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CORRELATOR POLARIMETRY
Jim Cordes
1995 November 27
This writeup is intended to clarify the relationship between polarimetry done with a cor­
relator and continuum polarimetery done with a single­channel 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 antisym­
metric components of the cross­correlation function of the LHCP and RHCP voltages.
Monochromatic, Linearly Polarized Signals
Consider a monochromatic, linearly polarized signal at RF:
~
E(t) = c cos(!t)“x + s cos(!t)“y: (1)
where c = cos ü and s = sin ü and ü is the position angle. The corresponding components
of circular polarization are
E L;R (t) = E x (t) \Sigma E y (t \Gamma t 1=4
) (2)
where t 1=4
j ú=2! 0 is a delay equal to one quarter cycle at the center frequency ! 0 . The
IF signals are (with implicit filtering off of the upper sideband)
L IF ; R IF = E L;R (t) cos(!LO1 t) (3)
and the baseband voltages are (again with implicit filtering)
`(t); r(t) = L IF ; R IF (t) cos(!LO2 t): (4)
The autocorrelation functions (acfs) of the baseband fields are
h`(t)`(t + Ü )i = 1
2
cos(ffi!Ü )[c 2 + s 2 + 2cs cos(!t 1=4
)] = 1
2
cos(ffi!Ü ) (5)
hr(t)r(t + Ü )i = 1
2 cos(ffi!Ü )[c 2 + s 2
\Gamma 2cs cos(!t 1=4
)] = 1
2 cos(ffi!Ü ) (6)
from which two of the Stokes parameter correlations may be defined:
I(Ü ) = h`(t)`(t + Ü )i + hr(t)r(t + Ü )i = cos(ffi!Ü )
V (Ü ) = h`(t)`(t + Ü )i \Gamma hr(t)r(t + Ü )i = 0:
(7)
The cross correlation function (ccf) between ` and r yields L(Ü ) j Q(Ü ) + U(Ü ):
L(Ü ) = 2h`(t)r(t + Ü )i = [(c 2
\Gamma s 2 ) cos(ffi!Ü ) \Gamma 2cs sin(ffi!Ü ) sin(!t 1=4
)];
= cos 2ü cos ffi!Ü \Gamma sin 2ü sin ffi!Ü
= cos(ffi!Ü + 2ü)
(8)
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where ffi! j ! \Gamma !LO1 \Gamma !LO2 is the baseband frequency of the monochromatic signal.
Lines 1­2 of Eq. (8) explicitly show symmetric and antisymmetric parts of L(Ü ). When
transformed to the frequency domain, they correspond to Q(~!) and U(~!), respectively.
The position angle is found in the usual way to be
ü ~
! = \Gamma
`
1
2
'
tan \Gamma1 U(~!)
Q(~!):
(9)
Linearly Polarized Continuum Signal
Now consider a continuum signal, again with 100% polarization, analyzed in a total band­
width B. We assume there is no Faraday rotation across the bandwidth (see below). Stokes
parameters may be found by integrating the monochromatic result over frequency because
the SP's are variance­like quantities and the frequency components are statistically inde­
pendent (variances add). Performing integrals like
I(Ü ) =
Z B
0
dffi!I m (Ü )
L(Ü ) =
Z B
0
dffi!L m (Ü );
(10)
where I m ; Lm are the monochromatic results from Eq. (7)­(8) [that depend on the fre­
quency ffi!], we find that
I(Ü ) = sin BÜ
Ü
L(Ü ) = cos 2ü
`
sin BÜ
Ü
'
+ sin 2ü
`
cos BÜ \Gamma 1
Ü
'
V (Ü ) = 0:
(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 we have defined them in Eq. (5) ­ (8).
Elliptically Polarized Signals
A monochromatic signal with arbitrary elliptical polarization is handled as above. General
expressions that include differential timing delays, Faraday rotation, and LO phase offsets
may be found in Eq. (C3)­(C4) in my 1988 memo, Polarimetry with the 40 MHz Correlator.
Similarly, noise­like signals with arbitrary polarization may be analyzed by integration of
the monochromatic results, as we did above.
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Single Channel Polarimetry
By single channel polarimetry, I mean a system where only a single lag of the auto­and­
cross correlations is computed. This might be effected with analog multipliers rather than
a digital correlator. In this case, one must explicitly calculate the ccf between the RHCP
and LHCP components with a 90 ffi 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)
2hER (t)E L (t + Ü )i = cos 2ü cos !Ü + sin 2ü sin !Ü (12)
2hER (t)E L (t + Ü \Gamma t 1=4 )i = cos 2ü sin !Ü \Gamma sin 2ü cos !Ü: (13)
At zero lag (Ü = 0), the unshifted correlation (Eq. 12) gives cos 2ü while the shifted
correlation (Eq. 13) gives sin 2ü; both cos 2ü and sin 2ü are needed to solve for the position
angle without ambiguity.
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