# Dipole Antenna - Effective Area

1. Apr 29, 2012

### Axis001

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Aeff = G* λ2/4*pi

3. 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.

2. Apr 30, 2012

### rude man

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.)

3. Apr 30, 2012

### Axis001

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.

4. Apr 30, 2012

### rude man

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!