D. S. Weile Radiation

Transcription

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

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4 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0

5 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are

6 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are Partial differential equations.

7 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are Partial differential equations. Coupled.

8 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are Partial differential equations. Coupled. Vector valued.

9 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are Partial differential equations. Coupled. Vector valued. Second order.

10 Maxwell Redux Maxwell s Equation s are: 1 E = jωb = jωµh 2 H = J + jωd = J + jωɛe 3 D = q 4 B = 0 How can we solve them? They are Partial differential equations. Coupled. Vector valued. Second order. Linear and spatially invariant.

11 Vector Potential Solving for the fields directly from Maxwell s Equations is extremely difficult. Therefore, we introduce a vector potential A which makes the solution simpler. In the quantum electrodynamic theory, this A is the primary variable of interest. Because the divergence of B vanishes, we can write it in terms of a Magnetic Vector Potential B = A H = 1 µ A

12 Solving Maxwell s Equations This relation can be substituted into Faraday s Law, resulting in E = jω A or (E + jωa) = 0 What can we say about a curl free vector?

13 Solving Maxwell s Equations This relation can be substituted into Faraday s Law, resulting in or E = jω A (E + jωa) = 0 What can we say about a curl free vector? It is the gradient of a scalar! We thus define a Electric Scalar Potential Φ = E + jωa

14 Proof of a Vector Identity Consider an arbitrary closed contour C bounding a surface S. Using Stokes s Theorem Φ ds = Φ dl S Now, the contour integral is over a closed curve, parameterized by a function f(ξ) (say) between ξ = 0 and ξ = 1 with the property f(0) = f(1). C Thus C Φ dl = 1 0 Φ (f(ξ)) f (ξ)dξ

15 Proof of a Vector Identity By the chain rule, we have dφ dξ = Φ (f(ξ)) f (ξ) Therefore C 1 Φ dl = dφ 0 dξ dξ = Φ (f(1)) Φ (f(0)) = 0 Since this is true for an arbitrary surface Φ = 0

16 On the Way to an Equation We can substitute our expressions of H and E into Ampère s Law giving ( ) 1 µ A = J + jωɛ ( jωa Φ) A = µj + ω 2 µɛa jωɛµ Φ = µj + k 2 A jωɛµ Φ Now, we can define a new vector operator, the The Vector Laplacian 2 A = ( A) A

17 The Vector Laplacian What is the meaning of this strange definition? For a scalar, 2 Φ = Φ This of course makes no sense for a vector. The point is that 2 A = u x 2 A x + u y 2 A y + u z 2 A z

18 The Vector Laplacian What is the meaning of this strange definition? For a scalar, 2 Φ = Φ This of course makes no sense for a vector. The point is that 2 A = u x 2 A x + u y 2 A y + u z 2 A z This holds only in Cartesian coordinates; it has no meaning in other coordinate systems.

19 The Lorenz Gauge Substituting this definition we find ( A) 2 A = µj + k 2 A jωɛµ Φ This is unfortunately still coupled. On the other hand, while we have chosen the curl of A, its divergence is still unassigned. A choice of divergence is called a gauge condition. Given the form of this equation we choose the Lorenz Gauge A = jωɛµφ

20 The Wave Equation Substituting in the Lorenz gauge results in the Wave Equation for A 2 A + k 2 A = µj Finally, we can find an equation for Φ as well: or Wave Equation for Φ q = ɛe = ɛ ( jωa Φ) = ɛ 2 Φ jω 2 ɛ 2 µφ ( ) = ɛ 2 Φ + k 2 Φ 2 Φ + k 2 Φ = q ɛ

21 The Problem, Restated Potential Equations 2 A + k 2 A = µj 2 Φ + k 2 Φ = q ɛ Fields H = 1 µ A E = jωa Φ ( ) A = jωa jωɛµ = jω [A + 1k ] 2 ( A)

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23 Setting Up the Solution We will concentrate on solving for the field due to a z-directed current, so our equation becomes Scalar Wave Equation 2 A z + k 2 A z = µj z To solve this equation we note that it is Linear, and Spatially invariant. How can we use this?

24 The Implication of Linearity Since the equation is linear, if 2 A z1 + k 2 A z1 = µj z1 2 A z2 + k 2 A z2 = µj z2 then 2 (aa z1 + ba z2 ) + k 2 (aa z1 + ba z2 ) = µ (aj z1 + bj z2 ) This linearity also applies to infinite sums and integrals. Suppose the symbol differentiates with respect to r, and that ζ is a parameter. Suppose that for ζ L ζ ζ U 2 g (r, ζ) + k 2 g (r, ζ) = s (r, ζ).

