Cylindrical Radiation and Scattering

Transcription

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

2 Outline Cylindrical Radiation 1 Cylindrical Radiation

3 Outline Cylindrical Radiation 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

4 Outline Cylindrical Radiation 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

5 Two-Dimensional Sources Two-dimensional radiation is created by sources independent of z. The simplest such source (akin to a Hertzian Dipole) is an infinite filament. Such a current radiates TM z. Consider a filamentary current I on the z-axis. Since its radiation is 1 Independent of φ, 2 Independent of z, and 3 Outwardly traveling, the magnetic vector potential must be of the form A z = µch (2) 0 (kρ)

6 Radiation from a Filament Now, we must have Now 2π lim ρ 0 0 H φ ρ dφ = I H = 1 µ A implies H φ = C ρ For small ρ, we have [ H (2) 0 (kρ) ] = kch (2) 1 (kρ). H (2) kρ 1 (kρ) 2 + j 2 π kρ

7 Radiation from a Filament Substituting this into our equation, so finally and we have the kc lim ρ 0 2π Magnetic Vector Potential 0 ( kρ 2 + j ) 2 ρ dφ = I π kρ C = 1 4j A z = µi 4j H(2) 0 (kρ) 2jC π 2π = I

8 Fields of a Filament Cylindrical Radiation Given A z we can easily compute Filamentary Fields E z = k 2 I 4ωɛ H(2) 0 (kρ) H φ = ki 4j H(2) 1 (kρ) Using large argument approximations, we can find Far Fields E z = ηki j 8πkρ e jkρ H φ = ki j 8πkρ e jkρ

9 Observations Cylindrical Radiation Near the source, E and H are not in phase and have a complex relationship. Far from the source, E and H Are in phase, Have ratio η, and Decrease as ρ 1 2. The total radiation (per unit length) must be independent of radius (which can be proven directly). Therefore, we can use the far field expression to compute the power: P = 2π 0 E z H φρdφ = 2π 0 ( ) ) j ηki (ki 8πkρ e jkρ j 8πkρ ejkρ ρ dφ = ηk 2 I 2 8πkρ 2πρ = ηk 4 I 2

10 Two-Dimensional Notation To proceed further, it is useful to define ρ = xu x + yu y ρ = x u x + y u y The distance between these points is ρ ρ = (ρ ρ ) (ρ ρ ) = ρ 2 + (ρ ) 2 2ρ ρ = ρ 2 + (ρ ) 2 2ρρ cos(φ φ )

11 Radiation Due to a Filament The radiation due to a current I at the origin we have seen is A z = µi 4j H(2) 0 (kρ). Therefore, if the current is located at ρ, we have Filamentary Radiation A z (ρ) = µi 4j H(2) 0 (k ρ ρ ) Similarly, a filamentary magnetic current K at ρ radiates F z (ρ) = ɛk 4j H(2) 0 (k ρ ρ )

12 Outline Cylindrical Radiation 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

13 Green s Function Cylindrical Radiation If a current density J z (ρ) is independent of z, we can think of it as a bundle of filaments with current J z (ρ)ds. Then we have the General Formulas for Two-Dimensional Radiation A z (ρ) = µ 4j F z (ρ) = ɛ 4j J z (ρ )H (2) 0 (k ρ ρ )ds M z (ρ )H (2) 0 (k ρ ρ )ds From here, the fields can be computed in the usual way.

14 Far Fields Cylindrical Radiation In the far field, we can make the standard approximation Similarly, for x, ρ ρ ρ ρ cos(φ φ ) H (2) 0 (x) 2j πx e jx. Combining these (and remembering to use a simpler approximation in the denominator) we have The Far Field A z (ρ) = µ e jkρ 8jπkρ F z (ρ) = ɛ e jkρ 8jπkρ J z (ρ )e jkρ cos(φ φ ) ds M z (ρ )e jkρ cos(φ φ ) ds

15 Far Fields Cylindrical Radiation We can compute the far fields using the definitions of A and F. The results are similar to three-dimensional results. Relationship Between E and H E φ = ηh z E z = ηh φ Far Electric Fields E φ = jωηf z E z = jωa z

16 Outline Cylindrical Radiation 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

17 Plane Wave Expansion We often want to express plane waves in cylindrical wave functions. The result must be Thus Finite at the origin, and 2π-periodic. e jx = e jρ cos φ = To find the a n, use orthogonality 2π 0 e jρ cos φ e jmφ dφ = n= n= a n J n (ρ)e jnφ. a n J n (ρ) 2π 0 e j(n m)φ dφ

18 Plane Wave Expansion Using orthogonality 2π 0 e jρ cos φ e jmφ dφ = 2πa m J m (ρ) The integral on the right-hand side is well-known J n (x) = jn 2π 2π 0 e jx cos φ e jmφ dφ

19 Plane Wave Expansion Substituting this in, we find a m = j m and finally The Plane Wave Expansion of Cylindrical Waves e jx = e jρ cos φ = j n J n (ρ)e jnφ n= How might this formula be modified for waves travelling in other directions?

