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Chapter 5: Problem 16

A monostatic free-space \(10 \mathrm{GHz}\) pulsed radar system is used to detect a fighter plane having a radar cross section, \(\sigma,\) of \(5 \mathrm{~m}^{2}\). The antenna gain is \(30 \mathrm{~dB}\) and the transmitted power is \(1 \mathrm{~kW}\). If the minimum detectable received signal is \(-120 \mathrm{dBm},\) what is the detection range?

Short Answer

Expert verified
The detection range is approximately 4.99 km.

Step by step solution

01

Convert Units

Convert the transmitted power from kilowatts to watts. Given that the transmitted power is 1 kW, convert it to watts: \[ P_t = 1 \text{kW} = 1000 \text{W} \] Convert the antenna gain from decibels to a linear scale: \[ G = 10^{\frac{30}{10}} = 10^3 \]
02

Convert Minimum Detectable Signal to Watts

Given the minimum detectable received signal is -120 dBm, convert it to watts: \[ P_{min} = 10^{\frac{-120}{10}} \text{mW} = 10^{-12} \text{W} \]
03

Use the Radar Range Equation

The radar range equation for a monostatic radar is given by: \[ P_r = \frac{P_t G^2 \lambda^2 \sigma}{(4 \pi)^3 R^4} \] Rearrange the formula to solve for the detection range, R: \[ R = \left(\frac{P_t G^2 \lambda^2 \sigma}{(4 \pi)^3 P_r} \right)^{\frac{1}{4}} \] Substitute the values: \[ P_t = 1000 \text{W}, \, G = 10^3, \, P_r = 10^{-12} \text{W}, \, \sigma = 5 \text{m}^2, \, \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{m/s}}{10 \text{GHz}} = 0.03 \text{m} \] \[ R = \left(\frac{1000 \times (10^3)^2 \times (0.03)^2 \times 5}{(4 \pi)^3 \times 10^{-12}} \right)^{\frac{1}{4}} \]
04

Calculate Detection Range

Calculate \(R^{4}\): \[ R^{4} = \frac{1000 \times (10^6) \times 0.0009 \times 5}{(4 \pi)^3 \times 10^{-12}} \] This simplifies to: \[ R^4 = \frac{4.5 \times 10^{9}}{8.247 \times 10^{-6}} = \frac{4.5 \times 10^{9}}{8.247 \times 10^{-6}} \approx 5.46 \times 10^{14} \] Taking the fourth root: \[ R = (5.46 \times 10^{14})^{\frac{1}{4}} \approx 4.99 \times 10^3 \text{meters} \]

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Key Concepts

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

Radar Cross Section
Radar Cross Section (RCS) is a measure of how detectable an object is by radar. It represents the size of an equivalent perfect reflector that would return the same amount of radar energy back to the radar system as the target does. In simpler terms, RCS measures how big or small a target appears to radar waves. The unit for RCS is square meters (m虏).
The fighter plane in the exercise has an RCS of 5 m虏, which means it reflects radar waves as if it were a perfect sphere with a cross-sectional area of 5 m虏. A larger RCS means the object is easier to detect, while a smaller RCS makes the object harder to detect.
Antenna Gain
Antenna Gain describes how well an antenna can direct or focus energy in a particular direction compared to a theoretical antenna that radiates energy uniformly in all directions (an isotropic antenna). It is usually measured in decibels (dB).
The given antenna gain in the exercise is 30 dB. To use this value in calculations, it's converted to a linear scale using the formula: \( G = 10^{\frac{30}{10}} = 10^3 \). This means the antenna focuses energy 1000 times better than an isotropic antenna.
High antenna gain is crucial for extending the radar's detection range because it allows more focused and effective transmission and reception of radar signals.
Minimum Detectable Signal
The Minimum Detectable Signal (MDS) represents the weakest signal strength that can be reliably detected by the radar system. It is essential because it defines the radar's sensitivity.
In the exercise, the MDS is given as -120 dBm. Decibels relative to a milliwatt (dBm) are a logarithmic measure of power. To use this value in calculations, it must be converted to watts: \( P_{min} = 10^{\frac{-120}{10}} \text{mW} = 10^{-12} \text{W} \).
Knowing the MDS helps in determining how far a radar can detect an object, as lower (more negative) values indicate higher sensitivity.
Radar Range Equation
The Radar Range Equation is a fundamental formula used to determine the maximum range at which a radar system can detect a target. This equation considers various parameters like transmitted power, antenna gain, RCS, and MDS.
For a monostatic radar, the equation is given by: \[ P_r = \frac{P_t G^2 \ \ \lambda^2 \sigma}{(4 \pi)^3 R^4} \], where:
  • \( P_r \) is the received power
  • \( P_t \) is the transmitted power
  • \( G \) is the antenna gain
  • \( \lambda \) is the wavelength
  • \( \sigma \) is the radar cross section
  • \( R \) is the detection range
In the exercise, values are substituted into the rearranged formula to solve for \( R \): \[ R = \/left(\frac{1000 \times (10^{3})^2 \times (0.03)^2 \times 5}{(4 \pi)^3 \times 10^{-12}} \/right)^{\frac{1}{4}} \]This equation helps in calculating the radar detection range precisely.
Unit Conversion in Radar Calculations
Unit conversion is vital in radar calculations to ensure consistency and accuracy. Different parameters might be expressed in various units, requiring careful conversion before plugging them into formulas.
In the exercise, we see multiple conversions:
  • Transmitted power from kW to watts: \( 1 \text{kW} = 1000 \text{W} \)
  • Antenna gain from dB to a linear scale: \( 10^{\frac{30}{10}} = 10^3 \)
  • MDS from dBm to watts: \( 10^{\frac{-120}{10}} = 10^{-12} \text{W} \)
  • Frequency to wavelength: \( \lambda = \frac{c}{f} = \/frac{3 \times 10^{8} \text{m/s}}{10 \/text{GHz}} = 0.03 \text{m} \)
These conversions are essential to maintain uniformity across all calculations, reducing errors and ensuring the correct application of the radar range equation, ultimately giving an accurate detection range.

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Most popular questions from this chapter

GLONASS is the Russian satellite navigation system with one of two open signals called the L1OF band at \(1600.995 \mathrm{MHz}\). The system uses DSSS encoding and BPSK modulation and each GLONASS satellite transmits on a different frequency. The symbol rate is 511,000 chips \(/ \mathrm{s}\), the bandwith of the transmitted signal is approximately \(540 \mathrm{kHz}\), and there are 50 information bits per second. (a) What is the system's processing gain? (b) What is the noise in \(\mathrm{dBm}\) received in the \(540 \mathrm{kHz}\) bandwidth? (c) If the required system minimum effective SNR is \(6 \mathrm{~dB}\), what is the minimum acceptable power, in \(\mathrm{dBm}\), of the received signal? Assume that the receiver is noiseless.

A \(4 \mathrm{kHz}\) bandwidth voice signal is coded by a vocodor as an 8 kbit/s data stream. Coding increases the data stream to \(64 \mathrm{kbit} / \mathrm{s}\). What is the processing gain that can be achieved at the receiver if QPSK modulation is used with a modulation efficiency of 1.4 bit \(/ \mathrm{s} / \mathrm{Hz}\) ?

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 \(10 \mathrm{~km} ?\)

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.

If a received RF signal has an SIR of \(-5 \mathrm{~dB}\) and the processing gain (calculated from bit rates) that can be achieved for the modulation and coding used is \(15 \mathrm{~dB}\), what is the \(E_{b} / N_{0}\) after processing? There are 4 bits per symbol.
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