 a warm welcome to the 25th session in the third module of signals and systems. We have now gained some command on the sample and hold. The natural question is what about other kinds of interpolation? We had just raised the question in the end of the previous session. Let us now set out to answer that question. So, for example, suppose I did linear interpolation, by the way why would we do linear interpolation? Is it something that we do all the time in our childhood? What I mean by childhood is when we go to high school and you know we perform experiments and then we mark some points. We smoothly join the points you know we could join the points by a smooth curve or sometimes if we just want to use a scale we join the points with a straight line is it not? It is actually connect the dots as they say that is what interpolation is. So, very often we do this in high school, do not we? So, we have some independent variable here and you have some dependent variable and you mark some instances. Of course, let us assume the instances as is the case here to be uniformly spaced. At these instances we have certain values of the dependent variable. So, what is one way in which we get a feel of the curve? We just join them by straight lines, join the dots or you know fill in the gaps. Now, the question is how is this expressible in terms of systems? So, systemic description of this, what is it? Can I use the same kind of description that I had for the sample and hold? Can I think of this as a process of first doing an ideal sampling followed by some linear shift invariant operation? Give it a thought, let me give you the answer for the moment. You know, so what are you really doing here? You are passing the signal xt through an ideal sampler as usual and let us again take the sampling interval to be ts. Let us call the output xst as before and let us now pass it through a linear shift invariant system whose impulse response is, of course, we will call it xt, but the only difference is we have a different xt now. The xt is a triangle. You see what I am saying? Let us again reconstruct. We will call it xtl for linear, linearly interpolated. Let us sketch this for a typical situation. As usual, let us take a signal, a smooth signal if you please. Mark the sampling instance. Identify the samples. Let me draw xt first and showing xt in red and now what would happen when you pass it through the linear shift invariant system? Every impulse is going to get replaced by the impulse response and then you have these discrete impulses. So, you would essentially have these impulse responses duly shifted and scaled and then of course, add it together. Let us draw the shifted and scaled ones. Let us begin by drawing them in green. How would we express the shifted and scaled impulses? We would essentially have them to be x at the points n ts multiplied by the impulse response shifted to n ts for all in tj n. So, for n equal to minus 1, you have this one here. For n equal to minus 2, of course, it is outside the scope of this drawing. For n equal to 0, you have it here and similarly for n equal to 1, 2 and so on. So, these are the shifted and scaled impulse responses. I have shown them to you in green. Now, let us add the green responses. You see, look at it very carefully. Let us focus our attention for example, between minus ts and 0. Let us focus our attention here. What happens when you add two straight lines? When you add two straight lines, the result has to be a straight line. In fact, that is not at all difficult to see. We can prove it formally. Let us do that for a minute. Very simple. If you have a 1 t plus b 1, that is how you express a linear segment in time t and add it to a 2 t plus b 2. It is simply a 1 t plus a 2 t plus b 1 plus b 2. That is also a linear segment. Now, when you know that you are going to get a linear segment by addition, how do you need to find out what that linear segment is going to be? Look for two points. It is very easy to draw a linear segment. If I know two points of the linear segment, I know the whole segment. So, now, let us go back to the previous drawing that we had. As I said, let us focus on this part minus ts to 0 here. When you add these two segments here, which are now emboldening, what are you going to get? At this extreme, at minus ts, you are essentially going to get 0 plus the sample value at minus ts. At 0, you are going to get the sample value at 0 plus 0. You are essentially going to get these two points. So, I will show in red what you expect to get. In red, the linear segment that you will get is just a linear segment that joins them. And the same argument can be used for any other pair of subsequent points. So, you will agree with me that the consequence of what I have done is to join these points by straight lines like this. So, now, x LT is what I am going to show in emboldened or maybe I will say red with some crosses, crossed red like this. I am going to cross this. So, you see the effect, the whole idea of linear interpolation is really just again passing the ideally sampled output through a linear shift invariant system. How would we now characterize this linear shift invariant system in the frequency domain? Again, let us find out the Fourier transform. Again, instead of finding the Fourier transform directly, let us see if we can relate it to the previous one. So, suppose we sample, suppose we look at the whole circuit first. The only thing is we keep the hold only between minus ts by 2 and plus ts by 2. So, you know, you remember the hold was earlier between 0 and ts. So, now we hold, we keep that same ht, but move it backwards. In fact, we now call it a zero order hold and zero order refers to the fact that this is essentially a polynomial of zero order in t. Suppose we had this impulse response. Let us call it h0t to emphasize the zero order. Let us convolve h0t with itself. What you expect to get? In fact, I leave this to you as an exercise. Evaluate this convolution and verify, verify essentially that it gives you the same impulse response as what you need for a linear hold, for a linear what I call a first order hold. So, you know, this is called a first order hold. Now, it is a first order polynomial in the independent variable. So, zero order hold when convolved with itself gives you a first order hold and a first order hold essentially has piecewise linear impulse response over the time axis. Now, once we see this impulse response in this perspective, it is very easy to see the effect in the frequency domain. When you convolve a function with itself, you are multiplying the Fourier transform by itself. So, you can immediately write down the frequency response of the first order hold. It is the square. It is t s e raised to the power minus j omega t s by 2. Now, you know, please remember, this was the frequency response of the zero order hold, but as running from 0 to t s. And I need to multiply it by e raised to the power plus j omega t s by 2 because we have shifted it backwards by t s by 2. So, it is this response which is now squared. And as you can see, this would give us t s squared times essentially sine omega t s by 2 by omega t s by 2 the whole squared. Let us draw the magnitude. In fact, we do not really need to worry about magnitude in phase here because anyway the phase is 0. You know, it is essentially just the square. So, on the cycles per second frequency axis, you have essentially the square of a sinc function. Now, the sinc function starts with 1 and then it goes down, is not it? What is important is that the fall-off is going to be faster. So, you know, the sinc function has a maximum at 0. It takes the value of 1 at 0. So, I am essentially plotting the sinc part, sinc f t s part. And sinc f t s is always less than 1 in magnitude. So, squaring means reduction bringing below the sinc function. When you square a number which is less than 1, it goes down. So, it would look something like this. It will fall even faster. What is the good part of this? The good part of this, I mean, there is a good part and a bad part. The good part of this is that the side loops are now suppressed. So, you have lower side loops, the weaker side loops as compared to the original 0 order hold or the sample and hold circuit. That is the good part. So, you know, all those other aliases beyond the first one are going to get suppressed more. What is the bad part? The bad part is that, now look at this. The region between minus 1 by 2 t s and 1 by 2 t s also suffers a steeper drop. So, you know, unfortunately the equalization that I need to do needs to be a little steeper now. So, you have gained that you have lost some of the aliases a little more strongly, but you lose because your equalization has to be steeper. There is less flatness in the original signal band. The signal has got the original signal band has got more severely distorted. We will see more in some of the subsequent sessions. Thank you.