20 The Addition Theorem We are also interested in expanding the field of a filament with respect to a different center. We have seen that a filamentary current I located at ρ = ρ radiates a field with A z = µψ with ψ(ρ, φ) = I 4π H(2) 0 (k ρ ρ ). We can think of the current generating this field as a current sheet, confined to the ρ = ρ cylinder, with J z (φ) = Iδ(φ φ ) ρ where the denominator ensures that 2π 0 J z (φ)ρdφ = I

21 The Addition Theorem We can expand the field inside and outside the tube of current: an J n (kρ)e jnφ for ρ < ρ n= ψ(ρ, φ) = a n + H (2) n (kρ)e jnφ for ρ > ρ n= Since E z ψ, ψ must be continuous at ρ. Thus This is solved if we let a n J n (kρ ) = a + n H (2) n (kρ ). an = H (2) n (kρ )a n a n + = J n (kρ )a n

22 The Addition Theorem We now have ψ(ρ, φ) = Now H φ = ψ ρ = n= n= k k a n H (2) n (kρ )J n (kρ)e jnφ for ρ < ρ a n J n (kρ )H (2) n (kρ)e jnφ for ρ > ρ n= n= a n H (2) n (kρ )J n(kρ)e jnφ for ρ < ρ a n J n (kρ )H (2) n (kρ)e jnφ for ρ > ρ

23 The Addition Theorem From boundary conditions, we have H φ (ρ = ρ + ) H φ (ρ = ρ ) = J z We thus have k a n [J n (kρ )H (2) n (kρ ) H (2) n (kρ )J n(kρ )]e jnφ = J z n= The standard Wronskian formula gives J n (kρ )H (2) n (kρ ) H (2) n (kρ )J n(kρ ) = 2j πkρ

24 The Addition Theorem Therefore, plugging in, 2j πρ n= a n e jnφ = Iδ(φ φ ) ρ We can now find the a n using orthogonality: j π n= a n 2π 0 e jnφ e jmφ dφ = I 2π 0 4ja m = Ie jmφ a m = I 4j e jmφ δ(φ φ )e jmφ dφ

25 The Addition Theorem We thus find that the field of a filament can be expanded as A z = µψ with I 4j H (2) n (kρ )J n (kρ)e jn(φ φ ) for ρ < ρ n= ψ(ρ, φ) = J n (kρ )H (2) n (kρ)e jn(φ φ ) for ρ > ρ I 4j n= Equating our earlier expression, this gives The Addition Theorem H (2) 0 (k ρ ρ ) = n= n= J n (kρ)h (2) n (kρ )e jn(φ φ ) for ρ < ρ J n (kρ )H (2) n (kρ)e jn(φ φ ) for ρ > ρ

26 Outline Cylindrical Radiation from a Circular Cylinder from a Wedge 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

27 from a Circular Cylinder from a Wedge TM by a Circular Cylinder Consider a PEC cylinder of radius a. Let it be excited by a z-polarized incident wave Ez i = E 0 e jkx jkρ cos φ = E 0 e Using our plane wave expansion, we may write E i z = E 0 The total field is of course n= E z = E i z + E s z ; j n J n (kρ)e jnφ it must vanish on the surface of the cylinder.

