 A warm welcome to the fifth session in the fourth module of signals and systems. We have been looking at the definitions of two very important general transforms in the last few sessions, the Laplace transform and the Z transform. The Laplace transform deals with a more general class of continuous independent variable systems and the Z transform deals with the more general class of discrete independent variable systems. What we would like to do in this session is to consider a few examples of the Laplace and the Z transforms for different continuous independent variable signals and discrete independent variable signals. So, let us take a few examples. Let us for example, consider the continuous independent variable signal x of t given by e raised to the power of 2 t u t plus e raised to the power 3 t u t. Now, when we find its Laplace transform, it would be given by integral of x t into e raised to the power minus s t d t over all t and of course, the integral is linear. So, when we look at this integral, it can be split into two integrals which can each be simplified. Now, each of the integrals has its own condition for convergence. So, if you look at this integral, it would converge if and only if the real part of 2 minus s is less than 0 or that means, the real part of s is greater than 2. In contrast, this would converge if and only if the real part of 3 minus s is less than 0 meaning the real part of s is greater than 3. If you want both of them to converge at once, both of these must be satisfied and therefore, we can now put down the condition for convergence of this Laplace transform to be the intersection. The real part of s greater than 2 intersection with real part of s greater than 3 which essentially is real part of s greater than 3. So, we can put down the region of convergence in the s plane as follows, you have sigma equal to 3 here. We draw a vertical line passing through sigma equal to 3 of course, with omega equal to 0 and we identify this region, this is the region of convergence. Now, what is the Laplace transform expression? We need to specify that too, here we have already specified the region of convergence, but we also need to specify the specific expressions, we need to evaluate each of these integrals. So, we have to evaluate this integral and this integral and that is very easy to do. You notice that if the region of convergence is where we are operating, then we can use a very standard result, it is of the form e raise the power 2 minus s into t dt which is a very simple indefinite integral to evaluate, it will just be e raise the power 2 minus s into t divide by 2 minus s and the same thing would hold for e raise the power 3 minus s dt. So, let us write down the expression of the Laplace transform and here of course, in the region of convergence, both of the terms at t equal to or t tending to infinity that is the more correct way of saying it, both terms go towards 0 and therefore, what remains is minus 1 by 2 minus s plus minus 1 by 3 minus s which is of course, 1 by s minus 2 plus 1 by s minus 3. So, this is the Laplace transform, this is the Laplace transform expression. Of course, we can simplify this in various ways, for example, we are perfectly at liberty to combine these terms. The Laplace transform of e raise the power of 2tut plus e raise the power 3tut is 2s minus 5 in the numerator divided by s minus 2 into s minus 3 in the denominator with the region of convergence real part of s greater than 3. Now, what is important here? We could of course, rewrite the expression in any way that we desire, but the expression has a meaning together with the region of convergence and here it will be interesting for us to see what could be other regions of convergence. In fact, here we are talking about real part of s greater than 3, but remember that came as an intersection of two regions real part of s greater than 2 and real part of s greater than 3 together. So, 2 and 3 are important values and if you took any one of them individually. So, let us look at each of the terms individually, e raise the power of 2tut it is very easy to see I leave it to you has a Laplace transform given by 1 by s minus 2 with the region of convergence real part of s greater than 2 and e raise the power 3tut correspondingly has the expression 1 by s minus 3 with the region of convergence real part of s greater than 3. So, you know we could have variants now for example, if you replaced this term here if you made this e raise the power of 2t u minus t and put a minus sign then the expression would be the same, but the region of convergence would change to real part of s less than 2 and the same thing could be argued for e raise the power of 3t u minus t with a minus sign. So, much so for the Laplace transform now let us take a corresponding example of the z transform. So, and here I could go a little faster because much of the reasoning is very similar. So, let us take x of n given by 2 raise to the power of n u n plus 3 raise to the power of n u n and let us query its z transform of course, the z transform is going to be and we can evaluate each of them individually. Now these are very easy to check for their convergence they are both geometric series and of course, you would require that both their common ratios have a magnitude less than 1 for convergence. So, for convergence we would want both need to have common ratios less than 1 in magnitude which means mod 2z inverse needs to be less than 1 and mod 3z inverse needs to be less than 1. Implying mod z needs to be greater than 2 and mod z also needs to be greater than 3. We need the intersection of these 2 as before you know in the case of the Laplace transform also. And what is their intersection of course, the intersection is mod z greater than 3 and how does this region look in the z plane? This region can be sketched like this in the z plane. You need to draw a circle center at the origin pardon my drawing, but it is essentially suggestive with a radius of 3 and the region exterior to the circle one can make similar inferences in this case. For example, suppose we were to take the other sequence other means instead of 2 raise the power of n u n you took the sequence moving in the left direction. So, minus 2 raise the power of n for n going from minus 1 to minus infinity then you know you would have a different circumstance the same thing can be argued for 3 and so on. So, I leave it to you to make those little changes in the context of the z transform. In the next session we are now going to see properties of both these transform. Thank you.