 What we have been strategically doing all this while is taking simple examples of the rational Z transform where we are immediately able to decompose into partial fractions. But the trouble comes when you for example have a double pole at a point, a repeated pole. Let us take an example. In with a simple example of just one repeated pole and that is it. So you have 1 by 1 minus half z inverse the whole square and of course you have two possible regions of convergence. You know there are, there is only one singularity here that is half. So you could either have this as the region of convergence or you could have this as the region of convergence to mod z less than half or mod z greater than half and we need to be able to find the corresponding sequence for each of these regions of convergence. Now we invoke here the property, we can do it in two ways. We can either treat 1 by 1 minus half z inverse the whole square as a product of two z transforms each of which is 1 by 1 minus half z inverse. But you know this is specific to this case. Suppose you had 1 by 1 minus half z inverse the whole to the power 6, what exactly would you do? You know so we need to evolve a general strategy and if you take a product of z transforms you are going to start convolving sequences, convolution is in general a more complex operation than what we are going to do. We will invoke the property of taking the derivative of the z transform. So we invoke the fact that minus z times dxz dz corresponds to n times xn where the z transform of xn is xz, we invoke this property now and we take the specific xn equal to half raised the power of nu n where upon we know what the z transform is, we have dealt with the z transform before. So xz would then be 1 by 1 minus half z inverse with mod z greater than half of course we can also take xn to be minus half raised the power of nu minus n minus 1 where upon xz would be the same expression but with mod z less than half as the region of convergence. So we will keep both of these in mind, it is not really terribly a cause of worry whether we have this region of convergence or this region of convergence after we take the derivative. So let us take the derivative, so the derivative of xz as you see here is minus 1 1 minus half z inverse raised to the power minus 2 times minus half minus 1 z raised to the power minus 2 is that correct? Minus 1 1 minus half z inverse squared or this raised to the power minus 2 and then the derivative of what is inside that is minus half times minus 1 z raised to the power minus 2. So if you multiply this by minus z, you get minus z in the numerator as well. So let us simplify this, this is very clearly, you have 4 minus signs that makes it a plus, you have half z inverse 1 by 1 I mean divided by 1 minus half z inverse squared what you get, you see what we have here is what we wanted except for the factor of one, except for the factor of half z inverse in the numerator. Now having a half is not a problem, you can remove the half, you know I said the z transform is a linear operator so that is not a problem, you can just remove the half by multiplying by 2. What do you know about a z inverse? When you want to remove a z inverse, you need to multiply by z, multiplication by z is equivalent to advancing the sequence by one step that means replacing n by n plus 1. So it is very clear to us that n times half to the power of n, u n would have the z transform half z inverse divided by 1 minus half z inverse the whole squared with the region of convergence mod z greater than half, this follows in a straightforward way from what we have done so far and you know in continuation n times half raised to the power of n, u minus n minus 1 negative of this as the z transform same thing but mod z less than half, simple. Let us call these sequences, let us give these sequences a name, so let depending on the region of convergence either you have the first or the second sequence for half z inverse by the same denominator. So let us say for either case of the z transform expression half z inverse divided by 1 minus half z inverse squared, let us call the sequence what we are saying is you see the strategy is the same whether you have the first region of convergence of the second, the strategy is multiply by 2 and then advance. So for 1 by 1 minus half z inverse the whole squared the sequence becomes 2 times g n plus 1, simple. Now this is a general strategy that we can use, in fact I leave it to you as an exercise to prove the following or rather to work out the following. So I leave it to you as an exercise, use inductive logic, inductive reasoning and the forth the strategy just gone by the inverse z transform of 1 by 1 minus alpha z inverse the whole to the power m, m is a positive integer and for 2 cases, case 1 mod z greater than alpha, case 2 mod z less than alpha of course, mod alpha if you please. Now you see once we have done this we are more or less in good shape for dealing with any sequence any z transform that has multiple poles in the denominator. For example, suppose we happen to have multiple poles, suppose we had a triple pole at alpha equal to half. So you have a term of the form, let me write it down. So consider for example, say 1 minus half z inverse plus 2z to the power minus 2 divided by 1 minus 1 third z inverse the whole to the power 3. Now you can always rewrite this, you can rewrite it as 1 by 1 minus 1 third z inverse to the power 3 plus minus half z inverse by the same thing plus 2z to the power minus 2 by the same thing. And if you know the inverse z transform for this with the given region of convergence, you know it for this as well, all that you do is multiply that inverse z transform by minus half and then delay it by 1 step. You know it for this 2, multiply that inverse z transform by 2 and delay it by 2 steps. So you see, when you have a partial fraction expansion now, you have to be careful. If you have multiple poles, a partial fraction expansion needs to allow a degree less than the number of repetitions of that pole or number of occurrences of that pole in the numerator just less than. So a term, typical term in the partial fraction expansion, when you have a triple pole as you have seen at z equal to 1 third would have a second degree term in the numerator.