 A warm welcome to the sixth session in the fourth module of signals and systems. We are now getting familiar with the Laplace transform and the Z transform. We now need to put down certain properties, you know we could keep taking examples, but it will be easier for us to write down certain properties, to investigate certain properties that these transforms obey, thereby leading us to deal with a much wider class of signals both continuous and discrete independent variable. So, let us begin with the first very important property of linearity. Now, when we say linearity what we really mean is think of the Laplace transformer or the Z transformer as an operator. So, you are saying you have this machine called a Laplace transformer. It takes the signal x t and generates capital X of s with a region of convergence and let us write region of convergence to be script R capital X. So, this means the region of convergence of the Laplace transform capital X of s, this is how we should read it, region of convergence of the Laplace transform capital X as a function of s. So, we do not explicitly need to write as a function of s here, you have just said capital X. Now similarly, let us take a sequence and talk about its Z transform as being the output of a Z transformer. It produces capital X of z with a region of convergence script R of X again. Of course, here we are talking about two different kinds of regions of convergence. You have seen the kind of region of convergence that a Laplace transform has and you have seen the kind of regions of convergence that a Z transformer has. Now, what linearity really means is these transformers behave linearly. That means if I were to give alpha times x 1 t plus beta times x 2 t to the Laplace transformer, the output would be alpha x 1 of s plus beta x 2 of s and what is important and somewhat tricky is the region of convergence. What would be the region of convergence of this in general? Well, it is difficult to make a general statement, but what can be said with certainty is that the ROC will definitely include the intersection. It could expand beyond the intersection. In fact, let us prove this, a general proof first. In fact, the same thing can be said for the Z transform. Let us first state it and then prove both of them together in parallel, so to speak. So what we are talking about is the linearity of the Z transform and that amounts to saying alpha times x 1 n plus beta times x 2 n when subjected to Z transformation results in alpha times x 1 z plus beta times x 2 z and once again what can we say about the region of convergence? All that we can say is that the region of convergence will definitely include the intersection. More than that could be a problem to say. We will now first prove both of these. Consider the Laplace transform of the sum or linear combination. Of course it is, so I have alpha x 1 t plus beta x 2 t and now let me split, let me use distributivity. Now this is easy to see. So in fact, now we can very clearly see that this essentially is capital X 1 of S by expression and the same thing holds for this, this is capital X 2 of S by expression. Now the only issue is to talk about its region of convergence. So all that we can say as we can see from this sum here is that if for a given S both these integrals converge, which integrals are we talking about these two integrals here? This one and this one, if for a particular S both of these integrals converge then the sum or that linear combination also converges and therefore this S which is included in the intersection must also be in the region of convergence of that linear combination. So this proves what we wanted to prove that at least the intersection is included if not more. Now where does that if not more come from? Let me give you an example, very simple example. Let us take the following sum, you know we dealt with this in one of the previous sessions. Let X 1 of t, we erase the power 2tut and X 2 of t, we erase the power 3tut, we have already calculated their Laplace transform, we have been doing this is not it in the previous sessions. So the Laplace transform of X 1 t is 1 by S minus 2 with the region of convergence real part of S greater than 2 and similarly the Laplace transform of X 2t is 1 by S minus 3 and the region of convergence real part of S greater than 3. Now let us take any linear combination. So for example, let us take 2 times X 1t plus 3 times X 2t, what would be its Laplace transform? The Laplace transform would of course be according to this principle that we have derived 2 by S minus 2 plus 3 by S minus 3 and of course the ROC is at least the intersection of real part of S greater than 2 and real part of S greater than 3, which is of course the second term is not it, the intersection is essentially the real part of S greater than 3 and this also happens to be the ROC here that can be verified. In contrast, if you took the following example, you took X 1 of t to be erase the power 2tut plus erase the power 3tut and you took X 2 of t to be erase the power 2tut minus erase the power 3tut and now you added them. So you took X 1t plus X 2t that would be essentially 2 times erase the power 2tut. Now the interesting thing is that if you look at the regions of convergence of course you can look at their Laplace transforms and the regions of convergence. So if you looked at the first Laplace transform it would be X 1 of S given by 1 by S minus 2 plus 1 by S minus 3, you know you can go back and see that you are summing there and you are taking a difference. So the first Laplace transform would be 1 by S minus 2 plus 1 by S minus 3 and the second one would be 1 by S minus 2 minus 1 by S minus 3 and both have the regions of convergence given by real part of S greater than 3. Now when I take their intersection, what do you notice? Their intersection is of course real part of S greater than 3 nothing very great about that but then if you look at the Laplace transform of the sum, the sum is also linear combination. So Laplace transform of X 1t plus X 2t is essentially the Laplace transform of 2 erases the power of 2tut which is very easily calculated to be 2 by S minus 2 and the region of convergence happens to be real part of S greater than 2 which of course includes real part of S greater than 3 but has expanded beyond. So it is very interesting, the region of convergence could expand in this process of linear combination beyond the intersection. Now let me carry you a few steps for the same reasoning in the context of the Z transform. Let me first complete the proof for the Z transform. So the Z transform of a linear combination is essentially this expression which can now be broken down and you can very easily see that this essentially corresponds to capital X 1 of Z and this essentially corresponds to capital X 2 of Z. So this becomes alpha times capital X 1 of Z plus beta times capital X 2 of Z. Now again if a particular Z belongs to the intersection I am bringing in the symbol for intersection now you know this is the symbol for the intersection of 2 sets or 2 regions in this case. Then it also allows that linear combination to converge and so we have proved that at least the intersection is there in the region of convergence. Now I am going to leave it to you as an exercise to construct example similar to what I did in the Laplace transform. The region of convergence could be exactly the intersection or the region of convergence could expand beyond the intersection. It is not too difficult do the exercise and that will give you some confidence with the Z transform. We will meet again in the next session. Thank you.