Radar Systems - Range Equation

Radar range equation is useful to know the range of the target theoretically. In this chapter, we will discuss the standard form of Radar range equation and then will discuss about the two modified forms of Radar range equation.

We will get those modified forms of Radar range equation from the standard form of Radar range equation. Now, let us discuss about the derivation of the standard form of Radar range equation.

Derivation of Radar Range Equation

The standard form of Radar range equation is also called as simple form of Radar range equation. Now, let us derive the standard form of Radar range equation.

We know that power density is nothing but the ratio of power and area. So, the power density, $P_{di}$ at a distance, R from the Radar can be mathematically represented as −

$$P_{di}=frac{P_t}{4pi R^2}:::::Equation:1$$

Where,

The above power density is vapd for an isotropic Antenna. In general, Radars use directional Antennas. Therefore, the power density, $P_{dd}$ due to directional Antenna will be −

$$P_{dd}=frac{P_tG}{4pi R^2}:::::Equation:2$$

Target radiates the power in different directions from the received input power. The amount of power, which is reflected back towards the Radar depends on its cross section. So, the power density $P_{de}$ of echo signal at Radar can be mathematically represented as −

$$P_{de}=P_{dd}left (frac{sigma}{4pi R^2} ight ):::::Equation:3$$ Substitute, Equation 2 in Equation 3.

$$P_{de}=left (frac{P_tG}{4pi R^2} ight )left (frac{sigma}{4pi R^2} ight ):::::Equation:4$$

The amount of power, $P_r$ received by the Radar depends on the effective aperture, $A_e$ of the receiving Antenna.

$$P_r=P_{de}A_e:::::Equation:5$$

Substitute, Equation 4 in Equation 5.

$$P_r=left (frac{P_tG}{4pi R^2} ight )left (frac{sigma}{4pi R^2} ight )A_e$$

$$Rightarrow P_r=frac{P_tGsigma A_e}{left (4pi ight )^2 R^4}$$

$$Rightarrow R^4=frac{P_tGsigma A_e}{left (4pi ight )^2 P_r}$$

$$Rightarrow R=left [frac{P_tGsigma A_e}{left (4pi ight )^2 P_r} ight ]^{1/4}:::::Equation:6$$

Standard Form of Radar Range Equation

If the echo signal is having the power less than the power of the minimum detectable signal, then Radar cannot detect the target since it is beyond the maximum pmit of the Radar s range.

Therefore, we can say that the range of the target is said to be maximum range when the received echo signal is having the power equal to that of minimum detectable signal. We will get the following equation, by substituting $R=R_{Max}$ and $P_r=S_{min}$ in Equation 6.

$$R_{Max}=left [frac{P_tGsigma A_e}{left (4pi ight )^2 S_{min}} ight ]^{1/4}:::::Equation:7$$

Equation 7 represents the standard form of Radar range equation. By using the above equation, we can find the maximum range of the target.

Modified Forms of Radar Range Equation

We know the following relation between the Gain of directional Antenna, $G$ and effective aperture, $A_e$.

$$G=frac{4pi A_e}{lambda^2}:::::Equation:8$$

Substitute, Equation 8 in Equation 7.

$$R_{Max}=left [ frac{P_tsigma A_e}{left ( 4pi ight )^2S_{min}}left ( frac{4pi A_e}{lambda^2} ight ) ight ]^{1/4}$$

$$Rightarrow R_{Max}=left [frac{P_tGsigma {A_e}^2}{4pi lambda^2 S_{min}} ight ]^{1/4}:::::Equation:9$$

Equation 9 represents the modified form of Radar range equation. By using the above equation, we can find the maximum range of the target.

We will get the following relation between effective aperture, $A_e$ and the Gain of directional Antenna, $G$ from Equation 8.

$$A_e=frac{Glambda^2}{4pi}:::::Equation:10$$

Substitute, Equation 10 in Equation 7.

$$R_{Max}=left [frac{P_tGsigma}{left (4pi ight )^2 S_{min}}(frac{Glambda^2}{4pi}) ight ]^{1/4}$$

$$Rightarrow R_{Max}=left [frac{P_tG^2 lambda^2 sigma}{left (4pi ight )^2 S_{min}} ight ]^{1/4}:::::Equation:11$$

Equation 11 represents another modified form of Radar range equation. By using the above equation, we can find the maximum range of the target.

Note − Based on the given data, we can find the maximum range of the target by using one of these three equations namely

Equation 7

Equation 9

Equation 11

Example Problems

In previous section, we got the standard and modified forms of the Radar range equation. Now, let us solve a few problems by using those equations.

Problem 1

Calculate the maximum range of Radar for the following specifications −

Peak power transmitted by the Radar, $P_t=250KW$

Gain of transmitting Antenna, $G=4000$

Effective aperture of the receiving Antenna, $A_e=4:m^2$

Radar cross section of the target, $sigma=25:m^2$

Power of minimum detectable signal, $S_{min}=10^{-12}W$

Solution

We can use the following standard form of Radar range equation in order to calculate the maximum range of Radar for given specifications.

$$R_{Max}=left [frac{P_tG sigma A_e}{left (4pi ight )^2 S_{min}} ight ]^{1/4}$$

Substitute all the given parameters in above equation.

$$R_{Max}=left [frac{ left ( 250 imes 10^3 ight )left ( 4000 ight )left ( 25 ight )left ( 4 ight )}{left ( 4pi ight )^2 left ( 10^{-12} ight )} ight ]^{1/4}$$

$$Rightarrow R_{Max}=158:KM$$

Therefore, the maximum range of Radar for given specifications is $158:KM$.

Problem 2

Calculate the maximum range of Radar for the following specifications.

Operating frequency, $f=10GHZ$

Peak power transmitted by the Radar, $P_t=400KW$

Effective aperture of the receiving Antenna, $A_e=5:m^2$

Radar cross section of the target, $sigma=30:m^2$

Power of minimum detectable signal, $S_{min}=10^{-10}W$

Solution

We know the following formula for operating wavelength, $lambda$ in terms of operating frequency, f.

$$lambda =frac{C}{f}$$

Substitute, $C=3 imes 10^8m/sec$ and $f=10GHZ$ in above equation.

$$lambda =frac{3 imes 10^8}{10 imes 10^9}$$

$$Rightarrow lambda=0.03m$$

So, the operating wavelength,$lambda$ is equal to $0.03m$, when the operating frequency, $f$ is $10GHZ$.

We can use the following modified form of Radar range equation in order to calculate the maximum range of Radar for given specifications.

$$R_{Max}=left [frac{P_t sigma {A_e}^2}{4pi lambda^2 S_{min}} ight ]^{1/4}$$

Substitute, the given parameters in the above equation.

$$R_{Max}=left [ frac{left ( 400 imes 10^3 ight )left ( 30 ight )left ( 5^2 ight )}{4pileft ( 0.003 ight )^2left ( 10 ight )^{-10}} ight ]^{1/4}$$

$$Rightarrow R_{Max}=128KM$$

Therefore, the maximum range of Radar for given specifications is $128:KM$.