Genel Elektromanyetik Dalga REFERENCE BOOKS FOR ELECTROMAGNETIC WAVE THEORY AND MICROWAVES. 1) D.CHENG, “ FIELD & WAVE ELECTROMAGNETICS” ADDISON-WESLEY 2) UMRAN S. INAN, AZIZ S. INAN , “ENGINEERING ELECTROMAGNETICS” , ADDISON-WESLEY 1999 3) R.COLLINS, “FOUNDATIONS OF MICROWAVE ENGINEERING” Mc GRAW-HILL 1992 4) D.POZAAR, “MICROWAVE ENGINEERING” ADDISON-WESLEY 1990 FOUR FUNDAMENTAL ELECTROMAGNETIC FUNCTIONS • E r ) , ( t r r - Electric Field Intensity – (V/m) • D r ) , ( t r r - Electric Flux Density (Displacement vector) – (C/m 2 ) • H r ) , ( t r r - Magnetic Field Intensity – (A/m) • B r ) , ( t r r - Magnetic Flux Density – (Wb/m 2 ) THREE TYPES OF ELECTRIC CURRENTS • c J r ) , ( t r r - Conduction Current Density – (A/m 2 ) = c J r E r ss s s = s Conductivity (S/m) • u J r ) , ( t r r - Convection Current Density – (A/m 2 ) = u J r u r rr r r = r Free Electric Charge Density – (C/m 3 ), = u r Velocity Vector – (m/s) • d J r ) , ( t r r - Displacement Current Density – (A/m 2 ) = d J r t D ¶ ¶ r MAXWELL’S EQUATIONS These four fundamental equations of electromagnetics on the basis of three separate experimentally established facts, namely, Coulomb’s law, Ampere’s law (or the Biot-Savart law), Faraday’s law, and the principle of conservation of electric charge. The validity of Maxwell’s equations is based on their consistency with all of our experimental knowledge to date concerning electromagnetic phenomena. The physical meaning of the equations is better perceived in the context of their integral forms, which are listed below together with their differential counterparts: 1. Faraday’s law is based on the experimental fact that time-changing magnetic flux induces electromotive force: ? C . E r ? ? ? ? -- - - = = = = s l d r s d t B r r . ¶¶ ¶ ¶ ¶¶ ¶ ¶ t B E ¶¶ ¶ ¶ ¶¶ ¶ ¶ -- - - = = = = ·· · · r r where the contour C is that which encloses the surface S, and where the direction of the line integration over the contour C (i.e., dl) must be consistent with the direction of the surface vector ds in accordance with the Right-Hand rule. 2. Maxwell’s second equation is a generalisation of Ampere’s law, which states that the line integral of the magnetic field over any closed contour must equal the total current enclosed by the contour. Maxwell’s second equation expresses the fact that time-varying electric fields produce magnetic fields. The first term of this equation (also referred to as the conduction-current term) is Ampere’s law, which is a mathematical statement of the experimental findings of Oersted, where as the second term, known as the displacement–current term, was introduced theoretically by Maxwell in 1862 and verified experimentally many years later in Hertz’s experiments. ? s ds ds dl s s . . . ? ? ¶ ¶ + = t D J H r r r t D J H ¶ ¶ + = · r r r where the contour C is that which encloses the surface S. 3. Gauss’s law is a mathematical expression of the experimental fact that electric charges attract or repel one another with a force inversely proportional to the square of the distance between them (i.e.,Coulomb’s law): ? ? ? ? ? ? ? ? = = = = s v dv s d rr r r r r . D rr r r = = = = D . r where the surface S encloses the volume V. The volume charge density is represented with rr r r r to distinguish it from the phasor from rr r r used in the time-harmonic form of Maxwell’s equations. 4. Maxwell’s fourth equation is based on the fact that there are no magnetic charges (i.e., magnetic monopoles) and that, therefore, magnetic field lines always close on themselves: ? ? ? ? = = = = s s d 0 . B r r 0 B . = = = = r where the surface S encloses the volume V. This equation can actually be derived from the Biot-Savart law, so it is not completely independent. CONSERVATION OF ELECTRIC CHARGE s d r ) t , r ( r r J Total charge in volume V is, Q = ? V V . d rr r r ? = = - V S S . J t r r d I d dQ 0 Lim ? D V = = = = V d Q d 0 Lim ? D V V J v S D ? S d r r J r . = Conservation of electric charge J . t r = - d dr ( CONTINUITY EQUATION ) t D J J H u c ¶ ¶ + + = · r r r r FOR DC CASE: J H r r = · u c J J J r r r + = = = u c J J r r constant, Then the Ampere Law is u c J J H + = · r r ) J J ( H ( u c + = · r r ) 0 J r = ? continuity equation FOR TIME VARIABLE CASE: t D J H ¶ ¶ + = · r r r ? ¶ ¶ + = ¶ ¶ + = · ) D ( t J ) t D J ( ) H ( r r r r r t J ¶ ¶ - = r r DERIVATION OF DISPLACEMENT CURRENT(J d ) If the free charge density of a region is varying with time there will be current flow through a surface enclosing a volume of the region. If ? ? ? ? = = = = V dv ) t ( ) t ( Q r is the free charge in volume v Then ? ? ? ? = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ S ds t ) t ( Q J r where J is the free current density out of surface S of volume v. ? ? ? ? ? ? ? ? -- - - = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ V V . J ds dv t ) t ( r r applying Gauss’ theorem. t ) t ( ¶¶ ¶ ¶ ¶¶ ¶ ¶ = = = = r J r A statement of conservation of charge at a point. When the charge density remains constant, as in steady current flow, 0 = = = = J r , as expected for conservation of steady constant around a circuit. A changing charge density must cause a related change in its electric field. By Gauss’ law r = = = = D r J D r r -- - - = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ t t r or 0 ) t ( = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ + + + + D J r r , a statement of “total” current density conservation, with t ¶¶ ¶ ¶ ¶¶ ¶ ¶ D r the “displacement current density”, applicable to time dependent charge distributions and field. The magnetic force law as used so far has been expressed in terms of steady currents. In the form of Ampere’s law at a point this is J H r r = = = = x This gives 0 ) x ( ”” ” ” = = = = J H r r since 0 ) A x ( = = = = r Maxwell realised that Ampere’s law had to be extended to apply to all current flow circumstances and postulated that t x ¶¶ ¶ ¶ ¶¶ ¶ ¶ + + + + = = = = D J H r r r In integral form this generalised Ampere’s law is then ? ? ? ? ? ? ? ? ¶¶ ¶ ¶ ¶¶ ¶ ¶ + + + + = = = = S ds ). t ( dl D J . H C r r r Example: Consider a certain type of humid soil with the following properties: ss s s =10 -2 S-m -1 , ee e e r =30, and mm m m r =1. Find the ratio of the amplitudes of conduction and displacement current at 1kHz, 1MHz, and 1GHz. Solution: Using | | | |J c | | | | max /| | | |J d | | | | max =ss s s /(ww w w .ee e e ), we have f 10 x 6 m F 30x8,85x10 m - S 10 . f 2 1 6 1 12 - -1 2 -- - - -- - - -- - - @@ @ @ -- - - = = = = p we s Thus we have ss s s /(ww w w .ee e e )@@ @ @ 6000, 6, and 0,006, respectively, at 1kHz, 1MHz, and 1GHz CAPACITOR E r a A C e = = = = = dt A d A a C dt d a C dt dQ C I d ) E . . ( . . E . . . r r e e dt A d ) E . . ( r e = ds . ) ds ( dt d 2 2 S S ? ? ? ? ? ? ? ? = = = = d J . D r r If we consider as an open surface (S 1 )? ? ? ? c C S I s d l d = = ? ? r r r r . J . H 1 c ) t ( .E J c r r s = = = = t ¶ ¶ = D J d r r S 2 ? ? ? ? d S I s d . s d . l d . 2 = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ = = = = ? ? ? ? ? ? ? ? ? ? ? ? r r r r r r t E t D H e I c =I d Example: Consider a parallel-plate capacitor consisting of two metal plates of 50 cm 2 area each,separated by a porcelain layer of thickness a=1 cm(for porcelain e r =5,5 cm and s =10 -14 S/m).If a voltage f t =110 2 cos(120p t) V is applied across the capacitor plates,find (a)J c (b)J d and (c )the total current I through the capacitor. Solution: The electric field in the porcelain layer is E(t) m V t a Q t 01 , 0 ) 120 cos( 2 110 p = = = 1,1 2 x 10 4 cos(120p t) V/m The conduction (J c ) and the displacement current (J d ) densities are then a) J c =s .E(t)=(10 -14 S/m)[(1,1 2x10 14 cos(120p t)V/m)] =1,1 2x10 -10 cos(120p t) A/m -2 b) J d = t t E t t E t D r ¶ ¶ = ¶ ¶ = ¶ ¶ ) ( ) ( 0 e e e =(5.5x 8.85 x10 -12 F-m -1 )[-1.1 2x 10 4 x120p sin(120p t) V-(ms) -1 ] @@ @ @ -(2.86x 10 -4 )sin(120p t) A-m -2 Note that the conduction current is in phase with the electric field, whereas the displacement current is 90 o out of phase. Also note that the amplitude of the conduction current is ~5x10 5 times smaller than the displacement current, as expected since porcelain is an excellent insulator. c) The total current of I can be found by integrating the total current density over the cross-sectional area of the capacitor plates as A ) J J ( ds ). J J ( I d c d A c + + + + = = = = + + + + = = = = ? ? ? ? since J c and J d are both approximately uniform and perpendicular to the plates throughout the dielectric region. Note also that ‰ J c ‰ max <<‰ J d ‰ max , so that for all practical purposes we can write @@ @ @ = = = = A J I d [-2,86x10 -4 sin(120p t)A/m 2 ] (50x10 -4 m 2 ) @@ @ @ -1,43x10 -6 sin (120p t) A=1,43 sin(120p t) m A Note that we could have easily obtained this result by using the familiar current-voltage relationship of an ideal capacitor that is , dt / ) t ( d . C ) t ( I f = = = = where . pF 3 , 24 a / A . C @@ @ @ = = = = e CONSTITUTIVE RELATIONS E . D r r e = and H . B r r m = are the two constitutive relations in magnestatic case. E r and B r are related to the medium independent quantities, D r and H r , in material media, where ,in general 0 e „ e and m„m 0 . The volume charge density rr r r r and conduction-current density J r are typically sources from which electric fields or magnetic fields orginate ,respectively.In conducting media(s„ 0),any present electric field E r must be accompanied by a volume current density of E J s = r . SIMPLE MEDIUM e , ss s s , mm m m are macroscopic parameters