http://www.cv.nrao.edu/course/astr534/dapertures... Two-Dimensional Aperture Antennas The field pattern of a two-dimensional aperture The method we used to show that the field pattern of a one-dimensional aperture is the one-dimensional Fourier transform of the aperture field illumination can simply be generalized to the more realistic case of a two-dimensional aperture: Z 1 f(l; m ) / g(u; v)e ÀiÙ(lu+mv) dudv (3C1) where m is the y-ais analog of l on the -ais, and v Ñ y= Z 1. In words, The electric field pattern of a two-dimensional aperture is the two-dimensional Fourier transform of the aperture field illumination. The Uniformly Illuminated Rectangular Aperture A two-dimensional rectangular aperture with side lengths D and D y. Dividing lengths in the u Ñ = v Ñ y= l Ñ sin m Ñ sin y is the angle from the (y; z) plane and y is the angle from the (; z) plane. aperture plane by the wavelength yield the normalized coordinates and. The direction from the origin to any distant point can be specified by and, where The two-dimensional counterpart of a uniformly illuminated one-dimensional aperture is a uniformly y y illuminated rectangular aperture with sides D and D. If the illumination g(; ) is constant over the aperture, the integrals over u and v in the Fourier transform are separable and ld md y f(l; m ) / sinc sinc ; 1 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... where sinc() Ñ sin(ù)=(ù) : Squaring the electric field pattern gives the relative (normalized to unity at the peak) power pattern P n(l; m ) = sinc ld md y sinc Given the relative power pattern, we can calculate the absolute power gain G in any direction by invoking energy conservation: Z +1 Z +1 Defining the temporary variable a: Z G dê = 4 Ù = G 0 P n (l; m)dl dm Z " # +1 Z sin(ùld =) " # +1 sin(ùmd y=) 4Ù = G 0 dl dm ÙlD = ÙmD = y D µ gives, for, ÙlD ÙD a Ñ ; so da = dl Z " # +1 ÔZ sin(ùld =) 1 sin a dl Ù da ÙlD = a = ÙD D since we can look up the definite integral in square brackets; its value is Ù. 4Ù = G 0 : D D y Thus the peak power gain is G 0 = 4ÙD D y and the power pattern of a uniformly illuminated rectangular aperture with side lengths D and D is y 4ÙD D y ld md y 4ÙD D y D y D G = sinc sinc Ù sinc sinc y (3C when and are much smaller than one radian. y In general, the peak power gain of an aperture antenna is proportional to the geometric area Ageom (Ageom = D D y in this case) of the aperture. The constant of proportionality is 4Ù= for a uniformly illuminated aperture and somewhat less for any other illumination pattern. of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... Using the relation G Ae = 4Ù we find that the on-ais effective collecting area is G ma(a e ) = 4Ù 0 = 4Ù D D y = D 4Ù D y = A geom The peak effective area of an ideal uniformly illuminated aperture equals its geometric area, independent of wavelength. With any other illumination taper, the effective area is smaller than but proportional to the geometric area. It is useful to define the aperture efficiency Ñ A as the ratio of the effective area to geometric area: Ñ A Ñ ma(a e ) A geom (3C3) Thus Ñ A = 1 for an ideal uniformly illuminated aperture and Ñ A < 1 otherwise. D µ Large ( ) waveguide horns are nearly uniformly illuminated unblocked apertures, so their actual gains and effective collecting areas can be calculated accurately. This makes them useful for measuring the absolute flu densities of strong sources such as Cas A and Cyg A and defining the practical flu-density scales used by radio astronomers (see Baars et al. 1977, A&A, 61, 99). The "Little Big Horn" at Green Bank, WV in 1959. It was used to measure the absolute flu density of the strong source Cas A. An ecess noise of T Ù 3:5 K was found but not recognized as important. The Uniformly Illuminated Circular Aperture Most apertures associated with reflectors and lenses are circular. The power pattern of a uniformly illuminated circular aperture is known as the Airy pattern. This linked interactive plot shows how the Airy pattern behaves as a function of wavelength and aperture size. 3 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... The Circular