 we are tempted to consider s equal to 1 minus z inverse. What it means is we are approximating the derivative by taking the difference between the current sample and the past sample. Now we do not need to go too far to see that this is not adequate as far as our conditions go. In fact all that we need to do is to write z in terms of s. So z inverse is clearly 1 minus s and therefore z is 1 by 1 minus s if we take this transformation to be true and then we only need to substitute s equal to j omega and then we have 1 by 1 minus j omega does not even have a magnitude of 1. So mod of 1 by 1 minus j omega is not 1 for all omega. So we have failed on a very important count anyway and in fact this transformation you know we should have this is if this is z then the imaginary axis should have gone to the unit circle. So we are failing on that count anyway and anyway this is not going to be so you know a simple approximation like this is not going to help us. In fact let us use some more cues to lead us to a transformation of choice. Let us do it the other way. Let us put z equal to erase the power j omega in the same transformation and see what we get. We get 1 minus erase the power minus j omega and that can be rewritten as erase the power minus j omega by 2 times 2j sine omega. So you know there are two counts on which you know of course we were pessimistic a minute ago and we noted that the transformation was not alright but then you know to move to a solution you cannot remain pessimistic you have to find some way out of the tricky situation in which you are and we could now take an optimistic view and say that you know if you what you would have liked here is that this be essentially imaginary. So this is the good part of the substitution and this is the bad part you know when you put when you took a point on the unit circle you should have gone to a point on the imaginary axis but this would have put you on a point in the imaginary axis it is this which is creating a trouble here. So the one thing is we want to do away with this term now how do you do away with this term you cannot do away with this term but by multiplying by its complex conjugate or dividing by the same term. So it is clear that we cannot be happy with just a polynomial in z or z inverse we need a rational a proper rational function in z or z inverse to replace s and that rational function in z should be such that it cancels out this factor that is the first observation the second observation is that when omega goes from minus pi to plus pi even if this term were to be absent you would be taken only from minus 2 to plus 2 here not from minus infinity to plus infinity from that count also it is inadequate in fact this is the problem with any sinusoid or co-sinusoid as you might want to call it just a sine or cosine term can never take you all over the real axis among the trigonometric functions the sine and cosine functions do not take you all over the all over the real axis which function takes you all over the real axis the tangent function or the cotangent function. So somehow we need a cotangent or a tangent function here now how do we get a tangent function you cannot get a tangent function by subtracting z's. So you can get a tangent function if you divide that is another you see if you divide a sine function by a cosine function or a cosine function by a sine function you get a tangent function that is another reason why you would want some kind of proper rational function of z to replace s. And in fact, if we take tan tangent, we have the answer, tan omega by 2, when omega goes from minus pi to plus pi, tan omega by 2 would indeed run all the way from minus to plus infinity. So, if we can go from here, you see, we should make, we should try to get tan here, tan omega by 2 here. And how can we get a tan omega by 2 by dividing sin omega by 2 by cos omega by 2? So, let us do that. And in fact, once we allow for that division, we are also likely to do away with this trouble. So, we will be killing two birds with one stone in the proverbial sense, not literally. Indeed, to get tan omega by 2, you know what you are saying is you want j tan omega by 2. And j tan omega by 2 can be written as 2j sin omega by 2 by 2 cos omega by 2. And nothing stops you from multiplying by e raised to power minus j omega by 2 in the numerator and the denominator. And in fact, there we have the answer. This is nothing but 1 minus e raised to power minus j omega divided by 1 plus e raised to power j omega. Please check that so easy. And therefore, we have the answer to the transformation that we want. We are therefore, asking for s equal to 1 minus z inverse by 1 plus z inverse. And now, let us check that this transformation indeed does all the jobs that we wanted to do. Of course, one job we have already verified. Let us start with that thing that we have already done. So, let us start with z equal to e raised to power j omega. We will just formality, but we will just complete it. And we have seen that you know s becomes j tan omega by 2 here. So, the good thing is omega going from