Re: plotting transfer function in octave 5.2: How to fix error: set: "dataaspectratio' must be finite

[Sergei Steshenko]

On 07/08/2020 18:31, shall689 wrote:
> Hello Torsten,
>
> > If you consider the complete control loop in discrete time (the system
> > has to be discretized), then you do not need any transfer function for
> > your hold device. Instead, it is considered in the calculation of the
> > discrete time system representation. If, e.g., the used D/A has a zero-
> > order hold, than the discretization method makes use of the fact that
> > the input u is constant during a sampling interval.
> Yes, if I use the ZOH method when discretizing, the ZOH will be included.
>
> What if there are two feedbacks, i.e a voltage and a current feedback (see
> diagram below)?  Should the ZOH  be placed in both loops as shown in the
> second diagram?  My system is sampling all feedbacks at the same time, so
> there is essentially only one delay; however, I think two zero order holds
> would add two delays.
>
> Stephen
>
> <https://octave.1599824.n4.nabble.com/file/t372348/TwoLoopControlDiagram.jpg>
>
> <https://octave.1599824.n4.nabble.com/file/t372348/TwoLoopControlDiagramWithZOH.jpg>
>
>
>
> --
> Sent from: https://octave.1599824.n4.nabble.com/Octave-General-f1599825.html
"I think two zero order holds would add two delays" - before deciding on 
adding delays think about the following. One delay stage adds 2 * pi * 
delay / frequency phase shift, where 'frequency' is the frequency at 
which you want to know phase shift.

In a control system magnitude and phase plots the most important area is 
the area in which magnitude response is about 0db - this is where one 
looks at phase margin. Let's call this frequency frequency_0db. So, if 
frequency_0db is much less than sample rate, then one sample delay 
introduces negligible phase shift.

--Sergei.