 A warm welcome to the second session of the third module in the course of Signal and Systems. In the first session, we had introduced the basic theme of this module. We said that we want to bring parity between continuous and discrete independent variables. We also said that there were issues there, issues of whether there would be something lost in that effort to bring a parity. And then we also asked whether work would require to be done to bring parity. Now, in this session, what we will do is to answer in some very simple contexts all these questions that we have raised. What I mean by that is we shall also bring in one more notion here which will enable us to say whether the context is simple or difficult in the light of module 3. What is it that makes it easy for us to reconstruct with sampling and difficult for us to reconstruct with sampling? In fact, the central idea that makes it either easy or difficult to go from continuous to discrete independent variable is what is called a priori or information beforehand. The a priori information that we have or what we know about the signal beforehand makes it easier or more difficult. Let me first explain with an example, a very simple example. Let us take a very simple trivial constant signal. So, I have the signal let us say x t is equal to alpha, alpha is a constant independent of time. And I ask the same question, can I sample the signal and reconstruct it? And if so, how? The answer is very simple. We need only one sample anywhere, in other words take t equal to any t 0 and of course except t 0 is equal to alpha. So, if you sample, you know everything about the signal, so it is very interesting. If you are given a constant signal, you do not need to represent that signal as a function of a continuous independent variable at all. All that you need to do is to take one sample and that tells you everything about the signal. Now, here this is almost obvious. I started with what I might call a quote unquote obvious example, almost begging to show its truth. But what this illustrates is that there is always this tussle between a priori information. What was the information that we had before we started the process of sampling here? The information that we had before was that the signal was a constant. It is a very great deal of information. In fact, let me go one step further. Suppose I knew the signal is 0 everywhere, then how many samples do I need to represent the signal? None at all. Now, that is a trivial signal and a trivial situation. But now let me take a non-trivial signal, but a trivial situation. So, let us take a situation where we know the signal is, we know this, we need no samples at all. So, you know the a priori information or information beforehand leaves us with no need for sampling. There we are, a non-trivial signal. I mean it is a simple but non-trivial signal, it is a sinusoid. But I know all that there is to know about the signal. I know the frequency of the sine wave, I know the amplitude of the sine wave, I know the phase, the starting phase of the sine wave. So, I know everything there is to know about that sine wave. I do not need to make any measurements at all. So, if I say how many samples would I need to take to reconstruct the signal perfectly? My answer is I do not need to do any sample. A priori information I have is quite enough. Now let me take a slightly different example, where I reduce the a priori information. For variety let us take a different kind of signal, let us take an exponential signal. We can later go to a sinusoid as well, let us take an exponential signal. So, if we have an exponential signal, there you are, there is a priori information there. You know the signal is exponential. X of t is of the form some a0 e raised to the power alpha t, where a0 and alpha are constants. Now, we make only two measurements on this signal. I measure x at t equal to t1 and I get a0 e raised to the power alpha t1 and I do the same at t equal to t2. Now, we can ask the question, how do we reconstruct, that is very easy here, very simple. Let us write the two equations down, divide one by the other and of course, we know t2 and t1. Therefore, if we take the logarithm of this, the natural logarithm ln as it is often called. So log natural x t2 divided by x t1 is alpha into t2 minus t1 and all that you need to do to get alpha is divide by t2 minus t1. In fact, you could write log natural x t2 minus log natural x t1 divided by t2 minus t1 gives you alpha. So, we have an equation for alpha ready here. Now, how do we get a? That is very easy. All that I need to do is to look at any one of the samples. So, x at t1 is a0 e raised to the power alpha t1. Now, you know alpha. So, therefore, a0 is simply x t1 into e raised to the power minus alpha t1. So, I know a02, very simple. So, we have a beautiful situation here, where if I have a non-trivial, so it is not really a trivial signal. It is a non-trivial signal, it is an exponential. What I mean by a trivial signal is things like a constant signal or a zero signal, but this is not quite trivial. But I have the a priori information about the nature of the signal. I know the signal is an exponential and I need it to make only two measurements. In fact, the beauty is those two measurements could be anywhere in time except of course at the extremities and anyway it does not make sense practically to talk about measurements at extremities. By extremities, I mean t tending to infinity or minus infinity or something like that. You know, that is not really a practical thing to do. So, any finite t1 and t2 gives me a very sensible process to reconstruct x t. Now, let us try to make the problem a little more complicated. Suppose I know the signal has exponentials and only exponentials in it, but I do not know how many. You know, for example, I could assume the signal is of the following form. Let me write it down. So, it could be let us say x of t is of the form a1 erases the power alpha 1 t plus a2 erases the power alpha 2 t plus am erases the power alpha mt. And we will consider two situations, m known, m unknown. Now, if m is known, the problem is still quite tractable. So, with this a priori information, we can in fact make two m measurements on x t. And hopefully, these two m measurements should give us a clue to the two m unknowns that there are. There is only one issue. It is a very highly non-linear equation in these unknowns. So, let me make the problem even simpler here. Let us go back to this problem. So, I have m known and I also have a1 to am known. Now, things are somewhat easier. In fact, now I put this as a question for you. Write down the result of m measurements. So, you know, it is like this. When you write down, in fact, let me do this for you to an extent. When we write down this result of m measurements, how does it look? It looks like a1 erases the power alpha 1 and then we will have to say t1, of course. And then the same thing goes up to alpha m. Remember, it is t1 here, not tm. It is, of course, equal to x at t1 and keep doing this until you get x tm. Now, it is interesting. We could simply say put, you know, now what are the unknowns here? The unknowns are alphas. So, my exercise, now I will leave this to you as an exercise. Exercise or maybe a bit of a challenge, actually. Can we solve these for the alphas? How? If so, I am putting this to you as a challenge. Let us begin with triggering some thought in your minds. But let me do the simpler of the questions. Let me take the other context where I know the alphas, but I do not know the a's. We consider alpha known and a unknown. In fact, here this makes the problem very easy. So, I could go back to the previous set of equations. The equations are the same. They do not change. These are all known now. If the alphas are known, all these are known. And the unknowns are these. And the beauty is this is a set of m linearly independent equations in m unknown. So, it is a very simple situation to deal with. I have a set of simultaneous linear equations involving the a's and I can solve it routinely by any. And of course, you need to ask whether this is a set, you know, I said it is linearly independent, but I again ask you why I said that. This is a special kind of equation that we get here. The coefficients are not arbitrary. They are really sample versions of an exponential. So, that assures us that the equations are independent provided. Of course, that we have separate alpha. The alphas are distinct. So, we have seen in this discussion two examples of a relatively simple problem of sampling and reconstruction. We shall see more in the next session. Thank you.