minus pi towards plus pi takes you from capital omega going all over from minus to plus infinity. Minus pi goes to minus infinity, 0 goes to 0 and plus pi goes to plus infinity. And this is monotonically increasing all that we wanted out of the mapping between the unit circle and the imaginary axis has been satisfied. Is that correct? As we move from minus pi to plus pi, there is also this mapping is 1 to 1. There is for every, you see so I mean here you have j tan omega by 2. So, essentially what you are saying is omega, the capital omega is tan small omega by 2. That is the mapping that you are establishing here. And this is 1 to 1. That is the beauty of it. You know for every small omega, there is a unique capital omega. For every unique capital omega, there is a unique small omega. In spite of the fact that small omega is on a finite interval and capital omega is on an infinite interval, we still have a 1 to 1 mapping. This is the beauty of this mapping. It is 1 to 1. It is on 2 both ways. So, it is invertible. And of course, it is monotonically increasing as we wanted it to be. Well, now about the left and the right half plane. So, let us write the inverse mapping. That is easier to deal with. So, where S is 1 minus Z inverse by 1 plus Z inverse. So, of course I can write 1 plus Z inverse times S is 1 minus Z inverse. So, I have S plus S Z inverse is 1 minus Z inverse. And that tells me that S plus 1 times Z inverse is 1 minus S. And therefore, Z is 1 plus S by 1 minus S. Now, we can say a lot about the real and imaginary part. Let S be equal to sigma plus j omega. So, Z is of course, obviously 1 plus sigma plus j omega divided by 1 minus sigma minus j omega. And therefore, we are interested in the magnitude of Z. It is the magnitude of Z that concerns us. Let us see what happens to the magnitude of Z. The magnitude of Z is of course, the magnitude of the numerator which is 1 plus. In fact, let us take magnitude of Z squared. It is the square of the magnitude of the numerator 1 plus sigma squared plus omega squared divided by the square of the magnitude of the denominator. That is easy to see. And of course, you have an omega squared common. So, the omega has no contribution to the magnitude. I mean it does not affect the nature of the magnitude. What makes the numerator and denominator different is the factors 1 plus omega squared and 1 minus omega squared. Indeed, when sigma is greater than 0, then 1 plus omega squared is bound to be greater than 1 minus omega squared. That is very easy to see. And nothing changes if you add omega squared to both sides. So, numerator is greater. So, numerator by denominator is greater than 1 clearly. Is that correct? For sigma greater than 0, sigma plus 1, the whole squared must be greater than sigma 1 minus sigma the whole squared. And in that case, the numerator is strictly greater than the denominator. And therefore, mod Z is more than 1. In contrast, when sigma is less than 0, then 1 plus sigma squared is strictly less than 1 minus sigma squared. And nothing changes if you add omega squared to both sides. And therefore, it is very clear that the numerator by the denominator is strictly less than 1. And therefore, mod Z is less than 1. Is that correct? So, we have what we wanted. We have a satisfactory mapping of between the imaginary axis and the unit circle. The outside of the unit circle corresponds to the right half of the S plane. That is the real part of S being greater than 0. And the interior of the unit circle corresponds to the left half of the Z plane. There is a 1 to 1 there. In fact, it is not even too difficult to see that it is 1 to 1. You have a 1 to 1 mapping between S and Z. Not just on the imaginary axis, but everywhere. You have got an inverse. So, you know, you can go from S to Z or you can go from Z to S uniquely. So, we have luckily arrived at the transformation that we want. This transformation meets our requirements. S is 1 minus Z inverse by 1 plus Z inverse. Now, as a point for further reflection, I put the following question to you. How do you interpret this as an approximation of derivatives in terms of shifts? And the hint is to expand this in terms of a power series. Of course, you have a 1 minus Z inverse, but 1 by 1 plus Z inverse can be expanded as a power series. And that would give you some insight on what you mean on what are we trying to do in terms of approximating derivatives by shifts. It is not too simple, but it is interesting. Of course, that is besides the main point of the discussion. We have got what we wanted. We are happy. We have got a rational function. The rational function does all that we wanted to. Now, this transformation which we have now so satisfactorily constructed is called a bilinear transformation. It is called bilinear because it is a ratio of 2 degree 1 functions of the complex variable. This is an example of a bilinear transformation. The set of bilinear transformations has been studied in depth in complex analysis. In general, a bilinear transformation looks like this. E Z plus B by C Z plus D equal to S is the general