[SciPy-user] signal.lti

Ryan Krauss ryanlists at gmail.com
Mon Sep 11 17:51:52 CDT 2006


I thought about this over the weekend.  I was wrong about the matrices
not being correct.  The A matrix really is just 0.  So, passing in the
numerator and denominator polynomials does correctly create a system,
you just can't take its step response.

I will play with other systems for now.

Is anyone maintaining this code?  I am going to be using it more the
systems modeling portion of my mechatronics class and I am trying to
convert students from Matlab.  I will also be teaching Dynamic System
Modeling in the spring which is basically transfer functions and
impulse and step responses.  I would be willing to invest some time in
this if no one is doing it.  But if someone already knows the code and
can easily modify it, that would be great.

Ryan

On 9/8/06, Travis Oliphant <oliphant at ee.byu.edu> wrote:
> Ryan Krauss wrote:
>
> >I am having a rough time trying to use the lti portion of
> >scipy.signal.  For starters, I just wanted to generate a step response
> >for an integrator (1/s).
> >
> It turns out your starting-point is a special-case not handled by the
> current code (I checked this on old SciPy and it raised errors as
> well).   The current code is "general-purpose" but does not handle
> lonely integrators.  Thus, any time your denominator ends in a zero, you
> will have trouble.  This case needs to be dealt with separately but so
> far isn't (the error could be more informative of course).  You would be
> better of trying a simple first-order system as your starting point.
>
> 1/(s+1)
>
> Which works fine for me with current SciPy.
>
> from pylab import *
> from scipy import *
>
> plot(*signal.step((1,[1,1])))
>
>
>
>
>
> -Travis
>
>
>
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