The central concept in the digital processing of analog signals is that the sampled signal must be a unique representation of the analog signal. For a sinusoid x(t) = cos(2πf0t + θ), a sampling rate S > 2f0 ensures that the digital frequency of the sampled signal falls in the principal period −0.5 ≤ F ≤ 0.5 and permits a unique correspondence with the underlying analog signal. More generally, the sampling theorem says that for a unique correspondence between an analog signal and the version reconstructed from its samples (using the same sampling rate), the sampling rate must exceed twice the highest signal frequency fmax. The critical rate S = 2fmax is called the Nyquist rate or Nyquist frequency, and ts= 2fmax1 is called the Nyquist interval. For the sinusoid x(t) = cos(2πf0t + θ), the Nyquist rate is SN = 2f0 = T2 and corresponds to taking two samples per period (because the sampling interval is ts= T2).
REVIEW PANEL 3.16
The Sampling Theorem: How to Sample an Analog Signal Without Loss of Information For an analog signal band-limited to fmax Hz, the sampling rate S must exceed 2fmax.
S = 2fmax defines the Nyquist rate. ts= 2fmax1 defines the Nyquist interval.
For an analog sinusoid: The Nyquist rate corresponds to taking two samples per period.
Consider an analog signal x(t) = cos(2πf0t + θ) and its sampled version x[n] = cos(2πnF0+ θ), where F0= f0/S. If x[n] is to be a unique representation of x(t), we must be able to reconstruct x(t) from x[n]. In practice, reconstruction uses only the copy or image of the periodic spectrum of x[n] in the principal period −0.5 ≤ F ≤ 0.5, which corresponds to the analog frequency range −0.5S ≤ f ≤ 0.5S. We use a lowpass filter to remove all other replicas or images, and the output of the lowpass filter corresponds to the reconstructed analog signal. As a result, the highest frequency fH we can identify in the signal reconstructed from its samples is fH = 0.5S.
REVIEW PANEL 3.17
The Highest Frequency Cannot Exceed Half the Sampling or Reconstruction Rate Upon sampling or reconstruction, all frequencies must be brought into the principal period.
Whether the frequency of the reconstructed analog signal matches x(t) or not depends on the sampling rate S. If S > 2f0, the digital frequency F0= f0/S is always in the principal range−0.5 ≤ F ≤ 0.5, and the reconstructed analog signal is identical to x(t). If S < 2f0, the digital frequency exceeds 0.5. Its image in the principal range appears at the lower digital frequency Fa= F0− M (corresponding to the lower analog frequency fa= f0−MS), where M is an integer that places the digital frequency Fabetween −0.5 and 0.5 (or the analog frequency fabetween −0.5S and 0.5S). The reconstructed analog signal xa(t) = cos(2πfat + θ) is at a lower frequency fa = SFa than f0 and is no longer a replica of x(t). This phenomenon, where a reconstructed sinusoid appears at a lower frequency than the original, is called aliasing. The real problem is that the original signal x(t) and the aliased signal xa(t) yield identical sampled representations at the sampling frequency S and prevent unique identification of x(t) from its samples!
REVIEW PANEL 3.18
Aliasing Occurs if the Analog Signal cos(2πf0t + θ) Is Sampled Below the Nyquist Rate If S < 2f0, the reconstructed analog signal is aliased to a lower frequency |fa| < 0.5S. We find fa as fa= f0− MS, where M is an integer that places fain the principal period (−0.5S < fa≤ 0.5S).
3.6 Aliasing and the Sampling Theorem 53
EXAMPLE 3.8 (Aliasing and Its Effects)
(a) A 100-Hz sinusoid x(t) is sampled at 240 Hz. Has aliasing occurred? How many full periods of x(t) are required to obtain one period of the sampled signal?
The sampling rate exceeds 200 Hz, so there is no aliasing. The digital frequency is F = 100
240 = 125.
Thus, five periods of x(t) yield 12 samples (one period) of the sampled signal.
(b) A 100-Hz sinusoid is sampled at rates of 240 Hz, 140 Hz, 90 Hz, and 35 Hz. In each case, has aliasing occurred, and if so, what is the aliased frequency?
