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A free-space \(2 \mathrm{GHz}\) pulsed monostatic radar system transmits a \(2 \mathrm{~kW}\) pulse and has a minimum detectable received signal power of \(-90 \mathrm{dBm}\). What is the antenna gain required to be able to detect a target with a radar cross section of \(10 \mathrm{~m}^{2}\) at \(10 \mathrm{~km} ?\)

Short Answer

Expert verified
The antenna gain required is approximately 74.25.

Step by step solution

01

Convert units if necessary

Convert the threshold received power from \texttt{dBm} to watts using the formula: \( P_{r_{\text{min}}} = 10^{\frac{-90}{10}} \times 10^{-3} = 10^{-12} \text{ W} \)
02

Identify known values

Given data:- Transmitted power: \(P_t = 2 \text{ kW} = 2000 \text{ W}\)- Minimum detectable signal power: \(P_r = 10^{-12} \text{ W}\)- Radar cross section: \(RCS = 10 \text{ m}^2\)- Target distance: \(R = 10 \text{ km} = 10,000 \text{ m}\)- Frequency: \(f = 2 \text{ GHz}\)
03

Apply the Radar Range Equation

Use the radar range equation for a monostatic radar system:\[ P_r = \frac{P_t G^2 \text{RCS} \lambda^2}{(4\pi)^3 R^4} \]Where \(G\) is the antenna gain and \(\lambda\) is the wavelength.
04

Calculate the wavelength

The wavelength (\(\lambda\)) can be found using the speed of light (\(c\)) and frequency (\(f\)):\( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{2 \times 10^9} = 0.15 \text{ m} \)
05

Isolate the antenna gain (G)

Rearrange the radar range equation to solve for the antenna gain (\(G\)):\[ G = \left( \frac{P_r (4\pi)^3 R^4}{P_t \text{RCS} \lambda^2} \right)^{\frac{1}{2}} \]
06

Plug in the known values

Substitute the known values into the rearranged equation:\[ G = \left( \frac{10^{-12} (4\pi)^3 (10,000)^4}{2000 \times 10 \times (0.15)^2} \right)^{\frac{1}{2}} \]
07

Simplify and solve

Perform the calculations step by step:1. Calculate \((4\pi)^3 = 248.05 \)2. Compute the distance term: \( (10,000)^4 = 10^{16} \)3. Compute the wavelength squared: \( (0.15)^2 = 0.0225 \)4. Multiply together: \( 10^{-12} \times 248.05 \times 10^{16} = 2.4805 \times 10^6 \)5. Divide by the denominator: \( \frac{2.4805 \times 10^6}{2000 \times 10 \times 0.0225} = 5.51 \times 10^3 \)6. Take the square root: \( \sqrt{5.51 \times 10^3} \approx 74.25 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radar Range Equation
The radar range equation is a critical concept in understanding radar system performance. It calculates the received power of a radar signal as it reflects off a target.
This equation combines various parameters such as transmitted power, antenna gain, target distance, and radar cross-section.
The general form of the radar range equation for a monostatic radar system is:
\[ P_r = \frac{P_t G^2 \text{RCS} \lambda^2}{(4\pi)^3 R^4} \]
Where:
- \(P_r\): Received power
- \(P_t\): Transmitted power
- \(G\): Antenna gain
- \(\text{RCS}\): Radar cross-section of the target
- \(\lambda\): Wavelength of the signal
- \(R\): Distance to the target
This equation helps in designing and optimizing radar systems for various applications.
Antenna Gain
Antenna gain is a measure of how well an antenna focuses energy in a particular direction. It is an important parameter in radar systems as it affects the strength of the received signal.
The higher the gain, the more efficiently the antenna transmits and receives signals.
In the radar range equation, the gain \(G\) appears squared, making it a significant factor in determining the received signal power.
A higher antenna gain allows the radar system to detect targets at greater distances.
Antenna gain is typically expressed in decibels (dB).
For example, in our original problem, the required gain to detect a target at 10 km with the given parameters is calculated using:
\[ G = \left( \frac{P_r (4\pi)^3 R^4}{P_t \text{RCS} \lambda^2} \right)^{\frac{1}{2}} \]
Plugging in the known values gives us a gain of approximately 74.25.
Signal Power Conversion
Signal power conversion is a key aspect of radar system calculations, especially when dealing with different units.
In our provided exercise, the received signal power is given in dBm (decibel-milliwatts), which needs to be converted to watts for use in the radar range equation.
The conversion formula is:
\[ P_{r_{\text{min}}} = 10^{\frac{-90}{10}} \times 10^{-3} = 10^{-12} \text{ W} \]
Here, -90 dBm is converted to \(10^{-12} W\), making it easier to plug into the radar range equation.
Understanding these conversions is essential because radar systems often use various units to express power levels, and precise conversions ensure accurate calculations.
Wavelength Calculation
Calculating the wavelength \(\lambda\) is crucial for many radar system calculations. The wavelength connects the frequency of the radar signal and the speed of light.
The formula to find the wavelength is:
\[ \lambda = \frac{c}{f} \]
Where:
- \(c\): Speed of light (approximately \(3 \times 10^8 \text{ m/s}\))
- \(f\): Frequency of the radar signal
In the original problem, the radar operates at a frequency of 2 GHz (or \(2 \times 10^9 \text{ Hz}\)).
So, the wavelength calculation is:
\[ \lambda = \frac{3 \times 10^8}{2 \times 10^9} = 0.15 \text{ m} \]
This calculated wavelength is then used in the radar range equation to determine the necessary antenna gain.
Understanding how to calculate wavelength helps in configuring radar systems to operate effectively under different conditions.

