 Now you see what we need to do is to complete the process of design. We have n with us and we have epsilon and we have agreed we will choose epsilon to be the very most that it can be and that is square root of d1 right. We cannot you know unless you are willing to compromise on order it does not make sense to choose epsilon any different from square root of d1. Having agreed to that we now want to put down the poles as usual we need to find out the discrete system function. The next step in design is to obtain the poles. Now before I proceed to this I wish to make an observation about the transition band of filters. The transition band is characterized by no specification at all on the magnitude. Of course it is expected that the magnitude would move smoothly from the pass band to the stop band and normally the transition band does show a monotonically decreasing character of magnitude but that is not specified. So what characterizes the transition band is unspecified magnitude and not specifically desired magnitude at all neither of course we definitely do not ask that the magnitude response be 0 all over the transition band we cannot ask for. If you are asking for that we are in fact asking for an ideal filter or something like an ideal filter and anyway it serves no purpose to make the magnitude response 0 all over the transition band. Further we are not even asking really that it be monotonically decreasing although that is how it often is. Of course it depends on increasing or decreasing will depend on whether the pass band follows the stop band or the stop band follows the pass band. So you see whatever it be we normally do observe in most of the common designs that there is a smooth movement in monotonic fashion from the stop band to the pass band or the pass band to the stop band but this is not specified right. So even if one comes up with a design where it is non-monotonic that is acceptable for the transition band. In fact the sole characteristic of the transition band is nothing is asked either of magnitude or phase. Whatever emerges as a consequence of satisfying the pass band requirements and stop band requirements is accepted in the transition band. Well so much so then for finding the poles of the Chebyshev filter. Now how we would find the poles is to write down again the analytic continuation and we know h analog s into h and log minus s as was the case before for the Chebyshev filter can be obtained by replacing j omega by s. In other words omega needs to be replaced by s by j. So we have this product would be essentially the squared magnitude analytically continued and that is 1 plus epsilon squared C n squared s by j omega p and the poles are obtained by putting the denominator equal to 0 and let us solve that. Now here you must remember that these poles are complex. So to solve this 1 plus epsilon squared C n squared now you know there will be several poles indexed by some integers. So let us call that integer k index of the pole k is the pole index as was the case in the Butterworth filter you run it over the set of integers. The kth pole is satisfies 1 plus epsilon squared C n squared s k by j omega p is equal to 0 and therefore of course C n squared s k by j omega p is equal to minus 1 by epsilon squared. Now we could take both the positive and the negative square root on both sides right but we will see that it is adequate to take any one of them. Once you are going to run k over the integers it would take care of the case of positive and negative by running it over sufficient number of consecutive integers right. So we will go back to expanding C n. So C n x if you recall essentially cos of n times cos inverse x. So what we have for the kth pole is that C n s k by j omega p is plus or minus it does not matter but one could say you know 1 by epsilon right. Well plus or minus 1 by epsilon multiplied by j. So let us keep the plus minus and later we can choose any one of them as I said that we could be happy with keeping plus j by epsilon or minus j by epsilon. By running k over sufficient over a sufficient set of integers, sufficient set means 2n of them. We need consider only one of either plus or minus only and therefore we can now write down this equation cos n cos inverse s k by j omega p is plus j by epsilon. Now let us put cos inverse s k by j omega p equal to a k plus j b k. So you should remember this is a complex argument and therefore we will need to have complex solutions to it. Now we are working with entirely complex cosines, sines and everything. We can now take the cosine of both sides and solve this. So we have s k by j omega p is the cosine of a k plus j b k and that can be expanded in the standard way in which we expand trigonometric functions. This is cos a k cos j b k minus sin a k sin j b k. But we recall that cos j b k is nothing but cos hyperbolic of b k. Remember a k and b k are now real. So cos j b k is essentially the cos hyperbolic of b k and sin j b k is minus j times the hyperbolic sin of b k. Is that right? So there we go. We have this is equal to cos a k cos b k plus j sin a k sin of b k and we equate this to 1 by epsilon times j. From where we can equate the real part in the imaginary part separately. Is that right? So we equate the real part of cos to 0 here and the imaginary part to 1 by epsilon cos a k cos b k is equal to 0. Now cos b k cannot possibly be 0. So the only possibility is that cos a k is 0. Cos b k as we have seen before must be greater than 1. So therefore cos a k is 0. And if cos a k now a k is of course you know real argument so there is no problem. If cos a k is 0 then sin a k is plus or minus 1. That is very clear. Yes please. Yes. Oh yes, yes, yes he is absolutely correct. Yes, there is a question. Yes, that is correct. Yes, I need to make a correction. I am very glad that somebody pointed this out. Yes, so you see we have taken, we need to write down