 Hello, welcome all. Today in this lecture we are going to discuss the cascade form structure for the given IIR filter. This is the learning outcome. At the end of this session students will be able to draw the cascade form realization structure for a given IIR system. So, let us consider a high order IIR system with system function. Let us consider a high order IIR system with system function with system function h of z is equal to summation k equal to 0 to m b suffix k z to the power minus k divided by 1 plus summation k equal to 1 to n a suffix k z to the power minus k. So, let us call this one as equation number 1. The system can be factor the system can be factor into a cascade of second order subsystem, second order subsystem such that h of z can be expressed as h of z is equal to product of k equal to 1 capital k h k of z. Let us call this one as equation number 2, where k is an integer part of n plus 1 divided by 2. So, h k of z is equal to b k naught plus b k 1 z to the power minus 1 plus b k 2 z to the power minus 2 divided by 1 plus a k 1 z to the power minus 1 plus a k 2 z to the power minus. Now, equation 2 can be represented as equation 2 can be represented as x of n which is equal to x 1 of n. Now, this is the input point h 1 of z from this we are going to give the input to h 2 of z. From this, this process will continue till it reaches the k th stage. This is the h k of z and here we will get the output y of n which is nothing, but the output of the k th system y k of n here. So, here this will be the output of y 1 of n here we will write y 1 of n. So, that is the output of the first subsystem and this is equal to x 2 of n. So, similarly here we will get y 2 of n and output of this part will be x 3 of n. So, in that way here we will get y k minus 1 of n and here the input for this system will be x k of n. Now, realize each h k of z in form 2 cascade all structures to get the cascade form realization of the given IIR system. So, now I will show you the direct form 2 structure for one of the subsystem here. So, let me take that subsystem has h k of z. So, I will show the direct form 2 structure for this h k of z subsystem here. So, now this is the direct form 2 structure for h k of z. We will take one example here to understand how to draw the cascade form realization for the given system. Example, realize the system with difference equation y of n 3 by 4 y of n minus 1 minus 1 by 8 y of n minus 2 plus x of n plus 1 by 3 x of n minus 1 in cascade form in cascade form solution. From difference equation we get h of z is equal to y of z divided by x of z which is equal to 1 plus 1 by 3 z inverse divided by 1 minus 3 by 4 z inverse plus 1 by 8 z to the power minus 2. Now, which it can be written as we know that h of z can be written as h 1 of z into h 2 of z. So, now pause the video for some time and find out the values for what is the expression for h 1 of z and h 2 of z. I hope you have found the expressions for h 1 of z into and h 2 of z. From the above expression h 1 of z is equal to 1 plus 1 by 3 z inverse divided by 1 minus 1 by 2 z inverse and h 2 of z is equal to 1 divided by 1 minus 1 by 4 z inverse. So, this is h 1 of z and h 2 of z. Now, we realize the direct form 2 for h 1 of z and h 2 of z as shown below. So, h 1 of z is equal to this can be realized as we will put one delay unit here. So, here the coefficient will be 1 divided by 3 and here it will be 1 by 2. So, this is the realization for h 1 of z. Similarly, we realize the direct form 2 for h 2 of z. So, here it will be input z inverse. So, here it will be 1 by 4 and here we will get the output of the system. Now, we cascade these two direct form 2 realizations to get the cascade form realization for the given IIR system and at this point we will get one of the output that is y 1 of n and here it will be the input x of n which is equal to x 1 of n and here it will be the input x 2 of n and here we will get y 2 of n which is equal to y of n. This completes the solution.