25 The Implication of Linearity If this is the case then suppose ζu 2 A z + k 2 A z = µ J z (ζ)s(r, ζ) dζ ζ L we could then conclude that A z (r) = µ ζu ζ L J z (ζ)g(r, ζ) dζ This is called the principle of superposition. The function g is called a Green s function. The role of ζ is generally taken by r, the location of the source.

26 The Implication of Spatial Invariance Because free space is invariant, if 2 A z (r) + k 2 A z (r) = µj z (r) then, if r is any fixed location in space, 2 A z (r r ) + k 2 A z (r r ) = µj z (r r ) Notice that spatial invariance is a less common occurrence than linearity. Maxwell s equations in free space have been shown linear with centuries of experimental evidence. Spatial invariance, on the other hand, does not exist in a waveguide!

27 The Delta Function In one dimension, given a function f (x) continuous at x = 0 we can define The Delta Function f (x)δ(x) = f (0) The delta function is not really a function, but a generalized function. That is, it is a mapping from the set of continuous functions to R that is linear and continuous (on the function space). The integral notation is useful since all linear functionals can be written this way. The delta function can be conceived as an infinitely tall, thin pulse.

28 The Delta Function in Multiple Dimensions It helps to define a Three-Dimensional Delta Function δ(r) = δ(x)δ(y)δ(z) so that f (r)δ(r) dx dy dz = f (0) Similarly, if we let r = x u x + y u y + z u z, we have Shifted Three-Dimensional Delta Function δ(r r ) = δ(x x )δ(y y )δ(z z )

29 The Delta Function in Multiple Dimensions It helps to define a Three-Dimensional Delta Function δ(r) = δ(x)δ(y)δ(z) so that f (r)δ(r) dx dy dz = f (0) Similarly, if we let r = x u x + y u y + z u z, we have Shifted Three-Dimensional Delta Function δ(r r ) = δ(x x )δ(y y )δ(z z ) What are units of δ(x)? Of δ(r)?

30 Green s Function The equation we wish to solve is The Helmholtz Equation Now, we can write an 2 A z + k 2 A z = µj z Impulsive Formulation of the Current J z (r) = J z (r )δ(r r ) dv This expression envisions the current as a superposition of shifted and weighted impulsive currents.

31 Green s Function Suppose we can find a Green s Function 2 g(r) + k 2 g(r) = δ(r) which solves the problem for a single impulse at the origin. Then, because of linearity and spatial invariance, the solution to the Helmholtz equation is written as a

32 Green s Function Suppose we can find a Green s Function 2 g(r) + k 2 g(r) = δ(r) which solves the problem for a single impulse at the origin. Then, because of linearity and spatial invariance, the solution to the Helmholtz equation is written as a Convolution A z (r) = J z (r )g(r r ) dv

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34 Finding the Green s Function Consider the equation 2 g(r) + k 2 g(r) = δ(r) By symmetry, g(r) is only a function of distance from the origin; that is g(r) = g(r) where r is the radial coordinate in spherical coordinates. The equation thus reduces to ( 1 d r 2 r 2 dg ) + k 2 g = δ(r) dr dr

35 Finding the Green s Function Away from the origin, this reduces to ( d r 2 dg ) + r 2 k 2 g dr dr = 0 r 2 d2 g dg + 2r dr 2 dr + r 2 k 2 g = 0 r d2 g dr 2 + 2dg dr + k 2 rg = 0 d 2 (rg) dr 2 + k 2 (rg) = 0 Therefore, for constants C + and C we have g(r) = C + r e jkr + C e jkr r

36 Finding the Green s Function It should be immediately obvious that Why? C = 0

37 Finding the Green s Function It should be immediately obvious that Why? How do we find C +? We C = 0 Plug back into the original differential equation, and Integrate over a small sphere of radius ɛ (S ɛ ) including the origin. (Why do we do this?) Thus, lim 2 g dv + k 2 ɛ 0 S ɛ S ɛ g dv = S ɛ δ(r)dv

38 Finding the Green s Function We examine this term by term. lim δ(r)dv = ɛ 0 S ɛ

39 Finding the Green s Function We examine this term by term. lim δ(r)dv = 1 ɛ 0 S ɛ

40 Finding the Green s Function We examine this term by term. lim δ(r)dv = 1 ɛ 0 S ɛ lim ɛ 0 S ɛ e jkr r dv =

41 Finding the Green s Function We examine this term by term. lim δ(r)dv = 1 ɛ 0 S ɛ lim ɛ 0 S ɛ e jkr r ɛ π dv = lim ɛ π 0 re jkr sin θ dφ dθ dr =

42 Finding the Green s Function We examine this term by term. lim δ(r)dv = 1 ɛ 0 S ɛ lim ɛ 0 S ɛ e jkr r ɛ π dv = lim ɛ π 0 re jkr sin θ dφ dθ dr = 0