that describe the relationships among macroscopic field quantities,but they are based on the microscopic behaviour of the atoms and molecules in response to the fields.If these parametres are simple constants,we call the medium “simple material media”.In these case constitutive relations are as follows: E . D r r e = , H . B r r m = and E . J c r r s = PROPERTIES OF SIMPLE MEDIUM • HOMOGENOUS(=uniform): e , s , m are not the function of space . • LINEAR: e , s and m do not depend on the magnitudes of E r and B r • STATIONARY(= time-invariant):e , s , m do not change with time. • ISOTROPIC:e , s and m do not depend on the field(E r andB r ) directions. These properities are for a simple medium.Otherwise if media is nonlinear [(e , s , m ) may depend on the magnitudes of E r and B r ] , inhomogeneous [(e , s , m ) may depend on spatial coordinates {x,y,z}], time-variant [(e , s , m ) depend on time)] and anisotropic [(e , s , m ) may depend on the orientations E r and B r ] and we call it “complex material medium”. EXAMPLES FOR COMPLEX MEDIA: (1) Anisotropic medium with respect to ee e e : E D z y x r r ? ? ? ? ? ? ? ? ? ? = e e e 0 0 0 0 0 0 ? ? ? ? = = = = D r D r {E r } y y y E . D r r e = z z z E . D r r e = (2) An example of bianisotropic medium: H E D z y x z y x r r r ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + + + + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = = = = ee e e ee e e ee e e ee e e ee e e ee e e 0 0 0 0 0 0 0 0 0 0 0 0 ? ? ? ? = = = = D r D r {E r ,H r } SOME SPECIAL IMPORTANT CASES: 1) STATIC CASE 0 = ¶ ¶ t ? ? ? ? time-invariant 0 J c = r , 0 = = = = u J r ? ? ? ? no net charge movement Let us substitute these conditions into the Maxwell equations : • 0 ) r ( E = · r r • 0 ) r ( H = · r r • r = ) r ( D r r (Static charge) x x x E . D r r e =• 0 ) r ( B = r r We can define E r as ) r ( ) r ( E r r r y - = : y Potential function POISSON EQUATION ¸ - = r y 2 LAPLACE EQUATION 0 2 = y NOTE: If there is no charge movement we can say 0 H , B = r r 2) STATIONARY (= DC) CASE 0 = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ ?? ? ? t (No time-derivation) ? 0 J c „ r (There is a charge movement) The physical concepts are only related to space coordinates. So the Maxwell equations are: • 0 ) r ( E = · r r • u c J J ) r ( H r r r r + = · ( 0 D = ¶ ¶ t r ,No time-derivation) • r = ) r ( D r r • 0 ) r ( B = r r From the equations we can write J J J u c r r r = + J J J ) r ( H u c r r r r r = + = · , E J c r r s = J )) r ( H ( r r r = · 0 )) r ( H ( = · r r Then, 0 J = r By using the equation E J c r r s = and J H r r = · we can say: 1) 0 = s There is no coupling between H r and E r . 2) 0 „ s There is a coupling between H r and E r . 3) QUASISTATIONARY CASE ? ? ? ? There is a time variation ? ? ? ? Displacement current can be neglected w.r.t the conduction current. The physical concepts are related to both time, space and d J r , So the Maxwell equations are: c d J J t r r << „ ¶ ¶ 0 ) , ( ) , ( ) , ( ) , ( ) , ( t r J t r J t r H t t r B t r E u c r r r v r r r r r r + = · ¶ ¶ - = · ds t r J ds t r J dl t r H ds t r B t dl t r E c c c s u s c c s c + + + + = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ -- - - = = = = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ) , ( ) , ( ) , ( ) , ( ) , ( r r r r r r r r So, there is a coupling between ) , ( t r E r r and ) , ( t r H r r . 0 ) , ( ) , ( ) , ( = = t r B t r t r D r r r r r r Quasistationary case applicates on “low frequency circuits” and “electric machines”. 4) FREE SPACE CASE 0 e e = , 0 m m = , 0 = s , 0 = u J r , 0 = c J By using these equations we can write; t t r E t r H t t r B t t r H t r E ¶ ¶ =¸ · ¶ ¶ - = ¶ ¶ - = · ) , ( ) , ( ) , ( ) , ( ) , ( 0 0 r r r r r r r r r r m 0 ) , ( 0 ) , ( = = t r B t r D r r r r (The right side of equation is zero. Because, there is no charge) t t r E t r H t t r H t r E ¶ ¶ ? ¶ ¶ ? ) , ( ) , ( ) , ( ) , ( r r r r r r r r NOTE: Propagation does not need a material to occur . Generally; ) , ( ) , ( ) , ( t r J t r J t r H u c r r r r + = · AmpØre Law )) , ( ) , ( ( )) , ( ( t r J t r J t r H u c r r r r + = · ) , ( 0 t r J r v = 0 ) , ( ) , ( „ ¶ ¶ - = t t r t r J r r v r r So we should think about another current and this must be d J r , displacement current (if there is a charge). 5) SINUSODIAL STEADY CASE z z y x x a t) , r ( E a t) , r ( E a t) , r ( E t) , r ( E y r r r r r r r r + + = For instantaneous expressions: 1 ) ( E ) ( E 1 q j x e r r r r = The phasoral form of E x only related to r r , time is eliminated. So; } ) ( E { Re Im t j x e r ww w w r r = = = = t) , r ( E x } ) ( E { ) ( 1 Re Im 1 qq q q ww w w + + + + = = = = t j e r r We also can write these equations for y E and z E } ) ( E { ) ( 2 Re Im 2 qq q q ww w w + + + + = = = = t j e r r r t) , r ( E y } ) ( E { ) ( 3 Re Im 3 qq q q ww w w + + + + = = = = t j e r r r t) , r ( E z )] ( sin[ ) ( E 1 1 r t r r r r qq q q ww w w + + + + = = = = t) , r ( E xAs a result . ELECTRIC FIELD IN PHASORAL FORM z z y y x x a a r a r r r r r r r r r E ) ( E ) ( E ) ( E + + = } ) ( E { Re Im t j e r ww w w r r r r = = = = EE E E t) , r ( NOTE: Time derivation has to be sinusoidal ,since time variation is sinusoidal. So the problem can be defined as finding out ). ( B ), ( H ), ( D ), ( E r r r r r r r r r r r r So we must express Maxwell and Continuity Equations in their phasor forms: t D J H ¶ ¶ + = · r r r } ) ( D Re{ } ) ( J Re{ } ) ( H Re{ jwt jwt jwt e r t e r e r r r r r r r ¶ ¶ + = · } ) ( D Re{ } ) ( J Re{ ]} )) ( H Re{[( jwt jwt jwt e r jw e r e r r r r r r r + = · So the phasoral forms of Maxwell equations are resulted as: 1) Ampere Law: D J H r r r ww w w j + + + + = = = = ·· · · ? ? ? ? + + + + ? ? ? ? = = = = ? ? ? ? c s s d D j c s s d J l d C H r r r r r r . . . ww w w 2) Faraday Law: B E r r jw - = · ? ? - = c c s s d jw l d r r r r . B . E 3) Gauss Law (for electrics): ) ( D r r r r = ? ? = v s v dv s d . . D r r r 4) Gauss Law (for magnetic): 0 B = r ? = v s s d 0 . B r r NOTE: } ) ( { ) , ( Re Im jwt e r t r r r r r = 5) Continuity Law: r jw - = J r ? ? - = v s v dv jw s d . . J r r r And simple medium equations are; E . D r r e = H . B r r m = E . J c r r s = EXAMPLE: AIRCRAFT VHF COMMUNICATION SIGNAL The electric field component of an electromagnetic wave in air used by an aircraft to communicate with the air traffic control tower can be represented by m V 8 ) 10 5 , 7 cos( . 02 , 0 . ) , ( z t a t z E y b - · = r r Find the definition of b , , , D B H r r r .z SOLUTION: The electric field signal is sinusoidal so we should use phasoral forms. * m V . 02 , 0 . ) ( E z j y e a z b - = r r 1) E . H H E B E 0 0 r r r r r r · = ? ? ? ? ? - = · - = · m m w j jw jw 2) H . 1 E E H E 0 H D J H 0 0 0 r r r r r r r r r · ¸ = ? ? ? ? ? ? ? ¸ = · ¸ + = · + = · jw jw jw jw 0 J J J u c = + = r r r (b) H 0 ) E ( (a) E 0 ) H ( 0 0 ? - = = · ? ¸ = = · r r r r m jw jw 3) 0 E = r 4) 0 H = r We can reach rd 3 and th 4 equations by using (a) and (b) . By using st 1 equation we can find ) ( H z r : E . ) ( H 0 r r · = m w j z x z j x y y z y x z y x a e j a z a a a z r r r r r r = ¶ ¶ - = = · - ¶ ¶ ¶ ¶ ¶ ¶ b b . 02 , 0 . E 0 E 0 ) ( E x z j x z j a e w a e j w j z r r r . . 02 , 0 ). 02 , 0 .( ) ( H 0 0 b b m b b m - - - = = By using nd 2 equation we can find b : ) ( H . 1 ) ( E 0 z jw z r r · ¸ = y z j y x x z y x z y x a e j w a z a a a z r r r r r r . ) ( 02 , 0 . H 0 0 H ) ( H 0 b b m b - ¶ ¶ ¶ ¶ ¶ ¶ - - = ¶ ¶ = = · y z j a e w j z r r . . 02 , 0 ) ( H 0 2 b m b - = · y z j a e w j jw z jw z r r r . . 02 , 0 . 1 ) ( H 1 ) ( E 0 2 0 0 b m b - ¸ = · ¸ = y z j a e w z r r ¸ = - b m b . . 02 , 0 ) ( E 0 0 2 2 This is also equal to z j e b - . 02 , 0 so to find b we can write 0 0 0 0 2 2 02 , 0 02 , 0 e m b e m b b b w e w e z j z j = ? = - - m rad s m 8 s rad 8 5 , 2 10 3 10 5 , 7 = · · = = c w b When we put values into the equations m rad 5 , 2 = b m A 5 , 2 5 , 2 7 8 . 1 , 53 10 4 . 10 5 , 7 5 , 2 . 02 , 0 ) ( H m p x z j z j x a e e a z r r r - - · - = · · - = ( ) { } m A 8 ). 5 , 2 10 5 , 7 cos( 1 , 53 . H Re ) , ( m x jwt a z t e z t z H r r r - · - = = m V 8 ). 5 , 2 10 5 , 7 cos( 02 , 0 ) , ( y a z t t z E r r - · = 0 0 ) , ( ) , ( , ) , ( ) , ( m e t z H t z B t z E t z D r r r r = = • If we think about 0 = z then wt 0 p /2 p 3p /2 2p E(0,t)=0,02.cos(7,5.10 8 t) 0,02 V/m * 0 0 > = z z so m V 0 8 ) 5 , 2 10 5 , 7 cos( 02 , 0 ) , ( z t t z E y - · = r So second status follows first one 0 5 , 2 z rad behind. * 0 0 < = z z So, at that time this status goes in front for 0 5 , 2 z rad according to 0 = z status. = b Phase Constant. Example: Consider a certain type of humid soil with the following properties: ss s s =10 -2 S-m -1 , ee e e r =30, and mm m m r =1. Find the ratio of the amplitudes of conduction and displacement current at 1kHz, 1MHz, and 1GHz. Solution: Using | | | |J c | | | | max /| | | |J d | | | | max =ss s s /(ww w w .ee e e ), we have f 10 x 6 m F 30x8,85x10 m - S 10 . f 2 1 6 1 12 - -1 2 -- - - -- - - -- - - @@ @ @ -- - - = = = = p we s Thus we have s /(w .e )@@ @ @ 6000, 6, and 0,006, respectively, at 1kHz, 1MHz, and 1GHz ELECTROMAGNETIC BOUNDARY CONDITIONS In order to solve electromagnetic problems involving contiguous regions of different constitutive parameters, it is necessary to know the boundary conditions that the field vectors E (Electric Field Intensity), D (Electric Flux Density), B (Magnetic Flux Density) and H (Magnetic Field Intensity) must satisfy at the interfaces. Boundary conditions are derived by applying the integral form of Maxwell equations, which are; ? ? ? ? ? ? - = S d dt B d dl E c s c . . ? ? ? ? ? ? ? ? ? ? ? + + = S d J S d J dt D d dl H c c c s u s c s c . . . dV S d D v s v . . ? ? = ? ? r 0 S d . v = ? ? ? B to a small region at an interface of two media. The integral equations are assumed to hold for regions containing discontinuous media. The application of the integral form of a divergence equation to a shallow pillbox at an interface with top and bottom faces in the two contiguous media gives the boundary conditions for the normal components : 1) ) ) ) ? D n a . S ? n a ? 