Suppose that the continuous-time signal *u*(*t*) has the follow relationship with its frequency spectrum *U*(*f*)

$$ \begin{array}{c}\hfill U(f)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}u(t){e}^{-j2\pi\;f\kern0.24em t}dt}\hfill \\ {}\hfill u(t)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}U\left((f)\right){e}^{j2\pi\;f\;t}df}\hfill \end{array} $$

(1)

Provided that the sampling period is *T*
_{s}, sampling from *u*(*t*)can produce the following sequence.

$$ x(n)=u\left(n{T}_{\mathrm{s}}\right) $$

(2)

The sequence has the following spectrum according to the sampling theorem [12, 13].

$$ \begin{array}{c}X\left(\omega \right)={\displaystyle \sum_{n=-\infty}^{\infty }x(n){e}^{-j\omega n}=\frac{1}{T_{\mathrm{s}}}{\displaystyle \sum_{m=-\infty}^{\infty }U\left(\frac{\omega }{2\pi {T}_{\mathrm{s}}}-\frac{m}{T_{\mathrm{s}}}\right)}}\\ {}={f}_{\mathrm{s}}{\displaystyle \sum_{m=-\infty}^{\infty }U\left(\frac{f_{\mathrm{s}}\omega }{2\pi }-m{f}_{\mathrm{s}}\right)}\end{array} $$

(3)

In the equation, *f*
_{s} = 1/*T*
_{s}, representing the sampling frequency.

The familiar result leads us to obtain the spectrogram naturally by making left and right shift of the analog spectrum diagram on the frequency axis infinite time. That is to say, to show digital signal spectrum derived from sampling, more curves should be drawn. It is convenient to handle the sampling problem of the baseband signal in this way because the positive and negative frequency bands of this kind of signal spectrum are mutually connected. Moreover, it is enough to measure the serious degree of aliasing effect by making the right shift of an analog spectrum, and the contribution of further shift terms can be ignored. If we discuss the undersampling problem of the modulating signal, shift terms of the analog spectrum involved will be many. Furthermore, the positive and negative frequency bands of the spectrum for analog signal are mutually separate; consequently, the shifts of the two parts must be taken into account thus making the method of analog spectrum shift no longer an ideal tool for analyzing.

Writing Eq. (3) in an expanded form, we obtain

$$ \begin{array}{l}X\left(\omega \right)=\cdots +{f}_{\mathrm{s}}U\left[\frac{f_{\mathrm{s}}}{2\pi}\left(\omega +2\pi \right)\right]+\\ {}\begin{array}{ccc}\hfill \hfill & \hfill \hfill & \hfill \hfill \end{array}{f}_{\mathrm{s}}U\left(\frac{f_{\mathrm{s}}\omega }{2\pi}\right)+{f}_{\mathrm{s}}U\left[\frac{f_{\mathrm{s}}}{2\pi}\left(\omega -2\pi \right)\right]+\cdots \end{array} $$

(4)

In Eq. (4) we focus on one term equivalent to Eq3. (3) in which *m* = 0. For simplicity, the symbol *V*(*ω*) is introduced to express it.

$$ V\left(\omega \right)={f}_{\mathrm{s}}U\left(\frac{f_{\mathrm{s}}\omega }{2\pi}\right) $$

(5)

Although *X*(*ω*) is periodic with period of *2π*, *V*(*ω*) is not, and its shape is consistent with the analog spectrum *U*(*f*) . On the horizontal axis*,* however, there is a flexible scale factor, that is

$$ \omega =\frac{2\pi }{f_{\mathrm{s}}}f $$

(6)

Meanwhile, there is an amplitude factor *f*
_{s}. In this paper, we call *V*(*ω*) the scaled analog spectrum in the later parts. When using the symbols in Eq. (5), Eq. (4) can be expressed as

$$ \begin{array}{c}X\left(\omega \right)=\cdots +V\left(\omega +2\pi \right)+V\left(\omega \right)+V\left(\omega -2\pi \right)+\cdots \\ {}={\displaystyle \sum_{k=-\infty}^{\infty }V\left(\omega +2k\pi \right)}\end{array} $$

(7)

