Question

Consider a radar with a PRF of 5 kHz. What is the maximum unambiguous range, Rua, of this radar, in km? If a target is located at a range of 50 miles, how many pulses will the radar have transmitted before the first echo from the target arrives? What will be the apparent range of the target in kilometers? (The apparent range is the range corresponding to the time delay from the most recent pulse transmission time to the arrival of the target echo from a previous pulse once steady state is achieved.)

Answer 1

Answer:

RA = 20.45 Km

Explanation:

We need to define our variables

$\lambda = 5 KHz\\\Omega = \frac{1}{\lambda}= \frac{1}{5*10^3} =2*10^{-4}s$

For $\lambda = PRF$(Pulse repetition frequency)

And $\Omega = PRT$ (Pulse repeition time)

We proceed to calculate the Maximum unambiguous range of radar,

$R_{max} = \frac{C*(\lambda-\epsilon)}{2}$

Where $\epsilon=$Pulse Width.

We do not have the Pulse Width, so we can think that can be omited.

$R_{max} = \frac{C*\Omega}{2} =\frac{3*10^8*2*10^{-4}}{2}\\    Rmax = 30,000 m = 30 km.$

converting the miles to kilometers of our target, we have,

$R= 50x1.609 = 80.45km.$

So the number of the pulses can be calculated as follow,

$N= \frac{R}{R_{max}} = \frac{160.9/2}{30} =2.68$

Therefore, N=3.

Presently,X MOD Y $\rightarrow \frac{X}{Y}* Y$

Evident range of target, $RA = R MOD [\frac{C*PRI}{2}]$

$RA = 80.45 MOD [(\frac{3x108x2x10^{-4}}{2}) *10^{-3}]\\    RA = 80.45 MOD 30\\    RA = 20.45 Km$