To avoid aliasing, the sampling rate must exceed 200 Hz. If S = 240 Hz, there is no aliasing, and the reconstructed signal (from its samples) appears at the original frequency of 100 Hz. For all other choices of S, the sampling rate is too low and leads to aliasing. The aliased signal shows up at a lower frequency. The aliased frequencies corresponding to each sampling rate S are found by subtracting out multiples of S from 100 Hz to place the result in the range −0.5S ≤ f ≤ 0.5S. If the original signal has the form x(t) = cos(200πt + θ), we obtain the following aliased frequencies and aliased signals:
1. S = 140 Hz, fa= 100 − 140 = −40 Hz, xa(t) = cos(−80πt + θ) = cos(80πt − θ) 2. S = 90 Hz, fa= 100 − 90 = 10 Hz, xa(t) = cos(20πt + θ)
3. S = 35 Hz, fa= 100 − 3(35) = −5 Hz, xa(t) = cos(−10πt + θ) = cos(10πt − θ)
We thus obtain a 40-Hz sinusoid (with reversed phase), a 10-Hz sinusoid, and a 5-Hz sinusoid (with reversed phase), respectively. Notice that negative aliased frequencies simply lead to a phase reversal and do not represent any new information. Finally, had we used a sampling rate exceeding the Nyquist rate of 200 Hz, we would have recovered the original 100-Hz signal every time. Yes, it pays to play by the rules of the sampling theorem!
(c) Two analog sinusoids x1(t) (shown light) and x2(t) (shown dark) lead to an identical sampled version as illustrated in Figure E3.8C. Has aliasing occurred? Identify the original and aliased signal. Identify the digital frequency of the sampled signal corresponding to each sinusoid. What is the analog frequency of each sinusoid if S = 50 Hz? Can you provide exact expressions for each sinusoid?
0 0.05 0.1 0.15 0.2 0.25 0.3 −1 −0.5 0 0.5 1 Time t [seconds] Amplitude
Two analog signals and their sampled version
Figure E3.8C The sinusoids for Example 3.8(c)
Look at the interval (0, 0.1) s. The sampled signal shows five samples per period. This covers three full periods of x1(t) and so F1= 35. This also covers two full periods of x2(t), and so F2= 25. Clearly, x1(t) (with |F1| > 0.5) is the original signal that is aliased to x2(t). The sampling interval is 0.02 s.
54 Chapter 3 Discrete Signals So, the sampling rate is S = 50 Hz. The original and aliased frequencies are f1 = SF1= 30 Hz and f2= SF2= 20 Hz.
From the figure, we can identify exact expressions for x1(t) and x2(t) as follows. Since x1(t) is a delayed cosine with x1(0) = 0.5, we have x1(t) = cos(60πt −π3). With S = 50 Hz, the frequency f1= 30 Hz actually aliases to f2= −20 Hz, and thus x2(t) = cos(−40πt−π3) = cos(40πt +π3). With F =3050 = 0.6 (or F = −0.4), the expression for the sampled signal is x[n] = cos(2πnF − π
3).
(d) A 100-Hz sinusoid is sampled, and the reconstructed signal (from its samples) shows up at 10 Hz. What was the sampling rate S?
If you said 90 Hz (100 − S = 10), you are not wrong. But you could also have said 110 Hz (100 − S = −10). In fact, we can also subtract out integer multiples of S from 100 Hz, and S is then found from the following expressions (as long as we ensure that S > 20 Hz):
1. 100 − MS = 10 2. 100 − MS = −10
Solving the first expression for S, we find, for example, S = 45 Hz (with M = 2) or S = 30 Hz (with M = 3). Similarly, the second expression gives S = 55 Hz (with M = 2). Which of these sampling rates was actually used? We have no way of knowing!
3.6.1
Reconstruction Using a Different Sampling Rate
There are situations when we sample a signal using one sampling rate S1but reconstruct the analog signal from samples using a different sampling rate S2. In such situations, a frequency f0in the original signal will result in a recovered frequency of fr = f0(S2/S1) if S1> 2f0 (no aliasing) or fr = fa(S2/S1) if S1 < 2f0, where fais the aliased frequency. In other words, all frequencies, aliased or recovered, are identified by their principal period.
REVIEW PANEL 3.19
Aliased or Reconstructed Frequencies Are Always Identified by Their Principal Period Sampling: Unique digital frequencies always lie in the principal period −0.5 < Fa≤ 0.5.
Reconstruction at SR: Analog frequencies lie in the principal period −0.5SR< fa≤ 0.5SR. The reconstructed frequency is fR= SRFa= SRfSa.
EXAMPLE 3.9 (Signal Reconstruction at Different Sampling Rates)
A 100-Hz sinusoid is sampled at S Hz, and the sampled signal is then reconstructed at 540 Hz. What is the frequency of the reconstructed signal if S = 270 Hz? If S = 70 Hz?
1. If S = 270 Hz, the digital frequency of the sampled signal is F = 100
270= 1027, which lies in the principal period. The frequency fr of the reconstructed signal is then fr= SF = 540F = 200 Hz.
2. If S = 70 Hz, the digital frequency of the sampled signal is F = 100
70 = 107, which does not lie in the principal period. The frequency of the principal period is F = 10
7 − 1 = −37, and the frequency fr of reconstructed signal is then fr= 70F = SF = −30 Hz. The negative sign simply translates to a phase reversal in the reconstructed signal.