8.2 SIGNAL [RECONSTRUCTION](#page-13-0)

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8.2 SIGNAL RECONSTRUCTION

The process of reconstructing a continuous-time signal x(t) from its samples is also known as interpolation. In Sec. 8.1, we saw that a signal x(t) bandlimited to B Hz can be reconstructed (interpolated) exactly from its samples if the sampling frequency fs exceeds 2B Hz or the sampling interval T is less than 1/2B. This reconstruction is accomplished by passing the sampled signal through an ideal lowpass filter of gain T and having a bandwidth of any value between B and fs −B Hz. From a practical viewpoint, a good choice is the middle value fs/2 = 1/2T Hz or π/T rad/s. This value allows for small deviations in the ideal filter characteristics on either side of the cutoff frequency. With this choice of cutoff frequency and gain T, the ideal lowpass filter required for signal reconstruction (or interpolation) is

$$H(\omega) = T \operatorname{rect}\left(\frac{\omega}{2\pi f_s}\right) = T \operatorname{rect}\left(\frac{\omega T}{2\pi}\right)$$

\n(8.4)

The interpolation process here is expressed in the frequency domain as a filtering operation. Now we shall examine this process from the time-domain viewpoint.

TIME-DOMAIN VIEW:ASIMPLE INTERPOLATION

Consider the interpolation system shown in Fig. 8.5a. We start with a very simple interpolating filter, whose impulse response is rect(t/T), depicted in Fig. 8.5b. This is a gate pulse centered at the origin, having unit height, and width T (the sampling interval). We shall find the output of this filter when the input is the sampled signal x(t) consisting of an impulse train with the nth impulse at t = nT with strength x(nT). Each sample in x(t), being an impulse, produces at the output a gate pulse of height equal to the strength of the sample. For instance, the nth sample is an impulse of strength x(nT) located at t = nT and can be expressed as x(nT)δ(t−nT). When this impulse passes through the filter, it produces at the output a gate pulse of height x(nT), centered at t = nT (shaded in Fig. 8.5c). Each sample in x(t) will generate a corresponding gate pulse, resulting in the filter output that is a staircase approximation of x(t), shown dotted in Fig. 8.5c. This filter thus gives a crude form of interpolation.

The frequency response of this filter H(ω) is the Fourier transform of the impulse response rect(t/T). Thus,

$$h(t) = \text{rect}\left(\frac{t}{T}\right) \qquad \text{and} \qquad H(\omega) = T \text{sinc}\left(\frac{\omega T}{2}\right) \tag{8.5}$$

The amplitude response |H(ω)| for this filter, illustrated in Fig. 8.5d, explains the reason for the crudeness of this interpolation. This filter, also known as the zero-order hold (ZOH) filter, is a poor form of the ideal lowpass filter (shaded in Fig. 8.5d) required for exact interpolation.†

We can improve on the ZOH filter by using a first-order hold filter, which results in a linear interpolation instead of a staircase interpolation. A linear interpolator, whose impulse response is a triangle pulse (t/2T), results in an interpolation in which successive sample tops are connected by straight-line segments (see Prob. 8.2-3).

^† Figure 8.5b shows that the impulse response of this filter is noncausal, and this filter is not realizable. In practice, we make it realizable by delaying the impulse response by T/2. This merely delays the output of the filter by T/2.

Figure 8.5 Simple interpolation by means of a zero-order hold (ZOH) circuit. (a) ZOH interpolator. (b) Impulse response of a ZOH circuit. (c) Signal reconstruction by ZOH, as viewed in the time domain. (d) Frequency response of a ZOH.

TIME-DOMAIN VIEW: AN IDEAL INTERPOLATION

The ideal interpolation filter frequency response obtained in Eq. (8.4) is illustrated in Fig. 8.6a. The impulse response of this filter, the inverse Fourier transform of H(ω) is

$$h(t) = \text{sinc}\left(\frac{\pi t}{T}\right)$$

For the Nyquist sampling rate, T = 1/2B, and

$$h(t) = \text{sinc}\left(2\pi Bt\right)$$

This h(t) is depicted in Fig. 8.6b. Observe the interesting fact that h(t) = 0 at all Nyquist sampling instants (t = ±n/2B) except at t = 0. When the sampled signal x¯(t) is applied at the input of this filter, the output is x(t). Each sample in x(t), being an impulse, generates a sinc pulse of height equal to the strength of the sample, as illustrated in Fig. 8.6c. The process is identical to that depicted in Fig. 8.5c, except that h(t) is a sinc pulse instead of a gate pulse. Addition of the sinc

