Closed claudiavr closed 1 year ago
@claudiavr , please check if this equation is correct.
$SNR{total} = \sqrt{\sum{(\frac{Zi}{\sigma{Zi}})^2}}$ $\sigma{theor} = 100\% \cdot K \cdot \frac{1}{SNR{total}}$
In principle, I think the $Zi$ terms carry the photometry SNR and, summing $N{WP}$ terms with same SNR we get $SNR{total} = \sqrt{N_{WP}} \cdot SNR$.
The sigma_theor expression seems correct for me.
The sum expression for the SNR_total is correct, but I think the "i" parcels are not correct, if Z_i = (f_ord - f_ext)/(f_ord + f_ext), where:
f_ord is the flux of the ordinary beam (already subtracted from the sky) in the "i" image; f_ext is the flux of the extraordinary beam (already subtracted from the sky) in the "i" image.
Each parcel should be:
parcel_i = (f_ord + f_ext)/{f_ord + f_ext+ 2 npix[f_sky_ord + f_sky_ext + 2 B N_R^2 + 4(0.289* G)^2]},
where
f_sky_x is the sky counts per pixel in the "i" image in ord or ext beams.
Reopening to fix the expression.
Sugiro utilizarmos a expressao do erro teorico da polarizacao do link abaixo. Ela inclui o ruido de Poisson das contagens da fonte e ceu, ruido de leitura e ganho. Essa expressao deve ser adaptada para uso em medidas do seguinte modo:
http://www.inpe.br/etc/docs/calculo_erro_polarizacao.pdf