 And now we would like to look at the corresponding transformation for band pass filters. Now for band pass filters, we need to think a little more, you know for band pass filters it is very clear that you have multiple stop bands, there is even though there is only one pass band, there are multiple stop bands. How on earth could you get multiple stop bands from a single stop band? You cannot do it by a single element. In fact here we can probably take a queue from what we do in LC networks again, remember we are going to choose an impedance of an LC network. Now what simplest form of an LC network can give us an impedance which has multiple mappings that is you know you want multiple frequencies to have the same magnitude that is the kind of mapping you want. You want the stop band to go to multiple places, how on earth can that happen? That can only happen if you have a second order system, at least it cannot be less than second order, but we can probably make do with second order. Now what we mean by second order is you need an inductance and a capacitance to come together. So let us consider a series LC network and consider its impedance, we have an L and a C and of course we have L s plus 1 by C s. So that gives us LC s squared plus 1 divided by C s. Now we can divide by LC in the numerator and denominator and that gives us s by L which we will write as s squared plus omega naught squared by B s where omega naught squared is 1 by LC and B is a positive quantity equal to 1 by L. Let us consider this transform. Now omega naught squared which is 1 by LC has a very important physical significance. In fact all of us would probably be aware that for an LC network that corresponds to what is called the resonant frequency of the network, the resonant angular frequency of the network. In fact it is that frequency at which this impedance vanishes. So you see the interesting thing is that B also has a significance but that significance we may not all be aware of and let us not be too worried about it for the moment. We will see it in the long run. At the moment we will just regard B as a positive quantity, it is the reciprocal of 1 of L, 1 by L. So let us consider as a candidate a cell to be replaced by FBPS where FBPS is s squared plus omega naught squared divided by B s and let us go through the exercise first of checking for the, now we will do it the other way. We will first check property 1 then property 2 and then we will look at property 3. So of course property 1 is obvious again. This is a rational transformation, rationality is going to be maintained. So if the original low pass filter is rational, this has the transform filter has no choice but to be rational because you are replacing SL by a rational function of S. So rationality is maintained anyway, there is no problem with that. Now we need to look at the second property stability. So all that we need to do is to substitute S equal to, in fact you know we can do something simpler. Let us divide this into 2 parts. Let us write this as S by B plus omega naught squared by B into 1 by S and now let us substitute S equal to sigma plus j omega and then we have SL becomes sigma plus j omega divided by B plus omega naught squared by B into 1 by S which is sigma minus j omega divided by sigma squared plus omega squared. And therefore if you write SL as sigma L plus j omega L then sigma L is 1 by B times sigma plus omega naught squared by B into sigma by sigma squared plus omega naught squared by plus omega squared, the real part we are taking the real part out. This is the real part of this quantity and this is the real part of this quantity. So we have extracted them. Now you see it is obvious, if sigma is positive then this has no choice but to be positive and neither does this because 1 by B is positive, omega naught squared by B is positive. If sigma is positive, sigma squared plus omega squared cannot possibly be 0, it has to be positive anyway and therefore this is bound to be positive and so is this and therefore sigma L is positive. On the other hand, if sigma is negative then this is bound to be negative and this is definitely bound to be non-zero because sigma squared plus omega squared cannot possibly be 0 if sigma is negative, it is bound to be positive. So this is positive, this is positive, so this whole thing would become negative. So negative quantity plus the negative quantity would give a negative quantity and therefore sigma and sigma L necessarily have the same sign strictly. So if sigma L is positive, if sigma is positive then sigma L is positive, sigma is negative, sigma L is negative and in fact right from here we can also see that if sigma is 0 which means if you are going to the imaginary axis then both of these quantities are 0 and therefore sigma L is 0 as well and therefore we also know what to expect on the imaginary axis. The imaginary axis goes to the imaginary axis which is not a surprise because anyway you are dealing with an LC impedance. So stability has been obeyed and therefore we have guaranteed that if you have a stable analog filter design that would continue to remain stable when you go to the band pass domain. Now the question is why should we call this band pass and for that we need to look at the frequency, sinusoidal frequency transformation. So now let us look at what happens when S equal to j omega L only, S equal to j omega only. So of course that gives us S L which is equal to, it will give us j omega L and j omega L is going to be omega naught squared minus omega squared by j B omega. That is j omega squared minus omega naught squared by B omega and this is very interesting. Let us see how this behaves. Now let us consider some critical points on the frequencies. Which critical points we need to look at? Well let us take a band pass. So a typical band pass filter would have a characteristic like this. It would have two edges of the pass span. Let us