 Welcome to the 10th lecture on the subject of digital signal processing and its applications. We continue today with our discussion of the discrete time Fourier transform. We have just introduced the idea and introduced the term in the previous lecture. But we have promised that we will look at it in more detail in the lecture today and we do so. Let us recapitulate a few ideas that we had begun with in the previous lecture. We had said that if we take an arbitrary sequence Xn, now not necessarily the impulse response of a linear shift invariant system. Any sequence, then its dot product from inner product with the sequence e raised to the power j omega n where omega as you know is the normalized angular frequency. Its dot product or inner product with the sequence e raised to the power j omega n was given a name of course if this converged. So, when you take a dot product here unlike in the case of the finite dimensional space where a dot product of two sequences is bound of two vectors is bound to have a non convergent value of finite value. Unlike there the case of finite dimensional spaces in infinite dimensional spaces we do not have this guarantee. So, we cannot rest assured that this dot product will converge because there is an infinite summation involved. But where if it converges we call it the discrete time Fourier transform of the sequence. So, its dot product is called the discrete time Fourier transform abbreviated by DTFT of Xn. In fact, we are given a name to the discrete time Fourier transform of the impulse response. We have called it the frequency response of the system. The DTFT of the impulse response of an LSI system if it converges. If it converges the DTFT of the impulse response of an LSI system if it converges is called the frequency response of that system. In fact, we have used Hm normally to denote the impulse response and H omega to denote the frequency response. Now, we employ a combination of these ideas. So, we understand that X omega is like the projection of the sequence Xn on the sequence e raised to the power j omega n. And therefore, we have agreed that you could possibly represent or reconstruct Xn from its projections as you can do in the case of vectors. Omega is unique so to speak only over any interval of 2 pi to know why this is the case. You see 2 pi denotes the sampling frequency on the normalize scale. So, the maximum frequency component that could have been present in the original signal is not more than half the sampling frequency if you have taken care to avoid LSI. And if you have not taken care anyway it is indistinguishable now. So, you need to deal only with the frequencies from minus pi to pi if you talk about the original phases or in fact if you look at the discrete time Fourier transform it is very easy to see that it is going to be periodic with the period 2 pi. Let us take a minute to prove that. That is another way of saying this. You see the dt of t of Xn is easily defined as summation over all integers n Xn e raised to the power minus j omega n. And we use capital X of omega to denote this. Let us consider X of omega plus 2 pi. Only X of omega plus 2 pi is obviously going to be summation n going from minus to plus infinity Xn e raised to the power minus j omega plus 2 pi n which is of course summation n going from minus to plus infinity Xn e raised to the power minus j omega n e to e raised to the power minus j 2 pi n. But this is identically 1. e raised to the power minus j 2 pi n is identically 1. And therefore this is the same as X of omega. So X omega plus 2 pi is identically equal to X of omega for all omega. And of course in particular for omega between minus pi and pi. Now this is of course a mathematical way of demonstrating this. But the physical interpretation is that uniqueness is only over the region from 0 frequency to half the sampling frequency. And for every rotating phasor with frequency omega you have a counter rotating phasor with frequency minus omega. They come together to form a silent wave. So you have uniqueness only over the region minus pi to pi. Beyond that there is periodicity. So there is uniqueness only over any continuous interval of 2 pi. That is what we are saying. Contiguous means an unbroken interval of 2 pi. Is that right? Now of course this is true for any discrete time Fourier transform. And therefore we expect that we should be able to reconstruct Xn from its components which are X omega's over this unique interval. In particular you could take the unique interval from minus pi to pi. So what are we saying in mathematical language? We are saying that we expect X of n should get reconstructed by taking these components. How do you reconstruct a vector from its components? You multiply the components by unit vectors in the direction of each of those components and add up these. So if you have a 3-dimensional vector and if its components in the X, Y and Z directions are 1, 2 and 4 then how do you reconstruct the 3-dimensional vector 1 times a unit vector in the X direction plus 2 times a unit vector in the Y direction plus 4 times a unit vector in the Z direction. So you multiply each component by unit vector in the direction of that component and add overall such components. Now here there is a slight difference. Here the components omega run from minus pi to pi and these components are not discrete. They are continuous. There is a continuum of components. Now if you have a continuum of components you cannot add. What should you do? You should integrate. And of course we asked for a unit vector. So I do not know whether e to the power j omega n is a unit vector or not. So I have to make a provision that if it is not a unit vector I should allow for a constant to divide or multiply that vector essentially a constant multiplying that vector to make it a unit vector. Hopefully the constant can be independent of omega. So what I am saying is I multiply the component by the so called unit vector and the unit vector is here is the power j omega n times some constant kappa 0 let us call it. Integrate over omega for omega run from minus pi to pi. I expect that this should be true. So I already have a physical interpretation. Now I need to prove this mathematically. The physical interpretation is I expect that I should be able to reconstruct the sequence here infinite dimensional vector by taking the product of components multiplied by the so called unit vectors and integrating over all such components over the region of uniqueness of omega. But now I need to prove this mathematically. So let us look at this expression. Let us in fact forget about the constant. Let us consider summation or rather let us just consider integral minus pi to pi x omega e raised to the power j omega m d omega. So we will worry about kappa 0 afterwards. Now x omega of course we know is summation k running from minus to plus infinity x k e raised to the power minus j omega k. I am intentionally using a