'> Date: Fri, 17 Oct 2008 17:05:25 -0400
'> From: Fred Schwab <email@example.com>
'> To: firstname.lastname@example.org
'> Subject: [Alma-feic] Holography horn scans 122 and 123
'> Hi All,
'> Josh sent me the data for Scans 122 and 123 earlier this week, and I've had a look at these data now in Mathematica, doing my own transform to far-field. In each case I formed the linear combination (Beam1 - i Beam2)/2. On my Web page, http://www.cv.nrao.edu/~fschwab, I've put a link (the second bulleted item) to a transcript of the Mathematica run. First I show the results for the co-polar (Scan 122) beam combination, then the cross-polar (Scan123) one. The plots and textual content should be pretty much self-explanatory. I get a ratio of integrated cross-polar to co-polar ff power which is equal to ~0.00306, i.e., about -25.1 dB.
'> These are 101 x 101 scans. In the computed far-field phase plots there appear to be significant aliasing artifacts, so I wonder if finer sampling ought to have been used.
'> I believe that these results agree qualitatively with those given by Geoff and Josh in their memo dated 2008-10-09. (I believe they were looking only at single beam, rather than the Beam1 and Beam2 combination that I used.) In the final plot of my printout I compare a 45-deg cross-polar cut with Sri's Green Bank data, and I see only about the same level of agreement that Geoff and Josh showed.
'> I wasn't entirely sure what beam combination to use. In the header portion of the near-field data files it says that the z-offsets for Beam1 and Beam2 are 0.798 mm and 0 mm, respectively. This is the opposite of the case for Band 9 Scans, where Beam2 was the one offset by 1/4 wavelength.
NSI suggested we put the offsets the other way around, as that is what their software expects. This was an attempt to solve the 1/4 wavelength problem with their software, it did not help. We can do either way and should agree on which way around (or do both + and - 90 degrees, one gives the combined beam and the other shows the reflections).
'> Also, I'm puzzled that the value 0.798 mm is one-quarter wavelength at approximately 93.9 GHz. Why does that value appear rather than a number like 0.7205 mm, which corresponds to 1/4 wavelength at the
holography frequency of 104.02 GHz?
Sorry, this seems to be my mistake, I put in the wrong spacing. We will have scripts in the future to insure that this does not happen again. Josh only analyzed one beam.
> - Fred Schwab
- 28 Oct 2008