1 B Medium 1 1 1 1 , , e m s h d d h 2 ? B t a ? Medium 2 2 2 2 , , e m s 0 n a ? 0 . = ? ? ? S d B v s Let us apply this rule to the differential cylinder which has h d height. D S 0 S . a . S . a . n 1 n 1 = y D + D - D ? ? ? ? B B ? y D is flux coming out of side surfaces. To find B (Magnetic Flux Density) between two media, if h d is made 0, the flux through side surfaces also would be zero. 0 0 h ? y D ? ? d 0 0 S . a . S . a . n 2 n 1 = + D - D ? ? ? ? B B n 2 n 1 n 2 n 1 a . a . B B B B = = ? ? ? ? The normal component of a B (Magnetic Flux Density) field is continuous across an interface. They are equal too. 2 ) ) ) ) ? D n a . S ? n a ? 1 D Medium 1 1 1 1 , , e m s h d d h Medium 2 2 2 2 , , e m s 2 ? D dV S d D v s v . . ? ? = ? ? r D S Lets apply this rule to the differential cylinder again and obtain the limit when d goes to zero. ( ) 0 h ? d h 0 n 2 n 1 . lim a . a . h d r = - ? d ? ? ? ? D D s n 2 1 a . r = ? ? ? ? ? ? - ? ? ? D D [ ] 3 m cm ? r [ ] m h ? d [ ] 2 s m c = r s h 0 . lim h r = d r ? d s n 2 n 1 r = - ? ? D D Free space charge density( r ) is transformed to ( s r )free space charge density since h dimension is lost when d goes to zero ( ) 0 h ? d . The normal component of a D (Electric Flux Density) field is discontinuous across an interface where a surface charge exists, the amount of discontinuity being determined by equation. 3) dV . dt d S d . J v sv ? ? r - = ? ? ? ? ? ? ? ? d r - = - ? d ? ? ? ? h 0 n 2 n 1 . lim dt d a . J a . J h s n 2 1 . dt d a . J J r - = ? ? ? ? ? ? - ? ? ? The application of the integral form of a curl equation to a flat closed path at a boundary with top and bottom sides in the two touching media yields the boundary conditions for the tangential components. 4) ) ) ) Medium 1 1 1 1 , , s m e ? 1 E ? n a h d 2 / h d 2 / h d ? 0 n a w D ? 2 E Medium 2 2 2 2 , , s m e ? ? ? ? ? ? - = S d dt B d dl E c s c . . + D - D ? ? ? ? 2 t 2 1 t 2 a . w . a . w E E Addition coming from side dimensions = from S D the flux which dt dB passed through. = ? ? ? ? D d + D d - 0 n h 2 0 n h 1 a . w . 2 . dt d a . w . 2 . dt d B B For Medium 1 For Medium 2 0 h ? d while h d goes to zero. The addition coming from sides and 0 h = d 0 . . . . 1 2 1 1 = D - D ? ? ? ? t t a w E a w E t 2 t 1 t 2 1 0 a . E E E E = = ? ? ? ? ? ? - ? ? ? 0 a n 2 1 = · ? ? ? ? ? ? - ? ? ? E E The tangential component of an E (Electric Intensity) field is continuous across an interface. 5) ) ) ) Medium 1 1 1 1 , , s m e ? 1 D ? n a h d 2 / h d 2 / h d ? 0 n a w D ? 2 D Medium 2 2 2 2 , , s m e ? ? ? ? ? ? ? ? ? + = c s s c c c S d J S d dt D d dl H . . . To find passing conditions of H (Magnetic Field Intensity) in rectangle frame to take the limit while h d goes to 0, dt dB is not finished and the equation below is left true. h c 0 t 2 t 1 . J lim a a h d = - ? d ? ? ? ? H H s 2 1 n J a = ? ? ? ? ? ? - · ? ? ? H H The tangential component of an H (Magnetic Field Intensity) field is discontinuous across an interface where a surface current exists. SPECIAL CASES 1) Interface Between a Dielectric and a Perfect Conductor A perfect conductor is one with an infinite conductivity. In the physical world we only have ‘good’ conductors such as silver, copper, gold and aluminum. In order to simplify the analytical solution of field problems, good conductors are often considered as the perfect conductors in regard to boundary conditions. In the interior of a perfect conductor, the electric field is zero (otherwise it would produce an infinite current density) and any charges the conductor will reside on the surface only. The interrelationship between (E, D) and (B, D) through Maxwell’s equations ensures that B and H are also zero in the interior of a conductor in a time-carrying situation. ? n a Perfect Dielectric 0 , , 1 1 1 = s m e Perfect Conductor ¥ = s m e 2 2 2 , , Medium 1 Medium 2 0 t 1 = E 0 t 2 = E 0 n 1 = B 0 n 2 = B ? ? ? = · s 1 n J a H 0 t 2 = H s 1 n . a r = ? ? D 0 n 2 = D 0 n 1 = H 0 n 2 = H 0 t 1 = D 0 2 = t D a) 0 n 2 n 1 = = B B b) s n 2 n 1 r = - D D , s n 1 n 2 0 r = ? = D D c) 0 a 2 1 1 n = ? ? ? ? ? ? - · ? ? ? E E , 0 t 1 = E , 0 t 2 = E d) ? ? ? ? = ? ? ? ? ? ? - · s 2 1 1 n J a H H , ? ? ? = · ? = s 1 1 n t 2 J a 0 H H Medium 1 Medium 2 2) ) ) ) Interface Between Two Lossless Linear Media A lossless linear medium can be specified by a permittivity e and a permeability m with s =0.There are usually no free charges and no surface currents at the interface between two lossless media. We set: 0 J , 0 s = = s ? ? n a 0 , , 1 1 1 = s m e 0 , , 2 2 2 = s m e 0 a 2 1 n = ? ? ? ? ? ? - · ? ? ? E E t 2 t 1 E E = 0 a 2 1 n = ? ? ? ? ? ? - · ? ? ? H H t 2 t 1 H H = 0 . a 2 1 n = ? ? ? ? ? ? - ? ? ? B B n 2 2 n 1 1 . . H H m = m 0 . a 2 1 n = ? ? ? ? ? ? - ? ? ? D D n 2 2 n 1 1 . . E E e = e 3) Interface Between Two Lossy Linear Media In this case second special cases (interface between two lossless linear medium) are valid. Mediu Medium APPLICATION Perfect insulator 0 = s ? Perfect conductor ¥ = s ? The E and H field of a certain propagating mode ( ) 10 TE in a cross section of an a and b rectangular waveguide are : y y E . a E ? ? = and z z x x H . a H . a H ? ? ? + = where jBz 0 o y e . a x sin . H . a . . jw E - ? ? ? ? ? ? p p m - = jBz o x e . a x sin . H . a . jB H - ? ? ? ? ? ? p p = jBz 0 z e . a x cos . H H - ? ? ? ? ? ? p = 0 = s ¥ = s 1 0 ,e m y x 0 y = Bottom surface 0 x = Left y=b Top surface x=a Right surface We can ignore these jBz e - terms while solving these problem, TE 10 ? T: Transverse E: Electric 10: The numbers of combination. Where m , w , H o and B are constants. Assuming the inner walls of the waveguide are perfectly conducting, determine for the four inner walls of the waveguide: (a) the surface charge densities. (b) the surface current densities. SOLUTION: Equation (1), (2a), (2b) must satisfy Maxwell Equation and Boundary Condition. Volume V is closed with the perfect conductor planes 0 y , a x , 0 x = = = , b y = and it is a dielectric medium which has parameters o m and 1 e . (a) ? Free space charge density at 0 x = perfect conductor plane ( ) s 1 n 0 x s D . a r = = r ? ? = ? ? = x n a a 0 E . D 0 x 1 1 0 x 1 = ? ? ? ? ? ? e = = ? = ? ( ) 0 a . E . a D . a 0 x y 1 . 1 x 0 x 1 x o x s = ? ? ? ? ? ? e = = r = ? ? = ? ? = ( ) 0 0 x s = r = ? Free space charge density at a x = perfect conductor plane ( ) s 1 n a x s D . a r = = r ? ? = ? ? - = x n a a 0 E . D a x 1 1 a x 1 = ? ? ? ? ? ? e = = ? = ? ( ) 0 a . E . a D . a a x y 1 . 1 x a x 1 x a x s = ? ? ? ? ? ? e - = = r = ? ? = ? ? = ( ) 0 a x s = r = ? Free space charge density at 0 y = perfect conductor plane ( ) s 1 n 0 y s D . a r = = r ? ? = ? ? = y n a a 0 E . D 0 y 1 1 0 y 1 = ? ? ? ? ? ? e = = ? = ? ( ) ( ) ( ) 0 y y 1 y 0 y y 1 y 0 y s E . a . E . . a = ? = ? = e = e = r ( ) ? ? ? ? ? ? p p m e - = r = a x sin . H . a . . w . . j 0 o 1 0 y s ?Free space change density at b y = perfect conductor plane ( ) s 1 n b y s D . a r = = r ? ? = ? ? - = y n a a 0 E . D b y 1 1 b y 1 = ? ? ? ? ? ? e = = ? = ? ( ) ( ) ( ) b y y 1 y b y y 1 y b y s E . a . E . . a = ? = ? = e = e = r ( ) ? ? ? ? ? ? p p m e - = r = a x sin . H . a . . w . . j 0 o 1 b y s (a) ? Surface current density at 0 x = perfect conductor plane ( ) ( ) 0 x n 0 x s H a J = ? ? = ? · = ? ? = x n a a ( ) ( ) ? ? ? ? ? ? + · = ? = ? = ? z 0 x z x 0 x x x a . H a . H a ( ) ? = - = y 0 x z a . H ( ) ? ? - = ? ? ? ? ? ? p - = y o y 0 a . o cos . H a . a 0 . cos . H ? - = y o a . H ? Surface current density at a x = perfect conductor plane ( ) ( ) a x n a x s H a J = ? ? = ? · = ? ? - = x n a a ( ) ( ) ? ? ? ? ? ? + · - = ? = ? = ? z a x z x a x x x a . H a . H a ( ) ? = = y a x z a . H ( ) ? ? p = ? ? ? ? ? ? p = y o y 0 a . cos . H a . a a . cos . H . o y H a ? = - ?Surface current density at 0 y = plane ( ) ( ) 0 y n 0 y s H a J = ? ? = ? · = ? ? = y n a a ( ) ( ) ? ? ? ? ? ? + · = ? = ? = ? z 0 y z x 0 y x y a . H a H a ( ) ( ) ? = ? = + - = x 0 y z z 0 y x a . H a . H ? ? ? ? ? ? ? ? p + ? ? ? ? ? ? p p - = x 0 z 0 a . a x cos . H a . a x sin . H . a . jB ? Surface current density at b y = plane ( ) ( ) b y n b y s H a J = ? ? = ? · = ? ? - = y n a a ( ) ( ) ? ? ? ? ? ? + · - = ? = ? = ? z b y z x b y x y a . H a H a ( ) ( ) ? = ? = - = x b y z z b y x a . H a . H ? ? ? ? ? ? ? ? p - ? ? ? ? ? ? p p = x 0 z 0 a . a x cos . H a . a x sin . H . a . jB ENERGY CONSERVATION IN THE ELECTROMAGNETIC FIELDS POYNTING THEOREM t ¶ B ¶ - = E · r r t D J u ¶ ¶ + + E = H · r r r r s ( ) H ? ? ? ? ? ? ? ? ¶ B ¶ - = E · H r r r r t ( ) 1 ( ) E ? ? ? ? ? ? ? ? ¶ ¶ + + E = H · E r r r r r r t D J u s ( ) 2 Let us subtract first (1) equation from second (2) equation: [ ] n n s n n d J d d t t D d V u V V V E + E + ? ? ? ? ? ? ¶ B ¶ H + ¶ ¶ E = E · H - H · E ? ? ? ? r r r r r r r r r r 2 ( ) 3 E E r r Left side of the integrand of above ( ) 3 equation equals to: ( ) n n d ? E · H r r dv S d r u J r V H E r r s m e ( ) ( ) n d S d V S POYNTING ? ? E · H = H · E - r r r 3 2 1 r r H · E = R r r v ( ) 4 ? EM POWER FLUX DENSITY VECTOR: 2 m W 2 m V 2 m A (t-Domain) INSTANTANEOUS EM ENERGY CONSERVATION IN THE SIMPLE MEDIA POYNTING VECTOR H ^ r r P ; E ^ r r P Let us arrange ( ) 3 equation and write again: ( ) 4 3 42 1 4 4 4 4 4 3 4 4 4 4 4 2 1 4 43 4 42 1 4 43 4 42 1 r r 4 3 42 1 r r r LOSS JOULE POWER THERMAL V VOLUME V IN ENERGY EM TOTAL V DENSITY ENERGY NETIC ELECTROMAG VOLUME V TO J BY POWER SUPPLIED TOTAL V u S IN ENTERING TOTALPOWER W S dV dV dt d dV J S d P u ? ? ? ? E + ? ? ? ? ? ? ? ? ? ? H + E = E - - 2 2 2 ), ( 2 1 2 1 s m e ( ) 5 This equation (5) is the expression of ELECTROMAGNETIC ENERGY CONSERVATION IN t-DOMAIN IN THE SIMPLE MEDIA. (5) Equation is also known as POYNTING THEOREM IN t-DOMAIN At an instant t; 3 2 2 1 m J W e E = e ; 3 2 2 1 m J W m H = m ( ) 1 . 