 So we will first look at the Butterworth possibility now, we will look at the design of analog low pass filters using the Butterworth approximation. Now you see in the Butterworth approximation you want the pass pan and the stop pan both to be monotonic. So obviously if you want the pass pan to be monotonic and the stop pan to be monotonic, the only sensible way of doing it is to let the magnitude decrease starting from 0 frequency all the way down to the edge of the pass pan and then again starting from the edge of the stop pan all the way down to infinity. That seems to be the most sensible way to do it, which means you need, you see now that is another interesting thing. Since we are dealing only with magnitudes, we shall not design the system function of the analog filter directly. What we will design instead is what is called the magnitude complex system function. So in general we introduce the idea of what is called a magnitude squared system function. And a magnitude squared system function is essentially the analog system function or the field system function of the analog filter which we call h analog of s multiplied by h analog of minus s and we will explain why that is the case. You see you want to take the magnitude squared of h analog j omega, is not it? What you want is what we want to design is mod h analog j omega the whole squared. Well I mean you could say you want to design mod h analog omega j omega that is also alright but then I mean it is much easier to do squared as we will see in a minute, very equivalent because it is all non-negative. So you see mod h analog j omega whole squared can also be rewritten as h analog j omega multiplied by h analog j omega complex conjugate. But remember we are going to have real coefficients in h analog of s, there are going to be real coefficients in s. So when we take the complex conjugate the real coefficients remain unchanged but s gets complex conjugated. So you know we could have written s conjugate or you see since you are talking about j omega the complex conjugate of j omega is minus j omega. So even if we write minus s there we would get the complex conjugate. So in fact what we are saying is we want h analog s into h analog minus s evaluated over s equal to j omega. That is because taking the complex conjugate amounts to replacing j omega minus j omega. Now you may wonder why did we not write h analog s complex conjugate. You see what we want to ultimately design is what is called an analytic function of s. Now if we use a complex conjugate on s then what we get is a non-analytic function. The moment you complex conjugation does not allow for derivatives in the first place. Forget about any higher order kinds of differentiation. So we cannot use s complex conjugate but we can use minus s. Minus s will retain the analyticity and therefore we choose minus s over s conjugate. Anyway so we have chosen to design h analog s into h analog minus s and of course what we have to do is design this compound system and then we have to identify which of its zeros and poles correspond to h analog s and which of its zeros and poles correspond to h analog minus s but that is not a difficult job at all. We know that we want the filter to be stable and the analog domain in the s plane if stability is the case in fact not just stability we want it to be positive real. And positive realness means all the zeros and all the poles must be on the left half side of the s plane. So once we have the collection of zeros and poles corresponding to this product h analog s into h analog minus s it is very easy to identify what is h analog s. Take all the zeros and poles in the left half plane and put them into h analog s automatically all the minus s go into h analog minus s that is easy. So you see if you want a monotonic response for you see you want h analog j omega mod squared to monotonically decrease in past time and stop time. So it must have its maximum at omega equal to 0 and then it must steadily decrease as you move from 0 to infinity that is the kind of response that we want. Now of course the response at 0 should not blow up without bound it must be finite in fact it is most convenient to make it 1 and then you want the response to die. So what can be and be such as a function of omega remember we are taking h analog j omega into h analog minus j omega. So therefore the product is a magnitude squared function it is necessarily going to be a real function of omega. So I need to look for a real function of omega and the simplest real function of omega which monotonically decreases as you go from 0 to infinity is essentially a function where the denominator monotonically increases and you want it to be rational. So denominator must of course be a rational function in fact why rational the simplest thing to do is to choose the denominator to be a polynomial. So what simpler polynomial can you have than a polynomial which has a constant term and another term which increases monotonically with omega. In other words what we are envisaging is a magnitude squared function that looks like this 1 by 1 plus now you want it to be more negative. So you have omega by some parameter positive parameter to hold to the power 2 n now let us understand this. You see why we have chosen this as we anticipated at omega equal to 0 this is 1 and therefore the response starts at 1 as omega goes from 0 to infinity this monotone the denominator monotonically increases from 0 to infinity and therefore this fraction monotonically decreases from 1 towards 0 and further this omega c why have we introduced omega c and why have we introduced n. You see n controls the rate at which the response drops and omega c controls the point at which the response reaches a certain value. So for example at omega equal to omega c this is 1 and therefore this magnitude squared reaches what is called the half power point. The magnitude squared becomes half and therefore that is called the half power point if you were to put in a sinusoid in the unlocked domain into the filter then that sinusoid would come out with half power at omega equal to omega c. The point of half power changes as you change omega c and how fast the response drops