 Welcome all to the video lecture on structure for FIR filter. In this video session, we are going to continue with the lattice structure for FIR filter. At the end of this session, students will be able to draw the lattice realization structure for a given FIR system. So, in the last video lecture, we have discussed that how to draw the lattice structure for m equal to 1. Now, let us consider an FIR filter, consider an FIR filter for which m is equal to 2, then y of n is equal to x of n plus alpha 2 of 1 x of n minus 1 plus alpha 2 of 2 x of n minus 2. So, let us give this as equation number 6. So, we will cascade two lattice stages to obtain y of n. So, x of n is split into two parts. So, this will be the f naught of n and this will be the g naught of n. So, here we will place a delay unit z inverse and this will be taken to the plus addition unit, here we take it to addition unit. So, this will be k 1 and here it will be k 1. So, at this output we are getting f 1 of n and here we will be getting g 1 of n here. So, this is again carried to the further to the next summation unit. Now, here g 1 of n will be passed through the delay unit z inverse. So, here we will get the output of stage 2 that will be f 2 of n is equal to y of n and here it will be g 2 of n. So, here it will be k 2 coefficient and here it will be k 2 coefficients. So, this is the two stage lattice structures output from first stage is f 1 of n is equal to x of n plus k 1 x of n minus 1 and g 1 of n is equal to k 1 x of n plus x of n plus x of n minus 1. So, let us call this one as equation number 7 here. Output form second stage is f 2 of n is equal to f 1 of n plus k 2 g 1 of n minus 1. Similarly, g 2 of n is equal to k 2 f 1 of n plus g 1 of n minus 1. So, let us rename this one as equation sorry let us give the this expression as equation number 8. Now substitute substitute f 1 of n and g 1 of n from equation 7 in f 2 of n of equation 8 we get f 2 of n is equal to f 2 of n is equal to x of n plus k 1 plus k 1 x of n minus 1 plus k 2 into bracket k 1 x of n minus 1 plus x of n minus 2. So, after simplifying this one it can be written as x of n f 2 of n is equal to x of n plus k 1 into bracket 1 plus k 2 x of n minus 1 plus k 2 x of n minus 2. So, let us give this one as equation number 9 here. So, now pause the video for some time and find out the output from second stage. Now equation 9 is identical equation 9 is identical to the equation 6 if coefficients alpha 2 of 2 is equal to k 2 and alpha 2 of 1 is equal to k 1 into bracket 1 plus k 2. So, let us give this one as equation number 10 here. Similarly or we can say that k 2 is equal to alpha 2 of 2 here and k 1 is equal to alpha 2 of 2 divided by 1 plus alpha 2 of 1. So, let us give this one as equation number 11 here. So, again coefficients k 1 and k 2 of lattice can be obtained from alpha m of k of the direct form realization of the direct form realization. So, now pause the video for some time and find out the output from second stage. This is the equation 6 here it is written as y of n is equal to x of n plus alpha 2 of 1 x of n minus 1 plus alpha 2 of 2 x of n minus 2. Similarly, g 2 of n is equal to alpha 2 x of n plus k 1 of 1 plus k 2 x of n minus 1 plus x of n minus 2. So, let us give this one as equation number 12 here. Now, from two stage lattice structure we have found f 2 of n and g 2 of n. Now, for m minus 1 stage filter f naught of n is equal to g naught of n which is equal to x of n. So, for m minus 1 stage filter f m of n is equal to f m minus 1 of n plus k m g m minus 1 of n minus 1 where m is equal to 1, 2 and so on up to m minus 1. Similarly, g m of n is equal to k m f of m minus 1 f of m minus 1 of n plus g m minus 1 of n minus 1 where m is equal to 1, 2 and up to m minus 1 here. So, now this is the lattice structure for m minus 1 stages here x of n is the input and this will be the output here y of n. So, here it will be f naught of n and here it will be g naught of n. So, f naught of n is directly taken to the addition unit and g naught of n is passed through the unit delay system. So, at this end we will get g naught of n minus 1. So, here it will be f 1 of n output and here the output will be g naught g 1 of n and at the output of unit system delay it will be g 1 of n minus 1 and here it will be f 1 of n. Now, here at this stage it will be f 2 of n output of the second stage is mentioned as f 2 of n and output of this one is g 2 of n. This structure is can be drawn up to m minus 1 stages here. So, here it will be f of m minus 2 of n and here it will be g m minus 2 of n. So, at the output end we will get f of m minus 1 of n which is equal to y of n and here it will be g of m minus 1 of n. So, the reflection coefficients that here on stage 1 it will be k 1 and at stage 2 it will be k 2 and at stage m minus 1 it will be k m minus 1. Now, this completes the m minus 1 stage lattice structure here and here y of n is equal to f of m minus 1 of n. So, these are the references. Thank you.