 Now we share a very important problem or challenge you may look at it in either sense which nature poses whether it is in discrete time or continuous time the ideal system that we want to realize are irrational we shall see that very soon the ideal system when you put down ideal specifications for what we want to do the corresponding system turns out to be irrational and the whole game that we are going to play for several lectures all over the subject of filter design is to approximate this irrational ideal by rational system. The rational system you see nature in engineering this is probably a frequent occurrence what we want is what we cannot realize what we can realize often not what we want but we have to draw a compromise between the two that is often the case in engineering and technology and discrete time processing is no exception. Anyway let us look a little more at rational system and specifically linear constant coefficient difference equation. Now you see we have a linear constant coefficient difference equation let us take for example the linear constant coefficient difference equation that we had a minute ago y n is 5 by 6 y n minus 1 minus 1 by 6 y n minus 2 plus x n minus 1 by 4 x n minus 1. Now I want to make a few remarks about solving such difference equation by what I mean by solving is you are provided with the value of x n over a certain interval of time. So let us take an example let us assume assume that x n is known known for n equal to let us say N1 to n equal to N2 in particular N1 might be 0 and N2 might be say 20 or 100. So what we are saying is in a sense is that you know the input over a certain interval of the natural domain okay the input is known over an interval of the natural domain the output is known for some pre-specified points example suppose N1 is equal to 0 N2 is equal to 20 we continue this we are given y of minus 1 and y of minus 2 so you know the what we call the initial value and we wish to obtain y of n for n equal to N1 to N2 namely 0 to 20 how would we proceed in this case well the approach to proceeding is to for a moment assume that this difference equation lasted forever. So the approach is assume the difference equation lasted forever assume the Lwcd held forever find the corresponding system function what do you mean by the corresponding system function the ratio of yz to xz and what is that corresponding system function here you know it it is 1 minus 1 4th z inverse divided by we began with 1 minus 1 3rd z inverse into 1 minus half z inverse identify the poles identify the poles identify what are called the system poles now where is system poles here they are 1 3rd and half now of course we are only going to deal with the solution of such equations when the input also has a rational or you know the input is also a sum of what are called poly x terms. So we shall assume the input is also a combination of what are called poly x terms I explain what a poly x term poly x refers to polynomial multiplied by exponential. So for example you could take the poly x term example of a poly x term 2 n squared plus 5 n plus 1 multiplied by the 1 5th raise the power of n this is the poly x term this is the exponential part of the term and this is the polynomial polynomial in the natural domain variable. So poly comes from here and x comes from here a polynomial in the natural variable multiplied by an exponential is a poly x term now of course you can see that a pure exponential is also a poly x term with the polynomial of degree 0 a pure polynomial is also a poly x term with the exponential factor equal to 1 a constant is a very special case of a poly x term with the degree of the polynomial equal to 0 and the exponential factor equal to 1 and so on. So all these are covered in the class of poly x term. So when you have a poly x input as we call it that means an input which comprises of a sum of poly x term we can solve this linear constant coefficient difference equation very easily what we do is to identify what we call the input poles now now here for example the pole corresponding so the pole corresponding to this poly x term is 1 5th but with a multiplicity of 3 that means 1 5th taken 3 times why is it taken 3 times because there are 3 coefficients in the polynomial the polynomial as a degree 2 so it has 3 degrees of freedom the constant part of the polynomial the first degree part and the second degree part. So as many as are the degrees of freedom in the polynomial so many times is that pole repeated if the polynomial is of degree 0 of course that pole occurs with multiplicity 1 that means it occurs only once if the polynomial is of degree 1 that pole occurs twice is that polynomial is of degree 2 as we see here that pole occurs 3 times and so on so forth. Now of course you must remember even if you have just an n square term and a 0 degree term you must still assume that you have a repetition 3 times the pole is repeated 3 times the highest power of n that matters right so the essentially the multiplicity of the pole is 1 more than the highest degree of highest power of n in the polynomial is that correct