ECE 6340 Intermediate EM Waves. Fall Prof. David R. Jackson Dept. of ECE. Notes 6

Transcription

1 ECE 6340 Intermediate EM Waves Fall 016 Prof. David R. Jackson Dept. of ECE Notes 6 1

2 Power Dissipated by Current Work given to a collection of electric charges movg an electric field: ( qe ) ( ρ S E ) W = = v Note: The magnetic field never does any work: ( B) 0 q v = Power dissipated per unit volume: This could be conduction current or impressed current. W Pd = = ρv E = ( ρv v ) E= J E S t t ρ v E S = v t J v = ρ v v q= ρ S Note: We assume no crease ketic energy, so the power goes to heat.

3 Power Dissipated by Current (cont.) Power dissipated per unit volume for ohmic current (we assume here a simple lear media that obeys Ohm s law): ( ) ( ) c d = = σ = σ = σ = σ P J E E E E E E E Note: E E = E E Hence, we have P d = σe 3

4 Power Dissipated by Current (cont.) Return to the general result: P J E d = This is the power dissipated per unit volume. From this we can also write the power generated per unit volume due to an impressed source current: P = J E s i 4

5 Poyntg Theorem: Time-Doma From these we obta B E = M t D H = J + t B H ( E) = M H H t D E ( H ) = J E + E t Subtract, and use the followg vector identity: H ( E) E ( H ) = ( E H ) 5

6 Poyntg Theorem: Time-Doma (cont.) We then have B ( E H ) = M H J E H E t D t Now let J = J +σ E i M = M i so that i i D ( E H ) = J E σ E M H E H t B t 6

7 Poyntg Theorem: Time-Doma (cont.) Next, use E D = ε E t E t Assume: D= εe (simple lear media) and E E t t t = ( E E) = E Hence E ε ε 1 E D E E = = t t t And similarly, H B = t t 1 µ H 7

8 Poyntg Theorem: Time-Doma (cont.) Defe S E H We then have i 1 1 S = J E σ E M H ε E + µ H t i Next, tegrate throughout and use the divergence theorem: ( i i ) 1 1 S nˆ ds = J E M H d σ E d ε + µ d t E H S Note: We are assumg the volume to be stationary (not movg) here. 8

9 Poyntg Theorem: Time-Doma (cont.) ( i i ) 1 1 S nˆ ds = J E M H d σ E d ε + µ d t E H S Interpretation: W W e m P d s 1 = ε d = 1 = µ d = = σ d = ( i i ) P = S nˆ ds = power flowg out of f E E H S stored electric energy stored magnetic energy dissipated power P = J E M H d = source power (see next figure) S 9

10 Poyntg Theorem: Time-Doma (cont.) ( i i ) 1 1 S nˆ ds = J E M H d σ E d ε + µ d t E H S P = P P W + W t ( ) f s d e m or P = P + P + W + W t ( ) s d f e m 10

11 Poyntg Theorem: Time-Doma (cont.) P = P + P + W + W t ( ) s d f e m Power flow out of surface E ε, µ, σ H source 11

12 Poyntg Theorem: Note on Interpretation S = E H Does the Poyntg vector really represent local power flow? i 1 1 S = J E σ E M H ε E + µ H t i Consider: S = S + A A = Arbitrary vector function Note that S = S This new Poyntg vector is equally valid! They both give same TOTAL power flowg out of the volume, but different local power flow. 1

13 Note on Interpretation (cont.) Another dilemma : A static pot charge is sittg next to a bar magnet. S = E H 0 N S q Is there really power flowg space? Note: There certaly must be zero net power out of any closed surface: P = P + P + W + W t ( ) s d f e m 13

14 Note on Interpretation (cont.) Bottom le: We always get the correct result if we assume that the Poyntg vector represents local power flow. Because In a practical measurement, all we can ever really measure is the power flowg through a closed surface. Comment: At high frequency, where the power flow can be visualized as "photons" movg space, it becomes more plausible to thk of local power flow. In such situations, the Poyntg vector has always given the correct result that agrees with measurements. 14

15 Complex Poyntg Theorem Frequency doma: Hence ( ) E = M jωµ H i H = J + jωε E ( ) * i * H E M H j H = ( ) * i* * ωεc E H = E J j E i c ωµ Subtract and use the followg vector identity: Hence ( A B) = B ( A) A ( B) ( E H * ) H * ( E ) E ( H *) = * i * i* * ωµ ωεc E H = M H E J j H + j E Generalized lear media: ε may be complex. ε = ε j σ c ω 15

