 Welcome to the 26th lecture on the subject of digital signal processing and its applications. We have been discussing frequency transformations in the previous lecture. By analog frequency transformations we mean transformations from an analog filter to an analog filter. In fact more specifically a transformation that takes us from a low pass analog filter to an analog filter of appropriate nature whether it be high pass or band pass or band stop. In the previous lecture we had looked at the transformation that takes us from the low pass to the high pass filter and from the low pass filter to the band pass filter. We now need to look at a transformation that takes us from a low pass filter to a band stop filter and that is going to be a little more difficult than the other two. But in order of difficulty the high pass transformation is the easiest. The band pass is slightly more difficult and the band stop even more difficult. So today we now proceed to look at the band stop transformation. So what we are trying to do is to get a transformation let us call it HBS of S which would replace a low pass complex frequency variable SL so that when we replace SL by this transformation we would get a band stop filter with specified characteristics. What do we mean by specified characteristics? For a band stop filter the specifications look like this. Of course we always show only the positive frequency side. You have a pass band and another pass band and a clear stop band. So unlike the band pass filter I am sorry this should be the other way around I am sorry. Let me redraw this because I said you have a one pass band, you have another pass band and a clear stop band. So unlike the band pass filter we have just one stop band and two pass bands. So maybe in a sense it is complementary or dual to the band pass filter. In the band pass filter you have one pass band and two stop bands. Here you have one stop band and two pass bands. Now we want a transformation again which satisfies the three characteristics that we mentioned the last time. The first is that it must maintain rationality. The second is it must preserve stability and finally it must take the imaginary axis to the imaginary axis in a manner that carries with it the kind of behavior that you desire. That is you know the frequency axis must map to the frequency axis in a way that a low pass character transforms to a band stop character. Now then here we see there are two pass bands and therefore it is I mean it is intuitively clear that people require a second order transformation. We cannot make do, it is not going to be a monotonic transformation. There is going to be some point in the transformation where you have a movement from one behavior to another and that is going to happen only if the transformation is at least second order. It cannot be just a first order transformation. And once again we take recourse to our intuition of looking for functions that take you from the imaginary axis to the imaginary axis and there again we have LC networks to come to our aid. And last time we saw that the series LC network comes to our aid in the band pass filter and therefore we expect that if we use a parallel LC network it would come to our aid in a band stop filter. So let us see how the impedance of a parallel LC network would look. So the impedance would be of the form 1 by the sum of the admittances and the admittances are Cs plus 1 by Ls. And this is easy to simplify. We can easily simplify this to get Ls in the numerator LCs squared plus 1 and that can be simplified once again. So 1 by C times S divided by S squared plus 1 by LC. And once again we note that 1 by LC is the resonant frequency of the network and this is a positive concept. And therefore we can rewrite this expression in the form sum BS divided by S squared plus omega naught squared. And we are not terribly worried about the specific meaning of B and omega naught squared now. That will in fact as was the case with the band pass filter it will become obvious when we actually go through the transformation. Now one thing we need to do is to note that this is the reciprocal of S squared plus omega naught squared by BS. And of course it is rational. So the movement from rational to rational is guaranteed. There is no problem. So we do not need to worry about the fact that after transformation the resultant filter would remain rational. There is no problem on that count. But of course we need to worry about stability. Now stability actually follows from a very simple argument. In fact we will show that if a particular transformation is stable. In other words if it takes the real part of S to I mean if it takes if it preserves the sign of the real part of S. If it takes the left half plane to left half plane and the right half to the right half plane then the same thing happens to the reciprocal. So we will say H BPS which is S squared plus omega naught squared by BS is stable. In other words the real part of S and real part of H BPS have the same sign. Of course strictly the same sign. By strictly I mean if the real part is 0 the other real part is 0 as well. So imaginary axis goes to imaginary axis. Now of course the fact that the imaginary axis would go to the imaginary axis is obvious because we have taken an LC network. So when you take its impedance I mean when you put S equal to j omega it is definitely going to be imaginary. So we do not need to worry about that. So the imaginary axis going to the imaginary axis is not a problem. But what we need to check is this whether the sign of the real part