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From: | dastew |
Subject: | RE: Time delayed transfer function |
Date: | Sun, 11 Oct 2009 11:44:41 +0000 |
As far as I know, in octave, you must have the same sample rate to work in 'Z" space. If you were to write up a simulator in C++ for your system, you will find that If your feedback is sampling faster than the main part, then you would just use every second or third sample (depending on the sample rates). If it was slower then you would use the same sample twice etc. If you are working at a graduate level and are developing some math to handle variable sample rate math , then let me know how it works. Doug Date: Sun, 11 Oct 2009 12:35:47 +0200 Subject: Re: Time delayed transfer function From: address@hidden To: address@hidden CC: address@hidden Hi Doug,
Forward path transfer function (open loop) looks as follow: G(s)=210/(s+0.003899*s^2+0.0000023*s^3). In hardware realization it samples at 20kHz. The feed-back sensor is CCD with variable sample rate - from 300Hz to 2kHz. I assume that feedback path gain is 1 so transfer function of close loop will be: G(cl)=G(s)/(1+G(s)*exp^-Ts). Actually what I need is apply sinusoid signal by G(cl) and compare input and output at different values of "T" - the result will be tracking error of my system. The problem is if I trying different discretization for feed-forward and feed-back paths the "feedback" function return error. Thanks, Yuri. On Sat, Oct 10, 2009 at 11:08 PM, <address@hidden> wrote:
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