Electromagnetic scattering. Graduate Course Electrical Engineering (Communications) 1 st Semester, Sharif University of Technology

Similar documents
U V. r In Uniform Field the Potential Difference is V Ed

P235 Midterm Examination Prof. Cline

Ferromagnetism. So that once magnetized the material will stay that way even in the absence of external current it is a permanent magnet.

Scattering and bound states

Dr. Naser Abu-Zaid; Lecture notes in electromagnetic theory 1; Referenced to Engineering electromagnetics by Hayt, 8 th edition 2012; Text Book

ma x = -bv x + F rod.

Lecture #8-3 Oscillations, Simple Harmonic Motion

13 Harmonic oscillator revisited: Dirac s approach and introduction to Second Quantization

The accelerated expansion of the universe is explained by quantum field theory.

An Approximate Model for the Theoretical Prediction of the Velocity Increase in the Intermediate Ballistics Period

In this chapter we will start the discussion on wave phenomena. We will study the following topics:

SOLUTIONS. PROBLEM 1. The Hamiltonian of the particle in the gravitational field can be written as, x 0, + U(x), U(x) =

Problem Set 2. Chapter 1 Numerical:

Four-vector, Dirac spinor representation and Lorentz Transformations

Now multiply the left-hand-side by ω and the right-hand side by dδ/dt (recall ω= dδ/dt) to get:

Fourier Series Summary (From Salivahanan et al, 2002)

2 Q 10. Likewise, in case of multiple particles, the corresponding density in 2 must be averaged over all

Scattering by a Blade on a Metallic Plane

Department of Physics Preliminary Exam January 3 6, 2006

Reading from Young & Freedman: For this topic, read the introduction to chapter 25 and sections 25.1 to 25.3 & 25.6.

Problem T1. Main sequence stars (11 points)

A GENERAL FORM FOR THE ELECTRIC FIELD LINES EQUATION CONCERNING AN AXIALLY SYMMETRIC CONTINUOUS CHARGE DISTRIBUTION

Dispersion. February 12, 2014

Lecture 8 Symmetries, conserved quantities, and the labeling of states Angular Momentum

3.8 Three Types of Convergence

PH 221-2A Fall Waves - I. Lectures Chapter 16 (Halliday/Resnick/Walker, Fundamentals of Physics 9 th edition)

Lecture 16: Scattering States and the Step Potential. 1 The Step Potential 1. 4 Wavepackets in the step potential 6

ME Machine Design I. FINAL EXAM. OPEN BOOK AND CLOSED NOTES. Friday, May 8th, 2009

( ') ( ) 3. Magnetostatic Field Introduction

Mechanics Physics 151

Lectures 8 & 9: The Z-transform.

A simple model for Jeans instability in the presence of a constant magnetic field arxiv:astro-ph/ v1 12 Feb 2004

72. (30.2) Interaction between two parallel current carrying wires.

lecture 37: Linear Multistep Methods: Absolute Stability, Part I lecture 38: Linear Multistep Methods: Absolute Stability, Part II

Seismic Analysis of Structures by TK Dutta, Civil Department, IIT Delhi, New Delhi.

Announcement. Grader s name: Qian Qi. Office number: Phys Office hours: Thursday 4:00-5:00pm in Room 134

Singularity Extraction for Reflected Sommerfeld Integrals over Multilayered Media

Some Perspective. Forces and Newton s Laws

2. A crack which is oblique (Swedish sned ) with respect to the xy coordinate system is to be analysed. TMHL

Part A Here, the velocity is at an angle of 45 degrees to the x-axis toward the z-axis. The velocity is then given in component form as.

Potential Energy 4/7/11. Lecture 22: Chapter 11 Gravity Lecture 2. Gravitational potential energy. Total energy. Apollo 14, at lift-off

CS Lecture 13. More Maximum Likelihood

On the Radiation from a Short Current-Carrying Straight Wire Oriented Perpendicular to a Stratified Medium

CHAPTER 4 TWO STANDARD SHORTCUTS USED TO TRANSFORM ELECTROMAGNETIC EQUATIONS 4.1 THE FREE-PARAMETER METHOD

Research and Experiments on Electromagnetic Field Induced by Two. Coaxial Solenoid Coils of Axially Mag-lev Driving Needle

Moment of Inertia. Terminology. Definitions Moment of inertia of a body with mass, m, about the x axis: Transfer Theorem - 1. ( )dm. = y 2 + z 2.