call them omega p1 and omega p2. Yes please, there is a question. So the question is what are we practically doing when we make this transformation? Now the answer is all this is still in the phase of design. So we are not realizing it as yet. So all this is still design and we are computing the filter. We have not yet reached a point where we have obtained the filter coefficients. We are yet designing on paper. After we have completed the design on paper, then we would translate it into a realization. So as yet all this design is in calculation. So we have specifications for the discrete time filter. We have converted them into specifications for the desired kind of analog filter. Now we are going into the specifications of the desired kind of low pass filter. Now we will see in a minute. We will design that low pass filter. We will then convert s by using s equal to by using the bilinear transform and there we get a discrete time filter that we would try to realize. So realization is after completing the design. So there we have a band pass filter and we assume as usual that we have tolerances. Now we are quite satisfied with letting the pass band be between 1 and 1 minus delta 1 and the stop band be not more than delta 2 in magnitude and of course you know the nature can be specified. Now you see you will have to decide on the nature of the stop bands here. There are two stop bands. You cannot have different natures for the two stop bands. Either both of them must be monotonic or both of them must be equilateral. We do not at the moment have a facility to allow different kinds of natures for the two stop band. Anyway we do have a facility to allow different tolerances in the two stop band. All that we need to do is if the tolerances are different consider the stronger of the tolerance. So for example if one of the stop bands is the tolerance which forces it to be less than 0.1 and the other one must be less than 0.09 then it is 0.09 that you must choose in your design. Choose the stronger or the more stringent one and then you can put that one if you are satisfying 0.09 you are of course satisfying 0.1. So you will have to do that. Take the more stringent. Having done that we can agree on a set of specifications like this. Let us put omega naught somewhere in between here. We expect omega naught to be somewhere in between here. And now let us see where this maps. You see let us make a mapping of what are called some critical frequencies. So we have omega mapping to omega l. The critical frequencies are actually 0, omega naught and plus infinity or rather 0 plus. 0 plus means a very small positive frequency which goes almost to 0. Now it is very clear and the mapping of course as you know is omega l is omega squared minus omega naught squared by B omega. This is the mapping. It is very easy to see that omega equal to 0 plus. You see this is already tricky. Now if it is 0 plus then this is of course 0. This is minus omega naught squared by B into 0 plus. So it is minus infinity because you have a quantity in the denominator which is very small and positive. Here you have a fixed negative quantity. It is fixed negative quantity divided by a very small positive quantity gives you an infinite negative quantity. What about omega naught? That is very easy to see. Omega naught maps to 0. That is very easy to see. And as you go towards plus infinity now let us see what happens. You see omega l can also be written as omega by B minus omega naught squared by B omega. So as omega goes to plus infinity this quantity goes to 0 and this quantity goes to plus infinity. So therefore omega equal to plus infinity goes to plus infinity. Now what happens in between? We have only taken the critical points. What happens in between? We need to see that too. So for that let us write down this expression here. Omega l is omega B minus omega naught squared by B omega. So omega l is omega by B minus omega naught squared by B omega. So D omega l you see this D omega l D omega is 1 by B plus omega naught squared by B times omega squared. So it is strictly greater than 0 when omega runs from 0 plus to infinity. That is obvious because when omega is positive then this is definitely positive. What it means is that as you increase omega from 0 plus towards infinity you expect omega also to monotonically strictly increase and therefore going back to the previous table here. As you go from 0 plus towards plus infinity here 0 plus is at minus infinity on omega l omega naught is at 0 and plus infinity goes to plus infinity. So as you move omega from 0 plus towards plus infinity omega l would move monotonically from minus infinity to plus infinity and omega naught would come to 0. Now you need to see the situation is like this 0 plus would go to minus infinity. Somewhere in between you will have a mapping of this omega is 1 then you would have a mapping of omega p 1 then you would have 0 here then you would have a mapping of omega p 2 and then you would have a mapping of omega s 2 and finally plus infinity. It is going to follow the same sequence. You see we have full control on choosing omega naught squared and B to meet the specifications that we want. What are the specifications? The first thing is that I you know I must you know you want a symmetric magnitude response. So this is going to map to some point on the negative axis and this is going to map to some point on the positive axis. You want them to be equal and opposite. So what you want first is that omega p 1 squared minus omega naught squared divided by B omega p 1 should be equal to minus of omega p 2 squared minus omega squared divided by B times omega p 2 for magnitude symmetry for symmetric pass span edge. Otherwise the low pass filter would have one negative pass span edge and a different positive pass span edge. We cannot allow that. Now we can solve this. It is very easy to solve this. One can easily solve this. In