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Lecture 28 Antennas and Radiation and the Hertzian Dipole Maxwell's

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Lecture 28 Antennas and Radiation and the Hertzian Dipole Maxwell's Lecture 28 Antennas and Radiation and the Hertzian Dipole In this lecture you will learn: • Generation of radiation by oscillating charges and currents • Hertzian dipole antenna ECE 303 – Fall 2005 – Farhan Rana – Cornell University Maxwell’s Equations and Radiation r r r r ∂ µ H(rr,t ) ∇. µo H()r ,t = 0 ∇ × E()rr,t = − o ∂t r r r r r r ∂ ε E(rr,t ) ∇. εo E()r ,t = ρ(r ,t ) ∇ × H()rr,t = J ()rr,t + o ∂t Time-varying currents as the “source” or the “driving term” for the wave equation: r r r r ∂ µ ∇ × H(rr,t ) ∂ µ J(rr,t ) 1 ∂2E(rr,t ) ∇ × ∇ × E()rr,t = − o = − o − ∂t ∂t c2 ∂t 2 r r r 1 ∂2E()rr,t ∂ µ J()rr,t ⇒ ∇ × ∇ × E()rr,t + = − o c2 ∂t2 ∂t Maxwell’s equation predict outgoing radiation from sinusoidally time-varying currents (and charges - recall that current and 911 charge densities are related through the continuity equation: r ∂ρ(rr,t ) ∇. J()rr,t + = 0 ∂t ECE 303 – Fall 2005 – Farhan Rana – Cornell University 1 Electro- and Magneto-quasistatics and Potentials Electroquasistatics Magnetoquasistatics r r r r ∇. µ H(rr,t ) = 0 ∇. εo E()r ,t = ρ ()r ,t o r r r ∇ × E()rr,t = 0 ∇ × H(rr,t ) = J(rr,t ) Since: Since: r r r r ∇ × E()r ,t = 0 ∇. µo H(r ,t ) = 0 One could write: One could write: r r r r r r r E()r ,t = −∇φ ()r ,t µo H(r ,t ) = ∇ × A(r ,t ) Scalar potential Vector potential ECE 303 – Fall 2005 – Farhan Rana – Cornell University Electrodynamics and Potentials - I r r r r ∂ µ H(rr,t ) ∇. µo H()r ,t = 0 ∇ × E()rr,t = − o ∂t r r r r r r ∂ ε E(rr,t ) ∇. εo E()r ,t = ρ(r ,t ) ∇ × H()rr,t = J ()rr,t + o ∂t Vector Potential One still has: r r ∇. µo H()r ,t = 0 Therefore, one can still introduce a vector potential: r r r r µo H()r ,t = ∇ × A(r ,t ) Faraday’s law then becomes: r r ∂ µ H(rr,t ) ∇ × E()rr,t = − o ∂t r r ∂ ∇ × A()rr,t ⇒ ∇ × E()rr,t = − ∂t r r ⎡ r r ∂ A()r ,t ⎤ ⇒ ∇ × ⎢E()r ,t + ⎥ = 0 ⎣ ∂t ⎦ ECE 303 – Fall 2005 – Farhan Rana – Cornell University 2 Electrodynamics and Potentials - II Scalar Potential r r ⎡ r r ∂ A()r ,t ⎤ Since: ∇ × ⎢E()r ,t + ⎥ = 0 ⎣ ∂t ⎦ One may introduce a scalar potential as follows: r r ∂ A(rr,t ) E()rr,t + = −∇φ()rr,t ∂t r r ∂ A()rr,t ⇒ E()rr,t = − − ∇φ()rr,t ∂t Using the vector and scalar potentials, the expressions for E-field and H-field become: r r r r µo H()r ,t = ∇ × A(r ,t ) r r ∂ A(rr,t ) E()rr,t = − − ∇φ()rr,t ∂t ECE 303 – Fall 2005 – Farhan Rana – Cornell University Choosing a “Gauge” in Electromagnetism Non-uniqueness of the vector potential The vector potential A is not unique – only the curl of the vector potential is a well defined quantity (i.e. the B-field): r r r r µo H(r ,t ) = ∇ × A(r ,t ) Demonstration: suppose we change the vector potential - such that the new vector potential is the old vector potential plus the gradient of some arbitrary function r r r r r Anew ()r ,t = A(r ,t ) + ∇ψ (r ,t ) Then: 0 r r r r r ∇ × Anew ()r ,t = ∇ × A(r ,t ) + ∇ × ∇ψ (r ,t ) The new vector potential r r r r is just as good as it will ⇒ ∇ × Anew ()r ,t = ∇ × A ()r ,t give the same B-field Making the vector potential unique A vector field can be uniquely specified (up to a constant) by specifying the value of its curl and its divergence To make the vector potential A unique, one needs to fix the value of its divergence – a process that usually