43 Finding the Green s Function We examine this term by term. lim δ(r)dv = 1 ɛ 0 S ɛ lim ɛ 0 S ɛ e jkr Finally, note that r ɛ π dv = lim ɛ π 0 re jkr sin θ dφ dθ dr = 0 g = u r dg dr = u r C + jkre jkr + e jkr r 2

44 Finding the Green s Function Therefore lim 2 g dv = lim g dv ɛ 0 ɛ 0 S ɛ S ɛ 2π π jkɛe = lim g ds = lim [ u jkɛ + e jkɛ r C + ɛ 0 S ɛ ɛ ɛ 2 ] ɛ 2 u r sin θ dθ dφ = 4πC + lim ɛ 0 [ jkɛe jkɛ + e jkɛ] = 4πC + So finally we have 4πC + = 1

45 The Green s Function and the Magic Formula We thus have The Scalar Green s Function g(r) = e jkr 4πr or g(r) = e jk r 4π r Therefore, g(r r ) = e jk r r 4π r r The magnetic potential is thus given by the Magic Formula A z (r) = µ 4π J z (r ) e jk r r r r dv

46 The Solution to Maxwell s Equations Stacking all components of the current gives The Magnetic Potential A(r) = µ 4π J(r ) e jk r r r r dv From this we can compute The Fields H = 1 µ A E = jω [A + 1k ] 2 ( A)

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48 An Approximation Far away from an antenna, the radiated field simplifies considerably. r R = r r r u r r Define R = r r. As usual, r is the integration variable (source point) and r is a parameter (field point). The figure demonstrates that R r r u r

49 The Same Approximation, Taylor Approach For small x, Therefore 1 + x = x + O ( x 2) r r = (r r ) (r r ) = r 2 + (r ) 2 2r r ( r = r 1 + r r 1 2u r r ] r r [1 u r r r = r u r r ) 2 2u r r r

50 Approximating the Green s Function We can use this approximation to R in the exponent, but need not in the denominator. Why?

51 Approximating the Green s Function We can use this approximation to R in the exponent, but need not in the denominator. Why? This gives rise to the Far Field Magnetic Vector Potential A(r) = µ e jkr J(r )e jkur r dv 4π r Note: The integrand is a function of only θ and φ, not r. This allows us to talk about antenna patterns. The radial dependence is a Decay at the rate r 1 (Why?), and Wave travel in the radial direction (e jkr )

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53 The Magnetic Field We can write the expression from the previous slide as Now A(r) = µ e jkr N(θ, φ) = µf (r)n(θ, φ) 4π r and H(r) = 1 A = f (r) N(θ, φ) + N(θ, φ) f (r) µ N(θ, φ) r 1 + O(r 2 ) Since we are ignoring terms that decay faster than r 1, H(r) = N(θ, φ) f (r)

54 The Magnetic Field Thus we have The Far Magnetic Field f (r) = u r f (r) r H(r) = jk e jkr 4πr u r jkre jkr e jkr = u r 4πr 2 jke jkr u r = jkf (r)u r 4πr J(r )e jkur r dv = jk µ u r A(r)

55 The Electric Field The far electric field can be found with the same technique. Recall E(r) = jω [A(r) + 1k ] 2 A(r) = jωµ {f (r)n(θ, φ) + 1k } 2 [f (r)n(θ, φ)] Now [f (r)n(θ, φ)] = f (r) N(θ, φ) + f (r) N(θ, φ)

56 The Electric Field The far electric field can be found with the same technique. Recall E(r) = jω [A(r) + 1k ] 2 A(r) = jωµ {f (r)n(θ, φ) + 1k } 2 [f (r)n(θ, φ)] Now [f (r)n(θ, φ)] = f (r) N(θ, φ) + f (r) N(θ, φ) f (r) N(θ, φ)

57 The Electric Field The far electric field can be found with the same technique. Recall E(r) = jω [A(r) + 1k ] 2 A(r) = jωµ {f (r)n(θ, φ) + 1k } 2 [f (r)n(θ, φ)] Now [f (r)n(θ, φ)] = f (r) N(θ, φ) + f (r) N(θ, φ) f (r) N(θ, φ) = jkf (r)u r N(θ, φ) because N r 1, and We showed it two slides ago.

