**4. Fundamentals of radar overview**

This part discusses the FMCW radar working principle, radar sensor hardware overview, signal processing, and radar signal waveform raw data measurement through a digital oscilloscope.

### **4.1 Principle for FMCW radar**

The principle of operation of FMCW radar is simple. This radar sensor sends continuous waves with increasing frequency and receives them back after reflecting by an object or target. It is used to find the range and other information from a target using a frequency modulation technique on a continuous signal. The radar transmitter continuously transmits this modulated signal as a continuous wave (CW). The frequency modulation used by the radar can take many forms, such as triangular, saw-tooth, sinusoidal, or some other shape. The characteristic of a radar sensor is low transmitting power, ease of modulation, simple processing and ability to measure both range and velocity (Doppler) simultaneously. The radar signal processing can be used for real-time object recognition, target tracking, parameter deduction, and sometimes even signal classification of multi-target positions under all weather situations like foggy weather, dusty environment, rainy weather, etc. The advantages of FMCW radar against the other types of radar are low peak power, less sensitivity to clutter and accurate short-range measurement, which means that it is easier to integrate, a simpler algorithm for digital signal processing and cheaper to manufacture [24–26]. A continuous carrier modulated periodic function like a saw-tooth wave is transmitted to provide range data as shown in **Figure 1**, a frequency-time relation in the FMCW radar where the red line denotes the transmitted signal, the blue line indicates the received signal, υ<sup>o</sup> denotes the central frequency, υ<sup>w</sup> denotes frequency bandwidth for sweep and *tw* denotes period for the sweep. The modulation waveform has a linear saw-tooth sample, as shown in **Figure 1**. A signal obtained is having some delay time.

*Intelligent Mine Periphery Surveillance Using Microwave Radar DOI: http://dx.doi.org/10.5772/intechopen.100521*

**Figure 1.** *Saw-tooth frequency modulation.*

In the FMCW radar system, the frequency modulated signal received at a voltagecontrolled oscillator (VCO) is transmitted from the transmitter *TX* terminal and the reflected signals from the targets are received at the receiver RX. These *TX* and *RX* signals are multiplied in a mixer, and thus beat signals are generated. The beat signals are then passed through a low pass filter, and an output signal is thus obtained. In this method, the frequency of the input signal varies with time at the VCO.

**Figure 2** shows the block diagram of the FMCW radar system. Here, D/A denotes the digital to analog converter, VCO denotes the voltage controlled oscillator, BPF indicates the band-pass filter, and A/D means analog to the digital conversion, FFT means fast Fourier theorem, *TX* denotes the transmitter unit, and *RX* denotes the receiver unit [26].

The frequency of the transmitted signal *VT(υ; s)* at *TX*, the transmitter unit is represented as:

$$V\_T(\nu, s) = A e^{\frac{j2\pi\nu}{s}} \tag{1}$$

Where υ indicates the frequency at a particular time, *s* indicates the target's distance from the transmitter where *s* = 0, *A* denotes amplitude of the transmitted signal, and *c* represents the speed of light. The frequency reflected signal *VR (υ, s)* at the receiver unit *RX* is expressed as:

**Figure 2.** *Block diagram of the FMCW radar system.*

$$\mathbb{P}(V\_R(\mathbf{u}, \mathbf{s}) = \sum\_{k=1}^{K} A a\_k \chi\_k e^{j q \eta\_k} e^{j \frac{2\pi n}{\epsilon} (2d\_k - s)} \tag{2}$$

Where *φ<sup>k</sup>* and *γ<sup>k</sup>* are reflective coefficients for the phase and amplitude of the *k*th target, respectively. The *α<sup>k</sup>* denotes the amplitude coefficient for transmission loss from the target and *dk* denotes the distance between the transmitter and the *k*th target.

