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Radiation Integrals and Auxiliary Potential Functions Ranga Rodrigo June 23, 2010 Lecture notes are fully based on Balanis [?]. Some diagrams and text are directly from the books. Contents 1 The Vector Potential for an Electric Current Source J 4 2 Solution of the Inhomogeneous Vector Potential Equation 6 3 Far-Field Radiation 9 Introduction • In the analysis of radiation problems, the usual procedure is to specify the sources and then require the ﬁelds radiated by the sources. This is in contrast to the synthesis problem where the radiated ﬁelds are speciﬁed, and we are required to determine the sources. • It is a very common practice in the analysis procedure to introduce auxil- iary functions, known as vector potentials, which will aid in the solution of the problems. • The most common vector potential functions are the A (magnetic vector potential) and F (electric vector potential). 1 Integration path 1 Sources J , M Radiated ﬁelds E , H Integration Differentiation path 2 path 2 Vector potentials A, F The Divergence The basic deﬁnition of divergence is S A · dS ∇ · A = lim . ∆v→0 ∆v The expansion of divergence in Cartesian coordinates is ∂A x ∂A y ∂A z ∇· A = + + . ∂x ∂y ∂z The Curl The basic deﬁnition of curl is C A · dl ∇ × A = lim i n. ∆S→0 ∆S max The expansion of curl in Cartesian coordinates is ix iy iz ∂ ∂ ∂ ∇× A = ∂x ∂y ∂z . Ax Ay Az Vector Identities div curl = 0 ∇·∇× A = 0 curl grad = 0 ∇ × ∇φ = 0 2 Maxwell’s Equations in Integral Form d E · dl = − B · d S. (Faraday’s law) C dt S d H · dl = J · dS + D · d S. (Ampère’s circuital law) C S dt S D · dS = ρd v. (Gauss’ law for the electric ﬁeld) S V B · d S = 0. (Gauss’ law for the magnetic ﬁeld) S Auxiliary equation d J · dS = − ρd v. (Law of conservation of charges) S dt V Other relations D = εE , B H = , µ Maxwell’s Equations in Differential Form ∂B ∇×E = − . ∂t ∂D ∇×H = J + . ∂t ∇ · D = ρ. ∇ · B = 0. Auxiliary equation ∂ρ ∇· J = − . ∂t Maxwell’s Equations in Free Space 3 ∂B ∇×E = − . ∂t ∂D ∇×H = . ∂t ∇ · D = 0. ∇ · B = 0. 1 The Vector Potential for an Electric Current Source J • The vector potential A is useful in solving for the EM ﬁeld generated by a given harmonic electric current J . • The magnetic ﬂux B is always solenoidal; that is, ∇ · B = 0. Therefore, it can be represented as the curl of another vector because it obeys the vector identity ∇ · ∇ × A = 0. where A is an arbitrary vector. Thus we deﬁne B A = µH A = ∇ × A. (1) We can substitute this into Maxwell’s curl equation ∇ × E A = − j ωµH A , ∇ × E A = − j ω∇ × A. This can also be written as ∇ × E A + j ω × A = 0. From the vector identity ∇ × (−∇φe ) = 0, it follows that E A + j ωA = −∇φe , or E A = −∇φe − j ωA. 4 The scalar function φe represents an arbitrary electric scalar potential which is a function of position. Taking the curl of both sides of Eq. 1 and using the vector identity ∇ × ∇ × A = ∇(∇ · A) − ∇2 A, reduces it to ∇ × (µH A ) = ∇(∇ · A) − ∇2 A For homogeneous medium, µ∇ × H A = ∇(∇ · A) − ∇2 A Using Maxwell’s equation, ∇ × H A = J + j ω E A , µJ + j ωµ E A = ∇(∇ · A) − ∇2 A Substituting 11, ∇2 A + k 2 A = −µJ + ∇(∇ · A + j ωµ φe ) where k 2 = ω2 µ . In 1 the curl of A was deﬁned. Now we are at liberty to deﬁne the divergence of A, which is independent of its curl. Let ∇ · A = − j ω µφe , which is known as the Lorentz condition. Substituting, ∇2 A + k 2 A = −µJ . In addition, 11 leads to 1 E A = −∇φe − j ωA = − j ωA − j ∇(∇ · A). (2) ωµ Once A is known, H A can be found from Eq. 1 and E A from Eq. 2. E A can just as easily be found from Maxwell’s equation, ∇ × H A = J + j ω E A , with J = 0. It will be shown later how to ﬁnd A in terms of the current density J . It will be a solution to the inhomogeneous Helmholtz equation of Eq. 11. • Although magnetic currents appear to be physically unrealizable, equiv- alent magnetic currents arise when we use the volume or the surface equivalence theorems. • The ﬁelds generated by a harmonic magnetic current in a homogeneous region, with J = 0 but M = 0, must satisfy ∇ · D = 0. 