 So, welcome to the 31st session on Signals and Systems and here we continue the story with which we left in the past session. So, recall there was this polyglot who had come to the King's court and who claimed to know several languages and had challenged the courtiers to find out his mother tongue. And of course, the courtiers had to find out a scheme by which they tricked that polyglot into revealing his mother tongue. So, to cut a long story short, what the courtiers did was to look for a time when this visitor or guest as you might call him, a little bit of a troublesome guest was asleep, fast asleep and kept a bucket of cold water ready. As soon as they noticed that this gentleman was comfortable in his sleep, they threw the bucket of cold water on him and you can imagine what happened. He rose suddenly with great discomfort and you can guess in which language he uttered curses his mother tongue. Well, that is in a lighter way, but the message that goes out from there is that in order to reveal something that is very special about a situation or about a person, you need to put the person into a situation where that particular thing becomes important or to put the person into a quote unquote troublesome situation as you might call it. The troublesome situation here is being awoken very rudely with a bucket of cold water. We are going to do something similar to the linear shift invariant system which disobeys or which pretends to disobey absolute summability or absolute integrability and yet claims to be stable. So, in other words, the way we are going to go about proving the necessity of absolute integrability or absolute summability for the stability of a linear shift invariant system is to subject that linear shift invariant system to such a troublesome input. We could force it to reveal if it is stable, it would force it to reveal the absolute summability of its impulse response and therefore, the necessity would be established. So, let us get down to business. Once again, let us begin with the discrete case. So, in the discrete case, we have y of n is equal to summation k going from minus to plus infinity. Well, this time we could write h k or x k first, but we will write h k x of n minus k and remember what we want is to assume that we have a bounded output when the input is bounded. You see, so what do you mean by establishing necessity? Let me explain. See, when you say you want to establish necessity, what are you doing formally? You are saying that you are guaranteed that a bounded output emerges when there is a bounded input and you are showing that as a consequence, if this is to be true for all bounded inputs including the quote unquote troublesome input, then you have no choice but for the impulse response to be absolutely summable in the case of discrete system and absolutely integrable in the case of continuous variable systems. So, let us take the particular expression for the output that we have here and if the output is bounded, we know that the every sample of the output needs to be bounded. In other words, every absolute value of the output sample needs to be strictly less than infinity. So, we can take any particular instance of the output sample. Let us do that here. So, consider specifically n equal to 0, you know you can focus on any particular sample that you desire. So, y of 0 is summation k going from minus to plus infinity h k x of minus k and now let us give the troublesome input that we spoke about. The troublesome input is as follows, put x of minus k is equal to the complex conjugate of h k divided by the magnitude of h k if h k is not equal to 0 and 0 if h k equal to 0. Now, you know I need to explain this. I have written down a formalism there but that does not really convey what I am doing. What I am doing is, I am literally pinching the input everywhere to conform to the impulse response. So, I am taking each term in the summation for y 0 and I am looking at the corresponding point in the impulse response. If that point is 0, I simply make the input 0 then anyway it has no consequence. If that value is non-zero then I put a corresponding input point which will come together with the impulse response to create an absolute sum. So, let us first work it out algebraically and then I will explain it with an informal explanation. So, let us take this input but before that let us take an example to understand this particular formalism. So, the example is as follows. You notice that I have written complex conjugate remember we are allowing complex impulse responses to be most general. So, let us take the following impulse response h n you remember the notation for finite sequences right. So, I will put at 0 the value j 1 plus j at 1 2 plus j at 2 and say 4 minus j at minus 1 and let us stop there. So, this is h n what would be the corresponding x n now take each point here that point is mapped to the corresponding negative point in x n. So, 0 goes to 0 of course but you take minus the complex conjugate divided by the magnitude look at minus 1 and bring it to plus 1. So, take the complex conjugate and divide by the magnitude now let me continue this here take plus 1 and you have 1 minus j divided by mod of 1 minus j and finally, 2 minus j divided by mod of 2 minus j. So, this is the input that we are going to take let us calculate the output in this case the corresponding y 0 is going to have k ranging only from minus 1 to plus 2 h k times h of minus k. So, essentially let us begin with minus 1. So, h of minus 1 which is 4 minus j times x of minus k. So, x of plus 1 4 plus j divided by mod 4 plus j then h of 0 comes next and that is j and x of 0 comes next and that is minus j divided by mod of minus j plus h of 1 comes next and x of minus 1 comes with it and finally, you have h of 2 that is 2 plus j and x of minus 2 comes with it and that is 2 minus j by mod of 2 minus j. Now, what do we have here if you look at it it is very simple it is simply mod 4 minus j the whole squared divided by mod 4 minus j plus mod j the whole squared divided by mod j well I am just writing it down formally you could just simplify this. So, mod 1 plus j divided by I mean mod 1 plus j the whole squared divided by mod 1 plus j plus mod 2 plus j squared divided by mod 2 plus j and that is essentially the absolute sum of h namely mod 4 minus j plus mod j plus mod 1 plus j plus mod 2 plus j have I done here I have given it this troublesome input which forces the absolute sum to come out and of course, you notice that the input is bounded it has to be because it has a finite number of samples and all the samples are bounded and if the output needs to be bounded that absolute sum of the impulse response has to be bounded because the output at 0 is equal to the absolute that is what we have done we have forced the output at a particular point to be equal to the absolute sum of the impulse response. So, we have chosen a particular input which forces the absolute sum of the impulse response to come out at a particular location on the output. Now, you understand very well what is going on you see what was going on when we made this definition here when we made this definition what we were trying to do is to give it that particular troublesome input. So, we were saying essentially choose an input which forces the absolute sum to come out at n equal to 0 in y and of course, here we have taken a little word of caution we have looked at the points where the impulse response is nonzero input a corresponding nonzero point at the negative of that value in the input and wherever the impulse response is 0 anyway that term in the summation is absent. So, we might as well put that quantity equal to 0 in the input. Now, I have shown you how to prove that this condition is sufficient is necessary for the discrete case. In fact, now let us complete the proof let us complete the formal proof let us go back to what we have here. So, now given this particular input we now need to establish in general what happens. So, what happens in general well in general with this choice x of minus k equal to h k complex conjugate divided by mod h k when h k is nonzero and equal to 0 when h k is 0 y of 0 becomes equal summation over all k h k h k complex conjugate divided by mod h k which is summation k going from minus to plus infinity mod h k and this must be bounded this must be finite if y is to be bounded. Notice you know note see y must be bounded because x is bounded note that x is bounded here. In fact, you can find the bound the bound is equal to 1 if you look at the magnitude of this this has a magnitude of 1 and this has a magnitude of 0. So, either x has a magnitude of 1 in its samples or 0. So, clearly x is bounded if the input is bounded you have no choice but the output being bounded if the output is bounded y of 0 must be finite and if y of 0 is finite you have no choice as you see there but that the absolute sum must be finite. And therefore, so essentially this has come out as a consequence of this troublesome input so we have proved for the discrete case that this condition of absolute summability is necessary. Now, the exercise for you which I am going to leave to you is to prove that this condition is necessary in the continuous variable case which can be done by a very similar argument to this and I would like you to do it. Thank you.