Dear Janesh,
Re:
> Dear Franklin Bretschneider,
>
> Thank you so much for your reply and explanation about the filter using the
> stats and signal package.
>
> I decided to opt the filter method in signal package. I have a simple
> question about the cut off frequency here.
>
> I have 30 minute collected tidal data and I want to use the 48 hour low pass
> filter to my data to remove the fluctuations and then get only the residuals.
> What should be the cutoff frequency in my case ? I have tried to figure out
> cut off frequency with the following rationale : .
>
> The parameters for butter filter are n, Wn and type. In the help, W is
> defined as critical frequencies of the filter. W must be a scalar for
> low-pass and high-pass filters, and W must be a two-element vector c(low,
> high) specifying the lower and upper bands. For digital filters, W must be
> between 0 and 1 where 1 is the Nyquist frequency.
>
> A value of 1 corresponds to half the sampling frequency. In my case the
> sampling frequency is 2 hr^-1. Hence a value of 0.01 corresponds to a
> frequency cutoff of .01*1 = .01 hr^-1 or 100 hrs time. Using unitary method,
> if 100 hours cut off frequency is 0.01 then 48 hours cut off frequency is
> 0.01/100*48 = 0.0048 hr^-1 . Is that correct ?
>
> Thank you so much
>
> Janesh
>
With digital filters (called "z-plane" in the signal package), the cut-off
frequency must indeed be given as a fraction of the Nyquist freqency.
With your tidal data, taken at 2 samples per hour, the nyquist is 1 per hour.
So, a cutoff interval of 48 h means a cutoff frequncy of 1/48 f(nyq) , or
0.020833. This must be fed into the butterworth function.
Note btw that at the cut-off frequency, the amplitude is still rather high
(about 0.707), so the tidal signal (period about 12.5 h) will not be attenuated
much.
I hope this helps. It's always best to check with a simple example, such as an
f= 1/48 sine, which after your filter should be reduced to an amplitude of
0.707). Then check with a simulated tidal signal (say, a sine of about 12:25 h
period).
With such simple tests, I often find my own programming errors.
Best wishes,
Franklin
--
Franklin Bretschneider
--
Dept Biology
Kruyt Building W711
Padualaan 8
3584 CH Utrecht
The Netherlands
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