Engr227 EM Slides 10 Antennae 2024 (4) PDF

title slide Lecture 10 Antennae Electromagnetism (Engr. 227) Lecture 10 Slide 1 Marconi broadcasts across the Channel In 1873 Maxwell proposes the electromagnetic theory of light....

title slide Lecture 10 Antennae Electromagnetism (Engr. 227) Lecture 10 Slide 1 Marconi broadcasts across the Channel In 1873 Maxwell proposes the electromagnetic theory of light. Guglielmo Marconi was the first person to broadcast a message across the English Channel in 1899. Hertz had only demonstrated short range radio links twelve years previously in 1887. In the next few years he created radio links across Europe. He received a Nobel prize for Physics for this achievement in 1909 Electromagnetism (Engr. 227) Lecture 10 Slide 2  Electromagnetic Waves Electromagnetic waves are a solution Maxwell’s equations (see lecture 6) ∂B ∂D ∇⋅ D = ρf ∇ ⋅ B =0 ∇×E = − ∇×H = j+ ∂t ∂t kz =ωt + nπ For an electromagnetic wave the electric and magnetic fields act as sources for each other giving them a variation in space as well as time. In free space the electric and magnetic fields are perpendicular to each other and the direction of propagation. The fields of a wave propagating in the z direction vary in time and space as E Previously in lecture 6 =E ˆi E o cos ( ωt − k z z ) B =−ˆj o cos ( ωt − k z z ) c This makes the wave travel with maxima at k z z =ωt + nπ where n = +/- 0,1,2,3… Electromagnetism (Engr. 227) Lecture 10 Slide 3  Creating Electromagnetic Waves + + + If electric charge changes its location with time or flows between two locations, the electric field associated with that charge (spreading to infinity) cannot change immediately. The effect associated with charge movement cannot travel faster than the speed of light. A oscillatory charge movement gives an oscillating wave travelling in free space at a velocity of c = 3x108 m/s. A single sharp movement of the charge gives a pulse. Electromagnetism (Engr. 227) Lecture 10 Slide 4  Electric Dipoles Charging an object and moving it around a evacuated ring quickly is done in particle accelerators but requires lots of effort. A simpler way of generating electromagnetic radiation is to move positive and negative charge with respect to each other in an electrical circuit. For instance with two oppositely charged spheres connected by wires to an AC generator. Every half cycle the spheres they swap their charge. This is known as a Hertzian electric dipole. An electric dipole radiates perpendicular to the current. Defining q as the maximum charge occurring on one sphere and L as the separation of the centres of the spheres then the strength of a Hertzian dipole (where the separation between the charges is less than a half wavelength) is known as the dipole moment p and is given by p = qL Electromagnetism (Engr. 227) Lecture 10 Slide 5 Electric field pattern of a dipole antenna https://it.wikipedia.org/wiki/Radiazione_di_dipolo_elettrico Electromagnetism (Engr. 227) Lecture 10 Slide 6  Poynting Vector and Radiated Power The power density of an electromagnetic wave is given by the “Poynting Vector, S” which is given as S = E x H Where E is the electric field strength (V/m) and H is the magnetic field strength (A/m), hence the units are Watts/m2 The Poynting vector for a short dipole antenna on a sphere of radius r is worked out as sin θ d 2 p sin θ E θ ( radiation = ) = ω2 p 2 4πεo r c dt 2 4πεo r c 2 z r 2 sin θ dθ dφ µ o sin θ d 2 p µ o sin θ 2 Bφ ( radiation = ) = ωp 4πr c dt 2 4πr c r dθ 1 µo 2  ωp  2 2 θ r =S = E θ Bφ   k sin θ (10.1) µo εo  4 πr  x φ 2π ω k where k is the wavenumber = = λ c r sin θ r sin θ dφ The power flow is then calculated by y integrating over the surface of a sphere Electromagnetism (Engr. 227) Lecture 10 Slide 7  Power Flow The power flow of an antenna can be calculated by integrating the Poynting vector over the surface of a sphere z π r 2 sin θ dθ dφ ∫ 2 P = S.2 πr sin θ dθ (10.2) 0 r dθ 2 2 2 2 2 2 p Z0 ω k I L Z0 k This gives=P = θ r 12π 12π x φ where p ω = q Lω = I L as charge multiplied by angular frequency gives the peak AC current I. r sin θ 2 r sin θ dφ Evaluation of Zo L y then gives P ≅ 40π2 I 2   (10.3) λ (by taking c = 3 × 108) This is independent of radius as the power flow through any closed sphere surrounding the dipole is constant. We can then define a radiation resistance as, 2P R rad = (10.4) I2 Electromagnetism (Engr. 227) Lecture 10 Slide 8  Radiation Resistance The current in the antenna see’s a resistance as the power dissipated in the air (i.e. radiation is emitted). 2 The radiation resistance for a short electric dipole is 2P L R rad= 2= 80π2   (10.5) where I is the peak current. I λ The equivalent circuit for the antenna is, Antenna EM Radiation Z= S R S + jXS Rrad Transmitter Raerial jXaerial Capacitive for the dipole antenna prior to resonance (X is negative) For maximum power transfer, the source should be matched to the antenna and the antenna reactance should be zero. Feed power will be split between Rrad which is transmitted and Rariel which is losses in the antenna. Electromagnetism (Engr. 227) Lecture 10 Slide 9  Dipole capacitance The electric dipole has a capacitance associated with it. This can be calculated as approximately the capacitance of two charged spheres or radius, a, separated by a −1 distance, d. If a