A 2 m long center-fed dipole antenna operates in the AM broadcast band at 1 MHz. The dipole is made of copper wire with a radius of 1 mm. (a) Determine the radiation efficiency of the antenna. (b) What is the antenna gain in dB

a) the radiation efficiency of the antenna is 0.3018 ≈ 30%

b) the antenna gain in dB is - 3.4 dB

Explanation:

first we find the type of antenna given the data in the question;

given that, frequency f is 1 MHz

wavelength λ = c / f

c is the speed of light ( 3 × 10⁸), f is frequency 1MHz

so  λ = (3 × 10⁸) / 1000000

λ = 300 m

therefore the length is less than λ/50

l/λ = 1/300 < 1/50      {since l = 1m}

so the antenna is Hertzian Dipole

a)

we find the radiation resistance and loss of resistance to get the radiation efficiency.

the antenna has a radiation α is 1 mm and is made of copper.

therefore its permeable

Цc ≈ Ц0 ≈ 4π × 10⁻⁷ H/m and ∝c= 5.8 × 10⁷ S/m.

R_road = 80π² ( l/λ)²

we substitute 2m for l and 300 for λ

R_road = 80π² ( 2/300)²

R_road = 0.0359 ≈ 3.51 × 10⁻² Ω

Next is resistance loss

R_loss = l/2πa √(πfЦc / ∝c)

we substitute; l = 2, f = 1MHz = 1000000, ∝c = 5.8 × 10⁷,  Цc = 4π × 10⁻⁷, α = 0.001

R_loss = 2/2π(0.001) √((π×1000000×4π × 10⁻⁷) / 5.8 × 10⁷ )

=  318.3098 × 0.00026089

= 0.08304 ≈ 8.3 × 10⁻² Ω

Now to get the radiation efficiency, we say;

E = R_road / ( R_road + R_loss)

we substitute

E = 0.0359 / ( 0.0359 + 0.08304 )

E = 0.0359 / 0.11894

E = 0.3018 ≈ 30%

therefore the radiation efficiency of the antenna is 0.3018 ≈ 30%

b)

to find the antenna gain, we say

G = ED

we know that the directivity of a Hertzian Dipole D = 1.5

so we substitute

G = 0.3028 × 1.5

G = 0.4527

to find the gain in decibels;

G_dB = 10log (0.4527) dB

G_dB  = - 3.4 dB

therefore the antenna gain in dB is - 3.4 dB