 Welcome all. In this lecture, we are going to discuss the parallel form structure for the given IIR filter system. This is the learning outcome. At the end of this session students will be able to draw the parallel form realization structure for a given IIR system. A parallel form realization of IIR system can be obtained by performing a partial expansion of H of z is equal to C plus summation k equal to 1 to k A suffix k divided by 1 minus p suffix k z to the power minus 1. So, let us call this one as equation number 1, where p suffix k they are the poles and A suffix k they are the coefficients in the parallel partial fraction expansion. And C is the constant and it is defined as b n divided by A n. The equation 1 can be written as H of z is equal to C plus A 1 divided by 1 minus p 1 z to the power minus 1 plus A 2 divided by 1 minus p 2 z to the power minus 1 plus and so on A k divided by 1 minus p k z to the power minus 1. So, let us call this one as equation number 2. Now, H of z can be written as C plus H 1 of z plus H 2 of z plus and so on till H k of z. So, let us call this one as equation number 3. We know that H of z is nothing but Y of z divided by X of z. So, now let us call this one as equation number 4. So, let us use equation number 4 in equation number 3 here. So, this can be written as Y of z divided by X of z is equal to C plus H 1 of z plus H 2 of z plus and so on H k of z. So, let us call this one as equation number 5. This can be simplified as Y of z is equal to C plus X of z H 1 of z plus X of z H 2 of z plus so on X of z H k of z. So, let us call this one as equation number 6. Now, apply inverse z transform on equation number 6. So, let us get Y of n is equal to C here it will be X of z there will be a 1 X of z term here C into X of n plus X of n into H 1 of n can be written as Y 1 of n. Similarly, X of n into H 2 of n can be written as Y 2 of n and so on this is Y k of. So, let us call this one as equation number 5 sorry equation number 7. Now, equation 4 can be realized as can be realized as can be realized as X of n. So, this will be the input to the system. So, here it will be H 1 of z here it will be H 2 of z and this continues and here it will be H k of z. So, here we are going to get the output Y of n here. So, we will connect these systems as shown in this figure. So, this is the parallel form structure realization for the given I i r system. Let us take one example here. So, let us take this example realize the system given by difference equation Y of n is equal to minus 0.1 Y of n minus 1 plus 0.72 Y of n minus 2 plus 0.7 X of n minus 0.252 X of n minus 2 in parallel form. The system function is given as H of z is equal to 0.7 minus 0.252 z to the power minus 2 divided by 1 plus 0.1 z to the power minus 1 minus 0.72 z to the power minus 2. We know that H of z can be written as C plus H 1 of z plus H 2 of z ok. Now pause the video for some time and find out the values for C H 1 of z and H 2 of z ok. I hope you have found out the values for C H 1 of z and H 2 of z. Now C is equal to 0.35 and H 1 of z is equal to 0.206 divided by 1 plus 0.9 z to the power minus 1. H 2 of z is equal to 1.144 divided by 1 minus 0.8 z to the power minus 1. Now H 1 of z and H 2 of z can be realized in direct form 2 as shown below ok. So, H 1 of z so this is H 1 of z and H 2 of z is equal to 0.144 divided by 1 minus 0.8 z to the power minus 1. Now H 1 of z and H 2 of z can be realized in direct form 2 as shown below ok. So, H 1 of z so this is H 1 of z and H 2 of z is equal to 0.5 and here the input will be x of n. So, here the coefficient will be minus 0.9 and here the coefficient will be 0.206 and here we will get the output y 1 of n ok. So, similarly H 2 of z is equal to x 1 of n input to the system z inverse is equal here the coefficient will be 0.8 and here the coefficient will be 0.144 ok. So, now we will put them in the places of H 1 of z and H 2 of z. Now the realization of H of z can be given as x of n here it will be 0.35, 0.206 z inverse. So, here it will be minus 0.9. So, here it will be 0.8, 0.144. So, this will be the final solution for the given problem statement. These are the references. Thank you.