 Hello everyone, I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering, WIT, Swollapur. Welcome you all for a video lecture on Cascade Form Structures, the Learning Outcome. At the end of this session, student will be able to draw the cascade realization structure for a given FIR system. Now, I will see the cascade structures. Now, consider a transfer function h of z is equal to 1 minus 2 z raise to minus 1 into 3 minus z raise to minus 1. Now, let us think of the zeros of this transfer function. Now pause this video for a minute and think on what are the zeros? Okay, you might have already written the zeros. Now, there are two zeros. The first one, this term becomes 0 when the value of z is equal to 2. So, therefore, z 1 is equal to 2. For the second, this term to become 0, the value of z must be equal to 1 upon 3. So, that z raise to minus 1 is equal to 3 and 3 minus 3 becomes 0. So, therefore, z 2 is equal to 1 by 3. Now, let us consider multiplying these two terms. We get 1 into 3 as 3 minus 6 z raise to minus 1 minus z raise to minus 1 plus 2 z raise to minus 2 which gives us 3 minus 7 z raise to minus 1 plus 2 z raise to minus 2. Now, to understand the cascade form structure, let us consider this h of z represented as h 1 z into h 2 z. So, we know that here when two systems are connected in cascade, the corresponding systems will draw. Let us, the system is h n as input. The first system is unity impulse response h 1 n which is given to the second system whose unity impulse response is h 2 n and the output of this is y of n. The equivalent system is the impulse response of the equivalent system will be h 1 n into sorry convolution h 2 n. So, according to community property, this can be written as h 2 n convolution with h 1 n and so, this input x of n output y of n and once again this can be written as h 2 n cascade connection of two systems and h 1 n. So, what does it mean? If you have two systems connected in cascade with impulse responses h 1 n and h 2 n, then you can change the order of the two systems and you can take it as first system h 2 n with h 1 n. So, as we have seen h of z is equal to 1 minus 2 z raise to minus 1 into 3 minus z raise to minus 1. Now, let us realize the system, this is 1 minus 2 z raise to minus 1. Comparing it to our standard equation, the value of b 0 is 1, value of b 1 is equal to minus 2. Similarly, for this system the value of b 0 is 3 and value of b 1 is equal to minus 1. So, therefore, if I am considering x n input b 0 which is equal to 1. So, this delay gives me x of n minus 1. If I am adding these two, I will get the first system. Now, this is given as an input to second system. So, this is 3 and this is minus 1. Adding together I will get y of n at the output. So, we have this is h 1 z and this is h 2 z. So, here we have connected these two systems in cascade. The order of this can be interchanged and gives you a equivalent system. Now, we use the same for the generalized transfer function. So, now, let us discuss cascade form structures for the system transfer function h of z is equal to summation h of z is equal to summation k equal to 0 to m minus 1 b k z raise to minus k. So, in this case this particular transfer function can be written as a product of second order subsystems as h of z is equal to product of k equal to 1 to capital K h k z where this h k z is equal to b k 0 plus b k 1 z raise to minus 1 plus b k 2 z raise to minus 2. Where this k equal to 1 to capital K h k z where this h k z is equal to b k 0 plus b k 1 z raise to minus 1 plus b k 2 z raise to minus 2. Where this k equal to 1 to up to capital K here k this capital K is integer part of m by 2. Because suppose for example, we have m is equal to phi. So, this m minus 1 will be 4 the order of the term will be 4 order of the polynomial will be 4 and therefore, there will be 2 subsystems. So, phi divided by 2 the it is integer part is 2. So, these this h k z is a second order subsystem. Now, while forming these subsystems the coefficient b 0 may be equally divided into k odd number of subsystems like b 0 equal to b 1 0 into b 2 0 into b 3 0 dot dot dot b k 0 or it can be taken into one of the system. Now, when we are combining the 0's to form a second order subsystem you can combine complex conjugate 0's. So, that the coefficients b k i in the second order system whereas, the real 0's can be combined in any arbitrary manner. So, with this the cascade structure for our h of z will look like the input x of n. This will go to the first system whose transfer function is h 1 z. Now, here I can consider this as x 1 n. So, your x of n will be equal to x 1 n. The output y 1 n of this will go as input to second order subsystem. So, this will be second system. So, likewise this will continue this will go to a system may be h k minus 1 z the output of that will go to a final system h k z and this output will be y of n. So, this y capital k n will be equal to y of n whereas, here the input to output of the system will be y of k minus 1 n will be equal to input here will be equal to x of k n. So, we will have k subsystems which are connected in cascade whereas, individual systems in this the realisation will be because now individual system is basically h k z is equal to b k 0 plus b k 1 z raise to minus 1 plus b k 2 z raise to minus 2. So, if I have x of n. So, this will be z raise to minus 1 gives me x of n minus 1 further z raise to minus 1 gives x of n minus 2. So, this is b k 0 here it will be b k 1 and here this is b k 2. So, if I add all these terms together I will get the realisation structure for this. So, this is the direct form realisation of the second order subsystem. So, every subsystem in this will be system like this. Now depending on the value of m say for example, m is equal to 7. So, when m is equal to 7 our equation here it will be k equal to 0 to 6. So, there will be the order of this will be 6. So, the number of terms will be 3. So, in that case all the three subsystems subsystems will have coefficients b k 0 b k 1 and b k 2. But if I consider m is equal to 6 then from this equation k equal to 0 to m minus 1 that is k equal to 0 to 5 b k z raise to minus k there the order of the system will be 5. And so you have for all three systems for out of those three for one of the system your b k 2 will be 0. So, this is how given systems can be realised using cascade realisation. Let us consider another example on realisation. Let us assume x z is equal to 1 minus 1 by 2 z raise to minus 1 1 plus 1 by 4 z raise to minus 2 into 1 plus 1 by 3 z raise to minus 1 plus 3 by 8 z raise to minus 2. So, now we have two subsystems. So, the realisation structure will be here x n you have to draw two second order systems. So, z raise to minus 1 z raise to minus 1 this will be b k 0 that is 1 this will be minus 1 by 2 1 by 4 add the terms together. So, this is h 1 side this will go as an input to second system. So, z raise to minus 1 z raise to minus 1 this is the b k 0 of this is 1 this will be 1 by 3 and this will be 3 by 8. So, if I add this to I will get y of and so this is h 2 z. So, this is how given a transfer function we can draw the cascade form structure for the FIR system. Thank you.