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Most popular questions from this chapter

A 10 GHz bistatic radar has a minimum detectable received signal power of \(-150 \mathrm{dBm},\) an antenna gain of \(26 \mathrm{~dB},\) and a required range of \(100 \mathrm{~km}\). What is the transmitted pulse power in dBm needed to detect a (a) conventional fighter aircraft having an RCS of \(5 \mathrm{~m}^{2}\) ? (b) a stealth aircraft with an RCS of \(0.05 \mathrm{~m}^{2}\) ?

A new communication system is being investigated for sending data to a printer. The system will use GMSK modulation and a channel with \(25 \mathrm{MHz}\) bandwidth and an information bit rate of \(10 \mathrm{Mbit} / \mathrm{s}\). The modulation format will result in a spectrum that distributes power almost uniformly over the 25 MHz bandwidth. [Parallels Example 5.3] (a) What is the processing gain? (b) If the received RF SIR is \(6 \mathrm{~dB},\) what is the effective system SIR (or \(\left.E_{b, i} / N_{o, i}\right)\) after DSP? Express your answer in decibels.

Describe the following concepts. (a) Clusters in a cellular phone system. (b) Multipath effects in a central city area compared to multipath effects in a desert.

The L1 band of the global positioning system (GPS), is centered at \(1.57542 \mathrm{GHz}\) and has two overlapping spread-spectrum encoded signals. The stronger of these is the coarse acquisition \((\mathrm{C} / \mathrm{A})\) signal with an information bit rate of 50 bits \(/ \mathrm{s}\) and a transmission rate of 1.023 million chips per second using BPSK modulation with an RF bandwidth of \(2.046 \mathrm{MHz}\). In ideal conditions the \(\mathrm{C} /\) A signal received has a power of \(-130 \mathrm{dBm}\). A GPS receiver has an antenna noise temperature is of \(290 \mathrm{~K}\). (a) What is the processing gain? (b) What is the noise in \(\mathrm{dBm}\) received in the \(2.046 \mathrm{MHz}\) bandwidth? (c) What is the SNR in decibels? (d) If a C/A signal is received from each of 10 satellites (so there are 9 interfering signals), what is the total interference power for one satellite's \(\mathrm{C} / \mathrm{A}\) signal? (e) With respect to just one of the \(\mathrm{C} / \mathrm{A}\) signals, what is the SINR (signal to interference plus noise ratio) at the receiver? (f) If the receiver does not contribute noise, what is the effective SNR of the despread bitstream from each satellite? (g) If the required minimum effective SNR is \(6 \mathrm{~dB}\), what is the minimum acceptable power, in \(\mathrm{dBm}\), of the GPS signal received from one satellite?

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