58 The Electric Field [f (r)n(θ, φ)] jk [f (r)n r (θ, φ)]

59 The Electric Field [f (r)n(θ, φ)] jk [f (r)n r (θ, φ)] = jk [f (r) N r (θ, φ) + N r (θ, φ) f (r)]

60 The Electric Field [f (r)n(θ, φ)] jk [f (r)n r (θ, φ)] = jk [f (r) N r (θ, φ) + N r (θ, φ) f (r)] jkn r (θ, φ) f (r)

61 The Electric Field [f (r)n(θ, φ)] jk [f (r)n r (θ, φ)] = jk [f (r) N r (θ, φ) + N r (θ, φ) f (r)] jkn r (θ, φ) f (r) k 2 u r f (r)n r (θ, φ) = k 2 u r u r f (r)n(θ, φ) Therefore E(r) = jωµ {f (r)n(θ, φ) + 1k } 2 [f (r)n(θ, φ)] = jω [A(r) u r u r A(r)]

62 Relationship Between Far Electric and Magnetic Field Notice that 1 η u r E(r) = ɛ jω µ u r [A(r) u r u r A(r)] = jω ɛµ u r A(r) µ = jk µ u r A(r) = H(r)

63 Far Field Summary Far Field Vector Potential A(r) = µ e jkr 4π r J(r )e jkur r dv Far Fields E(r) = jω [A(r) u r u r A(r)] H(r) = 1 η u r E(r) What do these last two equations say about the nature of the field far from an antenna?

64 Outline

65 The Hertzian Dipole The simplest possible antenna is a Hertzian dipole; its current is (Units?) J(r) = Ilδ(x)δ(y)δ(z)u z

66 The Hertzian Dipole The simplest possible antenna is a Hertzian dipole; its current is J(r) = Ilδ(x)δ(y)δ(z)u z (Units?) While its fields can be computed exactly, we will use the far field formulas for practice. A(r) = µ e jkr 4π r = u z Il µ e jkr 4π r Now u z Ilδ(x )δ(y )δ(z )e jkur r dx dy dz E(r) = jω [u θ A(r)] u θ jω [u φ A(r)] u φ (How did I come to this conclusion?)

67 Hertzian Fields Since u z u θ = sin θ and u z u φ = 0, we have E θ (r) = jω sin θa z (r) We therefore have the Hertzian Dipole Radiation Fields E θ (r) = jkilη e jkr 4πr sin θ H φ (r) = jkil e jkr 4πr sin θ

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69 The Poynting Vector The Poynting vector in the far field is radial, and proportional to r 2. In this case, it is given by S r (r) = 1 2 Re { E θ (r)hφ (r)} = I 2 l 2 k 2 η 32π 2 r 2 sin2 θ The most important thing to compute given this is the Power Radiated P f = π 2π 0 0 S r (r)r 2 sin θ dφ dθ

70 The Power Radiated by a Hertzian Dipole For the Hertzian dipole, this becomes P f = I 2 l 2 k 2 η π 2π 32π 2 sin 3 θ dφ dθ 0 0

71 The Power Radiated by a Hertzian Dipole For the Hertzian dipole, this becomes P f = I 2 l 2 k 2 η 32π 2 = I 2 l 2 k 2 η 16π π 2π 0 π 0 0 sin 3 θ dθ sin 3 θ dφ dθ

72 The Power Radiated by a Hertzian Dipole For the Hertzian dipole, this becomes P f = I 2 l 2 k 2 η 32π 2 = I 2 l 2 k 2 η 16π = I 2 l 2 k 2 η 16π π 2π 0 π 0 π 0 0 sin 3 θ dθ sin 3 θ dφ dθ sin θ(1 cos 2 θ) dθ

73 The Power Radiated by a Hertzian Dipole For the Hertzian dipole, this becomes P f = I 2 l 2 k 2 η 32π 2 = I 2 l 2 k 2 η 16π = I 2 l 2 k 2 η 16π = I 2 l 2 k 2 η 16π π 2π 0 π 0 π 0 0 sin 3 θ dθ sin 3 θ dφ dθ sin θ(1 cos 2 θ) dθ [ cos θ + cos3 θ 3 = I 2 l 2 k 2 η 4 16π 3 = I2 l 2 k 2 η 12π ] π 0

74 Outline

75 Radiation Resistance The input impedance of an antenna is the ratio between the (phasor) voltage and current at the terminals. The imaginary part of this impedance is very hard to calculate in general, since it involves knowing the near field of the antenna. (Why?) The real part is easy to calculate and is known as the radiation resistance. The Radiation Resistance R rad = 2P f I 2 For a Hertzian Dipole, this works out to R rad = η (kl)2 6π. What does it mean that it is proportional to (kl) 2?

76 Directivity An isotropic radiator radiates the same power in every direction. The directivity of an antenna in a given direction is the ratio of the power radiated in that direction to the power radiated in that direction by an isotropic radiator radiating the same total power. An isotropic radiator radiating P f radiates a Poynting vector S isotropic r = P f 4πr 2 The directivity of an antenna is its maximum directivity over the entire sphere.

77 Directivity A general formula is thus Directivity D(θ, φ) = 4πr 2 S r (θ, φ) P f The directivity of a Hertzian dipole in the θ direction is thus D(θ, φ) = 4πr 2 I 2 sin 2 θl 2 k 2 η 12π 32r 2 π 2 I 2 l 2 k 2 η = 3 2 sin2 θ and its overall directivity is D = 3 2