 Now, we shall look at the possibility of an equi-ripple pass band and a monotonic stop band. Is that right? You see, we move on to what is called the Chebyshev approximation of filter design. Chebyshev is a very well-known mathematician. He has made some very important contributions to theory of polynomials and rational approximations. So, his work has given a lot of insight into the behavior of polynomials and into the choice and the use of polynomials and rational functions. In particular, one class of polynomials that he proposed are called the Chebyshev polynomials, which are actually related to trigonometric functions, but are not trigonometric functions on their own. And we are going to employ those polynomials in our design here. What we want to do is to have an equi-ripple pass band and a monotonic stop band. So, we expect the pass band to have a magnitude something like this and we expect of course, the stop band to have a magnitude something like this. Now, if you want an equi-ripple behavior in the pass band, you want this essentially to correspond, you know, where do you get equi-ripple behavior from? Equi-ripple essentially one way to get equi-ripple behavior is to use the sinusoidal function because the sinusoids oscillate in equi-ripple manner over cycles. And what is a function which is sinusoidal in a certain range and then becomes monotonic later? Now, here we have to use a little beyond high school mathematics. We have to extend our ideas of trigonometry to complex functions. So, in fact, if we choose the function cos theta for example, cos theta can be written in complex form is e raised to the power j theta plus e raised to the power minus j theta by 2. And at the same time, the hyperbolic cosine or the hyperbolic function cos x is defined as e raised to the power x plus e raised to the power minus x by 2 is the close relationship between them cos hyperbolic. Now, theta in principle can be complex and so can x. So, one does not have to restrict this definition only to reals and when we do that, we get this whole class of functions that is equi-ripple in a certain region and then becomes monotonic. In fact, let us define the function C n of x which is cos of n times cos inverse x. Now, interestingly we will show that it can also be written as well in terms of the hyperbolic cosine. So, let cos inverse x be equal to theta remember we are now not restricting ourselves to real x at all although we will later on take the special cases of real x in different regions, but we are allowing x to be complex. And therefore, let cos inverse x be equal to theta and we are agreeing that all quantities are complex and therefore, x is of course, cos theta and therefore, we could expand x in terms of theta it is e raised to the power j theta plus e raised to the power minus j theta by 2, but this is also cos of j theta and therefore, j theta is cos inverse of x. Now, go back to C n x n theta which is also cos j n theta and therefore, that is also equal to cos of n times cos inverse x because j theta is cos inverse x. So, therefore, it is a very interesting observation that we have that cos of n times cos inverse x is also equal to cos of n times cos inverse x. And now, we will write down this expression for a few values of the integer n. So, let us take capital N equal to 0 and capital N equal to 1 now those are easy. So, C 0 x is cos 0 which is 1, C 1 x is cos of cos inverse x and that is simply x. So, as you see both C 0 and C 1 are polynomials you see C 0 is an even polynomial by an even polynomial I mean a polynomial with only even powers of the argument and C 1 is an odd polynomial. So, it is a polynomial with only odd powers of the argument now we shall generalize this by using a simple step of induction. We shall write down C n plus 2 x and C n x. So, let us write down C n plus 2 x plus C n x and that gives us cos of n plus 2 times cos inverse x plus cos of n times cos inverse x. And we can make use of the trigonometric identity when you add 2 cosines. So, this is 2 times quote the product of 2 cosines the first one is the average of these 2 and the second is this minus this by 2. So, you have this is equal from trigonometric identities to 2 times cosine of n plus 1 cos inverse x you see n plus 2 plus n times cos of cos inverse x this minus this by 2. And therefore, we have a very interesting recursive relation between the C n's C n plus 2 times x plus C n x is equal to 2 x times C n plus 1 x. And therefore, we have a recursion C n plus 2 x is 2 x C n plus 1 x minus C n x this is the recursion which takes you to successive C n x starting from 0 and 1. So, by induction we can now construct the C n x beginning with the 0 and 1 case. So, if I know C 0 and if I know C 1 I can construct C 2 if I know C 1 C 2 I can construct C 3 I can I can keep doing this. And this also makes it very clear that the C n's are all polynomials because you see by induction if C n is a polynomial in x and if C n plus 1 is a polynomial in x then this expression must be a polynomial in x because you are multiplying 2 x by a polynomial in x and subtracting another polynomial of x. And therefore, when you subtract 2 polynomials you must get a polynomial. In fact, you can also say something about the order of the polynomial. You see we have already seen by induction that C 0 is of degree 0 and C 1 is of degree 1. So, in general if we assume C n to be of degree n then you have a degree n polynomial here and you have a degree n plus 1 polynomial here. So, multiplying by x increases the degree by 1. So, it would become a degree n plus 2 polynomial coming from here and a degree n polynomial coming from there and therefore, the overall polynomial is of degree n plus 2. So, therefore, if we have proved in the inductive step that C n is of degree n and C n plus 1 is of degree n plus 1 then it follows by mathematical induction that C n plus 2 must be of degree n plus 2. Much more follows from this recursion. We have seen that C 0 is odd and C 1 is even. Let us assume that by again as an inductive basis that this is true for a particular n and n plus 1. You have to assume in the basis step here it is true for 2 successive values of the integer and it is of course, true for 0 