Figure 8.6 Ideal interpolation for Nyquist sampling rate.

pulses generated by all the samples results in x(t). The nth sample of the input x(t) is the impulse x(nT)δ(t −nT); the filter output of this impulse is x(nT)h(t −nT). Hence, the filter output to x(t), which is x(t), can now be expressed as a sum

$$x(t) = \sum_{n} x(nT)h(t - nT) = \sum_{n} x(nT)\operatorname{sinc}\left[\frac{\pi}{T}(t - nT)\right]$$

For the case of Nyquist sampling rate, T = 1/2B, this expression simplifies to

$$x(t) = \sum_{n} x(nT) \operatorname{sinc}(2\pi Bt - n\pi)$$

\n(8.6)

Equation (8.6) is the interpolation formula, which yields values of x(t) between samples as a weighted sum of all the sample values.

EXAMPLE 8.3 Bandlimited Interpolation of the Kronecker Delta Function

Find a signal x(t) that is bandlimited to B Hz, and whose samples are

$$x(0) = 1$$

and $x(\pm T) = x(\pm 2T) = x(\pm 3T) = \cdots = 0$

788 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE

where the sampling interval T is the Nyquist interval for x(t), that is, T = 1/2B.

Because we are given the Nyquist sample values, we use the interpolation formula of Eq. (8.6) to construct x(t) 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. (8.6) survives. Thus,

$$x(t) = \operatorname{sinc}\left(2\pi Bt\right)$$

This signal is illustrated in Fig. 8.6b. Observe that this is the only signal that has a bandwidth B Hz and the sample values x(0) = 1 and x(nT) = 0(n = 0). No other signal satisfies these conditions.

8.2-1 Practical Difficulties in Signal Reconstruction

Consider the signal reconstruction procedure illustrated in Fig. 8.7a. If x(t) is sampled at the Nyquist rate fs = 2B Hz, the spectrum X(ω) consists of repetitions of X(ω) without any gap between successive cycles, as depicted in Fig. 8.7b. To recover x(t) from x(t), we need to pass the sampled signal x(t) through an ideal lowpass filter, shown dotted in Fig. 8.7b. As seen in Sec. 7.5, such a filter is unrealizable; it can be closely approximated only with infinite time delay in the response. In other words, we can recover the signal x(t) from its samples with infinite time delay. A practical solution to this problem is to sample the signal at a rate higher than the Nyquist rate (fs > 2B or ω*^s* > 4πB). The result is X(ω), consisting of repetitions of X(ω) with a finite bandgap between successive cycles, as illustrated in Fig. 8.7c. Now, we can recover X(ω) from X(ω) using a lowpass filter with a gradual cutoff characteristic, shown dotted in Fig. 8.7c. But even in this case, if the unwanted spectrum is to be suppressed, the filter gain must be zero beyond some frequency (see Fig. 8.7c). According to the Paley–Wiener criterion [Eq. (7.43)], it is impossible to realize even this filter. The only advantage in this case is that the required filter can be closely approximated with a smaller time delay. All this means that it is impossible in practice to recover a bandlimited signal x(t) exactly from its samples, even if the sampling rate is higher than the Nyquist rate. However, as the sampling rate increases, the recovered signal approaches the desired signal more closely.

THE TREACHERY OF ALIASING

There is another fundamental practical difficulty in reconstructing a signal from its samples. The sampling theorem was proved on the assumption that the signal x(t) is bandlimited. All practical signals are timelimited; that is, they are of finite duration or width. We can demonstrate (see Prob. 8.2-20) that a signal cannot be timelimited and bandlimited simultaneously. If a signal is timelimited, it cannot be bandlimited, and vice versa (but it can be simultaneously nontimelimited and nonbandlimited). Clearly, all practical signals, which are necessarily timelimited, are nonbandlimited, as shown in Fig. 8.8a; they have infinite bandwidth, and the spectrum X(ω) consists of overlapping cycles of X(ω) repeating every fs Hz (the sampling frequency), as

