 A very warm welcome to the 18th lecture on the subject of wavelets and multirate digital signal processing. We build further in this lecture, the idea of uncertainty and what is called the time bandwidth product. The product of sigma t squared and sigma omega squared as we defined them the last time. Now, we shall just recall a few ideas from the previous lecture to put our discussion in perspective. So, what we intend to talk about today is what is called the time bandwidth product. Essentially, as I said the product of the time variance and the frequency variance and we wish to build a lower bound on that product based on some very fundamental principles that could be explained when we look at functions as vectors. So, with that background let us recall some of the definitions that we had brought out in the previous lecture. Last time we had agreed to confine ourselves to a function x t belonging to L 2 R and we said we could define its time and frequency variances. In fact, the time variance was first defined by choosing what is called the time center. The time center t 0 was defined according to this. Essentially, treating x t or mod x t squared as a mass in terms of t and then looking for the center of mass that is the interpretation of this. And we assume this center was a finite number. I mean that is a reasonable assumption in almost all cases that we will talk about. From the time center we defined the time variance. The time variance was given by t minus t 0 the whole squared mod x t squared d t integrated over all time divided by the integral of mod x t squared. Essentially, this time variance was indeed the variance of a density which we constructed out of that mass mod x t squared divided by the norm of x the whole squared. Now, similarly we could talk about the frequency center and the frequency variance. So, we define the frequency center to be capital omega 0 given by minus to plus infinity integral omega mod x cap omega the whole squared divided by the norm x cap squared. Again we assume the frequency center is finite which is of course, always almost always the case there is no problem. And we also noted that for a real x t this frequency center is 0. So, we said if x t is real then mod x cap omega squared is an even function of omega and that implies that omega naught is 0 because of symmetry. Now, we could similarly define the frequency variance. The frequency variance sigma omega squared so to speak is essentially integral omega minus omega naught squared modulus squared of x cap omega divided by a similar integral of mod x cap squared. So, once again the frequency variance is like a variance on the density constructed out of the modulus squared of the Fourier transform. Now, let us see what happens when we shift a function in time as we expect when we shift a function in time there is no change in the magnitude of the Fourier transform. In fact, all that happens is that the center shifts. So, effect of shifting in time so let us consider y of t to be x of t minus t 1. Now, we can easily write down the integrals corresponding to y t and they are not very difficult to evaluate. I shall not go through all the details of working, but I shall straight away put down some of the important results here. In fact, I put it as an exercise. We can show that the time center of y would essentially be you see I will take an example. Suppose for example, you have a function centered at the point t equal to 5. So, t 0 is equal to 5. Now, if you shift the function backward by 5 units in time you would get a function centered at 0 that is easy to visualize and if one just writes the integrals down carefully that is easy to prove. So, essentially what it means is that when you shift backwards by the center you are bringing it to be centered around 0 and with that we can generalize saying that the time center of y is essentially t 0 plus t 1. So, for example, if t 1 is equal to minus t 0 then the center of y would be 0. Anyway, so one can show the time center of y is 0. The time variance of y is equal to the time variance of x. Again, this is a very easy result to show just amounts to writing the integrals down carefully and making a transformation of variable. So, I would not go into the full proof here. Similarly, we can now analyze what happens in frequency. What happens to the Fourier transform? Let us look at that first. So, if y of t is equal to x of t minus t 1 the Fourier transform y cap omega is e raise the power minus j omega t 1 times the Fourier transform of x and very clearly mod y cap omega the whole square is equal to mod x cap omega the whole square that is very easy to see here. So, you see under a shift in time the magnitude squared of that matter the magnitude of the Fourier transform is unchanged. The mass as seen on the frequency axis remains unchanged in the Fourier domain. So, you can shift as much as you desire and it has no effect on the Fourier transform magnitude squared. As you can see it affects the phase, but not the magnitude and the phase is not of consequence to us as far as the center and the variance are concerned. So, anyway with that we have understood what happens when we shift. Now, let us take the dual operation of shifting namely modulation. So, in fact, let us look at it as a shift in the frequency domain. So, let us consider y t to be a modulated version of x t. So, let us consider y t equal to e raise the power j omega 1 t times x t. Now, of course it is very easy to see the Fourier transform y cap omega is x cap omega minus omega 1. You will recall that this idea is used in amplitude modulation of that matter in frequency modulation in communication engineering. When we shift a signal on the frequency axis we are modulating in time. Now, of course here we are talking about modulation by a complex exponential. If you modulate by a sine wave we are essentially modulating by two complex exponentials and adding up these two modulated signals just to put the idea of modulation in perspective. Anyway, coming back to this. So, when