16 Complex Poyntg Theorem (cont.) Defe 1 S = E H * (complex Poyntg vector) Note: 1 ( * Re S = Re E H ) = E H = S Then 1 1 S = E J + M H + j ω ε E µ H ( i * i *) ( ) * c Next use ε = ε jε ε = ε + jε * c c c c c c µ = µ j µ Next, collect real and imagary parts the last term on the RHS. 16

17 Complex Poyntg Theorem (cont.) S = E J + M H + j E H E + H ( i* i * ) ( ) ( ω ε ) c µ ω ε c µ or S = E J + M H + j E H E + H 4 4 ( i* i * ) ( ω ) ε c µ ω ε c µ Next, tegrate over a volume and apply the divergence theorem: 1 ( i* i *) 1 1 ˆ S n ds = E J + M H d + jω ε c E µ H d 4 4 S 1 1 ωε c E + ωµ H d 17

18 Complex Poyntg Theorem (cont.) Fal form of complex Poyntg theorem: 1 ( i* i * E J + M H ) d = S nˆ ds S jω µ H ε c E d ωε c E + ωµ H d 18

19 Complex Poyntg Theorem (cont.) Interpretation of P s : 1 ( i* i * P ) s = E J + M H d Re P s ( i i ) = E J + M H d = Ps We therefore identify that Ps complex source power [ A] Re Ps Im P s real power (watts) from the sources [ W] imagary power (vars) from the sources [ AR] 19

20 Complex Poyntg Theorem (cont.) Interpretation of P f : 1 ( * P ) ˆ f = E H n ds S Re P f S ( ) ˆ = E H n ds = P f We therefore identify that Pf complex power flowg out of S Re P Im P f f real power (watts)flowg out of S [ W] imagary power (vars)flowg out of S [ AR] 0

21 Complex Poyntg Theorem (cont.) Interpretation of energy terms: ε c E 4 ( E E ) * = εc 1 1 ε c Re ( * E E ) 1 1 = ε c E E = ε c E Note: The real-part operator may be added here sce it has no effect. Note: We know this result represents stored energy for simple lear media (ε c = ε), so we assume it is true for generalized lear media. Hence Similarly, 1 1 ε E d = ε E d = 4 c c e 1 1 µ H d = µ H d = 4 W W m Note: The formulas for stored energy can be improved for dispersive media (discussed later). 1

22 Complex Poyntg Theorem (cont.) Interpretation of dissipation terms: Recall: σ εc = ε j ω ε is real for simple lear media 1 1 ( * ) 1 ( * ωε Re ) c E = ωε c E E = ωε c E E = ωε E E = ωε E c c For simple lear media: Hence σ ωε c = ω = σ ω 1 ωε c E d = σ d = E P e d Note: This formula gives the correct time-average power dissipated due to electric losses for simple lear media. We assume the same terpretation holds for generalized lear media (ε is complex).

23 Complex Poyntg Theorem (cont.) Interpretation of dissipation terms (cont.): Similarly, 1 ωµ H d = P m d This is the time-average power dissipated due to magnetic losses. Note: There is no magnetic conductivity, and hence no magnetic conduction loss, but there can be magnetic polarization loss. 3

24 Complex Poyntg Theorem (cont.) Summary of Fal Form 1 ( i* i * E J + M H ) d = S nˆ ds S ωε c E + ωµ H d jω µ H ε c E d 4 4 ( )( ) P = P + P + j ω W W s f d m e 4

25 Complex Poyntg Theorem (cont.) ( )( ) P = P + P + j ω W W s f d m e P abs We can write this as Ps = Pf + Pabs where we have defed a complex power absorbed P abs : ( P ) Re abs = Pd ( P ) = ( ω)( W W ) Im abs m e 5

26 Complex Poyntg Theorem (cont.) Ps = Pf + Pabs This is a conservation statement for complex power. Re( abs ) d P = P Im ( P ) = ( ω)( W W ) abs m e Watts ARs P f Complex power flow out of surface E ARS consumed H Power (watts) consumed ε µ c, Source 6

27 Example L I + - Ideal (lossless) ductor Denote: P = complex power absorbed = W + jq abs abs abs Calculate P abs usg circuit theory, and verify that the result is consistent with the complex Poyntg theorem. Note: W abs = 0 (lossless element) 7

28 Example (cont.) L I + - Ideal ductor 1 1 Pabs = I = ( ZI) I 1 = 1 = jω LI * * ( jω ) LI I * P abs Note: = jq abs (lossless element) 8

