 So, what are the specifications then for the low pass filter? Pass pan edge 1, top pan edge of mode AS1 and mode SL2. The one which is closer is important that puts us more stringent transition band. This is carried as they are. Pass pan tolerance pick the more stringent of the 2 pass pan tolerances if they are different. Of course, if they are the same there is no problem just take that tolerance as it is. The stop pan tolerance carried as it is. Stop pan nature carry as they are. Equal or monotonic as they are. Naturally you cannot have different kinds of natures for the 2 pass pans or 2 stop bands in the band pass and band stop filter they have to be carried as they are. Finally, yes please there is a question. So, the question is when we design we are ultimately going to design with the same tolerance. Yes we are going to are we going to design with the same tolerance. Yes indeed we are going to design with the same tolerance but that tolerance could come from initially different tolerances. That means that obviously when you have different tolerances one of the tolerances is more stringent than the other. There is a well ordering principle between tolerances which is not there between natures. One nature is not better than another nature or more difficult than another nature and secondly one nature cannot transform into another nature. You see unlike tolerances. Tolerances if you have definitely obeyed the tolerance of 0.1 you obeyed the tolerance of 0.12 but that is not true with natures. So, you can possibly have different tolerances and satisfy both of them by taking the more stringent one but you cannot do that for natures. So, the nature must be the same but the tolerances can be different. Yes there is a question. So, the question is can omega that is a good question. Can omega naught fall outside omega s1 to omega s2 that is very interesting. Now you see omega naught is the geometric mean of omega p1 and omega p2. So, all that we can say is that it is between omega p1 and omega p2 but then of course you can yes in principle choose omega s1 and omega s2. So, that it is just on one side of this geometric mean in principle yes it is possible but in practice that would be very rare but then there is trouble. If you do that then you have the whole all this transformation goes haywire. That means if we do indeed put omega s1 and omega s2 just on one side of the geometric mean then we have trouble in this transformation. That is a good question. So, you know can omega s1 and omega s2 you know of course you are free to choose it anywhere between omega p1 and omega p2 and then omega naught squared is omega p1 times omega p2. So, it is the geometric mean but if you choose omega s1 and omega s2 on one side of the geometric mean between omega p1 and omega p2 then this design procedure falls flat. You cannot do this because both the mapping of omega p1 and omega s1 and omega s2 will come on the same side of the low pass frequency axis. So, then in fact I leave this to you as you know a food for thought. What would you do? Can you change your strategy a little bit if indeed this pathological situation arises right? So, I put this is a good question that has been asked we will put that as a challenge. In the pathological case omega s1 and omega s2 being on the same side of omega naught what can one do? One would need of course to modify this procedure. Now that we are anyway discussing challenges I would like to put one more challenge before the class. Another challenge is for the same low pass specifications the Chebyshev filter order is always less than equal to the Butterworth filter order but a yield by the same low pass specifications is only tolerance. Tolerance is same and the passband stopband edges are the same of course the edges are different. That means you are asking for an equi-repro passband or a monotonic stopband and putting the same passband edge and stopband edge and the same tolerance. Then you ask for a monotonic passband and a monotonic stopband with the same passband edge, same stopband edge and same tolerances and you find the order in both these cases. And the challenge is to show that the Chebyshev order will always be less than or equal to the Butterworth order. So it is very interesting. There is a reason therefore why we would choose one nature or the other. I told you that filter design is essentially a game of approximating the ideal and approximations always involve compromises. Nothing comes for free. So it might seem that when you choose an equi-repro passband you should always be doing better why at all should you use a monotonic passband if the order is more. The answer lies in the phase response what you have brushed under the carpet. So phase response of the Chebyshev filter happens to be worse in many ways and worse in what way? Worse in the sense of group delay and phase delay. You would like the phase delay and the group delay to be as constant as possible over the passband at least over the passband anyway. And that poses a problem. So you know in fact it would be interesting. It is not easy to prove this. I mean you know it is little involved but one should in one the designs that one would do in this course you would when you actually go through a design for the Butterworth of the Chebyshev filter it would be definitely worthwhile to plot the phase response. For