Gaussian Illumination Pattern A good approimation for "real" radio telescopes with practical feeds and circular apertures is a circular Gaussian illumination pattern. The normalized Gaussian function g(u) in one dimension is defined by 1 u g(u) Ñ p ep À ; Ù Û Û where Û is the rms (root mean square) width defined by ÔZ 1 Z 1 Û Ñ u g(u)du g(u)du 1= R 1 (u)du g = 1 and "normalized" means. g = ep[àu =(Û )] This bell-shaped curve is a plot of the (not normalized) Gaussian function. For the circular Gaussian illumination pattern the field pattern is where Û is the rms radius of the illumination pattern in wavelengths. The function f(l; m separable: where u + v u v ep À = ep À ep À Û Û Û Z 1 Z 1 u v f(l; m ) / ep À ep À ep [ÀiÙ(lu + mv)]dudv ; Û Û f(l; m ) / f(l) Â f(m); ) is 4 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... Z 1 u f(l) Ñ ep À ep(àiùlu)du Û and Z 1 v f(m) Ñ ep À ep(àiùmv)dv Û Each integral is the one-dimensional Fourier transform of a Gaussian. The Fourier transform of the Gaussian is the Gaussian (derivation). Given this Fourier f() = ep(àù ) F (s) = ep(àùs ) = 1 a = ( p ÙÛ) transform of the Gaussian with ÙÛ, we can use the similarity theorem with to get so the field pattern of an aperture with circular Gaussian illumination is and the normalized power pattern is f(l) / ep(àùl  ÙÛ ) f(m) / ep(àùm  ÙÛ ) f(l; m ) / ep[àù Û (l + m )] P n(l; m ) = [ f(l; m)] / ep[à4ù Û (l + m )] : For a reflector many wavelengths across, so and. Thus the power pattern produced by circular Gaussian illumination is also a circular Gaussian. The half-power beamwidth HPBW is the solution of Û µ 1 l Ù m Ù y Ô ep À4Ù Û HPBW 1 = 4Ù Û HPBW = ln() 4 p ln() HPBW = radians; ÙÛ where Û is the rms width of the illumination pattern in wavelengths. Since a Gaussian etends to Æ1 However, the Gaussian illumination g(u; v) falls off eponentially and is negligible for u + v µ Û. In practice, the value of Û is chosen so that g(u; v) is down quite a bit (e.g., 15 or 0 db) at the, this calculation formally assumes that the aperture itself is infinite. edge of the actual finite reflector. Tapering (or grading) the illumination this way: (1) broadens the beamwidth HPBW slightly, () reduces the aperture efficiency Ñ A slightly, (3) reduces the sidelobe level significantly, and (4) reduces spillover (wasted power, increased noise from ground pickup) significantly compared with uniform illumination. 5 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... Spillover is illumination etending beyond the aperture. Here the relative field strength at the edge is Î. Using Gaussian tapering as an eample, let g(u ; ) Î Ñ g(0; 0) ma 0 be the tapered field strength at the edge of the reflector. Thus Î = 0:1 (field taper) corresponds to Î = 0:01 (power taper), or a 0 db edge taper. This corresponds to an aperture of radius Ù :15Û. For small Î our previous calculation, which assumed an infinite aperture so Î = 0, will slightly underestimate the beamwidth of a finite reflector but still be fairly accurate because only a small fraction of the illumination will "spill over" the edge of the reflector. For a circular reflector with diameter D, uma = D and g(u ma ; 0 ) u Î = = ep ma À : g(0; 0) Û Then = 0 Recall that the beamwidth for Î is u ma À ln (Î) = = D : Û 8 Û p ln() HPBW = ; ÙÛ so for a finite aperture, HPBW Ù p ln() q À8 ln(î) ÙD 6 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... HPBW Ù p À8 ln() ln(î) Ù D Eample: What is the half-power beamwidth of a circular aperture with a Gaussian illumination tapering to Î = 0:1 at the edge? p À8 ln() ln(0:1) HPBW Ù :14 Ù D Ù 1 D Eample: Estimate the beamwidth in arcsec of the GBT (D GHz. HPBW Ù 1 :14 c D = 1 00 m) as a function of frequency in HPBW Ù 1 :14 rad  0665 arcsec=rad  3:00  10 8 m s (GHz)  10 9 Hz=GHz  100 m HPBW Ù 705 arcsec (GHz) This estimate is slightly low, as epected. The measured beamwidth of the GBT is about 740 arcsec / (GHz), or about 1:=D. A good approimation for the half-power beamwidths of most single-dish radio telescopes is HPBW Ù 1 :=D (3C4) Reflector Accuracy Requirements Real radio telescopes don't have perfectly paraboloidal reflectors. Small