 A warm welcome to the ninth session of the fourth module in signals and systems. We had come upon a very important property of the Laplace transform towards the end of the previous session, namely, when we convolve two signals, the Laplace transforms are multiplied. Let us formally state this first. But the difficult question is what ROC? We had not answered this in the last session. You see, what we saw? I mean, let me put down just one important step here. That is what we need to really look at carefully. What we saw essentially was that the Laplace transform of yt was this product, product of two integrals. Now, suppose there is an s which belongs to the intersection of the two regions of convergence. This product will certainly converge for that s. So, one thing we can say with certainty, the region of convergence includes this intersection. So, if you ask me, what is the region of convergence of xt convolved with ht? The region of convergence of the convolution is at least the intersection. It can be more. Let us take an example. Let us consider the convolution of two exponential sequences. So, all this is quite familiar to us. The only thing we need to worry is what happens to limits when you take this product and this product will allow the limits to go only for tau greater than equal to 0 and t minus tau greater than equal to 0. It has a meaning for 0 less than equal to tau less than equal to t and this is a null region for t less than 0. So, in fact, we can write this convolution as follows. This integral multiplied by ut and now expand this. So, you could take e raised to the power 3t common or if you like e raised to the power 3t ut common and multiplied by integral from 0 to t e raised to the power 2 minus 3 into tau d tau. Let us simplify this. So, you can easily see this can be simplified to e raised to the power of 3t minus e raised to the power 2t into ut. That is very interesting. The convolution of these two exponentials is actually a linear combination of those exponentials. This is something that I specifically wanted to point out here. This is a very important property of exponentials. When you have two exponentials with distinct parameters, if you convolve them, you actually get a linear combination of them. So, the Laplace transform can be found by this root. Let us evaluate it by this root. We can easily evaluate the Laplace transform by using linearity. So, it is the Laplace transform of e raised to the power of 3t ut minus the Laplace transform of e raised to the power of 2t ut, which is 1 by s minus 3 with a region of convergence, real part of s greater than 3 and 1 by s minus 2 with a real part of s greater than 2. But then we can take their intersection which happens to be real part of s greater than 3 and that can be taken as the region of convergence. We can simplify this. Lo and behold, you get the product of the Laplace transforms of the individual exponentials. We illustrated this with an example. When you convolve two signals, the Laplace transforms are multiplied. This is a very important property. In fact, this has serious implications for linear shift invariant systems. Let us see what it implies. For a linear shift invariant system, where you have an input and impulse response, both of which have Laplace transforms. That means, if you can ensnare both of them by an appropriate complex exponential, by a decaying complex exponential, then you can apply this property. So, consider a linear shift invariant system. Let us call it script s. It has an impulse response of h t. It has an input x t and the output y t is the convolution of the two. Now, as usual, let us take their Laplace transforms if they exist. Let us assume that x of t has a Laplace transform capital X of s with the region of convergence script r subscript capital X and similarly for h of t. What is the Laplace transform of y t? Very simple. We have answered it already. y t has a Laplace transform x s times h s and the region of convergence is at least the intersection of the two, if not more. Now, you know this whole business of at least if not more requires some explanation. We need to say when this more can come. But before we embark on that issue, let us embark on a much more important issue namely a new definition that we want to make, a new way to characterize a linear shift invariant system, where you are dealing only with signals with Laplace transforms. So, you are guaranteed that the input output, the impulse response, all of them have a Laplace transform. Look back at what you are saying here. What are we saying here? We are saying y of s as an expression is x of s times h of s and therefore, y of s divided by x of s is independent of the input and therefore, it characterizes the linear shift invariant system completely to the extent that we are dealing with signals with Laplace transforms. Why are we saying it characterizes the LSI system completely? You see, in the first module, you have seen that the linear shift invariant system is completely characterized by its impulse response. What is h s? It is the Laplace transform of the impulse response and if I can invert that Laplace transform, then I know the impulse response, I know everything about the linear shift invariant system. Essentially, due to invert has to be put in, it is not really something that we have done so far very formally. You have to put them in inverted commas. This word has to be put in inverted commas because if we invert h s, meaning if we can go back to h t from h s, we have not indicated how we would do that. You have just been going one way by and large. We have been going from the signal to its Laplace transform or from the signal to its z transform. We have not really talked about how you would invert it. We will do it once we realize why we should be doing it. So, if we invert h s, meaning given h s and the region of convergence r h, if we can calculate the Laplace inverse or you can go back to h t by whatever mechanism, then we know everything about the system because h t tells us everything about the linear shift invariant system. That is what we mean by this h s characterizing that system completely. So, if we can see, when you say it characterizes the system completely, it does so only when the input and the output have Laplace transforms and the impulse response also has a Laplace transform and then we would like to give this quantity and name. We will call it the system function. The system function therefore, is essentially the Laplace transform of the impulse response and remember the system function is always associated with the region of convergence. You cannot dissociate the system function from its region of convergence. So, it is always h s together with the region of convergence. This constitutes the system function. For example, suppose we had a linear shift invariant system script s with an impulse response h t equal to e raise to the power 2 t u t, then the system function of this LSI system would be 1 by s minus 2 with real part of s greater than 2 and it would always be true that if x t were given to the system, and y t came out and if both of them had Laplace transforms, this had a Laplace transform of x of s with a region of convergence r sub x and the same here, then y s divided by x s would be equal to this quantity. Now, we have to be a little careful here. What about the regions of convergence of y and s? Let us take the simple situation first that their regions of convergence have some non-null intersection and that non-null intersection also has a non-null intersection with r h. In that case, the matter is simple. At least you have some non-null part, non-empty part, which is in the intersection of all the three. Could it happen that there is a null intersection? I am not going to answer that question. Immediately I would like to reflect upon it. You might later be able to answer this question. But at the moment, let us take a situation where there is a non-null intersection all over. So, at least we are saved the bother of having to wonder. There is no agreement in the regions of convergence at all between two transforms and we are trying to find out a ratio of two Laplace transforms. Let us not get into that situation right now, but you might run to that situation later. We will see more of this in the next session. Thank you.