 construct Hz equal to 1 minus half Z inverse the whole square divided by 1 minus 1 third Z inverse into 1 minus 1 fifth Z inverse multiplied by 3 if you like using the lattice structure. Let us expand this. So we would need to expand this because it is not in the pole form or pole 0 form that the lattice structure can be realized. We need to explicitly as polynomials and therefore we need to expand this. Let us do so. So Hz is clearly 3 times 1 minus 2 times half Z inverse that is Z inverse plus 1 fourth Z raised to power minus 2 divided by 1 minus 1 third plus 1 fifth Z inverse plus 1 by 5 3 Z times Z to the power minus 2. So that can be expanded as 3 minus 3 Z inverse plus 3 by 4 Z raised to power minus 2 divided by 1 minus now 5 and 3 is 8. So 8 by 15 Z inverse plus 1 by 15 Z raised to power minus 2. Is that correct? Now let us start by realizing the denominator by Anz. So of course very clearly now here I have conveniently chosen the numerator degree to be less than or equal to the denominator degree. Of course if it were not I would first need to carry out a long division right. So we first realize Anz. Clearly N is equal to 2. So A2z is 1 minus 8 by 15 Z inverse plus 1 by 15 Z raised to power minus 2. And clearly the coefficient of Z raised to power minus 2 in Z inverse is 1 by 15 and this should be equal to K2. Therefore K2 is 1 by 15. Now we know the relation between A2 and A1 just for the sake of clarity though we have done it in general just for the sake of clarity let us write down that relationship explicitly. We know that A2z is A1z plus K1 times Z inverse A1 till day Z and A2 till day Z is Z inverse A1 till day Z plus I am sorry this should be K2 plus K2 A1z. Is that correct? And therefore if we take A2z minus K2 times A2 till day Z then this term would vanish and it would leave us with 1 minus K2 squared times A1z. I am just redoing that step to make it explicit though we have done it in general for N plus 1. You do not need to do this every time. And therefore A1z is A2z minus K2A2 till day Z divided by 1 minus K2 squared and this is very easy to write. Now A2z is 1 minus 8 by 15 Z inverse plus 1 by 15 Z raise the power minus 2 minus K2 which is 1 by 15 times. Now A2 till day Z is essentially this written in reverse order. So we have 1 by 15 minus 8 by 15 Z inverse plus Z raise the power minus 2 divided by 1 minus 1 by 15 the whole squared. The question is what is the difference between A1 person has asked what is the difference between A2z and A2 till day Z? They are very different. The coefficients are very different. You need to look very carefully to observe that they are very different. The coefficients are in reverse order. One needs to pay careful attention to understand the example. They are very different. The coefficients are in reverse order. It is 1 minus 8 by 15 and 1 by 15 and here it is 1 by 15 minus 8 by 15 and 1. Alright. Now if you look at this difference 1 by 15 Z to the power minus 2 is eliminated because you have 1 by 15 Z to the power minus 2 minus 1 by 15 Z to the power minus and that is expected. That has to happen because the degree must go down by 1. So as expected the degree decreases by 1 because the Z to the power minus 2 term is eliminated and that leaves you with 1 by 1 minus 1 by 15 the whole squared minus 8 by 15 into 1 minus 1 by 15 Z inverse divided by 1 minus 1 by 15 the whole squared. So in fact we can even simplify this further. Is that correct? We get this, don't we? We can simplify this. So you see is this expression correct? Is this correct? Yes. Now we can simplify because we can note that this is of the form 1 minus x squared and this is 1 minus x. So you can write 1 plus x here and you can cancel this. You can take it separately you know either way whatever you like. So you get 1 minus 8 by 15 divided by 1 plus x 1 plus 1 by 15 times Z inverse and that leaves you with 1 minus 8 by 16 Z inverse 1 minus half Z inverse. As expected this is again, this is A1Z. Is that correct? It should be noted as expected that the coefficient of Z to the power 0 in A1Z again comes out to be 1 as expected. And therefore the coefficient of the highest power of Z namely Z inverse in A1Z is essentially K1 which means K1 is minus half and once you have K1 you have completed the lattice for the denominator but now we need to write the lattice for the numerator. Essentially what we have is a lattice of the form 1. We have an E2 there and let us go back to the structure, the generic structure and then draw it for this specific case. This is the generic structure. We need a minus Kn plus 1 and a plus Kn plus 1 and we need a Z inverse every time you take one step. So you need a Z inverse here and you need a Z inverse here. A minus, so you see remember it is E2 here and E2 till day here. So we need to keep