and 1. So, suppose C n and C n plus 1 obey this property that they are odd when n is odd and even when n is even. We will prove that n plus 2 also obeys the property. So, you see let n be odd then n plus 2 will also be odd. Now, this is then assumed to be an odd polynomial and this happens to be an even polynomial. But an even polynomial multiplied by x makes it an odd polynomial again and an odd polynomial minus an odd polynomial must give you an odd polynomial. So, if C n is odd then and C n plus 1 is even then you go back to C n plus 2 being odd and conversely if n is even and then C n is assumed to be an even polynomial then C n plus 1 would be an odd polynomial multiplying an odd polynomial by x makes it even and therefore, an even polynomial minus an even polynomial would give you back an even polynomial. And therefore, by mathematical induction C n is an odd polynomial for odd n and an even polynomial for even n. All these are very interesting conclusions because we have seen. So, let us summarize these conclusions. C n x is an odd polynomial for odd n and even polynomial for even n. These are called the Chebyshev polynomials. In fact, now we can use two different expressions for the same polynomial. You see we will just repeat that once again and the two different expressions are useful in different regions. So, same C n x can be written as cos of n times cos inverse x and cos of n times cos inverse x. This is useful for x between 0 and 1 because then cos inverse x becomes real and this is useful when x is greater than 1 because cos for a real argument is always greater than 1 greater than or equal to 1. You see cos x for real x is of course, an even function. So, cos of x is equal to cos of minus x and therefore, we need only to look at the positive side of x. Cos x is always greater than 1. x greater than or equal to 0 real cos cos x is always greater than or equal to 1. In fact, cos of 0 is 1 and one can easily see that d dx of cos x is essentially e raise to the power of x minus e raise to the power minus x by 2. This is always greater than 0 for x greater than 0 and therefore, the derivative is always positive and therefore, cos x strictly increases beyond 1. So, you see for x greater than 1, we can use the expression involving hyperbolic cosine and for x less than 1, we can use the trigonometric expression and that gives us an insight into the behavior of the Chebyshev filter. Now, let us make use of this. You see what we want is to use that 0 to 1 region to create a pass band and the beyond 1 region to create the transition band and the stop band. So, let us do the following as we did for the Butterworth case. You see what you want is see up to between 0 and 1, this Chebyshev polynomial is going to oscillate or alternate. So, it is going to be equiple and it is not going to go beyond a certain range depending on what you must see. You see you can always multiply this Chebyshev polynomial by a constant by a tolerance to keep it within a certain range and beyond x equal to 1 of course, you can allow it to increase monotonically so that you go into the transition band and the stop band. The only catch is the Chebyshev polynomial itself is increasing, but you want the resultant expression to decrease. So, you can put it in the denominator, but if you put it in the denominator, you do not want to create a situation at 0, you wanted to take the value 1. So, we use the same strategy as we did for the Butterworth filter. We create the expression h analog j omega the whole squared for the Chebyshev case is equal to 1 by 1 plus epsilon squared C n squared times omega by omega p. Now, why are we doing this? You see for omega between omega p by the way is the pass span edge, why are we doing this? For omega between 0 and omega p, this argument is between 0 and 1 and therefore, you are employing the equilateral part of the Chebyshev response. For omega greater than omega p, this quantity becomes greater than 1. So, you are employing the monotonic part of the Chebyshev polynomial and of course, for omega greater than omega p, this quantity would monotonically increase and therefore, the denominator monotonically increases leading to a monotonic decrease of this expression and therefore, we have achieved what we wanted. An equilateral behavior in the pass span and a monotonically decreasing behavior starting from the edge of the pass span all through the transition band and then down into the stop band. Of course, the quantity epsilon squared allows you the pass span tolerance. So, we say epsilon squared because epsilon is assumed to be real and therefore, we take a square of a real number to ensure it is positive. That is what we really mean by saying epsilon squared. So, therefore, let us now write down the two requirements. This is the magnitude squared expression for a Chebyshev filter. Let us write down the two requirements on of course, here what is to be designed if you want to design a Chebyshev filter. So, design a Chebyshev filter design means the following. Obtain epsilon, obtain n and then obtain therefore, obtain h analog s and how would you obtain h analog s? You would have to identify the poles and then segregate those poles that correspond to h analog s by removing the poles that correspond to h analog minus s. So, let us obtain epsilon first. Epsilon is very easy. You see it is very clear that there would be an oscillatory behavior in the pass span. I am plotting h analog j omega the whole squared and then from the edge of the pass span and downwards, there would be a monotonic behavior and the oscillatory behavior essentially correct. See how much is the oscillation? How well when c n is equal to 0 of course, you reach this point and when c n is equal to 1 that is the maximum value it can take when the cosine takes the value of 1 you reach this point. And therefore, the pass span constraint you want this to be the tolerance. This should not go below 1 minus delta 1 and this should not go above delta 2 that is what you want. So, the pass span requires that 1 by 1 plus epsilon squared should be greater than 1 minus delta 1 the whole squared. Remember you are talking about the squared magnitude function and therefore, we have a