 A warm welcome to the 15th lecture on the subject of discrete time signal processing and its applications. And we proceed in this lecture to say more about the rational Z transform, how to invert it, how to deal with systems whose impulse response has a rational Z transform. What we can say about systems with the rational Z transform for the impulse response and so on, right. So we are going to spend quite a bit of time on the rational Z transform because that occurs very frequently in our use of LSI systems. Let us recall a couple of points that we talked about in the previous lecture. So in the previous lecture, we had talked about the rational Z transform. Let me just recapitulate what it means. The rational Z transform is a Z transform which is a ratio of two finite series in Z. An example can be 1 minus half Z inverse divided by 1 minus 1 fourth Z inverse 1 minus 1 third Z inverse. This is an example of a rational Z transform. Now I must always emphasize that a Z transform has associated with it a region of convergence. A Z transform is always an expression with a region of convergence. It is incomplete without either. So for example, if we look at this particular Z transform here, 1 minus half Z inverse by 1 minus 1 fourth Z inverse into 1 minus 1 third Z inverse, we have different possible regions of convergence based on the singularities. In fact, if you look at the denominator here, there are two singularities in the denominator. One singularity at Z equal to 1 fourth and one singularity at Z equal to 1 third. Now the points of singularity cannot be included in the region of convergence. So what we need to do is to identify the possible regions of convergence by excluding these points of singularity. Let us see how to do that. So we have the points of singularity to be 1 fourth and 1 third. Let us mark them in the Z plane. Use the following convention now. We will use crosses to denote the poles. Recall that the poles are the points where the denominator becomes 0 and we will use circles to denote the zeros. So we have two poles here and there is a 0 at the point Z equal to half. This is called the pole 0 plot for the Z transform that we saw a minute ago. Now these are singularities. The poles give you the singularities and therefore the regions of convergence must exclude these singularities. What possible regions of convergence can we have? Let us mark them. You have to identify concentric circles. So this is the Z plane again, complex plane. You have 1 fourth here, you have 1 third there. Let us draw concentric circles with the centre of the region passing through each of these singularities. If you want simply connected regions, that is if you have regions where any two points can be joined by a line completely within that region, then there are only three possible simply connected regions that these two concentric circles can give. Let us mark them with different shadings. So we have one like this. This is one possible region of convergence. The other like this, the third one like this and just marking it externally, you know. Let me write these same regions of convergence down analytically. So we have the three possible regions of convergence. Mark Z less than 1 fourth and that is essentially the region where we have shaded like this. Mark Z between 1 fourth and 1 third and that is the region that we have shaded like this. Finally, Mark Z greater than 1 third and that is the region that we have shaded like this. But that each of these regions of convergence would give different sequences corresponding to the same expression. In fact, we will see in a minute how we can use the partial fraction decomposition strategically to arrive at these three different possible sequences here. There we go. So we have 1 minus half Z inverse divided by 1 minus 1 fourth Z inverse times 1 minus 1 third Z inverse can be decomposed by partial fraction expansion into two terms. To obtain this term, I multiply this Z transform by 1 minus 1 fourth Z inverse and put Z equal to 1 fourth. So I get 1 minus 4 by 2 divided by 1 minus 4 by 3 and here I get by multiplying by 1 minus 1 third Z inverse and putting Z equal to 1 third 1 minus 3 by 2 divided by 1 minus 3 by 4. Of course, you can evaluate these. So you know let me call this whole thing. I do not really, I am not terribly worried what it is. Let us call it A and let us call this whole thing B. So you see we have two terms now. We have A by 1 minus 1 fourth Z inverse plus B by 1 minus 1 third Z inverse. And let us take each region of convergence in turn. So let us take mod Z less than 1 fourth. If mod Z is less than 1 fourth, of course mod Z is also less than 1 third. So both of these terms would need to be expanded by using powers of Z and not Z inverse, is not it? So you have to rewrite these. The strategy is when you have a region of convergence, look at how it behaves with respect to each term. Here you have two terms. Mod Z less than 1 fourth behaves the same way for both the terms. If mod Z is less than 1 fourth, it is also less than 