 Welcome all. In this lecture, we are going to discuss the lattice structure realization for the given FIR filters. This is the learning outcome. At the end of this session, students will be able to draw the lattice realization structure for a given FIR system. Let us begin the development by considering a sequence of FIR filters with the system function. Let us consider a sequence of FIR filters H of Z, sorry H of Z is equal to AM of Z where m is equal to 0, 1, 2 up to m minus 1. So, let us call this one as equation number 1, where by definition AM of Z is the polynomial. So, which can be written as AM of Z is equal to 1 plus summation k equal to 1 to m alpha m alpha m of k Z to the power minus k, where m is greater than or equal to and A naught of Z is equal to 1 from which we have from which we have Y of Z is equal to X of Z into 1 plus summation k equal to 1 to m alpha m of k Z to the power minus k. Because we know that H of Z is equal to Y of Z divided by X of Z. Now, taking inverse Z transform on both the sides, we get Y of n is equal to X of n plus summation k equal to 1 to m alpha m of k X of n minus k. So, let us call this one as equation number. Equation 3 represents an FIR system with system function H of Z is equal to AM of Z. Now, suppose that we have a filter. Now, suppose that we have a filter of order m equal to 1, the output of such filter is output of such filter is Y of n is equal to X of n plus alpha 1 of 1 X of n minus 1. So, let us call this one as equation number 4. This output can also be obtained from this output, can also be obtained from single stage lattice filter shown below. So, here it will be X of n summation at this point. Here, we will put a unit delay system Z inverse. So, this will be F naught of n and here it will be G naught of n. So, after this point, we will get G naught of n minus 1 and here it will be k 1 and here it will be k 2. And at this point, we will get F 1 of n is equal to Y of n and here we will get G 1 of n is equal to here we will get G 1 of n. Pause the video for some time and find out the expressions for F naught of n, G naught of n, F 1 of n and G 1 of n from the above figure. So, F naught of n is equal to G naught of n is equal to X of n because the input branch is divided into two parts and the upper part we call it as F naught of n and the lower part we call it as G naught of n and both are coming from the same line. So, F naught of n and G naught of n are equal to X of n. Now, F 1 of n can be written as so, F naught of n we get two lines one line from F naught of n and another one line from G naught of n minus 1 are added in the adder unit and then it is represented as F 1 of n. So, from this figure we can write it as F naught of n plus k 1 G naught of n minus 1. So, we know the values for G naught of n and F naught of n. So, we substitute those values in this expression here. So, it can be written as X of n plus k 1 X of n minus 1. So, let us call this one as phi u a equation number phi u a. Similarly, G 1 of n is equal to k 1 F naught of n plus G naught of n minus 1 which is equal to k 1 X of n plus X of n minus 1. So, let us call this one as phi u b. Now, we will compare equation 4 and phi by comparing equation 4 and phi we get alpha 1 of 0 is equal to 1 and alpha 1 of 1 is equal to k 1. These are the references. Thank you.