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Re: derivative of an input signal


From: Sergei Steshenko
Subject: Re: derivative of an input signal
Date: Tue, 7 May 2013 09:39:09 -0700 (PDT)


--- On Tue, 5/7/13, Przemek Klosowski <address@hidden> wrote:

> From: Przemek Klosowski <address@hidden>
> Subject: Re: derivative of an input signal
> To: address@hidden
> Date: Tuesday, May 7, 2013, 8:06 AM
> On 05/07/2013 08:41 AM, primi wrote:
> > i hav a low pass filter output..i need to find the
> derivative of it, becoz
> > the derivative converts
> > the slowly changing output of the LPF into a more
> defined and faster
> > changing waveform.
> > 
> > some one please help me.iam new to octave.
> > 
> > i tried diff(), gradient() and polyder()..bt i didint
> get the required
> > output..
> 
> Please provide more information about your problem: what
> data do you have initially, and why are the results you get
> not satisfactory?
> 
> Basically, if you have a vector X with evenly spaced samples
> of some signal x(t) for t=t0+dt*i where i=1...N, then
> diff(X)/dt will be an approximation to the time derivative
> of that signal. How good is this approximation is determined
> by how fast the signal changes with respect to the time
> interval between the samples. You say it's slow, so it
> should be OK.
> 
> If your samples are not evenly spaced in time and the time
> interval values are in vector dt (which then would be only
> 1..N-1 long), you can do diff(X)./dt. Again, it depends on
> what your input data is.
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> 

Derivative can also be trivially calculated with FFT.

Regards,
  Sergei.


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