Figure 8.7 (a) Signal reconstruction from its samples. (b) Spectrum of a signal sampled at the Nyquist rate. (c) Spectrum of a signal sampled above the Nyquist rate.

illustrated in Fig. 8.8b.† Because of infinite bandwidth in this case, the spectral overlap is unavoidable, regardless of the sampling rate. Sampling at a higher rate reduces but does not eliminate overlapping between repeating spectral cycles. Because of the overlapping tails, X(ω) no longer has complete information about X(ω), and it is no longer possible, even theoretically, to recover x(t) exactly from the sampled signal x(t). If the sampled signal is passed through an ideal lowpass filter of cutoff frequency fs/2 Hz, the output is not X(ω) but Xa(ω) (Fig. 8.8c), which is a version of X(ω) distorted as a result of two separate causes:

• 1. The loss of the tail of X(ω) beyond |f | > fs/2 Hz.
• 2. The reappearance of this tail inverted or folded onto the spectrum. Note that the spectra cross at frequency fs/2 = 1/2T Hz. This frequency is called the folding frequency.

^† Figure 8.8b shows that from the infinite number of repeating cycles, only the neighboring spectral cycles overlap. This is a somewhat simplified picture. In reality, all the cycles overlap and interact with every other cycle because of the infinite width of all practical signal spectra. Fortunately, all practical spectra also must decay at higher frequencies. This results in insignificant amount of interference from cycles other than the immediate neighbors. When such an assumption is not justified, aliasing computations become little more involved.

Figure 8.8 Aliasing effect. (a) Spectrum of a practical signal x(t). (b) Spectrum of sampled x(t). (c) Reconstructed signal spectrum. (d) Sampling scheme using anti-aliasing filter. (e) Sampled signal spectrum (dotted) and the reconstructed signal spectrum (solid) when anti-aliasing filter is used.

The spectrum may be viewed as if the lost tail is folding back onto itself at the folding frequency. For instance, a component of frequency (fs/2) + fz shows up as or “impersonates” a component of lower frequency (fs/2) − fz in the reconstructed signal. Thus, the components of frequencies above fs/2 reappear as components of frequencies below fs/2. This tail inversion, known as spectral folding or aliasing, is shown shaded in Fig. 8.8b and also in Fig. 8.8c. In the process of aliasing, not only are we losing all the components of frequencies above the folding frequency fs/2 Hz, but these very components reappear (aliased) as lower-frequency components, as shown in Figs. 8.8b and 8.8c. Such aliasing destroys the integrity of the frequency components below the folding frequency fs/2, as depicted in Fig. 8.8c.

The aliasing problem is analogous to that of an army with a platoon that has secretly defected to the enemy side. The platoon is, however, ostensibly loyal to the army. The army is in double jeopardy. First, the army has lost this platoon as a fighting force. In addition, during actual fighting, the army will have to contend with sabotage by the defectors and will have to find another loyal platoon to neutralize the defectors. Thus, the army has lost two platoons in nonproductive activity.

DEFECTORS ELIMINATED: THE ANTI-ALIASING FILTER

If you were the commander of the betrayed army, the solution to the problem would be obvious. As soon as the commander got wind of the defection, he would incapacitate, by whatever means, the defecting platoon before the fighting begins. This way he loses only one (the defecting) platoon. This is a partial solution to the double jeopardy of betrayal and sabotage, a solution that partly rectifies the problem and cuts the losses to half.

We follow exactly the same procedure. The potential defectors are all the frequency components beyond the folding frequency fs/2 = 1/2T Hz. We should eliminate (suppress) these components from x(t) before sampling x(t). Such suppression of higher frequencies can be accomplished by an ideal lowpass filter of cutoff fs/2 Hz, as shown in Fig. 8.8d. This is called the anti-aliasing filter. Figure 8.8d also shows that anti-aliasing filtering is performed before sampling. Figure 8.8e shows the sampled signal spectrum (dotted) and the reconstructed signal Xaa(ω) when an anti-aliasing scheme is used. An anti-aliasing filter essentially bandlimits the signal x(t) to fs/2 Hz. This way, we lose only the components beyond the folding frequency fs/2 Hz. These suppressed components now cannot reappear to corrupt the components of frequencies below the folding frequency. Clearly, use of an anti-aliasing filter results in the reconstructed signal spectrum Xaa(ω) = X(ω) for |f | < fs/2. Thus, although we lost the spectrum beyond fs/2 Hz, the spectrum for all the frequencies below fs/2 remains intact. The effective aliasing distortion is cut in half owing to elimination of folding. We stress again that the anti-aliasing operation must be performed before the signal is sampled.

