8.1 THE [SAMPLING](#page-13-0) THEOREM

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8.1 THE SAMPLING THEOREM

We now show that a real signal whose spectrum is bandlimited to B Hz [X(ω) = 0 for |ω| > 2πB] can be reconstructed exactly (without any error) from its samples taken uniformly at a rate fs > 2B samples per second. In other words, the minimum sampling frequency is fs = 2B Hz.†

To prove the sampling theorem, consider a signal x(t) (Fig. 8.1a) whose spectrum is bandlimited to B Hz (Fig. 8.1b).‡ For convenience, spectra are shown as functions of ω as well as of f (hertz). Sampling x(t) at a rate of fs Hz ( fs samples per second) can be accomplished by multiplying x(t) by an impulse train δ*^T* (t) (Fig. 8.1c), consisting of unit impulses repeating periodically every T seconds, where T = 1/fs. The schematic of a sampler is shown in Fig. 8.1d. The resulting sampled signal x(t) is shown in Fig. 8.1e. The sampled signal consists of impulses

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^† The theorem stated here (and proved subsequently) applies to lowpass signals. A bandpass signal whose spectrum exists over a frequency band fc − (B/2) < |f | < fc + (B/2) has a bandwidth of B Hz. Such a signal is uniquely determined by 2B samples per second. In general, the sampling scheme is a bit more complex in this case. It uses two interlaced sampling trains, each at a rate of B samples per second. See, for example, [1]. ‡ The spectrum X(ω) in Fig. 8.1b is shown as real, for convenience. However, our arguments are valid for complex X(ω) as well.

Figure 8.1 Sampled signal and its Fourier spectrum.

spaced every T seconds (the sampling interval). The nth impulse, located at t = nT, has a strength x(nT), the value of x(t) at t = nT.

$$\bar{x}(t) = x(t)\delta_T(t) = \sum_n x(nT)\delta(t - nT)$$

Because the impulse train δ*^T* (t) is a periodic signal of period T, it can be expressed as a trigonometric Fourier series like that already obtained in Ex. 6.9 [Eq. (6.25)],

$$\delta_T(t) = \frac{1}{T} [1 + 2\cos\omega_s t + 2\cos 2\omega_s t + 2\cos 3\omega_s t + \cdots] \qquad \omega_s = \frac{2\pi}{T} = 2\pi f_s$$

Therefore,

$$\bar{x}(t) = x(t)\delta_T(t) = \frac{1}{T}[x(t) + 2x(t)\cos\omega_s t + 2x(t)\cos 2\omega_s t + 2x(t)\cos 3\omega_s t + \cdots]$$

(8.1)

To find X(ω), the Fourier transform of x(t), we take the Fourier transform of the right-hand side of Eq. (8.1), term by term. The transform of the first term in the brackets is X(ω). The transform

778 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE

of the second term 2x(t) cos ωst is X(ω − ωs) + X(ω + ωs) [see Eq. (7.32)]. This represents spectrum X(ω) shifted to ω*^s* and −ωs. Similarly, the transform of the third term 2x(t) cos 2ωst is X(ω − 2ωs) + X(ω + 2ωs), which represents the spectrum X(ω) shifted to 2ω*^s* and −2ωs, and so on to infinity. This result means that the spectrum X(ω) consists of X(ω) repeating periodically with period ω*^s* = 2π/T rad/s, or fs = 1/T Hz, as depicted in Fig. 8.1f. There is also a constant multiplier 1/T in Eq. (8.1). Therefore,

$$\overline{X}(\omega) = \frac{1}{T} \sum_{n=-\infty}^{\infty} X(\omega - n\omega_s)$$

\n(8.2)

If we are to reconstruct x(t) from x(t), we should be able to recover X(ω) from X(ω). This recovery is possible if there is no overlap between successive cycles of X(ω). Figure 8.1f indicates that this requires

$$f_s > 2B \tag{8.3}$$

Also, the sampling interval T = 1/fs. Therefore,

$$T < \frac{1}{2B}$$

Thus, as long as the sampling frequency fs is greater than twice the signal bandwidth B (in hertz), X(ω) consists of nonoverlapping repetitions of X(ω). Figure 8.1f shows that the gap between the two adjacent spectral repetitions is fs − 2B Hz, and x(t) can be recovered from its samples x(t) by passing the sampled signal x(t) through an ideal lowpass filter having a bandwidth of any value between B and fs − B Hz. The minimum sampling rate fs = 2B required to recover x(t) from its samples x(t) is called the Nyquist rate for x(t), and the corresponding sampling interval T = 1/2B is called the Nyquist interval for x(t). Samples of a signal taken at its Nyquist rate are the Nyquist samples of that signal.