is present. Now let H BPS you see let S be equal to sigma plus j omega. Whereupon let H BPS give sigma BP plus j omega BP then 1 by H BPS which is H BS of S is the reciprocal of this. The reciprocal of this is sigma BP minus j omega BP divided by sigma BP squared plus omega BP squared. The complex conjugate divided by the mod squared. And now you can see very clearly that the real part of H BLS is simply sigma BP divided by sigma BP squared plus omega BP squared. Which is of course sigma BS if you wish to call it that. And now it is very obvious that when sigma BP is not 0 this denominator cannot possibly be 0. And if the denominator is not 0 it must be positive because it is a sum of 2 squares. And then sigma BP and sigma BS must have the same sign. It is obvious now. So we have proved stability for the band stop transformation as well. So now we have a very strong candidate. All that we need to do is to look at what it does to the sinusoidal frequencies. And let us indeed embark upon that question. So what does it do when you substitute S equal to j omega? H BS of j omega is GB omega divided by omega naught squared minus omega squared. And therefore this is at the point g omega L. It is of course imaginary as you can see. Where omega L is b omega divided by omega naught squared minus omega squared. And here we have a little more work to do. You see it is very clear that we now have a point of singularity somewhere on the positive real axis. Omega equal to omega naught is a point of singularity. Singularity means it diverges. The function is undefined, that point. And therefore we expect something unusual to happen at that point. We will see in a minute that that is exactly what the case is. So let us now consider the band stop transformation. So you know you have, let us map the critical points. You know the critical points are going to be a little more difficult to map now. You see let us try and visualize what it is that we desire from this transformation. If you look at the band stop, if you look at the band pass transformation. What happened there? And let me draw it once again for you. This is band stop. In the band pass transformation what happened was you had these two stop bands and a pass band in between. And what effectively happened was you had an omega naught somewhere in between here. Where it is the geometric meaning of omega p1 and omega p2. And this was omega p1 and this was omega p2. And the effect of this transformation was to bring omega naught to 0. This to minus infinity and plus infinity went to plus infinity. So in fact a pair you know this entire thing went to one side of the frequency axis in the low pass filter. And this entire thing went to the other. Now in a way in the band stop transformation we have exactly the reverse situation. So we have two such pairs of structures. This pair and this pair if you wish to call it that. And therefore what we expect is that one of these pairs would go on to the positive side of the frequency axis and the other would go on to the negative side. By a pair I mean a segment or the entire stop band and a segment or the entire pass band. So a stop band, pass band pair. In the band pass filter it was a segment of the pass band and the entire stop band. Now here we expect it to be a segment of the stop band and the entire pass band. That will go to one side and the other side of the frequency axis in the low pass filter. And therefore the infinity is going to come somewhere in between and as expected that infinity is going to come to omega equal to omega naught. That is why it is a point of singularity. So at omega equal to omega naught there is going to be a jump. It is going to jump from plus infinity to minus infinity. Let us see indeed what happens. So let us take inspiration from the band pass filter. Let as in the band pass filter omega naught squared be omega p1 omega p2 and b be equal to omega p2 minus omega p1. So here maybe it is not quite appropriate to call b the bandwidth. It is like a combination of the stop band and the transition bands. It is not exactly a bandwidth. It is something slightly different. But it is indeed a segment of the real axis from the frequency axis. That is of course true. If we do this then we can now map the critical points. 0 it is very clear map to 0. In fact let us take 0 plus. By 0 plus we mean a very small positive frequency which goes towards. So 0 plus of course goes to 0 plus. As you know from 0 plus the first point that you encounter of importance is omega naught. Now omega naught is a point of singularity. So we cannot simply use omega naught itself. At omega naught itself the expression diverges so we cannot use it. But we can use omega naught minus and omega naught plus. So let us use omega naught minus. By omega naught minus we mean a value just before omega naught. Now omega a as you know is b omega divided by omega naught squared minus omega squared. This is the frequency transformation. So to omega naught minus the denominator is slightly positive. And of course the numerator is finite and positive. And therefore this is going to go to plus infinity. In contrast when you go to omega naught plus the denominator is slightly negative. And of course the numerator is clearly positive. And therefore this goes to minus infinity. So now we understand what we mean by omega equal to omega naught being a point of singularity. Not only is the function discontinuous at that point there is a huge jump. So in fact we will see its derivative is also discontinuous. And that is why there is a jump. A function could be discontinuous but its derivative might turn out to be continuous. In that case you do not have an infinite jump. But here we have an infinite jump. So it is more than just discontinuity. It is discontinuity in the derivative. The seamless kind of singularity. And of course when you go to plus infinity and you see now plus infinity is tricky. We must divide the numerator and denominator by omega i. So omega L is B divided by omega naught squared by omega minus omega. That is how we write it. And therefore as omega tends to plus infinity, omega naught squared by omega tends to 0. 