 A warm welcome to the 15th session in the third module of signals and systems. We have explained in the previous two sessions the Nyquist theorem, what is required to be able to reconstruct a signal, a band limited signal perfectly from examples. We have also given a constructive proof namely put down a system which will give us that reconstruction and the system essentially retained all frequencies up to a little beyond f m on the positive side, little before minus f m on the negative side. So, we said keep all the frequencies from minus f m minus delta to plus f m plus delta and throw out all the rest and that would reconstruct the signal from example, because all the copies would be destroyed and the original would be retained. Now we also derived the impulse response of that system, let us quickly recapitulate the impulse response itself. The impulse response of the reconstructing system is given by 2 f c times sinc 2 f c t. Now let us identify what the difficulties with these systems are for different values of f c. One problem is the stability of this system. So, is the system stable? In fact, I am going to leave this as a challenge for you. The challenge is show that this system is unstable and how do you propose to show it is unstable? It is not too easy, not too difficult that is why I am calling it a challenge. How would you show it is unstable? Let us call this impulse response h t if you like. What would make this system unstable? Stability or instability is determined by the integral of mod h t, the absolute integral of h t. So, you need to convince yourself whether this absolute integral, let us call it h integral you know h tilde if you like. It is a constant of course. Now we do not know whether it is a finite or an infinite quantity. If it is infinite, it has to be non-negative of course. If it is infinite, the system is unstable. If it is finite, the system is stable and the challenge essentially before you is to show that this is the case. Again, how do you show it is the case? A simple way is to show that this integral diverges by taking pieces of the integral and showing that each piece is bigger than a certain quantity and all these quantities sum to a divergence series. So, let me give you a hint how you could go about it. I want you to solve this problem on your own. So, I am just going to give you enough hints so that you can solve it. Those of you who take the challenge can proceed and come up with an answer and the hints are as follows. So, let me now draw the parts of this. So, you know how does this h t look? First let us catch that. This is how the h t looks. All these are where the sink takes integer values and therefore, if you were to construct mod h t, it would look like this. So, whereas this is h t itself in black here, mod h t is in green. The black and the green coincide here, but they are different here. They coincide here and they are different here and so on. Now, what you need to do to show that the system is unstable is to take each of these segments here. Step number one, step number two, lower bound the area, each of these areas. Step number three, show that the sum of the areas is divergent. There are various ways of doing it. I am sure many of you will come out with your own variants, but the fact is that when you take these different absolute parts of the sum, you know, these different, the positive and negative lobe, the main lobe and the positive and negative side lobes and make them all positive and add up all of these, they diverge. So, why is this bad? You know, this is one of those peripheral examples. We said, take a system which, you know, h f is a Fourier transform of the impulse response. It is a frequency response of a system. It is one of those peripheral examples where the system is actually unstable, but it still has a frequency response. So, the impulse response is not absolutely integrable, but it has a Fourier transform. Where is the seeming contradiction in this? The seeming contradiction is that if the system is unstable, you are not guaranteed that its impulse response has a Fourier transform, but it might just and this is one of those might just cases. In fact, in some sense, this is an idealized system, a system that has a one frequency response up to a certain point and 0 everywhere else. There is a sharp brick wall response, you know, brick wall. Let me draw that frequency response. You will understand why it is called a brick wall. The frequency response looks like this minus f c to plus f c 1 and 0 outside. So, this is like a brick wall here, you know. So, these are called brick wall responses and brick wall responses are the problem here. Brick wall responses give you unstable systems. Now, this is the point which I am not formally proving at this stage. In fact, I am again, this is I am not asking you to do this challenge if you cannot because it is a bit difficult, but some of you who might really be motivated might actually want to take this as an even bigger challenge. Can you show that when there is this brick wall character in a frequency response, you have trouble with stability. Anyway, that was an aside. It is beyond the scope of the current discussion, but it is just to excite some of you who might just have that insight, but anyway. So, we have this unstable system which we are trying to use and why are we not very happy with using unstable systems because although the fact that there is a frequency response says that any sinusoid is going to give you a finite amplitude sinusoid coming out, there could be bounded inputs which give you unbounded outputs when you try to reconstruct. So, the process of reconstruction of a signal from its samples could be a situation where you have bounded inputs, but unbounded outputs not a thing that we like at all could be I am not saying see an unstable system does not necessarily have to give unbounded outputs every time a bounded input or an unbounded input is given to it nothing can be said. One thing that can be said for certain is that there are definitely some bounded inputs for which there are unbounded outputs, but beyond that nothing can be said about which bounded inputs give you unbounded outputs and which unbounded inputs give you unbounded outputs that is system to system specific. In this case, our main grudge is that the system is unstable. Now, why did we ask for that little margin? There is another reason for that. The reason is that you can ask the system to become stable by relaxing this brick wall business. You do not have to have that sharp wall if you have a margin. Let us explain what we mean. So, what we are saying is what you really wanted was you had the original signal going all the way from minus f m to plus f m on the frequency axis and you had the sampling frequency of f s and f s minus f m came somewhere here. What you really wanted to do was to isolate this from its copies. Now, we have this margin here which I can also interpret as a do not care region. What I mean by a do not care region is I do not worry about what the frequency response is in that region. Anyway, I am not interested what the system h f does there. Why? Because I have ensured whatever I wanted from the signal is up to f m. I mean f m on this side on the positive side minus f m on the negative side. So, whatever be my frequency response h f in that marginal region makes no difference and we could we can now make that response not brick wall. So, for example, we could have a response that looks something like this. Let me sketch it on the very same graph. So, you could have a response which is one all the way up to f m and falls off smoothly after that up to some f c and minus f c on this side not brick wall. What is the advantage of doing this? The moment you have a spectrum a frequency response like this. Let me show it to you again this one where the frequency response is now continuous. Notice there is a big difference between this one and the brick wall response. This response the green one which I am showing here is a continuous function a continuous spectrum a continuous frequency response continuous as in the sense of functions as opposed to the brick wall response. The brick wall response is discontinuous at the boundary at f c and minus f c. It is the discontinuity in the frequency response which created that instability. As I said the proof of this statement is a little beyond the scope of this discussion, but it is again an open challenge for any of you who want to take it to show that indeed when you have such discontinuities in the frequency response then instability is a consequence. But if we accept that as an advanced statement that we are just taking for granted at this stage then moving from the brick wall response to smoothly dropping response is going to save us from dealing with an unstable system to reconstruct and that is possible because of the margin that we introduced. How large can the margin be? Well the margin can be as large as the region between f m and f s minus f m allows. Let us look at the margin here this is the margin here. Now remember you are guaranteed that f c can at least go to f s by 2 f s by 2 is in the middle here between f s I mean here you know in between these exactly. So f s by 2 is the maximum possible f m that you can have or just beyond the maximum possible f m that you can have. So essentially you could have an h f that looks like this flat up to f m drops up to f s by 2 drops smoothly this is a continuous frequency response. Now it is a little more difficult to obtain the inverse Fourier transform of this. We will see more about this frequency response in the next session. Thank you.