11 3. The Sampling Theorem We have seen that a time-limited function can be reconstructed from its Fourier coe!cients. The reader will probably have noticed that there is symmetry between frequency and time domains. That is to say, apart from the assignmen t of the sign of the exponent of exp (2lvw) > the v and w domains are essentially equivalent. For many purposes, it is helpful to use not w> v with their physical connotations, but abstract symbols like t> u= Taking the lead from this inference, let us interc hange the w> v domains in the equations (2.6, 2.13), making the substitutions w $ v> v $ w> W $ 1@w> ˆ{ (v) $ { (w) =.We then have, ˆ{ (v)= 0>v 1@2w (3.1) X4{ (pw) sin ( (p w@w)){ (w)=  (p w@w)p= 4 X = 4{ (pw) sin ((@w)(w pw)) = (3.2)(@w)(w pw)p= 4 This result asserts that a function bandlimited to the frequency interval |v|  1@2w (meaning that its Fourier transform vanishes for all frequencies outside of this baseband ) can be perfectly reconstructed by samples of the function at the times pw= This result (3.1,3.2) is the famous Shannon sampling theorem. As such, it is an interpolation statement. It can also be regarded as a statement of information content: all of the information about the bandlimited continuous time series is contained in the samples. This result is actually a remarkable one, as it asserts that a con tinuous function with an uncountable infinity of points can be reconstructed from a countable infinity of values. Although one should never use (3.2) to interpolate data in practice (although so-called sinc methods are used to do numerical integration of analytically-defined functions), the implications of this rule are very important and can be stated in a variety of ways. In particular, let us write a general bandlimiting form:12 1. F R E Q UENCY DO M A IN FO RM ULATIO N ˆ{ (v)=0>v (3.3) vf If (3=3) is valid, it su!ces to sample the function at uniform time intervals w 1@2vf (Eq. 3.1 is clearly  then satisfied.). Exercise. Let w =1={ (w) is measured at all times, and found to vanish, except for w = p =0> 1> 2> 3 and the values are [1> 2> 1> 1] = Calculate the values of { (w) at intervals w@10 from 5 w 5 and   plot it. Find the Fourier transform of { (w) = The consequence of the sampling theorem for discrete observations in time is that there is no purpose in calculating the Fourier transform for frequencies larger in magnitude than 1@(2w)= Coupled with the result for time-limited functions, we conclude that all of the information about a finite sequence of Q observations at intervals w and of duration, (Q 1)w is contained in the baseband |v|  1@2w> at frequencies vq = q@ (Q w) = There is a theorem (owing to Paley and Wiener) that a time-limited function cannot be band-limited, and vice-versa. One infers that a truly time-limited function must ha ve a Fourier transform with non-zero values extending to arbitrarily high frequencies, v= If such a function is sampled, then some degree of aliasing is inevitable. For a truly band-limited function, one makes the required interchange to show that it must actually extend with finite values to w = ±4= Some degree of aliasing of real signals is therefore inevitable. Nonetheless, such aliasing can usually be rendered arbitrarily small and harmless; the need to be vigilant is, however, clear. 3.1. Tapering, Leakage, Etc. Suppose we have a continuous cosine { (w)= cos (2s1w@W1) = Then the true Fourier transform is 1 ˆ{ (v)= { (v s1)+  (v + s1)} = (3.4)2 If it is observed (continuously) over the interval W@2  w  W@2> then w e have the Fourier transform of { (w)= { (w)  (w@W ) (3.5) and which is found immediately to be ½ ¾ W sin (W (v s1)) sin (W (v + s1))ˆ{ (v)= (3.6) + 2 (W (v s1)) (W (v + s1)) The function sinc(Ts) =sin (W v) @ (W v) > (3.7) plotted in Fig. (2), is ubiquitous in time series analysis and worth some study. Note that in (3.6) there is a “main-lobe” of width 2@W (defined by the zero crossings) and with amplitude maximum W .To each side of the main lobe, there is an infinite set of diminishing “sidelobes” of width 1@W between zero crossings. Let us suppose that s1 in (3.4) is chosen to be one of the special frequencies vq = q@W > W = Q w> in particular, s1 = s@W= Then (3.6) is a sum of two sinc functions centered at vs = ±s@W= Avery important feature is that each of these functions vanishes identically at all other special frequencies vq>q 6= s= If13 3. THE SA M P LING THE O REM 1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -5 -4 -3 -2 -1 0 1 2 3 4 5 s Figure 2. The function, sinc(vW )= sin(vW ) @ (vW ) > (solid line) which is the Fourier transform of a pure exponential centered at that corresponding frequency. Here W =1= Notice that the function crosses zero whenever v = p> which corresponds to the Fourier frequency separation. The main lobe has width 2> while successor lobes have width 1, with adecayrateonlyasfastas 1@ |v| = The function sinc2 (v@2) (dotted line) decays as 21@ |v|, but its main lobe appears, by the scaling theorem, with twice the width of that of the sinc(v) function. we confine ourselves, as the inferences of the previous section imply, to computing the Fourier transform at only these special frequencies, we would see only a large value W at v = vs and zero at every other such frequency. (Note that if we conv ert to Fourier coe!cients by division by 1@W > we obtain the proper values.) The Fourier transform does not vanish for the continuum of frequencies v 6= vq, but it could be obtained from the sampling theorem. Now suppose that the cosine is no longer a Fourier harmonic of the record length. Then computation of the Fourier transform at vq no longer produces a zero value; rather one obtains a finite value from (3.6). In particular, if s1 lies halfway between t wo Fourier harmonics, q@W  s1  (q +1)@W , |ˆ { (vq+1)|{ (vq)| > |ˆwill be approximately equal, and the absolute value of the remaining Fourier coe!cients will diminish roughly as 1@ |q p| = The words “approximately” and “roughly” are employed because there is another sinc function at the corresponding negative frequencies, which generates finite values in the positive half of the v axis. The analyst will not be able to distinguish the