6 ( ) dV W W W W V m e m e ? + = + ( ) 2 . 6 JOULE LOSS: ( ) dV V ? E 2 s (6.3) ( ) E E r r s 2 m A 2 m V ? 3 m W (i) JOULE LOSS FOR THE UNIT VOLUME IN t-DOMAIN: 2 E s : 3 m W Note: E r and H r is coupled with required MAXWELL EQUATIONS (ii) TOTAL POWER OUTGOING FROM THE CLOSED S SURFACE: ( ) t S d P S Y = ? r r If a vector F whose divergence ( ) 0 = F r is zero is added to the P r , what happens; { } { } dV F S d F V S ? ? + H · E = + H · E r r r r r r r 0 ( ) dV V ? H · E = r r + ( ) dV F V ? r ( ) S d E S r r r H · = ? ( ) t Y = ? This has the physical meaning! ? This is measurable Here ( ) t r P , r r is the function gives the Power Flux that is measurable. ELECTROMAGNETIC ENERGY CONSERVATION IN t-DOMAIN AND APPLICATIONS OF P r APPLICATION OF RECEIVER AND TRANSMITTER P r P r S d r S H r (1) RECEIVER (2) TRANSMITTER P r E r 1. Let’s apply Poynting Theorem to Receiver and Transmitter dV dV dt d S d P V V S ? ? ? E + ? ? ? ? ? ? H + E = - 2 2 2 2 1 2 1 s m e r r ( ) ? ? ? ? R + E + ? ? ? ? ? ? H + E = E - v s v v u s d dv dv dt d dv J r r 4 43 4 42 1 r r 2 2 2 2 1 2 1 s m e Power given by u J r source to volume V e m s V R u J r e m V 2. Let’s explain the heat on the wire using R r whose resistance is R ohm. I r r r p j 2 I = H z=L v z=0 a x y z L 2 S 1 S 3 S R r R r 3 2 1 S S S S + + = ( ) a a r p j 2 I = = H If a voltage is applied to the wire, L V z = E The conduction current is given in z direction. H · E = R D r r r ; z z a r r E = E ; j j a r r H = H j j a a z z r r r · H E = R . ? ( ) r z a r 3 2 1 r - H E = R R j ( ) ( ) ( ) ( ) r S r S S z r S a ds a s d a ds a s d r r r r 4 4 3 4 4 2 1 r r r r - R - R - - - R - = R - ? ? ? ? 3 2 1 0 Independent from space coordinates. I = R I = I = = R V V aL a L V aL P p p p 2 2 2 Let’s apply “Poynting Theorem” to a wire which has stationary (DC) current: ( ) 0 = + m e W W dt d Because H r and E r are stationary. 2 m WEXAMPLE 4 Electromagnetic power flux density produced by an uniform plane wave in a lossy media: ( ) m V z t x b w - E = E cos 0 (1) ( ) z j x e z b - E = E 0 ( ) m A z t y b w h - E = H cos 1 0 (2) ( ) z j y e z b h - E = H 0 1 Characteristic impedance a) ( ) 2 ? , m W t r = R r r b) ? ) ( = R r ort r r c) 3 ? ) , ( ), , ( m J t r W t r W m e = r r d) ? ) ( ), ( = r W r W ort ort m e r r e) Provide the Poynting Theorem in t-domain. SOLUTION ( ) ( ) z t a t r z b w h - E = H · E = R 2 2 0 cos , r r r r r ( ) ( ) [ ] 2 2 2 0 cos 1 2 , m W z t a t r z b w h - + E = R r r r (a) ( ) ( ) [ ]dt z t a T dt t r T r p p T z p T p ort b w h - + E = R = R ? ? 2 2 0 0 0 cos 1 2 1 , 1 ) ( r r r r r } r r e m e m e m h p = = 120 0 0 ? ? ? ? ? ? ? ? = 0 0 1 e m c r r c e m w me w b = =2 2 0 2 ) ( m W a r z ort h E = R r r r (b) 2 0 ~ E ort P ( ) 3 2 2 0 cos 2 1 2 1 ) , ( m J z t t r W e b w e e - E = E E = r r r 3 2 0 4 1 ) ( m J r W ort e E = e r ( ) ( ) z t t r W m b w e m m m - ? ? ? ? ? ? E = H H = 2 2 2 0 cos 2 1 2 1 , r r r ( ) ( ) ( ) t r W z t t r W e m , cos 2 1 , 2 2 0 r r = - E = b w e 3 2 0 4 1 ) ( m J r W ort m E = e r ( ) ? ? ? ? ? ? E + ? ? ? ? ? ? H = H · E - 2 2 ? 2 1 2 1 e m dt d dt d r r The Complex Poynting Theorem v(t)=Re{ } t j j e e V v w q i(t)=Re{ } t j j e e I i w q p(t)= ) t ( i ) t ( v ........................Instant power p(t)= ) t cos( ) t cos( I V i v q + w q + w ?p(t)= 2 I V [ ] ) t 2 cos( ) cos( i v i v q + q + w + q - q Here ? q - q = = P T 0 i v P AV ) cos( 2 I V dt ) t ( p T 1 P or [ ] ) t ( p ) t ( p D = q = cos I V 2 1 P AV i v q - q = q (Phase between current and voltage) So, complex power is: P= * I V 2 1 = q q - q = j ) ( j e I V 2 1 e I V 2 1 i v P= REACT AV P P + { } * AV I V Re 2 1 P = { } * REACT I V Im 2 1 P = p(t) t AV P ) t , r ( H ) t , r ( E ) t , r ( P r r r r r r · = , [ ] 2 m W , power that pass through 2 m P r represents complex power density: ) r ( P j ) r ( P H E 2 1 ) r ( P REACT AV * r r r r r r r r + = = { } * AV H E Re 2 1 P r r r · = { } * REACT H E Im 2 1 P r r r · = For the cartesian coordinate system: = AV P ( ) ( ) { } z * z y * y x * x z z y y x x a H a H a H a E a E a E Re 2 1 r r r r r r + + · + + [ ] y z z y x H E H E 2 1 P - = Let’s start from the beginning, using Maxwell’s equations: ( ) * * u * E j J H r r r we - s + = · H j E r r wm - = · ( ) * * * H E E H H E r r r r r r · - · = · ( ) ( ) * u * * * J E E E j H H j H E r r r r r r r r - we - s - wm - = · We should take the integral of the last equation in the volume V: ( ) [ ] ? ? ? ? - s - e - m w - = · v v * u v * * * s * dv J E dv E E dv E E H H j s d H E r r r r r r r r r r r ? ? ? ? ? ? - = s + ? ? ? ? ? ? ? ? ? v * u v * s dv J E Re 2 1 dv E E 2 1 s d P Re r r r r r r ......equation for the real part(Average) ( ) ? ? ? ? ? ? - = w - w w + ? ? ? ? ? ? ? ? ? v * u v eav mav s dv J E Im Im 2 1 dv 2 1 s d P Im r r r r ....equation for the imaginar part(Reactive) ELECTROMAGNETIC WAVE PHENOMENA IN THE SIMPLE MEDIA EQUATION OF THE ELECTRIC FIELD WAVES IN THE SIMPLE MEDIA ? ? ? ? ? ? + • + ¶ ¶ • ¶ ¶ - = - ? ? ? ? ? ? ? ? ? ? ? ? + • + ¶ ¶ • ¶ ¶ • - = - • · ¶ ¶ • - = · · = • = • = • - ¶ ¶ • - · = ¶ ¶ • + · u J E t E t E u J E t E t E E H t E H E J E t E H t H E u r r r r r r r r r r r r r r r r r r r s e m e r s e m m e r s e m 2 2 ) ( ) ( ) 4 ( 0 ) 3 ( ) 2 ( ) 1 ( 0 ) 5 ( 1 2 2 2 t u J t E t E E ¶ ¶ • + • = ¶ ¶ • • - ¶ ¶ • • - r r r r m r e m s m e *This equation (5) is 2nd order, linear, partial differential equation according to time and space . *The right side contains only the source terms (non-homogeneous equation). *Since EM phenomena occurs in a linear media, the Superposition Principle can be employed in determining the EM solution vectors. EQUATION OF THE MAGNETIC FIELD WAVES IN THE SIMPLE MEDIA ( ) ( ) ( ) ( ) 0 ) ( 2 = • · + ? ? ? ? ? ? ¶ ¶ • - • + ? ? ? ? ? ? ¶ ¶ • - ¶ ¶ • = - • · + • · + · ¶ ¶ • = · · H J t H t H t H H J E E t H u u r r r r r r r r r r m s m e s e ) 6 ( 2 2 2 u J t H t H H r r r r ·· · · -- - - = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ • • • • • • • • -- - - ¶¶ ¶ ¶ ¶¶ ¶ ¶ • • • • • • • • -- - - mm m m ss s s mm m m ee e e u J term source t H loss to due term r r · - ¶ ¶ • • - : : m s * The most important terms in the electromagnetic wave phenomena are: t D AND t B ¶ ¶ ¶ ¶ r r * The charecteristical terms of the wave equations are :(1) and (2) * We should do s =0 in the equations of (5) and (6) if we want to investigate the properties of the EM Wave Phenomena. THE EM WAVE EQUATIONS IN THE LOSSLESS SIMPLE MEDIA: { } 2 2 2 t ¶ ¶ • • - m e => Operator for H E D B r r r r , , , waves (source terms) * EM Waves in the media where the sources do not exist: { } 2 2 2 t ¶ ¶ • • - m e ? ? ? ? ? ? ? ? ? ? • ) , ( ) , ( t r H t r E r r r r = 0 homogeneous EM wave equation in t-domain (8) * In the cartesian coordinate system: (9) x X a t z E t z E r r • = ) , ( ) , ( => Under this assumption the differential equation provided by the ) , ( t z E x r is: ) 10 ( 0 2 2 2 2 = ¶ ¶ • • - ¶ ¶ t E z E x x r r m e 0 ) , ( ) , ( ) , ( ) , ( ) , ( ) , ( 2 2 2 2 2 2 2 2 = ? ? ? ? ? ? ? ? ? ? ? • ? ? ? ? ? ¶ ¶ • • - ¶ ¶ + ¶ ¶ + ? ? ? ? ? ¶ ¶ t r H t r E t r H t r E t r H t r E t z y x z z y y x x r r r r r r r r r r r m eSolution of the above equation is : ) ( ) ( ) , ( 2 1 ut z g ut z g t z E x + + - = (11) * 1 g and 2 g are the arbitrary functions of only the (z-ut) and (z+ut) respectively . e.g: ) ( ) ( ) ( ut z e A ut z ut z g - - • • - = - etc... m e • = D 1 u (m/s) => is the propagation velocity of the EM waves in the z-direction. * 1 g (z-ut) is the wave component that propagates in the (+z) direction with the (u) speed. * 2 g (z+ut) is the wave component that propogates in the (-z) direction with the (u) speed. * The forms of the 1 g and 2 g functions are determined by the Boundary and the Initial Conditions THE D’ALEMBERT SOLUTION * The case fills the whole space z ) , 0 ( , ) , ( ¥ ¸ ¥ -¥ ¸ t Let us denote the x or y-component of the EM Waves with u(z,t): ) 2 ( ) ( ) , ( ) 1 ( ) ( ) 0 , ( 0 0 z U t z u t z u z u t = = = = ¶¶ ¶ ¶ ¶¶ ¶ ¶ = = = = = = = = is given as the data ,so we can determine the forms of the 1 g and 2 g functions in terms of the given data as follows: 1 0 ) ( 2 1 0 0 1 2 1 0 2 1 2 1 ). ( 2 1 ) ( 2 1 ). ( 2 1 ) ( 2 1 ) ( ) 4 ( ) ( ) ( ) ( ) , ( ) 3 ( ) ( ) ( ) ( ) , ( ) ( ) ( ) , ( c d U u z u g c d U u z u z g if z U z g u z g u o z u t z u z g z g o z u ut z g ut z g t z u z o z z + + = - - = ? = ¢ + ¢ - = ¶ ¶ = + = ? + + - = ? ? l l l l [ ] l l d U u ut z u ut z u t z u ut z ut z ). ( 2 1 ) ( ) ( 2 1 ) , ( 0 0 ? + - + + + - = PROPERTIES OF THE PLANE WAVES A plane wave cannot have a wave component in the propagation direction: 0 = • E r , 0 = • H r no ( ) sources 0 ) , ( ) , ( ) , ( 0 0 = ¶ ¶ + ¶ ¶ + ¶ ¶ t z z E t z y E t z x E z y x 4 3 4 2 1 4 3 4 2 1 ) ( ) , ( ) ( ) , ( t M t z H t K t z E z z = = where K(t) and M(t) are wave components propagating ın the z- direction,since they are not function of (z-ut) and (z+ut). ? • = c u m 1 The phase speed of the EM waves in the lossless media.