An anti-aliasing filter also helps to reduce noise. Noise, generally, has a wideband spectrum, and without anti-aliasing, the aliasing phenomenon itself will cause the noise lying outside the desired band to appear in the signal band. Anti-aliasing suppresses the entire noise spectrum beyond frequency fs/2.

The anti-aliasing filter, being an ideal filter, is unrealizable. In practice, we use a steep cutoff filter, which leaves a sharply attenuated spectrum beyond the folding frequency fs/2.

SAMPLING FORCES NONBANDLIMITED SIGNALS TO APPEAR BANDLIMITED

Figure 8.8b shows that the spectrum of a signal x(t) consists of overlapping cycles of X(ω). This means that x(t) are sub-Nyquist samples of x(t). However, we may also view the spectrum in Fig. 8.8b as the spectrum Xa(ω) (Fig. 8.8c), repeating periodically every fs Hz without overlap. The spectrum Xa(ω) is bandlimited to fs/2 Hz. Hence, these (sub-Nyquist) samples of x(t) are actually the Nyquist samples for signal xa(t). In conclusion, sampling a nonbandlimited signal x(t) at a rate fs Hz makes the samples appear to be the Nyquist samples of some signal xa(t), bandlimited to fs/2 Hz. In other words, sampling makes a nonbandlimited signal appear to be a bandlimited signal xa(t) with bandwidth fs/2 Hz. A similar conclusion applies if x(t) is bandlimited but sampled at a sub-Nyquist rate.

VERIFICATION OF ALIASING IN SINUSOIDS

We showed in Fig. 8.8b how sampling a signal below the Nyquist rate causes aliasing, which makes a signal of higher frequency (fs/2) + fz Hz masquerade as a signal of lower frequency (fs/2) − fz Hz. Figure 8.8b demonstrates this result in the frequency domain. Let us now verify it in the time domain to gain a deeper appreciation of aliasing.

We can prove our proposition by showing that samples of sinusoids of frequencies (ωs/2)+ω*^z* and (ωs/2)−ω*^z* are identical when the sampling frequency is fs = ωs/2π Hz.

For a sinusoid x(t) = cos ωt, sampled at intervals of T seconds, x(nT), its nth sample (at t = nT) is

x(nT) = cosωnT n integer

Hence, samples of sinusoids of frequency ω = (ωs/2)±ω*^z* are†

$$x(nT) = \cos\left(\frac{\omega_s}{2} \pm \omega_z\right) nT = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT \mp \sin\left(\frac{\omega_s}{2}\right) nT \sin\omega_z nT$$

Recognizing that ωsT = 2πfsT = 2π, and sin(ωs/2)nT = sinπn = 0 for all integer n, we obtain

$$x(nT) = \cos\left(\frac{\omega_s}{2}\right) nT \cos\omega_z nT$$

Clearly, the samples of a sinusoid of frequency (fs/2)+fz are identical to the samples of a sinusoid (fs/2)−fz. ‡ For instance, when a sinusoid of frequency 100 Hz is sampled at a rate of 120 Hz, the apparent frequency of the sinusoid that results from reconstruction of the samples is 20 Hz. This follows from the fact that here, 100 = (fs/2)+fz = 60+fz so that fz = 40. Hence, (fs/2)−fz = 20. Such would precisely be the conclusion arrived at from Fig. 8.8b.

^† Here we have ignored the phase aspect of the sinusoid. Sampled versions of a sinusoid ^x(t) ⁼ cos(ω*^t* ⁺ θ ) with two different frequencies (ωs/2) ± ω*^z* have identical frequency, but the phase signs may be reversed

depending on the value of ^ωz. ‡ The reader is encouraged to verify this result graphically by plotting the spectrum of a sinusoid of frequency (ωs/2) + ω*^z* (impulses at ±[(ωs/2) + ωz]) and its periodic repetition at intervals ωs. Although the result is valid for all values of ωz, consider the case of ω*^z* < ωs/2 to simplify the graphics.

