 A warm welcome to the 15th session of the fourth module in signals and systems. We have now embarked on the discussion of a new property of the Laplace transform. In some sense, the dual of what we have been discussing in the last few sessions. We took a derivative of the Laplace transform and we asked ourselves what would happen in the time domain and it was simple. Taking a derivative in the Laplace domain in terms of s was equivalent to multiplying by the time variable in the time domain. So let me put down this property once again formally so that we can now put down the property that we are going to talk about with reference to it. So the differentiation property as we have learned it. If x t corresponds to capital X of s with the region of convergence r x, what we have answered is what happens when we take the derivative. So d x s d s in fact we have said that d x s d s we have taken a derivative in the Laplace domain with almost the same region of convergence except possibly for the extremities. The derivative corresponds to minus t times x t that is the property we have learned. So in some sense derivative in one domain corresponds to multiplication by the independent variable in the other domain. Here you know the independent variable is t for time. So the question that we want to ask is is there a dual of this property that is what we want to now answer. What we mean by that is can we reverse the roles of the Laplace variable and the natural domain independent variable. Now to answer this question we have to take recourse to the formal inversion of the Laplace transform that we talked about in the previous session. Let us review that formal inversion. We said that essentially if you have a Laplace transform x of s you multiply x of s with e raised to the power of s t integrate with respect to s. But let this integral go from any you know essentially on a vertical line in the region of convergence. So how does a vertical line in the region of convergence look essentially let us sketch it. So if you had this region of convergence this is the s plane and you had some line beyond which or before which you had a region of convergence. So let us say this is the region of convergence the one which I am shading in red here. What we are doing essentially is to choose say a line like this, this green line and you would essentially move from the lower most or the bottom most end of that green line to the top most this way. That is what you mean by this integral with respect to s. Remember you know in the complex plane you integrate with respect to a contour. So here the contour of integration is the vertical line in the s plane in the region of convergence. So you could think of it as sigma minus j infinity to sigma plus j infinity that is one way to understand it. So you could say integral from sigma minus j infinity to sigma plus j infinity provided sigma brings you into the region of convergence. And this needs to be divided by 2 pi j that is a matter of constants. So this is called the formal inverse that means it is a formula which tells you the inverse of the Laplace transform x of t is equal to this. Now this formal inversion is useful not for inversion it is interesting it is inversion is not very conveniently done by this formal inversion sometimes it might be. You have seen that inversion is actually done more strategically by using experience. So we saw a few typical examples of a Laplace transform and you could use that experience to invert. This formal inversion gives you many insights into the Laplace transform and one of them is what we are now going to derive. So the formal inversion tells us that x of t is essentially 1 by 2 pi j integral on a vertical line in the region of convergence capital X of s e raised to the power of s t d s and let us now take the derivative of both sides as we did previously too except it was done in the Laplace transform not in the inverse Laplace transform. So d x t d t would be d d t for all this business. Let us call this vertical line in the ROC as capital L. Now this operator d d t would act only on this expression and therefore if this is meaningful we can take the d d t into the integral and let us do that now let us take the d d t into the integral here. What would you get? You would get d d t of x t is 1 by 2 pi j integral over the same L provided of course there are some conditions we will see integral over the same L x s d d t e raised to the power of s t d s. So this says something very interesting it says that if I look at the right hand side this is the formal Laplace inverse of s times x s. And therefore if you look at it what we are saying is that d x t d t has the Laplace transform s times x s and hopefully L is included. So you know multiplication by s is normally not going to change the region of convergence too much. So in most contexts of the application of this property here we are talking about the same region of convergence. Again it is the extremities which could be an issue but keeping those little nitty gritty details aside let us say essentially that you have chosen a line in the region of convergence that line carries over to the region of convergence when you replace s x of s by s times x of s which happens for most of those contours except possibly for some extreme contours. So with this then what do we have? We have now a kind of dual you have multiplied by the independent variable in the Laplace domain here that is s, s times x s and it has corresponded to taking a derivative with respect to the variable in the other domain this is a kind of dual. So this gives you what is called a dual differentiation property it says taking a derivative in time is equivalent to multiplication by s and r x remains almost the same. So this in fact gives us now an insight into why the rational Laplace transform is so important every time you take a derivative in time you are multiplying by s. The other way of looking it is every time you are multiplying by s you are taking a derivative in time if you multiply once by s it is the first derivative in time if you multiply once again by s it is the second derivative in time and this can continue. So in fact now let us take an example what does a rational Laplace transform auger for the system which it is describing let us take an example suppose we had h of s this is the system function given by s plus 3 divided by s plus 2 into s plus 4 we can simplify this and now we can interpret this y of s is essentially x of s times this system function. So let us expand that let us multiply out by cross multiplication you get s square plus 6 s plus 8 into y s is equal to s plus 3 times x s and now we can interpret this using the dual differentiation property we have the second derivative of y t plus 6 times the first derivative plus 8 times y t itself is equal to the first derivative of x t plus 3 times x t this is essentially obtained term by term now this is essentially what is called a linear constant coefficient differential equation and do you remember at the end of the first module in signals and systems that I told you it is this class of linear shift invariant systems which is most typical in natural engineering systems. Now we see the rational Laplace transform playing a very important role it is essentially the fact that this kind of a linear constant coefficient differential equation describes many natural engineering systems not just electrical engineering system mechanical engineering system civil engineering systems aerospace systems so many others many systems can be described or at least approximated in a certain region of operation by using linear constant coefficient differential equations and the rational Laplace transform is a very appropriate tool to understand these linear shift invariant systems that is why we are so keen on looking at the rational Laplace transform. Now we need to complete one detail we said a rational Laplace transform is a ratio of finite series in s and in a series you can have both positive powers of s and negative powers of s what happens when you have a negative power of s we need to complete that detail so let us do that what happens when we have a negative power of s let us take the same dual differentiation property again it says that if x of t has the Laplace transform capital X of s with the region of convergence rx then dx t dt has the Laplace transform s times x s with essentially the same rx so how do we go from here to here now we want to go the other way you essentially want what you call a division by s a negative power of s so you would go from here to here by using what is called indefinite integration or more appropriately by what is called a running integral what is a running integral let us see essentially we are saying that if now we will just change the variables the ways you know can become confusing so if g of t has the Laplace transform g of s with the region of convergence rg then integral running integral this is essentially the running integral of g the running integral of g has the Laplace transform 1 by s times gs or s inverse gs and essentially it is the same rg but perhaps with some problems with the boundaries in fact just go back and see what I am trying to say let us interpret it in that light so we are saying to go from here to here you would essentially note that x of s is s inverse times s times x of s and this is what we are calling g of s now that is what you should interpret this to mean so this is g of s here and s inverse g of s corresponds to the running integral essentially the inverse operation of difference here we shall see more about this in the next session. Thank you.