Electromagnetic Radiation

Electromagnetic Radiation - NUS/ECE EE4101 Electromagnetic...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 1 Electromagnetic Radiation 1. Radiation Mechanism When electric charges undergo acceleration or deceleration, electromagnetic radiation will be produced. Hence it is the motion of charges (i.e., currents) that is the source of radiation. Yet not all current distributions will produce a strong enough radiation for communication. We will first study some typical current distributions and the radiation fields that they produce.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 2 2. Vector and Scalar Potentials From Maxwell’s fourth equation: 0 = Β ( ) 0 = × A For any vector function A , A B × = So we can write:
Background image of page 2
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 3 () A B E × = = × ω j j From Maxwell’s first equation: Then, ( ) 0 = + × A E j For any scalar function φ , 0 ∇×∇ = So we can write: j + =−∇ EA That is, j = −∇ −
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 4 From Maxwell’s second equation: () 22 1 j ω j ω j j ωω kj ω εφ ω μ με φ με μεφ ×= + ∇×∇× = + −∇ − ∇∇ −∇ = ∇ + ∇+ = + + HJ D AJ A AA J A J A We can further specify the divergence of A according to Lorentz’s gauge as: j ω ⋅= A k ω =
Background image of page 4
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 5 Using Lorentz’s gauge, we have: 22 k μ += AA J Now, the first, the second, and the last Maxwell’s equations have been satisfied. To satisfy the third one, put into the third equation, j φ ω =−∇ − EA () j k ρ ε φω φφ ⋅= ∇⋅ = ∇⋅ −∇ − = D E A
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 6 A and φ are called vector and scalar potentials and they satisfy the following inhomogeneous Helmholtz equations: 22 k k k ρ φφ ε μ ωμε ∇+ = += = AA J Note that each component of A is governed by the same scalar equation as that for . Hence it suffices to solve only one scalar equation, namely, the inhomogeneous Helmholtz equation.
Background image of page 6
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 7 3. Solutions to the Vector and Scalar Potentials Solutions to the vector and scalar potentials are (see Supplementary Notes): () ' ' 1 ' 4 ' 4 jkR v jkR v e 'd v R e v R φ ρ πε μ π = = ∫∫∫ RR AR JR R' R R R = field point R ’ = source point R = | R - R ’|
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Hon Tat Hui Electromagnetic Radiation NUS/ECE EE4101 8 1 1 j μ ωε =∇ × = ∇× HA EH Once the potentials are known, the electric and magnetic fields can be found from:
Background image of page 8