B. Spherical Wave Propagation

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1 11/8/007 Spheical Wave Popagation notes 1/1 B. Spheical Wave Popagation Evey antenna launches a spheical wave, thus its powe density educes as a function of 1, whee is the distance fom the antenna. We can detemine the powe adiated by an antenna if we know the powe density of the popagating spheical wave it poduces. HO: Total Radiated Powe To descibe how an antenna distibutes adiated powe, we fist need to undestand the concept of a solid angle measued in units of steadians. HO: The Steadian We find that an antenna neve adiates powe equally in all diections. Instead, it adiates powe moe in some diections and less in othes. The mathematical desciption of this distibution is called adiation intensity. HO: Radiation Intensity

11/8/007 Total Radiated Powe 1/6 Total Radiated Powe So, we know that an antenna (located at the oigin) will poduce a adiated powe density of the fom: ˆ W ( ) = U ( θφ, ) Q: But this is the powe

2 11/8/007 Total Radiated Powe 1/6 Total Radiated Powe So, we know that an antenna (located at the oigin) will poduce a adiated powe density of the fom: ˆ W ( ) = U ( θφ, ) Q: But this is the powe density a function of spatial position (i.e., a function of, θ, φ ). Is thee any way to detemine the total powe adiated by an antenna? A: The powe density function gives us all we need to know to detemine the powe adiated by an antenna ( P ad ). To see why, fist conside some apetue (i.e., window) defined by suface aea S. Say that at some (pehaps lage) distance away fom this suface, thee exists a adiating antenna. S This adiating antenna poduces a popagating electomagnetic wave at all points thoughout the entie univese! The entie univese includes ou apetue (suface S) and thus thee must be electomagnetic enegy popagating though this apetue (i.e., fom one side of suface S to the othe).

3 11/8/007 Total Radiated Powe /6 W ( ) S Q: But an electomagnetic wave contains enegy, and thus enegy must be passing though this apetue. Can we detemine the ate of this enegy flow though suface S? A: No poblem! If we know the powe density W ( ), we can always detemine the powe flowing though some apetue S using a suface integation: W S ( ) P = ds Hopefully this makes sense to you! We simply integate the powe density (in W/m ) flowing though the suface S (in m ) to detemine the ate of enegy flow (in Watts) though the entie suface S. Q: So the value P is the powe adiated by the antenna?

11/8/007 Total Radiated Powe 3/6 A: Absolutely not! Although the powe P does depend on the powe density poduced by the antenna, it also depends on the location, oientation, and size of apetue S.

4 11/8/007 Total Radiated Powe 3/6 A: Absolutely not! Although the powe P does depend on the powe density poduced by the antenna, it also depends on the location, oientation, and size of apetue S. Fo example, if the apetue size appoaches zeo, the powe P flowing though the apetue likewise appoaches zeo. This of couse does not mean that the powe adiated by the antennas is zeo it is likely vey lage. Howeve, thee ae sufaces S wee the suface integation does tell us pecisely the adiated powe! Conside now a closed suface one that completely suounds the adiating antenna. It tuns out that integating the powe density acoss this closed suface will tell us the powe adiated by the antenna: ad W S ( ) P = ds Q: Yikes, why does this wok? What s so special about a closed suface? A: Essentially, this woks because of consevation of enegy.

5 11/8/007 Total Radiated Powe 4/6 Remembe, an antenna popagates electomagnetic enegy outwad fom the antenna. This enegy flows in all possible diections (although not unifomly in all diections!). If we integate the powe density acoss a suface that completely suounds the antenna, then we ae captuing the enegy flowing outwad in all possible diections. Thee ae no holes in a closed suface! Ou answe thus descibes the total powe flowing outwad fom the antenna. By consevation of enegy, this must be equal to the powe being adiated by the antenna! Thee is one impotant caveat to this statement. The volume suounded by closed suface S must be lossless. If thee is lossy mateial in this volume, then some of the adiated powe will be absobed by this mateial, and thus will not popagate though closed suface S. In this case we will find: P > ds ad W S ( ) if volume is lossy But, we will assume that the volume is not lossy, that it is essentially fee-space (e.g., a clea atmosphee). Q: But what should the closed suface S be? Doesn t the integation depend on the size/shape of closed suface S?

6 11/8/007 Total Radiated Powe 5/6 A: Although this would seem to be the case, it is in fact not. Any and all closed sufaces that suound the antenna will in fact povide the same answe fo the suface integation. Afte all, thee is only one coect value of P ad! Q: Since it doesn t matte, can I just choose any closed suface S that suounds the antenna. A: Theoetically yes, but being efficient (i.e., lazy) enginees, we might choose a suface S that makes the suface integation as easy as possible. This closed suface is simply a sphee, centeed at the oigin (i.e., centeed at the antenna location). To see how this simplifies things, conside a spheical suface S, with adius a. This suface is thus descibed as: = a 0 θ π 0 φ π and ds = ˆ d θ d φ. Given that the adiated powe density has the fom: ˆ W ( ) = U ( θφ, ) we find that the suface integal is:

