 So, warm welcome to the 22nd lecture on the subject of digital signal processing in its applications. We briefly recapitulate what we did in the previous lecture and then go on to the details of the current one. We have been discussing the whole theme of discrete time filter design for the past few lectures. So far, we have come to the following point, we have agreed that we cannot meet the ideal specifications ever for the piecewise constant standard piecewise constant filters. We have also identified that we need to relax the ideal specifications by introducing tolerances and a transition band. We have also agreed subsequently that having made these relaxations, a realization is always possible and the realization is possible either with infinite impulse response filters or with finite impulse response filters both are possible. We first decided to look at infinite impulse response filter design and that is because it takes a cue from standard analog design. So, we can take advantage of the standard analog designs that are available in the literature to design discrete time filters by an analog to discrete time transformation. And we identified what we call the bilinear transformation, subsequently we said that having made a transformation from a discrete time filter to a corresponding analog filter that is to be designed. One needs to then convert from an arbitrary kind of analog filter for example, a non low pass filter to a low pass filter. So, there is a transformation in the analog domain itself which we will study later. So, we will first look at low pass filter design and then we look at how we can generalize to other kinds of filters in the analog domain. And therefore, we were looking at the design of low pass analog filters and there again we agreed there are 4 possibilities depending on whether the pass band or the stop and the stop band are monotonic or non-monotonic and we said if it is non-monotonic then the best choice is equi-report. So, yesterday we looked at the case of monotonic stop band and monotonic pass band and that corresponds to what is called the Butterworth filter design. And the Butterworth filter design we have identified how to design the 2 parameters that characterize the Butterworth filter. The 2 parameters that characterize the Butterworth filter are as follows. The parameter called the order n and what is called the half power frequency or omega c. We have also written down the equations for obtaining the order and the half power frequency. Both of them have a range associated. So, the order has a minimum that is required and of course, you can take any order more than the minimum but the most appropriate choice is to take the integer just above the quantity that we get. You see because the order corresponds to the use of resources, the more the order the more the resources that you need to employ and therefore, you would like to use as small in order as possible but sufficient to meet the specifications. And therefore, you choose the ceiling or the integer just above a quantity that we calculated yesterday. Having chosen the order we can then choose the half power frequency again in a range and that range comes again because of the operation of ceiling. Now having chosen the half power frequency we then have the complete analog system function. So, the analog system the analog magnitude function more appropriately. So, the analog magnitude function is 1 by 1 plus omega by omega c to the power 2 n and this n is known and omega c is known the complete analog magnitude function is known. Now we need to complete the design by identifying h analog s. So, once we know n and omega c we know h analog s into h analog minus s which is 1 by 1 plus s by j omega c you see how do you get the analog system function you get it by replacing omega by s by j. You see s is equal to j omega on the imaginary axis when we wanted to find the frequency response we replaced s by j omega and therefore, to go back to the system function we need to replace omega by s by j. Now of course, this is acceptable if the function can be continued all over the complex plane as we have done here. In other words you know the function as restricted to the imaginary axis and you are assuming that it can be subjected to what is called analytic continuation. That means, you may continue that function all over the complex plane with the same expression. If that is acceptable then this replacement is valid. Now here this replacement is valid because this continuation is acceptable it is an analog it is an analytic expression in the analog domain and therefore, it can be continued all over the complex plane. Anyway once we have this now it is very easy to identify the poles and zeros of this analog system function all that we need to do of course, we can see very clearly there are no zeros at all and the poles can be obtained by equating the denominator to 0 and that is very easy to solve s by j omega c to the power 2 n is equal to minus 1 and you see you need to take the 2 nth root of both sides. So, you need to write minus 1 in a way in which you can identify all the 2 nth roots and therefore, you should write minus 1 as e raise the power j 2 k plus 1 times pi and odd multiple of pi any odd multiple of pi in the argument creates minus 1. Now you need to write this because you need to identify that when we take the 2 nth root on both sides you are going to have 2 n possible solutions and to identify these distinct solutions you need to identify the multiplicity in the phase of minus 1. So, now let us solve this equation. So, therefore, s by j omega c is e raise the power j 2 k plus 1 by 2 n times pi simple and therefore, s is equal to j omega c times e raise the power j. Now you can see the angle it is 2 k plus 1 by 2 n multiplied by pi. So, of course, you could write this in the form in its polar form clearly this contributes this and this contributes a magnitude of 1 and this really contributes the magnitude and therefore, the polar form. So, in fact, we should you know we should identify there are roots indexed by k. So, we should write s k here you know there are the roots are indexed by the integer k s k is omega c times e raise the power j 2 k plus 1 by 2 n times pi plus pi by 2 the plus pi by 2 comes from this term j here this adds the phase of pi by 2. So, this is the magnitude and the phase. So, it is very clear that all the s k's have a magnitude of omega c and