This discussion again shows that sampling a sinusoid of frequency f aliasing can be avoided if the sampling rate fs > 2f Hz.

$$0 \le f < \frac{f_s}{2} \qquad \text{or} \qquad 0 \le \omega < \frac{\pi}{T}$$

Violating this condition leads to aliasing, implying that the samples appear to be those of a lower-frequency signal. Because of this loss of identity, it is impossible to reconstruct the signal faithfully from its samples.

GENERAL CONDITION FOR ALIASING IN SINUSOIDS

We can generalize the foregoing result by showing that samples of a sinusoid of frequency f⁰ are identical to those of a sinusoid of frequency f⁰ + mfs Hz (integer m), where fs is the sampling frequency. The samples of cos 2π(f⁰ +mfs)t are

$$\cos 2\pi (f_0 + m f_s) nT = \cos (2\pi f_0 nT + 2\pi mn) = \cos 2\pi f_0 nT$$

The result follows because mn is an integer and fsT = 1. This result shows that sinusoids of frequencies that differ by an integer multiple of fs result in identical set of samples. In other words, samples of sinusoids separated by frequency fs Hz are identical. This implies that samples of sinusoids in any frequency band of fs Hz are unique; that is, no two sinusoids in that band have the same samples (when sampled at a rate fs Hz). For instance, frequencies in the band from −fs/2 to fs/2 have unique samples (at the sampling rate fs). This band is called the fundamental band. Recall also that fs/2 is the folding frequency.

From the discussion thus far, we conclude that if a continuous-time sinusoid of frequency f Hz is sampled at a rate of fs Hz (samples/s), the resulting samples would appear as samples of a continuous-time sinusoid of frequency fa in the fundamental band, where

$$f_a = f - mf_s \qquad -\frac{f_s}{2} \le f_a < \frac{f_s}{2} \qquad m \text{ an integer} \tag{8.7}$$

The frequency fa lies in the fundamental band from −fs/2 to fs/2. Figure 8.9a shows the plot of fa versus f , where f is the actual frequency and fa is the corresponding fundamental band frequency, whose samples are identical to those of the sinusoid of frequency f , when the sampling rate is fs Hz.

Recall, however, that the sign change of a frequency does not alter the actual frequency of the waveform. This is because

$$\cos(-\omega_a t + \theta) = \cos(\omega_a t - \theta)$$

Clearly the apparent frequency of a sinusoid of frequency −fa is also fa. However, its phase undergoes a sign change. This means the apparent frequency of any sampled sinusoid lies in the range from 0 to fs/2 Hz. To summarize, if a continuous-time sinusoid of frequency f Hz is sampled at a rate of fs Hz (samples/second), the resulting samples would appear as samples of a continuous-time sinusoid of frequency |fa| that lies in the band from 0 to fs/2. According to Eq. (8.7),

$$|f_a| = |f - mf_s| \qquad |f_a| \le \frac{f_s}{2} \qquad m \text{ an integer}$$

Figure 8.9 Apparent frequencies of a sampled sinusoid: (a) fa versus f and (b) |fa| versus f .

The plot of the apparent frequency |fa| versus f is shown in Fig. 8.9b.† As expected, the apparent frequency |fa| of any sampled sinusoid, regardless of its frequency, is always in the range of 0 to fs/2 Hz. However, when fa is negative, the phase of the apparent sinusoid undergoes a sign change. The frequency belts in which such phase changes occur are shown shaded in Fig. 8.9b.

Consider, for example, a sinusoid cos(2πft +θ ) with f =8000 Hz sampled at a rate fs = 3000 Hz. Using Eq. (8.7), we obtain fa = 8000 − 3 × 3000 = −1000. Hence, |fa| = 1000. The samples would appear to have come from a sinusoid cos(2000πt − θ ). Observe the sign change of the phase because fa is negative.‡

In the light of the foregoing development, let us consider a sinusoid of frequency f = (fs/2)+fz, sampled at a rate of fs Hz. According to Eq. (8.7),

$$f_a = \frac{f_s}{2} + f_z - (1 \times f_s) = -\frac{f_s}{2} + f_z$$

Hence, the apparent frequency is |fa| = (fs/2) − fz, confirming our earlier result. However, the phase of the sinusoid will suffer a sign change because fa is negative.

