 A warm welcome to the 14th session in the fourth module on signals and systems. We are dealing with the specific class of Laplace transforms. Let us explain that class in some greater depth now. We are essentially dealing with what we call rational Laplace transforms. What do we mean by rational? Racial is essentially a ratio of two finite length series in the Laplace variable s which means ratio of a numerator finite length series in s to a denominator finite length series in s. Now, let us take an example in fact all the examples that we have been discussing all this while are really such rational Laplace transforms. So, for example, if we had something like 1 by s plus 2 into s plus 3, it would be expanded in the following way in which case we essentially have the numerator finite series is simply 1 and the denominator finite length series is s square plus 5 s plus 6. Now, of course, we say series because in principle we can have both positive and negative powers of s. Now, we have seen how to handle a rational Laplace transform. In fact, we discussed what is called the method of partial fraction expansion and in that process we can identify individual simple terms which we can invert. So, let us make a note of that. So, with partial fraction expansion in this example also, now this is the form of a partial fraction expansion and we need to fill up these numerators here. How would you fill up this term? You would essentially multiply both sides by s plus 2 and put s plus 2 equal to 0 or s equal to minus 2 and that could give us essentially 1 divided by 1 here. So, you get 1 and when we want to get this term, we multiply both sides by s plus 3 and put s plus 3 equal to 0 that means s is minus 3. That gives you 1 by minus 1 which is minus 1. So, essentially this can be written as 1 by s plus 2 minus 1 by s plus 3 that is easy to check as well. Now, this is only the expression for the Laplace transform. There are three possible regions of convergence. Let us write down each of these regions of convergence and the corresponding inverse Laplace transform. So, we will make a table, region of convergence, inverse Laplace transform. The first region is where real part of s plus 2 is less than 0 or real part of s is less than minus 2 and of course, thereby well when real part of s is less than minus 2, we have to worry about what is happening to minus 3. We could take real part of s plus 3 also to be less than 0 in which case we have real part of s less than minus 2, real part of s less than minus 3. Essentially this means real part of s is less than minus 3 that is the first region. Now, in this region the inverse Laplace transform, now we know the expression, the expression is common. The inverse Laplace transform looks at both terms and takes the left sided corresponding sigma. So, 1 by s plus 2 would lead to minus e raised to the power minus 2 t u minus t that is important u minus t and the same can be done for 1 by s plus 3 with the negative sign. And therefore, the inverse Laplace transform is e raised to the power minus 3 t minus e raised to the power minus 2 t times u minus t. Now, let us take the other two regions as well. The second region is where the real part of s lies between these two possibilities minus 2 and minus 3. So, it is so to speak to the right of minus 3 and to the left of minus 2. And therefore, the corresponding inverse Laplace transform would be minus e raised to the power minus 2 t u minus t minus e raised to the power minus 3 t u t. And finally, you have the third region where of course, you want it to be greater than real part of s to be greater than minus 2 and real part of s to be greater than minus 3 which can be combined into one condition real part of s is greater than minus 2. And there the inverse Laplace transform would be e raised to the power minus 2 t u t minus e raised to the power minus 3 t u t. So, in fact, if you look at these three possibilities that we have here, in the first region of convergence, the overall signal is also left sided. It is all a u minus t here. So, it is left sided. In contrast, when I look at the second region, it is infinite in extent on both sides, essentially it is both sided so to speak. So, it has matter on the left and right, not confined to left or right. And finally, this is right sided. So, this gives us some insight into the kinds of region of convergence that we can have. You could have a region of convergence which is infinite on one side, either on to the left side or to the right side. And then we have one kind of signal associated or we could have a region of convergence which lies in a strip between two vertical lines. And then there is a neither left nor right. It is essentially it is double sided. It extends infinitely in both directions. Of course, if you have multiple factors with essentially the same factor being raised to a higher power. So, for example, if there are repeated roots of the denominator, we know how to deal with them. So, we deal with them by expanding as many terms in the partial fraction expansion as the amount of repetition. So, example, if