writing the others but you have a minus K2 there and the K2 here is K1 here and the K1 here and this is E0 and this is E0 till day and of course this is E1 and this is E1 till day. So we need to take C0 times this plus C1 times this plus C2 times this and add them. So X is given here and Y is tupped off here and we now need to obtain C0, C1 and C2. We know the values. We know K1 is equal to minus half and K2 is 1 by 15 and please note that we had assumed a stable causal system to begin with. In fact we explicitly wrote the system in terms of its poles. You see that is the reason why I wrote the system in factored form in the numerator and denominator first. You see if you look back let me put that system function back here before you. The system function has poles which are both inside the unit circle and therefore the system is stable and as expected the denominator polynomial yields both lattice coefficients less than 1 in magnitude as was the necessary and sufficient condition. This is a verification not a proof but a verification. Now we need to determine the C0, C1, C2. So let us begin. Now we know what these are. We know what the A's are. A2 of course is 1 minus 8 by 15 Z inverse plus 1 by 15 Z raised to the power minus 2. A2 till day is therefore 1 by 15 minus Z to the power 0 if you like minus 8 by 15 Z inverse plus Z raised to the power minus 2. A1 Z is 1 minus half Z inverse and therefore A1 till day Z is minus half plus Z inverse and of course A0 and A0 till day are identically 1. Therefore what we have is C2 times A2 till day Z plus C1 times A1 till day Z plus C0 A0 till day Z. Now A0 till day Z is 1. This is equal to the numerator and the numerator is 1 minus Z inverse plus 1 by 4 Z raised to power minus 2. Now let us expand this relationship. C2 times A2 till day which is 1 minus 8 I am sorry it is 1 by 15 minus 8 by 15 Z to the power minus 1 plus Z raised to power minus 2 plus C1. A1 till day which is minus half plus Z inverse plus C0 is equal to 1 minus Z inverse plus 1 by 4 Z raised to power minus 2. As expected the coefficient of Z raised to power of minus 2 can be compared on both sides. It comes only from here and therefore we have C2 is equal to 1 by 4. Therefore we have C1 into minus half plus Z inverse plus C0 is equal to 1 minus Z inverse plus 1 by 4 Z raised to power of minus 2 minus C2 which is 1 by 4 times 1 by 15 minus 8 by 15 Z inverse plus Z raised to power minus 2. Subtracting the term with C2 from both sides. Now after you have subtracted as expected you see that 1 fourth Z to the power minus 2 is annihilated. So 1 fourth Z to the power minus 2 minus 1 fourth Z to the power minus 2. So you get a degree 1 term only let us write that down. So therefore C1 into minus half plus Z inverse plus C0 is equal to 1 minus 1 by 4 into 15 plus Z inverse into minus 1 plus 8 by 4 into 15 and the Z to the power minus 2 term is annihilated and is killed is annihilated. And therefore C1 is now very clear as well. This is essentially C1 now the coefficient of Z to the power minus 1 is C1. So once you have C1 you can subtract this term from the side again and find C0 finally. But we can continue this we can now subtract C1 times minus half plus Z inverse from the right hand side and obtain C0 and I leave that to you as an exercise. So this is illustrated the process of obtaining a lattice realization of an HZ and this lattice realization can be used to realize either an FIR system function or an IRR system function. In the case of an IRR system function the lattice structure gives an insight into the stability. In the case of an FIR system function the lattice structure is just another structure which is regular. And in the assignment on filter design you are encouraged to realize both the FIR and the IRR structures using the lattice form. Yes, now you see what I have done now what I, yeah that is a good point. You see what I did here was to ignore the gain of 3 and one there are 2 ways in which you can do this either you could multiply all the C's by 3 or you could put 1 gain of 3 in the end. So I had taken the numerator without the gain of 3 here but you could either incorporate that gain of 3 in all the C's so you could multiply each of the C's by 3 or you could just put 1 gain of 3 finally either way is fine that is a good point somebody is going to point that out. Is that correct? That completes the lattice structure and for the moment we are done with the subject of realization in the next lecture we shall begin on a very important theme in discrete signal processing namely the discretization of the frequency axis. If you want to understand the frequency domain behavior we need to discretize the frequency axis how do we do it and how do we do it efficiently. We shall discuss this beginning with the next lecture. Thank you.