1 third. So you know you do the same thing to both the terms in this particular case. So you have A times Z divided by Z minus 1 fourth plus plus B times Z by Z minus 1 third. Which you can also rewrite as minus A Z by 1 fourth minus Z plus minus B Z by 1 third minus Z. Or if you like, you could even multiply numerator and denominator by 4. So let us rewrite this. So you have minus 4 A Z by 1 minus 4 Z plus minus 3 B Z by 1 minus 3 Z. And the key is because mod 4 Z is less than 1 and mod 3 Z is also less than 1, you can expand these two terms in the following way. This becomes minus 4 A Z times summation n going from 0 to infinity 4 Z to the power of n plus minus 3 B Z summation n going from 0 to infinity 3 Z to the power of n. And from here, we can obtain the inverse Z transform. We can obtain the sequence. Compare terms, compare powers of n. So what you need to do is to look at the general expression for the Z transform and compare powers of Z raise the power of, powers of Z raise the power of n in both of them. So I will leave it to you to complete this. In fact, I shall write down the answer and leave it to you to prove it. So I have exercise, show that the sequence is minus A 1 fourth raise the power of n u minus n minus 1 minus B 1 third raise the power of n u minus n minus 1. Now I will take the second possible region of convergence and there I am going to have a little bit of trouble. You see the second possible region of convergence is not said between 1 fourth and 1 third. Now this is a little tricky. Well this is the case of course the partial fraction expansion does not change that holds. So you have A by 1 minus 1 fourth Z inverse plus B by 1 minus 1 third Z inverse but then mod Z is greater than 1 fourth but mod Z is less than 1 third. So there is a different way by which we treat this term and this term. So here you would have to do a different kind of expansion. You would have to expand it by keeping the powers of Z inverse in the first term. So we would have to expand A into 1 fourth Z inverse to the power n summed n going from 0 to infinity plus. Now here I will take minus 3 B Z times summation n going from 0 to infinity 3 Z to the power of n. And from here we will be able to recognize by comparing term by term. Interestingly this term would give us positive powers of Z inverse. This would give us negative powers of Z inverse. Positive powers of Z or negative powers of Z inverse. Positive powers of Z inverse or negative powers of Z. However you like to look at it. But in the expression for the Z transform the positive powers of Z inverse give us the samples at positive n. And the negative powers of Z inverse give us the samples at negative n. So there we go. Again we leave it to you as an exercise to recognize these terms and show that they come together to form A times 1 fourth raise the power of n u n minus B times 1 third to the power of n u minus n minus 1. Early we take the third possibility. The third possibility is where mod Z is greater than 1 third. So of course if mod Z is greater than 1 third it is also greater than 1 fourth. Mod Z is greater than 1 third. So of course the expansion proceeds as A times 1 fourth to the power of n you know I mean essentially 1 fourth to the power of n Z raised to the power minus n if you please summation n going from 0 to infinity. You recall you must always keep the powers of Z there. So 0 to infinity B times 1 third to the power of n Z raised to the power of minus n. So this is easy you know in this case the inverse is very easy to calculate. This essentially corresponds to A times 1 fourth raised to the n u n plus B 1 third raised to the power of n u n. Notice that in the first region of convergence we got what is called a left sided sequence. A sequence which is nonzero only for or only towards the left of a chosen point. In this case you could take that chosen point to be n equal to 0. So in the first region of convergence we have left sided sequence. What is a left sided sequence? A left sided sequence is a sequence which is nonzero to the left of a chosen point. Typically that chosen point is 0. Now the sequence that we got in the third region of convergence is what is called a right sided sequence. It is nonzero to the right of a chosen point and that of course chosen point happens to be n equal to 0. In the second tightly tricky case we have what is called a double sided sequence. So it is neither left sided nor right sided. It is double sided. In fact it is doubly infinite. It is infinite in extent to the left and it is infinite in extent to the right. Of course you can have sequences that are finite in extent to the left and finite in extent to the right and of course they have to be finite length sequences. So finite length sequence is you know you can view it as neither left sided nor right sided if you please. But it is definitely not doubly infinite. Anyway so much so. Now this gives us an idea how we deal with the region of convergence and it also convinces us of the importance of the region of convergence.