 A warm welcome to this discussion session, we have been having these discussion sessions where I invite my teaching associates to join me and participate in the discussion as the teaching associates essentially represent all of you who are participating in this course and they bring before us here some of the questions that have been discussed hotly or some of the questions which require some explanation or some of the discussions which require more elaboration or some other questions which they themselves have found to be important to answer. So, I am very happy to have with me today Siddhant Rana Day and Siddhant is one of my teaching associates he has been very active some of you must have seen him answer some questions on the forum and he has some very interesting points to discuss today. So, without further ado let me ask Siddhant to put forth his points one by one for discussion. On the discussion forums we had a question where the there was a system defined in the following way. So, where y of t was defined to be equal to x of t square u of t. So, there was a lot of discussion on this in the discussion forums and so many of us gave actually incorrect answers or misleading answers. So, and we learned it later that we were in fact wrong. So, I would just like to ask sir to clarify it. Yes, this was a very good question in fact some of my teaching associates had framed this question and it is one of those questions which illustrate you know that you can solve a problem both algebraically and graphically and there could be advantages in one approach to the other and what I am going to do first is to explain the meaning of the system. What does the system do? You know very often in the domain of signals and systems it is important not only to be able to write the algebra or write the algebraic description but also to be able to interpret it properly. Interpretation is critical in many concepts and signals and systems otherwise it just remains a mass of symbols you know it does not really give you a meaning and the meaning is what really gives beauty to the subject. Now, for example, here let us look at the description it says y of t is in fact let me put u t first followed by x of t squared. Now, let us forget about the u t for the moment suppose we took y of t. So, I will call it y 1 t y 1 t is x of t squared. Now, what does this really mean? See it is interesting you have to think of a mapping here. So, query any t for example, take t equal to plus 5 and take t equal to minus 5 it would say y 1 at plus 5 is x at plus 5 squared and similarly y 1 at minus 5 is also x at minus 5 squared but which is the same. So, you know when you write an equation like this. So, y 1 at minus 5 is x at minus 5 the whole squared which is also x at 25. So, when we write a system description like this you must interpret it as follows you are saying for this t calculate the output as follows from x and as follows comes here. So, you know you pick any t and your reading of of course, in this case you are luckily your equation is not implicit in the sense that you know it is not recursive. So, then if it were recursive then there is trouble in fact then you do not even know if you properly have solutions or not you know but here it is very clear definition is clear there is no ambiguity about the definition and I do this. So, it says at a given t pick up the value of x at the square of that value of t. Now, what is the problem with this system? This system can only take positive values in the argument here this argument can only be positive. So, it will always pick only non negative values is not it only non negative value greater than equal to 0 values are taken. What is the implication of that? It would mean that for example, if I had an x t like this only this part would ever figure in the output the negative part of t is lost forever is that right. So, now let us complete the system description by multiplication by u t let us now bring in u t. Now, you are multiplying by u t and u t is of course 1 for t greater than equal to 0 and 0 for t less than 0. So, now you are saying y of t is u t times x t squared meaning the first thing you do is whenever t is negative y t is automatically 0 because u t is 0 and then when t is greater than equal to 0 y t becomes essentially x of t squared and t to t squared is an invertible mapping for t greater than equal to 0 is that right. You see there is a 1 to 1 mapping as long as t is non negative the mapping is 1 to 1. So, there is a point in x which corresponds to the to the point in y of course, what it does is in some sense to compress and expand. So, between 0 and 1 of course, all the values between 0 and 1 get mapped between 0 and 1 and all the values beyond 1 get mapped beyond 1 and 1 is the point of transition 1 goes to 1. But before that you know there is a movement backwards and after that there is a movement forward. So, if you have a number greater than 1 it moves to a number it gets pushed you know. So, for example, y 1 of 2 is x of 4 you know on the other hand y 1 of half is x of 1 fourth you see