 I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering, Walchand Institute of Technology, Solapur. Today, we will see how to design a FIR filter using frequency sampling method, the learning outcome. At the end of the session, student will be able to explain finite impulse response filter design using frequency sampling method. To design a filter, the desired frequency response is given. Generally, it is represented as hd e raised to j omega. So, set of samples is determined from this desired frequency response and these are known as the DFT coefficients. After sampling, the IDFT of these samples gives us the filter coefficients. Generally, this hd e raised to j omega is sampled at m equal points or equally spaced points with omega k equal to 0 to m minus 1 at uniformly spaced around the unit circle. There are two types of design. One is known as a type 1 design which includes the sample present at omega equal to 0 and another design is known as type 2 design where the frequency sample at omega equal to 0 is omitted. We will concentrate on type 1 design. So, as I said earlier, the samples of hd e raised to j omega are taken at omega k equal to 2 pi k by m where k is equal to 0 to m minus 1. So, the spacing between two samples is 2 pi by m. These samples are given by or represented as h bar k equal to hd e raised to j omega where omega equal to omega k with k ranging from 0 to m minus 1. So, it is equal to hd now replace omega by omega k. So, gives us hd e raised to j 2 pi k by m where k is 0 to m minus 1. Now, from these samples, now these samples are actually known as the DFT coefficients. The filter coefficients hn can be computed from these samples by taking the IDFT of h bar k. So, this is written as h of n is equal to 1 upon m k equal to 0 to m minus 1 h bar k e raised to j 2 pi n k upon m where n is equal to 0 to m minus 1. So, this gives us the m minus sorry m coefficients of our filter. If the numbers which are obtained by the previous equation are real numbers, then these will represent the coefficients of our filter. Now, for getting these numbers as real, all the complex sums must appear as a complex conjugate pairs. So, the term h bar k e raised to j 2 pi n k upon m should be matched with the term with exponential e raised to minus j 2 pi n k upon m. So, that after addition n this term will become a real term. Therefore, therefore the matching terms are h bar k e raised to j 2 pi n k upon m and h bar of m minus k e raised to j 2 pi n m minus k by n. So, nothing on why this term rather e raised to j 2 pi n m minus k by m is the matching term for this. Pause the video for a minute and write down your answer. See as I told you this term e raised to j 2 pi n k m minus k upon m is equal to e raised to j 2 pi n m upon m into e raised to minus j 2 pi n k upon m. So, this is equal to e raised to j 2 pi n into e raised to minus j 2 pi n k upon m. So, this is equal to e raised to minus j 2 pi n k upon m because this term is equal to 1. So, this is the reason why the term with e raised to j 2 pi n k upon m and this term that is h of m minus k e raised to j 2 pi n m minus k by m are the matching terms. So, these terms are complex conjugate if h bar 0 is real and in case when m is odd h bar of m minus k is equal to h bar star k where k equal to 1 to m minus 1 by 2 in case when m is odd and if m is even in that case h bar m minus k is equal to h bar star k where h the star represents the complex conjugate k equal to 1 to m by 2 minus 1 and in case of m even this h bar of m by 2 must be equal to 0. So, if these conditions are satisfied in that case these terms are complex conjugate of each other and when you are adding these terms they will turn to be either a sine terms or a cosine terms and the resultant coefficients h of n will all be real. So, with this the filter coefficients h of n are given by h of n is equal to 1 upon m h bar 0 plus 2 into k equal to 1 to m minus 1 by 2 real part of h bar k e raised to j 2 pi n k upon n when m is odd and h of n is equal to 1 upon m h bar 0 plus 2 into k equal to 1 to m by 2 minus 1 real part of h bar k e raised to j 2 pi n k upon m. Of course, these equations we get from the previous equations where m is even. So, once we obtain these filter coefficients the system function of this f r filter can be determined by taking the z transform that is h z is equal to n equal to 0 to m minus 1 h of n z raised to minus n. So, this gives us the filter transfer function. So, this is how from the desired frequency response by sampling we obtain the coefficients which are known as the DFT coefficients and by taking IDFT we get the impulse response of our finite impulse response filter reference. Thank you.