29 Example (cont.) 1 Qabs = Im Pabs = ω LI 1 Re * = ωl II ( ) = ωl i = ω Li ( m ) = ω W 1 9

30 Example (cont.) Sce there is no stored electric energy the ductor, we can write Q = W W ω ( ) abs m e Hence, the circuit-theory result is consistent with the complex Poyntg theorem. Note: The ductor absorbs positive ARS. 30

31 Example We use the complex Poyntg theorem to exame the put impedance of an antenna I J s P f S Antenna Model: I Z =R +j X P 1 1 = I = ( Z I ) I 1 = Z I * * 0 31

32 Example (cont.) so Z P = Real part: I R = Re I P ( ) Re( P ) = P = P rad (no losses vacuum surroundg antenna) = Re P rad Note: The far-field Poyntg vector is much easier to calculate than the near-field Poyntg vector. Hence R = Re( ˆ) I S S n ds 3

33 Example (cont.) R = Re( S nˆ ) ds I S In the far field (r ) Im S = 0 (This follows from plane-wave properties the far field.) Hence R = ( S nˆ ) ds I S 33

34 Example (cont.) Im Imagary part: X = ( P ) where I ( ) ( ) Im P = Im Prad + ω Wm We Hence X I S Im S nˆ ds The RHS is zero! W m, W e 34

35 Example (cont.) We can say that X = I ( ω W W ) m e Hence 4ω 1 1 X = H E d I µ 0 ε0 4 4 However, it would be very difficult to calculate the put impedance usg this formula! 35

36 Dispersive Material The permittivity and permeability are now functions of frequency: ( ) ( ) ε = ε ω µ = µ ω (A lossy dispersive media is assumed here. It is assumed that the fields are defed over a fairly narrow band of frequencies.) The formulas for stored electric and magnetic energy now become: W e W m = = E H ω ω ( ωε ) ( ωµ ) Note: The stored energy should always be positive, even if the permittivity or permeability become negative. Reference: J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 63). 36

37 Momentum Density ector The electromagnetic field has a momentum density (momentum per volume): p = D B In free space: p = µε E H ( ) 0 0 p = 1 c S or [(kg m/s)/m 3 ] = [kg/(s m )] Reference: J. D. Jackson, Classical Electrodynamics, Wiley, 1998 (p. 6). 37

38 Momentum Density ector (cont.) Photon: From physics, we have a relation between the energy E and the momentum p of a sgle photon. E = pc E hf 34 = h = [J s] (Planck s constant) Calculation of power flow: S z Energy = = A t = Ec N z ( ) ( ) ( ) p p p E Ac t N = pc N = pn c = p c Hence A t p z = p 1 c S z A Photons movg: = c t S = S N p photons per unit volume ˆ z z This is consistent with the previous momentum formula. v 38

39 Momentum Density ector (cont.) Example: Fd the time-average force on a 1 [m ] mirror illumated by normally cident sunlight, havg a power density of 1 [kw/m ]. Note: the fields vary susoidally, but the frequency is arbitrary. A = 1 m z F z p c z 1 c 1 = S z = (1000) c c ( c pz )( Ac t) ( c = = pz )( Ac ) = (1000) ( 1) c = t c c F z = [N] 39

40 Solar Sail Sunjammer Project NASA has awarded $0 million to L Garde, Inc. (Tust, CA), a maker of flatable space structures, to develop a solar sail, which will rely on the pressure of sunlight to move through space when it takes its first flight as soon as 014. A smaller version of L'Garde's solar sail unfurled a vacuum chamber Ohio 005. This one was about 3,400 square feet, a quarter the size of the sail the company plans to loft to space as soon as 014. Photo courtesy NASA and L'Garde. 40

41 Solar Sail (cont.) 41

42 Maxwell Stress Tensor This gives us the stress (vector force per unit area) on an object, from knowledge of the fields on the surface of the object. References: D. J. Griffiths, Introduction to Electrodynamics, Prentice-Hall, J. D. Jackson, Classical Electrodynamics, Wiley, J. Schwger, L. L. DeRaad, Jr., K. A. Milton, and W.-Y. Tsai, Classical Electrodynamics, Perseus,

43 Maxwell Stress Tensor T T T xx xy xz = yx yy yz T T T T T T T zx zy zz 1 1 Tij = ε 0EE i j + µ 0 Hi H j δ ij ε 0 E + µ 0 H (for vacuum) i, j= xyz,, δ ij 1, = 0, i = i j j (Kronecker delta) 43