Figure 8.10 shows how samples of sinusoids of two different frequencies (sampled at the same rate) generate identical sets of samples. Both the sinusoids are sampled at a rate fs = 5 Hz (T = 0.2 second). The frequencies of the two sinusoids, 1 Hz (period 1) and 6 Hz (period 1/6), differ by fs = 5 Hz.

^† The plots in Figs. 8.9 and 5.17 are identical. This is because a sampled sinusoid is basically a discrete-time sinusoid.

^‡ For phase sign change, we are assuming that the signal has the form cos(2πft ⁺ θ ). If the form is sin(2πft + θ ), the rule changes slightly. It is left as an exercise for the reader to show that when fa < 0, this sinusoid appears as −sin(2π|fa|t − θ ). Thus, in addition to phase change, the amplitude also changes sign.

Figure 8.10 Demonstration of aliasing.

The reason for aliasing can be clearly seen in Fig. 8.10. The root of the problem is the sampling rate, which may be adequate for the lower-frequency sinusoid but is clearly inadequate for the higher-frequency sinusoid. The figure clearly shows that between the successive samples of the higher-frequency sinusoid, there are wiggles, which are bypassed or ignored, and are unrepresented in the samples, indicating a sub-Nyquist rate of sampling. The frequency of the apparent signal xa(t) is always the lowest possible frequency that lies within the band |f | ≤ fs/2. Thus, the apparent frequency of the samples in this example is 1 Hz. If these samples are chosen to reconstruct a signal using a lowpass filter of bandwidth fs/2, we shall obtain a sinusoid of frequency 1 Hz.

EXAMPLE 8.4 Apparent Frequency of Sampled Sinusoids

A continuous-time sinusoid cos(2πft + θ ) is sampled at a rate fs = 1000 Hz. Determine the apparent (aliased) sinusoid of the resulting samples if the input signal frequency f is (a) 400 Hz, (b) 600 Hz, (c) 1000 Hz, and (d) 2400 Hz.

The folding frequency is fs/2 = 500. Hence, sinusoids below 500 Hz (frequency within the fundamental band) will not be aliased and sinusoids of frequency above 500 Hz will be aliased.

(c) Since f = 1000 Hz can be expressed as 1000 = 0 + 1000, we see that fa = 0. Hence, the aliased frequency is 0 Hz (dc), and there is no phase sign change. The apparent sinusoid is y(t) = cos(0πt ±θ ) = cos(θ ). This is a dc signal with constant sample values for all n.

⁽a) Since f = 400 Hz is less than 500 Hz, there is no aliasing. The apparent sinusoid is cos(2πft +θ ) with f = 400.

⁽b) Since f = 600 Hz can be expressed as 600 = −400 + 1000, we see that fa = −400. Hence, the aliased frequency is 400 Hz and the phase changes sign. The apparent (aliased) sinusoid is cos(2πft −θ ) with f = 400.

796 CHAPTER 8 SAMPLING: THE BRIDGE FROM CONTINUOUS TO DISCRETE

(d) Here, f = 2400 Hz can be expressed as 2400 = 400 + (2 × 1000) so that fa = 400. Hence, the aliased frequency is 400 Hz and there is no sign change for the phase. The apparent sinusoid is cos(2πft +θ ) with f = 400.

We could have found these answers directly from Fig. 8.9b. For example, for case (b), we read |fa| = 400 corresponding to f = 600. Moreover, f = 600 lies in the shaded belt. Hence, there is a phase sign change.

DR ILL 8.3 A Case of Identical Sampled Sinusoids

Show that samples of 90 Hz and 110 Hz sinusoids of the form cosωt are identical when sampled at a rate 200 Hz.

DR ILL 8.4 Apparent Frequency of Sampled Sinusoids

A sinusoid of frequency f⁰ Hz is sampled at a rate of 100 Hz. Determine the apparent frequency of the samples if f⁰ is (a) 40 Hz, (b) 60 Hz, (c) 140 Hz, and (d) 160 Hz.

ANSWERS

All four cases have an apparent frequency of 40 Hz.