goes by the names “gauge fixing” or “fixing the gauge” or “choosing a gauge” r ∇. A(rr,t ) = ?? ECE 303 – Fall 2005 – Farhan Rana – Cornell University 3 Gauges in Electromagnetism Coulomb Gauge: r ∇. A()rr,t = 0 • This gauge is commonly used in electro- and magnetoquasistatics • This gauge is not commonly used in electrodynamics Lorentz Gauge: r 1 ∂φ(rr,t ) Relates divergence of the vector ∇. A()rr,t = − 2 ∂t potential to the time derivative of c the scalar potential • This gauge is commonly used in electrodynamics ECE 303 – Fall 2005 – Farhan Rana – Cornell University Vector Potential Wave Equation Using the vector and scalar potentials, the expressions for E-field and H-field were: r r r r ∂ A(rr,t ) µ H()rr,t = ∇ × A(rr,t ) E()rr,t = − − ∇φ()rr,t o ∂t Ampere’s Law becomes: r r r ∂ ε E(rr,t ) ∇ × H()rr,t = J ()rr,t + o ∂t 2 r r r r r r r 1 ∂ A()r ,t 1 ⎡∂φ()r ,t ⎤ ⇒ ∇ × ∇ × A()r ,t = µoJ ()r ,t − − ∇ c2 ∂t2 c2 ⎣⎢ ∂t ⎦⎥ Remembering that: r r r r r 2 r r r r 1 ∂φ()r ,t ∇ × ∇ × A()r ,t = ∇[∇. A ()r ,t ]− ∇ A(r ,t ) and ∇. A()r ,t = − c2 ∂t We finally get: 2 r r 2 r r 1 ∂ A(r ,t ) r r ∇ A()r ,t − = −µoJ()r ,t c2 ∂t2 This is the vector potential wave equation with current as the driving term ECE 303 – Fall 2005 – Farhan Rana – Cornell University 4 Scalar Potential Wave Equation Using the vector and scalar potentials, the expressions for E-field and H-field were: r r r r ∂ A(rr,t ) µ H()rr,t = ∇ × A(rr,t ) E()rr,t = − − ∇φ()rr,t o ∂t Gauss’ Law becomes: r r r ∇. εo E()r ,t = ρ ()r ,t ∂ ∇. A()r ,t ρ()rr,t ⇒ − ∇2φ()rr,t − = ∂t εo Remembering that: r 1 ∂φ(rr,t ) ∇. A()rr,t = − c2 ∂t We finally get: 2 r r 2 r 1 ∂ φ(r ,t ) ρ(r ,t ) ∇ φ()r ,t − 2 2 = − c ∂t εo This is the scalar potential wave equation with charge as the driving term ECE 303 – Fall 2005 – Farhan Rana – Cornell University Scalar and Vector Potential Wave Equations Given any arbitrary time-dependent charge and current density distributions one can solve these two wave equations to get the potentials: 2 r r 2 r 1 ∂ φ()r ,t ρ(r ,t ) Scalar potential wave equation ∇ φ()r ,t − 2 2 = − c ∂t εo 2 r r 2 r r 1 ∂ A()r ,t r r Vector potential wave equation ∇ A()r ,t − = −µoJ()r ,t c2 ∂t2 And then find the E- and H-fields using: r r r r µo H()r ,t = ∇ × A ()r ,t r r ∂ A()rr,t E()rr,t = − − ∇φ()rr,t 911 ∂t ECE 303 – Fall 2005 – Farhan Rana – Cornell University 5 Superposition Integral Solution of the Scalar Wave Equation We know that Poisson equation in electrostatics: rr − rr' ρ(rr) ∇2φ()rr = − εo rr r Has the solution: r ' ρ(rr') φ()rr = dv' ∫∫∫ r r r 4π εo r − r ' ρ(r ') The wave equation (which looks somewhat similar to the Poisson equation): 2 r r 2 r 1 ∂ φ(r ,t ) ρ(r ,t ) ∇ φ()r ,t − 2 2 = − c ∂t εo Has the solution: Retarded Potential: • The potential at the observation r r r r ρ(r ',t − r − r ' c) point at any time corresponds to the φ()r ,t = ∫∫∫ dv' charge at the source point at an 4π ε rr − rr' o earlier time • Electromagnetic disturbances travel at the speed c ECE 303 – Fall 2005 – Farhan Rana – Cornell University Superposition Integral Solution of the Vector Wave Equation We know that the vector Poisson equation in rr − rr' magnetostatics: 2 r r r r ∇ A()r = −µoJ(r ) rr r Has the solution: r ' r r r r µo J(r ') A()r = ∫∫∫ dv' r 4π rr − rr' J(r ') The wave equation (which looks somewhat similar to the vector Poisson equation): 2 r r 2 r r 1 ∂ A(r ,t ) r r ∇ A()r ,t − = −µoJ()r ,t c2 ∂t2 Retarded Potential: Has the solution: • The potential at the observation