Radiation from Antennas

Radiation from Antennas Ranga Rodrigo University of Moratuwa November 20, 2008 Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 1 / 32

Summary of Last Week s Lecture Radiation From Antennas Radiation From Antennas We learned that, when we derived the electric magnetic field due to the infinite plane sheet of time-varying, spatially uniform current density, the current sheet gives rise to uniform plane waves radiating away from the sheet to either side of it. The infinite current sheet was an idealized, hypothetical source. We are now in a position to learn the principles of radiation from physical antennas. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 2 / 32

Hertzian Dipole Hertzian Dipole Introduction The Hertzian dipole is an elemental antenna consisting of an infinitesimally long piece of wire carrying an alternating current I(t). To maintain the current flow in the wire we postulate two point changes terminating the wire at its two ends, so that the law of conservation of charge is satisfied. Thus if I(t) = I 0 cos ωt, (1) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 3 / 32

Hertzian Dipole Introduction Then and dq 1 = I(t) = I 0 cos ωt, dt (2a) dq 2 = I(t) = I 0 cos ωt, dt (2b) Q 1 (t) = I 0 sin ωt, ω (3a) Q 2 (t) = I 0 ω sin ωt = Q 1(t). (3b) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 4 / 32

Hertzian Dipole Introduction To determine the electromagnetic field due to the Hertzian dipole, we consider the dipole to be situated at the origin and oriented along the z-axis, in perfect dielectric medium. We shall use an approach based upon the magnetic vector potential for the static case and then extend it to the time-varying current element. For a current element of length dl = dli z situated at the origin carrying a current I, the magnetic field at a point P(r, θ, φ) is given by A = µidl 4πr = µidl 4πr i z. (4) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 5 / 32

Hertzian Dipole Introduction z A θ i θ A i r P A ri r r i θ dl = dli z θ y φ x Figure 1: Magnetic vector potential due to an infinitesimal current element. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 6 / 32

Hertzian Dipole Retarded Potential Retarded Potential If the current in the element is now assumed to be time varying in the manner I = I 0 cos ωt, we might expect the corresponding vector potential to be that in 4 with I replaced by I = I 0 cos ωt. This leads to fields inconsistent with Maxwell s equations. The reason is that time-varying electric and magnetic fields give rise to wave propagation, according to which the effect of the source current at a given time is felt at a distance r from the origin after a time delay of r/v p, where v p is the velocity of propagation of the wave. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 7 / 32

Hertzian Dipole Retarded Potential Conversely, the effect felt at a distance r from the origin at time t is due to the current which existed at the origin at an earlier time (t r/v p ). Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 8 / 32

Hertzian Dipole Retarded Potential Retarded Potential Thus for the time varying current element I 0 dl cos ωti z situated at the origin, the magnetic vector potential is given by A = µi ) 0dl (t 4πr cos ω rvp i z = µi (5) 0dl 4πr cos(ωt βr)i z where we have replaced ω/v p by β, the phase constant. The result given by 5 is known as the retarded magnetic vector potential in view of the phase-lag factor βr contained in it. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 9 / 32

Potential Functions Magnetic Vector Potential Gradient, Laplacian, and the Potential Functions Let s take a brief look at magnetic vector potential and related concepts. For any vector A, A = 0. It then follows from Gauss law for the magnetic field in differential form, B = 0, that the magnetic flux density vector B can be expressed as the curl of another vector A; that is B = A. (6) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 10 / 32

Gradient Potential Functions Gradient, Laplacian, and the Potential Functions The vector A in 6 is known as the magnetic vector potential. Substituting 6 into Faraday s law in differential form, E = B/ t, and rearranging, we then obtain or E + ( A) = 0 t ( E + A ) = 0 (7) t Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 11 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions If the curl of a vector is equal to the null vector, that vector can be expressed as the gradient of a scalar, since the curl of the gradient of a scalar function is identically equal to the null vector. The gradient of a scalar, say, Φ denoted Φ (del Φ) is defined in such a manner that the increment dφ in Φ from a point P to a neighboring point Q is given by dφ = Φ dl (8) where dl is the differential length vector from P to Q. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 12 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions Applying Stokes theorem to the vector Φ and a surface S bounded by closed path C, we then have ( Φ) ds = s = = 0. c c Φ di dφ (9) for any single-valued function Φ. Since 9 holds for an arbitrary S, it follows that Φ = 0. (10) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 13 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions To obtain the expression for the gradient in the Cartesian coordinate system, we write dφ = Φ Φ Φ dx + dy + x y z dz ( Φ = x i x + Φ y i y + Φ ) z i z (dxi x + dyi y + dzi z ) Then comparing with 8, we observe that (11) Φ = Φ x i x + Φ y i y + Φ z i z (12) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 14 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions The gradient of a scalar function Φ at a point is a vector having magnitude equal to the maximum rate of increase of Φ at that point and is directed along the direction of the maximum rate of increase, which is normal to the constant Φ surface passing through that point; that is, Φ = dφ dn i n, (13) where dn is a differential length along i n. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 15 / 32