7 11/8/007 Total Radiated Powe 6/6 π π 1 W S ( ) ds = U ( θ, φ) ˆ ˆ dθdφ = 0 0 ππ 0 0 U ( θφ, ) dθdφ Thus, we find that we can always detemine the adiated powe by integating ove the adiation intensity function poduced by the antenna: ππ P U ( θ, φ) dθdφ ad = 0 0

8 11/8/007 The Steadian 1/5 The Steadian Q: What the heck is a steadian? A: Fist, let s examine what a adian is! Recall fo a cicle (a two-dimensional object), we find: = φ whee: φ = ac length [ m] φ = angle adians = adius [ m] An ac length that entiely suounds the cicle would be the cicle s cicumfeence (C ). The angle φ that subtends this ac length is of couse be π, and inseting these values into the equation above gives: C = π φ = π = C Look familia?

9 11/8/007 The Steadian /5 Q: But wait! A cicle is a twodimensional stuctue, yet we (along with ou antennas) live in a thee-dimensional wold. A: Tue enough! The 3-dimensional equivalent of a cicle is a sphee. Conside a small patch on the suface of the sphee, with suface aea A. Each patch is subtended by a cone, whose point begins at the cente of the sphee. The lage the suface aea of the patch, the lage the cone. We say this cone foms a solid angle Ω, and the size of this cone is expessed in Steadians!

10 11/8/007 The Steadian 3/5 Q: How do we detemine the size of a solid angle in steadians? A: The size Ω of a solid angle that subtends a section of suface aea on a sphee with adius is: A Ω= whee A the aea of the subtended suface (i.e., the aea of the patch ) and Ω is expessed in steadians. Be caeful! The units of A and must be the same! Fo example if A = 100 m then must be expessed in metes. If =5 kilometes then A must be expessed in km. Now, we can ewite the above equation into its moe common fom: A =Ω Note that neithe the shape of the subtended patch of suface, no the shape of the esulting solid angle, mattes in the above elationship. In othe wods the subtended patch could be a cicle, tiangle, squae, o any othe shape. The esult above would be the same!

11 11/8/007 The Steadian 4/5 A 1 A Ω 1 Ω A 3 Ω 3 Theefoe, if A 1 = A = A 3, then Ω 1 =Ω =Ω 3. Even though they have diffeent shapes, they have pecisely the same size! Q: What if the solid angle gets so lage that it subtends the entie suface of the sphee? What is the size of the solid angle then??? A: Recall the suface of an entie sphee has aea: A = 4π We likewise know that: A =Ω Equating these two, we find that a solid angle that subtends the entie suface of a sphee must have a size of 4π steadians!

of 4π steadians subtends the entie suface of a sphee.

12 11/8/007 The Steadian 5/5 Thus, we conclude that a plana angle of π adians subtends the entie cicumfeence of a cicle, but a solid angle of 4π steadians subtends the entie suface of a sphee. Note this means that a solid angle of π steadians subtends the suface of a hemisphee.

13 11/8/007 Radiation Intensity 1/7 Radiation Intensity We found that the powe density of a spheical wave poduced by an antenna located at the oigin has the fom: W ˆ (, θφ, ) = U ( θφ, ) and that the total adiated powe fom an antenna located at the oigin is: ππ P U ( θ, φ) sinθ dθdφ ad = 0 0 Q: This adiation intensity U ( θ, φ ) seems to be vey impotant. What does it indicate? Does it have any physical meaning? A: It tuns out that an antenna does not (in fact, it cannot) adiate powe unifomly in all diections. Rathe, an antenna distibutes powe unequally moe powe in some diections and less powe in othes. The adiation intensity U ( θ, φ ) descibes this unequal distibution of powe it tells how the adiated powe is distibuted as a function of adiation diection. Q: What ae the units of the adiation intensity function?

14 11/8/007 Radiation Intensity /7 A: Radiation intensity is expessed in units of Watts/steadian! To see why, conside the (impossible) case whee an antenna does distibute adiated powe equally in all diections. We call this isotopic adiation. The intensity in this case is thus a constant: U ( θ, φ ) = U0 Although this isotopic intensity function is physically impossible, it will help illustate the physical meaning of intensity. We find that the adiated powe fo this case is: ππ P = U ( θ, φ) sinθ dθdφ ad = 0 0 ππ ππ 0 0 = 4π U U sinθ dθdφ = U sinθ dθdφ = U ( π )( ) 0 We can eaange this esult to detemine that the intensity poduced by an isotopic adiato is:

15 11/8/007 Radiation Intensity 3/7 U ( θφ, ) U0 = = Pad 4π Watts steadian Fo isotopic adiato only! Q: What s up with that 4π in the denominato? Does it have any significance? A: Absolutely! Look again at the units of intensity Watts/Steadian. Compae this to the expession fo isotopic adiation intensity: U 0 Pad = 4π Watts steadian In othe wods, the antenna adiates P ad Watts unifomly thoughout a solid angle of 4π steadians. Q: 4π steadians! Isn t that the size of a solid angle that subtends an entie sphee? A: You bet! 4π steadians is the lagest possible solid angle, one that includes all possible diections θ and φ. Now, say that an antenna adiates all its powe unifomly thoughout a one hemisphee (and thus no powe into the othe hemisphee).