( 0 0 1 m e • = c ) e m s e m e m e m , , 0 ; ; 1 0 = £ • = • • • = c u c u p r r o r r p THE RELATIONSHIP BETWEEN THE COMPONENTS OF AN UNIFORM EM PLANE WAVE PROPAGATING IN x-DIRECTION: z y x z y x x X E E E z y x a a a E t H E H E H E ¶ ¶ ¶ ¶ ¶ ¶ = · ¶ ¶ • - = · = = = • = • r r r r r r r r r r , 0 ; 0 ; 0 m ) 1 ( 0 ? ? ? ? ? ? ? ? • ¶ ¶ + • ¶ ¶ • - = • ¶ ¶ + • ¶ ¶ - ? y y z z z y y z a t H a t H a x E a x E r r r r m z y x z y x H H H z y x a a a H t E H ¶ ¶ ¶ ¶ ¶ ¶ = · ¶ ¶ • = · r r r r r r , e ) 2 ( ? ? ? ? ? ? ? ? • ¶ ¶ + • ¶ ¶ • = • ¶ ¶ + • ¶ ¶ - ? y y z z z y y z a t E a t E a x H a x H r r r r r r e Using (1) and (2) ,we can obtain the following simultaneous equations: ) 3 ( a t E x H y z ¶ ¶ • = ¶ ¶ - e ) 3 ( b t E x H z y ¶ ¶ • = ¶ ¶ e ) 3 ( c t H x E y z ¶ ¶ • = ¶ ¶ m ) 3 ( d t H x E z y ¶ ¶ • - = ¶ ¶ m *The wave component does not exist in the propogation direction: Let us have the y- electric wave component propagating in the +x direction: e m • = - = 1 , ) ( ) , ( 1 u ut x g t x E y ) 4 ( ' ' ; ; ) 4 ( ' ) ( 1 1 1 1 b g x E g g ut x a g u t E y y = ¶ ¶ ¶ ¶ = - = • - = ¶ ¶ z z Then we put the equations (4a) and (4b) into the statements (3a) and (3d): ) 5 ( ' 1 ) 5 ( 1 1 b g t H a g u x H z z m e - = ¶ ¶ • • = ¶ ¶ From the equation (5a): ? + • • = c dx g H g z 3 2 1 1 ' 1 m e ) 6 ( ) , ( ) , ( t x E t x H y z • = m e The relation in (6) is the magnetic wave accompanying to the given electric wave . W = · • = e m h h ; ) ( 1 E a H n r r r ,where n a r is the unit vector in the propagation direction. In our case it is equal to x a r ) ( ) , ( 1 ut x g t x E y - = * ) ( , , direction n propagatio a H E n - r r r constitutes the right hand ortogonal system.So: ) ( ) , ( ) ( ) , ( 2 1 ut x g t x H ut x g t x E z y + • - = + = m e * The relationship between the y z H and E can be found by using the 3b and 3d: ? + + - = ) ( ) ( ) , ( 2 1 ut x f ut x f t x H y r ( ) ) ( ) ( ) , ( 2 1 ut x f ut x f t x E z + + - - • = e m * Here if z y y y a ut x E H then a ut x E E r r r r • * - = • - = m e ) ( ) ( [ ] x y a ut x E H E P H E r r r r r r r • - • = · = = • 2 ) ( 0 m e TRANSVERSE ELECTROMAGNETIC WAVES: The plane waves mentioned here are arbitrary if they change with time. SINUSOIDAL PLANE WAVES: ? = • ? ? ? ? ? ? ? ? ? ? ¶ ¶ • • - ) 1 ( 0 , 2 2 2 H E t r r e m [ [ [ [ ] ] ] ] [ [ [ [ ] ] ] ] ) 2 ( . cos ) ( cos ) , ( 0 r k t z k y k x k t t r E o z y x r r r r r r -- - - EE E E = = = = + + + + + + + + -- - - • • • • EE E E = = = = ww w w ww w w ) 3 ( vector n propogatio a k a k a k k z z y y x x r r r v + + = Then we look for this solution:The given solution can be examined in t-domain or w-domain. The transformation of equation (1) in w domain: { } ) 4 ( 0 ) ( ) ( 2 2 = ? ? ? ? ? • • • + r H r E r r r r e m w => right sideless wave equation in w-domain The fasors according to the given solution: ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) { { { { } } } } ( ( ( ( ) ) ) ) ( ( ( ( ) ) ) ) 0 Re 0 0 = = = = EE E E = = = = EE E E • • • • EE E E = = = = EE E E ? ? ? ? • • • • EE E E = = = = -- - - r and e r e r r E jkr t j r r r r r r r r r ww w w SINUSOIDAL ELECTROMAGNETIC WAVE * EXAMPLE: x n a t kz a t z E t r E r r r r ) cos( ) , ( ) , ( 0 w - • E = • = We investigate this sinusoidal electric wave: ) cos( ) , ( 0 t kz t z E x w - • E = 1) At z=0 plane ) cos( ) , 0 ( 0 t t E x w • E = At z = 0 ) cos( t w = ) 2 cos( p w n t + ,n=0,1,2... = =>T= w p 2 (Sinusoidal) 2) At t w =0 wavelength k n n z k kz kz z E x ? = = + = E = p l l 2 ... 2 , 1 , 0 ), ( cos cos cos ) 0 , ( 0 E 0 2 p 2 3p p p 2 t w ) cos( ) , 0 ( 0 t E t E x w = 0 = t w 3) t w = 2 p w p p w p w p 2 2 ) sin( ) 2 cos( ) 2 , ( 0 0 = ? = • E = - • E = t t kz kz z E k * t w = p kz kz z E x cos ) cos( ) 2 , ( 0 0 • E - = - • E = p w p w p = ? t The instantaneous equiphase plane at t, K t kz = - w .And at t t D + , K t t z z k = D + - D + ) ( ) ( w so the propagation speed of the equiphase plane K in the (+z) direction is: E 0 2 p 2 3p p p 2 kz ) cos( ) , ( 0 kz E t z E x = w p w = t 0 = t w k t z Lim u t p w = D D = ? D 0 . SINUSOIDAL UNIFORM PLANE WAVE IN THE LOSSLESS SIMPLE MEDIA: 1) 0 ) , ( ) , ( 2 2 2 = ? ? ? ? ? • ? ? ? ? ? ? ¶ ¶ • • - t r H t r E t r r r r e m EM waves in t-domain (sourceless,right sideless) 2) { } ? ? ? ? ? ? ? • • • - ) ( ) ( ) ( ) ( 2 2 r B r D r H r E r r r r r r r r e m w EM waves in w -domain= =>HELMHOLTZ EQUATION 3) ) cos( ) , ( 0 r k t t r E r r r r - • E = w Basic uniform plane wave definition 4) { } t j r k j e r t r E then e r w • E = • E = E - ) ( Re ) , ( ) ( . 0 r r r r r r r r r z z y y z x n p a k a k a k a k k wavenumber u k here r r r r r r r r r r r + + = • = = • = = E • • • + E w e m w e m w 0 ) ( ) ( ) 5 2 2 0 ) ( ) ( 2 2 = E + E r k r r r r r ( ) [ ] ) ( 0 2 2 z k y k x k j z y x e r + + - • E = E r r r using E E E 2 ) ( - = · · and replacing k j r - instead of operator, We have ? ( ) r k r r r r r E - = E 2 2 ) ( *Direction cosinus: z n y n x n a a C a a B a a A r r r r r r • = • = • = cos cos cos r r r r k k w k e m e m e m e m w • = • • • • = • = 0 0 0 ( ) r a jk n e r c k r r r r r . 0 0 0 0 * - • E = E = • = w e m w phase r a k amplitude n : . : 0 r r r E EQUAL PHASE SURFACES: plane equalphase M z C y B x A M CONST r a n ? = + + = = ) cos( ) (cos ) (cos r r * : p u speed of the equal phase plane. *the phase at t-time: ( ) [ ] ( ) ( ) 0 0 0 0 0 : 0 1 0 . 0 . . . ) ( . ) ( . e m h e m h e m h h m e m w wm h e m w w w w w = • = = = H • E · H = E E · = E · - • = - = E · E · = H • = = ? = D D - ? = D - D D D = ? = D + - D + = - ? D r r n n n n r r p n n n t p n n NOTE a a a jk j H j a c k u t r a k r a k t a t r Lim u M r r a k t t M r a k t r r r r r r r r r r r r r r r r r r r r r r r r r r r EXAMPLE: AN AM BROADCAST SIGNAL is given as : ( ) ( ) m V x t a t x E z b p + = 6 10 5 , 1 cos 10 , r r Then find out that : a)the propagation direction, b) ? ) , ( = t r H r r c) ? ) ( ?, ) , ( = = r P t r P av r r r r Solution: a) Propagating in (-x) direction b) MHz f s rad 75 , 0 2 10 . 5 , 1 6 = = ? • = p w p w m k s rad u k p 400 005 , 0 2 2 005 , 0 10 . 3 10 5 , 1 8 6 = = = = • • = = = p p p l p p w b ? ? ? ? ? ? + • • = • • = • = x t a t x H t x E a t x H a t x H y y z y y y 400 2 . 10 . . 5 , 1 cos . 10 120 1 ) , ( ) , ( ) , ( ) , ( 6 0 0 p p p m e r r r r r c) ? ? ? ? ? ? • + • - = • · • = x t a t x H a t x E a t x P x y y z z 400 2 10 . . 5 , 1 cos 2 , 1 1 ) , ( ) , ( ) , ( 6 2 p p p r r r r{ } 2 2 2 1 * Re 2 1 * Re 2 1 ) ( m W a a x P n n av r r r r r r r r • E • • = ? ? ? ? ? ? E · • · E • = H · E • = m e m e REFLECTION AND TRANSMISSION OF THE ELECTROMAGNETIC WAVES Normal Incidence onto the Interface of Two Lossy Dielectrics If a uniform plane wave is normally incident on a dielectric boundary from a medium with 1 1 1 , , m e s to the other medium with 2 2 2 , , m e s the wave reflection and transmission can be explained using following schematic : Polarization A plane wave ) , ( t r r r E propagating in the n a r direction can be described in the following general form : ) , ( t r r r E = ( , ) h r t E r h a r + ( , ) v r t E r v a r where the components can be named as ( , ) h r t E r h a r horizontal ( , ) v r t E r v a r vertical Definition: The geometrical place (contour) drawn by the tip point of the ) , ( t r r E vector at a chosen r r point during ) , 0 ( ¥ ¸ t is the polarization of the wave. Types of Polarization (1) Linear Polarization If r r =0 is chosen for the convenience, we have the equation of ) , 0 ( t E r as: ) , 0 ( t E r = h E cos ) ( h t y w - h a r + v E cos ) ( v t y w - v a r where if p y y 2 n v h = - … ... 3 , 2 , 1 , 0 m m m = n , the parametric equations of the contour drawn by the tip of the ) , 0 ( t E r in the t-domain can be obtained as follows: (0, ) h t E = h E cos( ) h t w - y (0, ) v t E = v E cos( 2 ) h t n w - y + p = v E cos( ) h t w - y Thus the relation between the components of ) , 0 ( t E r at every instant t is ( ) . ( ) v v h h t const t = = E E E E or ( ) v h h t E E E which is a line passing through the origin with the slope v h E E in the ( ) h t E - ( ) v t E plane. For the case of (2 1) h v n - = y y + p … ... 3 , 2 , 1 , 0 m m m = n , the parametric equations of the contour drawn by ) , 0 ( t E r in the t-domain are : (0, ) h t E = h E cos( ) h t w - y ) , 0 ( t v E = v E cos (2 1) ( ) h t n w - y + + p = - v E cos( ) h t w - y Thus, this results in ( ) v t = E q ( ) . ( ) v v h h t const t = - = E E E Eor - ) (t h h v E E E equation is a line passing through the origin with the slope (- h v E E ) in the ) (t h E , ) (t v E plane which is (2) Circular Polarization For the right hand circular polarization, the following conditions must be satisfied: 0 E E E = = h v and 2 ) 1 2 ( p y y + = - n v h In this case the contour drawn by the tip of ) (t E is: ) (t E = [ ] ) sin( ) cos( h v h h t a t a o y w y w - + - E r r So the parametric equations of the contour are: = E ) (t v ) (t v E ) (t h E ( ) h t E = o E cos( ) h t w - y ) (t v E = o E sin( ) h t w - y Thus the relation between the components of ) (t E is ( 2 2 ) ( ) 1 h v o o t t ? ? ? ? ? ? + = ? ? ? ? ? ? ? ? ? ? E E E E 2 2 ) ( 2 ) ( o v h t t E E E = + which is a circle centred at the origin with the radius o E in the ) (t h E - ) (t v E plane: Tip of the ) (t E rotates in the mathematical positive direction the right -hand circular polarization. Similiarly, the left -hand circular polarization occurs if the following conditions are met: 0 E E E = = h v and 2 ) 1 2 ( p y y + - = - n v h o E o E o E - o E - ) (t v E ) (t h E Under these conditions the equation of ) (t E is: ) (t E = [ ] ) sin( ) cos( h v h h t a t a o y w y w - - - E r r The parametric equations of the contour are: ) (t h E = o E cos ) ( h t y w - ) (t v E = o E - sin ) ( h t y w - Thus we have 1 2 2 ) ( ) ( = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? E E E E - o v o h t t ? 