you know in fact in the class different people would perhaps look at Butterworth designs and Chebyshev designs. And therefore it would be worthwhile to plot the phase response in different cases and compare the phase responses. It would also be worthwhile to take the same specifications and compare the orders that one is getting for the Butterworth filter and Chebyshev filter to verify the truth of what one has been saying here in this challenge. So that would be an interesting thing to do. Anyway so much so then for yes please the question. So that is an interesting question. In fact maybe the question is you know when you when you took the band pass filter you had used omega the passband edges to define the omega not the the center frequency omega not squared. So why would you why can't you do the same thing for the band stop filter. That is a very good question. In fact maybe that is just an answer to a challenge that I posed. So anyway I would certainly urge you to reflect on what has been suggested. The suggestion is that instead of using omega p1 omega p2 geometric mean as omega not as omega not could you use the geometric mean of omega s1 and omega s2 as omega not. How would the whole design change and that perhaps is a very interesting alternative to consider. Anyway now we are well equipped then to design band stop filters as well and in fact now let us summarize the process of any kind of discrete band filter design. The overview with hindsight as they say. Hindsight means having now traversed the territory of discrete filter design discrete piecewise constant filter design. First the first step is discrete time unnormalized specifications. This is what is given to you. This must first be transformed into normalized specifications by dividing by the sampling frequency dividing by fs and multiplying by 2 pi. You get the normalized specs. The next step is to go from the normalized specifications to the corresponding analog filter specifications. So normalized discrete time specs in omega that is. We use to binary a transform, binary a frequency transform. Omega is time small omega by 2 and that gives you the analog filter specifications of appropriate nature in omega. Of course here the analog filter has the same nature. So high pass goes to high pass, band pass to band stop, band pass to band pass and band stop to band stop and of course low pass to low pass. The next step is to use the frequency transformation. The analog filter specs use the frequency transform to design to get the specs for the corresponding analog low pass filter. So you know how to do that for the band pass filter and the band stop filter. In our method of design we have put the passband edge at 1 and the stopband edge is chosen to be the more stringent that emerges from the frequency transformation. We have of course put the more stringent of the tolerances on the passband and the stopband and nature is carried as it is. So we have designed that analog low pass filter. Let us call it H analog low pass filter with the frequency variable SL. SL is the complex frequency variable in the analog low pass and of course now we know what to do. In fact we have two roots that we can follow. So we have H analog low pass filter as a function of SL. Now we can either make SL equal to HS. This could be either band pass, band stop or high pass whichever one we desire. So we make a frequency transformation in the analog domain and then we replace S by 1 minus Z inverse by 1 plus Z inverse the bilinear transform and we get the discrete H as a function of Z. This is one root that we can follow and that is the root that we have discussed so far. But you see there is another root. You could in principle replace SL by 1 minus ZL inverse by 1 plus ZL inverse and get something in between. So make a bilinear transformation first and then replace ZL inverse by some H discrete Z. And come there. You could also do that in principle. That means for each of these three kinds of filters the high pass, the band pass and the band stop you could conceive of a discrete transformation taking you from ZL inverse to Z or Z inverse as you like. Z inverse Now the exercise for you is to work out this discrete transformation which DZ for all three cases high pass, band pass and band stop. And a corollary to that question is observe carefully the nature of this transformation. Think of this transformation. Of course we expect that transformation to be rational. It could not be irrational. Now if it is a rational transformation you could think of it as a rational system in its own right. What properties does that system have? It is a question I leave to you as an exercise. With captain we are now all set to carry out filter designs. We are in a position to design the filters that we have been assigned whether band pass, band stop or high pass right from the discrete filter specifications right up to the discrete system function. So of course the infinite impulse response filter can now be designed. We shall now therefore actually we should now individually proceed to design the filters that we have been assigned and from the next lecture we shall commence our discussion on the design of finite impulse response filters. So with that now we are well equipped and we all urge one another to design the infinite impulse response filters that we are required. Thank you.