Dipole Antenna - Effective Area

[Axis001]

Homework Statement

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

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

Homework Equations

A_eff = G* λ²/4*pi

The Attempt at a Solution

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

A_physical = 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.

 

[rude man]

Homework Statement

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

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

Homework Equations

A_eff = G* λ²/4*pi

The Attempt at a Solution

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

A_physical = 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.)

 

[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.

 

[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!

 

[DeeNos]

The effective area of a dipole antenna is a measure of its ability to capture electromagnetic energy from an incoming wave. It is defined as the ratio of the power received by the antenna to the power density of the incident wave. In this problem, we are given the lengths of two different dipoles, one with a length of λ/60 and the other with a length of λ/2. We are also given the radius of the wires used for these dipoles.

To calculate the effective area, we can use the equation Aeff = G* λ2/4*pi, where G is the antenna gain and λ is the wavelength of the incoming wave. The gain of a short dipole is 1.5, while the gain of a half-wave dipole is 1.64. Using these values, we can calculate the effective area for each dipole:

Aeff(short dipole) = 1.5 * λ2/4*pi = 0.047L2
Aeff(half wave dipole) = 1.64 * λ2/4*pi = 0.052L2

Next, we can calculate the physical area of each dipole using the given radius of 1 cm for the wires:

Aphysical(short dipole) = π*(a2 + a*L) = π*(0.01 m)2 + (0.01 m)*(λ/60) = 0.000314L
Aphysical(half wave dipole) = π*(a2 + a*L) = π*(0.01 m)2 + (0.01 m)*(λ/2) = 0.0157L

Comparing the effective area to the physical area, we can see that the effective area is much smaller than the physical area for both dipoles. This is because the effective area takes into account the directional characteristics of the antenna, while the physical area is simply the geometric area of the wires.

In conclusion, we have calculated the effective area and physical area for a short dipole and a half-wave dipole, and found that the effective area is significantly smaller than the physical area. This highlights the importance of considering the antenna gain when designing and using antennas.