N-Point. DFTs of Two Length-N Real Sequences

Magnetorheological Fluid for Levitation Migration Technology

Pearson Physics Level 30 Unit VI Forces and Fields: Chapter 12 Solutions

Plasma-Wall Interaction: Sheath and Pre-sheath

The Weierstrass Approximation Theorem

Supporting Information for Supression of Auger Processes in Confined Structures

PHYSICS 110A : CLASSICAL MECHANICS MIDTERM EXAM #2

NB1140: Physics 1A - Classical mechanics and Thermodynamics Problem set 2 - Forces and energy Week 2: November 2016

Physics 201, Lecture 15

Causality and the Kramers Kronig relations

- 3 - A COUSTICS. k=ω /c the wavenumber in m - ¹ c = the speed of sound in the medium in ms - ¹

Water a) 48 o b) 53 o c) 41.5 o d) 44 o. Glass. PHYSICS 223 Exam-2 NAME II III IV

Generalized r-modes of the Maclaurin spheroids

Motion of Charges in Uniform E

Quantum Ground States as Equilibrium Particle Vacuum Interaction States

Electromagnetic Waves

Work, Energy and Momentum

Math 1600A Lecture 3, Section 002

Lecture 12: Waves in periodic structures

The Transactional Nature of Quantum Information

Unification of Electromagnetism and Gravitation. Raymond J. Beach

We last left off by talking about how the area under a force vs. time curve is impulse.

PHYS 102 Previous Exam Problems

Analytical solution for the electric potential in arbitrary anisotropic layered media applying the set of Hankel transforms of integer order

IN A SENSE, every material is a composite, even if the

THE ROCKET EXPERIMENT 1. «Homogenous» gravitational field

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.323: Relativistic Quantum Field Theory I

Exam 3 Solutions. 1. Which of the following statements is true about the LR circuit shown?

Electromagnetics I Exam No. 3 December 1, 2003 Solution

Kernel Methods and Support Vector Machines

which is the moment of inertia mm -- the center of mass is given by: m11 r m2r 2

e = n 1 ( ) 3 [ m 3] = n [ m 3] n

Momentum. February 15, Table of Contents. Momentum Defined. Momentum Defined. p =mv. SI Unit for Momentum. Momentum is a Vector Quantity.

PERIODIC STEADY STATE ANALYSIS, EFFECTIVE VALUE,

Massachusetts Institute of Technology Quantum Mechanics I (8.04) Spring 2005 Solutions to Problem Set 4

Pattern Recognition and Machine Learning. Artificial Neural networks

Data-Driven Imaging in Anisotropic Media

Physics 139B Solutions to Homework Set 3 Fall 2009

Donald Fussell. October 28, Computer Science Department The University of Texas at Austin. Point Masses and Force Fields.

Name: Partner(s): Date: Angular Momentum

Figure 1: Equivalent electric (RC) circuit of a neurons membrane

Optical Properties of Plasmas of High-Z Elements

i ij j ( ) sin cos x y z x x x interchangeably.)

7. Renormalization and universality in pionless EFT

Effects of an Inhomogeneous Magnetic Field (E =0)

Physically Based Modeling CS Notes Spring 1997 Particle Collision and Contact

Measuring orbital angular momentum superpositions of light by mode transformation

Newton's Laws. Lecture 2 Key Concepts. Newtonian mechanics and relation to Kepler's laws The Virial Theorem Tidal forces Collision physics

Chapter VI: Motion in the 2-D Plane

Q ESTIMATION WITHIN A FORMATION PROGRAM q_estimation

P032 3D Seismic Diffraction Modeling in Multilayered Media in Terms of Surface Integrals

a a a a a a a m a b a b

ANALYSIS OF REFLECTOR AND HORN ANTENNAS USING MULTILEVEL FAST MULTIPOLE ALGORITHM

General Properties of Radiation Detectors Supplements

Transcription:

Electroagnetic scattering Graduate Course Electrical Engineering (Counications) 1 st Seester, 1388-1389 Sharif University of Technology

Contents of lecture 5 Contents of lecture 5: Scattering fro a conductive wedge The scalar wave equation The vector solutions Introducing a source Solving the equations The scattering proble Special cases Scattering fro a conductive wedge 2

