The centre of a half-wave dipole is placed at the origin of the standard spherical coordinate system with the wire parallel to the z axis. This centre-fed dipole is in resonance at 0.9 GHz and the total radiated power is 200 W.(a) Calculate the effective height vector in the xy plane (θ = 90°).(b) Find the electric field vector (peak value) radiated by the dipole in the far-field region.(c) A second similar dipole is also placed in the xy plane parallel to the first at a distance of 1500 m. Neglecting losses and assuming the free-space condition, calculate the voltage U (peak value) induced in the dipole, i.e. the voltage in the equivalent receiving circuit.(d) Based on the voltage U, calculate the maximum rms power available at the matched load
Question
The centre of a half-wave dipole is placed at the origin of the standard spherical coordinate system with the wire parallel to the z axis. This centre-fed dipole is in resonance at 0.9 GHz and the total radiated power is 200 W.(a) Calculate the effective height vector in the xy plane (θ = 90°).(b) Find the electric field vector (peak value) radiated by the dipole in the far-field region.(c) A second similar dipole is also placed in the xy plane parallel to the first at a distance of 1500 m. Neglecting losses and assuming the free-space condition, calculate the voltage U (peak value) induced in the dipole, i.e. the voltage in the equivalent receiving circuit.(d) Based on the voltage U, calculate the maximum rms power available at the matched load
Solution
This is a complex problem that involves several steps. Here's how you can solve it:
(a) The effective height of a half-wave dipole is given by the formula h_eff = λ/2π, where λ is the wavelength. The frequency is given as 0.9 GHz, so first we need to convert this to Hz: 0.9 GHz = 0.9 x 10^9 Hz. The speed of light c is approximately 3 x 10^8 m/s. The wavelength λ is given by the formula λ = c/f, so λ = 3 x 10^8 m/s / 0.9 x 10^9 Hz = 0.333 m. Substituting this into the formula for h_eff gives h_eff = 0.333 m / 2π = 0.053 m. The effective height vector in the xy plane is then 0, since the dipole is parallel to the z axis.
(b) The electric field radiated by a dipole in the far-field region is given by the formula E = √(90*P/(r^2)), where P is the total radiated power and r is the distance from the dipole. We are not given a specific distance, so we cannot calculate a specific value for E. However, we can say that E is proportional to the square root of the power and inversely proportional to the distance from the dipole.
(c) The voltage induced in a dipole is given by the formula U = E*h_eff, where E is the electric field and h_eff is the effective height of the dipole. Again, without a specific value for E, we cannot calculate a specific value for U. However, we can say that U is proportional to the product of the electric field and the effective height of the dipole.
(d) The maximum rms power available at the matched load is given by the formula P = U^2 / R, where U is the voltage and R is the resistance of the load. Without specific values for U and R, we cannot calculate a specific value for P. However, we can say that P is proportional to the square of the voltage and inversely proportional to the resistance of the load.
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Time left 0:01:01Question 11Not yet answeredMarked out of 15.00Flag questionTipsQuestion textThe electric and magnetic fields of a short dipole (length L << λ) that is placed at the origin and oriented parallel to the z axis of the standard spherical coordinate system can be calculated in the general case at the point r,θ,φ from the following formulaeE⃗ =Erur+Eθuθ𝐸→=𝐸𝑟𝑢𝑟+𝐸𝜃𝑢𝜃 whereEr=j2ωμILcosθe−jkr4πr(1jkr+1(jkr)2)𝐸𝑟=𝑗2𝜔𝜇𝐼𝐿cos𝜃𝑒−𝑗𝑘𝑟4𝜋𝑟(1𝑗𝑘𝑟+1(𝑗𝑘𝑟)2)Eθ=jωμILsinθe−jkr4πr(1+1jkr+1(jkr)2)𝐸𝜃=𝑗𝜔𝜇𝐼𝐿sin𝜃𝑒−𝑗𝑘𝑟4𝜋𝑟(1+1𝑗𝑘𝑟+1(𝑗𝑘𝑟)2) andH⃗ =Hϕuϕ=uϕjkLIsinθe−jkr4πr(1+1jkr)𝐻→=𝐻𝜙𝑢𝜙=𝑢𝜙𝑗𝑘𝐿𝐼sin𝜃𝑒−𝑗𝑘𝑟4𝜋𝑟(1+1𝑗𝑘𝑟)ω is the angular frequency, I is the amplitude of the electric current in the dipole, L is the length of the dipole, and k = 2π/λ is the wave number.(a) Simplify the expression for the wave impedance of a short dipole η=|E⃗ |/|H⃗ |𝜂=|𝐸→|/|𝐻→| as much as possible when θ = 90°. (Keep in mind part b)(b) Using the software of your choice, plot the wave impedance η as a function of the distance-to-wavelength ratio r/λ along the axis perpendicular to the antenna (θ = 90°). Choose, e.g., 0.05≤r/λ≤3 and 0 Ω≤η≤1000 Ω.(c) Mark on the plot i) the level where η = η0 =377 Ω, ii) the distance r/λ = 1/(2π), and iii) the distance r/λ at which η = 0.997η0.(d) What is the reactance of a short dipole like? Which field (E or H) dominates in the distance range r/λ = 1/(2π)? How is this related to the reactance of a short dipole?(e) What distance can be considered as the outer boundary of the reactive near fields of a short dipole?(f) What is the distance of the far field region of a short dipole?(g) How would you assume that the wave impedance curve changes for a small loop antenna? What is the reactance of a small loop, inductive or capacitive?
The question is in relation to the antenna analysis. Which of the statement below is correct?Select one:a.A half wave dipole antenna has a maximum directive gain of 2.15 dBb.A long antenna is required to reduce radiation resistancec.A half wavelength dipole has a maximum directive gain of 2.15 dBid. Image theory is use to increase antenna radiated power and directivitye.A half-wave length dipole antenna has a radiation resistance of 50 Ohms
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