 Hello everyone, I am Prashant S. Malge, assistant professor, Department of Electronics Engineering, Walton Institute of Technology, Solaapur. Today, we will discuss decimation in time fast Fourier transform algorithm, the learning outcome. At the end of the session, student will be able to draw and explain this signal flow graph for decimation in time FFT algorithm. So, previously we have already seen the divide and conquer approach for computation of DFT, in which a endpoint DFT can be computed in terms of smaller size DFTs. So, let us consider n is 2 raise to v. So, basically n is the power of 2 and we want to compute this DFT by using divide and conquer approach. Let us use column wise mapping for input xn and row wise mapping for the output xk that is DFT. By choosing m equal to n by 2 and l equal to 2, it splits our input signal x of n into 2 data sequences of n by 2 points. As the first sequence is F1n which is equal to x of 2n, second signal is F2n which is x of 2n plus 1 with n equal to 0 to n by 2 minus 1. It means what? It is decimated into two sequences. One is all even valued samples of xn and second is a odd valued samples of xn. So, it means the with l equal to 0, the first column rather the first 0 sorry first row will have all even samples of x of n and second row will have all odd samples of. So, now let us consider this xk. As we know xk is represented as n equal to 0 to n minus 1 xn w n kn where w n is basically a 2 del factor which is e raise to minus j 2 pi by n. So, in terms of that this is written as w n kn where k equal to 0 to n minus 1 gives you n points of dft xk. So, now this can be written in terms of even samples of xn and odd samples of xn. So, this summation can be divided into two parts where all even valued samples are taken in first summation and odd valued summations are second in second summation. So, this can be written as x of 2 m now this m will be 0 to n by 2 minus 1. So, now n to be replaced by 2 m. So, it will be w n 2 m k. So, this gives you even and for odd this n I have to substitute as 2 m plus 1. So, m equal to 0 to n by 2 minus 1 x of 2 m plus 1. So, w n now once again n is to be replaced by 2 m plus 1 k. So, you will get this. So, now if you write this w 2 n. So, w 2 n is basically e raise to minus j because w n is e raise to minus j 2 pi by n w 2 n will be e raise to minus j 4 pi by n which is equal to e raise to minus j 2 pi by n by 2 which is w n by 2. So, w 2 n can be also be written as w n by 2. So, with this notation now this once again x k can be represented as now I am representing x 2 m as we have know already seen that it is the f 1 m. Now this w 2 m k I can write as w m k n by 2. So, w m k n by 2 and now here this can be written as w 2 m k n which is once again w m k n by 2 and the remaining factor is w k n. So, w k n is taken as here and this x of 2 m plus 1 as we have seen previously it is f 2 m. So, now this DFT is now represented as this. Now if you look at this and if you compare with this this is as if it is a n by 2 point DFT of signal f 1 m. So, this I will represent it as f 1 k where f 1 k is a n by 2 point DFT. Similarly, if you take this second part is a n by 2 point DFT of f 2 n right. So, n by 2 point DFT of f 2 n. So, now that will represent it as f 2 k. So, this x k can be written as f 1 k plus w k n f 2 k where k equal to 0 to n minus 1 gives us n DFT n points of DFT right. Now making use of some symmetry property of w k n and periodicity property we know that DFT is periodic with period size of the DFT. So, f 1 k is a n by 2 point DFT. So, f 1 k plus n by 2 will be f 1 k f 2 k plus n by 2 will be f 2 k and w k plus n by 2 as per symmetry property. So, w n k plus n by 2 equal to minus w k n. So, with these this x k can be written as x k equal to f 1 k plus w k and f 2 k for k equal to 0 to n by 2 minus 1 and k plus n by 2. So, if I take x of k plus n by 2 will be f 1 k plus n by 2 which is equal to f 1 k w k plus n by 2 n which is equal to minus w k n f 2 k plus n by 2 which is equal to f 2 k. Therefore, this x k which is n by n point DFT now can be represented in terms of n point n by 2 point DFT is f 1 k and f 2 k right. So, we have represented n point DFT x k in terms of n by 2 point DFT is f 1 k and f 2 k. Now, let us recall the number of complex multiplication and complex additions required for computing n point DFT because finally, we are comparing computations required for direct computation and your DIT fFT. Pause and video for a minute and think on this ok. So, as we know the number of computations required for computation of n point DFT by using direct computations is n square and for n point