We are saying that the Nyquist rate 2B Hz is the minimum sampling rate required to preserve the information of x(t). This contradicts Eq. (8.3), where we showed that to preserve the information of x(t), the sampling rate fs needs to be greater than 2B Hz. Strictly speaking, Eq. (8.3) is the correct statement. However, if the spectrum X(ω) contains no impulse or its derivatives at the highest frequency B Hz, then the minimum sampling rate 2B Hz is adequate. In practice, it is rare to observe X(ω) with an impulse or its derivatives at the highest frequency. If the contrary situation were to occur, we should use Eq. (8.3).†

^† An interesting observation is that if the impulse is because of a cosine term, the sampling rate of 2B Hz is adequate. However, if the impulse is because of a sine term, then the rate must be greater than 2B Hz. This may be seen from the fact that samples of sin 2πBt using T = 1/2B are all zero because sin 2πBnT = sinπn = 0. But, samples of cos 2πBt are cos 2πBnT ⁼ cosπ*ⁿ* ⁼ (−1)n. We can reconstruct cos 2πBt from these samples. This peculiar behavior occurs because in the sampled signal spectrum corresponding to the signal cos 2πBt, the impulses, which occur at frequencies (2n ± 1)B Hz (n = 0,±1,±2,…), interact constructively, whereas in the case of sin 2πBt, the impulses, because of their opposite phases (e±jπ/2), interact destructively and cancel out in the sampled signal spectrum. Hence, sin 2πBt cannot be reconstructed from its samples at a rate 2B Hz. A similar situation exists for signal cos(2πBt+θ ), which contains a component of the form sin 2πBt. For this reason, it is advisable to maintain sampling rate above 2B Hz if a finite-amplitude component of a sinusoid of frequency B Hz is present in the signal.

The sampling theorem proved here uses samples taken at uniform intervals. This condition is not necessary. Samples can be taken arbitrarily at any instants as long as the sampling instants are recorded and there are, on average, 2B samples per second [2]. The essence of the sampling theorem was known to mathematicians for a long time in the form of the interpolation formula [see later, Eq. (8.6)]. The origin of the sampling theorem was attributed by H. S. Black to Cauchy in 1841. The essential idea of the sampling theorem was rediscovered in the 1920s by Carson, Nyquist, and Hartley.

EXAMPLE 8.1 Sampling at, Below, and Above the Nyquist Rate

In this example, we examine the effects of sampling a signal at the Nyquist rate, below the Nyquist rate (undersampling), and above the Nyquist rate (oversampling). Consider a signal ^x(t) ⁼ sinc² (5πt) (Fig. 8.2a) whose spectrum is ^X(ω) ⁼ 0.2(ω/20π ) (Fig. 8.2b). The bandwidth of this signal is 5 Hz (10π rad/s). Consequently, the Nyquist rate is 10 Hz; that is, we must sample the signal at a rate no less than 10 samples/s. The Nyquist interval is T = 1/2B = 0.1 second.

Recall that the sampled signal spectrum consists of (1/T)X(ω) = (0.2/T)(ω/20π ) repeating periodically with a period equal to the sampling frequency fs Hz. For the three sampling rates fs = 5 Hz (undersampling), 10 Hz (Nyquist rate), and 20 Hz (oversampling), we see that

fs (Hz)	T = 1 (s) fs	TX(ω) 1	Comments
5	0.2	ω 20π	Undersampling
10	0.1	2 ω 20π	Nyquist rate
20	0.05	4 ω 20π	Oversampling

In the first case (undersampling), the sampling rate is 5 Hz (5 samples/s), and the spectrum (1/T)X(ω) repeats every 5 Hz (10π rad/s). The successive spectra overlap, as depicted in Fig. 8.2d, and the spectrum X(ω) are not recoverable from X(ω); that is, x(t) cannot be reconstructed from its samples x(t) in Fig. 8.2c. In the second case, we use the Nyquist sampling rate of 10 Hz (Fig. 8.2e). The spectrum X(ω) consists of back-to-back, nonoverlapping repetitions of (1/T)X(ω) repeating every 10 Hz. Hence, X(ω) can be recovered

Figure 8.2 Effects of undersampling and oversampling.

from X(ω) using an ideal lowpass filter of bandwidth 5 Hz (Fig. 8.2f). Finally, in the last case of oversampling (sampling rate 20 Hz), the spectrum X(ω) consists of nonoverlapping repetitions of (1/T)X(ω) (repeating every 20 Hz) with empty bands between successive cycles (Fig. 8.2h). Hence, X(ω) can be recovered from X(ω) by using an ideal lowpass filter or even a practical lowpass filter (shown dashed in Fig. 8.2h).†

DR ILL 8.1 Nyquist Sampling

Find the Nyquist rate and the Nyquist sampling interval for the signals sinc(100πt) and sinc(100πt)+sinc(50πt).

ANSWERS

The Nyquist sampling interval is 0.01 s and the Nyquist sampling rate is 100 Hz for both signals.

FOR SKEPTICS ONLY

Rare is the reader who, at first encounter, is not skeptical of the sampling theorem. It seems impossible that Nyquist samples can define the one and the only signal that passes through those sample values. We can easily picture infinite number of signals passing through a given set of samples. However, among all these (infinite number of) signals, only one has the minimum bandwidth B ≤ 1/2T Hz, where T is the sampling interval. See Prob. 8.2-15.