28 from a Circular Cylinder from a Wedge TM from a Circular Cylinder We may expand the scattered field as

29 from a Circular Cylinder from a Wedge TM from a Circular Cylinder We may expand the scattered field as E s z = E 0 n= j n a n H (2) n (kρ)e jnφ The total field on the surface of the cylinder is thus E z = E 0 n= [ ] j n J n (ka) + a n H (2) n (ka) e jnφ = 0 Therefore a n = J n(ka) H (2) n (ka)

30 from a Circular Cylinder from a Wedge TM from a Circular Cylinder, Current From here we can find any information about the scattering we want. For instance J z = H φ ρ=a = 1 E z jωµ ρ Now 1 jωµ = E 0 jωµ E z ρ = E 0 ρ=a jωµ n= n= ρ=a [ ] j n J n(ka) + a n H (2) n (ka) e jnφ [ ] j n H (2) n (ka)j n(ka) + J n (ka)h (2) n (ka) e jnφ H (2) n (ka)

31 from a Circular Cylinder from a Wedge TM from a Circular Cylinder, Current Using the Wronskian relation, we find The Surface Current J s (φ) = 2E 0 ωµπa n= j n e jnφ H (2) n (ka) For a thin wire, the first term dominates and we can even write E 0 I = 2π jωµ log ka

32 from a Circular Cylinder from a Wedge TM from a Circular Cylinder, Scattered Field Of course, the exact scattered field is given above: E s z = E 0 n= j n a n H (2) n (kρ)e jnφ Using the far field formula for Hankel Functions, we find the Scattered Far Field Ez s = E 0 e jkρ 2 πkρ a n e jnφ We may treat the other polarization in the same way.

33 from a Circular Cylinder from a Wedge due to Filamentary Excitation We can also consider the scattering due to a current I located at ρ. In this case, the incident field is E i z = k 2 I = k 2 I 4ωɛ 4ωɛ H(2) 0 (k ρ ρ ) H (2) n (kρ )J n (kρ)e jn(φ φ ) n= for ρ < ρ We can write the scattered field in the form E s z = k 2 I 4ωɛ n= c n H (2) n (kρ )H (2) n (kρ)e jn(φ φ )

34 from a Circular Cylinder from a Wedge due to Filamentary Excitation From the preceding, it is obvious that The final solution is thus the c n = J n(ka) H (2) n (ka) Total Field from Filamentary E z = k 2 I 4ωɛ n= [ ] H (2) n (kρ > ) J n (kρ < ) + c n H (2) n (kρ < ) e jn(φ φ ) Here ρ < = min(ρ, ρ ) ρ > = max(ρ, ρ )

35 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why?

36 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why? The coefficients c n are the same as the a n from the last problem. Why?

37 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why? The coefficients c n are the same as the a n from the last problem. Why? How can we assure they are the same for every problem?

38 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why? The coefficients c n are the same as the a n from the last problem. Why? How can we assure they are the same for every problem? It is often said this problem is a generalization of the plane wave scattering problem. Why?

39 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why? The coefficients c n are the same as the a n from the last problem. Why? How can we assure they are the same for every problem? It is often said this problem is a generalization of the plane wave scattering problem. Why? How can we recover the plane wave solution from this one?

40 Observations Cylindrical Radiation from a Circular Cylinder from a Wedge Our answer is symmetrical with respect to the substitution ρ ρ and φ φ. Why? The coefficients c n are the same as the a n from the last problem. Why? How can we assure they are the same for every problem? It is often said this problem is a generalization of the plane wave scattering problem. Why? How can we recover the plane wave solution from this one? How else might we approach the solution to this problem?

41 Outline Cylindrical Radiation from a Circular Cylinder from a Wedge 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

42 from a Wedge from a Circular Cylinder from a Wedge Now we look at filamentary scattering from a wedge. y The wedge is composed of two half planes φ = α and φ = 2π α. The filament carries current I and is located at (ρ, φ ). ρ α φ x

43 A Different Approach from a Circular Cylinder from a Wedge We can write the Total Electric Field E z = ν a ν J ν (kρ < )H ν (2) (kρ > ) sin[ν(φ α)] sin[ν(φ α)] The dependence on ρ and φ is clear enough. How can I just write a dependence on ρ and φ? Since E z (α) = E z (2π α) = 0, ν = ν m = mπ 2(π α)

44 Expansion of the Current from a Circular Cylinder from a Wedge The current can be thought of as a cylindrical sheet current (A/m) with an impulsive distribution: J z = I δ(φ φ ) ρ This current can be expanded in a Fourier series in the usual way: J z = I (π α)ρ sin ν m (φ α) sin ν m (φ α) m=1 From boundary conditions, this current is related to the field by J z = H φ (ρ + ) H φ (ρ )

45 from a Wedge from a Circular Cylinder from a Wedge From the usual equations we can find k jωµ m=1 H φ = k jωµ m=1 a νm H (2) ν m (kρ )J ν m (kρ) sin ν m (φ α) sin ν m (φ α) ρ < ρ a νm H (2) ν m (kρ )J νm (kρ) sin ν m (φ α) sin ν m (φ α) ρ > ρ Using the Wronskian, we can write J z = 2 ωµπρ a νm sin ν m (φ α) sin ν m (φ α) m=1 Equating the two expressions for J z we find a νm = ωµπi 2(π α)