different variable of summation here to distinguish it from the index n and I substitute that. So I have integral minus pi to pi x omega e raised to the power j omega m d omega is summation minus pi to pi summation k going from minus to plus infinity x k e raised to the power minus j omega k. The whole multiplied by e raised to the power j omega n and integrated with respect to omega. Now here I have a finite integral and of course I am assured that this infinite summation has converged. So I can bring that finite integral in and make it at only an omega. So you notice it is only this and this that depend on omega and I can make the integral act on them and bring the summation outwards. So I have, I will just remark here that it is of course an important technical point when you can make such interchanges of integrals, particularly when the integrals or summations are infinite in length. But we shall not dwell on those technicalities here. Let us take it that in this context it is acceptable and one of the justifications is that we have assured ourselves of the convergence of x omega. That is a good factor for it. Anyway, so this is equal to summation k going from minus to plus infinity x k integral from minus pi to pi e raised to the power j omega n minus k d omega. And it is this step we need to study. What is this? Well this is very easy to evaluate. In fact we shall evaluate it mathematically but let us evaluate it with some intuition first. What are we asking here? We are asking for the integral of a rotating number. Note here, see now for a moment you have to reverse the rule of omega n n, n minus k. Now you see you are integrating with respect to omega. So it is omega which is changing here not n and k. Then omega goes from minus pi to pi. Suppose n minus k is equal to 1, then you are going through 1 complete cycle. If n minus k is equal to 2, you are going through 2 complete cycles. Now of course n minus k can be 2 or minus 2. It depends on whether you are going clockwise or anticlockwise. But in any case you are always completing n minus k cycles when n minus k is not equal to 0. But when n minus k is equal to 0 what are you doing? You have e raised to the power j omega 0 which is 1. So you are essentially integrating 1 from minus pi to pi which simply becomes 2 pi. So when n minus k is not 0, the integral must go down to 0 because you are starting from a point and coming back to the same point. In fact each time you are going through for each positive value, for each particular value in the cycle you are going through the negative value as well. So we expect that this integral is going to vanish when n minus k is not equal to 0. But when n minus k is equal to 0 that integral is going to be equal to 2 pi. Now it is very easy to show this mathematically. In fact it is very easy to see that e raised to the power alpha x dx integrated is of course e raised to the power alpha x by alpha. This is the indefinite integral. So it is very clear that e raised to the power j omega n minus k e omega is going to be equal to e raised to the power j pi into n minus k minus e raised to the power minus j pi into n minus k divided by j into n minus k when n minus k is not equal to 0. This one raised to the power of n minus k by j minus n minus k which is j into n minus k which is automatically 0. So for n minus k not equal to 0 or n not equal to k that integral vanishes. To contrary when n is equal to k or n minus k is equal to 0 integral e raised to the power j omega n minus k d omega from minus pi to pi is simply 2 pi. It is 1 d omega. And therefore all that we need to do is to go back to that expression that we had that summation. You see what we have said now is that in this summation for n given n all these integrals for k not equal to n vanish and believe only the integral when k is equal to n and for k equal to n you get 2 pi here. And therefore we are saying that in fact we have answered 2 questions at once. We have suddenly answered what is kappa 0 as well. So we have said integral from minus pi to pi x omega e raised to the power j omega n d omega is 2 pi times x of n. And now it is very obvious that x n is therefore 1 by 2 pi integral from minus pi to pi x omega e raised to the power j omega n d omega. Now what is the, you see we have this mathematically we already made an interpretation of it in the beginning of the class in terms of vectors. It is a very interesting correlation that we see. And in fact as I said we have already answered the question what is kappa 0. Kappa 0 must be 1 by 2 pi. We did not need to work very hard to arrive at kappa 0. Now this is called the inverse discrete time Fourier transform. So we say the inverse discrete time Fourier transform of x omega which is periodic with period 2 pi is 1 by 2 pi integral minus pi to pi x omega e raised to the power j omega n d omega. Of course we assume again that this should converge. But the good thing is that if mod x omega is finite and if x omega is continuous then we expect that convergence will happen. There should not be too much of trouble. It is only in pathological cases that we would have trouble. So if you are given a discrete time Fourier transform then going back to the sequence is you know possibly not too difficult. I mean the ability to go back does not seem to be too much of a problem. But the other way is a problem. A sequence may not have a discrete time Fourier transform. In fact let me immediately give you an example of a sequence which does not have a discrete time Fourier transform. Let us take the sequence x n is equal to 2 raised to the power of n for n greater than equal to 0 and 0 for n less than 0. Obviously this summation k going from minus to plus infinity xk e raised to the power minus j omega k becomes summation k going from 0 to infinity 2 raised to the power of k e raised to the power minus j omega k which is summation k going from 0 to infinity 2 e raised to the power of minus j omega the whole raised to the k. Now this is a geometric progression with common ratio 2 raised to the power minus j omega. And obviously this common ratio has a magnitude greater than 1 not 2 raised to the power j omega is 2 which is greater than 1. So the geometric progression does not converge and therefore this x n does not have a discrete time Fourier transform. So you do not have to go very far to see examples of sequences that do not have a discrete time Fourier transform. So of course we need to confine ourselves if you want to deal with the discrete time Fourier transform we need to confine ourselves to sequences that do for the moment at least. Now this also illustrates an example of an impulse response which would not correspond to a frequency response. Of course it is obvious because this particular impulse response also would correspond to an unstable system. Impulse response is not absolutely summable. So this like this the Emissary system if it had this impulse response would be unstable.