Cartesian Coordinates

Transcription

1 Cartesian Coordinates Daniel S. Weile Department of Electrical and Computer Engineering University of Delaware ELEG 648 Cartesian Coordinates

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4 Separation of Variables Away from sources, in Cartesian coordinates, we have The Helmholtz Equation 2 ψ x ψ y ψ z 2 + k 2 ψ = 0 To apply separation of variables, we assume and plug in: ψ(x, y, z) = X(x)Y (y)z (z) 2 X x 2 YZ + X 2 Y y 2 Z + XY 2 Z z 2 + k 2 XYZ = 0

5 Separation of Variables We now divide by XYZ : 1 2 X X x Y Y y Z Z z 2 + k 2 = 0 We now make the following key observation:

6 Separation of Variables We now divide by XYZ : 1 2 X X x Y Y y Z Z z 2 + k 2 = 0 We now make the following key observation: Justification of Separation of Variables Each term in the equation depends on one variable and not the others. Since only one term depends on x, it cannot vary with x since then the other terms would need to vary with x to keep the equation true. Since this is a contradiction, every term in the equation must be constant.

7 Separation of Variables We thus have 1 2 X X x 2 = kx Y Y y 2 = ky Z Z z 2 = kz 2 subject to the separation equation k 2 x + k 2 y + k 2 z = k 2 We have also seen that solutions to these equations are of the form h (k x x) cos (k x x), sin (k x x), e jkx x, e +jkx x

8 Separation of Variables A function of the form ψ kx k y k z (x, y, z) = h x (k x x) h y (k y y) h z (k z z) (subject to the separation equation) is called an elementary wave function. In problems with boundaries, the allowed values of k i are determined by the boundary conditions and are called eigenvalues. The functions themselves are called eigenfunctions. Superpositions of these functions also work: ψ = A kx k y h x (k x x) h y (k y y) h z (k z z) k x k y ψ = f (k x, k y ) h x (k x x) h y (k y y) h z (k z z) dk x dk y k x k y

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10 The General Form A general Cartesian elementary wave function is of the form ψ(x, y, z) = e jkx x e jky y e jkzz A more telling version of this expression defines k = k x u x + k y u y + k z u z r = xu x + yu y + zu z so that General Plane Wave Expression ψ(r) = e jk r

11 Interpretation of the General Form In the time domain, the wave becomes ψ(r) = cos(ωt k r) Thus, the wave is traveling in the k direction with phase velocity v p = ω k This is of course the smallest phase velocity that might be assigned, and it is along the direction perpendicular to the equiphase planes.

12 Complex Propagation Constant Suppose k = β jα. The strange notation allows us to define γ = jk = α + jβ. If the propagation material itself is lossless, k 2 = ω 2 µɛ must be real. This implies is real, and so either α = 0 or k 2 = k k = β 2 α 2 j2α β α β = 0. What does this mean about the equiphase planes and the equiamplitude planes?

13 Complex Propagation Constant Suppose k = β jα. The strange notation allows us to define γ = jk = α + jβ. If the propagation material itself is lossless, k 2 = ω 2 µɛ must be real. This implies is real, and so either α = 0 or k 2 = k k = β 2 α 2 j2α β α β = 0. What does this mean about the equiphase planes and the equiamplitude planes? What type of wave has nonzero α?

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15 The Rectangular Waveguide Let us consider TM modes in an a b waveguide along the z-direction. (We assume a > b, and a is along the x-direction.) The Setup A = µψu z ψ = h x (k x x)h y (k y y)e jkzz Now, from our earlier work, E z = 1 ŷ ( ) k 2 kz 2 ψ The most important thing to note is that E z ψ.

16 TM Modes This immediately tells us that we need Boundary Conditions Now ψ(x = 0, y, z) = ψ(x = a, y, z) = 0 ψ(x, y = 0, z) = ψ(x, y = b, z) = 0 h x (k x x) = A x cos(k x x) + B x sin(k x x) h x (0) = 0 A x = 0 h x (k x a) = 0 k x = mπ a for m = 1, 2,...