therefore, all these roots lie on a circle with radius omega c a circle centered at the origin with radius omega c in the complex plane. In fact, just to get a feel of how the angles change of course, they all have the same magnitude, but just to get a feel of how the angles change let us enumerate the various angles for the case of n odd and n even. So, let us take the case of n equal to 3 and n equal to 4 that will enable us to see how these angles are distributed. So, for the case n equal to 3 let us write down the angles pole angles the pole angles are of course, given by pi by 2 plus pi by 2 n k pi by n and if n is equal to 3 then pi by 2 n is pi by 6. So, you can of course, you see all that you need to do is to identify that there are 2 n distinct poles here. So, if you run the integer k over 2 n consecutive values you will be covering all the poles. So, you could without any loss of generality start from k equal to 0 and in this case it could run from k equal to 0 to k equal to 5. So, in general you see you could have started with k equal to 1 it does not matter, but you need to run over consecutive 2 n values. So, let us begin with k equal to 0 your pi by 2 plus pi by 6 then you have and then subsequently move in steps of pi by 3 that is what this says each time you increment k by 1 you are adding an angle of pi by 3 let us catch these poles on the complex plane this is the complex plane this is the circle of radius omega c this is the first pole located you see this is the real axis and this is the imaginary axis. So, we could call this sigma k and capital omega k to denote the real part of the pole and the imaginary part of the pole. So, the first angle as you see is pi by 2 plus pi by 6. So, pi by 2 and then another angle of 30 degrees here. So, here we are this is the first pole this angle is pi by 6 subsequently we move in steps of pi by 3. So, when you move in the first step of pi by 3 you would reach here move the second step of pi by 3 and you would take the complex conjugate of this move in one more step of pi by 3 and you reach here and then of course, you have one more there. So, as you can see there are 6 poles and they are located at angles 0 pi pi by 3 2 pi by 3 and you know you can of course, calculate the remaining ones. Now, it is very interesting and quite satisfying to see that the poles are symmetric with respect to the real axis and also the imaginary axis. The symmetry with respect to the real axis is to be expected that is because you want the poles to occur in complex conjugate pairs. You want the coefficients of h and log s to be real and therefore, the poles must occur in complex conjugate pairs and they do occur indeed in complex conjugate pairs. Here of course, you do not need a complex conjugate wherever there are real poles they occur singly and wherever there are complex poles they occur in conjugate pairs that is to be expected. Now, in addition there is a symmetry with respect to the imaginary axis. Now, why do we have a symmetry in fact the symmetry more than the imaginary axis is with respect to the point 0 the origin. Now, why is that symmetry there because you have already created h and log s into h and log minus s. So, for every pole at s there is a pole at minus s and therefore, there is a symmetry with respect to the origin and that symmetry reflects as a symmetry with respect to the real axis and a symmetry with respect to the imaginary axis. Now, of course, as we expect there are no poles on the imaginary axis. If there were a pole on the imaginary axis stability would be affected the filter could never be stable. Just as the discrete system if there is a pole on the unit circle the filter cannot possibly be stable and therefore, here too there is no pole on the imaginary axis. And all that we need to do now to identify h and log s is to take away the poles in the left half plane. So, these poles these 3 poles give us the poles corresponding to h and log s and therefore, of course naturally these 3 poles are the poles corresponding to h and log minus s. Now, you see let us write down therefore, the system function corresponding to h and log s. In fact, let us call these poles since we have indexed them with k this is s 0, this is s 1, this is s 2. So, h and log s is equal to something divided by s minus s 0 s minus s 1 into s minus s 2. Now, what must come in the numerator of course, there are no zeros. So, you need a constant you need to put a constant. Now, what constant need you to put? You see the constant should be such that when you put s equal to j omega and when omega is equal to 0. In other words when you put s equal to 0 this should evaluate to 1. Now, when you put s equal to 0 what you have here is the product we write product like this product k going from 0 to 2 of minus s k because you want to we want this to evaluate to 1 at s equal to 0 and it is very easy to identify this product. You see if you look at it minus s 1 is of course, omega c minus s 0 into minus s 2 is mod s 0 squared because they are complex conjugates and therefore that is omega c the whole squared. And therefore, this product is essentially equal to omega c cubed very simple. In fact, I leave it to you as an exercise to work out the corresponding set of poles for capital N equal to 4. I shall straight away draw them I leave it to you as an exercise to work out these poles. In this case you would have 2 poles in each quadrant and they would have a separation of pi by 4 which is 45 degrees. So, this I will just mark some of them this is pi by 8 this angle is pi by 8 and this angle would be pi by 4. In fact, all the poles will be separated by an angle of pi by 4 and of course, you can draw the others. So, this together would amount to h and log s and here too you can number the poles. So, you can call this s 0 then s 1 and so on. And therefore, we can now write down h and log of s that essentially takes these 4 poles because they are in the left half plane. So, h and log s would be s minus s 0 s minus s 1 and the product of minus s k in the numerator which is essentially omega c to the power 4. I leave it to you as an exercise to complete the details of this. Now, having written down h and log s our job is almost done. The next step of course is to replace s by 1 minus 0 inverse by 1 plus 0 inverse and then we have the discrete system function and that completes the Butterworth design. So, much so for the design of Butterworth low pass filters.