 Hello everyone, I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering Walsh and Institute of Technology, Solaapur. Today, we will discuss decimation in frequency that is DIF, Fast Fourier Transform Algorithm, the learning outcome. At the end of the session, student will be able to draw and explain the signal flow graph for decimation in frequency FFT algorithm. So, we have seen the divide and conquer approach for computation of n point DFT in terms of smaller size DFTs. Here, we will make use of the divide and conquer approach. Let us assume m equal to 2 and l equal to n by 2 to compute n point DFT using divide and conquer approach. Basically, here it this implies the column wise storage of input data sequence, because there are two columns. So, 50 percent of the data points or sequence input sequence will be stored in first column and 50 percent in the second column. Now, as we know this is a n point DFT which is represented as x k equal to n equal to 0 to n minus 1 x of n w n k n, where w n is the twirl factor which is represented as e raised to minus j 2 pi by n. So, now, this can be written as in terms of first n by 2 points and second n by 2 points of the input sequence. So, n equal to 0 to n by 2 minus 1 x n w n k plus n. Now, once again here by using n equal to 0 to n by 2 minus 1 only, but x is now n plus n by 2. So, which gives you second 50 percent of your sequence x n. So, now, n is replaced by n plus n by 2. So, this will be n plus n by 2 k, where summation now we k is kept to 0 to n by 2 minus 1. So, now, this can be written as so this part will remain as it is, but the second part you have w n 1 term is n k and second term is w n n by 2 k. So, this w n n by 2 k is taken outside and here in the summation it is written as n equal to 0 to n by 2 minus 1 x of n plus n by 2 w n n k. Now, about this particular term w n n k by 2 gives us equal to as we know w n is e raise to minus j 2 pi by n raise to n k by 2. So, this gives you multiplication of these two terms which gives you e raise to minus j 2 pi n k by 2 n 2 2 2 gets cancelled n gets cancelled. So, this will be e raise to minus j pi k and this e raise to minus j pi k can be represented as e raise to minus j pi raise to k and as we know e raise to minus j pi is minus 1. So, it will be minus 1 raise to k. So, this term is minus 1 raise to k. So, with that this x k can be now represented as summation is of course, limits are same. So, taken this and common first part you have x n, second may we have minus 1 raise to k in 2 x of n plus n by 2 and this factor common w n k n. So, it is w n k n. So, this DFT can be written as now by using this equation right. So, this x k is actually for k equal to 0 to n minus 1 total you have n points. Now, in the next step let us divide this x k into two parts one with all even numbered samples x k and order number samples x k. So, split x k into even numbered and odd numbered samples as x 2 k where x of 2 k with k equal to 0 to n by 2 minus 1 you will give you all even numbered samples of x k. So, n equal to 0 to n by 2 minus 1 this minus 1 raise to when k is even for 2 k. So, for 2 k this will always be equal to 1 plus 1 and this w n 2 k n because now k is to be replaced by 2 k this comes out to be w n 2 k which give which can be written as w n k n by 2 right. Same way for x of 2 k plus 1. So, I have to replace k by 2 k plus 1. So, minus 1 raise to 2 k plus 1 is always equal to minus 1 because 2 k plus 1 is odd number. So, minus 1 raise to 3 minus 1 raise to 5 every time it is minus 1. So, this gives you negative and if you replace this k by 2 k plus 1 here as it is shown. So, the first part is here and second is w n k 2 k plus 1 will be w n 2 k n 2 k n and second is w n n. So, this is w n n. So, this can be written as this. So, from this we can have this w n n term and finally, this whole bracket into w n k n by 2 right. So, in both cases k runs from 0 to n by 2 minus 1. So, now if I represent the sequence x of n plus n x of n by n plus n by 2 as maybe g 1 n and second is sequence x 2 sorry g 2 n as x of n minus x of n plus n by 2 into w n n. Then with these representation now x 2 k can be represented as the DFT of g 1 n and x 2 k plus 1 as a DFT of g 2 n. So, with this x 2 k is represented as n equal to 0 to n by 2 minus 1 g 1 n w n k n by 2 and x 2 k is equal to n equal to 0 to n by 2 minus 1 g 2 n w n k n by 2 where k equal to 0 to n by 2 minus 1. So, these are the 2 n by 2 point DFTs computed of g 1 n and g 2 n where g 1 n and g 2 n are the sequences found from x of n right. So, therefore, this can be represented as a structure like this where the first 4 samples are g 1 n, second are for g 2 n. So, this is for a sample representation for a 8 point DFT. So, suppose this is as input sequence x 0, 1, 2, 3, 4, 5, 6 as we know g 1 n is equal to or now g 1 0 will be equal to g x of 0 plus x of 4 right because it is x of n plus x of n plus n by 2 now