Firstly, only two terms in Eq. (7) are taken into consideration that is *V*(*ω*) and *V*(*ω* + *2π*). In Fig. 1a, as for a length of amplitude magnitude of one signal’s scaled analog spectrum in the interval (−*π*, *π*], |*V*(*ω*)| is drawn by a solid line. At the same time, |*V*(*ω* + 2*π*)| is drawn by a dotted line. Suppose that ω_{1} is any point in the interval (−*π*, *π*], and its corresponding amplitude magnitude |*V*(*ω*
_{1})| is expressed by *P*, while |*V*(*ω*
_{1}) + 2*π*| is expressed by *P*
_{1}. Their corresponding complex number *V*(*ω*) and *V*(*ω* + *2π*) should be added together according to Eq. (7). Otherwise, the value |*V*| whose abscissa is *ω*
_{1} + 2*π* in the interval (*π*, 3*π*] also equals to |*V*(*ω*
_{1} + 2*π*)|, and it can be expressed by Q_{1}. Suppose that we cut down this strip of paper on which the scaled analog spectrum in the interval (−*π*, *π*] is drawn and stick it on a cylinder whose cross section is a unit circle in the following way. Fix the origin of coordinates (0,0) on a point of R on the cylinder, and make the vertical coordinate direction parallel to the axis of the cylinder. Furthermore, place the horizontal coordinate *ω* along the unit circle of the cylindrical cross section, which is perpendicular to the ordinate axis. The value of the horizontal coordinate ω is equal to the arc length from the origin point to this abscissa point. Consequently, it is equal to the angle between the vector \( \overline{OR} \) and the vector from the center of a unit circle O to this abscissa point. For the purpose of clarity, the abscissa that is fixed on the cylindrical surface is called circumferential coordinate, and the ordinate is called axial coordinate. Obviously, this paper just fully covers the cylinder surface because the width of the paper along the horizontal axis is exactly 2*π*. Next, choose the strip of paper on which the scaled analog spectrum in the interval (−*π*, 3*π*] is drawn, and fix it on the former paper in the same way to make them coincide with each other. Suppose that the pieces of paper are transplant, then the situation shown in Fig. 1b can appear, and *P*
_{1} overlaps together with *Q*
_{1} by now.

Thinking about the other terms in Eq. (7), we repeat the above steps. The transparent piece of paper in the interval ((2*k* − 1)*π*, (2*k* + 1)*π*] displays the amplitude magnitude of *V*(*ω* + 2*kπ*), which is any term in Eq. (7). As the value of *k* increases, the whole scaled analog spectrum |*V*(*ω*)| can be continuously expressed on the cylindrical surface. Therefore, the way of using a single curve and twining the cylindrical surface many layers replaces the way of using each shift term to represent the digital signal spectrum. By now, the aliasing effect is shown by twining the cylindrical surface with the same curve: more than one axis coordinate which are not equal to zero appear on the same circle coordinate of the curve. As a result of the addition of their corresponding complex number value, the shape of the digital spectrum is no longer similar with the spectrum of the analog signal used for sampling.

As a three-dimensional space curve, the expression on the cylindrical surface for the amplitude magnitude of the scaled analog spectrum is a three-dimensional vector type, and *ω* is used as its parameter.

$$ r\left(\omega \right)=\left[ \cos \omega, \kern0.5em \sin \omega, \kern0.5em \left|V\left(\omega \right)\right|\right] $$

(8)

As we see from Eqs. (5) and (6), the scale analog spectrum *V*(*w*) is proportional to the analog spectrum *U*(*f*). The spectrum on the cylindrical surface of the sampled digital signal can be derived from the spectrum of the analog signal directly. If we are only interested in the position and shape of the spectrum, then

$$ r(f)=\left[ \cos \frac{2\pi f}{f_{\mathrm{s}}},\kern0.5em \sin \frac{2\pi f}{f_{\mathrm{s}}},\kern0.5em \left|U(f)\right|\right] $$

(9)

The following example can demonstrate it. Suppose that an analog signal has the spectrum *U*(*f*), as depicted in Fig. 2a. Obviously, it contains a baseband signal whose bandwidth of double sideband is *B*
_{0} = 10*KHz* and two modulating signals. Among the two modulating signals, the carrier frequency and modulation bandwidth of one is *f*
_{1} = 20*KHz*, *B*
_{1} = 5*KHz*, and of the other is *f*
_{2} = 40*KHz*, *B*
_{2} = 8*KHz*. If the sampling frequency is 30 KHz, its spectrum on the cylindrical surface can be drawn according to Eq. (9), as illustrated in Fig. 2b.

It can be seen from the figure above, at such sampling rate, although in digital frequency domain, the two modulating signals are separate from the baseband signal, they overlaps and cannot be separated by filtering. We will return back to use this example in the fourth section, and use the arc distance of frequency band to obtain the appropriate sampling rate, so as to separate the signals.

The meaning of Nyquist criterion is expressed by the cylindrical surface spectrum in Fig. 3. What Fig. 3a expresses is a baseband signal whose highest frequency is below half the sampling frequency. Its cylindrical surface spectrum is depicted in Fig. 3b. Because digital signal spectrum cannot cover the whole cylindrical surface, there will not be aliasing at this time. The same analog signal is illustrated in Fig. 3c; however, the half sampling frequency is below the highest frequency of the spectrum. Then the digital spectrum enters into another cylindrical layer from its original “layer”, consequently, the aliasing occurs, as illustrated in phase 3D.