44 Maxwell Stress Tensor (cont.) Momentum equation: d T nˆ ds = F + p dt S d Total flow rate of momentum given to object from EM field Total force on object (rate of change of mechanical momentum) 1 p = E H c Rate of change of electromagnetic momentum side of region F 44

45 Maxwell Stress Tensor (cont.) d T nˆ ds = F + p dt S d In many practical cases the time-average of the last term (the rate of change of electromagnetic momentum side of region) is zero: No fields side the body Susoidal steady-state fields In this case we have Example : d dt d ( s ( ωt) ) = ωs ( ωt) cos( ωt) = ωs ( ωt) dt ( ( ωt) ) ω ( ωt) s = s = 0 F = T nˆ ds S The Maxwell stress tensor (matrix) is then terpreted as the stress (vector force per unit area) on the surface of the body. 45

46 Maxwell Stress Tensor (cont.) Example: Fd the time-average force on a 1 m mirror illumated by normally cident sunlight, havg a power density of 1 [kw/m ]. nˆ = zˆ x TEM z wave A = 1 m z F z = zˆ F S Note: The fields are assumed to be zero on the back side of the mirror, and no fields are side the mirror. F Txx Txy Txz 0 Txz = T nˆ = Tyx Tyy T yz 0 T = yz Tzx Tzy T zz 1 Tzz F z = Tzz 46

47 Maxwell Stress Tensor (cont.) nˆ = zˆ x TEM z wave A = 1 m z 1 1 Tzz = ε 0EE z z + µ 0 Hz Hz ε 0 E + µ 0 H 1 1 = ε 0 E + µ 0 H front side Assume: E = E xˆ, H = H x y yˆ E c c x / Hy = η0 47

48 Maxwell Stress Tensor (cont.) 1 1 F = ε E + µ H 1 = µ 0H y z 0 x 0 y front side (PEC mirror) F z = 1 µ 0 Hy 1 1 µ Re ( * HH) y y = µ Re 1 µ 0 H y 4 ( H ) y = 0 = F z We then have: 1 = µ = µ 0 4 H y H c y (The tangential magnetic field doubles at the shortg plate.) 48

49 Maxwell Stress Tensor (cont.) F 1 4 c c z = µ 0 Hy = µ 0 Hy Next, use F z = µ 0 H c y c c c 1 c c* 1 c Sz = Ex Hy = Ex Hy = Hy η0 c c ( Ex / Hy = η0 ) = ( c S / ) z µ η = π 0 0 ( ( ) ) / so F z = [N] 49

50 Force on Dielectric Object Here we calculate the electrostatic force on a dielectric object an electric field usg dipole moments. This is an alternative to usg the Maxwell stress tensor. Consider first the force on a sgle electrostatic dipole: + ( ( ) ( )) F = q E r E r + r = r r q r + p = q r r c For small dipoles: or ( ( )) F q r E r c ( ) F p E r c r q E Note that we have a gradient of a vector here. This is called a dyad. 50

51 Force on Dielectric Object (cont.) Dyadic representation: E = xˆ xe ˆ + ye ˆ + ze + yˆ xe ˆ + ye ˆ + ze + z xe ˆ + ye ˆ + ze x y z or ( x y ˆ z) ( x y ˆ z) ˆ ( x y ˆ z) E E x y E z E E x y E z E E x y E z E= xx ˆˆ + xy ˆˆ + xz ˆˆ + yx ˆˆ + yy ˆˆ + yz ˆˆ + zx ˆˆ + zy ˆˆ + zz ˆˆ x x x y y y z z z Hence we have: ( ˆ ˆ x y ˆ z) r E = x + y + z E = E E x y E z E E x y E E z E x y E z ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ x x + y + z + y x + y + z + z x + y + z x x x y y y z z z 51

52 Force on Dielectric Object (cont.) For a dielectric object: P ε r = εχ 0 e E ( 1 χ ) = + e ε= εε 0 r ( c ) ( ) ( ) ( ) F = p E r df r = P r E r d ( ) ( ) F = P r E r d Sce the body does not exert a force on itself, we can use the external electric field E 0. ( ) ( ) F = P r E r d 0 5

53 Force on Dielectric Object (cont.) We then have: ( ) ( ) F = εχ E r E r d 0 e 0 We can also write: or or 0 e ( ) ( ) F = εχ E r E r d 1 F E r E r d ( ) ( ) ( ) = εχ 0 e 1 F E r d = εχ 0 e ( ( ) ) 53