r r r r point at any time corresponds to the r r µo J()r ',t − r − r ' c A()r ,t = ∫∫∫ r r dv' current at the source point at an 4π r − r ' earlier time • Electromagnetic disturbances travel at the speed c ECE 303 – Fall 2005 – Farhan Rana – Cornell University 6 Time-Harmonic Fields and Complex Wave Equations r r r r E()rr,t = Re[ E ()rr e j ω t ] H()rr,t = Re[ H ()rr e j ω t ] r r A()rr,t = Re[ A ()rr e j ω t ] φ()rr,t = Re[ φ ()rr e j ω t ] r r r r j ω t ρ()rr,t = Re[]ρ ()rr e j ω t J()r ,t = Re[ J ()r e ] Assuming time harmonic currents, charges, and fields, the wave equations: 2 r r 2 r r 1 ∂ A(r ,t ) r r ∇ A()r ,t − = −µoJ()r ,t c2 ∂t2 2 r r 2 r 1 ∂ φ(r ,t ) ρ(r ,t ) ∇ φ()r ,t − 2 2 = − c ∂t εo become: 2 r r 2 r r r r ∇ A(r ) + k A(r ) = −µoJ(r ) ω k = ρ(rr) c ∇2φ()rr + k 2φ ()rr = − εo ECE 303 – Fall 2005 – Farhan Rana – Cornell University Superposition Integral Solutions of the Complex Wave Equations For time harmonic signals: r r r r j ω t A()r ,t = Re[ A ()r e ] φ()rr,t = Re[ φ ()rr e j ω t ] r r r r j ω t ρ()rr,t = Re[]ρ ()rr e j ω t J()r ,t = Re[ J ()r e ] The solutions to the complex wave equations are found as follows: r r r r r ρ(r ',t − r − r ' c) r ρ(r ') − j k rr−rr' φ()r ,t = ∫∫∫ r r dv' φ()r = ∫∫∫ r r e dv' 4π εo r − r ' 4π εo r − r ' Where we have used: j ω ()t − rr−rr' c − j ω rr−rr' c j ω t ρ()rr',t − rr − rr' c = Re[]ρ()rr' e = Re[ρ()rr' e e ] And: r r r r r µ J()r ',t − r − r ' c r r r r r o r r µo J(r ') − j k r −r ' A()r ,t = ∫∫∫ r r dv' A()r = ∫∫∫ e dv' 4π r − r ' 4π rr − rr' Where we have used: r r j ω ()t − rr−rr' c r − j ω rr−rr' c j ω t J()rr',t − rr − rr' c = Re[]J()rr' e = Re[J()rr' e e ] ECE 303 – Fall 2005 – Farhan Rana – Cornell University 7 Wave Equations and Methods of Solution Suppose we need to find the radiation emitted by some collection of sinusoidally time varying charges and currents 911 Method 1 (1) We can solve these two equations: r 2 r r 2 r r r r 2 r 2 r ρ(r ) ∇ A()r + k A ()r = −µoJ(r ) ∇ φ()r + k φ ()r = − εo (2) And then find the E-field and the H-field through: r r r r r r r r r µo H()r = ∇ × A ()r E(r ) = − jω A(r ) − ∇φ(r ,t ) Method 2 (1) We can solve just one equation: 2 r r 2 r r r r ∇ A()r + k A(r ) = −µoJ(r ) (2) And then find the H-field and the E-field through: r r r r r r r r r r µo H()r = ∇ × A(r ) ∇ × H(r ) = J(r ) + jω εoE(r ) ECE 303 – Fall 2005 – Farhan Rana – Cornell University Hertzian Dipole Antenna - I • A Hertzian dipole is one of the simplest radiating elements for which analytical solutions for the fields can be obtained • A Hertzian dipole consists of two equal and opposite ± charge reservoirs located at a distance d from each other, as shown below: z q(t) d I(t) x -q(t) • The two charge reservoirs are electrically connected and a sinusoidal current I(t) flows between them • Consequently, the charge in the reservoirs also changes sinusoidally: q(t) -q(t) dq(t ) I(t) I()t = dt t ECE 303 – Fall 2005 – Farhan Rana – Cornell University 8 Hertzian Dipole Antenna - II r z Suppose we could write a current density J () rr for the Hertzian dipole q(t) d I(t) Then: x -q(t) r r r r ∫∫∫ J(r ) dv = ∫∫∫ J(r ) dx dy dz = zˆ I d • The integral represents the total “weight” or “strength” of the dipole • If the size of the dipole is much smaller than the wavelength then one may write: r J(rr) = zˆ I d δ (x )δ (y ) δ (z) = zˆ I d δ 3()rr The above expression will give the same strength for the dipole, i.e.
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