At the receiver where, *s* = 0, Eq. (2) can be rewritten as:

$$V\_R(\mathbf{u}, \mathbf{0}) = \sum\_{k=1}^{K} A a\_k \mathbf{y}\_k e^{j\rho\_k} e^{j\frac{2\pi\alpha}{\varepsilon}(2d\_k)}\tag{3}$$

To get the frequency of a beat signal, the transmitted signal's frequency in Eq. (1) is multiplied by the frequency of the received signal in Eq. (3) at the position *s* = 0.

Thus, the frequency of beat signal = *VT*(*υ*,*s*) � *VR* (*υ*, 0).

$$\mathbf{A} = \mathbf{A} \, e^{j\frac{2\pi\mathbf{x}}{\varepsilon}} \sum\_{k=1}^{K} A a\_k \mathbf{y}\_k e^{j\phi\_k} e^{j\frac{2\pi\mathbf{x}}{\varepsilon}(2d\_k)}.\tag{4}$$

The output signal *Vout* (*υ*, *0*) is generated by passing through a BPF is presented as:

$$V\_{out}(\mathbf{0}, \mathbf{0}) = \sum\_{k=1}^{K} A^2 a\_k \chi\_k e^{j\varphi\_k} e^{j\frac{4\pi d\_k}{c}} \tag{5}$$

The distance and displacement of the target are assumed from the generated output signal in Eq. (5) by the use of signal processing.

The distance spectrum of the output signal *P*(*x*) is calculated by using Fourier transform:

$$\begin{split} P(\mathbf{x}) &= \int\_{v\_0 - \frac{w\_\epsilon}{2}}^{v\_0 + \frac{w\_\epsilon}{2}} V\_{out} e^{-j\frac{4\pi w}{\epsilon}} dv \\ &= \int\_{v\_0 - \frac{w\_\epsilon}{2}}^{v\_0 + \frac{w\_\epsilon}{2}} \sum\_{k=1}^K A^2 a\_k \chi\_k e^{j\frac{4\pi w\_k}{\epsilon}} e^{-j\frac{4\pi w}{\epsilon}} dv \\ &= A^2 \sum\_{k=1}^K a\_k \chi\_k e^{j\eta\_k} \int\_{v\_0 - \frac{w\_\epsilon}{2}}^{v\_0 + \frac{w\_\epsilon}{2}} e^{j\frac{4\pi (d\_k - s)}{\epsilon}} dv \\ &= A^2 \sum\_{k=1}^K \left[ a\_k \chi\_k e^{j\eta\_k} e^{j\frac{4\pi w\_\epsilon (d\_k - s)}{\epsilon}} u\_\omega \sin c \left\{ \frac{2\pi \mathfrak{u}\_\omega (d\_k - s)}{\epsilon} \right\} \right] \end{split} \tag{6}$$

In this equation, the function of *Sin c*(*s*) denotes:

$$\text{Sim}\ (\text{cs}) = \frac{\text{Sins}}{\text{s}} \tag{7}$$

The amplitude value of the distance spectrum j j *P x*ð Þ in Eq. (6) is given as:

*Intelligent Mine Periphery Surveillance Using Microwave Radar DOI: http://dx.doi.org/10.5772/intechopen.100521*

$$\begin{split} |P(\mathbf{s})| &= A^2 \Big| \sum\_{k=1}^{K} a\_k \chi\_k e^{j\rho\_k} e^{j\frac{4\pi\mathbf{u}\_0(d\_k - s)}{c}} v\_w \text{sinc} \left\{ \frac{2\pi\mathbf{u}\_w(d\_k - s)}{c} \right\} \Big| \\ &\leq A^2 \mathbf{o}\_w \sum\_{k=1}^{K} \left[ a\_k \chi\_k \left| \text{sinc} \left\{ \frac{2\pi\mathbf{u}\_w(d\_k - s)}{c} \right\} \right| \right] \Big| \end{split} \tag{8}$$

This is possible when the phase components *ϕκ* <sup>þ</sup> <sup>4</sup>*πυo*ð Þ *dk*�*<sup>s</sup> <sup>c</sup>* for all of *κ* are equal.