we modulate with e raise the power j omega 1 t we essentially shift on the omega axis as we have done here. So, y cap omega is x cap omega minus omega 1 and of course it is again easy to work out the following exercise. The time center of y is equal to the time center of x. In fact, that is very easy to see mod y t squared is equal to mod x t squared. And so, as far as the time domain is concerned as far as the variable t is concerned there is no change in the mass that is placed on t. So, it is not surprising that the time center is unchanged as also the time variance which we will now show. The time variance of y is equal to the time variance of x. What about the frequency variance and the frequency center? That goes exactly dual to what happened when we shifted in time. So, frequency center the frequency center of y is now going to be the frequency center of x plus plus omega 1 and put a bracket there. And what happens to the frequency variance? Well, shifting on the frequency axis has no effect on the variance. Again, I leave it to you to show this by writing down the integral carefully. It is a simple exercise. I shall put down the final result. So, the frequency variance of y is equal to the frequency variance of x. So, far so good. So, we have taken two dual operations the time shift and the frequency shift. And in both of them we have had no change in the variances. And obviously, when there is no change in the variances there is going to be no change in the product of the variances. So, the sigma t squared sigma omega squared product which we shall now give a name. We call it the time bandwidth product. So, we introduce the term for time bandwidth product given by the product sigma t squared times sigma omega squared. Essentially, the time bandwidth product is the product of the time variance and the frequency variance. And this product is very important from a number of perspectives. As you can see, there is a strong invariance that this product exhibits. When you shift a function in time or when you shift it in frequency, in other words modulate it in time, this product is unchanged. Let us make a note of that. The time bandwidth product is unchanged when we shift in time or frequency. And remember shifting in frequency is modulation in time. Now, we would like to look at this product in a little greater depth. What happens to this product when we stretch or compress? You see in the wavelet transform or for that matter in all the discussion that we have had in some of the previous lectures, we have talked about moving along the frequency axis by a process of dilation. We saw that when we stretched or compressed a function, a band pass function, it was equivalent to moving that function on the frequency axis, moving in a constant quality factor fashion, keeping the ratio of the center frequency to the bandwidth a constant. Now, in this process of compression or expansion together called dilation, what is happening to this fundamental quantity, the time bandwidth product is the question that we would now like to ask and answer. So, the first thing that we observe is that you know if you look at the scaling, let us look at that process. Suppose, you have a function you multiplied by a constant, that is a trivial thing, we will settle that matter first. So, we had you see here we are talking about scaling the independent variable, when we stretch or compress, we are talking about scaling the independent variable. What happens when you scale the dependent variable? So, if you multiply the function by a constant, you must answer that question first. So, again it is very easy to show that if y t is equal to some constant, let us say c 0 times x t, c 0 is a constant, of course, unequal to 0, that is straight forward. So, if c 0 is a constant, it is very easy to see that in the process of defining the center or the variance, the centers or variance always have a ratio of mod x squared or mod x cap squared involved. Now, you know if y t is c 0 times x t, mod y squared of that matter mod y cap squared is going to be mod c 0 squared times mod x squared. I am sorry here it is just x. So, here to it is mod c 0 squared mod x cap squared. Now, when we take a ratio, this mod c 0 squared being non-zero is going to cancel from the numerator and denominator. And therefore, when we scale the dependent variable, there is no effect on the time center or the frequency center. Let us make a note of that. The mod c 0 squared term of course greater than 0 strictly cancels from the ratio. And therefore, scaling the dependent variable or essentially multiplication by a constant, multiplying the signal by a constant leaves the centers and variances unchanged. This is important. Scaling the dependent variable leaves the centers and the variances unchanged. And now the most difficult of them all and yet elegant and beautiful, what happens when we scale the independent variable. In other words, let us consider y t equal to x of alpha t and alpha is a real number alpha not equal to 0. Now, we know what happens to the Fourier transform in this. So, you will recall that when we scale the t variable by alpha, the Fourier variable capital omega is scaled by 1 by alpha, but there is also a 1 by mod alpha factor that emerges outside. So, what we are saying is if x t has the Fourier transform x cap omega, then x of alpha t with alpha a real number alpha not equal to 0 has the Fourier transform 1 by mod alpha x cap omega by alpha here it is alpha here it is mod alpha. It is adequate for us to see what happens to one of them the time or the frequency domain and the other one can be interpreted. Let us take the time domain for simplicity. So, consider the time domain. Let us see what happens to the time center let y t be equal to x of alpha t as usual alpha real alpha not equal to 0. And let us ask what is the norm of y t norm squared of y t in the L 2 r sense against the norm of x. So, the L 2 norm of y is this the L 2 norm squared I mean and of course, this is easily seen to be mod x alpha t