17 The TM solution Therefore, we have the TM Wave Potential subject to and ψ(x, y, z) = sin ( mπ ) 2 ( nπ + a b ( mπx ) ( nπy ) sin e jkzz a b m = 1, 2,... n = 1, 2,... ) 2 + k 2 z = k 2 From here, we can find all of the fields from our TM formulas.

18 TE Modes Now, let us consider TE modes. The Setup Now, from our earlier work, F = ɛψu z ψ = h x (k x x)h y (k y y)e jkzz E x = ψ y E y = ψ x

19 TE Modes This immediately tells us that we need Boundary Conditions ψ(x = 0, y, z) ψ(x = a, y, z) = x x ψ(x, y = 0, z) ψ(x, y = b, z) = y y = 0 = 0 Now h x (k x x) = A x cos(k x x) + B x sin(k x x) h x (k x x) x = k x A x sin(k x x) + k x B x cos(k x x) = k x h x(k x x) h x(0) = 0 B x = 0 h x(k x a) = 0 k x = mπ for m = 0, 1, 2,... a

20 The TE solution Therefore, we have the TE Wave Potential subject to and ψ(x, y, z) = cos ( mπ ) 2 ( nπ + a b ( mπx ) ( nπy ) cos e jkzz a b ) 2 + k 2 z = k 2 m = 0, 1, 2,... n = 0, 1, 2,... m = n = 0 excluded From here, we can find all of the fields from our TE formulas.

21 Cutoff From the formula for k z, it is obvious that cutoff wavenumber for the TE mn or TM mn mode is The Cutoff Wavenumber (mπ ) 2 ( nπ k cmn = + a b ) 2 In terms of this wave number we have the propagation constant { jβ = j k γ mn = jk z = 2 kc 2 for k > k c α = kc 2 k 2 for k < k c

22 Cutoff Frequency and Wavelength From the cutoff k and the formula k = 2πf µɛ = 2π/λ, we find the Cutoff Frequencies f cmn = 1 (m ) 2 ( n 2 + ɛµ a b ) 2 and Cutoff Wavelengths λ cmn = 2 ( m ) 2 ( a + n b ) 2

23 Alternate Expression for γ Since we can always write the k c k = f c f Propagation Constant as a Function of Cutoff Frequency ( ) 2 jβ = jk 1 fcf for f > f c γ mn = jk z = ( ) 2 α = k c 1 f f c for f < f c

24 TE Impedance Recall from earlier that for TE modes From this we have E x = ψ y H x = 1 E y = ψ x H y = 1 2 ψ jωµ x z = jkz ψ jωµ x 2 ψ jωµ y z = jkz ψ jωµ y jωµh x jωµh y ψ = jk z x ψ = jk z y = jk z E y = jk z E y This gives the TE Modal Impedance Z TE 0mn = E x H y = E y H x = ωµ k z

25 TM Impedance By precisely the same logic, we find the TM Modal Impedance Z TM 0mn = E x H y = E y H x = k z ωɛ In the application of these formulas, we must remember that k z is real above cutoff and imaginary below. Indeed, these formulas are not particular to rectangular cross section, as we will see.

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27 TM Modes Let us assume we form a cavity by placing metal walls on our waveguide at z = 0 and z = c. Because we need to satisfy the BCs on the other four walls, we can start with our expression for TM waveguide modes. ( mπx ) ( nπy ) ψ(x, y, z) = sin sin [A z cos(k z z) + B z sin(k z z)] a b We need the x- and y-directed electric fields to vanish at z = 0 and z = c. Both of these fields are proportional to ψ z, so

28 TM Modes Let us assume we form a cavity by placing metal walls on our waveguide at z = 0 and z = c. Because we need to satisfy the BCs on the other four walls, we can start with our expression for TM waveguide modes. ( mπx ) ( nπy ) ψ(x, y, z) = sin sin [A z cos(k z z) + B z sin(k z z)] a b We need the x- and y-directed electric fields to vanish at z = 0 and z = c. Both of these fields are proportional to ψ z, so We need h z(0) = h z(k z c) = 0.