$$|P(\mathbf{s})| = A^2 \mathbf{o}\_w \sum\_{k=1}^{K} \left[ a\_k \gamma\_k \left| \text{sinc} \left\{ \frac{2\pi \mathbf{o}\_w(d\_k - s)}{c} \right\} \right| \right] \tag{9}$$

When the number of a target to be 1, the distance spectrum in Eq. (6) becomes:

$$|P(\mathfrak{s})| = \left[A^2 a\_1 \chi\_1 e^{j\phi\_1} e^{j\frac{4\pi\mathfrak{o}\_0(d\_1 - s)}{c}} \mathfrak{o}\_w \text{sinc}\left\{\frac{2\pi\mathfrak{o}\_o(d\_1 - s)}{c}\right\}\right] \tag{10}$$

Its amplitude value of the distance spectrum is given as:

$$|P(\mathbf{s})| = A^2 \mathbf{o}\_w a\_1 \chi\_1 \left| \operatorname{sinc} \left\{ \frac{2\pi \mathbf{o}\_w (d\_1 - s)}{c} \right\} \right| \tag{11}$$

This Equation indicates the distance of the target is specified by the amplitude value of the distance spectrum.

The phase value of the distance spectrum, *P(s),* is represented as:

$$\mathbf{s} < \mathbf{P}(\mathbf{s}) = \phi\_1 + \frac{4\pi\mathbf{u}\_o(d\_1 - s)}{\mathcal{L}} = \theta\_1(\mathbf{s}) \tag{12}$$

Here, because *θ1(s)* satisfies - *π* ≤*θ*1ð Þ*s* ≤*π* The displacement of the target is:

$$\frac{c(-\pi - \rho\_1)}{4\pi \mathbf{u}\_o} \le d\_1 \le \frac{c(\pi - \rho\_1)}{4\pi \mathbf{u}\_o} \tag{13}$$

If the phase value satisfies *<sup>φ</sup>*<sup>1</sup> <sup>¼</sup> *<sup>π</sup>* 6, and *υ<sup>o</sup>* ¼ 24*:*15 GHz in Eq. (13). Thus, it can be rewritten as �0.0036 [m] ≤ d1 ≤ 0.0026 [m] with *υ<sup>o</sup>* ¼ 24*:*15 GHz. This means a slight displacement of the target within –0.0036 [m] ≤d1 ≤0.0026 [m] is generated by the phase value of the distance spectrum.

On the other hand, the maximum distance for measuring *dmax* can be calculated using the following equation:

$$d\_{\max} = \frac{c}{4\Delta\nu} [\mathbf{m}] \tag{14}$$

Where *c* denotes the speed of light and Δ*υ* denotes the frequency resolution of the distance spectrum and expressed in hertz (Hz), which can be calculated using the following equation:

$$
\Delta \mathbf{u} = \frac{\mathbf{u}\_w}{\mathbf{t}\_{w\big|\_{\mathbf{t}\_r}}} \mathbf{[Hz]} \tag{15}
$$

Where *tw* denotes the sweep time, *ts* denotes the interval time for sampling and *υ<sup>w</sup>* denotes the bandwidth of sweep frequency, *dmax* denotes the maximum distance is expressed in metre (m).

Now, when sweep time, *tw* = 1024 μs. Interval time for sampling, *ts* = 1μs, the bandwidth of sweep frequency, *υ<sup>w</sup>* = 200 MHz.

Then,

$$\begin{aligned} \Delta \mathbf{u} &= \frac{\mathbf{u}\_w}{t\_{w\big|\_{\mathbf{f}\_I}}} = \frac{200 \times 10^6 \text{Hz}}{1024 \times 10^{-6} \text{\AA}^{-1} \text{\AA}^{-6} \text{s}}\\ \Delta \nu &= \mathbf{1.95} \times \mathbf{10^5} [\text{Hz}] \end{aligned} \tag{16}$$

Using value of Δ*υ*, *dmax* can be calculated as follows:

$$d\_{\max} = \frac{c}{4\Delta\nu} = \frac{3 \times 10^8}{4 \times 1.95 \times 10^5} = 384 \,\mathrm{[m]}\tag{17}$$