n times. Right? So we need to write down n times. So I need to make a correction here. That is correct. So you see s by j omega p. Well, let us therefore make a correction here. Yes, let us put back this argument. So you have a k plus j b k here. So let me repeat that step. Yes. So in fact let us so we have cos n times a k plus j n b k is equal to plus j by epsilon. That is correct. So you are right. So we now need to so let us complete this working. Let us expand this on the left hand side. We have cos n a k times cos j n b k plus or rather minus j sin n a k sin or well you know I am skipping a step here but maybe I will write it down first. So sin n a k sin j n b k. This is the left hand side. Is that alright? Yes. So we need to correct it. Yes. And LHS will therefore evaluate to cos n a k as usual cos n b k plus j sin n a k sin n b k. That is correct. And this is equal to j by epsilon which is the RHS. Is this fine? Yes. So yes, we do need to introduce a correction there. I am glad that was pointed out. Yes. Is that clear? So we need of course now proceed to equate the left hand side and the right hand side and therefore the real and imaginary parts of the left and right side. So once again we would get cos n a k cos n b k is equal to 0 and sin n a k sin n b k is 1 by epsilon. As before we observed that cos n b k cannot be 0 and that means that cos n times a k is 0. Is that correct? Yes. Cos n a k is 0. Now if cos n a k is 0 then clearly sin n a k has no choice but to be either plus or minus 1. Sin squared of n a k needs to be 1 because cos squared plus sin squared is 1. So of course you have sin n a k again is either plus or minus 1. And here too we might take either the positive sign or the negative sign. And all that will happen the only change that will take place is that we need to run k over all the integers once again all the required integers to cover both the positive and negative sign. So here too we can be satisfied with taking one of them right and run k over a sufficient number of integers. So anyway what we have is well let me put because you know this we do not need to refer to it again and again. So here the situation is that this is either plus 1 or minus 1. Let us take it to be plus 1 in which case this becomes 1 by epsilon. Is that right? So we have sin n b k is 1 by epsilon which means b k is now clearly 1 by n sin inverse of 1 by epsilon that is interesting. And of course b k as you can see has nothing to do with k that is interesting. So b k is not indexed by the integers at all. So the index the integer index is going to act on a k not on b k how will it act on a k and yes. So here it is very good that student corrected because had we not made that correction on route we would have had trouble now in indexing a k yes. So it was very appropriate that that student interjected and made a correction on n. So we have cos n a k is equal to 0 which means n a k must clearly be an odd multiple of pi by 2 and that tells us that a k must be of the form 2 k plus 1 pi by 2 n and now we know the poles because we know a k and b k. I emphasize in fact maybe it is a good idea maybe it is good that you know it was serendipity that we made that mistake because it is very important to see that we do need a dependence on n when we satisfy the equation for the cosine part. It is a dependence on n which allows you to create multiple poles the dependence on k on the integer index is a consequence of cos n a k being 0 not just cos a k yeah anyway. So coming back to this we now have an expression for the poles and that we have done partly before let us put back that transparency. So you will recall that we have written down s k by j omega p is cos of cos inverse of this was equated to a k plus j b k. So s k by j omega p is cos a k plus j b k and of course we have expanded this. So I will just renumber this we will give this the number 23 okay. So now we know where the poles lie s k can now we calculate it is j omega p times cos a k cos j b k minus j omega p times sin a k sin j b k and now we can use the standard strategy of putting cos j b k equal to cos p k and sin j b k equal to minus j times the sin hyperbolic of b k and therefore we have s k is j omega p cos a k cos b k minus j times you see minus j into minus j omega p sin a k sin b k and once again remember neither sin b k nor cos p k have anything to do with k sin b k is essentially 1 by epsilon or b k or rather you know if you looked at it before we have written an expression for b k yeah. So b k is 1 by n sin inverse of 1 by epsilon so neither sin b k nor cos p k have anything to do with k right. So you might as well just write cos b and sin b there so we will just write b k is equal to b for all k so we can just call this cos b and we can call this sin b and clearly this is the real part and this is the imaginary part here this is the imaginary part and this is the real part of the pole so s k is the real part so minus omega p times sin a k sin b plus j omega p cos a k cos b and now what we need to do you see we want to find out this is the real part this is the imaginary part and the real part and we can call the real part s k as we did and the imaginary part capital omega k and we can now write down an equation that relates s k and omega k we are trying to find a contour a curve in the imaginary in the imaginary in the complex plane on which these poles lie so where is that contour well how do we obtain the contour the different poles are indexed by the case so if we eliminate k we get the contour and to eliminate it all that we need to do is to note that sin squared plus cos squared of a k must be equal to 1 is that right therefore sin squared a k plus cos squared a k must be equal to 1 and that means sigma k by omega p sin b the whole squared plus omega k by omega p times cos b the whole squared is equal to 1 and we know what contour this is you see if the two arguments