29 TM Modes Thus, we have the h z(0) = 0 B z = 0 h z(k z c) = 0 k z = pπ c Elementary Wave Functions for TM Cavity Modes subject to ψ(x, y, z) = sin A z = µψ ) sin ( mπx a ( nπy b m, n = 1, 2,... p = 0, 1, 2... ) cos ( pπz ) c

30 Resonant Frequencies and TE Elementary Wave Functions The Resonant Frequencies f rmnp = 1 ( m ) 2 ( n ) 2 ( p ) ɛµ a b c By similar effort Elementary Wave Functions for TE Cavity Modes subject to ψ(x, y, z) = cos ( mπx a F z = ɛψ ) cos ( nπy ) sin b ( pπz ) c m, n = 0, 1, 2,... p = 1, 2...

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32 Alternate Mode Sets The most important way of finding waveguide modes is by assuming them TE or TM to z. This characterization holds for non-cartesian coordinates as well. In Cartesian coordinates, other choices are possible and sometimes useful. Thus, we can create a set of TM x modes by choosing or a set of TE x modes with A = µψu x F = ɛψu x

33 TM x Modes Using the standard formulas for computing fields from potentials we can write ( ) E x = k 2 ψ H ŷ x 2 x = 0 E y = 1 2 ψ ŷ x y H y = ψ z E z = 1 2 ψ ŷ x z H z = ψ y From the formula for E z, it is easy to find the Wave Potential for TM x Waves in a Rectangular Guide ( mπx ) ( nπy ) ψ(x, y, z) = cos sin e jkzz a b

34 TE x Modes Similar methods give rise to TE x mode formulas: ( ) H x = k 2 ψ ẑ x 2 E x = 0 H y = 1 2 ψ ẑ x y E y = ψ z H z = 1 2 ψ ẑ x z E z = ψ y From the formula for E z, it is easy to find the Wave Potential for TE x Waves in a Rectangular Guide ( mπx ) ( nπy ) ψ(x, y, z) = sin cos e jkzz a b

35 Mode Impedance Despite the novel mode formulation, we can still compute the impedance of a mode relative to z. For TM x modes, we have Similarly, E x = 1 ( ) 2 ŷ x 2 + k 2 ψ jωɛe x = We thus have the TE x Mode impedance H y = ψ z = jk zψ Z TE 0mn = E x = k 2 H y ωɛk z [ k 2 ( mπ ) 2 a ( mπ ) 2 ] ψ a We can find modal impedance for TM x waves similarly.

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37 Partially Filled Waveguide As an example of the need for alternative modes, consider the partially filled guide y d b z 1,µ 1 2,µ 2 a x To find the modes in this, we will start with TM x modes.

38 TM x Modes in Partially Filled Waveguide We will expand A x differently in the two regions, taking into account the step change in µ. We thus write Note: ψ 1 = C 1 cos k x1 x sin nπy b e jkzz ψ 2 = C 2 cos [k x2 (a x)] sin nπy b e jkzz 1 The form of the solution is determined by previous slides. 2 The solutions in the y-direction must be the same since the boundary conditions have not changed. 3 The solutions in the z-direction must have identical k z to ensure continuity of fields. 4 The behavior of the solution at x = 0 and x = a is determined by the metal waveguide wall.

39 TM x Modes in Partially Filled Waveguide For these to be legitimate solutions, they must satisfy the usual separation conditions: k 2 x1 + ( nπ b ) 2 + k 2 z = k 2 1 = ω2 µ 1 ɛ 1 k 2 x2 + ( nπ b ) 2 + k 2 z = k 2 2 = ω2 µ 2 ɛ 2 Now we must enforce continuity at the boundary. This implies 1 E y1 (x = d) = E y2 (x = d) 2 E z1 (x = d) = E z2 (x = d) 3 H y1 (x = d) = H y2 (x = d) 4 H z1 (x = d) = H z2 (x = d) How many of these conditions do we actually need?

