InstrumentalWidthOfEIS
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I'm getting first level data from zero level after usual reduction by 'eis_prep'. However, when i am fitting the lines by xcfit, i'm getting the spectrum slightly displaced from the wavelength values given in the data window. What error this will make in the calculations?. Secondly, can we get the accurate width variation ? , and can we use these datas for line-width variation studies ?

--A.K. Srivastava, 02-July-2007


In response to your first query - do you get the same shift (and what is it) if you got with a simple gauss_fit?

In response to your second query we would need more information. Various work has been carried on line widths looking at the differences at different stages or positions of loops by Hara et al., and Doschek et al (both been submitted), so clearly line widths can be measured.What variation are you referring to?

--Hiro Hara, 09-Jul-2007


About the instrumental width, Doschek et al 2007 proposed a FWHM of 2.5 pixels (i.e. 56 mA) in orbit given that the FWHM is 1.956 pixels in laboratory and Brown et al. 2008 give us between 54mA and 57mA depending on the wave band and the wavelength.

Is there any instrumental profile available?

Is it a Gaussian profile? What is the width commonly used ?

--Celine Boutry, 16-Jan-2009


Hi Céline,

I think the answer may be in your question, actually.

The Doschek et al. article states a single number, but that is based on comparisons with older data. However, it is still consistent with the numbers derived from comparing the pre-launch laboratory EIS calibration data with the on-orbit EIS data by Brown et al. (2008). The Brown et al. numbers are inferred widths, but the reasoning is pretty logical; the fact that the Doschek et al. (2007) number falls within that range of 0.054nbsp;— 0.057nbsp;Å is comforting.

In summary, Brown et al. (2008) assume a Gaussian instrumental line profile, and they infer an instrumental width of 0.054 Å in the short-wavelength channel (170 — 210 Å) and 0.057 Å for the long-wavelength channel (250 — 290 Å).

NOTE! However, the instrumental width that is discussed by Brown et al. is not the FWHM of the instrumental width, but rather the (1/e)-1/2 half-width σ that naturally falls out of the Gaussian function:

G(x) = exp ( -x2 / 2 (σ2) )

The Gaussian width σ is related to the FWHM by the relation:

FWHM = 2 × (2 ln(2) )1/2 × σ

i.e.,

FWHM = 2.356 σ I hope this helps. And please let me know if I've slipped up in my maths!