16 11/8/007 Radiation Intensity 4/7 The adiation intensity in one hemisphee (with a solid angle of π steadians) is theefoe P π, and zeo in the othe: ad D ( θφ, ) P ad in one hemisphee π = W std 0 in the othe hemisphee Note that the antenna in this case places all its adiated powe in a solid angle half the size of the isotopic case. As a esult, the intensity in the solid angle is twice as lage as the isotopic case! O, if an antenna adiates all its powe unifomly thoughout a solid angle of Ω steadians, the adiation intensity within this solid angle will be P ad Ω, and zeo outside the solid angle: D ( θφ, ) P ad inside solid angle Ω Ω = W std 0 outside solid angle Ω Note that as the solid angle gets smalle, the intensity will incease (assuming ad P emains unchanged). As the antenna focuses its powe into a smalle and smalle cone, the intensity in that cone will get lage and lage.

17 11/8/007 Radiation Intensity 5/7 This is somewhat(?) analogous to compessing a fixed amount of gas into a smalle and smalle volume the pessue within the volume will get moe intense! These ae simple examples to illustate the meaning of adiation intensity U ( θ, φ ). Howeve, we find that the adiation intensity of eal antennas will be continuous functions of θ and φ. Fo example: D ( θ, φ) = cos θ sinφ Q: I m a bit confused. What s the diffeence between adiation intensity U (, ) W? θ φ and powe density ( ) A: Recall the two ae elated by the expession: W ˆ (, θφ, ) = U ( θφ, ) Fom this expession we note these diffeences: 1. Radiation Intensity U ( θ, φ ) is a scala quantity, while powe density W ( ) is a vecto quantity.. Radiation Intensity U ( θ, φ ) is a function of coodinates θ and φ only, while powe density W ( ) is a function of all thee spheical coodinates, θ, φ.

11/8/007 Radiation Intensity 6/7 Fom these obsevations we can conclude: Radiation Intensity U ( θ, φ ) is a desciption of how an antenna (located at = 0) behaves how it distibutes enegy acoss

18 11/8/007 Radiation Intensity 6/7 Fom these obsevations we can conclude: Radiation Intensity U ( θ, φ ) is a desciption of how an antenna (located at = 0) behaves how it distibutes enegy acoss diffeent diections defined by coodinates θ and φ. Powe density W ( ) is a desciption of the popagating (spheical) electomagnetic wave ceated by the antenna. It is defined at all points in the univese we can detemine the magnitude and diection of the powe density at any location in space a location denoted by coodinates, θ, φ. Finally, let s again conside the mythical isotopic adiato. We know that the intensity of such a adiato will be unifom acoss all diections: U ( θφ, ) U0 = = Pad 4π Watts steadian And thus the powe density ceated by this isotopic adiato will be: ˆ W (, θφ, ) = U ( θφ, ) ˆ = U0 P ˆ ad = 4π Pad = ˆ 4π Fo isotopic adiato only!

19 11/8/007 Radiation Intensity 7/7 Note that this makes pefect sense! Conside a sphee, centeed at the oigin, with adius. At the cente of this sphee is an isotopic adiato, and thus the powe density must be pecisely the same at evey point on the suface of this sphee. Q: Why is that? A: Remembe, the adiation of an isotopic adiato is independent of θ and φ (i.e., the same in all diections), and evey point on the sphee is pecisely the same distance fom the adiato (the distance ). The powe density thus must likewise be a constant acoss this entie spheical suface. Q: But what is the value of this constant? A: We simply take the powe flowing though this sphee (i.e., P ad by consevation of enegy), and divide it by the suface aea of the sphee. Recall the suface aea of this sphee is 4π! Pad (, θ, φ ) = ˆ 4π W Watts unit aea

20 11/8/007 Spheical Coodinates 1/ Spheical Coodinates * Geogaphes specify a location on the Eath s suface using thee scala values: longitude, latitude, and altitude. * Both longitude and latitude ae angula measues, while altitude is a measue of distance. * Latitude, longitude, and altitude ae simila to spheical coodinates. * Spheical coodinates consist of one scala value (), with units of distance, while the othe two scala values ( θ, φ ) have angula units (degees o adians). z P(, θ, φ ) y x

21 11/8/007 Spheical Coodinates / 1. Fo spheical coodinates, (0 < ) expesses the distance of the point fom the oigin (i.e., simila to altitude).. Angle θ (0 θ π ) epesents the angle fomed with the z-axis (i.e., simila to latitude). 3. Angle φ (0 φ < π ) epesents the otation angle aound the z-axis, pecisely the same as the cylindical coodinate φ (i.e., simila to longitude). z θ = 45 P(3.0, 45,60 ) P(0,θ,φ) =3.0 y φ = 60 x Thus, using spheical coodinates, a point in space can be unambiguously defined by one distance and two angles.

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50 woking pages fo Paul Richads class notes; do not copy o ciculate without pemission fom PGR 2004/11/3 10:50 CHAPTER7 Solid angle, 3D integals, Gauss s Theoem, and a Delta Function We define the solid angle,