 A warm welcome to the third session of module 4 of signals and systems. In the previous session, we had established a generalization of the discrete time Fourier transform for the context of discrete sequences. Just for variety we decided we would deal with the discrete case first this time. In this session, we shall look at the generalization for continuous time signals. So, now let us take again that example which we had used in the first discussion. We said that in order to be able to hold or capture this exponentially growing sequence, we would need to multiply by a decaying exponential. So, we multiply x t with e raised to the power minus sigma t and then we can of course question under what conditions this suitably modulated signal has a Fourier transform. So, when does this have a Fourier transform? Of course, we are talking about a continuous time Fourier transform. So, in other words what we are asking is when would this integral converge that is what we are asking. So, what we need to do is to write down this integral explicitly write it down in full expansion. So, let us do exactly that. Now, clearly for this integral to converge there is only one way. We have to ensure that the exponential part the exponentially changing part not the complex exponential which rotates not the phase of part. So, what we are saying here is this determines convergence. This only has a magnitude of 1 and of course, when would this particular term allow convergence when this is a decaying exponential for its domain of existence. So, e raised to the power 1 minus sigma t needs to be a decaying exponential in its domain and what would that mean? That essentially tells us that 1 minus sigma needs to be less than 0. So, simple. So, that e raised to the power 1 minus sigma t is a negative power of t. Remember, we are taking t going from 0 to plus infinity that is why we need it to be a negative power of t for convergence. So, sigma greater than 1. Now, again let us go back to the expression that we wrote in the previous part of this discussion. So, let us focus our attention on this expression as we wrote it. What are we doing here? We could think of sigma plus j omega as an overall complex constant s. Sigma is the real part of the complex constant and omega is the imaginary part of the complex constant. So, in effect what we are saying is you are writing an expression integral minus to plus infinity x t e raised to the power minus sigma t into minus j omega t which can be combined into e raised to the power minus s t where s is a complex constant equal to sigma plus j omega. This is the real part and this is the imaginary part and the real part is what determines convergence. The imaginary part has no role to play. In fact, the imaginary part is essentially a calculation of the Fourier transform. So, this amounts to calculating the Fourier transform of x t into e raised to the power minus sigma t. Now, let us calculate this quantity for this particular case. For this particular case integral 0 to infinity e raised to the power 1 minus sigma t into e raised to the power minus j omega t is convergent. In fact, provided as we said sigma is greater than 1 and you could write it in terms of s. Now, at infinity this term goes to 0 provided sigma greater than 1 and of course, at 0 it takes the value 1. So, essentially this evaluates to 1 by 1 minus s again sigma greater than 1. This is important. You cannot forget this requirement. So, here again if you look at it from the point of view of s as a complex number and then put down s in the complex plane, what are we talking about? Let us sketch the situation. So, again we have the complex s plane with its real part here, imaginary part here. You know the real part is sigma, the imaginary part is omega and this is sigma equal to 1. So, all complex numbers whose real part is equal to 1 are along this vertical line. Let us draw that vertical line in red and therefore, the real part of s or sigma being greater than 1 as required essentially means the region shown shaded in red here. In fact, this becomes what we call the region of convergence. Now, again region of convergence of what? We have not yet given this quantity a name. We have calculated it. Now, for the discrete case when we did something like this, we gave it a name based on the name of the complex variable. So, we called it the z transform. We could of course, call this the s transform, but that is not what is done commonly. It is named with respect provided and given to a very well known mathematician or scientist as you might call him Laplace. I am not quite sure how they pronounce it in his original mother tongue or in his original land, but that is how it is often pronounced in several parts of our country and I believe in some other countries too. So, Laplace. So, this transform that we encounter here is called the Laplace transform. So, we say the Laplace transform. These are all different ways of saying it real part of s greater than 1 or sigma greater than 1 or real in bracket s greater than 1. And once again here you might wonder why I have talked about the region of convergence in addition to the expressions. The Laplace transform is never just an expression. It is always an expression with a region of convergence. And like in the z transform, I wish to make it clear now why you need it, why you must put down that region of convergence along with the expression. And towards that objective, let us look at what happens if you just took the exponential going on the other side. So, you have this e raised to the power of t and you have restricted it to t greater than equal to 0. Suppose you restricted it to t less than equal to 0, what could happen? What would the Laplace transform look like? Let us find out. So, I am saying suppose you had this signal. This is t. This is t equal to 0. It starts at 1 for t equal to 0 and decays backwards. Essentially Laplace transform would be if we call this x 1 t in the spirit that we had for the discrete sequence. Then it is x 1 t multiplied by e raised to the power minus s t dt. And of course, this would be simplified because this only needs to be calculated from minus infinity to 0. Let us expand it. Everything else is the same here except that you now need this to be substituted. So, when we substitute t tending to minus infinity, we need this quantity to vanish. And when will that happen? That will happen if real part of 1 minus s is greater than 0, not less than 0. In other words, real part of s is less than 1. Very simple. So, you would have essentially a 0 corresponding to this limit. And when t is equal to 0, you get 1 here. So, in total what are we saying? We are saying that the Laplace transform of e raised to the power of t u minus t is 1 by 1 minus s with the real part of s being less than 1. By the way, you know, you will realize that I need to make a little correction here in the expression that I had derived for the Laplace transform of e raised to the power t u t because I had to worry about the sign. So, let us make that little change now. When you go back here, you will realize that here the 0 is at the bottom. So, therefore, when you put t equal to 0, that term would be subtracted. So, the only little change that we need to make here is that this is minus 1 by 1 minus s, which is essentially 1 by s minus 1, a small change, not a very serious 1. But now, we can make this correction and compare. Here, we have essentially the same expression, the negative of that expression. But the region of convergence is real part of s less than 1. How does it look? It is the inverted region here. So, the remaining part of the s plane from what we had previously. The same expression, essentially the same expression, but the opposite region of convergence and a totally different signal. We will see more in the next session. Thank you.