 a warm welcome to the 25th session of the fourth module in signals and systems. We have now reached a stage where we can see beyond just the transform in the inverse transform. So, in the previous session, we had identified a methodology to look at the causality of a linear shift invariant system by looking at the system function. And in fact, now we understand what the importance of the region of convergence is. The region of convergence contained the information about causality. In this session, we are going to take on the identification of the other property that we were talking about beyond linearity and shift invariance namely stability. So, the question that we are going to answer is how do we determine stability of linear shift invariant systems from the system function? Of course, we would look at both continuous and discrete independent variable. So, if we are looking at the continuous independent variable, we would be looking at the Laplace transform. So, system function would be a function of s. And if we are looking at the discrete independent variable, we would look at the z transform. So, system function would be a function of z. Let us take the continuous independent variable case first. So, let us review the context. We have a linear shift invariant system. Its impulse response is H of t. H of t is Laplace transformable. And H t has the Laplace transform given by capital H of s with the region of convergence R h. This is the expression. This is the region of convergence. We shall assume that H s is rational. So, therefore, we are actually looking at a restricted context. We are not trying to look at all possible continuous independent variable systems which have a Laplace transform for their impulse response. That is because more general systems could be a little tricky to deal with. But then, we are quite happy dealing only with this class because rational systems actually do cover a wide class of systems in nature. They cover all the systems which are described by differential equations. So, in that sense, we are doing well. So, let us now you see it is clear that the expression itself may not tell us about the stability. And the stability information is going to be contained in the region of convergence. In fact, let me explain that you immediately with an example that the region of convergence is going to be critical to the stability and not the expression. So, let us take an example. A very simple example with which we had started in the beginning of the discussion of the Laplace transform. So, let us take the expression H s to be common to two different situations. And once we do this, we have identified the pole. The pole is at s equal to minus 2. Let us sketch it in the S plane. And therefore, you have two possible regions of convergence. Let us identify them. This is the dividing line between those two regions of convergence. This is one region of convergence. And the other one I am going to show in green. Let us write down the inverse Laplace transform for each of these regions of convergence. So, if the region of convergence is the red one, then the inverse Laplace transform is e raised to power minus 2t ut. If the region of convergence is the green one, then the inverse Laplace transform is e raised to power minus 2t, u minus t with a negative sign. Let us sketch both of these. So, the green one looks something like this. It begins with minus 1 and then grows without bound. And the red one begins with 1 at 0, but decays. Of course, you must think of this as a decaying exponential. So, that is how it is looking. So, now the question is what is going to determine stability or otherwise? You must look at the absolute integral. So, for determining stability, we need to consider what is called the absolute integral, the impulse response. And you can see that the absolute integral for the red response converges and for the green one diverges. And therefore, this red one corresponds to a stable system and the green one corresponds to an unstable system. So, it is very interesting. You have taken the same expression 1 by s plus 2, but with a different region of convergence, you have once a stable system and once an unstable system. So, this illustrates very clearly why the region of convergence is central to stability or instability. Now, let us analyze this situation to understand what made the system stable or unstable. You see, let us go back to the situation that we had with just the one pole that we took. So, we had just one pole. And of course, the region of convergence cannot include the pole. So, it must either be to the left of the pole or the right of the pole. This is true universally for all the poles. And you also know that the poles are critical in determining the region of convergence. In fact, it is very simple. You draw vertical, for a rational system, you draw vertical lines to the poles and the regions in between these vertical lines are the possible regions of convergence. So, here you have essentially just one pole and you go to the left of the pole or the right of the pole. In fact, any given region of convergence is going to be to the left of a given pole or to the right of a given pole. It cannot lie on the pole. So, this idea of left of the given pole or right of the given pole is going to be applicable even if there are multiple poles. Also, it does not matter if the pole is repeated. Still, we can talk about the ROC being to the left of the pole or the right of the pole. Now, before we have also seen the implication in terms of the inverse. So, if I have a pole and if I take the region of convergence to the right of the pole, I get a right sided signal. If I take the region of convergence to the left of the pole, I get a left sided signal. So, let us review that. We must review some of the critical ideas that we have to use. Focus on a given pole. The region of convergence is either to its left or to its right. If to the left, we get a corresponding left sided poly-extr. And now, I can write one more statement in the same breath, so to speak. If to the right, we get a right sided poly-extr. So, you should read this as two sentences. You should read it as if to the left, we get a corresponding left sided poly-extr. If to the right, we get a corresponding right sided poly-extr by using the word respectively. This is called a multi-statement. We have encountered it before. Now, I say poly-extr because the pole might be repeated. Now, let us determine the role of any one pole in stability or instability. For stability, we want this to hold. Now, in a rational system, h t is going to be a sum of poly-extr, one corresponding to each distinct pole and possibly what are called singularity functions or generalized functions. So, if you look at the system function, we look at its numerator and denominator. Both of them can be expressed ultimately as finite series in S. And ultimately, we can write them down as polynomials in S. Now, you need to see and rewrite h s as follows. We need to write it as a quotient polynomial in S plus remainder polynomial in S divided by 1 plus d tilde S. That means, you have ensured that the constant term in the polynomial can be brought to 1. There is a constant term. And of course, the remainder polynomial is going to be of degree less than the denominator. So, once you do this, if you see a quotient polynomial, the moment this is non-zero. Now, you know, again, there is an issue here. If the numerator degree is greater than the denominator degree, so you have brought out terms like something times S plus something times S squared or a polynomial of degree other than 1, other than 0, that is other than just one term. The moment you have terms like S or S squared and so on in the quotient polynomial, it means you have differentiators present. There is a differentiation on a double differentiation operation. The moment you have a double differentiation operation, instability is implicit. So, if non-zero, here, the system is immediately unstable if degree greater than equal to 1. So, what we are saying is, the moment we have differentiators or integrators, the system is unstable. So, let us now assume that we do not have them and proceed. So, what I am saying is, if you find the system is unstable before you even investigate that remainder part, there is nothing more to be done. So, that kind of a system has been dispensed with immediately. Now, we shall in the next session look at what to do with the rest of it, if you have not been able to disqualify already. Meet you in the next session. Thank you.