8.4 DUAL OF TIME [SAMPLING: SPECTRAL](#page-13-0) SAMPLING

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8.4 DUAL OF TIME SAMPLING: SPECTRAL SAMPLING

As in other cases, the sampling theorem has its dual. In Sec. 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 fs > 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 fs > the spectral width of X(ω) in hertz ( fs > 2B).

We now prove the dual of the time-sampling theorem. This is the spectral sampling theorem, which 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 < T¹ and t > T² (assuming T² > T1). The signal width or duration is τ = T² −T¹ seconds.

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.

$$X(\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt = \int_{0}^{\tau} x(t)e^{-j\omega t}dt$$

\n(8.8)

Figure 8.15 Periodic repetition of a signal amounts to sampling its spectrum.

We now construct xT⁰ (t), a periodic signal formed by repeating x(t) every T⁰ seconds (T⁰ > τ ), as depicted in Fig. 8.15b. This periodic signal can be expressed by the exponential Fourier series

$$x_{T_0}(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}$$

where (assuming T⁰ > τ )

$$D_n = \frac{1}{T_0} \int_0^{T_0} x(t) e^{-jn\omega_0 t} dt = \frac{1}{T_0} \int_0^{\tau} x(t) e^{-jn\omega_0 t} dt$$

From Eq. (8.8), it follows that

$$D_n = \frac{1}{T_0} X(n\omega_0)$$

This result indicates that the coefficients of the Fourier series for xT⁰ (t) are (1/T0) times the sample values of the spectrum X(ω) taken at intervals of ω0. This means that the spectrum of the periodic signal xT⁰ (t) is the sampled spectrum X(ω), as illustrated in Fig. 8.15b. Now as long as T⁰ > τ , the successive cycles of x(t) appearing in xT⁰ (t) do not overlap, and x(t) can be recovered from xT⁰ (t). Such recovery implies indirectly that X(ω) can be reconstructed from its samples. These samples are separated by the fundamental frequency f⁰ = 1/T⁰ Hz of the periodic signal xT⁰ (t). Hence, the condition for recovery is T⁰ > τ ; that is,

$$f_0 < \frac{1}{\tau} \, \mathrm{Hz}$$

Therefore, to be able to reconstruct the spectrum X(ω) from the samples of X(ω), the samples should be taken at frequency intervals f⁰ < 1/τ Hz. If R is the sampling rate (samples/Hz), then

$$R = \frac{1}{f_0} > \tau \text{ samples/Hz}$$

804 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE

SPECTRAL INTERPOLATION

Consider a signal timelimited to τ seconds and centered at Tc. 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. (8.6), we obtain the spectral interpolation formula†

$$X(\omega) = \sum_{n=-\infty}^{\infty} X(n\omega_0) \operatorname{sinc}\left(\frac{\omega T_0}{2} - n\pi\right) e^{-j(\omega - n\omega_0)T_c} \qquad \omega_0 = \frac{2\pi}{T_0} \qquad T_0 > \tau \tag{8.9}$$

For the case in Fig. 8.15, Tc = T0/2. If the pulse x(t) were to be centered at the origin, then Tc = 0, and the exponential term at the extreme right in Eq. (8.9) would vanish. In such a case, Eq. (8.9) would be the exact dual of Eq. (8.6).

EXAMPLE 8.6 Spectral Sampling and Interpolation

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

$$X(0) = 1$$

and $X(\pm 2\pi n) = 0$ $n = 1, 2, 3, ...$

Find x(t).

We use the interpolation formula Eq. (8.9) (with Tc = 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. (8.9) survives. Thus, with X(0) = 1 and τ = T⁰ = 1, we obtain

$$X(\omega) = \operatorname{sinc}\left(\frac{\omega}{2}\right)$$

and $x(t) = \operatorname{rect}(t)$

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

^† This can be obtained by observing that the Fourier transform of xT⁰ (t) is 2^π % ⁿ Dnδ(ω − nω0) [see Eq. (7.22)]. We can recover x(t) from xT⁰ (t) by multiplying the latter with rect(t − Tc)/T0, whose Fourier transform is T⁰ sinc(ωT0/2)e−jωTc . Hence, X(ω) is 1/2π times the convolution of these two Fourier transforms, which yields Eq. (8.9).