 A warm welcome to the 17th session in the fourth module of signals and systems. In the previous session, we had looked at both rational and irrational systems and we again re-emphasized why rationality was an important virtue in the Laplace transform. Now we would like to understand more about rational systems and we have given a hint of what we would do towards the end of the previous session. Let us now do that. So let us consider as I said in general rational Laplace transform. So we have a numerator series and a denominator series. Now we will first make a simplification. We shall first reduce these to polynomials in S. So we will take an example. Suppose we have something like S inverse plus 2 S plus 1 divided by S to the power minus 2 plus S to the power minus 1 plus some constant, let us say 3 plus 2 S. We shall write this as S to the power minus 2 taken common from the denominator and S to the power minus 1 taken common from the numerator and then multiply by S. So we have 1 plus 2 S squared plus S divided by again multiply by S squared in the denominator. And now of course when you have powers of S like this you can adjust them either in the numerator or denominator. This will be S squared by S. Now this is indeed of the form a polynomial in S divided by a polynomial in S. So numerator polynomial divided by denominator polynomial. To be complete let us write down these two polynomials. The numerator polynomial is going to be S squared plus 2 S to the power 4 plus S cubed and the denominator polynomial is going to be S plus S squared plus 3 S cubed plus 2 S to the power 4 simple. Now let us look at what happens when we put each of these equal to 0. So let us put the denominator polynomial equal to 0 first. We get a set of roots for example in the example that we just discussed a few minutes ago. We are talking about this polynomial and we are saying put it equal to 0. So what would we have there? You would have S plus S squared plus 3 S cubed plus 2 S to the power 4 is equal to 0. So of course either S is 0 or the rest of it. Now of course we can in principle solve this. The question is what happens to this rational Laplace transform at these points where the denominator tends to 0? Of course when the denominator tends to 0 the entire function has an infinite magnitude. Let us sketch that situation. So remember this Laplace transform is a function of S. So you could think of it as lying upon the S plane. You have built something on the S plane. Let us look at the S plane as the base on which you construct the magnitude of this function. I am drawing a 3 dimensional or pseudo 3 dimensional plot now. So this is the S plane what you see here. This is the imaginary part of S and the real part of S. And on the vertical so to speak you are going to plot the magnitude of the Laplace transform. Now the magnitude will occur only on one side. It is always non-negative. So that is why I am saying that you could think of this whole structure as lying upon an S plane. You could think of the S plane as the base on which you are putting something because you do not have anything on the other side. In fact I would go to the extent of saying you can think of the magnitude as a function of S being a tent on the S plane. So the magnitude is like a tent on the S plane. And let us assume you have some place where the denominator goes to 0. Suppose it lies here. What would happen to this tent at that place? The tent would rise infinitely towards the sky at that place. You know you have a tent all over. Let me draw that tent notionally in blue everywhere else. And at that point the tent would seem to rise infinitely into the sky at this place. In contrast wherever you have a point where the numerator is equal to 0. Suppose there is a point here where the numerator is equal to 0. What would happen to the tent at that place? Let us show it notionally in blue again. The tent would seem to come and get held at that point. The tent would touch the ground at that point. So you see what I mean? There are places where the numerator is 0 where the tent touches the ground. It is as if you know you put a nail there to hold the tent. And there are places where the denominator is 0 and the tent is actually rising towards the sky there as if held up by a long stick you know to hold the tent up you need to do two things. You need to plug the tent down by nails or by something hard at certain places. And you need to hold the tent up with poles with long sticks. So that is what these are all about. These are the poles which hold up the tent. And these the places where the denominator or rather the numerator is 0 are the 0s which pin down the tent. So it is poles and 0s together which make a rational system function. If I know the poles and the 0s I know almost everything about the rational system function except for one little thing. Let me hint at what is still left with an example. So if we knew that such a tent or rather that such a rational system function or rational Laplace transform it has 1 0 at s equal to minus 1 and 2 poles 1 at s equal to minus 3 and 1 at s equal to minus 5. What is the Laplace transform? Laplace transform would take the form where we know up to this point s plus 1 all this is fine. What we do not know here is whether it is multiplied by a constant. So now you can again think of it in terms of the tent. You know where the tent is held up by poles and where the tent has been plugged to the ground with the 0s. What you do not know is how high the tent is in between. You can scale the tent up. You know you can plug it at the same places and you can hold it up at the same places. But then you can scale the tent up and down even with those poles and 0 locations. That scaling factor is not determinable by the poles and 0s. But otherwise the rational Laplace transform is completely characterized by its poles and 0s. Well that scaling factor would only scale everything by a constant. But what we will now see is that the poles and 0s are sufficient for us to obtain the inverse Laplace transform notionally. Notionally means except to within a constant. Once you later know the constant you can of course take care of it by multiplying by that constant. So we will now talk about inverting. Inverting a rational Laplace transform. Very simple. In fact you have done it before. You just need to put down the steps. Identify the poles, distinct poles. Make a partial fraction expansion. Now before you make a partial, now here I need to put in one more caveat, one more point that has to be considered. Before you make this partial fraction expansion you must ensure that the numerator degree is less than the denominator degree, less than. If it is equal to or greater you need to take a few terms common few terms outside. So if you have n of s, let us first write down the ratio of polynomialness. So the Laplace transform can be written numerator polynomial in s. Let us call it n of s divided by denominator polynomial in s. Let us call it ds. Now of course just to review ds equal to 0 gives the poles and ns equal to 0 gives the zeros. And it is this that we have identified for this keeping in mind possible repetitions. So when I say distinct poles point number 3, each distinct pole will now give a set of terms of the form some constant. Let us call it k 0 or you can even say k l divided by s plus alpha to the power l. So you know for example, if you have a pole at s equal to minus 2 and the pole occurs 3 times, then you would have 3 such terms k 1 divided by s plus 2 plus k 2 divided by s plus 2 the squared and plus k 3 divided by s plus 2 cubed. Now each of these terms can be inverted as per the region of convergence. What I mean by that is look at whether the region of convergence is to the right of that pole or to the left of that pole. If it is to the right of the pole, give it the corresponding right sided expression, right sided time domain function, time domain signal. And if it is to the left, give it the corresponding left sided expansion, left sided time domain signal. And of course, you know how to give the time domain signal, you can use it by successfully differentiating the Laplace transform 1 by s plus alpha. So we have dealt with this before, we know how to deal with terms like this. So everything fine except as I emphasized all this works provided the numerator degree is less than the denominator degree. Let me now come to that problem. Note that the numerator degree needs to be less than the denominator degree for this to be valid. If not you have n s by d s first do a long division. So you know n s long divided by d s and carry out the long division and bring a remainder and a quotient. The quotient is in s, let us call it q of s and the remainder is in s again r of s. That means you can write down n s by d s is q s plus r s by d s. Here degree r s is strictly less than the degree of d s and this is a polynomial in s. Now we have created a new problem, how do we invert this polynomial in s? The rest of it is okay. We will have to answer this question in the next session. Thank you.