squared d t. Now, it is an easy integral to evaluate all that we need to do is to make a transformation of variable. So, we have lambda is alpha t where upon d lambda is 0. So, this is the alpha d t and therefore, d t is 1 by alpha d lambda. Now, here in the integral you see let me just bring the integral back here the relevant integral. So, we are going to replace d t by d lambda by alpha and depending upon whether alpha is positive or negative if alpha is positive the limits still remain from minus to plus infinity. If alpha is negative the limits go from plus infinity to minus infinity, but then there is also a 1 by alpha negative there. So, 1 by alpha is negative and therefore, if you take the negative and the reversal of integral together all in all we always have the following. The norm of y in the L 2 R sense is always 1 by mod alpha times integral from minus to plus infinity x lambda mod squared d lambda. Where upon what we have concluded is norm y in the L 2 R squared sense squared is 1 by mod alpha times the norm of x in the L 2 R sense squared. Now, of course we can use the same reasoning I have illustrated the thought behind the reasoning and if we write down the time center and the frequency center expressions let us write down the time center as we said we are going to focus on time. So, let us write down the time center expressions the time center of y is going to be given by t times mod y t squared d t integrated divided by the integral of mod y t squared d t. Now, again I shall not repeat all the working that I have done to relate the norms of y and x. Essentially the central idea there is a transformation of variable put lambda equal to alpha t and do this all throughout the integral and one can straight away write down the final result. It is very easy to show by making that substitution that the time center of y is equal to 1 by mod alpha times the time center of x actually not mod alpha I am sorry it should be just 1 by alpha because we must take into account the sign as well not modulus you must know where it is modulus and where it is not here it is not modulus. So, for example, I will take the example of alpha equal to minus 1 if alpha is minus 1 then the time center is also reflected. So, if the time center of x was t 0 the time center of y becomes minus t 0. So, it is not mod alpha it is just alpha sometimes simple intuitive reasoning can also clarify certain points for us anyway here it is not modulus of alpha. Similarly, one could use the same transformation and show the time variance of y now here it is modulus. So, it will be 1 by modulus squared of the time variance of x and in fact if we now note that what we are doing in time is reversed in frequency. So, in time we have t being replaced by alpha t in frequency capital omega gets replaced by capital omega divided by alpha with a scaling of 1 by mod alpha, but please remember the mod alpha scaling is not going to affect either the time center or the time variance or the frequency center or the frequency variance. So, scaling the independent variable scaling the dependent variable has no effect we saw that a few minutes ago scaling the independent variable has an effect let us make a note of this. So, we have in the frequency domain x of alpha t which is essentially y t corresponds to 1 by mod alpha x cap omega by alpha and this has no effect no effect on the variances or the center as we saw a minute ago. It is essentially this that we need to look at carefully this one the 1 by alpha scaling and one can use a set of steps very similar to what we did in the time domain and arrive at the following conclusions. So, the conclusions are the frequency center of y is equal to alpha times the frequency center of x and as far as the frequency variance goes the frequency variance of y is mod alpha times the frequency variance of x. I leave it to you as an exercise to prove this it is easy to do by making the same kind of substitution of variable anyway. Now, let us look at the time band with product the time band with product of y is essentially the time variance of y multiplied by the frequency variance of y, but the time variance of y by this argument is clearly seen to be 1 by mod alpha squared times the time variance of x multiplied by. So, I am continuing this as a product here multiplied by the frequency variance of y is mod alpha squared times the frequency variance of x that is very interesting. The time variance is multiplied by 1 by mod alpha squared and the frequency variance is multiplied by mod alpha squared. So, when you take a product these two cancel and because they cancel you get just the time variance of x times the frequency variance of x which is essentially the time band with product of x. This is a very significant statement that we have made. What we have just shown is that the time band with product is unaffected by a scaling of the independent variable as well. It is very strongly invariant as we can see. It is a very serious statement. Let us write it down. The time band with product is invariant number 1 shift in time, number 2 shift in frequency or modulation in time, number 3 multiplication by a constant multiplying the function by a constant and finally, number 4 the most interesting of them all scaling the independent variable. Essentially what we have been calling dilation all this y dilation by alpha. So, it is invariant to dilation. So, the time band with product is something very fundamental about a function. What is it then variant to? What does it change with? It changes essentially with the shape. So, different shapes have a certain time band with product associated with them. That is interesting. Now, I put before you a question for thought. I like to put certain questions to challenge your imagination and this is one of them. Can two different shaped functions have the same time band with product? So, it is not to leave you entirely in the dark. I give you a hint. What would happen if you took the Fourier transform of a function? Can one employ the property