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Millimeter-wave radar generates a narrower beam than conventional microwave radar, making it less vulnerable to antiradar missiles than conventional radar. (a) Calculate the angular width \(2 \theta\) of the central maximum, from first minimum to first minimum, produced by a 220 GHz radar beam emitted by a \(55.0-\) cm-diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric "window.") (b) What is \(2 \theta\) for a more conventional circular antenna that has a diameter of \(2.3\) \(\mathrm{m}\) and emits at wavelength \(1.6 \mathrm{~cm} ?\)

Short Answer

Expert verified
(a) 0.006 radians; (b) 0.017 radians.

Step by step solution

01

Determine Wavelength from Frequency

First, calculate the wavelength of the radar using the speed of light, which is approximately \( c = 3 \times 10^8 \) m/s. Given the frequency \( f = 220 \) GHz for the millimeter-wave radar, the wavelength \( \lambda \) is calculated using the formula:\[ \lambda = \frac{c}{f} \]Substitute \( c = 3 \times 10^8 \) m/s and \( f = 220 \times 10^9 \) Hz:\[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{220 \times 10^9 \text{ Hz}} = \frac{3}{2200} \text{ m} = 0.00136 \text{ m} = 1.36 \text{ mm} \]
02

Use the Angular Width Formula for Millimeter-wave Radar

Using the formula to find the angular width \( 2\theta \) for a circular aperture in terms of the wavelength \( \lambda \) and the antenna diameter \( D \):\[ \theta = 1.22 \frac{\lambda}{D} \]For the first minimum to first minimum:\[ 2\theta = 2 \times 1.22 \frac{\lambda}{D}=2.44\frac{\lambda}{D} \] For the millimeter-wave radar, substitute \( \lambda = 0.00136 \text{ m} \) and \( D = 0.55 \text{ m} \):\[ 2\theta = 2.44 \frac{0.00136}{0.55} = 2.44 \times 0.00247 \approx 0.006 \text{ radians} \]
03

Calculate for the Conventional Antenna

For the conventional radar antenna with diameter \( D = 2.3 \) m and wavelength \( \lambda = 0.016 \) m:Using the formula for \( 2\theta \):\[ 2\theta = 2.44 \frac{\lambda}{D} = 2.44 \frac{0.016}{2.3} \]Calculate:\[ 2\theta = 2.44 \times \frac{0.016}{2.3} = 2.44 \times 0.00696 \approx 0.017 \text{ radians} \]

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Key Concepts

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

Angular Width Calculation
Understanding the angular width calculation is key to determining how focused a radar beam will be. The central maximum refers to the broadest part of the beam, and calculating this allows us to predict the beam's size and directionality. In radar applications, the angular width up to the first minimum on each side is crucial. The formula:
  • For angular width: \[ 2\theta = 2.44 \frac{\lambda}{D} \]
highlights the dependence on both wavelength \(\lambda\) and antenna diameter \(D\).
This means that a smaller wavelength or a larger diameter results in a narrower beam, while the opposite leads to a wider beam.
Radar systems adjust these factors based on their specific needs, allowing for precision in applications like aviation and weather forecasting.
Radar Beam Frequency
The frequency of a radar beam directly impacts its wavelength, which means it also affects the angular width. Millimeter-wave radar operates at very high frequencies, typically in the range of 30 GHz to 300 GHz.
In the given exercise, a frequency of 220 GHz is used.
This high frequency translates into shorter wavelengths, hence the term "millimeter-wave."
  • Higher frequencies (shorter wavelengths) result in more focused beams.
  • Lower frequencies (longer wavelengths) create broader beams.
This is why millimeter-wave radar is less vulnerable to interference and is used for detailed imaging applications, like in autonomous vehicles.
Circular Antenna Diameter
The diameter of the circular antenna plays a significant role in the angular width of the radar beam.
A larger diameter antenna focuses the beam more narrowly by gathering more of the radar signal at a given frequency or wavelength.
  • A 55.0 cm diameter is used in the millimeter-wave radar calculations, providing narrowed focus suitable for precision.
  • A 2.3-meter diameter antenna, like in the conventional setup, results in a wider beam for broader coverage.
Thus, the antenna diameter is a critical parameter engineers adjust according to the application’s requirements.
Increased antenna dimensions significantly impact the performance of the radar system by enhancing its directional properties.
Radar Wavelength
Wavelength is a central element in radar technology, defining how the system identifies distances and objects.
It is inversely related to frequency – as seen in the relationship:
  • \[ \lambda = \frac{c}{f} \]
where \(c\) is the speed of light and \(f\) is the frequency. For the millimeter-wave radar example, the calculated wavelength is 1.36 mm.
This is crucial since shorter wavelengths penetrate less matter but provide more precise measurements.
In comparison, the 1.6 cm wavelength used in the traditional radar system represents a compromise between range and resolution.
Your radar choice of wavelength depends on the target application's need for accuracy and range.
Speed of Light in Calculations
The speed of light is a fundamental constant used in these calculations and it is approximately equal to:
  • \[ 3 \times 10^8 \text{ m/s} \]
This value is crucial in determining the wavelength from the frequency. Understanding this relationship is vital, as radar systems rely on electromagnetic waves to detect objects.
The speed of light helps bridge the gap between the frequency at which a radar operates and the resultant wavelength.
The consistency of this value allows engineers to accurately design radar systems for various operational requirements.
In all radar equations, maintaining this parameter ensures precise calculations and enhances the reliability of radar performance across different conditions.

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Most popular questions from this chapter

The distance between the first and fifth minima of a single-slit diffraction pattern is \(0.35 \mathrm{~mm}\) with the screen \(40 \mathrm{~cm}\) away from the slit, when light of wavelength \(550 \mathrm{~nm}\) is used. (a) Find the slit width. (b) Calculate the angle \(\theta\) of the first diffraction minimum.

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(589 \mathrm{~nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first two diffraction minima on the same side of the central diffraction maximum?

A single slit is illuminated by light of wavelengths \(\lambda_{a}\) and \(\lambda_{b}\), chosen so that the first diffraction minimum of the \(\lambda_{a}\) component coincides with the second minimum of the \(\lambda_{b}\) component. (a) If \(\lambda_{b}=350 \mathrm{~nm}\), what is \(\lambda_{a} ?\) For what order number \(m_{b}\) (if any) does a minimum of the \(\lambda_{b}\) component coincide with the minimum of the \(\lambda_{a}\) component in the order number (b) \(m_{a}=2\) and (c) \(m_{a}=3 ?\)

Two yellow flowers are separated by \(60 \mathrm{~cm}\) along a line perpendicular to your line of sight to the flowers. How far are you from the flowers when they are at the limit of resolution according to the Rayleigh criterion? Assume the light from the flowers has a single wavelength of \(550 \mathrm{~nm}\) and that your pupil has a diameter of \(5.5 \mathrm{~mm}\).

Assume that Rayleigh's criterion gives the limit of resolution of an astronaut's eye looking down on Earth's surface from a typical space shuttle altitude of \(400 \mathrm{~km}\). (a) Under that idealized assumption, estimate the smallest linear width on Earth's surface that the astronaut can resolve. Take the astronaut's pupil diameter to be \(5 \mathrm{~mm}\) and the wavelength of visible light to be \(550 \mathrm{~nm}\). (b) Can the astronaut resolve the Great Wall of China (Fig. 36-40), which is more than \(3000 \mathrm{~km}\) long, 5 to \(10 \mathrm{~m}\) thick at its base, \(4 \mathrm{~m}\) thick at its top, and \(8 \mathrm{~m}\) in height? (c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth's surface?

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