Respuesta :
Answer:
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
The radiation efficiency of the antenna is 41.8 %. The antenna gain is -1.630 dB
In the field of antennas, the term "efficiency" is frequently used, which is the measurement of the electrical efficiency by which any radio antenna transforms the power of radio-frequency received at its ends into radiated power.
From the given parameters;
- The length of the antenna (l) = 2 m
- The frequency of the antenna = 1.0 × 10⁶ Hz
- The radius of the wire = 1.0 × 10 ³ m
- The radiated power [tex]\mathbf{P_r = 80 \ W}[/tex]
Using the relation of the wavelength in order to be able to calculate the radiation resistance, we have:
[tex]\mathbf{wavelength \ \lambda = \dfrac{c}{f}}[/tex]
[tex]\mathbf{wavelength \ \lambda = \dfrac{3\times 10^8 }{1 \times 10^6}}[/tex]
[tex]\mathbf{wavelength \ \lambda = 300 \ m}[/tex]
Now, the radiation resistance [tex]\mathbf{R_r}[/tex] is expressed as:
[tex]\mathbf{R_r = 80 \pi ^2 ( \dfrac{l}{\lambda})^2}[/tex]
[tex]\mathbf{R_r = 80 \pi ^2 ( \dfrac{2}{300})^2}[/tex]
[tex]\mathbf{R_r =35.09 \ m \Omega }[/tex]
The loss resistance [tex]\mathbf{R_l = \dfrac{1}{2\pi r } \sqrt{\dfrac{\pi F \mu __C}{\sigma _C} }}[/tex]
where:
- the magnetic constant = [tex]\mathbf{ \mu __C}[/tex]
- the conductivity of the annealed copper = [tex]\mathbf{\sigma _C}[/tex]
∴
[tex]\mathbf{R_l = \dfrac{1}{2\pi \times 10^{-3} } \sqrt{\dfrac{\pi \times (1 \times 10^6) \times (4 \pi \times 10^{-7 })}{ 5.8 \times 10^{-7}} }}[/tex]
[tex]\mathbf{R_l =41.52 \ m \Omega}[/tex]
Therefore, the radiation efficiency is the ratio of the radiation resistance by the total amount of both the radiation resistance and loss.
∴
[tex]\mathbf{\varepsilon = \dfrac{R_r}{R_r+ R_l}}[/tex]
[tex]\mathbf{\varepsilon = \dfrac{35.09}{35.09+ 41.52}}[/tex]
[tex]\mathbf{\varepsilon = 0.4580}[/tex]
[tex]\mathbf{\varepsilon = 45.8\%}[/tex]
Recall that:
- For short dipole, the antenna directivity D is usually 1.5
∴
The antenna gain G is
- G = ξ × D
- G = 0.458 × 1.5
- G = 0.687
Converting the antenna to decibel (dB), we have:
- = 10 log (0.687)
- = -1.630 dB
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