ECE 6340 Intermediate EM Waves. Fall Prof. David R. Jackson Dept. of ECE. Notes 17

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1 ECE 634 Intermediate EM Waves Fall 16 Prof. David R. Jacson Dept. of ECE Notes 17 1

2 General Plane Waves General form of plane wave: E( xz,, ) = Eψ ( xz,, ) where ψ ( xz,, ) = e j( xx+ + zz) The wavenumber terms ma be complex. Helmholtz Eq.: E+ E = Propert of vector Laplacian: ( ) ( Eψ xz,, ) E ψ ( xz,, ) = Hence E ψ + Eψ = so ψ + ψ =

3 General Plane Waves (cont.) ψ ψ + = This gives x z ( + ) ψ = or + + = x z (separation equation or wavenumber equation) 3

4 General Plane Waves (cont.) Denote = x ˆ + ˆ + z ˆ x z r = xx ˆ + x ˆ + xz ˆ Then ψ ( xz,, ) = e j r and = (wavenumber equation) Note: For complex vectors, this is not the same as saing that the magnitude of the vector is equal to. 4

5 General Plane Waves (cont.) We can also write = β jα Wavenumber vector so jβ r α r ψ ( xz,, ) = e e The β vector gives the direction of most rapid phase change (decrease). The α vector gives the direction of most rapid attenuation. To illustrate, consider the phase of the plane wave: Φ ( xz,, ) = β r= βxx β βzz Φ ( xz,, ) = β ˆ ˆ xx β βzzˆ = β Similarl, Ψ ( xz,, ) = α Ψ 5

6 General Plane Waves (cont.) A general plane wave: ψ ( xz,, ) = = β jα S = Re S 1 S = E H * e j r z S α β x In the most general case, all three vectors ma be in different directions. 6

7 General Plane Waves (cont.) Smbol for a plane wave: In the most general case, the vector ma be complex. In this case it is not possible to actuall visualize it as vector in 3D space. The blue arrow is still used as a smbol for the vector. z x 7

8 General Plane Waves (cont.) Next, loo at Maxwell s equations for a plane wave: E = jωµ H H = jωεc E = xˆ + ˆ + zˆ x z = xˆ( j ) + ˆ( j ) + zˆ ( j ) = j x z Hence = j j E = jωµ H j H= jωε E c 8

9 General Plane Waves (cont.) Summar of Maxwell s curl equations for a plane wave: E =ωµ H H = ωεc E 9

10 General Plane Waves (cont.) Gauss law (divergence) equations: D = ρ v B = j = ( ε E ) j = ( µ H) E= H = Reminder: The volume charge densit is zero in the sinusoidal stead state for a homogeneous source-free region. 1

11 General Plane Waves (cont.) Furthermore, we have from Farada s law E =ωµ H Dot multipl both sides with E. E H = Note : E ( E) = ( for an vectors E, ) 11

12 General Plane Waves (cont.) Summar of dot products: E= H = E H = Note: If the dot product of two vectors is zero, we can sa that the vectors are perpendicular for the case of real vectors. For complex vectors, we need a conjugate (which we don t have) to sa that the vectors are orthogonal. 1

13 Power Flow 1 S = E H * E = E ψ H 1 = E jωµ 1 = ( j E ) jωµ 1 = E ωµ 1 = ψ ( E) ωµ 13

14 Power Flow (cont.) so 1 ( ) [ ] * 1 * S = ψe E ψ ωµ 1 ( * * = ψ E ) E ωµ Note: µ is assumed to be real here. Use so that ( ) ( ) A ( B C) = B AC C AB ( ) ( ) ( ) * * * * * * = E E E E E E and hence S 1 = ψ ( ) ( ) E ωµ E E E * * * * 14

15 Power Flow (cont.) S 1 = ψ ( ) ( ) E ωµ E E E * * * * Assume E = real vector. (The same conclusion holds if it is a real vector times a complex constant.) ( ) * ( ) * * * = = = E E E (All of the components of the vector are in phase.) Note: This conclusion also holds if is real, or is a real vector times a complex constant. Hence S 1 = ψ ωµ E * 15

16 Power Flow (cont.) S 1 = ψ ωµ E * The power flow is: < S >=Re S so < S >= 1 ωµ ψ E Re (assuming that µ is real) Recall: = β jα 16

17 Then Power Flow (cont.) ψ < S > = E β ωµ Power flows in the direction of β. Assumption: Either the electric field vector or the wavenumber vector is a real vector times a complex constant. (This assumption is true for most of the common plane waves.) 17

18 Direction Angles First assume = real vector (so that we can visualize it) and the medium is lossless ( is real): z θ = β = direction of power flow x φ The direction angles (θ, φ) are defined b: = sinθcosφ x z = sinθsinφ = cosθ Note: + + x z = sin θ + = cos θ 18

19 Direction Angles (cont.) From the direction angle equations we have: cosθ = tanφ = z x Even when ( x,, z ) become complex, and is also complex, these equations are used to define the direction angles, which ma be complex. 19

20 Homogeneous Plane Wave Definition of a homogenous (uniform) plane wave: (θ, φ) are real angles In the general loss case (complex ): = xˆ + ˆ + z ˆ x z = xˆsinθcosφ+ ˆsinθsinφ+ zˆ cosθ = rˆ where rˆ xˆsinθcosφ+ ˆsinθsinφ+ zˆ cosθ ˆr = real unit vector pointing in the direction of power flow

