Consider a half-wave dipole that is radiating into free space when the origin is at the centre of the dipole and its axis is aligned with the z axis. The input power fed to the dipole is 300 W. If we assume that the dipole has an efficiency of 75%, calculate the power density at point P(r,θ,φ) of the spherical coordinate system with r = 300 m, θ=75° and φ=45°.
Question
Consider a half-wave dipole that is radiating into free space when the origin is at the centre of the dipole and its axis is aligned with the z axis. The input power fed to the dipole is 300 W. If we assume that the dipole has an efficiency of 75%, calculate the power density at point P(r,θ,φ) of the spherical coordinate system with r = 300 m, θ=75° and φ=45°.
Solution
To solve this problem, we need to use the formula for the power density (S) of a half-wave dipole antenna, which is given by:
S = (P * η * (cos^2(θ/2)) / (4 * π * r^2))
where:
- P is the input power fed to the dipole (300 W),
- η is the efficiency of the dipole (75% or 0.75),
- θ is the angle from the axis of the dipole (75°),
- r is the distance from the center of the dipole (300 m).
First, we need to convert the angle θ from degrees to radians because the trigonometric functions in the formula use radians. We can do this using the conversion factor π/180:
θ = 75° * π/180 = 1.309 radians
Then, we can substitute the values into the formula:
S = (300 W * 0.75 * (cos^2(1.309/2)) / (4 * π * (300 m)^2))
Solving this equation will give us the power density at the point P in W/m^2.
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