0 plus if you like. And of course omega tends to plus infinity. So this whole thing omega L tends to 0. Now again you may ask whether it is 0 plus or 0 minus. It should go to 0 minus. And therefore we need to write here 0 minus. Now we have the critical point mapping very clearly before us. Now this critical point mapping makes it clear that omega naught is a point where we have trouble. So it will be difficult for us to show both the segments all in one figure. So let us show them in two figures. That will be easier to show. So we will take the first segment of the band pass, the band stop transformation and omega naught somewhere here in between. Now remember omega naught is a geometric mean of omega p1 and omega p2. So it can be somewhere in between. And omega s2 is going to be somewhere beyond there. So you know we say omega naught plus or other omega naught minus. We go from 0 to omega naught minus. And 0 to omega naught minus is carried, so this is omega. This is carried 0 plus. 0 plus is carried to 0 plus on omega L. As you know, you know the expression, you have already worked this out. Omega p1 is going to go to plus 1. That we know from the band pass transformation. We have worked it out. The reciprocal of 1 is 1. So you see omega p1 will go to 1. Omega s1 is going to come to another point beyond plus 1. So it is called that omega LS1. And omega naught minus would go to plus infinity. Therefore we have this mapping for the segment between 0 plus and omega naught minus. This is the frequency mapping. And of course here too you must not forget that the dependent variable is carried as is from the band stop filter to the low pass filter. The tolerances are carried as they are. The nature of the passband and stopband are carried as they are. Here too you must not forget that we have two passbands and therefore in principle you could have had two different tolerances for the two passbands. But then when you put down the specifications for the low pass filter you need to choose the more stringent of the two. Whichever demands more that must be put. And the other would automatically be satisfied. If at all they happen to be different. If they are the same there is no issue at all. As far as nature goes you do not have the you know the luxury of having two different natures in the passband. The two passbands must be of the same nature. Either equiple or monotonic. And of course the stopband gets carried as it is. So that is for the segment from 0 plus to omega naught minus. Now what happens for the segment? From omega naught plus to plus infinity. We will just squeeze this part. Omega naught plus as you know goes to minus infinity. We have already worked out for the band stop transformation. That omega p2 goes to minus 1. This is easy to verify. And therefore omega s2 is going to come to a point. Omega sl2. And of course plus infinity is going to come to 0 minus. And therefore we have this specification being put down for the low pass filter. Plus is one little word of caution. If you look at the frequency transformation here. This or if you like this one. If you recall in the band pass transformation. It was omega squared minus omega naught squared divided by B omega. Here this is not exactly the frequency transformation is the negative reciprocal. And that is not surprising. Because the reciprocal of one by the reciprocal of an imaginary number. Has the reciprocal magnitude and the opposite sign. The sign has been reversed. So plus 1 and minus 1 have got reversed here. That is the point of observation. The reciprocal of an imaginary number is reciprocated in magnitude. And the sign is reversed. That should be noted. These are small. And that is why you know you have to be careful. That you know the plus 1 and minus 1 roles have got reversed. So omega p2 now maps to minus 1 not to plus 1. And omega p1 maps to plus 1 and not minus 1. That should be clearly noted. Anyway it is a minor point. But we have now understood the transformation. So now we have the totality of the transformation before us. And we can see now that you know we can put down the specifications on the low pass filter except for one important detail. We have got the specifications on the positive frequency side and the negative frequency side. But they have to be symmetric. Magnitude symmetric and phase anti-symmetric. Phase anyway we are not putting down anything at all. So it is only the magnitude which is of concern. Now again the symmetry is there anyway as far as the pass pan edge goes. The pass pan edge is plus 1 on the positive frequency side and minus 1 on the negative frequency side. The problem is with the stop pan edge. Omega SL2 and omega SL1 or omega LS whatever you know whatever you want to call it LS or SL. Now you see these the mapped stop pan edges may not be the relative of one another. Stop pan edges omega LS1 and omega SL2 may not be negative of one another. So we have to do what is harder to do here. We take the minimum of mod SL1 and mod SL2.