passengers as will often be the case have different strength? They have different builds and different strengths. What does that mean? For example, suppose we were to try and convolve the following two functions. Let us take an example. Convolve this function. The function that rises from 0 to 1 over an interval of t1. So, t1 to t1 plus capital T with another function that rises over an interval of t2. Let us call this x1 t and it is called as x2 t. Now, here you should think of it as again passengers and a train with passengers inside the train and outside the train. Passengers space very closely together into a continuum. The only thing is the passengers have different strengths. So, the tallest and strongest passengers are in the end and the weaker passengers are in the beginning. But you know, you have to be careful. Again, suppose again without loss of generality, let us take capital T2 to be greater than capital T1 because you can always interchange the order of the functions and the convolution is not affected. So, without loss of generality here, let T2 be greater than T1. And we can do the same. We could put the platform with the passengers which have a longer strip and the train with the passengers that have the shorter strip. And let us draw both the platform and the train. The platform would now look like. I do not need to carry out all the explanations again. I will just try to withdraw the platform and the train now. In fact, we must now write down proper expressions analytically for the passengers on the platform and on the train. So, here for this one, I would write the expression. Well, how much is this theory? It is take the difference from the point lambda at which you are from T2. So, lambda minus T2 divided by the entire interval which is covered. So, capital T2 into 1. That is the value anywhere here in this region. And here I need to take again I could write a similar expression there. So, I would have written so T minus lambda and so on and so forth. So, I can write a similar expression for this. I could write here T minus lambda minus T1 divided by T1. Now, for example, if you take lambda equal to this T minus T1 plus capital T1, you can see that gives you 1. And you take this value T minus T1 here for lambda, you get 0. Now, again here we have three regions as before. However, now the convolution integral is a little more complicated. So, regions are the same. Let us write down the regions. In fact, I will now calculate the convolution each of the three regions. Let us write down those three calculations. So, T1, you know the regions do not change. T is greater than equal to T1 plus T2, but less than equal to T1 plus T2 plus capital T1. And in that region, you essentially have the frame not quite having entered the platform. So, I will draw the platform and the phone in different colors here. I will superpose the frame which is the region of overlap. Let us mark that in green and let us calculate the product. This is the region of overlap. And we also need to write the expression everywhere. The expression for this is T minus lambda minus small t1 divided by capital T1. And we need to take the product to calculate the convolution. We take the product of this with this integrated over T2 to T minus T1, this region, integration region or integration interval. Let us write that down. We are saying essentially the convolution region 1 is the product lambda minus T2 divided by capital T2 times T minus lambda minus T1 divided by capital T1 d lambda integrated from T2 to T minus T1. So, essentially it is going to be a quadratic, but to be expected. When you multiply two line segments and integrate, you are going to get some kind of a quadratic here. Now, in fact, you know, the integrand is quadratic. When you integrate, it is going to be cubic. So, in fact, if you look at it, expand this, look at the expression here. The integrand is quadratic. So, you would have a lambda squared term and so on. So, the integral will be cubic. So, I mean, you know, it would require a little bit of algebra to work this out. And that calculation I am leading to do as an exercise. I am leading to you the key where you have to work a little hard, but it would explain to you several things about convolution when you do it patiently and carefully. Now, the same approach can be found for region 2 and region 3. In fact, the only change, the integrand is really the same in region 2. It is just the limits that change. However, here, one has to unlike the earlier example where the convolution remain constant all through region 2. The convolution would not remain constant here because there are passengers of different strength interacting with each other even all through region 2. Region 2 is the region where the train moves within the platform strip. And finally, there is the portion where the train starts leaving the platform strip and there again the expression is the same, but the region over which you integrate that expression, the integrand is different. So, I leave this as an exercise to you. I recommend that you complete this and sketch this. You expect that you would get three cubic segments in your sketch. It would be a good exercise to complete this and understand those sketches in some depth. Please do that exercise. Thank you.