so suppose you apply now let us take the simple case suppose you apply this poly x term as the input xn of course valid only in that region 0 to 20. Now a remark about how large or how small that region can be large is no problem it can go all over the integer x small is a problem you cannot make it smaller than the degree of the difference equation that means the highest pass sample that you need to deal with in fact here that you know you need to deal with 2 pass samples so you cannot just assume that this difference equation holds for 3 samples and then a solve it by the approach that we are using if the length over which this difference equation holds is large enough to cover the span of the difference span means how many samples you are involving at a time then it is valid to use the procedure that we are describing otherwise there is no solution but to operate the difference equation step by step so of course you know yn in terms of yn-1 yn-2 you know let us go back to that difference equation. So if you look at it you can always sit down and solve it term by term you know let us go back to this equation here so here you have yn is 5 by 6 yn-1-1 by 6 and so on so you know if you know y of minus 1 and if you know y of minus 2 and you know x of 0 and you know x of minus 1 then of course I can find y of 1 or y of 0 and so on. Once you know y of 0 you can go one step further you can find y of 1 so you can keep solving this difference equation step by step and obtain the output that you can do when the difference equation is valid only for too small an interval but that is not a practical way of doing it when it is valid for a very long time and what we are describing is the process for solving it when it is valid for a very long time if it is valid for a very small time it is not worth doing all that we are talking about right. Anyway so long enough means long time means long enough for all the samples to be covered samples that are involved now in this case we say the system poles are at one third and half the input poles and you know with each pole you write its multiplicity the input poles are at one fifth and with the multiplicity of three now we write down what are called the union poles we write down what are called the union poles now what we mean by the union poles are the poles of the system union with the poles of the input now how do you take now I said union not union means when you have a common pole you must bring them together now of course in this case there is no common pole right but suppose we also had one more we had a one third with a multiplicity of one here also in the input we introduce that too that means the input is a combination of two poly x term a one fifth to the power of n multiplied by a polynomial of degree 2 plus a one third to the power of n multiplied by a polynomial of degree 0 suppose the input is a combination of two poly x term what would be the union poles now look for the common pole which are the common poles one third one third is common now one third occurs with multiplicity one here and multiplicity one here so in the union it will occur with multiplicity 2 half occurs with multiplicity 1 one fifth occurs with multiplicity 3 this is the set of union is that clear now once you have the union poles you know the form of the output immediate you know the form of the output immediate what is the form of the output the form of y n now take each pole in turn one third occurring with a multiplicity of two contributes a 1 0 plus a 11 n times one third to the power n half with a multiplicity of one contributes a 2 0 into half raised to the power of n is that correct and one fifth with a multiplicity of 3 contributes a 3 0 plus a 3 1 n plus a 3 2 n squared into one fifth raised to the power of n this is the form of y is that is that clear is that clear yes any questions of this so far this is a very important point if you understand this half the job is done now all these a is a 1 0 a 1 1 a 2 0 a 3 0 a 3 1 a 3 all of them are unknown constant we have to determine the constant how do we determine the constant we determine the constant by imposing the constraints that are given to us right so the constants are determined are determined from the system constraints what do you mean by the system constraints one constraint is the Lwcd itself so put this y n back into the Lwcd and some constants will emerge on their own then take into account the conditions on why that are given to you some more of these constants will be constrained is that right so I give you a little exercise to help you understand this concept better the exercise is as follows obtain y n for n equal to 0 to 100 given that y n minus 5 by 6 y n minus 1 or you know we will do the same thing 5 by y n is equal to 5 by 6 y n minus 1 minus 1 by 6 y n minus 2 plus x n minus 1 by 4 x n minus 1 and x n is equal to 1 by 5 raised to the power of n for n equal to minus 1 up to 100 y of minus 1 is equal to 2 and y of minus 2 is equal to 4 okay so I leave this is an exercise for you to work out using the process that you just described now one remark about resume