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A pulse radar determines target by round trip time of a pulsed microwave signal. The frequency used by radar transmitter is 10GHz with transmitted power 2KW (Pulse power). The Antenna size of radar transmitted this signal is based on \(\lambda\ (\text{Wavelength})=\frac{C\ (\text{speed of light})}{f\ (\text{Frequency})}\) with a Gain (Gt) of 28dB is used to detect the target (aeroplane) having its cross section area as 12m2. The receiver has its capability as -90dBm as minimum detectable signal (Pmin). There is an isolation between trans and receive chain as (80-100dB) determine Radar maximum range.

Based on the paragraph answer following questions :

What is maximum range of this radar (Rmax) upto which target can be determined ?

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

8114 m

To determine the maximum range of the radar (\(R_{\text{max}}\)), we'll use the radar range equation. The equation for the maximum range of a radar system is given by:

\[R_{\text{max}} = \left( \frac{P_t \cdot G_t \cdot G_r \cdot \lambda^2 \cdot \sigma}{(4\pi)^3 \cdot P_{\text{min}}} \right)^{1/4}\]

where:

  • \(P_t\) is the transmitted power (2 kW or 2000 W)
  • \(G_t\) is the transmitter gain (28 dB)
  • \(G_r\) is the receiver gain (also taken as 28 dB for symmetry)
  • \(\lambda\) is the wavelength of the radar signal
  • \(\sigma\) is the radar cross-section of the target (12 m2)
  • \(P_{\text{min}}\) is the minimum detectable signal (-90 dBm)

First, let's convert given values:

  • Gain (\(G_t\) and \(G_r\)) in linear scale:
  • \(-90 \, \text{dBm}\) is equal to:
  • Frequency \(f = 10\,\text{GHz} = 10^{10}\,\text{Hz}\)

Substitute these values into the radar range equation:

\[R_{\text{max}} = \left( \frac{2000 \times 631 \times 631 \times (0.03)^2 \times 12}{(4\pi)^3 \times 10^{-9}} \right)^{1/4}\]

Calculate the range:

\[R_{\text{max}} = \left( \frac{2000 \times 398161 \times 0.0009 \times 12}{248.05 \times 10^{-9}} \right)^{1/4}\]

 

\[R_{\text{max}} = \left( \frac{8594846400}{248.05 \times 10^{-9}} \right)^{1/4}\]

 

\[R_{\text{max}} = \left( 3.48 \times 10^{19} \right)^{1/4}\]

 

\[R_{\text{max}} \approx 8114 \, \text{m}\]

Thus, the correct answer is 8114 m.

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