To summarize, for a given set of samples taken at a rate fs Hz, there is only one signal of bandwidth B ≤ fs/2 that passes through those samples. All other signals that pass through those samples have bandwidth higher than fs/2, and the samples are sub-Nyquist rate samples for those signals.

8.1-1 Practical Sampling

In proving the sampling theorem, we assumed ideal samples obtained by multiplying a signal x(t) by an impulse train that is physically unrealizable. In practice, we multiply a signal x(t) by a train of pulses of finite width, depicted in Fig. 8.3c. The sampler is shown in Fig. 8.3d. The sampled signal x(t) is illustrated in Fig. 8.3e. We wonder whether it is possible to recover or reconstruct x(t) from this x(t). Surprisingly, the answer is affirmative, provided the sampling rate is not below the Nyquist rate. The signal x(t) can be recovered by lowpass filtering x(t) as if it were sampled by impulse train.

^† The filter should have a constant gain between 0 and 5 Hz and zero gain beyond 10 Hz. In practice, the gain beyond 10 Hz can be made negligibly small, but not zero.

Figure 8.3 Effect of practical sampling.

The plausibility of this result becomes apparent when we consider the fact that reconstruction of x(t) requires the knowledge of the Nyquist sample values. This information is available or built into the sampled signal x(t) in Fig. 8.3e because the nth sampled pulse strength is x(nT). To prove the result analytically, we observe that the sampling pulse train pT (t) depicted in Fig. 8.3c, being a periodic signal, can be expressed as a trigonometric Fourier series

$$p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n) \qquad \omega_s = \frac{2\pi}{T}$$

Thus,

$$\overline{x}(t) = x(t)p_T(t) = x(t)\left[C_0 + \sum_{n=1}^{\infty} C_n \cos(n\omega_s t + \theta_n)\right]$$ $$= C_0 x(t) + \sum_{n=1}^{\infty} C_n x(t) \cos(n\omega_s t + \theta_n)$$

The sampled signal x(t) consists of C0x(t), C1x(t) cos(ωst + θ1), C2x(t) cos(2ωst + θ2), … . Note that the first term C0x(t) is the desired signal and all the other terms are modulated signals with spectra centered at ±ωs,±2ωs,±3ωs,…, as illustrated in Fig. 8.3f. Clearly the signal x(t) can be recovered by lowpass filtering of x(t), as shown in Fig. 8.3d. As before, it is necessary that ω*^s* > 4πB (or fs > 2B).

EXAMPLE 8.2 Practical Sampling

Demonstrate practical sampling by sampling signal ^x(t) ⁼ sinc2 (5πt) with the rectangular pulse sequence pT (t) illustrated in Fig. 8.4c. Sketch the original and sampled signals and their spectra, and discuss recovery of x(t) from its samples.

Figure 8.4 An example of practical sampling.

784 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE

The signal x(t) and its spectrum are shown in Figs. 8.4a and 8.4b, respectively.

The period of pT (t) is 0.1 second so that the fundamental frequency (which is the sampling frequency) is 10 Hz. Hence, ω*^s* = 20π. The Fourier series for pT (t) can be expressed as

$$p_T(t) = C_0 + \sum_{n=1}^{\infty} C_n \cos n\omega_s t$$

Hence,

$$\overline{x}(t) = x(t)p_T(t)$$

= C₁x(t) + C₁x(t) cos 20 $\pi$ t + C₂x(t) cos 40 $\pi$ t + C₃x(t) cos 60 $\pi$ t + …

Use of Eq. (6.14) yields ^{C0 = 1 4} and Cn ^{= 2 nπ} sin ^nπ 4 . Consequently, we have

$$\overline{x}(t) = x(t)p_T(t)$$

= $\frac{1}{4}x(t) + C_1x(t) \cos 20\pi t + C_2x(t) \cos 40\pi t + C_3x(t) \cos 60\pi t + \cdots$

and

$$\overline{X}(\omega) = \frac{1}{4}X(\omega) + \frac{C_1}{2}[X(\omega - 20\pi) + X(\omega + 20\pi)] + \frac{C_2}{2}[X(\omega - 40\pi) + X(\omega + 40\pi)] + \frac{C_3}{2}[X(\omega - 60\pi) + X(\omega + 60\pi)] + \cdots$$

where Cn = (2/nπ )sin(nπ/4). The sampled signal and its spectrum are shown in Figs. 8.4d and 8.4e, respectively.

The spectrum X(ω) consists of X(ω) repeating periodically at the interval of 20π rad/s (10 Hz). Hence, there is no overlap between cycles, and X(ω) can be recovered by using an ideal lowpass filter of bandwidth 5 Hz. An ideal lowpass filter of unit gain (and bandwidth 5 Hz) will allow the first term on the right-hand side of the foregoing equation to pass fully and suppress all the other terms. Hence, the output y(t) is

^y(t) ^{= 1 4} x(t)

DR ILL 8.2 The Role of Sampling Pulse Area

Show that the basic pulse p(t) used in the sampling pulse train in Fig. 8.4c cannot have zero area if we wish to reconstruct x(t) by lowpass-filtering the sampled signal.