Introduction An exaple of another canonical, exactly solvable scattering proble is that of an infinite, perfectly conducting wedge The wedge is assued to have an angle of Note that the wedge is infinitely extended and infinitely long Also, an iportant liiting case is when 0. (Why?) k i Scattering fro a conductive wedge 3

Scalar equation This tie the syste has no rotational syetry Nonetheless, the natural waves of the syste can still be found by solution of the wave equation in cylindrical coordinates As before, first consider the scalar wave equation k 2 2 0 y 2 2 1 1 2 k 0 2 2 2 In cylindrical coordinates x Scattering fro a conductive wedge 4

Scalar equation The solution should be found for the angles 0 2 Let us consider solutions of the type sin, k (,, ) f, ( )exp k jk cos Where the radial function f satisfies f, k 2 2 2 k f f, ( ) AJ k BY k k, k 0 k k k 2 2 Scattering fro a conductive wedge 5

Vector solution For the oent we do not ipose any boundary conditions Solutions of the vector wave equation can once again be found: M 1, k ˆ, k ˆ ˆ, k, k k k k N 1 jk ˆ ˆ k, k, k 2, k jk 2 k, k ˆ These vectors can be used to construct the electric field in the region outside the wedge Scattering fro a conductive wedge 6

Vector solution But, which kind of function should we choose to satisfy the boundary conditions? For the M vector to be a legitiate electric field its radial coponent should vanish on the wedge surfaces for all, k y 0 for 0,2 jk, f, ( )exp cos k k 0 2 2 x Scattering fro a conductive wedge 7

Vector solution What about the N vector? Now we should deand for all, k, k 0 for 0, 2 y jk, f, ( )exp sin k k 2 0 x 2 Scattering fro a conductive wedge 8

Vector solution solution (horiontal polariation): E, k ˆ, k M, k k k ˆ case 1 j H E N j, k j jk 2 k, k jk, k ˆ 2 ˆ k, k ˆ No electric field along, but agnetic field has a -coponent Scattering fro a conductive wedge 9

Vector solution solution (vertical polariation) 1, k jk, k ˆ 2 E N ˆ ˆ, k jk 2 k, k k 1 j H E M j j, k ˆ, k ˆ k k, k case No agnetic field along, but electric field has a coponent Scattering fro a conductive wedge 10

Vector solution Here, unlike the proble of a conducting cylinder, the solution extends to =0. To ensure the finiteness of the electroagnetic energy (not necessarily field) near the origin we should have f, ( ) J k k y, J ( k )exp jk cos k, J ( k )exp jk sin k 0 x Scattering fro a conductive wedge 11

Vector solution The total electric field which satisfies boundary conditions at surfaces of conducting wedge y and ensures correct behavior k i as 0 Is a superposition of these and vector solutions x Reeber: this representation is for the total field; for the scattering proble we cannot separate the incident and scattered waves in this representation Scattering fro a conductive wedge 12

Vector solutions in the scattering proble k i, If the incident wave has a certain, then we know fro the unifority of the syste in the direction that the total solution consists of waves with k k k k k k 2 2 i, i, i, Then, for exaple for the case, the total field can look like E, k i,, k ˆ i, a k k ˆ i, J ( k )cos exp jk, k i, i, Scattering fro a conductive wedge 13

Vector solutions in the scattering proble Or: E k i, a J ki, k ( )cos J ( ki, ) sin k ˆ ˆ exp jki, Note: electric field is not necessarily finite as 0, it is electric energy density which should reain integrable Scattering fro a conductive wedge 14

Vector solutions in the scattering proble For case or vertical polariation E a jk k k 1, ki, jk, ki, ˆ 2 ˆ 2 i,, k i, ˆ i, J ( k )sin exp jk, k i, i, E ki, ki, jki, a j J( k 2 i, )sin J ( k 2 i, )cos k k k 2 i, 2 k ˆ J ( ki, )sin exp jki, ˆ ˆ Scattering fro a conductive wedge 15

Vector solutions in the scattering proble So far, it is fine, but unfortunately it is not so easy to find the constant coefficients a to describe the scattering solution for a particular incident wave propagating in a certain direction For a cylinder we could separate the incident and scattered field fro the beginning and then apply the boundary condition on the cylinder surface Also, it was essential to expand the incident plane wave in Bessel functions in order to be able to solve the proble. This is not possible here because of the non-integer Bessel functions used. Scattering fro a conductive wedge 16