fact, one would obtain omega naught squared is omega p 1 into omega p 2 which is very interesting. Now this is not at all difficult equation to solve. It is a simple you know linear equation in omega naught squared. So you can easily solve it and you would obtain omega naught squared is omega p times omega p. This is very interesting. What it says is that the so called center frequency or the resonant frequency as we chose it in the beginning is the geometric mean of the pass span edges and this is not unfamiliar. In fact, if we take you know a typical LC network with a resistance added by the way, then you know the it is indeed true that the edges of the pass by the points of half power as known do turn out to have a geometric mean equal to the center frequency or the resonant frequency. This is not surprising. This is indeed a property of a band pass filter as known with LCR networks. So this is surprising, but true in this case as well. So omega naught squared is determined. Now what is it that would determine B? Actually nothing at all. But we can put something down to make our whole life easier. You remember in the high pass filter, we had put down the low pass pass span edges 1 and let us standardize that to make life easy. So we could put down the pass span edges 1 not just you see what we have said so far is that the pass span edge on the positive side and on the negative side must be the negative of one another. That is all right. But let us make that 1 and minus 1. That makes life easier. So let us put down. Let us ask for this is our choice. Let us ask for omega P1 to map to minus 1 or equivalently omega P2 to map to plus 1. That means omega P1 whole squared minus omega P1 omega P2 because now we have agreed that omega naught squared must be omega P1 omega P2 divided by B omega P1 should be equal to minus 1. That would of course very clearly give us B is omega P2 minus omega P1. And of course the same thing would follow if you put the condition on omega P2. So now we have an explicit value for B and for omega naught. So in fact our band pass transform is complete and we are now in a position to design band pass filters too. What do we need to do? Let us just put down the steps. We have this band pass filter. First step take omega naught squared equal to omega P1 omega P2. B equal to omega P2 minus omega P1. And now we know why we call this B. B stands for bandwidth. So in a way B is the length of the pass band. Having taken that the next step is to design a low pass filter with the following specs. You see omega S1 would transform to omega SL1. S1 omega SL1 is omega S1 squared minus omega naught squared divided by B omega S1. And similarly omega S2 would translate to omega S2 squared minus omega naught squared divided by B omega S2. Now omega S1 would definitely be negative and this would be positive. Take the smaller of mod omega SL1 and mod omega SL2. This is omega SL2 for stop band edge. You see what we are saying in effect is that the pass band edge is of course 1. But this would give us one stop band edge here and this would give us another stop band edge. Which of them should we take? They may not be equal. So naturally you must take the string end, the more the tougher condition there. Which is the tougher condition? The one which is closer. That is why we have taken a minimum here. So take the minimum of the modulus of SL1 and the modulus of omega SL2. Whichever is less that means is closer. Of course both of them will be more than 1 in magnitude. That is for sure. But the one which has a smaller magnitude should be chosen as the stop band edge. So of course now you know the pass band edge of the low pass filter. You know the nature of the pass band. You know the stop band edge of the low pass filter, the more stringent of the two. You know the nature of the stop band. You know the tolerance in the pass band. You know the tolerance in the stop band. So you can design the low pass filter using either the Butterworth or the Chebyshev or inverse Chebyshev or elliptic approximation and then transform the low pass variable SL using a squared plus omega naught squared by BS where you know what omega naught squared and BR. With that then you would, yes please, a question. Oh, the question is what happens to the minus infinity to 0 interval? Where the minus infinity to 0 interval O? You see it is very clear that minus omega, yeah that is a good question. So let us just look at, let us answer that. You see one thing that you need to check is what happens when you put S equal to minus j omega here. So we will see very quickly that this mapping must be odd. You see the mapping the frequency transformation is odd. So S equal to, so if omega goes to omega L then minus omega would go to minus omega L. That is very easy to see. The mapping must be odd. In fact you can see it here. Omega squared minus omega naught squared by B omega. When you replace omega by minus omega this whole thing is negated. So the mapping is odd, isn't it? So therefore whatever you are doing on the positive side of the frequency axis is mirrored on the negative side. And that anyway is required because you want magnitude symmetry. Is that clear? So with that then we have completed our design of the band pass filter as well. And now all of us are in a position to of course once you design the band pass analog filter you can transform it using S equal to 1 plus z inverse by 1 minus z inverse using the bilinear transform. You have the discrete time filter with you and then you can realize the discrete time filter as you desire. So now we have completed the design of band pass filters. And now all of us are well equipped to complete our assignment on the design of the band pass filter as well. And then we shall in the next lecture look at how we might design band stop filters and with that we shall be well equipped to complete the assignment given to us on filter design.