you had a factor of the form in the denominator, say s plus 3 to the power 3 with all other factors, the corresponding partial fraction expansion terms would look like this. And we know how to deal with these. We can obtain these, this all these are constants, you know the ones which I am showing here in brackets. And these constants can be obtained by successive differentiation. We have explained this, we talked about partial fraction expansion. I definitely recommend that you review some of your ideas of partial fraction expansion, something we learn in high school and one should review it for understanding the Laplace transform. Now the issue is why are we so keen on talking about rational Laplace transform? To answer that question, we would now like to put down one more property of the Laplace transform first. And this property is in some sense a dual to what we have been discussing, you know we have been talking about differentiating the Laplace transform. But now let us talk about differentiating the function, let us see what would happen. So, if I differentiated the function, how would I obtain the corresponding Laplace transform? In fact towards that objective, you would first need to put down a method to invert the Laplace transform formally. So, formal inversion, formal means a clear formula. In fact you can call this a property too, it is indirectly a property of the Laplace transform. You see what was the Laplace transform? It was essentially multiplying a signal by a decaying exponential where you have chosen sigma appropriately and then taking the Fourier transform. This would give you the Laplace transform. In fact, this tells us how you would invert. All that you need to ensure is that this s lies in the region of convergence for all omega which means the easiest thing to do is to take a vertical line in the region of convergence. That is the simplest thing to do. Take a vertical line in the region of convergence, strictly inside. Now the region of convergence would include many such vertical lines and the line would have a fixed sigma. Take the inverse Fourier transform first. So, you know essentially you are just trying to invert step by step. You know if you look at what you have done to go to the Laplace transform, we are saying go through the steps backwards. So, go backwards that is what we are saying. Take the inverse Fourier transform and then divide by erase the power sigma t because you have a fixed sigma here that is what we are saying. So, let us write that down. Take the inverse Fourier transform and then divide by erase the power sigma t. Remember erase the power of sigma t is never 0, so admissible. This division is admissible. Now how would you write this term? We need to write it down formally. So, let us do that. So, let X t have the Laplace transform capital X of s and we will go through these two steps. We will first take the inverse Fourier transform. You need to integrate after multiplying by erase the power j omega t with respect to omega. Omega would run from minus to plus infinity and you need to divide by 2 pi. Now after that you need to multiply this by erase the power minus sigma t. So, you know multiplication by erase the power minus sigma t is essentially division by erase the power sigma t. Now you know there is one little observation that we have to make. What we have called s here? You know you remember we said we multiply X t by erase the power sigma t and then take a Fourier transform. Let us expand this. Erase the power sigma t, erase the power minus j omega t dt integrated. So, here the s that we have taken I hope many of you must have noticed that the s which we have taken is actually you see you would have taken X t erase the power minus s t. If you want to take the Laplace transform, erase to the power minus s t dt. This is what the Laplace transform would have been. So, minus s has the value sigma minus j omega. So, s has the value minus sigma plus j omega. This little observation should be made at this step. When I mentioned it in the beginning, I was a little careless about the choice of s. In principle it was okay, but then this little change needs to be made. So, if we make that change, now what is the interpretation of this expression here? What are we doing here? Let us write it down. The s remember is this minus sigma plus j omega not sigma plus j omega. Now, you know if you look at it carefully, this integral is with respect to omega. So, this is essentially a product with a function dependent only on t can be taken right inside. So, you can push this inside. This is what we get and this is identified all as s. This is s and this is also s. The only problem is that we have a d omega and not a d s. But what is s really? s is minus sigma plus j omega. So, if you were to write d s, sigma is a constant. And therefore, d s is simply j d omega, where upon d omega is 1 by j d s. So, we could rewrite this as 1 by 2 pi j integral x s e raised to the power s t d s. And s is on the vertical line s equal to essentially minus sigma plus j omega in the r o. We will see more in the next session.