 Hello everyone, myself, K. R. Beradhar, assistant professor, department of ENTC, WIT, Saulapur. Today I am going to deal the topic called discrete Fourier transform. Let us start with the learning outcomes first. At the end of this session, students will be able to find the discrete time Fourier transform of the sequence, compare different methods of finding the DFT. These are the some of the contents. Start with definition of DFT, different methods of finding DFT, matrix method first one, DFT using FFT and lastly references. Recall once the conventional method of finding the DFT and think for a moment whether it goes suitable for conventional method or using two other methods mentioned earlier. Using conventional method definitely takes more number of steps and even it lengthier and it takes more time to find the DFT using. About mention using matrix method and using FFT, it is required very less time to compute the DFT of the sequence. So, this is the definition of DFT. Discrete Fourier transform for an endpoint sequence is defined as x of k is equal to summation equal to 0 to n minus 1 x of n e raised to minus j 2 pi k n divided by capital n. So, this is also written in terms of twiddle factor x of k is equal to summation equal to 0 to n minus 1 x of n W n raised to n k, where W n is the twiddle factor which is equal to e raised to minus j 2 pi divided by capital N, where k is equal to 0 1 2 3 up to n minus 1. Different methods of finding the DFT are conventional method already I mentioned, but other than these there are two methods to find the DFT that is DFT using matrix method, DFT using first Fourier transform algorithm also called the DFT using FFT. So, these two methods I will discuss by taking one example here. The example shown here is find the DFT of the sequence x of n is equal to 1 2 3 4. Let us take this example x of n is equal to 1 2 3 4. We need to find the DFT for this sequence using first method that is using matrix method. So, if you want to find the DFT using matrix method we need to form the matrix bar given value of n that is here n is equal to 4. How we are going to form the matrix is this is omega 4 raised to 0, omega 4 raised to 1, omega 4 raised to 2, since here n is equal to 0 1 2 and 3, here k is equal to 0 1 2 and 3. Now, we know that is a total factor omega n raised to nk, we keep on substituting the value of n and k, so in the first row k is equal to 0, all that is nothing but factors is going to becomes equal to 0 that is omega 4 raised to 0, omega 4 raised to 0 this is also omega 4 raised to 0, this is omega 4 raised to 0. Similarly, in the column also you are going to get omega 4 raised to 0 omega 4 raised to 0 omega 4 raised to 0. Since n is equal to 0, we are going to get that is omega all 0s. Here omega 4 raised to nk is there, n value is 1 k is also 1, this is also 1 omega 4 raise to, here n value is 2, k value is 1, that is 2 into 1 is 2. Similarly, this is omega 4 raise to 3, this all similarly, omega 4 raise to 2, omega 4 raise to 4, omega 4 raise to 6, this is omega 4 raise to 3, omega 4 raise to 6, omega 4 raise to 9. Sometimes this is also called omega r w. Now, we are going to form, find the values of w here. To find that one, we need to draw a circle, chop up with equally 4 points usual. This is a value is 1, this value is minus 1, this is j, this is minus j. Now, this assume that omega 4 raise to, this is 0, this is omega 4 raise to 1, this is omega 4 raise to 2, this is omega 4 raise to 3. Now, if I want to write these values here, omega 4 raise to 0 all that is equal to 1, this is 1, this is also 1, 1, this side 1, 1, 1. Similarly, omega 4 raise to 1 value is minus j, omega 4 raise to 2 is minus 1, omega 4 raise to 3 value is j. Similarly, omega 4 raise to 2 value 2 is minus 1, this is omega 4 raise to 