- Radar Systems Tutorial
- Radar Systems - Home
- Radar Systems - Overview
- Radar Systems - Range Equation
- Performance Factors
- Radar Systems - Types of Radars
- Radar Systems - Pulse Radar
- Radar Systems - Doppler Effect
- Radar Systems - CW Radar
- Radar Systems - FMCW Radar
- Radar Systems - MTI Radar
- Delay Line Cancellers
- Radar Systems - Tracking Radar
- Antenna Parameters
- Radar Systems - Radar Antennas
- Matched Filter Receiver
- Radar Systems - Radar Displays
- Radar Systems - Duplexers
- Phased Array Antennas
- Radar Systems Useful Resources
- Radar Systems - Quick Guide
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In this chapter, we will learn about the Doppler Effect in Radar Systems.

If the target is not stationary, then there will be a change in the frequency of the signal that is transmitted from the Radar and that is received by the Radar. This effect is known as the **Doppler effect**.

According to the Doppler effect, we will get the following two possible cases −

The

**frequency**of the received signal will**increase**, when the target moves towards the direction of the Radar.The

**frequency**of the received signal will**decrease**, when the target moves away from the Radar.

Now, let us derive the formula for Doppler frequency.

The distance between Radar and target is nothing but the **Range** of the target or simply range, R. Therefore, the total distance between the Radar and target in a two-way communication path will be 2R, since Radar transmits a signal to the target and accordingly the target sends an echo signal to the Radar.

If $\lambda$ is one wave length, then the number of wave lengths N that are present in a two-way communication path between the Radar and target will be equal to $2R/\lambda$.

We know that one wave length $\lambda$ corresponds to an angular excursion of $2\pi$ radians. So, the **total angle of excursion** made by the electromagnetic wave during the two-way communication path between the Radar and target will be equal to $4\pi R/\lambda$ radians.

Following is the mathematical formula for **angular frequency**, $\omega$ −

$$\omega=2\pi f\:\:\:\:\:Equation\:1$$

Following equation shows the mathematical relationship between the angular frequency $\omega$ and phase angle $\phi$ −

$$\omega=\frac{d\phi }{dt}\:\:\:\:\:Equation\:2$$

**Equate** the right hand side terms of Equation 1 and Equation 2 since the left hand side terms of those two equations are same.

$$2\pi f=\frac{d\phi }{dt}$$

$$\Rightarrow f =\frac{1}{2\pi}\frac{d\phi }{dt}\:\:\:\:\:Equation\:3$$

**Substitute**,$f=f_d$ and $\phi=4\pi R/\lambda$ in Equation 3.

$$f_d =\frac{1}{2\pi}\frac{d}{dt}\left ( \frac{4\pi R}{\lambda} \right )$$

$$\Rightarrow f_d =\frac{1}{2\pi}\frac{4\pi}{\lambda}\frac{dR}{dt}$$

$$\Rightarrow f_d =\frac{2V_r}{\lambda}\:\:\:\:\:Equation\:4$$

Where,

$f_d$ is the Doppler frequency

$V_r$ is the relative velocity

We can find the value of Doppler frequency $f_d$ by substituting the values of $V_r$ and $\lambda$ in Equation 4.

**Substitute**, $\lambda=C/f$ in Equation 4.

$$f_d =\frac{2V_r}{C/f}$$

$$\Rightarrow f_d =\frac{2V_rf}{C}\:\:\:\:\:Equation\:5$$

Where,

$f$ is the frequency of transmitted signal

$C$ is the speed of light and it is equal to $3\times 10^8m/sec$

We can find the value of Doppler frequency, $f_d$ by substituting the values of $V_r,f$ and $C$ in Equation 5.

**Note** − Both Equation 4 and Equation 5 show the formulae of Doppler frequency, $f_d$. We can use either Equation 4 or Equation 5 for finding **Doppler frequency**, $f_d$ based on the given data.

If the Radar operates at a frequency of $5GHZ$, then find the **Doppler frequency** of an aircraft moving with a speed of 100KMph.

Given,

The frequency of transmitted signal, $f=5GHZ$

Speed of aircraft (target), $V_r=100KMph$

$$\Rightarrow V_r=\frac{100\times 10^3}{3600}m/sec$$

$$\Rightarrow V_r=27.78m/sec$$

We have converted the given speed of aircraft (target), which is present in KMph into its equivalent m/sec.

We know that, the speed of the light, $C=3\times 10^8m/sec$

Now, following is the **formula for Doppler frequency** −

$$f_d=\frac{2Vrf}{C}$$

**Substitute** the values of 𝑉𝑟, $V_r,f$ and $C$ in the above equation.

$$\Rightarrow f_d=\frac{2\left ( 27.78 \right )\left ( 5\times 10^9 \right )}{3\times 10^8}$$

$$\Rightarrow f_d=926HZ$$

Therefore, the value of **Doppler frequency**, $f_d$ is $926HZ$ for the given specifications.

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