deviations from the best-fit paraboloid may be caused by permanent manufacturing errors, changing gravitational deformations as the reflector is tilted, thermal distortions resulting from solar heating, and bending by strong winds. There will be some shortest wavelength min below which these errors degrade the reflector performance so severely that the telescope becomes unusable. We can define the reflector surface efficiency Ñ s as the power gain of the actual reflector divided by the power gain of a perfectly paraboloidal reflector with the same size and illumination. Net we will calculate how Ñs 7 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... varies with the rms (root mean square) surface error in wavelengths, Ï=. The classic reference for this calculation is Ruze, J. 1966, Proc. IEEE, 54, 633. Deviations Ï of the actual reflector surface (thick curve) from the best-fit paraboloid (thin curve) degrade short-wavelength performance. Where the actual reflector surface deviates from the best-fit paraboloid by a distance Ï, the path length of the reflected wave will be in error by almost Ï and the phase error Î (radians) will be Ù 4ÙÏ Î = (Ï) = : An oversimplified eample would be a bumpy surface, half covered with small bumps of height and half covered with small dips of the same depth Ï. Then the contribution of each area Ï Ü element to the far (electric) field is reduced by a factor cos. Î Vectors sums of the electric fields produced by elements of perfect and imperfect apertures. Bumps in the imperfect aperture produce phase shifts. ÆÎ Î Ü 1 cos Î Ù 1 À Î =::: In the limit rad, and E(Î) Î Ù 1 À ::: E(0) so the relative power gain is G(Î) G(0) Ù Ô E(Î) E(0) Ù 1 À Î 4ÙÏ Ù 1 À 8 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... This rough estimate shows that the surface errors must be an order-of-magnitude smaller than the shortest usable wavelength, a severe requirement indeed. A more realistic calculation makes use of the fact the most error distributions are roughly Gaussian. Suppose that the surface errors have a Gaussian probability distribution P with rms Û: where the normalization factor in front of the eponential term ensures that Z 1 as required for any probability distribution. Then the relative field strength is obtained as the weighted sum over all possible Ï: Z 1 By substituting e iz = cos (z) + i sin(z) we can turn this integral into a more familiar one, the Fourier transform of a Gaussian: Note that the i sin(z) part drops out immediately because it is antisymmetric over the symmetric region of integration. To make this look even more like our usual form for a Fourier transform, let s Ñ =, Ñ Ï, and a Ñ ( p ÙÛ). Then Z E 1 Recall that and apply the similarity theorem to get P (Ï) = 1 Ï p ep À ; Ù Û Û P (Ï)dÏ = 1 Power is proportional to E so our final result for the reflector surface efficiency is E h i À dï E(0) Ù 4ÙÏ cos  1 Ï p ep Ù Û Û Z E 1 h i Ài À dï : E(0) Ù 4ÙÏ ep  1 ÙÏ p ep Ù Û ÙÛ h i (ÀiÙs) [ÀÙ(a) ]d : E(0) Ù ep ep f() = ep(àù ) has Fourier transform F (S) = ep(àùs ) Ô E h i ÀÙ E(0) = 1 s p ep jaj Ù Û a E h i [ÀÙ Û s ] E(0) = ep E h i À : E(0) = ep 8Ù Û 9 of 10 09/7/010 1:18 PM
http://www.cv.nrao.edu/course/astr534/dapertures... Ô 4ÙÛ Ñs = ep À (3C5) This is often called the Ruze equation. The surface efficiency Ñ s declines rapidly as the rms error in wavelengths Û= eceeds 1=16 Ù 0:06. Eample: A traditional criterion for the shortest wavelength min at which a radio telescope works reasonably well is Û Ù min 16 because the surface efficiency is only Ô Ù Ñs Ù ep À Ù 0:54 4 and falling eponentially at shorter wavelengths. The 100 m diameter GBT is intended to operate at frequencies as high as Ù 100 GHz, or mm. To meet this specification, the rms deviation from a perfect paraboloid must not eceed m, the thickness of two sheets of paper! Û Ù 3 mm=16 Ù 00 Ö min Ù 3 The power gain of a perfect paraboloidal reflector is proportional to. If the reflector surface has a Gaussian error distribution with rms Û, then its gain increases as at low frequencies, reaches a maimum at and falls eponentially at higher frequencies. = 4ÙÛ 10 of 10 09/7/010 1:18 PM