54 Force on Dielectric Object (cont.) A non-uniform electrical field will generate a net attractive force on a neutral piece of matter. The force is directed toward the region of higher field strength. 54

55 Force on Magnetic Object We have (derivation omitted) the magnetostatic force as: Hence, we have ( ) ( ) F = M r B r d 0 Recall: M = χ m H or or m ( ) ( ) F = χ H r B0 r d χ ( ) ( ) m F = B r B0 r d µ 1 ( ( ) ) m F = χ B r d µ µ = µµ 0 r 55

56 Foster's Theorem Consider a lossless system with a port that leads to it: Lossless system (source free) Z = jx S p ε = ε = ε c µ = µ Coaxial port z PEC enclosure Foster s theorem: dx dω > 0 R. E. Coll, Field Theory of Guided Waves, IEEE Press, Piscataway, NJ,

57 Foster's Theorem (cont.) The same holds for the put susceptance: Z = 1 Y jx = 1 jb X = 1 B dx 1 db = dω B dω db dω > 0 57

58 Foster's Theorem (cont.) Start with the followg vector identity: ( A B) = B ( A) A ( B) Hence we have (applyg twice, for two different choices of the (vectors) * * * H H H E = ( E) E ω ω ω * * * E E E H = H H ω ω ω ( ) Add these last two equations together: 58

59 Foster's Theorem (cont.) * * * * H E H H E + H = ( E) E ω ω ω ω * * E E + H ω ω ( H) Usg Maxwell's equations for a source-free region, * * * * H E H H E + H = ( jωµ H) E ω ω ω ω * * E E + H j E ω ω ( ωε ) 59

60 Foster's Theorem (cont.) * * * * H E H H E + H = ( jωµ H) E ω ω ω ω * * E E + H j E ω ω ( ωε ) Usg Maxwell's equations aga, and the cha rule, we have: * * * * * E H = ( H) = ( jωε E) = j ( ωε ) E jωε ω ω ω ω ω * * * * * H E = ( E) = ( jωµ H) = j ( ωµ ) H + jωµ ω ω ω ω ω We then have 60

61 Foster's Theorem (cont.) * * * * H E H * E E + H = ( jωµ H) E j ( ωε ) E jωε ω ω ω ω ω Simplifyg, we have or * * H E + H j H + j j E ω ω ω * ( ωµ ) ωµ ( ωε ) cancels * * H E E + H = E j ( ωε ) E + H j ( ωµ ) H ω ω ω ω * * E + H = j ( ωε ) E + ( ωµ ) H ω ω ω ω * * H E 61

62 Foster's Theorem (cont.) * * H E E + H = j ( ωε ) E + ( ωµ ) H ω ω ω ω Applyg the divergence theorem, * * H E E + H nˆ ds = j4 e + m ω ω W W S Hypothesis: the total stored energy must be positive. Therefore, * * H E Im E + H nˆ ds > 0 ω ω S 6

63 Foster's Theorem (cont.) * * H E Im E + H nˆ ds > 0 S ω ω The tangential electric field is only nonzero at the port. Hence we have: * * H E Im E + H nˆ ds > 0 ω ω S p Assume that the electric field (voltage) at the port is fixed (not changg with frequency). Then we have * E = 0 ω Lossless system (source free) ε = ε = ε c µ = µ Z = jx Coaxial port S p z 63

64 Foster's Theorem (cont.) Then * H Im E nˆ ds > 0 ω S p At the coaxial port: E = ˆ ρ E H ρ = ˆ φ H φ Lossless system (source free) nˆ = εc = ε = ε µ = µ zˆ Z = jx Coaxial port S p z Hence: * H φ Im Eρ ds > 0 S p ω Im ds > 0 ω * or ( EρHφ ) S p (sce the electric field is fixed and not changg with frequency) 64

65 Foster's Theorem (cont.) * Im ( EρHφ ) ds > 0 ω S p Im ( P z ) > 0 ω Lossless system (source free) Z = jx S p ω I z > * Im 0 ε = ε = ε c µ = µ Coaxial port z * Im Y > 0 ω ( I = Y ) z ω < * Im Y 0 65

66 Foster's Theorem (cont.) ω < * Im Y 0 Im[ Y ] > 0 ω ω B > 0 ω X > 0 66

67 Foster's Theorem (cont.) Example: Short-circuited transmission le Z Z 0 Z L = 0 l = tan ( ) 0 β X Z jz l β = ω LC Inductive Capacitive π/ 3π/ 5π/ β l 67