bilinear transformation, where A, B, C, D are complex constants in general. For those of you who wish to gain more insight into this transformation, I would recommend looking at any standard text on complex analysis, perhaps a first level engineering mathematics text. In reasonably comprehensive discussion on complex analysis, which is often a part of the syllabus of an engineering mathematics subject in the first initial years, the bilinear transformation is described in depth. One of the important properties of a bilinear transformation is that it takes straight lines and circles to straight lines and circles. That is the property, which is often pointed out in the context of bilinear transforms. As you can see, this transform is no exception. It takes the straight line, the imaginary axis into the unit circle. Of course, you can also think further to see what other straight line circle relationships there are in this bilinear transform. But that is typical of a bilinear transform. There is a straight line circle to straight line circle correspondence. Anyway, that was a remark to put you in perspective with complex analysis. But now, we have the tools that we want and we have agreed that we are going to use this bilinear transformation. So, what we now need to do is to put down. What is the process that we will now follow? Now we shall take the discrete time filter specifications. Of course, these are unnormalized. Normally, you would be given unnormalized specifications. What are these unnormalized specifications? To take an example, suppose you have a low pass filter, you may say the sampling rate is so much. Let us say it is 50 kilohertz just to make a point. The pass band edge is 20 kilohertz. Let us say the stop band edge is 30 kilohertz. Can it be? No, it cannot. So, this cannot be. Let us make it 22 kilohertz. If the sampling rate is 50 kilohertz, at best we can deal with the frequency of 25 kilohertz. So, stop band edge is 22 kilohertz and the tolerances of the pass band and stop band are 0.1. So, what we are saying in effect is that we want this kind of a frequency response. Of course, I am showing only the positive side. Now you see, pi corresponds to 25 kilohertz since the sampling rate is 50 kilohertz. Therefore, 20 kilohertz would essentially be 4 by 5 times pi or 0.8 times pi. This is the pass band edge and 22 kilohertz is 22 divided by 25. That is 0.885 and therefore, you have a stop band going from 0.88 to pi and a pass band going all the way from 0 to 0.8. And of course, you are allowing the response to vary between 1.1 and 0.9 here and from 0 to 0.1 there. This is the response that you want to realize. So, this is how you would translate the specifications into the normalized angular frequency domain or these are called the normalized specifications on the normalized angular frequency axis. Is that clear? Is that clear how we go from the unnormalized specifications to normalized specifications? Very easy. All that we are doing is really to normalize the frequency axis. Is that correct? So, here the change is, see it is very interesting. The S to Z transformation, this process of normalization are all transformations of the independent variable, not the dependent variable. The dependent variable is not being transformed. It is the independent variable which is being transformed and the dependent variable is being carried with the independent variable. Is that right? Now, once we have the normalized specifications, the next step. So, remember this can be similarly done for band pass, similarly for high pass filter, band pass filter, band stop filter. You can go to a set of normalized specifications and the next thing to do is to use the bilinear transform. The bilinear transform will take you from the normalized discrete specifications to analog specifications. That means you have specifications in omega and now you have here specifications in capital omega. And how would we carry out this movement from discrete small omega to capital omega? Capital omega is tan omega by 2. That takes you from omega to capital omega. So, the nature of the. So, here we now have a corresponding analog filter to be designed. So, what would you do next? Of course, you would design that analog filter. Take advantage of the known methods for analog filter design and design that analog filter. So, we will assume we design that analog filter. Let us call it H analog and that would be a function of S. It is a rational function of S. Rational, stable, causal. There are several different approaches and we will look at some of them. In fact, we will begin by looking at the low pass filter design. So, once we have an analog filter, we are done because now all that you need to do is to replace S by 1 minus Z inverse by 1 plus Z inverse to get a stable, causal, rational discrete filter and our job is done. So, this is the scheme of things. Therefore, now in the next lecture, we need to look at how we can design analog filters given the specifications of the analog filter. We will begin with the low pass analog filter as a case in point.