 A warm welcome to the 26th session of the fourth module in signals and systems. We were looking at continuous independent variable systems in the previous session where the impulse response had a Laplace transform and the Laplace transform was rational. So, essentially we are looking at rational linear shift invariant systems and we are trying to analyze whether the system is stable or not. Now, we identified the importance of the region of convergence. In fact, we saw the region of convergence was critical. And what we identified was that you need to look at two things to determine stability. You need to look at where the poles lie and where the region of convergence lies with respect to the poles. So, we said we could focus on each pole individually. Now, we also said that we should first take care of a basic disqualification if there is one. That is, if you have a system function and if you then do a long division and bring it into the form of quotient plus remainder by denominator. And if the quotient brings you integrators or differentiators into the system, there is immediate instability there. In fact, just for completeness, let me show you why the differentiator and integrator are inherently unstable. So, let us prove that. So, we can prove with a counter example. Let us consider a differentiator with input x t equal to sin of t squared. Clearly, the output is going to be 2 t times cos t squared. The mod x t is less than or equal to 1 for all t. In contrast, mod y t grows without bound in t. And therefore, a bounded input has produced an unbounded output. And therefore, the system is unstable. Now, let us take a corresponding proof for the integrator. Integrators are inherently unstable. Again, we give a proof with counter example. Let us take once again the running integrator. And let us give the unit step input to the system. Now, I will show the input in black. And I will show the output in green. The running integrator produces 0 all the while up to t equal to 0. And from there, it grows again without bound. So, at the point t equal to 1, you have essentially integrated an area of 1. So, it reaches 1 and then grows beyond without bound. In fact, you know what the output is y of t is going to be t times u t. So, you have a bounded input resulting in an unbounded output. And therefore, the system is unstable. So, you see what it means is the moment you have a differentiator or an integrator. There is inherent instability. So, we have to discard the question of stability there and then. And what is the system function of a differentiator? Let us write it down and that of an integrator true. The system function for a differentiator is simply s. And for an integrator or running integrator, of course, it is 1 by s. And of course, you can repeat this. So, if you have a cascade of differentiators. So, if you have a double derivative, for example, this would be have s squared as the system function. And similarly, you can have double integrators and so on. And all these are immediately unstable. So, let us do a wave with them. So, you know, let us assume that we do not have them and then look at the non-trivial discussion that follows. Otherwise, we have nothing to do. We do not even need to look at the quotient by the remainder and so on. That is not going to be able to overtake this instability created by the differentiator or the integrator. Now, let us assume they are not there and that means the degree of the numerator is less than the denominator degree. So, we have a remainder which is the numerator divided by denominator. And we know that we can decompose this into partial fractions. We can then invert. We have done all this. We do not have to review all the steps in gory detail, but we can invert each partial fraction term. Now, remember we are going to consider distinct poles. You see, because if we have this pole repeated, we do not call them distinct poles. We can club together all the terms corresponding to a given pole and create a poly-x term. The exponential parameter comes from the pole and the polynomial comes from the coefficients in the partial fraction expansion. Now, what we need to do is to look at any one poly-x term and see whether that poly-x term can be annihilated or can be affected by the presence of the others. In other words, if one of those poly-x terms has an exponential parameter that grows without bound, you know, and a growing x, see it is very clear. The region of convergence is either to the left of the pole or the right of the pole. If it is to the left of the pole, then you are going to get a left sided exponential term. Left sided does not immediately mean growing or decaying. It depends on the exponential term. It depends on the real part of the exponential term. For example, if the exponential term is e raised to power 2t, it would be growing to the right and decaying to the left. If the exponential term is e raised to power minus 2t, it would be growing to the left and decaying to the right. So, the question is, let us focus on one of these poly-x terms. Suppose, you discover that the exponential part of the poly-x term is a growing exponential. Forget about the polynomial part. In fact, we can show that the growing or decaying nature. So, let us put this down formally. What I am saying is, focus on one poly-x term. A simple lemma says the absolute integral of this poly-x term diverges for a growing exponential, converges for a decaying exponential. The polynomial term has no role to play. The exercise for you is to demonstrate this. So, you know, it does not matter if a pole is repeated. You only need to see whether it contributes a growing exponential or a decaying exponential. Now, the next question is, suppose you have a growing exponential coming from one of these exponential terms, poly-x terms. Can one of the other poly-x terms overcome that growth and, you know, render it null in the sense when I take the overall impulse response, which is the sum of such poly-x terms? If one of those poly-x terms is growing without bound, can one of the other poly-x terms bring it into control? The answer is no. That is the next subtle point here. So, lemma. The poly-x terms are linearly independent. That means one poly-x term cannot annihilate or overcome another. And more mathematically speaking, a linear combination of two poly-x terms cannot be identically zero. The exercise for you is to demonstrate this again. So, where are we now? We are clear that the moment you have one of those poly-x terms which grows without bound, your impulse response is going to be absolutely non-integral and the system is going to become unstable. So, let us put that down too. You are saying one exponentially growing term, poly-x term and that growing or decaying has only to do with the exponential part, not with the polynomial part. One exponentially growing term makes that system unstable. Now, when will a poly-x term grow? There are two possibilities. A poly-x term would be exponentially growing. The pole is of positive real part and the region of convergence is to the right of the pole or the pole is of negative real part and the region of convergence is to the left of the pole. Let us draw the diagram in the S-plane. So, both of these cause exponential growth. You have a pole in the left half plane, that means negative real part and then of course, the region of convergence has to be on one of the sides of the pole. So, this causes exponential growth. In contrast, if you have a pole in the right half plane, then the region of convergence being to the right of it causes exponential growth and therefore, what should you not have? Either of these, either of these red are forbidden and therefore, what should you have? If a pole is to the right of the imaginary axis, this is the imaginary axis with real part equal to 0. If a pole is to the right of the imaginary axis or in the right half plane, the region of convergence should be to the left and if a pole is in the left half plane, the region of convergence should be to the right for stability. In either case, the imaginary axis is included in the region of convergence. Do you see that? This is the beauty. For stability, the imaginary axis is included in either case in the region of convergence. So, in fact, we have a very simple condition. Look at the imaginary axis. If it is included in the region of convergence, it is stable. If it is not included, it is unstable. We will see more about this in the next session. Thank you.