Документ взят из кэша поисковой машины. Адрес оригинального документа : http://www.naic.edu/~astro/aotms/backends/95-01.pdf Дата изменения: Mon Apr 26 01:47:44 1999 Дата индексирования: Sat Dec 22 05:19:32 2007 Кодировка: Поисковые слова: п п п

CORRELATOR POLARIMETRY
Jim Cordes 1995 November 27

This writeup is intended to clarify the relationship between polarimetry done with a correlator and continuum polarimetery done with a single-channel system. First I derive the correlation functions for a monochromatic signal and then a continuum signal. These show explicitly how the Q and U Stokes parameters are related to the symmetric and antisymmetric components of the cross-correlation function of the LHCP and RHCP voltages.

Monochromatic, Linearly Polarized Signals
Consider a monochromatic, linearly polarized signal at RF: ~ E t= c cos!t^ + s cos!t^ x y: 1 where c = cos and s = sin and is the position angle. The corresponding components of circular polarization are EL;R t= Ex t Ey t , t1=4 2 where t1=4 =2!0 is a delay equal to one quarter cycle at the center frequency !0. The IF signals are with implicit ltering o of the upper sideband LIF ;RIF = EL;R t cos!LO1 t 3 and the baseband voltages are again with implicit ltering `t;rt= LIF ;RIF t cos!LO2 t: 4 The autocorrelation functions acfs of the baseband elds are 1 h`t`t + i = 2 cos ! c2 + s2 +2cs cos!t1=4 = 1 cos ! 2 hrtrt + i = 1 cos ! c2 + s2 , 2cs cos!t1=4 = 1 cos ! 2 2 from whichtwo of the Stokes parameter correlations may be de ned: I = h`t`t + i + hrtrt + i = cos ! V = h`t`t + i,hrtrt + i =0: The cross correlation function ccf between ` and r yields L Q + U : L = 2h`trt + i = c2 , s2 cos ! , 2cs sin ! sin!t1=4 ; = cos 2 cos ! , sin 2 sin ! = cos ! +2 1

5 6 7

8


where ! Lines 1-2 of transformed The position

! , !LO1 , !LO2 is the baseband frequency of the monochromatic signal. Eq. 8 explicitly show symmetric and antisymmetric parts of L . When to the frequency domain, they correspond to Q~ and U ~ , respectively. ! ! angle is found in the usual wayto be

!= ~

U ~ ! , 2 tan,1 Q~ : !

1

9

Linearly Polarized Continuum Signal
Now consider a continuum signal, again with 100 polarization, analyzed in a total bandwidth B. We assume there is no Faraday rotation across the bandwidth see below. Stokes parameters may be found byintegrating the monochromatic result over frequency because the SP's are variance-like quantities and the frequency components are statistically independentvariances add. Performing integrals like
I L

= =

Z Z

B B

0 0

d !I

m

d !Lm



10
;

where Im ;Lm are the monochromatic results from Eq. 7-8 that depend on the frequency ! , we nd that
I L V

= sin = 0
:

B B

 sin = cos 2



 cos + sin 2

B

, 1

11

When transformed to the frequency domain, Eq. 11 yield the Stokes parameters vs. frequency. In this case, the SPs are independent of frequency and it may be seen that a linearly polarized continuum source with arbitrary position angle is representable with correlation functions as wehave de ned them in Eq. 5 - 8.

Elliptically Polarized Signals
A monochromatic signal with arbitrary elliptical polarization is handled as above. General expressions that include di erential timing delays, Faraday rotation, and LO phase o sets may be found in Eq. C3-C4 in my 1988 memo, Polarimetry with the 40 MHz Correlator. Similarly, noise-like signals with arbitrary polarization may be analyzed byintegration of the monochromatic results, as we did above. 2


Single Channel Polarimetry
By single channel polarimetry, I mean a system where only a single lag of the auto-andcross correlations is computed. This might be e ected with analog multipliers rather than a digital correlator. In this case, one must explicitly calculate the ccf between the RHCP and LHCP components with a90 phase shift as well as without a phase shift. This may be seen by using Eq. 1-2 and calculating the cross correlations for a monochromatic signal 2hERtEL t + i = cos 2 cos ! + sin 2 sin ! 12 2hER tEL t + , t1=4i = cos 2 sin ! , sin 2 cos ! : 13 At zero lag = 0, the unshifted correlation Eq. 12 gives cos 2 while the shifted correlation Eq. 13 gives sin 2 ; both cos 2 and sin 2 are needed to solve for the position angle without ambiguity.

3