Examples of Antenna Array | Solved Examples based on Parameters of Antenna Array | Engineering Funda

Engineering Funda
27 Apr 202410:04

Summary

TLDRIn this video, the presenter tackles two complex antenna array problems, following up on a previous video where three similar questions were solved. The first problem involves calculating the distance between two half-wave dipole antennas, considering a phase shift and radiation angle. The second problem deals with finding the minimum height for a dipole above a conducting plane to achieve maximum far-field radiation at a specific angle. The presenter explains key concepts, such as phase shifts, array factors, and errors commonly made by students, providing step-by-step solutions for better understanding.

Takeaways

  • 📡 In this video, two interesting questions related to antenna arrays are solved, following a previous video on the same topic.
  • 🧐 In question 4, two half-wave dipole antennas are excited with a 3 MHz frequency, and the phase shift of the origin element is -π/2.
  • ✏️ Key concept: If the reference element has a phase shift, that phase is negative. This is crucial for solving antenna problems.
  • 📐 The distance between the antennas, denoted as 'd,' is derived using the formula for ψ: βd cosθ + α. At the maximum radiation point (θ = 60°), d is λ/2, or 50 meters.
  • 🔍 The wavelength (λ) is calculated using c/f, where c is 3×10^8 and f is 3 MHz, yielding 100 meters.
  • 💡 Question 5 discusses an infinitesimally small dipole (Hertzian dipole) placed at a height (h) above a conducting plane and determining the minimum height (h) for maximum radiation at a 60° angle.
  • 🧲 A Hertzian dipole has an α value of 0, while a dipole of λ/2 would have an α value of π. This is essential for solving the radiation pattern.
  • 🔗 The Array Factor for two elements is given by cos(ψ/2), where ψ = βd cosθ + α. Using this, the problem is solved for maximum value of h.
  • 🔢 The correct minimum value of h for maximum radiation at 60° is λ, with n = 1, meaning h = λ.
  • ⚠️ Common mistake: Students often assume α = π, which leads to incorrect results (h = 0.5λ), but for a Hertzian dipole, α = 0.

Q & A

  • Question 1: What is the primary focus of the video?

    -The primary focus of the video is to solve two interesting questions related to antenna arrays.

  • Question 2: What is the frequency given in question 4?

    -The frequency given in question 4 is 3 MHz.

  • Question 3: What phase shift is observed between the two half-wave dipole antennas in question 4?

    -The phase shift between the two half-wave dipole antennas is given as pi by two (π/2), but the reference element at the origin has a phase shift of -π/2.

  • Question 4: How is the distance 'd' between the two antennas calculated in question 4?

    -The distance 'd' is calculated using the formula psi = βd cos θ + α, where ψ = 0, β = 2π/λ, cos θ = cos 60°, and α = -π/2. After simplifying, the distance is found to be λ/2, which equals 50 meters.

  • Question 5: What concept does the speaker emphasize about the phase shift in antenna arrays?

    -The speaker emphasizes that if the reference element (origin) is given a phase shift, that shift should be treated as negative.

  • Question 6: In question 5, what is the minimum value of 'h' for the maxima in the far-field radiation pattern at 60 degrees?

    -The minimum value of 'h' is equal to the wavelength λ for the maxima in the far-field radiation pattern to occur at 60 degrees.

  • Question 7: What is the role of the array factor in solving question 5?

    -The array factor helps in determining the condition for maxima in the far-field radiation. It is expressed as cos(ψ/2), where ψ = βd cos θ + α, and the maxima occur when the argument of the cosine function is a multiple of π.

  • Question 8: How is the distance 'd' related to the height 'h' in question 5?

    -The distance 'd' between the dipole and its image is equal to 2h, where 'h' is the height of the dipole above the conducting plane.

  • Question 9: What common mistake do students make when solving for the minimum value of 'h' in question 5?

    -Students often mistakenly assign α = π for a Hertzian dipole, but for an infinitesimally small dipole (Hertzian dipole), α should be 0.

  • Question 10: Why is it important to distinguish between different types of dipoles in the problem-solving process?

    -It is important because the value of α depends on the type of dipole. For example, for a Hertzian dipole, α = 0, whereas for a dipole of λ/2, α = π. Incorrect assumptions about the type of dipole can lead to errors in calculations.

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Étiquettes Connexes
Antenna ArrayPhase ShiftRadiation PatternDipole AntennasEngineering ExamsFrequency CalculationCompetitive Exam PrepAntenna TheoryArray FactorHertzian Dipole
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