 A warm welcome to the 23rd session in the 4th module on Signals and Sets. In this session, we now look at the Laplace transform and the Z transform together from an important point of view. And that is, we focus now on the regions of convergence. We draw some parallels and there again we focus on rational Laplace and Z transforms. So, let us put down a context. You know, we have now seen how to deal with rational Laplace transforms and invert them. We have seen how to deal with rational Z transforms and invert them too. Let us put down the parallels. So, you see, both of them are associated with poles and zeros. We know how to invert. We now focus on the part where we have decomposed it as quotient plus remainder. Let us first look at the quotient part and there there is an important difference. So, the quotient part is a finite series in S or Z. And let us look at them separately. You know, they have to be dealt with very different. Now, for the Laplace transform, if there is a quotient part and it has anything other than a constant, one thing is very clear. You know, if there is a term like S, so a 0 S, that means there is a unit doublet or there is a differentiation operation coming into the picture. If you have a term of the form a 0 till day 1 by S, so negative power of S, there is an integrator coming into the picture in the system. So, the moment there is anything other than a constant in the quotient part, the system immediately becomes unstable. In contrast, let us look at the Z transform. If you have a quotient part, it essentially gives you a finite length sequence and if this Z transform that we are talking about is the system function of a system, this finite length sequence will add to the impulse response. Now, by itself, this finite length sequence can never change the stability or the instability of the system. So, you know, there is a big contrast here. The effect of the quotient in the Laplace transform, if that Laplace transform corresponds to the system function, immediately makes the system unstable. That is not true for the quotient part of the Z transform. However, there is this big dissimilarity as far as the quotient part is concerned. As far as the remainder part is concerned, the remainder divided by the denominator. Let us come to that now. So, both for the Laplace and the Z transform, if we consider the remainder divided by the denominator, we can make a partial fraction expansion of this. And there is a great similarity in the way we deal with this partial fraction. Each distinct pole gives a poly-X term. Now, for example, in the Laplace transform, if you had something like 1 by S minus alpha to the power m with other factors, you know, other factors here, it would give you a poly-X term of the form e raised to the power alpha t multiplied by a polynomial of degree m minus 1 in t. And if we multiplied either by ut or u minus t, depending on, now that is what we need to see, depending on the region of convergence, region of convergence to the left or right of S equal to alpha. See, we have dealt with this before. You know how to do this. You identify where S equal to the vertical line passing through S equal to alpha and see if the region of convergence is to the left of that or to the right of it. If it is to the left of it, then you have a left sided solution. So, u of minus t is the correct answer. And if the region of convergence is to the right of that, then you have the right sided solution. u of t is the answer. So, that is, in fact, true in the Laplace transform. Something very similar is true in the Z transform. So, let us see the Z transform. In the Z transform, suppose you had a term of the form, now I am taking the portion, tremendous everything in terms of Z inverse. So, 1 minus alpha Z inverse to the power m and there are other factors in the numerator, other factors in the denominator. On inversion, this would give you a poly x term of the form, a polynomial in n of degree m minus 1 multiplied by alpha raised to the power of n. And again, you have a u n or shifted version of this or u minus n or a shifted version of this. And this time, the choice is of course, multiplied by either of them. And this time, the choice between u n and u minus n is based on whether the region of convergence is to the exterior or the interior. So, you know, you draw the circle passing through alpha, circle passing through alpha with a center at the origin. And you look at the region of convergence. If the region of convergence is to the exterior of that circle, then you have the right sided sequence. If it is to the interior of that circle, you have the left sided sequence. So, here the choice is between exterior and interior. And of course, the region of convergence can never pass through a pole. The pole is a singularity. The pole is the point where the Z transform can never converge. So, it can never pass through the pole. That is true for the Laplace transform and the Z transform. The guideline