 A warm welcome to the third session of the third module in signals and systems. We are now growing more and more confident of how we can process a continuous time signal which is band limited in an equivalent discrete time framework. Let us take an example to illustrate the point now. So, let us assume we have a continuous time signal band limited to 10 kilohertz. We wish to process this using a simple RC circuit. Now, we would of course generate the output continuous time signal. And here we will choose the time constant as follows. Essentially 1 by tau is an indication of frequency. In fact, it is an indication of, you see it is a time constant here. So, this is an indication of Hertz frequency. Let us take this to be much less than 10 kilohertz. So, let us make the time constant say 3 kilohertz. Well equivalent to 3, that is 1 by tau is 3 kilohertz. Now, you see note I am writing equivalent here, I will explain what I mean by that. Let us find out what the equivalent discrete time processing paradigm would be here. Towards that objective, let us find out the frequency response. So, to find the frequency response, we could proceed in various ways, we could find the impulse response and take the Fourier transform or we could look at the input and the output phases and find their ratio. Of course, we are talking about the input and output voltages here, that is what we mean by the input and the output. So, now we can see what happens. This expression has a magnitude given by the positive square root of 1 by 1 plus omega squared tau squared. So, what is the significance of omega equal to 1 by tau at omega tau equal to 1 or capital omega is equal to 1 by tau, the magnitude falls to 1 by square root 2 or the power falls to half. So, if I were to give an input sinusoid at 0 frequency, it would come out with the same power, but at this frequency, angular frequency 1 by tau, it would come out with half the power, that is the significance. So, a lot of people think of this as a critical point, you know half power point. Of course, you could take any multiple, you see you could ask what is the value capital omega at which the power falls to 10 percent, you know we could take any such value. Let us get a feeling. So, in general if I want the power to fall by ratio k as compared to what it is at 0, what is the omega to which I need to go to? Let us find it out. We are essentially saying the magnitude squared which is 1 by 1 plus omega squared tau squared must be k or 1 plus omega squared tau squared is 1 by k. So, omega squared tau squared is equal to minus 1 minus 1 by k and therefore, omega is equal to 1 by tau into plus 1 by k minus 1. So, of course, suppose for example, k is 10 percent or 0.1. So, therefore, let us take k to be 0.1 for example, then we are talking about capital omega being 1 by tau into 1 by 0.1 that is 10 minus 1 that is 9, so 3 by tau. So, you know this 0.1 is also an interesting number, 1 by 0.1 that is 10 minus 1 9 square root of 9 is 3, so 3 by tau. So, 3 by tau has a significance at omega equal to 3 by tau we fall to 10 percent or 0.1 of the power and that is what I meant when I said take it equivalent you know you remember I said take it equivalent that is what I meant here I said take 1 by tau equivalent to 3 kilo hertz. Now, equivalent means it is the corresponding angular frequency you see you want you know at omega equal to 3 by tau is what you want, so f is 1 by 2 pi tau into 3. So, of course, what you are saying here is take 1 by tau is equal to actually 1 by tau needs to be equal to the corresponding angular frequency which is 2 pi into 3 kilo hertz that is what we mean and if we do that now you are saying that at 9 kilo hertz you have the power falling down to 10 percent at 3 kilo hertz you had the power falling down to 50 percent half power one tenth power. So, you know if you treat the half power point as a approximate cutoff of a low pass operation that you want to this is a low pass operation is not it. If you understand this let us look at the frequency response of this you have already seen that we can sketch this frequency response here we have done it before let us do it here again this magnitude has a sketch that looks like this I am showing only the positive omega psi essentially it is a low pass kind of response and you can think of various points on the low pass response the 1 by tau point is the half power point the 3 by tau point is of course, the one tenth power point and so on. So, that is that is what we are really talking about here now. So, essentially you could now think of a circumstance where you restrict this response. So, we can now sample that signal now it is band limited to 10 kilo hertz. So, sample at more than twice let us sample it at 24 kilo hertz. So, now what is the situation on the omega axis 2 pi into 12 corresponds to pi equal to omega. So, corresponds to small omega equal to pi and if you look at the RC circuit and draw the response of the RC circuit let us draw that on the same graph of course, this response would go all the way up to plus infinity, but we now need to truncate this and keep only this part. So, I will just shade that part in red I am keeping only this part of the RC response and reducing it to 0 here. Now, what would in principle we would have to do is the following now J tau capital omega and capital omega has to be normalized. How do you normalize? You divide by the sampling frequency that means you put omega is equal small omega I mean is omega divided by 1 by T s which is omega multiplied by T s. So, capital omega is small omega divided by T s this is how we obtain the expression for capital H of omega that is the frequency response of the discrete time system. But then this should be restricted it is not just for all small omega this is to be restricted. So, it is like this you see now note let us draw it on a separate graph again you have this capital H omega. Now, you make a change of variable replace omega by small omega divided by T s and that would give us the same in terms of small omega. Now, restrict this to omega from minus pi to pi that means chop it and then of course capital H of small omega is periodic in omega. So, you should think of this being repeated around every multiple of 2 pi all this business will have to be repeated around every multiple of 2 pi. So, this would be the corresponding discrete time system. Now, you would of course continue you can find the inverse DTFT and so on, but that is a fairly complicated thing to do. What can be an approximation to this system? We can approximate this system by treating it as a low pass filter where omega equal to 3 by tau is well into the stop band. Now, remember 3 by tau corresponds to 3 into 3 kilohertz that is 9 kilohertz and this corresponds to 3 by 4 times 12 kilohertz or omega equal to 3 by 4 of pi. So, 3 by 4 of pi is well into the stop band and on the other hand 1 by tau which is 3 kilohertz is just half the power. So, now you can approximate this by an ideal low pass filter which has a cutoff somewhere between these two depending on how you prefer. If you think 50 percent of the power is low enough you could put that as the cutoff. If you think your power should be as low as 10 percent then you of course you could put the cutoff at 3 by 4 pi, but 10 percent seems to be too low a power to be kept in the pass band. So, maybe we could choose something in between instead of 1 by tau which is 3 kilohertz we could put at 6 kilohertz perhaps. So, we could think of this not as this non-ideal low pass filter which comes from an RC circuit, but as an almost ideal low pass filter which cuts off at 6 kilohertz on the original frequency scale and 6 kilohertz on the original frequency scale means pi by 2 on the normalized omega scale. We shall talk further about this in the next session. Thank you.