2 2 ) ( 2 ) ( o v h t t E E E = + which is a circle centred at the origin with the radius o E rotated in the mathematical negative direction in the ) (t h E - ) (t v E plane: o E o E o E - o E - ) (t v E ) (t h E (3)Eliptical Polarization The necessary and sufficient conditins for eliptical polarization are h v E E „ and O v h y y y = - . For example the right- hand eliptical polarization occurs, if 2 p y = o , 2 E E = v , 1 E E = h .at the point r r =0, the equation of ) , 0 ( t E r is: ) , 0 ( t E r = 1 E cos ) ( h t y w - h a r + 2 E sin ) ( h t y w - v a r The parametric equations of the contour drawn by ) , 0 ( t E r are: ) (t h E = 1 E cos ) ( h t y w - ) (t v E = 2 E sin ) ( h t y w - Thus we have 1 2 2 2 1 ( ) ( ) = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? E E E E t t v h which is an ellipse, rotating in the mathematical negative direction in the ) (t h E - ) (t v E plane: We have the left- hand eliptical polarization, If 2 p y - = o , 2 E E = v , 1 E E = h at the point r r =0, the equation of ) , 0 ( t E r is : ) , 0 ( t E r = 1 E cos ) ( h t y w - h a r - 2 E sin ) ( h t y w - v a r The parametric equations of ) (t E are: ) (t h E = 1 E cos ) ( h t y w - ) (t v E = 2 E - sin ) ( h t y w - , thus we get the equation 1 2 2 2 1 ( ) ( ) = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? E E E E - t t v h . Iwhich is an ellipse being rotated in the mathematical negative direction in the ) (t h E - ) (t v E plane: 1 E 2 E 1 E - 2 E - ) (t v E ) (t h E 1 E > 2 E Theorem I : An EM plane wave with the elliptical polarization is the sum of two waves with the linearly polarized in the h a r and v a r directions. Theorem II A linearly polarized wave is the sum of the right -hand and left- hand polarized waves. For example, let us consider an EM wave propagating in the +z direction and polarized in the x- direction. It can be expressed as x o a z jkz e r r - E E = ) ( . ) ( ) ( ) ( z z z lc rc E E E + = r r r where ( ) jkz e y x o lc a j a z - - E E = r r r 2 ) ( 1 E 2 E 1 E - 2 E - 1 E > 2 E ) (t v E ) (t h E ( ) jkz e y x o rc a j a z - + E E = r r r 2 ) ( which are the left- hand and right- hand polarized waves, respectively. EM WAVE PROPAGATION AT THE LOSSY MEDIA Maxwell Equations for the lossy media are : E · = - w j H m H · = ( E + ) e s jw + u J H = 0 E = / r e Since we have the wave solutions for the lossless media, so those solutions can be used for the lossy media as well if we can obtain the wave equations for the lossy media in the same form as the lossless media : E · = - jw H m H · = j w ( ) j s e - E w ur + u J uur = jw E c e + u J where Complex dielectric constant : ' c e e - j ' ' e = ( ) w s e j - ? ' e e , '' s e w For the sourceless lossy medium 0 u J = r , 0 r = From Faraday’s law which is · E ur =-j w H m , we have ( ) · · E uur = -j w m · H ur and using the Ampere law · H ur =jw c E e r and and Gauss law E ur =0 we have ( ) E uur - 2 E ur = 2 c w me E ur and 2 E ur + 2 c w me E ur = 0 So the EM Waves propagating in the + z –directions can be expressed in the form of ) ( ) ( ) ( z a z a z E y y x x E + E = r r r r r ( H r z)= c y x c x y z a z a h h ) ( ) ( E - E r r r r where ' ' ' e e m w s e m e m h j j c c - = - = = , z jk o z jk o x c c e e z - - + E + E = E ) ( r , 1/ 2 1/ 2 [ ( )] [1 ] c c j jk j j j j j s g a + b = = w me = w m e - w s g = w me - we z o z o x e e z c g g - - + E + E = E ) ( r where ) ( ) ( o o o o - - + + E = E E = E + E = E o x z) ( e z j z j o z j z j e e e e e b a b a F - - - F E + { } ) ( Re ) , ( jwt x x e z a t z E = E r r f j { } ( , ) cos[ ] cos[ ] z z x o o o o z t a e t z e t z + -a + - a - E = E w - b + F + E w + b + F r r ( , ) cos[ ] cos[ ] z z o o y o c o c c c z t a e t z e t z + - -a + a - ? ? E E ? ? H = w - b + F - F - w + b + F - F ? ? h h ? ? ? ? r r EXAMPLE: Microwave energy is applied muscle tissue at f=915 MHz ; m s kas / 6 , 1 = s ; 51 = r e ; ; 1 = r m p U , , , , l b a g ? , h =? b a p p we s e m e m g b a we s me me g p j e e e e e j j j jw jk j j jw jw jk j j j j j o o r r c c c + = + = = = = - = - = = + = - = = = - - - 7 , 142 5 , 40 . 818 , 148 . 084 , 1 . . 92 , 136 . 175 , 1 . 92 , 136 10 . 85 , 8 . 51 . 10 . 915 . 2 6 , 1 1 10 . 3 51 10 . 915 . 2 . 1 1 13 , 74 987 , 17 90 . 66 , 31 2 12 6 8 6 044 , 0 7 , 142 2 2 = = = p b p l m 8 , 15 . . 7 , 48 ) 1 ( j r o c c e w j = - = = e s e m m e m h W U 6 6 2 .915.10 40,288.10 142,7 p w p = = = b m/s D f, tg 1 << d ? the medium is good dielectric so we can apply the Binomial series THE PENETRATION DEPTH: The wave comfort’s space go down 1/e. z o e z a - + + E = E . ) ( ? z=1/d Amplitude z d 2d EXAMPLE FM broadcast signal is propagating in the +y direction which is given by ( ) ( ) . . 10 . 92 , 2 68 , 0 3 m A z x y j a j a e y r r r + - = H - - p Find out f, l b , , p U and ) ( y E r . SOLUTION: lossless media so 0 , 0 = = a s . ( ) x z y j y s m p a j a e y a y U m MHz c f c c k j r r r r r r + = E ? = = = H · = E = = = = = = = ? = = = = = ? = - - p p p e m h h h p p b p l p b p w b w p w e m w b b g 68 , 0 3 0 0 0 8 0 0 0 . 10 . 92 , 2 . 120 ) ( 120 ) ( ) ( 10 . 3 94 , 2 68 , 0 2 2 102 2 2 . 68 , 0 SIMPLE EM WAVES PROPAGATION IN SIMPLE MEDIA 1)LOSSY MEDIA: = g propagation constant we s me w me w b a g j j j jk j c c - = = = + = 1 Sınusoidal E and H waves propagating in the +z direction : c c c c o z c x z j z c x z j z y j z t e t z e e z e e z j h we s e m h j j b w h h a b a b a — = ? ? ? ? ? ? - = - + - E - = H ? E - = H E = E - + - + - - + - - + 2 1 0 0 0 1 ) cos( ) , ( ) ( ) ( 2)GOOD DIELECTRIC: Electric currents in the lossy media: E = E = + = E + E = H · r r r r r r r r r we s we s j j j j j j d c d c d j r t j r c j r we s we s d = E E = = r r r r d c c j j tan Dominant current in a good dielectric is displacement type of current. Behaviour for good dielectric at an operation frequency w ? = c d tan we s << 1 2 2 2 w e s <<1 , so this term can be neglected. W ? ? ? ? ? ? + @ we s e m h 2 1 j c EXAMPLE: (Dry soil) Consider a certain type of dry soil with the following properties: 7 0 10 4 - = = p m m H/m, 0 3e e = , f =10 MHz, m S 5 10 - = s Find the b a , and skin depth. SOLUTION: c d tan = we s = 006 , 0 10 . 85 , 8 . 3 . . 2 10 12 5 @ - - p c d tan <<1 so dry soil is a good dielectric. K – - + – = – ? << 2 ! 2 ) 1 ( 1 ) 1 ( 1 : x n n nx x x NOTE n ( ) ? ? ? ? ? ? + - = ? ? ? ? ? ? ? ? + - = 2 2 2 2 tan 8 1 tan 2 1 1 8 1 2 1 1 c c j j j j d d me w g e w s we s me w g s rad / 10 . 2 7 p w = ( ) { } { } m rad m Np j j 3629 , 0 Im 10 Re 006 , 0 8 1 003 , 0 1 10 . 85 , 8 . 3 . 10 4 . 10 . 2 . 3 2 12 7 7 = = = = ? ? ? ? ? ? + - = - - - g b g a p p g skin depth= a 1 =920 m EXAMPLE:(Propagation in Galium Arsenide) f=10 GHz, 4 10 . 5 tan , 1 , 9 , 12 - = = = c r r d m e are given.Find the c p U h b a , , , . SOLUTION: 4 10 . 5 tan - = c d <<1 so the galium arsenide is a good dielectric in 10 GHz. c d tan = c d we s we s tan = ? m Np r c 188 , 0 1 . 2 tan 2 0 0 @ = ? ? ? ? ? ? ? ? = e e m d we e m s a ( ) ( ) m rad r c c c 23 , 752 neglected be can term this so , 1 tan tan 8 1 1 8 1 1 2 2 2 2 2 = = << ? ? ? ? ? ? ? ? ? ? ? ? ? ? + = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + = e w b d d me w e w s me w b m c U s m r p 3 7 10 . 35 , 8 2 10 . 35 , 8 - = = = = b p l e W + @ ? ? ? ? ? ? ? ? + = ? ? ? ? ? ? + @ - - 4 4 0 0 10 . 5 , 2 105 2 10 . 5 1 1 2 1 j j j r c e e m we s e m h 3)GOOD CONDUCTOR: E = E = + = E + E = H · r r r r r r r r r we s we s j j j j j j d c d c c c d j tan j sE s d = = = weE we r r r r Dominant current in a good conductor is conduction type of current. Behaviour : good dielectric at an operation frequency w ? = c d tan we s >>1 { } { } ( ) factor n attenuatio j j e e j j j j j j 2 Im Re 2 2 2 2 1 2 1 2 4 2 1 2 1 2 2 1 wms b a g b g a b a g we s me w g we s me w g we s me w we s me w g p p p = = ? = = = ? + = ? ? ? ? ? ? ? ? + ? ? ? ? ? ? = ? ? ? ? ? ? = ? ? ? ? ? ? ? ? = ? ? ? ? ? ? - = + - - W = @ ? ? ? ? ? ? - = = = = = - 4 2 1 1 2 2 2 p s wm s wm we s e m h ms p b p l ms w b w j c s m p e j j m f U EXAMPLE: For copper f:100 MHz , m mm depth skin m a d 28 , 3 00328 , 0 1 = = = = Skin depth is very short so EM waves cannot reached to the inside of copper. EXAMPLE: Comparasion of skin depths f=10 kHz, 7 0 10 4 - = = p m m H/m Sea water: 81 = r e , m S 4 = s , 1 10 . 84 , 8 tan 4 >> @ c d good conductor d m 52 , 2 1 @ = a Humid soil: 10 = r e , m S 2 10 - = s , 1 10 . 8 , 1 tan 3 >> @ c d good conductor d m 50 1 @ = a Dry soil: 3 = r e , m S 4 10 - = s , 1 9 , 59 tan >> @ c d good conductor d m 503 1 @ = a EXAMPLE: VLF waves propagation in the ocean. m S 4 = s , 81 = r e , 1 = r m . Sınusoidal E wave is propagating in the +z direction; ( ) m V z x z t e a t z ) 10 6 cos( , 3 0 b p a - E = E - r r Find ; a)the C P U h l b a , , , , and skin depth. b) ) , ( t z H r c)There is a submarine in the 100 m depth and it can measure m V / 1m and bigger than m V / 1m electric field magnitude.What are the magnitudes of E and H waves for the communication with the submarine? SOLUTION: a) Hz f s rad 3 3 10 . 