 A warm welcome to the 21st session of the fourth module of signals and systems. In the previous session, we have been looking at the differentiation property of the Z-transform and we have been looking at rational Z-transforms in some detail. In this session, we will continue with that discussion. So, we look at rational Z-transforms where we have repeated factors now. So, let us take an example first. Let us consider, for example, the Z-transform h of z is 1 by 1 minus half z inverse square into 1 minus 1 third z inverse with mod z greater than both of them. There is mod z greater than half and mod z greater than 1 third, which means mod z needs to be greater than half, which becomes the region of convergence. Now, the question is to find the inverse Z-transform of this expression with its region of convergence. How do we go about it? So, the first thing is to decompose into partial fractions and how would the partial fractions for this expression look? h z would be of the form sum a 1 divided by 1 minus half z inverse. So, this is the form we want. We want each factor to be explicit in all its powers. So, for example, you will see here that the repeated factor occurs with all its powers, a power of 1 and a power of 2. So, of course, now if you wish to make a partial fraction decomposition, we must equate this. So, you equate it to 1 by 1 minus half z inverse squared into 1 minus 1 third z inverse and let us find the easier ones first. So, we can find a 2 1. How do we find a 2 1? We will multiply both sides by 1 minus half z inverse the whole squared and then put z equal to half. So, that can be done. Let us do that. You would of course get 1 divided by 1 minus 1 third and z is half so that z inverse would be 2. So, 1 minus 2 third that makes it 1 third in the denominator and therefore, a 2 1 is equal to 3. Similarly, we can find a 2 and I will show you the working in green here. A 2 can be found by multiplying both sides by 1 minus 1 third z inverse then putting z equal to 1 third. So, we can do that, multiply both sides by 1 minus 1 third z inverse and then put z equal to 1 third that leaves you with 1 minus 1, 1 by 1 minus half into z is 1 third so z inverse is 3. So, we have this squared 1 minus 3 by 2 the whole squared and that is minus half squared so that is 1 fourth in the denominator. So, a 2 becomes equal to 4 and now there are different ways of finding a 1 1 once you have done this. So, let us put down these two first. We have h z here is already we have calculated most of it a 1 1 minus divided by 1 minus half z inverse plus we have calculated a 2 1 to be 3 and a 2 to be 4. Now, one easy way to find a 1 1 here simply subtract. So, you can do a subtraction 1 by 1 minus half z inverse minus these two term and here when we subtract we should see. So, in fact, the exercise which I am going to give you is to complete this work. Now, there is another way in which you can find a 1 1 let us explain that too and this is more general. So, we have h z is of this form we can multiply both sides by 1 minus half z inverse squared. Now, we can differentiate both sides with respect to z or rather with respect to z inverse. So, treat z inverse as a variable here and now what we do essentially see when you do that a 2 1 disappears and d d z inverse of 1 minus half z inverse into a 1 1 is minus half a 1 1 and then of course, here when we differentiate this with respect to z inverse it would still retain the 1 minus half z inverse factor. So, now when we put z inverse or rather 1 minus z inverse equal to 0 we would get minus half a 1 1 is essentially d d z inverse into 1 minus half z inverse h z all this evaluated with 1 minus half z inverse equal to 0 or z equal to half or z inverse equal to 2. So, we can find a 1 now the exercise for you is also to verify that you get the same answer. In fact, this method is more general because if you have a higher power of the factor 1 minus half z inverse for example, suppose you had 1 minus half z inverse to the power 3 for example, let me show you just the important steps. So, suppose we had you had a third power of 1 minus half z inverse in denominator then of course, you recognize that this would have three terms it would have the term of the form a 1 1 by 1 minus z inverse plus a 2 1 plus a 3 1 and then other terms. Now, again we can multiply both sides with 1 minus z 1 minus half z inverse cubed. So, if I multiply both sides here I multiply this by 1 minus half z inverse cubed then of course, this leaves just a 3 1 here this leaves you with a 2 1 into 1 minus half z inverse and this leaves you with a 1 1 into 1 minus half z inverse whole squared and now we can successively differentiate when you successively differentiate you would get the terms 1 by 1. So, for example, without any differentiation you get a 3 1 with the first differentiation you would get a 2 1 and with the second differentiation you would get a 1 1 that is how we could proceed anyway. This was meant to illustrate how we could make a partial fraction decomposition and just reviewing these ideas because you need to use them again and again. Now, having done that what is the next step the next step is to apply the differentiation property again and again. So, for example, you know how to deal with 1 by 1 minus half let us do that we know how to deal. Now, if you differentiate this with respect to z you would get of course, minus 2 and this becomes a power of 3 the denominator and then minus half there you are. So, now I take minus z times d dz. So, here of course, there is a minus 1 here too. So, it would be 2 times half times z to the power minus 1 because you are multiplying by minus z there are 4 negative signs and 1 minus half z inverse the whole cube. Now, this would correspond this overall operation here. So, this minus z d dz would correspond to multiplication by n and therefore, you could now reason out what happens when you have just 1 by 1 minus half z inverse cube. You know there is a z inverse factor here which you have to take care of. If you do not want the z inverse factor you need to advance the sequence by 1. So, essentially you are multiplying by z. So, you are advancing the sequence by 1 it is a shift is not it we have seen that before if you shift a sequence then it amounts to multiplying the z transform by a power of z an appropriate power of z integer power of z. Anyway this was to review how we could invert a rational z transform. So, now we have all the tools in fact, we were already experienced in inverting rational Laplace transform we have now seen how to invert a rational z transform. I shall say a little more about rational z transforms in the next session. Thank you.