4. So, omega 4 raise to 0 is, you know 4 raise to 1, omega 4 raise to 2, omega 4 raise to 3, this is omega 4 raise to 5 again, 5, 6, 7, 8, 9, 10. This will continue. Now, omega 4 is to 0 is already 3 is j, 4 is now 1, omega 4 is to 6. This is 4, 5, 6 is minus 1. Similarly, omega 4 is to 3 is what is the value omega 4 is to 3 is j. What about 6? This is 4, 5, 6 is minus 1. Similarly, omega 4, 9, 9 is again minus j. This is the matrix how we are going to form. So, after finding this matrix, we need to write this matrix as usual. This is 1, 1, 1, 1. This side all column become 1 and this is minus j. This is also minus j. This is j. This is j minus 1, minus 1, minus 1. This is 1. Now, we need to remember like this. So, what is the problem? Sequence has given sequence 1, 2, 3, 4. 1, 2, 3 and 4. Multiply. This is 4 by 4 matrix. This is 4 by 1 matrix. The inner dimensions are same so that you can multiply these two matrices. Now, this 1 will be all that is row elements will be multiplied with the column. Similarly, second row element will be multiplied with the second column and third row element will be multiplied with the column. Now, this becomes 1 plus 2. This 1 into 1 is 1. 1 into 2 is 2. 1 into 3 plus 3 plus 1 into 4 is equal to this is total 10. Similarly, second row will be multiplied with the column 1 minus 2 j minus 3 plus 4 j is equal to minus 2 plus 2 j. Similarly, 1 minus 2, this is plus 3 minus 4 is equal to 4 plus 3 plus 1 is 4, 4 plus 4 minus 6 is minus 2. Similarly, 1 plus 2 j minus 3 minus 4 j is equal to minus 2 minus 2 j. This is the final answer to find. This is the sequence for the given sequence 1, 2, 3, 4 d of t is x of k is equal to 10 minus 2 plus 2 j minus 2 minus 2 minus 2 j. Similarly, using second method is using FFT that is DFT using FFT. In this diagram method, we need to draw the butterfly structure for the given input, four inputs. This is x of 0, x of 2, x of 1, x of 3. So, what is this? This is 1, this is 3, this is 2, this is 4. Draw the butterfly structure here using DFT algorithm. Now, this will be minus 1, minus 1, this is minus 1, minus 1. The output will be x 0, x 1, x 2 and x 3. Here, omega 4 is to 0, omega 4 is to 0, this is omega 4 is to 0, omega 4 is to 1. For the given input, here 1, this is 3. So, 1 comes here and 3 will be multiplied with omega 4 is to 0, that is 1. 3 will be multiplied with 1, it is 3 only, it comes here. 3 plus 1 is equal to 4. Similarly, 1 comes from here. So, 3 comes and 3 will be multiplied with 1, 3 multiplied with minus 1, again minus 3. So, 1 minus 3 is equal to minus 2. Similarly, in this case, 2 comes here. So, this 4 will be multiplied with omega 4 is to 0 is 1, 4 will be multiplied with 1, again 4 comes here, 4 plus 2 is equal to 6. Similarly, 2 comes here, 2 and 4 will be multiplied with minus 4, 2 minus 4 is minus 2. So, at the output of the second stage, so this 4 comes here and that is 4 and 6 will be multiplied with 0, omega 4 is to 1, 6 plus 4 equal to 10. So, you are getting same answer as the previous case. This is minus 2. Similarly, this minus 2 comes will be multiplied with omega 4 is to 1 is minus j, that is 2 minus 2 multiplied by minus j is plus 2 j, minus 2 plus 2 j. Similarly, 4 comes from this one, 4 comes here and 6 will be multiplied with minus 1, minus 6, minus 6 is equal to minus 2. Similarly, this comes minus 2, minus 2, this is minus 2, minus j plus 2 j and multiplied with minus 1, minus 2 j, this is the answer. So, x of k is equal to, answer will be 10, minus 2 plus 2 j, minus 2, minus 2 j, this is the final. So, the same as the previous. So, but using these two methods, you can find the DFT very fast and accurate, whereas the conventional method requires very large time and more computational steps. This is the answer. These are references I referred. Thank you.