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Re: derivative of an input signal

From: Przemek Klosowski
Subject: Re: derivative of an input signal
Date: Tue, 7 May 2013 11:06:16 -0400
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:17.0) Gecko/20130402 Thunderbird/17.0.5

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() i didint get the required

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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