is that for the rational Laplace transform of the rational Z transform, you essentially identify the poles. In the Laplace transform, you draw vertical lines through these poles. And the possible regions of convergence are interspersed in between these vertical lines. In the Z transform, you again draw circles centered at the origin passing through the poles. And the possible regions of convergence are disks between these circles, the interleaving period, the interleaving region between these circles. So, I will give you an example for the Z transform as a variant now. If you have something like this, HZ is 1 by 1 minus 2 Z inverse. Let us take this for some variety. Now the ROC has to be specified. So, let us draw what is called a pole 0 plot. And to bring in some variety, let us also put some 0 there. Let us put 1 0 at Z equal to 1 fourth. So, we have a 0 at Z equal to 1 fourth. And we have pole at Z equal to 2 and Z equal to 1 third. The pole at Z equal to 2 is of multiplicity 2. And the pole at Z equal to 1 third is of multiplicity 1. And of course, the numerator degree is denominator is less than the denominator degree. So, we can straight away make a partial fraction expansion. We can straight away invert. What would the inverse look like? The inverse would be of the form some let us say a 1 0 plus a 1 1 n times 2 raised to the power of n. Now here we have to take a decision you know u n or u minus n and then plus simply some b 0 times 1 third to the power of n. And again we have to take a decision between u n or u minus n. Now how do we take the decision? So, let us draw the pole 0 plot here. This is the z plane and we draw circles passing through the pole. So, we likely draw the real and imaginary axis here and there is a pole at 2. So, we mark the poles with crosses. There is a pole at 1 third. The poles are marked with crosses and the zeros are marked with circles. Draw the circles with the center at the origin and passing through the poles. Now there are three possible regions of convergence. Let me mark each of them. This is ROC 1 shown green. This is ROC 2 shown in blue. And finally, this is ROC 3 shown in black. And ROC 1 is to the interior of both of the poles ROC 2 is to the exterior of the pole at 1 third and to the interior of the pole at 2. And finally, ROC 3 is to the exterior of both of the poles. Now when an ROC is to the exterior of the pole, you choose the right sided sequence. So, u n is the choice. Of course, u n is initially the choice, but then you may have to shift that. So, you have to make it u of n plus 1 or n plus 2 depending on some loose factor of z squared or something. If you have a factor of z squared to be taken care of, you must replace n by n plus 2. If you have a factor of loose factor of z inverse, you need to replace n by n minus 1. So, to that extent essentially there is a u n term. The moment you see the ROC being to the exterior of a pole and the moment the ROC goes to the interior of the pole, you have a mu minus n term again possibly shifted. So, you know if you go back to this expression, which we had in the previous expansion, the choice here, you know we have to make a choice between u n and u minus n here and u n and u minus n here. For ROC 1, you would choose a u minus n form for both of them. For ROC 2, you would choose the u n form for this and u minus n form for this. And for ROC 3, you would choose the u n form for both. So, for ROC 1, you would choose the u minus n form for both. For ROC 2, you would choose the u n form for one third to the power of n and the u minus n form for the 2 to the power of n term. And for ROC 3, you would choose the u n form for both. So, that makes it clear how we make a choice between the u n and u minus n term. So, now, you see we see a parallel, you have poly x terms and you also know how to obtain these poly x terms both for the Laplace transform and the z transform. Now, in the next session, we are going to answer an important question. You have been given a system function. Let it be the Laplace transform or let it be the z transform. That means a system function for a continuous independent variable system or for a discrete independent variable system. You have been given, of course, if you have been given a system function, it is inadequate without knowing the region of convergence. So, you have also been told about the region of convergence. So, in principle, you can find the impulse response. If you can find the impulse response, you can query the properties of the system. So, of course, a system is linear shift invariant, otherwise you could not have got to that extent of getting a system function. But now what more, what are the other properties that you need to query? Causality and stability. How should we determine causality and stability by looking at the system function? It is something that we are going to begin discussing in the next session and continue perhaps for some time. See you in the next session. Thank you.