 A warm welcome to the 20th session in the fourth module on signals and systems. We have been dealing with the rational Laplace transform for a while, we have also seen an example of how to use the Laplace transform in analyzing a system in the session just gone by. I have given you an exercise to work out and now I think it is high time we looked at the Z transform for some time and then come back the Laplace transform to see more about it. But we need to have some parallels I told you right in the beginning of this module that we would like to deal with the Laplace transform and the Z transform together because we would like to build parallels. So now let us ask what happens to the Z transform when you differentiate. So let us investigate the differentiation property of the Z transform I am sure it will take you a minute to get accustomed against the new context. So we have a sequence X of n and we take its Z transform remember we have used this symbolism to denote the fact that we are taking the Z transform capital X of Z with the region of convergence R X capital X of Z with the region of convergence script R X and we know that capital X of Z is summation n going from minus to plus infinity X n Z raised to the minus n. Now let us consider the derivative D X Z D Z with essentially the same ROC except again possibly for some boundaries what is this derivative D X Z D Z is going to be D D Z of the summation we will differentiate term by term if that derivative converges and now we can multiply both sides by minus Z. So we guess minus Z times D D Z of X Z is the summation on all n n X n times Z raised to the power minus n and as you can see this is essentially the Z transform of n times X n a new sequence we have a very elegant relationship here. So therefore, if X of n has the Z transform capital X of Z with the region of convergence R X then n times X of n has a Z transform minus Z D X Z D Z with essentially the same R X essentially the same ROC except for some boundaries. Now let us take an example which take our famous exponential sequence which we have been dealing with this time we will take for variety half raised to the power of n u n of course, Z transform is 1 by 1 minus half Z inverse with the region of convergence mod Z greater than half then n times half to the power of n u n would have a Z transform given by minus Z times D D Z of this expression with essentially the same region of convergence mod Z greater than half and we can simplify this this is the entire expression and you have 1 2 3 4 negatives which can all be combined together into a positive and you have a Z and a Z to the power minus 2 which combines into Z to the power minus 1 and you have a half. So, overall you have half Z to the power minus 1 divided by 1 minus half Z inverse the whole squared with mod Z greater than half as the Z transform of n times half to the power of n u n. Notice that here you have a square of the denominator in this with some numerator added. So, now this gives us the methodology for dealing with repeated factors in the denominator in the Z transform we have gone through this process for the Laplace transform. So, we do not need to discuss all of it with the same detail the principles are similar, but what we now have to do is to recognize the presence of these factors. So, you know in some sense now suppose I just have the plane factor 1 by 1 minus half Z inverse the whole squared how would I deal with that or the other way of asking is suppose I want to write this not as a function of Z inverse, but as a function of Z how do I deal with this. What about convolution in the Z domain? What is the effect of convolution in the Z domain? Let us review that and let us now establish certain properties there. We are now equipped to deal with repeated factors. Now let us see what happens when we convolve two sequences. So, if I have X of n with a Z transform given by capital X of Z with the region of convergence R X and H of n with a Z transform given by capital H of Z with the region of convergence R H. So, now let us consider a situation X n has the Z transform capital X of Z with the region of convergence R X and H n has the Z transform capital H of Z with the region of convergence script R H and we ask what is the Z transform of X convolve with H. We know what to expect it is the product and the region of convergence is at least the intersection. Now with this property and the property that we derive for differentiation we are in a position to handle all rational Z transforms. Just as we handled all rational Laplace transforms. Let us now put down a procedure. Let us take an example. Let mod alpha be less than mod beta and let us consider a Z transform Y of Z given by 1 by 1 minus alpha Z inverse into 1 minus beta Z inverse and the region of convergence is mod Z greater than the maximum of mod alpha mod beta. I am explicitly writing it like this which is of course, mod beta here. So, what it means is I can first decompose into partial fractions. Now you know here decomposition into partial fractions can be done in two ways one way which I am writing down now. So, I can express it as a sum of two terms and I can obtain this by multiplying both sides by 1 minus alpha Z inverse and putting 1 minus alpha Z inverse equal to 0 which gives me Z is equal to alpha. So, 1 by 1 minus beta alpha inverse and I can similarly obtain this term by putting 1 minus beta Z inverse equal to 0 after multiplying both sides by 1 minus beta Z inverse and that could give me essentially putting Z equal to beta and then 1 by 1 minus alpha beta inverse. So, this is the partial fraction expansion. Let us simplify and verify. Of course, one can verify that this is correct. So, you could now invert term by term. Of course, the region of convergence here is mod Z greater than mod beta which also means and therefore, I can invert each term here. When I invert this, I would get alpha times alpha raise the power of n u n and when I invert this, I get beta times beta raise the power of n u n and therefore, the overall inverse Z transform is 1 by alpha minus beta times alpha to the power of n plus 1 minus beta to the power n plus 1 into u n. We have obtained this by partial fraction expansion. We can also obtain the same expression using the convolution property. So, I have y of Z is 1 by 1 minus alpha Z inverse into 1 minus beta Z inverse which is a product of two terms and of course, mod Z is greater than mod beta, but if this is the case, then it is also true that mod Z is greater than mod alpha for this and mod Z is greater than mod beta for this. So, for each individual term, you have the corresponding outward region of convergence and we can invert each of them separately. What does that give you? When you invert them separately, so 1 by 1 minus alpha Z inverse with mod Z greater than mod alpha gives alpha raise the power of n u n and 1 by 1 minus beta Z inverse with mod Z greater than mod beta gives beta raise to the power of n u n and y n is their convolution. Now, u of k is equal to 1 only for k greater than equal to 0. So, together the product lies only between 0 and n. So, obviously, it has no meaning when n is less than 0, the product is going to be 0 and therefore, y of n now takes the value. The u n comes from the fact that only for n greater than equal to 0 can 0 less than equal to k less than equal to n be a non-trivial interval. Now, this is very easy to sum. It is a simple geometric progression. Now, I can multiply by beta in the numerator and denominator and I can now simplify. So, therefore, it is very clear that this sum is essentially this and y of n is indeed beta to the power n plus 1 minus alpha to the power n plus 1 divided by beta minus alpha. And you will see the similarity to the expression that we derived some time ago there. So, it shows you that both the approaches lead you to the same answer, the convolution approach and the multiplication of two Z transforms approach using partial fraction expansion. We shall see more about rational Z transforms in the coming sessions. Thank you.