Nyquist's criterion The greatest part of information sources are analog, like sound. Today's telecommunication systems are mostly digital, so the most important step toward communicating is a signal digitization. In this case, signal is a voltage or current representation of the real world phenomena. Word digital means discrete value of a signal at discrete periods of time (intervals). These discrete intervals are very important and defined by the Nyquist's criterion. The Nyquist's criterion says that a signal must be sampled at least twice as often as its highest frequency to enable correct reconstruction of the signal when transmitted. Real world signals are very irregular. For example speech or music. Frequency analysis of the music signal shows that there are components that have very high frequencies. Theoretically, a signal can have all of the frequencies in the spectrum, so that if one wants to take samples of such signal, respecting the Nyquist's criterion, sampling frequency goes to infinity. Practically that is unobtainable. But fortunately, music, speech, etc. have almost all of the components inside a bandwidth of around 0-4 khz for speech or 0-12 khz for music. Upper frequency is taken as a highest frequency of the signal, and all of the components above are being filtered out using a lowpass filter. According to Nyquist's criterion, sampling rate is e.g. 8 khz for speech. Every sample gets its 8 bits code word and that makes 8bit*8kHz = 64 kbit/s speech bit rate used as a basic rate in N-ISDN. Let's prove now the Nyquist's criterion. Knowledge in Fourier analysis and a bit of mathematical analysis is required to understand the following. We'll assume an real sampling signal made of pulses of the duration time thau distanced for period T ( Fig. 1-b and Fig. 3 ). Samples carrying signal ( Fig. 1-c ) will be produced by the multiplication of the original signal ( Fig. 1-a ) and the sampling signal ( Fig. 1-b ). Spectrum of the original signal Xi(t) is defined by the Fourier transformation as follows : Spectrum in Fig. 2 is a speech spectrum. One can see that almost whole spectrum is placed under the frequency of 4 khz.
To prove Nyquist's criterion, we'll have to find the spectrum of the signal shown in Fig. 1-c. First step toward this spectrum is spectrum of the sampling signal shown below in Fig. 3. The spectrum shown on the right is the real part ( positive frequencies ) of the spectrum of the real "red" signal on the left. Theoretically, we assume an ideal rectangular pulse sampling signal shown in Fig. 3 above in black. Pulse duration time is thau, and distance between two neighboring pulses is T. The spectrum of that signal is the same as the one shown above, but the spectral components have no width, they are delta-functions multiplied with the amplitude. Let's start finding the blue envelope of the spectrum. We'll use Fourier transformation for a periodical function. This complex function represents Fourier transformation. We distinguish amplitude and phase of every spectrum component ( for every n ). Absolute value of this function is amplitude and, if n is considered a real variable, this absolute value represents the spectrum envelope. Spectral components frequencies are discrete values given as a multiplication of the frequency f=1/t and integer number n (...,-
2,-1,0,1,2,...). That means that the distance between two neighboring spectral components is f [Hz]. Samples carrying signal ( PAM signal - Pulse-Amplitude Modulation signal ) is multiplication of the original signal and sampling signal. We'll multiply those in the form of the inverted Fourier transformation. Sampling signal spectrum has no complex part because we assume t1=thau/2. The part of the formula above, between integral sign and exponential function is Fourier transformation of the PAM signal ( samples carrying signal ). It is very important to note that the tranformation is distinguished on real and complex part. Real part is spectrum amplitude. Complex part is a key problem. Written in this form, integral represents Fourier transformation of the function X(t)*exp(j*n*w0*t). Fourier transformation has a very interesting property. If X'(jw) is transformation of X(t), than transformation of X(t)*exp(jxt) is X'(jw - jx). What impact does this property have to our spectrum? We draw the original signal Xi(t) spectrum in DSB ( Double Side Band ) form as shown in Fig. 4-a below. If this signal is multiplied with exp(jxt) before transformation, its spectrum looks like in Fig. 4-b.
If one takes this property in account, it is obvious that the PAM spectrum will be made of pairs shown in Fig. 4-b for every n*w0, n >= 0. That's because of sum from minus to plus infinity for n. The original spectrum will appear every f=1/t khz,infinite number of times, but the intensity of this original spectrum will be determined by the envelope function of the sampling signal. Every original spectrum is DSB and it is multiplied with spectrum envelope function of the sampling signal (Fig. 3). Now, we can draw the whole spectrum of the PAM signal (Fig. 5). Band width of the original signal is limited to B by a low pass filter. Figure 5 PAM spectrum is made of damped original spectrums. Each of these has its own band width determined by the spectrum itself or by the filter. The original signal can be reconstructed by the same filter and by an amplificator. Original spectrum will be cut from the PAM spectrum. But what if one part of an original spectrum enters ( interferes ) whit the neighboring original spectrum. In that case, the original spectrum cannot be reconstructed and the original information is lost. This is always the case when the original spectrum is infinite and unfiltered. In that case every original spectrum will be mixed with every other spectrum. That means that the first condition for the successful reconstruction is : original signal has to be frequency limited to bandwidth B before sampling. But the problem is still not solved because situation in Fig 6-a is possible. Two limited original spectrums can interfere if the distance between them is less than 2*B. And here is the essence of the Nyquist's criterion. Frequency distance between two neighboring original spectrums has to be at least twice larger than the original spectrum bandwidth B ( Fig 6-b and c ).
bsp; c a &n bsp; b &n Figure 6 Since the distance between two neighboring original spectrums is rate of sampling, signal must be sampled at least twice as often as its highest frequency. Next formula is formalization of that law. One has to be careful with the term bandwidth. It considers band width from frequency 0 Hz to B Hz. If one takes that in account, it is maybe better to say that sampling rate has to be at least twice faster than the highest frequency fh. Frequency fh is the upper low pass filter frequency. Nyquist's criterion is proved for the sampling of the electrical signal that is an image of the real world phenomena, but it can be applied elsewhere. One of the school examples is turning of the wheel in wrong direction while in a movie. Why does it happen? We can mark a point on the wheel, and watch that point only. Making movie is taking pictures every "t" time. If the point on the wheel makes exactly one turn inside the time "t", we'll see that the wheel is not turning at all. If we increase sampling rate, we'll see the point on the wheel turning in wrong direction. If we increase the sampling rate to be twice faster than a wheel rotation speed, we'll finally see the point on the wheel turning in right direction. Even in this case, Nyquist's criterion has to be respected. Of course, in real movies, details on the wheel determine the right sampling rate. This was just the first part of the story. The other one is about how many bits can be transimtted through an analog system in time in the presence of noise. Shannon law says something about it, and Shannon himself in his book "Communication in the Presence of Noise" (Proc. IRE, Vol. 37, 1949, 10-12).