40 TM x Modes in Partially Filled Waveguide For these to be legitimate solutions, they must satisfy the usual separation conditions: k 2 x1 + ( nπ b ) 2 + k 2 z = k 2 1 = ω2 µ 1 ɛ 1 k 2 x2 + ( nπ b ) 2 + k 2 z = k 2 2 = ω2 µ 2 ɛ 2 Now we must enforce continuity at the boundary. This implies 1 E y1 (x = d) = E y2 (x = d) 2 E z1 (x = d) = E z2 (x = d) 3 H y1 (x = d) = H y2 (x = d) 4 H z1 (x = d) = H z2 (x = d) How many of these conditions do we actually need? Two. Why?

41 Finding the Eigenvalues From our work with alternative mode sets, we find E y1 = 1 nπ C 1 k x1 jωɛ 1 b sin k x1x cos nπy b e jkzz E y2 = 1 nπ C 2 k x2 jωɛ 2 b sin [k x1 (a x)] cos nπy b e jkzz These must be equal at x = d so we have the E-Field Continuity Condition 1 ɛ 1 C 1 k x1 sin k x1 d = 1 ɛ 2 C 2 k x2 sin [k x2 (a d)]

42 Finding the Eigenvalues A similar condition can be found from the continuity of H z : H z1 = nπ b C 1 cos k x1 x cos nπy b e jkzz H z2 = nπ b C 2 cos [k x2 (a x)] cos nπy b e jkzz These must be equal at x = d so we have the H-Field Continuity Condition C 1 cos k x1 d = C 2 cos [k x2 (a d)]

43 Finding the Eigenvalues Dividing these two equations yields a transcendental equation for k z : TM x Eigenvalue Equation k x1 ɛ 1 tan k x1 d = k x2 ɛ 2 tan [k x2 (a d)] Once k z is found (and hence k x1 and k x2, either one of the previous equations may be used to find C 1 /C 2. TE x modes can be found similarly, and lead the the TE x Eigenvalue Equation k x1 µ 1 cot k x1 d = k x2 µ 2 cot [k x2 (a d)]

44 Partially Filled Waveguide Observations The resulting k z can be shown to be between those obtained for homogeneous guides filled with material 1 or 2. The TE x and TM x modes are no longer degenerate. Computing the cutoff frequency can be done without solving transcendental equations by setting k z = 0 and finding k x1 and k x2. Knowledge of the cutoff frequency is no longer sufficient using the usual formulas to find k z ; a transcendental equation must be solved at each frequency. The dominant mode is affected by the orientation of the inhomogeneity as well as the shape of the guide.

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46 The Slab Waveguide Waves can be guided without metal. (This should be known to everyone here, especially those studying optics. The simplest example of a dielectric waveguide is the slab waveguide: 0,µ 0 x a d,µ d z This problem is simple primarily because it is two dimensional. We assume no variation with y.

47 Slab Guide Setup It turns out that we can examine modes either TE x and TM x, or TE z and TM z. We choose the latter. Given the lack of y variation, the TM z equations are: E x = k z ψ ωɛ x E z = 1 jωɛ H y = ψ x ( k 2 k 2 z ) ψ Also, because of the symmetry of the problem, the solutions must either be odd or even functions of x.

48 Form of the Solution In the dielectric region we choose Odd Dielectric Mode ψd o = A sin ux e jkzz for x < a 2 In the air we choose Odd Air Mode ψ o a = Be vx e jkzz for x > a 2 ψ o a = Be vx e jkzz for x < a 2 Why are these expression of different forms?

49 Separation Equations As usual, all of this is subject to the separation equations, which say the length of the wavevector is the wavenumber. The guide wavenumber k z must be the same in both media. Separation Equations u 2 + k 2 z = k 2 d = ω2 µ d ɛ 0 v 2 + k 2 z = k 2 0 = ω2 µ a ɛ 0 Of course, E z and H y must be continuous.

