There's an important property of the DFT known as the shifting theorem. It states that a shift in time of a periodic x(n) input sequence manifests itself as a constant phase shift in the angles associated with the DFT results. (We won't derive the shifting theorem equation here because its derivation is included in just about every digital signal processing textbook in print.) If we decide to sample x(n) starting at n equals some integer k, as opposed to n = 0, the DFT of those timeshifted sample values is Xshifted(m) where
Equation (319) tells us that, if the point where we start sampling x(n) is shifted to the right by k samples, the DFT output spectrum of Xshifted(m) is X(m) with each of X(m)'s complex terms multiplied by the linear phase shift ej2pkm/N, which is merely a phase shift of 2pkm/N radians or 360km/N degrees. Conversely, if the point where we start sampling x(n) is shifted to the left by k samples, the spectrum of Xshifted(m) is X(m) multiplied by e–j2pkm/N. Let's illustrate Eq. (319) with an example.
3.6.1 DFT Example 2
Suppose we sampled our DFT Example 1 input sequence later in time by k = 3 samples. Figure 35 shows the original input time function,
Figure 35. Comparison of sampling times between DFT Example 1 and DFT Example 2.
We can see that Figure 35 is a continuation of Figure 32(a). Our new x(n) sequence becomes the values represented by the solid black dots in Figure 35 whose values are
Equation 320
Performing the DFT on Eq. (320), Xshifted(m) is
Equation 321
The values in Eq. (321) are illustrated as the dots in Figure 36. Notice that Figure 36(a) is identical to Figure 34(a). Equation (319) told us that the magnitude of Xshifted(m) should be unchanged from that of X(m). That's a comforting thought, isn't it? We wouldn't expect the DFT magnitude of our original periodic xin(t) to change just because we sampled it over a different time interval. The phase of the DFT result does, however, change depending on the instant at which we started to sample xin(t).
Figure 36. DFT results from Example 2: (a) magnitude of Xshifted(m); (b) phase of Xshifted(m); (c) real part of Xshifted(m); (d) imaginary part of Xshifted(m).
By looking at the m = 1 component of Xshifted(m), for example, we can doublecheck to see that phase values in Figure 36(b) are correct. Using Eq. (319) and remembering that X(1) from DFT Example 1 had a magnitude of 4 at a phase angle of –90 (or –p/2 radians), k = 3 and N = 8 so that
Equation 322
So Xshifted(1) has a magnitude of 4 and a phase angle of p/4 or +45°, which is what we set out to prove using Eq. (319).
URL http://proquest.safaribooksonline.com/0131089897/ch03lev1sec6
Amazon  