8.2-2 Some Applications of the Sampling Theorem

The sampling theorem is very important in signal analysis, processing, and transmission because it allows us to replace a continuous-time signal with a discrete sequence of numbers. Processing a continuous-time signal is therefore equivalent to processing a discrete sequence of numbers. Such processing leads us directly into the area of digital filtering. In the field of communication, the transmission of a continuous-time message reduces to the transmission of a sequence of numbers by means of pulse trains. The continuous-time signal x(t) is sampled, and sample values are used to modify certain parameters of a periodic pulse train. We may vary the amplitudes (Fig. 8.11b), widths (Fig. 8.11c), or positions (Fig. 8.11d) of the pulses in proportion to the sample values of the signal x(t). Accordingly, we may have pulse-amplitude modulation (PAM), pulse-width modulation (PWM), or pulse-position modulation (PPM). The most important form of pulse modulation today is pulse-code modulation (PCM), discussed in Sec. 8.3 in connection with Fig. 8.14b. In all these cases, instead of transmitting x(t), we transmit the corresponding pulse-modulated signal. At the receiver, we read the information of the pulse-modulated signal and reconstruct the analog signal x(t).

Figure 8.11 Pulse-modulated signals. (a) The signal. (b) The PAM signal. (c) The PWM (PDM) signal. (d) The PAM signal.

One advantage of using pulse modulation is that it permits the simultaneous transmission of several signals on a time-sharing basis—time-division multiplexing (TDM). Because a pulse-modulated signal occupies only a part of the channel time, we can transmit several pulse-modulated signals on the same channel by interweaving them. Figure 8.12 shows the TDM of two PAM signals. In this manner, we can multiplex several signals on the same channel by reducing pulse widths.†

Digital signals also offer an advantage in the area of communications, where signals must travel over distances. Transmission of digital signals is more rugged than that of analog signals because digital signals can withstand channel noise and distortion much better as long as the noise

^† Another method of transmitting several baseband signals simultaneously is frequency-division multiplexing (FDM) discussed in Sec. 7.7-4. In FDM, various signals are multiplexed by sharing the channel bandwidth. The spectrum of each message is shifted to a specific band not occupied by any other signal. The information of various signals is located in nonoverlapping frequency bands of the channel (Fig. 7.45). In a way, TDM and FDM are duals of each other.

Figure 8.12 Time-division multiplexing of two signals.

Figure 8.13 Digital signal transmission: (a) at the transmitter, (b) received distorted signal (without noise), (c) received distorted signal (with noise), and (d) regenerated signal at the receiver.

and the distortion are within limits. An analog signal can be converted to digital binary form through sampling and quantization (rounding off), as explained in the next section. The digital (binary) message in Fig. 8.13a is distorted by the channel, as illustrated in Fig. 8.13b. Yet if the distortion remains within a limit, we can recover the data without error because we need only make a simple binary decision: Is the received pulse positive or negative? Figure 8.13c shows the same data with channel distortion and noise. Here again, the data can be recovered correctly as long as the distortion and the noise are within limits. Such is not the case with analog messages. Any distortion or noise, no matter how small, will distort the received signal.

The greatest advantage of digital communication over the analog counterpart, however, is the viability of regenerative repeaters in the former. In an analog transmission system, a message signal grows progressively weaker as it travels along the channel (transmission path), whereas the channel noise and the signal distortion, being cumulative, become progressively stronger. Ultimately, the signal, overwhelmed by noise and distortion, is mutilated. Amplification is of little help because it enhances the signal and the noise in the same proportion. Consequently, the distance over which an analog message can be transmitted is limited by the transmitted power. If a transmission path is long enough, the channel distortion and noise will accumulate sufficiently to overwhelm even a digital signal. The trick is to set up repeaters along the transmission path at distances short enough to permit detection of signal pulses before the noise and distortion have a chance to accumulate sufficiently. At each repeater, the pulses are detected, and new, clean pulses are transmitted to the next repeater, which, in turn, duplicates the same process. If the noise and distortion remain within limits (which is possible because of the closely spaced repeaters), pulses can be detected correctly.† This way the digital messages can be transmitted over longer distances with greater reliability. In contrast, analog messages cannot be cleaned up periodically, and their transmission is therefore less reliable. The most significant error in digitized signals comes from quantizing (rounding off). This error, discussed in Sec. 8.3, can be reduced as much as desired by increasing the number of quantization levels, at the cost of an increased bandwidth of the transmission medium (channel).