46 from a Circular Cylinder from a Wedge Plane Wave from a Wedge We can generalize our solution to plane wave scattering by taking the limit as the limit as the source recedes. The original incident field was E i z = k 2 I 4ωɛ H(2) 0 (k ρ ρ ) Using the large argument approximation and the far field approximation of ρ ρ we write Ez i = ωµi 2j 4 πkρ e jkρ e jkρ cos(φ φ )

47 from a Circular Cylinder from a Wedge Plane Wave from a Wedge This can be written as E i z = E 0 e jkρ cos(φ φ ) where E 0 = ωµi 2j 4 πkρ e jkρ Why is E 0 dependent on ρ?

48 from a Circular Cylinder from a Wedge Plane Wave from a Wedge This can be written as E i z = E 0 e jkρ cos(φ φ ) where E 0 = ωµi 2j 4 πkρ e jkρ Why is E 0 dependent on ρ? For large ρ our solution is 2j E z = πkρ a n j νm J νm (kρ) sin ν m (φ α) sin ν m (φ α) m=1

49 from a Circular Cylinder from a Wedge Plane Wave from a Wedge Plugging in for I and a n we find The Total Field E z = 2πE 0 π α j νm J νm (kρ) sin ν m (φ α) sin ν m (φ α) m=1

50 Outline Cylindrical Radiation from a Circular Cylinder from a Wedge 1 Cylindrical Radiation 2 from a Circular Cylinder from a Wedge

51 2.5 Dimensions? Cylindrical Radiation from a Circular Cylinder from a Wedge 2.5 dimensions is a term of art, not an actual physical description. The geometry of the problem is assumed two-dimensional. The source, however, may be three dimensional. The problems are attacked by superposition (i.e. Fourier Methods). Suppose ( ) 2 x y z 2 + k 2 ψ = 0

52 The General Approach from a Circular Cylinder from a Wedge Since the boundaries are independent of z, we can define ψ in terms of A Superposition ψ(x, y, z) = 1 2π ψ(x, y, k z )e jkzz dk z Substituting into the Helmholtz equation, we find that [ ] 2 x y 2 + (k 2 kz 2 ) We define (as usual) k 2 ρ = k 2 k 2 z. ψ = 0

53 A Filament Cylindrical Radiation from a Circular Cylinder from a Wedge Suppose we have a filament of current I(z). We of course have a TM z field, A z = µψ. The wave function can be written in the form ψ = 1 2π f (k z )H (2) 0 (k ρρ)dk z. Here, then, the transform of the function is ψ = f (k z )H (2) 0 (k ρρ).

54 A Filament Cylindrical Radiation from a Circular Cylinder from a Wedge The azimuthal magnetic field is H φ = ψ ρ = k ρf (k z )H (2) 0 (k ρ ρ). Now, by Ampère s law, lim ρ 0 H φ dl = Ĩ For small ρ, H φ = k ρ f (k z ) d dx ( 2j π ln γx 2 ) x=kρ = 2j πρ f (k z)

55 A Filament Cylindrical Radiation from a Circular Cylinder from a Wedge Thus or Finally, we find the lim ρ 0 H φ dl = 2πρ 2j πρ f (k z) = Ĩ f (k z ) = Ĩ 4j 2.5-D Filament Solution ψ = 1 8πj Ĩ(k z )H (2) 0 ( ) ρ k 2 kz 2 e jkzz dk z

56 Another Way Cylindrical Radiation from a Circular Cylinder from a Wedge We can always make a line current out of dipoles. We thus have The Spatial Approach ψ = e jk ρ 2 +(z z ) 2 I(z ) 4π ρ 2 + (z z ) 2 dz Now, suppose we choose I(z ) = Ilδ(z ). Then ψ = Il e jkr 4πr. Also, we have Ĩ(k z) = Il, since the Fourier Transform of the delta function is unity.

57 A New Identity Cylindrical Radiation from a Circular Cylinder from a Wedge Plugging this transform into the 2.5-D solution, ψ = Il 8πj ( ) H (2) 0 ρ k 2 kz 2 e jkzz dk z Setting these two expressions to each other, we find An Identity e jkr r = 1 2j ( ) H (2) 0 ρ k 2 kz 2 e jkzz dk z