Potential Functions Electric Scalar Potential Returning now to 7 we write E + A t Gradient, Laplacian, and the Potential Functions = Φ (14) Where we have chosen the scalar to be Φ. Rearranging 14, we obtain E = Φ A t The quantity Φ in 15 is known as the electric scalar potential. (15) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 16 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions Electromagnetic Potentials The electric scalar potential Φ and the magnetic vector potential A are known as the electromagnetic potentials. The electric scalar potential is related to the source charge density ρ, whereas the magnetic vector potential is related to the source current density J. For the time varying case, the two are not independent since the charge and current densities are related through the continuity equation. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 17 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions For a given J it is sufficient to determine A, since B can be found from 6 and then E can be found by using Ampère s circuital law H = J + D/ t. For static fields, that is, for / t = 0, the two potentials are independent. Equation 6 remains unaltered, whereas 15 reduces to E = Φ. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 18 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions To proceed further, we recall that the Maxwell s equations in differential form are given by E = B t H = J + D t D = ρ B = 0 From 16d, we expressed B in the manner (16a) (16b) (16c) (16d) B = A (17) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 19 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions and then from 16a, we obtained E = Φ A t (18) We now substitute 18 and 17 into 16c and 16b, respectively, to obtain ( φ A ) = ρ (19a) t ε A µε ( Φ A ) = µj (19b) t t Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 20 / 32

Potential Functions Laplacian of a Scalar Gradient, Laplacian, and the Potential Functions We now define the Laplacian of a scalar quantity Φ, denoted 2 Φ (del squared Φ) as or 2 Φ = Φ (20) 2 Φ = 2 Φ x 2 + 2 Φ y 2 + 2 Φ z 2 (21) The curl of a vector is a vector. The divergence of a vector is a scalar. The gradient of a scalar is a vector. The Laplacian of a scalar is a scalar. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 21 / 32

Potential Functions Laplacian of a Vector Gradient, Laplacian, and the Potential Functions Next, we define the Laplacian of a vector, denoted 2 A as 2 A = ( A) A (22) Expanding the right side of 22 in Cartesian coordinates and simplifying, we obtain in the Cartesian coordinate system, 2 A = ( 2 A x )i x + ( 2 A y )i y + ( 2 A z )i z (23) Thus in the Cartesian coordinate system, the Laplacian of a vector is a vector whose components are the Laplacians of the corresponding components of A. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 22 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions It should however be cautioned that this simple observation does not hold in the cylindrical and spherical coordinate systems. Using 20 and 22, we now write 19a and 19b as 2 A 2 Φ + t ( A) = ρ ε (24a) ( A + µε Φ ) µε 2 A = µj t t2 (24b) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 23 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions Potential Function Equations Equations 24a and 24b are a pair of coupled differential equations for Φ and A. To uncouple the equations, we make use of a theorem known as Helmholtz s theorem which states that a vector field is completely specified by its curl and divergence. Therefore, since the curl of A is given by 17, we are at liberty to specify the divergence of A. We do this by setting A = µε Φ t which is known as the Lorentz condition. (25) Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 24 / 32

Potential Functions Gradient, Laplacian, and the Potential Functions This uncouples 24a and 24b to give us 2 Φ µε 2 Φ t 2 = ρ ε (26) 2 A µε 2 A = µj (27) t2 These are the differential equations relating the electromagnetic potentials Φ and A to the source charge and current densities ρ and J, respectively. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 25 / 32

Hertzian Dipole Let s get back to our discussion on magnetic vector potential due to the Hertzian dipole 1. To augment the reasoning behind the retarded magnetic vector potential, we consider differential equations for the electromagnetic potentials. 1 Please note that the discussion on magnetic vector potential was included for the sake of completion. There will be no examination questions on the derivation of 26 and 27. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 26 / 32

Hertzian Dipole For the magnetic vector potential, we obtained 2 A µε 2 A = µj (28) t2 which reduces to 2 A z µε 2 A z t 2 = µj z (29) for A = A z i z and J = J z i z. Equation 29 has the form of the wave equation, except in three dimensions and with the source term on the right side. Thus the solution for A z must be of the form of a traveling wave while reducing to the static field case for no time variations. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 27 / 32

Hertzian Dipole Fields due to Hertzian Dipole Fields due to Hertzian Dipole Expressing A in 5 in terms of its components in spherical coordinates, A = µi 0dl cos(ωt βr) (cos θi r sin θi θ ) (30) 4πr Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 28 / 32

Hertzian Dipole Fields due to Hertzian Dipole The magnetic field due to the Hertzian dipole is then given by H = B µ = 1 µ A i r i θ i φ = 1 r 2 sin θ r sin θ r µ = 1 r θ φ µr A r ra θ 0 [ r (ra θ) A ] r θ Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 29 / 32

Hertzian Dipole Fields due to Hertzian Dipole or H = I 0dl sin θ 4π [ cos(ωt βr) r 2 ] β sin(ωt βr) i φ (31) r Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 30 / 32

Hertzian Dipole Fields due to Hertzian Dipole Using Maxwell s curl equation for H with J set equal to zero in view of perfect dielectric medium, we then have E t = 1 ε H i r i θ i φ = 1 r 2 sin θ r sin θ r ε r θ φ 0 0 r sin θh φ 1 = εr 2 sin θ θ (r sin θh 1 θ)i r εr sin θ r (r sin θh φ)i θ Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 31 / 32

Reference Hertzian Dipole Fields due to Hertzian Dipole Nannapaneni Narayana Rao. Elements of Engineering Electromaganetics. Prentice Hall, 4th edition, 1994. Ranga Rodrigo (University of Moratuwa) Radiation from Antennas November 20, 2008 32 / 32