after omega c is controlled by n the larger the n the faster it drops. Now using just these two parameters omega c and n we shall be able to meet any magnitude specifications that we choose let us see how. So it is a thing that you have arrived at the magnitude specifications as follows. This I am talking about now the analog domain you have a pass point edge a stop point edge in the pass point you have agreed that you want the response to be between 1 plus delta 1 and 1 minus delta 1 and in the stop point you do not want it to be beyond delta 2 and the pass point edge is at omega p and the stop point edge is at omega s and this is the unlock frequency omega. Now we shall see a minute that we can meet this by using the Butterworth filter but we can you know do a little better in the Butterworth filter we do not even need this we do not even need 1 plus delta 1 we can remain between 1 and 1 minus delta 1 in the pass point. In fact it is very easy we will return the equations that we need to satisfy you see and is that at the pass point edge 1 by 1 plus omega p divided by omega c to the power 2 must be greater than or equal to 1 minus delta 1 the whole squared remember we are talking about squared magnitude and the second equation that we have is at the edge of the stop point at omega s it must be less than equal to delta 2 squared and that is very easy to solve these 2 equations. In fact these are all non-negative quantities so if we multiply both sides by a non-negative number then the you see remember we must not forget we are dealing with inequalities here so when we multiply both sides of inequality we have to be careful whether we are multiplying by a non-negative number or a negative number here luckily all of them happen to be non-negative anyway you know let us take the first inequality and let us note that so let us give them numbers let us call this inequality 1 and let us call this inequality 2 from inequality 1 we have 1 by 1 minus delta 1 squared minus 1 is greater than equal to omega p by omega c to the power 2n and the second inequality we take the reciprocal now when we take the reciprocal of course the inequality is reversed they are all non-negative quantities so if I take the reciprocal of both sides the inequality is reversed and of course luckily none of the sides is 0 so reciprocal is acceptable. So we take the reciprocal of the second and you get 1 by delta 2 squared is less than or equal to 1 plus omega s by omega c whole to the power 2n therefore we have 1 by delta 2 squared minus 1 is less than or equal to omega c by omega s by omega c to the power 2n three inequalities here you see what we need to do is to first eliminate one way to solve for the two quantities omega c and omega I mean omega c and n you see so what is the design parameters here omega c and n all right you see we can we let us give these quantities names so now let us observe these two equations that we have here let us write them down again so we proceed as follows we will obtain n first and then omega c so to obtain n first we need to do away with omega c first right from the two equally flash the equations before you again so we have these two equations here which we have converted back into these two equations here right now please note let us make a few observations about these quantities here 1 minus delta 1 squared is a quantity less than 1 so therefore this quantity is going to be greater than 1 for sure and minus 1 is therefore going to be positive on the other hand if you look at this quantity here delta 2 squared is again significantly less than 1 so 1 by delta 2 squared is significantly greater than 1 right and therefore this minus 1 is going to be positive right so both of these are positive now we can take the logarithm on both sides right so in fact you know you could the logarithm is a non-negative operation is that right so logarithm is a monotonic operation which works on non-negative arguments in fact it is what is called a strictly monotonic so you know there is no place where it stabilizes of plateaus so if I take the logarithm on both sides of an inequality of non-negative quantities that inequality is preserved so let me take the logarithm of both sides let us keep the name 1 by 1 minus delta 1 squared minus 1 to be d1 and 1 by delta 2 squared minus 1 to be d2 so we have and of course both of them are greater than 0 as we observed right so the first equation that we have is d1 is greater than equal to omega p by omega c to the power 2n and the second equation that we have is d2 is less than or equal to omega s by omega c to the power 2n now if we take the logarithm as we do we have 2n times log omega p minus log omega c is less than equal to log d1 and of course log d2 is less than equal to 2n times log omega s minus log omega c what we want to do is to subtract these two inequalities to get rid of log omega c in a way and the most convenient thing to do is to reverse this inequality by taking the negative on both sides so let us reverse the inequalities so we have log d1 or minus log d1 is less than or equal to 2n log omega c minus log omega p and log d2 is less than equal to 2n log omega s minus log omega c simple and now we can just add now we can just add these inequalities is that right and one does away with you see log d2 minus log d1 is log d2 by d1 and this is less than equal to 2n times log omega s by omega p and why we have chosen this form is because omega s is greater than omega p so this logarithm is positive omega s is of course the stopband is more than the path stopband edge is more than the path stopband edge and therefore omega s by omega p is a quantity greater than 1 so its logarithm would be positive and another question is what about d2 and d1 remember d2 is 1 by delta 2 squared minus 1 now delta 2 squared is much less or less definitely than 1 by delta 1 1 minus delta 1 squared so d2 is definitely greater than d1 d2 is 1 by delta 2 squared minus 1 delta 2 squared is lower than 1 minus delta 1 squared so 1 by delta 2 squared is more than 1 by 1 minus delta 1 squared and therefore d2 is greater than d1 as well and therefore we have positive quantities on both sides of this inequality here so we are in