Consider a radar with a PRF of 5 kHz. What is the maximum unambiguous range, Ru, 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?
Answers
RA = 20.45 Km
Explanation:
We need to define our variables
\begin{gathered}\lambda = 5 KHz\\\text{\O}mega = \frac{1}{\lambda}= \frac{1}{5*10^3} =2*10^{-4}s\end{gathered}λ=5KHzØmega=λ1=5∗1031=2∗10−4s
For \lambda = PRFλ=PRF (Pulse repetition frequency)
And \text{\O}mega = PRTØmega=PRT (Pulse repeition time)
We proceed to calculate the Maximum unambiguous range of radar,
R_{max} = \frac{C*(\lambda-\epsilon)}{2}Rmax=2C∗(λ−ϵ)
Where \epsilon=ϵ= Pulse Width.
We do not have the Pulse Width, so we can think that can be omited.
\begin{gathered}R_{max} = \frac{C*\text{\O}mega}{2} =\frac{3*10^8*2*10^{-4}}{2}\\ Rmax = 30,000 m = 30 km.\end{gathered}Rmax=2C∗Ømega=23∗108∗2∗10−4 Rmax=30,000m=30km.
converting the miles to kilometers of our target, we have,
R= 50x1.609 = 80.45km.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} =2.68N=RmaxR=30160.9/2=2.68
Therefore, N=3.
Presently,X MOD Y \rightarrow \frac{X}{Y}* Y→YX∗Y
Evident range of target, RA = R MOD [\frac{C*PRI}{2}]RA=RMOD[2C∗PRI]
\begin{gathered}RA = 80.45 MOD [(\frac{3x108x2x10^{-4}}{2}) *10^{-3}]\\ RA = 80.45 MOD 30\\ RA = 20.45 Km\end{gathered}RA=80.45MOD[(23x108x2x10−4)∗10−3] RA=80.45MOD30 RA=20.45Km
Answer:
Explanation:
\begin{gathered}\lambda = 5 KHz\\\text{\O}mega = \frac{1}{\lambda}= \frac{1}{5*10^3} =2*10^{-4}s\end{gathered}λ=5KHzØmega=λ1=5∗1031=2∗10−4s
For \lambda = PRFλ=PRF (Pulse repetition frequency)
And \text{\O}mega = PRTØmega=PRT (Pulse repeition time)
We proceed to calculate the Maximum unambiguous range of radar,
R_{max} = \frac{C*(\lambda-\epsilon)}{2}Rmax=2C∗(λ−ϵ)
Where \epsilon=ϵ= Pulse Width.
We do not have the Pulse Width, so we can think that can be omited.
\begin{gathered}R_{max} = \frac{C*\text{\O}mega}{2} =\frac{3*10^8*2*10^{-4}}{2}\\ Rmax = 30,000 m = 30 km.\end{gathered}Rmax=2C∗Ømega=23∗108∗2∗10−4 Rmax=30,000m=30km.
converting the miles to kilometers of our target, we have,
R= 50x1.609 = 80.45km.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} =2.68N=RmaxR=30160.9/2=2.68
Therefore, N=3.
Presently,X MOD Y \rightarrow \frac{X}{Y}* Y→YX∗Y
Evident range of target, RA = R MOD [\frac{C*PRI}{2}]RA=RMOD[2C∗PRI]
\begin{gathered}RA = 80.45 MOD [(\frac{3x108x2x10^{-4}}{2}) *10^{-3}]\\ RA = 80.45 MOD 30\\ RA = 20.45 Km\end{gathered}RA=80.45MOD[(23x108x2x10−4)∗10−3] RA=80.45MOD30 RA=20.45Km