Introducing a source The traditional way to treat this proble is by solving the proble in y Current line source presence of the sources Once the proble is solved, the 0 sources is oved to a point infinitely far away. The field generated by such a source is a far one field and 0 x has a plane-wave character. Therefore by taking this liit of the full solution the proble is solved. Scattering fro a conductive wedge 17

Introducing a source Let us consider this proble in the sipler case where k = 0. So, we are interested in waves which are unifor along. Fro the field expansion discussed, we expect the total electric field vector to look like ( ) a J ( k )cos J k E sin k ˆ ˆ E ˆ a J ( k)sin Here we used ki, 0 ki, k Scattering fro a conductive wedge 18

Introducing a source Let us return to Maxwell equations, with electric and agnetic line current sources along, and unifor along 0 y 0 E j H M H je J i i Magnetic current x i i i i J J ˆ M M ˆ Scattering fro a conductive wedge 19

Introducing a source With Maxwell equations in cylindrical coordinates, and fields unifor in -direction: E j H H j E E j H H j E E i E j H M H i H je J Scattering fro a conductive wedge 20

Introducing a source Two separate sets for horiontal and vertical polariations Vertical electric field Horiontal electric field E j H H j E E j H H j E H i H je J E i E j H M Horiontal agnetic field Vertical agnetic field Scattering fro a conductive wedge 21

Introducing a source 1 st set leads to: E E 2 2 i k E 2 2 jj 2 nd set gives: H H 2 2 i k H 2 2 jm Left hand side sae as the scalar wave equation in cylindrical coordinates for unifor fields along Scattering fro a conductive wedge 22

Introducing a source We would like to solve these equations in presence of the conducting wedge, for infinitely thin line sources I i 0 J 0 0 0 y M i 0 M 0 0 0 Boundary conditions: 0 0 x E E 0 for 0 E E 0 for 2 E E Scattering fro a conductive wedge 23

Solving the equations () Consider the case first. Solution in the region 0 which 0, 0 y satisfies boundary conditions at the wedge surfaces, and shows the right behavior at the origin is 0 x, ( )sin E a J k 2 Scattering fro a conductive wedge 24

Solving the equations () y Solution in the region which satisfies the wedge boundary conditions, and the radiation condition at the infinity is 0 0, 0 x (2), ( )sin E b H k 0 2 Scattering fro a conductive wedge 25

Solving the equations () y Next we have to atch the solutions at the source by deanding that 0 0, 0 0 x E 0, E 0, E j 0 E I0 0 0 0 Scattering fro a conductive wedge 26

Solving the equations () Resulting equations ( is shown explicitly) (2) b H k 0 a J k 0 1 1 ( )sin ( )sin (2) b H k 0 a J k 0 1 1 ( )sin ( )sin 2 j I 0 0 k0 Scattering fro a conductive wedge 27

Solving the equations () (2) b H k0 a J k 0 ( ) ( ) 2 j I b H ( k ) a J ( k ) sin (2) 0 0 0 0 2 k0 a I0 2 j H k 2 0 (2) W ( k ) 0 sin 0 W J ( k ) H ( k ) J ( k ) H ( k ) (2) (2) 0 0 0 0 2 j j J ( k0) Y ( k 0) J ( k 0) Y ( k 0) k 0 Scattering fro a conductive wedge 28

Solving the equations () Solution for 0 a I H ( k )sin 2 (2) 0 0 0 E I H k J k 0 0 0 2 0 (2), ( )sin ( )sin Scattering fro a conductive wedge 29

Plane wave solution () What if we had no wedge? Then the equation to be solved for this case is the sae, but has to be solved for a source in free space E E j I 2 2 0 k E 2 2 0 0 0 The solution is I E H0 kr 4 0 (2) 2 2, R 0 20 cos 0 Scattering fro a conductive wedge 30

Plane wave solution () Now, if we ove the source to a point far away in the free space proble then I I 2 E H0 kr jkr j 4 4 kr (2), 0 0 exp / 4 R 2 cos 2 2 0 0 0 cos 0 0 0 R y, exp cos E A jk 0 0 I 2 A exp jk j / 4 0 0 0 4 k0 0 0 x Scattering fro a conductive wedge 31

Plane wave solution () Apart fro its aplitude this is the field of a plane wave propagating towards the origin along the angle 0 A0 exp jk cos 0 A0 exp jki r ki k cos 0, k sin 0, 0 ki r cos, sin, 0 0 Scattering fro a conductive wedge 32