of duality somewhere? And with that slight hint which I am sure should take you rather far, I shall leave the rest of the reasoning to you to answer this question. Anyway, that would also bring forth one more kind of invariance. So, it is in some sense giving you two gifts all at once and answer to this question and one more kind of invariance at this time band with product exhibits. Anyway, so now what we have seen is that this time band with product is a very important quantity in the context of functions. It in some sense is characteristic of the shape, though I must remark not unique to the shape, but characteristic of the shape. Now, we saw what was the time band with product of this pulse in the previous lecture. Last time, we had noted that the time band with product of this pulse. Now, you will agree with me that I do not really need to write the limits here, nor do I need to give you the extent of this, nor even do I need to give you the height of this. Just the shape would do, because whatever be this extent, whatever be this height, the time band with product is all the same. But as you noted, the time bandwidth of this product tends to infinity and we had asked, well this is terrible. It told us why we were not content with the Haar. When you are dealing with Haar in terms of the scaling function or the wavelet function, you are essentially working with functions of an infinite time band with product. Having a large time band with product is bad. It means that I cannot localize nicely in time and frequency. So, the whole objective or the whole game is to see how small you can make this time band with product. So, that is the question that the uncertainty principle is going to ask and we shall formulate that question today. A basic fundamental question, how small can the time band with product be? The Haar, terrible infinity. Well, we shall have good news very soon. We shall be able to come up with functions whose time band with product is not bad at all. But then we will again see what nature often does to us poor scientists and engineers. You can reach within a certain level of this bound that we are going to come to soon, but reaching the bound itself is impossible. Anyway, that is just a prelude. Now, what we need to do is to establish that bound. And to establish that bound, let us first simplify our work as we should always do. Try and take away unnecessary trappings from the problem, identify the essence or the core of the problem and then try and solve the problem. So, what we are trying to establish is the fundamentally lower bound on the time band with product. We have already noted the invariance of the time band with product to a number of different operations namely shifting in time, shifting in frequency, multiplication by a constant and scaling the independent variable. Now, what we shall do therefore, is to do away with the requirement of time and frequency center. So, let us first to answer this question assume without loss of generality and this without loss of generality is because of all that I just said that the time center and the frequency center are both 0. After all, if the time center is not 0, we could always shift and make the time center 0 in time without affecting the time band with product. If the frequency center is not 0, we can always shift in frequency which means modulate in time and bring the frequency center to 0. Of course, if the function is real, we do not even need to do that. So, this is without loss of generality. Let us make the time center and the frequency center 0, where upon we then have the time band with product to be essentially the following. It is integral minus to plus infinity the whole squared of x multiplied by t squared d t divided by the norm of x squared times something similar in frequency. So, we will keep this here minus to plus infinity omega squared mod x cap omega squared d omega and here again divided by the norm of x cap the whole squared. Now, we have already made an observation about this and we will formally make that observation again. We have already seen that this essentially is minus to plus infinity j omega x cap omega mod whole squared d omega and that we have noted is essentially you see noting that if x t has the Fourier transform x cap omega then d x t d t has the Fourier transform g omega times x cap omega, where upon integral minus to plus infinity j omega x cap omega mod squared d omega is essentially integral from minus to plus infinity d x t mod d x t d t mod squared d t, but there is a factor of 2 pi. So, it is 2 pi times this and of course, we also see that norm of x cap squared is 2 pi times the norm of x squared and therefore, if we now put back all together here and I put all these together I keep these as they are in the time bandwidth product and I note that this is essentially the norm squared or the energy in the derivative divided by the energy in the function, the factor of 2 pi cancels from the numerator and denominator here. So, we could write down all in all thus the time bandwidth product can be written as essentially the energy that the L 2 norm of t x t whole squared divided by the L 2 norm of t x t L 2 norm of x multiplied by the L 2 norm of d x t d t whole squared divided by the L 2 norm of x the whole squared a very interesting result. I will just remind you that when we put back the expression for the time bandwidth product this is essentially the L 2 norm of t x t and that is how we come to this conclusion. So, here of course, there is a bit of abuse of notation we are talking about a function t x t and then taking its L 2 norm. So, this is a ratio of 2 products of L 2 norms and now we are all set to minimize this product. In fact, that would be our next step in the beginning of the next lecture. What is the minimum value of this product minimum over all x that can exist in L 2 R and that would give us a very fundamental bound in nature which is called the uncertainty bound. We shall derive that uncertainty bound in the coming lecture. Thank you.