good shape and now we have a very simple relationship and therefore terms are to be greater than or equal to half log d2 by d1 divided by log omega s by omega p and this is what governs the choice of what is called the order n is the rate at which the response decays and n has a direct relationship to how much of resource you are going to require to realize the filter that is not too difficult to see you see once you have n it tells you what is the power of s you see after all omega or g omega is s so a higher power of s means a higher power of z and a higher power of z means more delays more additions more multiplications so n has a direct bearing on the amount of resource that you require to realize the filter so n let us make a remark here n is a reflection of the resource requirement of the filter the more the n the more resources you need to invest in realizing the filter if you look just for a minute at the expression for n then you will see that it is very very intuitively clear why this is a reflection of resource let us see how n remember n is greater than or equal to some quantity now this quantity need not be an integer so what must you do you must put here what you call a ceiling operation a ceiling operation means the integer just above that for example suppose this quantity works out to be 6.4 then the ceiling of that is 7 if it works out to be 7.9 the ceiling is 8 if it works out to be 5.1 the ceiling is 6 so if it is a little bit above 5 the ceiling goes up by 1 step the ceiling is the integer just above so n must be greater than or equal to the ceiling of that quantity now let us look at the quantity itself for a minute and make some inferences you see it is very clear that n the requirement of n is going to be more if the numerator is more or if the denominator is less let us first look at the denominator when you have the denominator will be less the denominator would be less if omega s and omega p are close to one another if omega s is far away from omega p the denominator is more and therefore the requirement of order goes down so asking for a very sharp transition band means asking for a higher order now let us look at the numerator in the numerator d1 you see when would the numerator be more either when d2 is more or when d1 is less if d2 is more you are saying that 1 by delta 2 squared is more and that means delta 2 squared is less that means you are asking for smaller tolerance in the stop band if you are asking for smaller tolerance in the stop band you are demanding more and therefore you have to invest more resources on the other hand d1 is less than 1 by 1 minus delta 1 squared is less that means 1 minus delta 1 squared is more that means you are reducing the tolerance in the past and that means you are again asking for more and therefore you have to invest more engineering design whether it is in discrete time system design or meaning that any other branch of engineering is always a game of ask for more invest more when you ask for more you have to invest more and that is true for all the 3 kinds of asking here if you ask for a sharper transition band you have to invest more if you ask for a smaller stop band tolerance you have to invest more if you ask for a smaller pass band tolerance you have to invest more ask for more, invest more, simple. Now of course once you have completed the choice of n and you know now you understand why you had a 2n there. You had a 2n because you took h analog s into h analog minus s. So you know whatever poles and zeros you have in the original h analog are doubled when you go into h analog s into h analog minus s that is why you put 2n there and anyway you are considering the magnitude squared function. So you need to double, now this n is called the order, we will give it a name, n is called the order of the filter. More demand, more order, finding out omega c is very easy, finding out omega c is very easy. You could go back to the equation that we had here. Once you found omega, once you found n, finding omega c is very easy. You had this equation earlier. d1 is greater than equal to omega p by omega c to the power n and this is true. Now all that we need to do is to take omega c to the other side in both of them. We have from the first equation from this equation we would have omega c to the power 2n is greater than equal to omega p to the power 2n divided by d1 and from second equation taking omega c to the power 2n here and bringing this here you have omega c to the power 2n is less than equal to omega s to the power n by d2. Now you have a beautiful range in which omega c can lie. Of course, omega c is a positive quantity. So if you know omega c to the power 2n you can find out omega c by taking the 1 by 2nth root or you can use logarithms that is not a problem. But what we can see from here is that omega c to the power 2n must lie between omega s to the power 2n by d2 and omega p to the power 2n by d1 and any omega c satisfying this will do. Any omega c which satisfies this inequality will do. You may wonder why you got a range of omega c which is acceptable. That is because n took a sealing operation. So in n you took a sealing operation. So you introduce some amount of tolerance there and therefore you got a range of omega c which is possible. If that you know when you had the expression for n, if the right hand side happened to be an integer then you would get no tolerance here. You would get a strict value of omega c. But because you have got a sealing operation there you have introduced some tolerance. So therefore you have a range of omega c to choose from. Any omega c which satisfies this inequality will do in this range. Is that right? So yes, there is a question. Well, so the question is does omega c lie between omega p and omega s? Well that is not necessary. It depends on the tolerances that you have given. Omega c is the half power point. If your pass point tolerance is more than half power then of course omega c lies inside the pass point. So omega c is only the half power point. That is all that can be said. Anyway with this then we come to the end of this lecture. In the next lecture we will proceed further to complete the design of the Butterworth filter.