with the real if these coefficients have been equal if this had been equal to this we would have landed up with a circle but because these are unequal we get an ellipse father which is the major and which is the minor part axis of the ellipse now of course this ellipse is aligned with the axis in other words the major and minor axis are coincident with the vertical and horizontal yeah it is not an inclined ellipse the question is which is the major axis and which is the minor axis and to answer that question we need to decide which is greater is cos b greater or is sin b greater which one would be greater it is the course hyperbolic which is always greater for a real argument because cos squared b is 1 plus sin squared b and therefore cos squared b is always going to be greater than sin squared b and therefore in this it is very clear that the major axis is on the imaginary and the minor axis is on the real is that clear to everybody yeah so we get an ellipse this is the contour that we land up with this is the contour sigma by omega p sin b the whole squared plus omega by omega p cos b the whole squared is equal to 1 this is the contour and of course this point would be omega p times sin b and this would be omega p cos b and the other points can be determined of course one must mark the specific poles so you know ak and well sin ak and cos ak now need to be determined or you need to find out ak and then you know find out where these poles will lie precisely now the easiest thing to do is to you see how do you mark these poles the easiest thing to do is to draw two circles one with radius omega p sin b and the other with radius omega p cos b that right what I am saying is it will be easiest for us to draw two circles like this the real part can be marked by taking the inner circle here all that you need to do is to see where this angle the angle is 2k plus 1 pi by 2n for all that you need to do is to draw and draw a radial line making an angle of 2k plus 1 pi by 2n and see where it intersects the circle and that gives you the real part on the other hand the imaginary part can be obtained by using the larger circle and on the larger circle one takes the same radial line but then you know the imaginary part is obtained by the measurement coming from the larger circle so one has to be careful in marking the poles you see the one I must I must I must emphasize that when you draw a radial line here with angle ak ak is 2k plus 1 pi by 2n where it intersects this circle will give you the real part where it intersects this circle gives you the imaginary part but one must not straight away take the intersection of this arc with the ellipse to find the location of the pole no that is not what it is one must take the measurement of real imaginary part and mark it and it of course would lie on the ellipse right so I am just giving you a strategy to measure the real imaginary part but one must not use the radial line with angle ak to intersect with the ellipse and mark the pole there that is not correct alright anyway I leave it to you as an exercise this is an exercise actually mark the poles mark the poles on the ellipse for n equal to 2 I am sorry for n equal to 3 and n equal to 4 and you would get a feel of how the poles are located you would also observe that there are 2n values of k to be taken as was the case for the Butterworth filter you can take any consecutive 2n values so you could start with k equal to 0 and run all the way up to k equal to 2n minus 1 or you could start with 1 and run up to 2n it does not matter whatever it be after you mark all the 2n poles on the ellipse the poles in the left half plane would give you the poles corresponding to h and log s is that right so let us write that down the remaining steps are identical mark poles poles in LHP give you h and log s now there is one important observation here which does not happen in the Butterworth filter and that is what do you want the numerator to be in h and log s in other words what is the magnitude response when omega equal to 0 now there you have to be careful because the response for omega equal to 0 is not 1 identically here it depends on whether n is odd or even is that right so you see you recall the response at omega equal to 0 how would you determine it well we have 1 by 1 plus cn squared epsilon squared cn squared omega by omega p and cn is cos of n cos inverse now let us look at the 2 situations when you have cos inverse 0 cos inverse 0 of course can be taken to be pi by 2 however when n is even this becomes cos of an even multiple of pi by 2 which is either plus 1 or minus 1 when n is even this evaluates to plus minus 1 and therefore this evaluates to 1 so this response is 1 by 1 plus epsilon squared on the other hand when n is odd and of course you have an odd multiple of pi by 2 so this evaluates to 0 so this response is just 1 by 1 plus 0 that is 1 so you have to distinguish between n odd and n even to put down the response at omega equal to 0 and therefore when you put down the numerator in h and log s you must at s equal to 0 equate it to the expected response for omega equal to 0 and not identically 1 if n is odd it evaluates to 1 if n is even its square would evaluate to 1 by 1 plus epsilon squared and therefore the magnitude itself would evaluate to 1 by 1 plus epsilon squared out that right that care need to be taken when specifying the Butterworth the Chebyshev filter unlike the Butterworth and finally of course once you have h and log s the remaining process is common to the Butterworth filter replace s using the bilinear transform and get the discrete time system function that completes the design of the Chebyshev low pass filter now we are well equipped to proceed to see how we could design other kinds of filters either with the Butterworth approximation or the Chebyshev approximation by using what are called analog frequency transformation.