Current Distribution

[tennant]

Hello,

Was reading Michael's PRST-AB on modeling microbunch instability and see lots of figures showing the current distribution (current vs time). Is this column generated in a particular output file or did you compute this using sdds tools?

Thanks,
chris

[michael_borland]

Chris,

I think you are referring to the output from CSRCSBEND. Use the OUTPUT_FILE parameter to specify a filename, and you should see what you are looking for.

--Michael

[tennant]

Michael,

Okay, thanks - it's the linear density. I see it now.

As I was reading through your paper I'm able to follow until you describe the numerous steps you take to obtain the gain curve (p. 5). In particular, I am not familiar with NAFF. Any suggestions/help/examples you could provide to understand better what is involved with this analysis?

Thanks,
chris

[tennant]

I should've looked at the SDDS Toolkit manual first! I see a sddsnaff now. I don't suppose you have an example for all the steps you discuss in your paper (e.g. make histogram, remove ends, polynomial fit, naff, etc...)?

[michael_borland]

Chris,

I don't have the link for the original paper, but this paper describes the algorithm (see page 5) and gives references
cdsweb.cern.ch/record/301630/files/p183.pdf

The attached script is the one I used to perform the analysis. It expects a set of histogram files (produced with the HISTOGRAM element) with the run parameters encoded in the file rootname. See the comments for explanations.

--Michael

[headdoggy64]

Hello,

Just wondering if this procedure is valid.

1)

Code: Select all

 elegant2genesis bunch.out test.out -slices=200 -totalCharge=20e-12 

2)

Code: Select all

 sddsfft test.out test.tmp -columns=s,current 

to get FFT of current
3)

Code: Select all

 sddsfft test.out test2.tmp -columns=s,gamma 

to obtain FFT of modulations in energy

Also, why does the current after sddsfft still have the units of Amp?

Thanks,
Alex.

[michael_borland]

Alex,

What you are doing is valid, at least mathematically, since the spacing of the s points is uniform.

I'm not sure why it is surprising that the FFT of I(s) has units of Amps. Unlike a continuous Fourier transform, the discrete Fourier transform is just a sum of the samples times some exponentials. There's no change of units.

--Michael