DFT the number of complex additions required are n into n minus 1 ok. So, now, let us continue with this remember now we want to compare these computations for direct computation of f 1 k and f 2 k because these are n by 2 point DFT is require n by 2 square complex multiplications each. Now, n by 2 additional complex multiplications are required for multiplying w n k into f 2 k right. And then with the first step that is after dividing that n point DFT into n by 2 point DFTs the number of complex multiplications are for 2 n by 2 point DFTs 2 n by 2 square plus additional n by 2. So, which is n square by 2 plus n by 2. So, which is approximately for a large value of n n square by 2. So, the number of complex multiplications comes down from n square to n square by 2 and same way this is only with one step right. So, in the next step further decimation we can do. So, by decimating f 1 n and f 2 n in time results into n by 4 point DFTs. So, v 1 1 n as f 1 2 n. So, v 1 2 n as a f 1 2 n plus 1. So, basically v 1 2 and v 1 1 n these are two sequences we get by decimating your f 1 n in time. So, these two are n by 4 point DFTs. Similarly, from f 2 n we get v 2 1 and v 2 2 n right. So, these are also 2 n by 4 point DFTs. So, it means what now these n by 2 point DFTs f 1 k and f 2 k now can be written in terms of these n by 4 point DFTs. So, by computing n by 4 point DFTs we can now compute these n by 2 point DFTs f 1 k and f 2 k. So, this is how we can do it f 1 k equal to v 1 1 k w n by 2 k v 1 2 k where k equal to 0 to n by 4 minus 1 on same lines f 1 of k plus n by 4 equal to v 1 1 k minus w k n by 2 v 1 2 k and same for f 2 k and f 2 k plus n by 4 which is written in terms of your n by 4 point DFTs v 2 1 and v 2 2 k right. So, where v i j k are all n by 4 point DFTs of sequences v i j right. So, now if we compare once again a computation of v i j k these are all n by 4 point DFTs. So, they are required and how many are there v 1 1 k v 1 2 k v 2 1 k and v 2 2 2 k. So, there are 4 n by 4 point DFTs. So, they need 4 into n by 4 square complex multiplications with which we get additionally multiplications if we compare. So, 4 into n by 4 square means n square by 4 plus here you need n by 4 multiplications here you need n by 4 multiplications means n by 2 plus for f 2 k multiplied by w k and n by 2. So, those additional n multiplications. So, the total number of multiplications will be n square by 4 plus n right. So, once again if you compare for a large values of n. So, it negligible. So, it comes out to be n square by 4. So, with twice decimation it reduces from n square to n square by 4 almost one fourth of the total and the same way you can continue right. So, it means for n equal to 2 raise to v this decimation can be performed as v times v equal to log of n to the base 2. So, the total number of complex multiplication is reduced to n by 2 log of n to the base 2 because there are n by 2 multiplications for each step. So, the number of complex additions will be n into log of n to the base 2. Actually these calculations we can understand from this particular structure. Now, remember this structure is known as a butterfly structure where you have supposed to inputs a and b and here is a multiplication factor some today factor w dash n. So, this a comes here this b gets multiplied by w dash n. So, multiplication of this comes here as and here this is a summing point. So, it will be a plus w dash n b and here as it gets multiplied by minus 1. So, this will give you a minus w dash n b right. So, basically because as we have seen x k written as f 1 k plus some factor into f 2 k and k plus n by 4 as f 1 k minus some factor into f 2 k. So, this structure is going to help us to draw the algorithm. So, this one structure actually requires one multiplication and there are two additions. So, one complex multiplication and two complex additions right. So, using this you can draw the signal flow graph. Remember here the input sequence comes in the bit reversed order right with one decimation it divides into 0, 2, 4, 6 and second decimation 0 4 2 6 1 5 and 3 7. So, now the first stage the computation of x k in terms of f 1 k and f 2 k f 1 k computation in terms of v 1 1 k and v 1 2 k right. So, using those equations you can draw this particular structure right. So, here the multiplication factor will be. So, one of the if I say this is f 1 0 plus f 2 0. So, x of 0 will be equal to f 1 0 plus f 2 0 and x of 1 will be f 1 1 plus f 2 1 into this factor w n k that is w 1 8 and so on. So, similarly you can understand this particular structure and this is what the signal flow graph references. Thank you.