1 / 11

# Fundamental Concepts (2 sessions) - PowerPoint PPT Presentation

Fundamental Concepts (2 sessions) . Review of Electromagnetic Theory. Maxwell’s Equations: Constitutive Relations:. is magnetic conductive current density (in volts/square meter). Boundary Conditions: Constitutive parameters are σ, ε , μ .

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Fundamental Concepts (2 sessions) ' - lena

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

• Maxwell’s Equations:

• Constitutive Relations:

is magnetic conductive current density (in volts/square meter)

• Boundary Conditions:

• Constitutive parameters are σ, ε,μ.

• Linear Medium: σ, ε,μ are independent of E and H.

• Homogeneous Medium: σ, ε,μ are not functions of space variables or.

• IsotropicMedium: σ, ε,μ are independent of direction (scalars).

is magnetic resistivity

• When a medium is source-free: J = 0, ρv = 0

• In practice, only two of Maxwell’s equations are used:

• Since other two are implied.

• Also, in practice, it is sufficient to make tangential components of fields satisfy necessary boundary conditions.

• Since normal components implicitly satisfy their corresponding boundary conditions.

• Wave Equations:

• Altogether there are six scalar equations for Ex, Ey, Ez, Hx, Hy, Hz the form of:

• Time-varying Potentials:

• Time-Harmonic Fields:

• Source-free wave equation in phasor representation:

• General wave equation in phasor representation:

• Special Case 1: Poisson’s equation for static case (ω = 0):

• Special Case 2: Laplace’s equation for static case and source-free:

• Classification of EM Problems:

• ThisClassification help to answer the question of “What method is best for solving a problem”.

• Three independent items define a problem uniquely:

• (1)the solution region (problem domain) of the problemas R:

• (2) the nature of the equation describing the problem,

• (3) the associated boundary conditions as S.

• Classification of Solution Regions:

• There are two classifications:

• Solution region R is interior (inner, closed, or bounded)

• Solution region R is exterior (outer, open, or unbounded)

• If part or all of S is at infinity, R is exterior otherwise R is interior.

• For example, wave propagation in a waveguide is an interior problem.

• For example, wave propagation in free space (scattering of EM waves by raindrops, and radiation from a dipole antenna) are exterior problems.

• Solution region R could be linear,homogeneousandisotropic.

Ris the solution region

Sis the boundary condition

• Classification of Differential Equations:

• EM problems are classified in terms of equations describing them.

• Equations could be differential or integral or both defined as:

• For example:

• Another example:

• A second-order partial differential equation (PDE):

• or simply:

• PDE operator:

• In non-linear PDEs, coefficients are function of quantity

• Any linear second-order PDE can be classified as elliptic, hyperbolic, or parabolic:

• An elliptic PDE usually models an interior problem such as:

• A Hyperbolic PDE usually models an exterior problem as:

• A ParabolicPDE usually models an exterior problem such as diffusion (or heat) equation:

Laplace’s equation:

Poisson’s equations:

Elliptic problem

parabolic, or hyperbolic problem

• Nondeterministic Problems:

• Previous problems are deterministic, since quantity of interest can be determined directly.

• Another type of problem where quantity is found indirectly is called nondeterministicor eigenvalue.

• StandardEigen problem is of the form of:

• A more general version is generalized Eigen-problem having the form of:

• Only some particular values of λcalled eigenvaluesare permissible.

• Eigen-problems are usually encountered in vibration and waveguide problems.

• In these problems eigenvalues λcorrespond to physical quantities such as resonance and cutoff frequencies.

Where source term has been replaced by λ

Where M, like L, is a linear operator

• Classification of Boundary Conditions:

• Usually boundary conditions are of the Dirichlet and Neumann types.

• Dirichlet boundary condition:

• A good example is the charged metal plate.

• Because all points on a metal are at same potential, a metal plate can readily be modeled by a region of points with some fixed voltage.

• Neumannboundary condition:

• Mixed boundary condition:

• These conditions are called homogeneous boundary conditions.

• General ones are inhomogeneous:

• Dirichlet:

• Neumann:

• Mixed:

i.e., the normal derivative of vanishes on S

h(r) is a known function

• Some Important Theorems:

• Superposition Principle:

• If each member of a set of functions φn , n=1,2,…,Nis a solution to PDE:

• Then a linear combination of them also satisfies the PDE as:

• Uniqueness Theorem:

• This guarantees that solution a PDE with some prescribed boundary conditions is only one possible.

• If a set of fields (E,H) is found which satisfies simultaneously Maxwell’s equations and prescribed boundary conditions, this set is unique.

• Therefore, a field is uniquely specified by sources (ρv,J) within medium plus tangential components of E or H over boundary.

• To prove uniqueness theorem, suppose there exist two solutions:

• Uniqueness Theorem (cont.):

• We denote the difference of the two fields as:

• These must satisfy the source-free Maxwell's equations:

• Dotting both sides with ΔEgives:

• Integrating over volume:

• Therefore ΔE and ΔH satisfy the Poynting theorem just as E1,2and H1,2

• Only Etand Htcontribute to surface integral on the left side.

• Therefore, if E1tand E2tor H1tand H2t are equal over S, ΔEtand ΔHtvanish on S.

• Consequently, surface integral is identically zero, and hence right side must vanish also.

• It follows that ΔE=0 due to second integral on right side and hence also ΔH=0 throughout the volume.

• Thus E1=E2 and H1=H2, confirming that the solution is unique.

Using: