DUAL OF TIME SAMPLING: SPECTRAL SAMPLING

8.4 DUAL OF TIME SAMPLING: SPECTRAL SAMPLING

As in other cases, the sampling theorem has its dual. In Section 8.1 , we discussed the time-sampling theorem and showed that a signal bandlimited to B Hz can be reconstructed from the signal samples taken at a rate of f s > 2B samples/s. Note that the signal spectrum

exists over the frequency range (in hertz) of −B to B. Therefore, 2B is the spectral width (not the bandwidth, which is B) of the signal. This fact means that a signal x (t) can be reconstructed from samples taken at a rate f s > the spectral width of X ( ω) in hertz (f s > 2B).

We now prove the dual of the time-sampling theorem. This is the spectral sampling theorem, that applies to timelimited signals (the dual of bandlimited signals). A timelimited signal x (t) exists only over a finite interval of τ seconds, as shown in Fig. 8.15a . Generally, a

timelimited signal is characterized by x (t)= 0 for t < T 1 and t > T 2 (assuming T 2 >T 1 ). The signal width or duration is τ=T 2 −T 1 seconds.

Figure 8.15: Periodic repetition of a signal amounts to sampling its spectrum. The spectral sampling theorem states that the spectrum X ( ω) of a signal x (t) timelimited to a duration of τ seconds can be

reconstructed from the samples of X ( ω) taken at a rate R samples/Hz, where R > τ (the signal width or duration) in seconds. Figure 8.15a shows a timelimited signal x (t) and its Fourier transform X ( ω). Although X (ω) is complex in general, it is adequate for

our line of reasoning to show X ( ω) as a real function.

We now construct x T0 (t), a periodic signal formed by repeating x (t) every T 0 seconds (T 0 > τ), as depicted in Fig. 8.15b . This periodic signal can be expressed by the exponential Fourier series

where (assuming T 0 > τ)

From Eq. (8.16) it follows that

This result indicates that the coefficients of the Fourier series for x T0 (t) are (1/T 0 ) times the sample values of the spectrum X ( ω) taken at intervals of ω 0 . This means that the spectrum of the periodic signal x T0 (t) is the sampled spectrum X ( ω), as illustrated in Fig. 8.15b . Now as long as T 0 > τ, the successive cycles of x (t) appearing in x T0 (t) do not overlap, and x (t) can be recovered from x T0 (t). Such recovery implies indirectly that X ( ω) can be reconstructed from its samples. These samples are separated by the fundamental frequency f 0 = 1/T 0 Hz of the periodic signal x T0 (t). Hence, the condition for recovery is T 0 > τ; that is,

Therefore, to be able to reconstruct the spectrum X ( ω) from the samples of X (ω), the samples should be taken at frequency intervals

f 0 < 1/ τ Hz. If R is the sampling rate (samples/Hz), then

SPECTRAL INTERPOLATION

Consider a signal timelimited to τ seconds and centered at T c . we now show that the spectrum X ( ω) of x (t) can be reconstructed from the samples of X ( ω). For this case, using the dual of the approach employed to derive the signal interpolation formula in Eq. (8.11b) , we obtain the spectral interpolation formula [ † ]

For the case in Fig. 8.15 ,T c =T 0 /2. If the pulse x (t) were to be centered at the origin, then T c = 0, and the exponential term at the extreme right in Eq. (8.18) would vanish. In such a case Eq. (8.18) would be the exact dual of Eq. (8.11b) .

EXAMPLE 8.6

The spectrum X ( ω) of a unit duration signal x (t), centered at the origin, is sampled at the intervals of 1 Hz or 2π rad/s (the Nyquist rate). The samples are:

Find x (t). We use the interpolation formula (8.18) (with T c = 0) to construct X ( ω) from its samples. Since all but one of the Nyquist samples are

zero, only one term (corresponding to n = 0) in the' summation on the right-hand side of Eq. (8.18) survives. Thus, with X (0)= 1 and τ =T 0 = 1, we obtain

For a signal of unit duration, this is the only spectrum with the sample values X (0)= 1 and X (2 πn)= 0 (n ≠ 0). No other spectrum satisfies these conditions.

[ † ] This can be obtained by observing that the Fourier transform of x T 0 (t) is 2 πΣ n D n δ(ω−nω 0 ) [see Eq. (7.26) ]. We can recover x (t) from x T 0 (t) by multiplying the latter with rect (t −T c )/T 0 , whose Fourier transform is T 0 sinc ( ωT 0 /2)e −j ωtc . Hence, X ( ω) is

1/2 π times the convolution of these two Fourier transforms, which yields Eq. (8.18) .