21 Homogeneous Plane Wave (cont.) Hence we have = j r ( ) ˆ β = α = r ˆ rˆ The phase and attenuation vectors point in the same direction. The amplitude and phase of the wave are both constant (uniform) in a plane perpendicular to the direction of propagation. Note: A simple plane wave of the form ψ = exp (-j z) is a special case, where θ =. z = z ˆ Note: A homogeneous plane wave in a lossless medium has no α vector: α = rˆ = x 1

22 Infinite Current Sheet An infinite surface current sheet at z = launches a plane wave in free space. z Plane wave x J s ( x, ) Assume + s x jx ( x J Ae ),, = real The vertical wavenumber is then given b + + = x z

23 Infinite Current Sheet (cont.) Part (a): Homogeneous plane wave =.5 =.5 x 1 z =.5.5 =± We must choose z = +.77 ( outgoing wave) Then = ˆ ˆ x(.5) + (.5) + zˆ (.77) z S = β = Power flow x β = ˆ ˆ x(.5) + (.5) + zˆ (.77) α = 3

24 Infinite Current Sheet (cont.) Part (b) Inhomogeneous plane wave = = 3 x = 4 9 z = ± j 1 We must choose z = j 1 The wave is evanescent in the z direction. Then = ˆ ˆ ˆ x() + (3) + z( j 1) β = ˆ ˆ ˆ x() + (3) + z() α = ˆ z( 1) 4

25 Infinite Current Sheet (cont.) Power flow (in x plane) z β = ˆ ˆ x() + (3) α = zˆ ( 1) z cosθ = = j 1 so π θ = + j(1.956) [rad] x φ S = β β tanφ = = β x so φ = Note: Another possible solution is the negative of the above angle. The inverse cosine should be chosen so that sinθ is correct (to give the correct x and ): sinθ >. 5

26 Propagation Circle + + = x z ( ) = + z x Propagating waves: + < x ( so = real) z x Propagating (inside circle) = z x Evanescent (outside circle) = j + z x Free space acts as a low-pass filter. 6

27 Radiation from Waveguide z PEC WG E E π x ( x,,) = cos a b x a 7

28 Radiation from Waveguide (cont.) Fourier transform pair: j( x x ) E + + (,, z) E ( x,, z) e dx d = x 1 j( x x+ ) (,, ) = (,, ) x x ( π ) E x z E z e d d 8

29 Radiation from Waveguide (cont.) 1 j( x x+ ) (,, ) = (,, ) x x ( π ) E x z E z e d d E+ E = E E E E = x z Hence 1 ( π ) E + + = j( x x+ ) E x E E e dd x z 9

30 Radiation from Waveguide (cont.) Hence E E E + + E = z x Next, define z x We then have E z + E z = 3

31 Radiation from Waveguide (cont.) Solution: z E (,, z) = E (,,) e x x j z Note: We want outgoing waves onl. Hence 1 j ( x x+ + z z ) (,, ) = (,,) x x ( π ) E x z E e d d Power flow + < x (Homogeneous) + > x (Inhomogeneous) 31

32 Radiation from Waveguide (cont.) Fourier transform of aperture field: j( x x ) E + + (,,) = E ( x,,) e dx d x b/ a/ π x = cos e a b/ a/ + j( x+ ) a/ b/ π x j( ) j( ) e + x x + dx e d = cos a a/ b/ x dx d π a a cos x b E ( x,,) = bsinc π a x 3

33 Radiation from Waveguide (cont.) Note: z E (,, z) = E (,,) e j z = x x x x Hence E (,, ) x xz = For E z we have E Ez E = + = z ( j ) E + ( j ) E = z z This follows from the mathematical form of E as an inverse transform. 33

34 Radiation from Waveguide (cont.) Hence E z = E z In the space domain, we have E x z E e d d 1 j ( x x+ + z z ) z(,, ) = (,,) x x ( π ) z 34

35 Radiation from Waveguide (cont.) Summar (for z > ) z PEC E (,, ) x xz = E WG 1 j ( x x+ + z z ) (,, ) = (,,) x x ( π ) E x z E e d d E x z E e d d 1 j ( x x+ + z z ) z(,, ) = (,,) x x ( π ) z π a a cos x b E ( x,,) = bsinc π a x 35

36 Some Plane-Wave η Theorems Theorem #1 E E η H H = (alwas true) Theorem # If PW is homogeneous: E = η H (loss medium) E = η H (lossless medium) Theorem #3 If medium is lossless: β α = 36

37 Example Example ( ) ˆ j x x + + z z ˆ ψ (,, ) E = e = x z Plane wave in free space Given: x = = = j 3 z Note: It can be seen that + + = x z and E= Find H and compare its magnitude with that of E. Verif theorems 1 and 3. 37

38 Example (cont.) so E= jωµ H j E= jωµ H 1 H = E ωµ = ωµ [ ψ xz] (,, j 3) (,1,) (,, ) 38

39 Example (cont.) 1 H = [ zˆ () + xˆ ( j 3)] ψ η * * * x x z z x z H= H H = HH + HH = H + H 1 7 H = + j 3 ψ = ψ η η Hence E H = η 7 Note: The field magnitudes are not related b η! 39

40 Example (cont.) E E= η H H Verif: (Theorem #1) 1 H= [ zˆ () + xˆ ( j 3)] ψ xz,, η E= ˆ ψ ( xz,, ) ( ) At the origin (ψ = 1) we have: E E= H H = [ zˆ() + xˆ( j 3)] [ zˆ() + xˆ( j 3)] = ( 4 3) = η η η 4

41 Example (cont.) β α= Verif: (Theorem #3) β = Re = xˆ ( ) ( ) Hence ( ) zˆ ( ) α = Im = 3 β α= x = = = j 3 z 41

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