Dipole Antenna - Effective Area

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Discussion Overview

The discussion revolves around determining the effective area (Aeff) for short dipole antennas with specific lengths (L = λ/60 and λ/2) and comparing Aeff with the physical area. Participants explore the calculations involved and express confusion regarding the lack of a numerical wavelength value needed for a definitive answer.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the formula for effective area as Aeff = G * λ² / (4 * π) and calculates Aeff for both dipole types, expressing confusion over obtaining a numerical answer without a given wavelength.
  • Another participant confirms the formulas are correct but emphasizes the need for an actual wavelength value to compute numerical effective areas.
  • A participant mentions that the effective area of the λ/60 dipole does not significantly depend on L, noting that its gain leads to an effective area close to that of the half-wave dipole, despite the short dipole's low radiation resistance.
  • One participant expresses frustration over the wording of the homework statement, questioning its clarity and correctness.
  • A participant shares their attempt to solve for L using a polynomial equation, resulting in effective and physical area values that are close but seem too small, raising further concerns about the calculations.

Areas of Agreement / Disagreement

Participants generally agree on the formulas used but express disagreement and confusion regarding the lack of a numerical wavelength value and the clarity of the problem statement. The discussion remains unresolved regarding the effective area calculations.

Contextual Notes

The discussion highlights limitations due to missing assumptions, particularly the absence of a numerical wavelength value, which is critical for calculating effective areas. There is also uncertainty regarding the interpretation of the homework statement.

Axis001
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Homework Statement



Determine the effective area (Aeff) for a short dipole with L = λ/60 and λ/2 dipole. If the wires used for dipoles has radii a = 1 cm compare Aeff with the physical area.

G(short dipole) = 1.5
G(half wave dipole) = 1.64

Homework Equations



Aeff = G* λ2/4*pi

The Attempt at a Solution



Aeff = 1.64* λ2/4*pi = L2/pi * 1.64 = 0.052L2

Aphysical = pi*(a2 + a*L) = pi*10-4 + 1*10-2*L

What is confusing me is that I cannot seem to get a numerical answer for this problem even though it is apparently possible. I'm sure I just keep over looking something simple and making this easy problem harder than it has to be.
 
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Axis001 said:

Homework Statement



Determine the effective area (Aeff) for a short dipole with L = λ/60 and λ/2 dipole. If the wires used for dipoles has radii a = 1 cm compare Aeff with the physical area.

G(short dipole) = 1.5
G(half wave dipole) = 1.64

Homework Equations



Aeff = G* λ2/4*pi

The Attempt at a Solution



Aeff = 1.64* λ2/4*pi = L2/pi * 1.64 = 0.052L2

Aphysical = pi*(a2 + a*L) = pi*10-4 + 1*10-2*L

What is confusing me is that I cannot seem to get a numerical answer for this problem even though it is apparently possible. I'm sure I just keep over looking something simple and making this easy problem harder than it has to be.

All your formulas are correct. You need the actual value of λ to get a numerical answer for the effective areas for both antennas.

Interestingly, the actual effective area of the λ/60 dipole is practically speaking not a function of L. Since the gain is 1.5, that makes the effective area not much less than that of the half-wave one! (However, the short dipole has a teeny-tiny radiation resistance, going as (L/λ)2., making it more or les useless.)
 
That is what is baffling me is there is no provided value for wavelength but my professor insists that a numerical value is possible. Since the two areas should be equivalent I set up a polynomial equation with them and got a L of 0.1135 m. But when I solve for the areas I get 4.1077 x 10^-3 m for effective area and 3.9 x 10 -3 m for the physical area. The fact that they are so close makes since but the area values seems far to small.
 
Is the wording exactly what you posted? The wording is not in good English ...
"Determine the effective area (Aeff) for a short dipole with L = λ/60 and λ/2 dipole".
Other than that I'm baffled too!
 

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