Plane wave solution () So, in the original proble, if we take the sae liit, while keeping the aplitude constant, the resulting solution will be that of the total field for the incident plane wave along 0 0 Note: keeping aplitude constant eans that we k s should put 4 k I exp jk j / 4 A 2 0 0 0 0 0 0 Scattering fro a conductive wedge 33

Plane wave solution () Result: 4 E A j J k 0 0 2 1, sin ( )sin This way we have found the coefficients for the total k i electric field when the incident wave is a plane wave with ki k cos 0, k sin 0, 0 0 0 Scattering fro a conductive wedge 34

Plane wave solution () The corresponding agnetic field is A k 4 H j 0 J k j 2 0, sin ( )sin 1 A 4 H j 0 J k j 2 0, sin ( )cos 1 Scattering fro a conductive wedge 35

scattering In a siilar way, the scattering proble can be solved Now we solve the equation for the agnetic field 2 H H 2 2 M i 0 M 0 0 0 2 i k H jm With the boundary condition H 0 for 0, 2 Scattering fro a conductive wedge 36

scattering The overall solution can be written as, ( )cos H a J k 0 (2), ( )cos H b H k 0 0 0 At the line source, they should be atched by H 0, H 0, H j 0 H M 0 0 0 0 Scattering fro a conductive wedge 37

scattering Due to the analogy with the case, we conclude that if the incident agnetic field is 0 0 H A j i exp 0 ki r k i Then the total agnetic field is given by 2 H A j J k 1 0 0 0 2 0, cos ( )cos 2 1 Scattering fro a conductive wedge 38

scattering i E But what about the electric field? The (horiontal) incident field is k i E i 0 1 j H ˆ jk rˆ k A exp i i i 0 0 The total electric field is E 1 H H ˆ ˆ ρ j Scattering fro a conductive wedge 39

scattering E 2, j A j cos 0 0 2 0 J ( k) sin ( )cos k ρˆ J k ˆ Copare with the general expression found before Scattering fro a conductive wedge 40

Special cases Let us consider soe specific y cases. First when=. k i Corresponds to reflection of a plane incident wave by a flat, infinite perfect conductor 0 x Solution for vertical fields () 2, 4 sin ( )sin E A j J k 0 0 1 Scattering fro a conductive wedge 41

Special cases, 4 sin ( )sin E A j J k 0 0 1 We can use the relationship To arrive at exp j cos J0( ) 2 j cos J( ) 1, exp cos exp cos E A0 jk 0 A0 jk 0 Incident wave A exp 0 jki r Scattering fro a conductive wedge 42

Special cases The reaining (scattered) ter corresponds to: A0 exp jk cos 0 A0 exp jks r k s k cos 0, k sin 0, 0 y This is what one would have expected. k i k s The case is siilar except 0 for the sign of the reflected (agnetic) field x Scattering fro a conductive wedge 43

Special cases Another iportant case is the liit 0 k i This is the liit of a half-infinite plane which is quite iportant 0 0 Total fields:, 2 / 2 sin / 2 ( )sin / 2 E A j J k 0 0 / 2 1 0 0 / 2 0, 2 / 2 cos / 2 ( )cos / 2 H A j J k Scattering fro a conductive wedge 44

Nuerical exaples We plot the scattered field as k i function of the observation angle (at a certain distance) for a certain incoing wave 0 0 scattered field: 4 E A j J k 0 0 2 1, sin ( )sin,, exp k r E, A exp jk cos E E A j s, 0 i 0 0 Scattered field Scattering fro a conductive wedge 45

Nuerical exaples () k i 45, k 2 0 0 0 30 0 90 Scattering fro a conductive wedge 46

Nuerical exaples () k i 45, k 2 0 0 0 180 0 225 Scattering fro a conductive wedge 47

Nuerical exaples () k i 0, k 2 0 0 0 30 0 90 0 180 Scattering fro a conductive wedge 48

Nuerical exaples () For scattering, we plot the scattered agnetic field as function of the observation angle (at a certain distance) for a certain incoing wave scattered field: 2 H A j J k 0 0 2 0, cos ( )cos,, exp k r H, A exp jk cos H H A j s, 0 i 0 0 Scattering fro a conductive wedge 49

Nuerical exaples () k i 0, k 2 0 0 0 30 0 90 0 180 Scattering fro a conductive wedge 50