n by 2 is 4. So, 0 plus 4 is this and if you take g 1 of g 2 of 0 will be x of 0 minus this because of this negative sign here it is minus of this and after that multiply w n n ok. So, w n n means w 0 8 same way g 1 1 is equal to x 1 1 plus x of x of 1 plus x of 5 and g 2 1 will be x of 1 minus x of 5 into w 1 8 and so on. So, this is how these are computed. So, from input sequence these are constructed ok. Now, this is a n by 2 point sequence this is a n by 2 point sequence. So, now we can compute n by 2 point DFT that is 4 point DFT in this case of this sequence 4 point DFT of this sequence right and so on. So, same procedure is repeated for computing n by 2 point DFT is x 2 now this is x 2 k which is further decimated in terms of 2 n by 2 point DFT right. So, with the same process these smaller DFTs are computed for n equal to 2 raise to v that process actually involves v equal to log of n to the base 2 number of stages ok. So, you need to decimate it v number of times where v is given as log of n to the base 2 right. So, this is once decimated. Now, for the second state this 4 point DFT will be computed in terms of 2 point DFTs second decimation. Now, the 2 point DFT is in computed in terms of 1 point DFTs that is the third decimation and 1 point DFT of a sequence is the sequence itself. So, number of decimation will be thrice. So, for n equal to 8 it will be log of 8 to the base 2 is 3. So, it is 3 times decimation right. So, with this now let us draw the signal flow graph for 8 point DFT right. Before that let us recall this because we want to compare the multiplications required for direct computation with this FFT algorithm right. So, number of complex multiplications and additions required for n point DFT pause the video for a minute and think on this. So, you might be knowing for computation of an n point DFT by using direct computation it needs n square complex multiplications and n into n minus 1 complex additions right. So, now need this for comparing the multiplications and additions for direct computation with our DIF FFT algorithm right ok. So, the basic butterfly structure as we have seen here for computation of your G 1 n and G 2 n it will be like a plus b and a minus b into this that is x of n plus x of n plus n by 2 x of n minus x of n minus 2 n by 2 into a factor right. So, this particular structure actually needs one complex multiplication and two complex additions. So, it has total as we have seen there are v number of stages where v is given as log of n to the base 2 and every stage as we have seen in previous diagram there are four there were four butterfly structures in first stage where n was 8. So, number of butterfly structures are n by 2. So, the total number of complex multiplications will be n by 2 for each stage and total log of n to the base 2 stages. So, n by 2 log of n to the base 2 number of complex multiplications are required and of course, additions will be because there is only one. So, two additions are required base basically. So, n into log of n to the base 2 will be the total number of complex additions required for this. So, with this actually now we can draw this, but signal flow graph as we have seen for n equal to 8 this comes out to be x of n plus n by n plus 4 x of n minus n plus n by 2 means actually 4 and w n 8 for first stage. Second stage may you are now further competing in terms of n by 4 point 80. So, in terms of that you will get this now g 1 n only it is written for g 1 n further decimated into h 1 n and h 2 n in terms of 2 n by 4 minus 1. So, n by 4 minus 1 for n equal to 8 will be 0 1 right. So, it gives you this and this w 2 n 8 it gives. So, this actually can be written as w a n by 2 can be taken here. So, 2 n by n is 8. So, it will be w 2 n 8 right and in the next stage it will be further decimation of h 1 n in terms of h 1 1 n into i 1 1 1 n and i 1 1 2 n where these are n by 8 minus 1 means 1 point sequence right. So, there is a third stage of decimation right. So, with these equations and you can draw the structure like this as we are we have seen g 1 n construction g 1 n in terms of your input sequences this is a x of 2 k means even numbered samples of d f t if you are competing d f t of this, but further decimation of this you get in terms of h 1 n and further decimation gives you 1 terms of d f t and finally, this gives you 0 4 2 4 6. So, these are the d f t points right. So, the same structure repeating for n by 2 point d f t's then further n by 4 point d f t's gives you this structure right. So, if you look at here in this particular d i f f t your input sequence in the natural order and output sequence in the bit reversed order right. So, there are n by 2 point rather n by 2 butterfly structures for every stage and there are total 3 stages for n equal to 8. So, this is what is the signal flow graph for 8 point d f t computation by using d i f f t algorithm. So, reference thank you.