3 10 . 6 = = p w ) ( 1 10 . 96 , 2 10 . 85 , 8 . 81 . 10 . 6 4 tan 5 12 3 conductor good c >> = = = - p we s d m rad m Np or 218 , 0 2 = = = wms b a m e e j j f f U m j j c c s m p c 59 , 4 1 10 . 70 , 7 1 10 . 66 , 8 9 , 28 . 10 . 3 2 2 89 , 2 218 , 0 2 2 4 2 2 1 4 3 = = W = = @ ? ? ? ? ? ? - = = = = = = = = = - - a d h g wm we s e m h l l p p b w p b p l p f h b) m A z y z c y z t e a t z z t e a t z c ) 4 218 , 0 10 . 6 cos( . . . 13 . ) , ( ) 10 . 6 cos( . . ) , ( 3 218 , 0 0 3 0 p p j b p h h a - - E = H - - E = H - - r r r r c) m kA c m kV z e e 9 , 36 13 84 , 2 84 , 2 10 . 1 . 0 0 min 0 min 0 0 6 100 . 218 , 0 0 218 , 0 0 = E = E = H ? = E ‡ E ‡ E = E - - - h DISPERSION PROPERTY OF A MEDIUM PHASE VELOCITY Phase velocity is given by b w = p u where • If w b » b is proportional with w , such as the situation in the lossless simple media ( 0 = s ) : me w b = , 1 p const u = = me • ) (w b b = b is the nonlinear function of w , such as the situation in the lossy simple media ( 0 s „ ) : ( ) p p u u w = . GROUP VELOCITY Let us consider a signal packet consisting of the two frequencies of } { ) ( ), 0 0 ( w w w w D - D + : [ ]+ = D + - D + z t o x E a E t z ) ( ) ( 0 0 cos { ) , ( b b w w r [ ] } ) ( ) ( 0 0 0 cos z t E b b w w D + - D - or ) ( ) ( 0 0 0 ) , ( cos cos 2 z t z t E E t z b w b w - D - D = x a v At the z=0 plane thetime variation of the signal can be given by ) ( ) ( cos cos 0 ) , 0 ( 2 t t o t E E w w D = . where o w w <<< D . w w D - 0 w w D + 0 w The space variation of the signal at the instant t=0 is ) ( ) ( cos cos 0 ) 0 , ( 2 z z o z E E b b D = . where o b b <<< D dır. t w • p u is the velocity of the equiphase plane of the carrier, so we have K z t o o = - b w K z z t t o O = - D + D + ) ( ) ( b w which resulted in O O t p t z u b w = D D = ? D 0 lim • g u is the velocity of equiphase plane of the envelope, so we have T z z t t T z t = D + D - D + D = D - D ) ( ) ( b w b w which resulted in w b d d t z t g u 1 lim 0 = D D = ? D RELATION BETWEEN THE PHASE AND GROUP VELOCITIES w w w b d du d d p p g g u u u - = ? ? ? ? ? ? = - 1 1 z = ? = 0 w d du SBT p p u no dispersion 0 < w d dup normal dispersion 0 > w d dup anormal dispersion Example: A narrow-band signal is propagating within a simple lossy medium.Carrier frequency:550MHz; 1 ; 5 , 2 ; 2 , 0 = = = r r c tg m e d . Required : ? , , , , = p g u u l b a . Investigate dispersion of the medium. Solution: m S tg c 5 10 53 , 1 - · = ? = s we s d sn m d d u sn m u m rad m Np g p 8 1 8 2 3 10 907 , 1 10 888 , 1 0183 , 0 8 1 1 10 82 , 1 2 · @ ? ? ? ? ? ? @ · @ @ @ ? ? ? ? ? ? ? ? ? ? ? ? ? ? + @ · @ @ - - w b b w we s e m w b e m s a Since p g u u > , anormal dispersion exists. DISSIPATED POWER WITHIN THETHICK MATERIAL wITH FINITE DISSIPATED POWER WITHIN THETHICK MATERIAL wITH FINITE DISSIPATED POWER WITHIN THETHICK MATERIAL wITH FINITE DISSIPATED POWER WITHIN THETHICK MATERIAL wITH FINITE CONDUCTIVITY CONDUCTIVITY CONDUCTIVITY CONDUCTIVITY ( ) net C dis v P average J E dv • ? ? = ? ? ? r r ( ) 2 2 1 1 3 2 2 C C O z average J E J E W e m * • • - d = = sE r r r r • Net dissipated power within the V volume: x y z d b l EM WAVE 2 2 2 0 0 2 2 2 1 2 net b b z dis dzdydx P e ¥ - - - d = sE ? ? ? l l W 2 0 1 2 n e t d is b P = s E dl W • So dissipated power within the material with the unit area and infinite thickness: 2 0 1 1 1 2 0 2 4 net dis P J = sE d = d s W CURRENT DISTRIBUTION WITHIN THE GOOD CONDUCTOR Red : Real Distribution; Green:Equivalent Distribution giving the same total current J 0 J e JO 2 0 J d z EXAMPLE: Calculation of the resistance of the copper wire at dc and a high frequency with the length of L=1km and radius a=1mm 3 7 6 3 1 7 10 5,48 5,8 10 10 10 41,5 5,8 10 2 DC MHZ L R S R a - = = = W s · · P · = = W · · Pd • For a mm << = 0661 , 0 d , 2 a s P d is used. REFLECTION AND TRANSMISSION OF THE ELECTROMAGNETIC WAVES NORMAL INCIDENCE ONTO THE INTERFACE OF TWO LOSSY DIELECTRICS d a L If a uniform plane wave is normally incident on a dielectric boundary from a medium with 1 1 1 , , m e s to the other medium with 2 2 2 , , m e s the wave reflection and transmission can be explained using following schematic : s 1 , m 1 , e 1 s 2 , m 2 , e 2 + 1 x E 2 x E - 1 x E z<0 z=0 z>0 Let us consider that E + x1 is a linearly polarized electromagnetic plane wave in the x direction, and is incident in the normal direction onto the boundary plane at z=0 . The medium in the right is assumed to be effectively infinite in extent so that there is no reflected wave in that region. So let us express the incident, reflected and transmitted wave components of the electric and magnetic fields: Incident EM Wave Components: For the region (z<0) the incident electric and magnetic wave components can be given , respectively z x x e E E 1 10 1 g - + + = z c x y e E H 1 1 10 1 g h - + = Reflected Wave Components : In the region of (z<0) ,the reflected electric and magnetic wave components can be given , respectively : z x x e E E 1 10 1 g - - = z c x y e E H 1 1 10 1 g h = - - Transmitted Wave Components: In the region of (z>0) the transmitted electric and magnetic wave components can be given , respectively : z x x E E E 2 20 2 g - + = z c x y e E H 2 2 20 2 g h - + = So these EM field components have to satisfy the Boundary Conditions at (z = 0) 20 10 10 x x x E E E = + - + and 2 1 1 y y y H H H = - - + or 2 20 2 10 1 10 c x c x c x E E E h h h = - - + For an imperfect dielectric material : Just considering the effect of s to e will solve the “lossy dielectric” matter. If 1 << we s , using e m w g ¢ = j and binomial series leads; ? ? ? ? ? ? = e m s a 2 and; ( ) 2 1 2 2 2 8 1 ? ? ? ? ? ? ? ? + · = e w s me w b 1 1 1 1 1 c c j j jk me w b a g = + = = 2 2 2 2 2 c c j j jk me w b a g = + = = i c r c c c j 1 1 1 1 h h e m h + = = i c r c c c j 2 2 2 2 h h e m h + = = ( ) ( ) 1 2 1 2 10 10 c c c c x x E E h h h h + - = = G + - Reflection coefficient ? ? ? ? ? ? - = we s e e j 1 '+ = T 10 20 x x E E Transmission coefficient G + =1 T 2 1 1 2 c c c T h h h + = + - G = 10 10 x x E E Transmitted and Reflected Components of EM Wave: ( ) + - + F + - = 10 1 cos 1 10 z t e E a E z x x i b w a r r ( ) 1 1 1 10 10 1 cos c z t e E a H z c x y i h a b w h F - F + - = + - + r r [ ] ( ) z z c x x avi a e E H E P c r r r r F = · = - + 1 1 1 10 cos 2 1 * Re 2 1 2 2 h a h (W/m 2 ) ( ) r z x x r z t e E a E F + F + + G = + + 10 1 cos 1 10 b w a r r ( ) 1 1 1 10 10 1 cos c r z c x y r z t e E a H h a b w h F - F + F + - G - = + + r r [ ] ( ) ( ) z z c x r r avr a e E H E P c r r r r - F G = · = + 1 1 1 10 cos 2 * Re 2 1 2 2 2 h a h V Transmitted Power Density; ( ) T z x x t z t e E a E F + F + - T = + - + 20 2 cos 2 10 b w a r r [ ] ( ) ) ( z z c x t t avt a e E H E P c r r r v F T = · = - + 2 2 2 10 cos 2 * Re 2 1 2 2 2 h a h Complex Poynting Theorem ( ) z avt z avr avi a S P V in dissipated power thermal a S P P r r r r r D + D = D - . ). ( EXAMPLE 1: A uniform plane wave is normally incident on the boundary between two lossy dielectric media with the following parameters; Medium 1: ¸ r1 = 2 s 1 = 0,2 1/W m m r1 =1 Medium 2: ¸ r2 =4 s 2 = 0,1 1/W m m 2 =1 ƒ = 10 GHz 1 10 = + x E (dimensionless) calculate the reflection and transmission coefficient and draw the phasor diagram for the incident, reflected and transmitted EM phasors and graphics of the variations of amplitudes of electric fields by distance. P i av P r av P t av Z=0 +z Solution: Medium 1 Medium 2 18 , 0 1 1 = we s (lossy dielectric) 05 , 0 2 2 = we s (low loss dielectric) 1 1 1 97 , 297 579 , 26 9 , 84 299 b a g j j + = + = — = a 1 =0,0704 Np/m b 1 =941,89 rad/m 25 , 410 46 , 10 57 , 88 4 , 419 2 j + = — = g a 2 @ 0 b 2 = 2664,07 rad/m ( ) 2 1 4 2 1 10 3 2 1 1 1 - - - = ? ? ? ? ? ? ? ? - = j j r c we s e m h ( ) 2 1 4 2 1 10 5 , 7 4 1 2 2 - - - = ? ? ? ? ? ? ? ? - = j j r c we s e m h ( ) ( ) - — = + - = G 52 , 169 17 , 0 1 2 1 2 c c c c h h h h - — = G + = T 15 , 2 8321 , 0 1 Amplitude HOMEWORK: Find the reflected and transmitted powers (clue: use the following formulas) Incident power: ( ) z c x i e E av P 1 1 10 2 2 2 1 a h - + = r 1 10 = + x E ( ) z z c x avi a e E P c r r 1 1 10 cos 2 1 2 1 2 h a h F = - + Reflected Power ( ) z a c r 1 cos h F ?T?.?E x10 + ?.e -a 2 z ?E x10 + ??G ?.e a 1 z ?E x10 + ?e -a 1z s 1, m 1 , e 1 s 2 , m 2 , e 2 E PLANE E x10 - -169,52 E x10 + -2,15 z c x avr e E P 1 1 10 2 2 2 2 1 a h G - = + rTransmitted Power: z z c x avt a e E P c r r ) cos( 2 1 2 2 2 10 2 2 2 h a h F T = - + NORMAL INCIDENCE ONTO THE INTERFACE OF THE PERFECT TWO DIELECTRICS 0 0 numbers. real are and ? ? . ? ? ? 1 2 1 2 1 2 1 2 2 1 2 1 2 1 0 2r 2 2 0 1r 1r 1 1 1 > + - = G < + - = G < > = = = h h h h h h h h h h h h h h h m h m m r For the region of z < 0, For the region z>0, i E r iav P r i H r i E r i H r iav P r y z Z=0 z 2 1 h h t E r Pt t H r Medium 2 Medium 1 ( ) ( ) z j x t y z j x r z j x t x z j x r e E H a e E H e E E a e E E 2 1 2 1 10 2 10 1 10 10 1 , 1 , b b b b h h - + + - + + G – = G – = G – = G – = ) ( ) ( ) ( ) ( E , 0 1 1 1 10 1 1 x 1 2 z j z j x x x e e E z E z E z z b b h h G + = + = > ? > - + - + FINDING STANDING WAVE PARAMETERS Standing Wave Pattern : Variations of ) ( 1 z z E x - z j z j x e e E z 1 1 1 10 x ) ( E b b G + = - + z z z e e z j z j 1 2 1 2 2 2 1 2 2 cos 2 1 ( 2 sin ) 2 cos 1 ( 1 . 