it is a complicated system in that one can neither think of it as pushing things forward nor can you think of it as pushing things back. There is a non-linear transformation on the axis there. So, the system description is unambiguous now having understood and it is easier to see this graphically having understood the system description. One can then describe certain operations algebraically or graphically as one chooses and now Siddhant actually has thought a lot about this example if he is going to put forward some of his thoughts now in the discussion to follow. So, Siddhant go ahead tell us more about what you understood about this system. Suppose you give the input x 1 of t equal to let us say t to the system and so for that input you will get the output y 1 of t y 1 of t which is equal to x 1 of t square as per the definition of the system which in turn is equal to t square for t greater than equal to 0. It is always 0 for t less than equal to 0. So, this will be the output for such a system. So, now suppose you define another system. So, this was the question in the forum. So, suppose you say that your input is not x 1 of t but x 1 of t minus 4 and you are giving this input to the system. So, the important thing about x 1 of t minus 4 is how you interpreted. So, x 1 of t minus 4 does not mean that the input at time t minus 4 is x 1 of t minus 4, x 1 of t is essentially just a function. So, x 1 of t minus 4 is the value, you can treat it as the value of the function at some point. So, x 1 of t minus 4 does not mean that the value of the signal at time t minus 4 is x 1 of t minus 4. It actually means that the value of the signal at time t is x 1 of t minus 4. So, instead of writing x 1 of t minus 4, you should write it as let us say x 2 of t. So, this is a much better definition for this signal. So, this is equal to t minus 4 and if you give this function as an input to the system, you will get y 2 of t square. y 2 of t square is just essentially replacing t by t square in this expression here, in this expression. So, you are just replacing t with t square in this expression. So, this is t square minus 4 and just to check that things are all working fine. Let us define the signal x 3 of t as x 2 of t plus 4. So, this is essentially just t. So, this is equal to x 1 of t and this is just to check that the system we are not doing anything incorrectly. So, this leads to the output being y 3 of t square which is replacing t by t square in x of 3 in x 3 which is just t square. So, you can see that. So, if you do it like this, if you write x 2 of t instead of writing x 1 of t minus 4, you will end up with correct expressions. Yes, that is very correct. So, what is Siddhant's try to point out is that you know some of I remember see in this discussion there was this whole issue of whether one should meticulously go through the algebra or whether one should do it graphically. Now, let me explain what Siddhant is trying to say. He is very correct in that he is saying that if I take t minus. So, when he says x 1, if you had x 1 t, let us begin with x 1 t again. So, x 1 t being equal to t would lead to the corresponding output y 1 t given by t square u t. So, simple you know. So, it is t square for t greater than equal to 0 and 0 for t less than 0. Now, when you take x 2 t which is described in terms of x 1 as x 1 t minus 4, we know the interpretation. You see if replacing t by t minus 4 means you put there see at t equal to 4, it takes the value that the original signal took at 0. So, what is that 0 goes to 4? In other words, you are pushing the signal forward by 4 steps and then of course, everything else gets pushed by 4 steps too. So, now you know you have to be careful. So, one way is now I will ask you to you know try and attempt the same thing graphically. So, how does this look? You had t, t would look like this. So, this was this is x 1 of t which is essentially just a straight line passing at 45 degrees to the origin and x 1 of t minus 4 in the same graph is this pushing the whole thing forward by 4 steps. Now, what would this system do? This system would essentially take whatever is here either in x 1 t for the first output or in x 2 t for the second output and then make a non-linear transformation on it and what you need to verify now is that what Siddhan did algebraically is this done graphically. So, you essentially as I told you the mapping is very clear. There is a one to one mapping for all points. Anyway for t less than 0, the output is always 0. For t greater than 0, you essentially pick the point t squared from the output and put it on that value of y. Siddhan has explained to you how to deal with the problem algebraic. You know you can and as I told you in my email sent to all the participants and in the discussion 2 which I recorded as a response to all of you. I said you know you need the skill to also decide which might be more strategic in a given situation. So, this was a very good question put up by my teaching associate and it was one of those where you really need to decide which skill is more experienced.