50 Continuity of E z From our equations of TM propagation, we find that E z = E z = E z = A jωɛ d u 2 sin ux e jkzz for x < a 2 B jωɛ 0 v 2 e vx e jkzz for x > a 2 B jωɛ 0 v 2 e vx e jkzz for x < a 2 Enforcing continuity at x = a 2 gives the E-Field Continuity Condition A u 2 sin ua ɛ d 2 = B v 2 e va 2 ɛ 0

51 Continuity of H y By the same token H y = Au cos ux e jkzz for x < a 2 H y = Bve v x e jkzz for x > a 2 Enforcing continuity at x = a 2 gives the H-Field Continuity Condition Au cos ua 2 = Bve va 2

52 The Odd TM Characteristic Equation Dividing the E-field continuity condition by the H-field condition gives The Characteristic Equation ua 2 tan ua 2 = ɛ d va ɛ 0 2 Coupling this with the Separation Equations u 2 + k 2 z = k 2 d = ω2 µ d ɛ 0 v 2 + k 2 z = k 2 0 = ω2 µ a ɛ 0 we can find the k z s and cutoffs of the odd TM modes.

53 Even TM Modes Doing precisely the same for even modes, we can write ψ e d = A cos ux e jkzz for x < a 2 ψ e a = Be v x e jkzz for x > a 2 Enforcing continuity of E z and H y gives the Even TM Characteristic Equation ua 2 cot ua 2 = ɛ d va ɛ 0 2

54 TE Modes TE modes are completely dual to TM modes. The characteristic equations are Characteristic Equations The fields are given by ua 2 tan ua 2 ua 2 cot ua 2 = ɛ d va ɛ 0 2 = ɛ d va ɛ 0 2 (odd) (even) H x = k z ψ ωµ x 1 ( H z = k 2 k 2 ) z ψ jωµ E y = ψ x

55 Cutoff in Dielectric Waveguides Above cutoff, the wave propagates in the z-direction unattenuated. In free space, k x is imaginary. In the dielectric, k x is real. Below cutoff, k z is complex due to radiation out of the core. The condition for guidance is Guidance Condition k 0 < k z < k d 1 After all, if k z < k 0, there will be propagation away from the guide. 2 If k z > k d, there evanescence everywhere (and the equations cannot be satisfied).

56 Cutoff Computation 1 By the previous argument, cutoff occurs when v = 0.

57 Cutoff Computation 1 By the previous argument, cutoff occurs when v = 0. 2 This implies that k z = k 0.

58 Cutoff Computation 1 By the previous argument, cutoff occurs when v = 0. 2 This implies that k z = k 0. 3 This in turn implies that u = k 2 d k 2 0.

59 Cutoff Computation 1 By the previous argument, cutoff occurs when v = 0. 2 This implies that k z = k 0. 3 This in turn implies that u = k 2 d k 2 0. We thus have the equations ( ) a tan k 2 2 d k 0 2 ( ) a cot k 2 2 d k 0 2 = 0 (odd) = 0 (even) for both TE and TM modes.

60 Cutoff Computation The cutoff equations are solved when Cutoff Wavenumbers a k 2 2 d k 0 2 = nπ 2 n = 0, 1, 2,... Consistent with Murphy s law, odd mode numbers go with even modes! Cutoff Frequencies n f c = 2a ɛ d µ d ɛ 0 µ 0

61 Observations 1 The TE 0 and TM 0 modes have no cutoff regardless of a. 2 If the slab is thin, the guidance is weak in that v 0. 3 The mode number n is roughly the number of half wavelengths in the dielectric; this characterization is more accurate for large refractive indices. 4 If the slab is weakly guiding, the cutoff of higher order modes will be at very high frequency.

62 Graphical Method for Characteristic Equation Solution Eliminating k z from the separation equations, we have u 2 + v 2 = k 2 d k 2 0 = ω2 (ɛ d µ d ɛ 0 µ 0 ). Given this, the TE characteristic equations become µ 0 ua µ d 2 tan ua (ωa = 2 2 µ d ua µ 0 2 cot ua (ωa = 2 2 We now assume µ d = µ 0 and let ) 2 ( ua ) 2 (ɛd µ d ɛ 0 µ 0 ) 2 ) 2 ( ua ) 2 (ɛd µ d ɛ 0 µ 0 ) 2 x = ua 2 The right hand side is thus a circle of radius r = ωa ɛd µ d ɛ 0 µ 0 2

63 Graphical Method for Characteristic Equation Solution Why bother with graphical methods when we have computers?