Re: spectral Analysis

Thanks to Maria for valuable information. I wanted to filter my data above
1 hour (60 min) and the data is in the 4 minute time step. For that i tried
to 60 min low pass the data using Lanczos funtion as follows.

*****************************************************************************************
  ihp = 0 ; Low pass
  sigma = 1.0 ; Lanczos sigma

  nWgt = 1201
 * fca = 1./(4*60) * ; start freq
  fcb = -999 ; last freq
  wgt = filwgts_lanczos (nWgt, ihp, fca, fcb, sigma)
  xBPF = wgt_runave_Wrap ( x, wgt, 0 )
***************************************************************************************
 Is fca assignment meets my requirement otherwise please let me know the
right calculation.

Thanks in advance


On Fri, Jul 18, 2014 at 10:55 PM, Maria Gehne - NOAA Affiliate <
maria.gehne@noaa.gov> wrote:

> Hi Paul,
>
> when you do a spectral analysis the lowest frequency you can resolve is
> 1/N if N is the length of your time series. The highest frequency is
> 0.5/10mins in your case, so half a cycle in 10 minutes. Or in general half
> a cycle in the time interval. The frequency units will be 1cycle/time unit
> so in your case cycles/10min. To get to cycles/min divide the frequency
> vector by 10.
>
> It helps if you know certain signals in your data and what frequency they
> should be at. That's a good sanity check when computing spectra to make
> sure the frequency units are correct.
>
> Hope this helps,
> Maria
>
>
> On Fri, Jul 18, 2014 at 3:58 AM, Steeven Paul Yerraguntla <
> steevenpaul@gmail.com> wrote:
>
>> Dear NCL users,
>> I have plotted spectral variance to my data using Data
>> Analysis examples in the NCL website. The data i have used is of 10 min
>> interval and my doubt is that whether spectral frequency would be
>> cycles/min or cycles/10min. If it is not cycles/min then how can i get
>> the spectral Frequency in cycles/min. Hope i will get the answer soon.
>> Thanks in advance.
>>
>> regards,
>> Paul.
>>
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>

Received on Sat Jul 19 12:15:27 2014