5 • Therefore, E F can be expressed as the curl of the vector potential F by 1 EF = − ∇×F . (3) • By letting ∇ · F = − j ωµ φm (4) we can obtain ∇2 F + k 2 F = − M. Once F is known, E F can be found from Eq. 3 and H F with M = 0. It will be shown later how to ﬁnd F once M is known. It will be a solution to the inhomo- geneous Helmholtz equation of 4. Total Fields Now, we have developed equations that can be used to ﬁnd the electric and magnetic ﬁelds generated by an electric current source J and a magnetic cur- rent source M. The procedure requires that the auxiliary potential functions A and F generated, respectively, by J and M are found ﬁrst. In turn, the cor- responding electric and magnetic ﬁelds are then determined (E A ,H A due to A and E F ,H F due to F ). The total ﬁelds are then obtained by the superposition of the individual ﬁelds due to A and F (J and M). 1 1 1 1 E = E A +EF = − j ωA − j ∇(∇ · A) − ∇ × F = ∇×HA − ∇×F. ωµ jω 1 1 1 1 H = H A + HF = ∇ × A − j ωF − j ∇(∇ · F ) = ∇ × A − ∇×EF . µ ωµ µ j ωµ 2 Solution of the Inhomogeneous Vector Potential Equation The solution of the inhomogeneous vector wave equation is µ e − j kR A= J dv . 4π V R To derive it, let us assume that a source with current density J z , which in the limit is an inﬁnitesimal source, is placed at the origin of a x, y, z coordinate system. 6 z (x, y, z) θ (x , y , z ) φ y x Since the current density is directed along the z-axis (J z ), only an A z com- ponent will exist. Thus we can write ∇2 A + k 2 A = −µJ as ∇2 A z + k 2 A z = −µJ z . At points outside of the inﬁnitesimal source, ∇2 A z + k 2 A z = 0. (5) Point source: A z is not a function of direction (θ and φ) and A z = A z (r ), where r is the radial distance. 1 ∂ ∂A z (r ) ∇2 A z (r ) + k 2 A z (r ) = 2 ∂r r2 + k 2 A z (r ) = 0. r ∂r d 2 A z (r ) 2 d A z (r ) + + k 2 A z (r ) = 0. dr 2 r dr This differential equation has two independent solutions e − j kr A z1 = C 1 : radially outward traveling wave. r e + j kr A z2 = C 2 : radially inward traveling wave. r 7 Therefore, we choose e − j kr A z = A z1 = C 1 . r In the static case (ω = 0, k = 0), this simpliﬁes to C1 Az = . r which is the solution to the wave equation 6 when k = 0. Thus at points outside the source, the time-varying and static solutions differ only by e − j kr , the phase retardation factor. In the presence of the source (J z = 0), and k = 0, the wave equation reduces to ∇2 A z = −µJ z . (6) This equation is recognized to be Poisson’s equation, the most familiar form being that relates the scalar electric potential φ to the electric charge density ρ. ρ ∇2 φ = − . The solution is 1 ρ φ= dv , 4π V r where r is the distance from any point in the charge density to the observation point. Using a similar form µ Jz Az = dv . 4π V r This represents the static solution. The time-varying solution is µ e − j kr Az = Jz dv . 4π V r If we consider the current densities in x- and y-directions (J x , and J y ), µ e − j kr A= J dv . 4π V r If the source is removed from the origin and placed at a position represented by (x , y , z ), µ e − j kR A(x, y, z) = J (x , y , z ) dv . 4π V R 8 where (x, y, z) is the observation point and R is the distance from any point on the source to the observation point. In a similar fashion we can obtain e − j kR F (x, y, z) = M(x , y , z ) dv . 4π V R If J and M represent linear densities, µ e − j kR A(x, y, z) = J s (x , y , z ) ds . 4π S R e − j kR F (x, y, z) = M s (x , y , z ) ds . 4π S R For electric and magnetic current I e and I m µ e − j kR A(x, y, z) = I e (x , y , z ) dl . 4π C R e − j kR F (x, y, z) = I m (x , y , z ) dl . 4π C R 3 Far-Field Radiation The far-ﬁelds radiated by antennas of ﬁnite dimensions are spherical waves. For these radiators, a general solution to the vector wave equation is spherical components is A = a r A r (r, θ, φ) + a θ A θ (r, θ, φ) + a φ A φ (r, θ, φ). ˆ ˆ ˆ The amplitude variations of r in each component are of the form 1/r n . Neglect- ing the higher order terms, for large r , e − j kr A a r A r (r, θ, φ) + a θ A θ (r, θ, φ) + a φ A φ (r, θ, φ) ˆ ˆ ˆ . r Consider 1 E A = −∇φe − j ωA = − j ωA − j ∇(∇ · A). ωµ If we evaluate the second term in spherical coordinates, we will notice that higher powers of 1/r are produced. Neglecting them, EA − j ωA for θ and φ components. 9 Neglecting the higher order terms of 1/r n , the radiated E - and H -ﬁelds have only θ and φ components. They can be expressed as Er 0, Eθ − j ωA θ , Eφ − j ωA φ . Hr 0, ω Eθ Hθ +j Aφ = − , η η ω Eφ Hφ − j Aθ = + . η η Radial far-ﬁeld terms exist only for higher order terms of 1/r n . In a similar manner, the far-zone ﬁelds due to magnetic source M (potential F )can be written as Hr 0, Hθ − j ωF θ , Hφ − j ωF φ . Er 0, Eθ − j ωηF φ = ηHφ , Eφ + j ωηF θ = −ηHθ The corresponding far-zone E - and H -ﬁeld components are orthogonal to each other and form TEM (to r ) mode ﬁelds. 10