 Hello everyone. I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering, Walshchand Institute of Technology, Soolapur. Today we will discuss FR filter design using windows. Learning outcome. At the end of this session, student will be able to explain the procedure to design FR filters using windowing technique. Also, he will be able to state the frequency domain characteristics of different windows. Design procedure begins with the frequency response specification that is hd omega and corresponding impulse response hdn that is a desired frequency response hd omega. The relation between these is hd omega equal to summation n equal to 0 to infinity hdn e raise to minus j omega n, where hdn can be obtained by taking inverse Fourier transform of hd omega. When this hdn is obtained, we have seen previously that this hdn that is a desired impulse response is infinite in duration and so, such a filter is not practically realizable. So, we mean we must truncate it to some finite duration. So, truncation of hdn to a length m is equivalent to multiplying it by a rectangular window, where this window function is defined as wn equal to 1 for n equal to 0 to m minus 1 0 otherwise. Let us recall what is the equivalent operation in frequency domain for multiplication of two signals in time domain, because in this case actually we are obtaining impulse response of the filter by multiplying hdn that is a desired impulse response by window function. Pause the video for a minute, recall what is that operation. Okay, we know that the multiplication in time domain is equivalent to convolution in frequency domain. So, in this case the unit impulse response of the FIFilter as we have seen is obtained by multiplying this desired unit impulse response by window coefficient wn. As in case of a rectangular window the coefficients are all 1, your hn will be equal to hdn for n equal to 0 to m minus 1 and 0 otherwise. Now, hn is a finite duration and it is causal and so it is practically realizable. So, in frequency domain this operation is as I told you that it is equal to h omega equal to 1 upon 2 pi integration minus pi to plus pi hd theta w omega minus theta d theta. So, that is basically a convolution of hd omega and w omega where this w omega is n equal to 0 to m minus 1 wn e raise to minus j omega n. So, as we have seen this rectangular window function the Fourier transform is w omega where if you obtain this values. So, the summation comes out to be 1 m minus e raise to minus j omega m divided by 1 minus e raise to minus j omega which gives us equal to e raise to minus j omega m minus 1 by 2 sin of omega m by 2 upon sin of omega by 2. So, the magnitude of this particular function is given by this between minus pi to pi and phase parties equal to minus omega m minus 1 by 2 whenever it is positive and minus omega m minus 1 by 2 plus pi when sin of omega m by 2 is less than 0 that is negative. So, if you see this particular function if you plot it is a some sinc function. So, this is the frequency response for a rectangular window plotted. So, if you look at this is this is the window of sorry this particular rectangular window is of length 15. So, if you see from this particular equation when omega equal to 0 you obtain this value of magnitude is equal to m. So, this is equal to m at 0 and this is the width of the main lobe. So, this is the main lobe and these are all the side lobes. If you see this is a main lobe width and we can very easily obtain from that equation this main lobe width is equal to 4 pi by m. So, if you look at the effect of this truncation that is a truncated filter as we have seen this is a convolution of the desired frequency response. So, this is the desired frequency response of a filter and this sinc function is the frequency response of our window. So, here for convolution we are multiplying this w omega minus theta with this h d theta shift and that is as per convolution operation. If we look look at and if you plot this h omega you will see that it is will not be exactly equal to what the desired frequency response is, but you will get this repulse in the passband as well as in the stopband. So, as in case of an ideal frequency response we are expecting this sharp cutoff, but practically because of truncation we will get this transition width or the transition will not be sharp. So, it takes some time to take transition from passband to stopband. So, basically the width of the transition band it depends on the main lobe width whereas, the side lobes peak of the side lobes here it depends on the peak of the side lobes of your window function. So, the width of the main lobe is 4 pi by m as far as the rectangular window is concerned and we see that with the increase in m that is the length of the filter the main lobe becomes narrower because 4 pi by m. So, which will improve the transition characteristics, but the side lobes of this function are relatively high. So, which will increase the peak of the side lobes. So, if I am going for a better transition characteristics it is ambering the stopband characteristics that is the stopband ripple increases. The ripple in the passband and stopband basically they depends on the area under the side lobes. So, in case of a rectangular window as we know that it is a sharp because all the coefficients are equal to 1 between the window length. So, by debring this window smoothly to 0 at each end the height of the side lobes can be decreased and when a height of the side lobe is reduced which will improve the stopband characteristics, but the effect of this is that it increases the main lobe width and hence wider transition band. Requirements are conflicting. So, if I am going for increasing m which is improving the transition characteristics, but at the same time it is affecting the stopband characteristics. So, these are the other window functions. Hamming, Hanning, Blackman and Bartlett, Bartlett are triangular. If you look at now in case of a rectangular window it was constant throughout the length, but whereas now here they are smoother around the each side. So, which improves the side lobe characteristics or which improves the stopband characteristics. So, these are the frequency response of these windows for two different lengths that is m is equal to 1 sorry m equal to 31 and m equal to 61. This is for Hamming window, this is for Hamming window and this is for Blackman window. So, actually these window functions we can obtain from these equations. So, these windows are all defined for n equal to 0 to m minus 1. So, Bartlett window, Blackman, Hamming and Hanning. So, these are the frequency domain characteristics of all these window functions in case of. So, this is the approximate transition width and this is the peak side loops. In case of a rectangular window this is 4 pi by m, Bartlett, Hamming, Hanning it is 8 pi by m, Blackman it is 12 pi by m and these are the side loops. If you look at in case of rectangular it is a width is less compared to all other, but whereas the peak side loops also is less. So, wider transition region is compensated by much lower side lobes and thus lower side lobes because in this case in case of a Blackman has a lower side lobes which improves the characteristics both the ripples will be less in passband as well as a stopband. So, this is a overall design procedure for the windowing method. First select a suitable window function then specify an ideal response that is HD omega. Compute the coefficients of ideal filter that is HDN that is by taking the inverse Fourier transform of those. Then multiply ideal coefficients by the window functions to give filter coefficients. Once you get these filter coefficients evaluate the frequency response of the resulting filter. So, then if you are not satisfied with the characteristics you may iterate and you may change the length of the filter and once again you can evaluate. So, this is how we can design the FI filters by using windowing method. References. Thank you.