constraint on epsilon. You see if you look at it this does not involve n at all that is interesting. The pass span has very little to 2 with n seems so. So, let us write down 1 plus therefore, epsilon squared must then be less than in fact it could be less than or equal to it does not matter if it is equal to it could be less than or equal to 1 by 1 minus delta 1 the whole squared minus 1 and in fact, recall that this is the d 1 that we saw in the Butterworth filter epsilon is therefore, less than equal to the square root of d 1 and you could take it to be the positive square root of d 1. Now, you see it looks like you have a range for epsilon I mean in principle you could even choose epsilon equal to 0. So, nothing comes for free as I said it looks as if n has nothing to do with the pass span, but it does indirectly. If you choose epsilon smaller you will see that you will actually be pushing for a larger n. So, now let us of course, we need to obey this constraint epsilon is less than equal to square root of d 1, but it is in our interest to choose the largest possible epsilon which is essentially epsilon equal to square root of d 1 we will see that in a minute. So, having chosen epsilon the stopband constraint says 1 by 1 plus epsilon squared C n squared omega s by omega p is greater than equal to delta 2 square I am sorry less than equal to delta 2 square it must be within the stopband. So, at the edge of the stop you see because of monotonicity. So, if the requirement is obeyed at the edge of the stopband it is obeyed all over the stopband if you have ensured that at the edge of the stopband you have reached the value less than delta 2 square you are bound to remain within delta 2 square all over the stopband. So, you need to check only at the edge of the stopband. Now, we can solve this very easily, but here remember we have omega s by omega p is greater than 1 and therefore, we must make use of the cos hyperbolic or cos expression and there we have 1 by delta 2 squared is less than or equal to 1 plus epsilon squared C n squared omega s by omega p or in other words 1 by delta 2 squared minus 1 is less than equal to epsilon squared C n squared omega s by omega p and there again we land up with the familiar d 2 here d 2 from the Butterworth filter. So, remember it is very interesting the same quantities occur both in the Butterworth filter and the Chebyshev filter. Now, in place of C n we must make use of the hyperbolic cosine expression and therefore, we have cos hyperbolic n times cos hyperbolic inverse cos inverse omega s by omega p the whole squared is greater than or equal to d 2 by epsilon squared and here we can take the take the positive square root on both sides. You see if we take the positive square root is a monotonically increasing operation and you have if you have positive quantities on both sides. So, if you take the square root on both sides the inequality is preserved, but you must remember to take the positive square root that gives us cos of n times cos hyperbolic cos rather cos hyperbolic or cos n times cos inverse omega s by omega p is greater than equal to the positive square root of d 2 divided by epsilon and of course, cos inverse of both sides. You see it is valid to take cos inverse of both sides because cos inverse and cos are both monotonically increasing functions. When is it valid to take a function of both sides of an inequality? It is valid if the function is monotonically increasing. You see if a is greater than b then f a is greater than f b if f is a monotonically increasing function of its argument. If it is a monotonically decreasing function of the argument then the inequality is reversed. If the function is neither monotonically increasing nor monotonically decreasing then it is invalid to take the function of both sides of an inequality that is the beauty. You can operate a function on both sides of an equality without any concern, but when there is an inequality you have to be worried about whether the function is monotonic or not. If it is not monotonic then the inequality can either be destroyed entirely or reversed or preserved. So, here of course, cos inverse you see cos itself is a monotonically increasing function. So, its inverse is also going to be a monotonically increasing function. So, since it is a monotonically increasing function we can take cos inverse on both sides and that gives us n times cos inverse omega s by omega p is greater than equal to cos inverse of square root of d2 divided by epsilon and that tells us an inequality for n as we expect. So, it is cos inverse square root of d2 by epsilon divided by cos inverse omega s by omega p. As expected there is a minimum value of the order and that minimum value is given by this expression and of course, you must put here a ceiling because n cannot be non integral. So, you need to put there the integer just above that quantity. If this quantity works out to be 7.3 the ceiling is 8 and so on. Now, of course, naturally you would choose n equal to the ceiling you do not want to invest more resources than required, but this is what requires to be obeyed by the order. And now it is also very clear why epsilon it directly plays a role in the order. You see cos inverse is a monotonically increasing function of its argument. The smaller you choose epsilon the larger this argument and the larger the cos inverse and therefore, the larger the requirement on the order. So, it is in our interest to choose as large and epsilon as we can and that is square root of d1. So, let us make a remark therefore, it is in our interest to choose as large and epsilon as we can namely epsilon equal to square root of d1 and having made the choice then of course, we have a simple expression for the order and that order is n is equal to the ceiling of cos inverse square root of d2 divided by square root of d1 of course, positive divided by cos inverse omega s by omega p. So, much so then for the design of epsilon and n we shall proceed in the next lecture to remark on the behavior of this choice as we change delta 2 delta 1 and what happens once we have made this choice of n as far as the choice of system function is concerned. So, we will do that in the next lecture.