1 1 b b b b b G + G + = G + G + = G + = - + + + G + G + - G + G + = 10 1 1 1 1 2 10 2 cos 1 2 sin tan arg cos . 2 cos 2 1 . ) . ( j b b b w b z z z t z x E t z E x 4 ) 1 2 ( 2 ) 1 2 ( ) 1 2 ( 2 * 2 2 , 1 , 0 2 2 2 * 1 min 1 min min 1 1 max 1 max 1 max 1 l b p p b l b p b p b - = ? - = - = = ? = - - = = n z n z n Z n z n z n n Z Finding max. and min. with the alternatif way (Grafic solution): 1 G - = G + - - = - = G + = G + - - = = 1 1 ......; 2 , 1 , 0 , ) 1 2 ( 2 1 1 .....; 2 , 1 , 0 , 2 2 min 1 max 1 2 min 1 2 max 1 z j z j e n n z e n n z b b p b p b z j x x e E z E 1 1 2 10 1 ) ( b G + = + i jG r G z j e 1 2 1 b G + z 1 2b plane - G 1 0 1 1 1 0 £ G £ ? ¥ = ? = G ¥ < < = ? = G VSWR VSWR VSWR (Voltage Standing Wave Ratio) [ ] ) ( ) 1 ( ) ( ) ( ) ( ) ( 1 1 1 1 1 10 10 1 1 1 z j z j z j x z j z j x x x x e e e E e e E z E z E z E b b b b b - - + - + - + - G + G + = G + = + = z e E T z E j z j x x 1 10 1 sin 2 ) ( 1 b b G + = - + { } 4 4 3 4 4 2 1 4 4 4 3 4 4 4 2 1 STANDING MOVING i x jwt x x t z z t E T e z E t z w b j b w e sin sin 2 ) cos( ) ( Re ) . ( 1 1 10 1 1 G - + - = = = + G - G + = = 1 1 min 1 max 1 E E VSWR 0 : 1 < z medium 0 : 2 > z medium NORMAL INCIDENCE ONTO THE INTERFACE OF THE PERFECT CONDUCTOR WITH THE PERFECT DIELECTRIC 0 . . 1 1 1 0 1 = = = s b e m w g j j 0 , 1 1 . 2 2 2 1 2 2 2 2 2 ? ¥ ? ? ? ? ? ? ? ? ? ? ? ? ? - = h s e s e m h w j 1 1 2 1 2 - = + - = G h h h h G = -1 , T= 0 G = -1 ? no transmission, EM wave is reflected T= 0 ? there is no transmited EM wave 4 4 3 4 4 2 1 r r r r z j x z j z j m r i a e e E E E E 1 1 1 sin 2 1 ) ( . b b b - - - = + = y X 0 . 1 1 1 ? s m e Z<0 z=0 z ? Z>0 0 2 ? h ¥ ? 2 2 2 .s m e 1 h0 , / . cos . 2 ) ( 0 , / . sin . . 2 2 1 1 1 1 1 2 1 1 1 1 = = + = + = = - = - H m A a Em H a e e E r H i H H E m V o Em j E y z y z j z j m x z r r r r r r r r r r b h h b b b If we multiply with e jwt and take the real part, we get the instantaneous functions of the EM waves: STANDING WAVE EQUATIONS z z Em j H x E P v 1 1 1 2 1 1 cos . sin . Re 2 1 Re 2 1 1 b b h - = = * D r r 0 Re 2 / 1 = j e j 0 0 2 1 = = ? D D v v P P ... 2 , 1 , 0 2 min 2 2 min min min 0 sin 0 ) , ( maximum ) , ( H ) [0, t (1) 1 1 1 1 1 1 1 1 = - = ? - = - = - = ? - = = ? = ? ¥ ¸ n n Z n n Z n Z n z z t z E t x l l l p p b p p b b r m A a t z E t z H m V a t z E t z E y m x m / cos ). cos( . 2 ) , ( / sin ). sin( . 2 ) , ( 1 1 1 1 1 r r r r w b h w b = =... 2 , 1 , 0 4 ) 1 2 ( 4 ) 1 2 ( 0 ) , ( 2 ) 1 2 ( z points aximum t) (z, E (2) max 1 max 1 max 1 1 = + - = ? + - = = + - = ? n n z n z t z H n m l l p b Distance of the two succesive maximal (or minimal) points are 2 l . Distance of the two sucessive max.(or min.) points are 4 l . J S = n a r x 1 H r Z n a a r r - = NORMAL INCIDENCE ONTO THE INTERFACE OF THE PERFECT CONDUCTOR WITH THE PERFECT DIELECTRIC 2 3l - 2 l - l - z Em E 1 1 sin 2 b = 4 3l - l - z Em H 1 1 1 cos 2 b h = 2 l - 4 l - Hz Ez z z y X 0 . 1 1 1 ? s m e Z<0 z=0 z ? Z>0 0 2 ? h ¥ ? 2 2 2 .s m e 1 h 0 . . 1 1 1 0 1 = = = s b e m w g j j 0 , 1 1 . 2 2 2 1 2 2 2 2 2 ? ¥ ? ? ? ? ? ? ? ? ? ? ? ? ? - = h s e s e m h w j 1 1 2 1 2 - = + - = G h h h h G = -1 , T= 0 G = -1 ? no transmission, EM wave is reflected T= 0 ? there is no transmited EM wave 4 4 3 4 4 2 1 r r r r z j x z j z j m r i a e e E E E E 1 1 1 sin 2 1 ) ( . b b b - - - = + = 0 , / . cos . 2 ) ( 0 , / . sin . . 2 2 1 1 1 1 1 2 1 1 1 1 = = + = + = = - = - H m A a Em H a e e E r H i H H E m V o Em j E y z y z j z j m x z r r r r r r r r r r b h h b b b If we multiply with e jwt and take the real part, we get the instantaneous functions of the EM waves: STANDING WAVE EQUATIONS m A a t z E t z H m V a t z E t z E y m x m / cos ). cos( . 2 ) , ( / sin ). sin( . 2 ) , ( 1 1 1 1 1 r r r r w b h w b = =z z Em j H x E P v 1 1 1 2 1 1 cos . sin . Re 2 1 Re 2 1 1 b b h - = = * D r r 0 Re 2 / 1 = j e j 0 0 2 1 = = ? D D v v P P ... 2 , 1 , 0 2 min 2 2 min min min 0 sin 0 ) , ( maximum ) , ( H ) [0, t (1) 1 1 1 1 1 1 1 1 = - = ? - = - = - = ? - = = ? = ? ¥ ¸ n n Z n n Z n Z n z z t z E t x l l l p p b p p b b r ... 2 , 1 , 0 4 ) 1 2 ( 4 ) 1 2 ( 0 ) , ( 2 ) 1 2 ( z points aximum t) (z, E (2) max 1 max 1 max 1 1 = + - = ? + - = = + - = ? n n z n z t z H n m l l p b Distance of the two succesive maximal (or minimal) points are 2 l . Distance of the two sucessive max.(or min.) points are 4 l . 2 3l - 2 l - l - z Em E 1 1 sin 2 b = Ez z J S = n a r x 1 H r Z n a a r r - = sin ( 1 1 ni x j io y i r a j io y i e E a E e E a E q b b - - = ? = r r r r r r ) cos sin ( 1 1 1 ) sin cos ( ) ( 1 i i z x j i z i x io i i ni i e a a E H E a H q q b q q h h + - + - = ? · = r r r r r r ) cos sin ( 1 1 r r nr z x j ro y r r a j ro y r e E a E e E a E q q b b - - - = ? = r r r r r r ) cos sin ( 1 1 1 ) sin cos ( ) ( 1 r r z x j r z r x ro r r nr r e a a E H E a H q q b q q h h + - + - = ? · = r r r r r r ) cos sin ( 2 2 t t nt z x j to y t r a j to y t e E a E e E a E q q b b - - - = ? = r r r r r r ) cos sin ( 2 2 2 ) sin cos ( ) ( 1 t t z x j t z t x to t t nt t e a a E H E a H q q b q q h h + - + - = ? · = r r r r r r * For z = 0; ) 0 ( ) 0 ( ) 0 ( ) 0 ( 2 1 2 1 = = = = = = z H z H z E z E t t t t t r i x j to x j ro x j io e E e E e E q b q b q b sin sin sin 2 1 1 - - - = + t r i x j t to x j r ro x j i io e E e E e E q b q b q b q h q q h sin 2 sin sin 1 2 1 1 cos ) cos cos ( 1 - - - - = + - * r i q q = (Snell’s Law of Reflection) 2 1 1 2 1 2 sin sin u u w u u w t i = = = b b q q 1 2 2 2 1 1 sin sin , n n u c n u c n t i = ? = = q q 4 3l - l - z Em H 1 1 1 cos 2 b h = 2 l - 4 l - Hz z By using Snell’s Law of Reflection, equation ( ) and ( ); to ro io E E E = + to t ro io i E E E q h q h cos 1 ) ( cos 1 2 1 - = + - - ) cos / ( ) cos / ( ) cos / ( ) cos / ( 1 2 1 2 i t i t io ro E E q h q h q h q h + - = = G ^ ) cos / ( ) cos / ( cos / 2 1 2 2 i t t io to E E q h q h q h + = = T ^ Reflection and transmission coefficients are related by the following equation: ^ ^ T = G + 1 If medium 2 is perfect conductor, 0 2 = h . We have 0 , 1 = T - = G ^ ^ . The angle ^ B q is called the BREWSTER ANGLE of no reflection for the case of perpendicular polarization. ^ ^ ? ? ? ? ? ? ? ? - = - = ? = B t t t B q h h q q q h q h 2 2 2 1 2 1 2 sin 1 sin 1 cos cos cos ( ) ( ) 2 2 1 1 2 2 1 2 / 1 / 1 sin m m e m e m q - - = ^ B For nonmagnetic media, 0 2 1 m m m = = , the right side of the above equation becomes infinitive, and ^ B q does not exist. In the case of 2 1 e e = and 2 1 m m „ ,the equation reduces to ( ) 2 1 / 1 1 sin m m q + = ^ B which does have a solution whether 2 1 / m m is greater or less than unity. PARALLEL POLARIZATION t z t x nt r z r x nr i z i x ni a a a a a a a a a q q q q q q cos sin cos sin cos sin r r r r r r r r r + = - = + = ) cos sin ( 1 1 1 1 i i ni z x j io y i r a j io y i e E a H e E a H q q b b h h + - - = ? = r r r r r r ) cos sin ( 1 1 ) sin cos ( ) ( i i z x j i z i x io i ni i i e a a E E a H E q q b q q h + - - = ? · = r r r r r r ) cos sin ( 1 1 1 1 r r nr z x j ro y r r a j ro y r e E a H e E a H q q b b h h + - - - = ? - = r r r r r r ) cos sin ( 1 1 ) sin cos ( ) ( r r z x j r z r x ro r nr r r e a a E E a H E q q b q q h + - + = ? · = r r r r r r ) cos sin ( 2 2 2 2 t t nt z x j to y t r a j to y t e E a H e E a H q q b b h h + - - = ? = r r r r r r ) cos sin ( 2 2 ) sin cos ( ) ( t t z x j t z t x to t nt t t e a a E E a H E q q b q q h + - - = ? · = r r r r r r Continuity requirements for the tangential components of E and H at z = 0 lead again to Snell’s laws of reflection, as well asto the following two equations: to ro io t to i ro io E E E E E E 2 1 1 ) ( 1 cos cos ) ( h h q q = + = + Solving for ro E and to E in terms of io E , we obtain the FRESNEL’S EQUATIONS. i t i t io ro II E E q h q h q h q h cos cos cos cos 1 2 1 2 + - = = G i t i io to II E E q h q h q h cos cos cos 2 1 2 2 + = = T i t II II q q cos cos 1 T = G + II G goes to zero when the angle of incidence i q equals to BII q . ( ) ( ) 2 2 1 1 1 1 2 2 / 1 / 1 sin e e e m e m q - - = BII If 0 2 1 m m m = = , a reflection-free condition is obtained when the angle of incidence in medium1 equals to the Brewster angle , BII q . ( ) 2 1 / 1 1 sin e e q + = BII Homework: The dielectric constant of pure water is 80. (a) Determine the Brewster angle for parallel polarization, BII q , and the corresponding angle of transmission. (b) A plane wave with perpendicular polarization is incident from air on water surface at i q = BII q .Find the reflection and transmission coefficients. TOTAL REFLECTION 2 2 1 1 2 1 sin sin r r o o r r o o i t w w e m e m e m e m b b q q = = 2 1 0 2 1 sin sin r r i t e e q q m m m = ? = = *For 2 1 e e > - that is, when the wave in medium 1 is incident on a less dense medium 2. In that case t q > i q . Since t q increases with i q , an interesting situation arises when t q = 2 / p . c q = i q (critical angle) and 1 2 1 1 2 sin sin r r c r r c e e q e e q - = ? = *For i q > c q ; 1 sin sin 1 cos , sin 2 2 1 2 1 2 - – = - = > i r r t t r r i j q e e q q e e q In medium 2, the unit vector nt a in the direction of propagation of a typical transmitted wave. t z t x nt a a a q q cos sin + = Both t E and t H vary spatially in accordance with the following factor: ) cos sin ( 2 2 t t nt z x j R a j e e q q b b + - - = i x i q e e b b q e e b a sin ) / ( , 1 sin ) / ( 2 1 2 2 2 2 1 2 2 = - = For i q > c q , a wave exists along the interface(in x direction), which is attenuated exponentially in medium 2 in the normal direction(z direction). This wave is tightly bound to the interface and is called a SURFACE WAVE.