Re: octave-control: stepinfo

[Doug Stewart]

On Mon, Feb 9, 2015 at 12:30 AM, Aidan Macdonald <address@hidden> wrote:

On Sun, Feb 8, 2015 at 7:59 PM, Doug Stewart <address@hidden> wrote:

On Sun, Feb 8, 2015 at 10:30 PM, Doug Stewart <address@hidden> wrote:

On Sun, Feb 8, 2015 at 10:29 PM, Aidan Macdonald <address@hidden> wrote:

On Sun, Feb 8, 2015 at 6:54 PM, Doug Stewart <address@hidden> wrote:

On Sun, Feb 8, 2015 at 9:48 PM, Aidan Macdonald <address@hidden> wrote:

We always bottom post on this list

 

Aidan Plenert Macdonald

4178 Decoro Street, San Diego, CA 92122

Cell: 805 418 0174

On Sun, Feb 8, 2015 at 6:31 PM, Doug Stewart <address@hidden> wrote:

On Sun, Feb 8, 2015 at 8:41 PM, Aidan Macdonald <address@hidden> wrote:

Hi,

I wrote some code for stepinfo in the octave-control package. I tried to match the guidelines found here. 

I attached the code. It doesn't match the Matlab standard perfectly, but I was wondering how much I should do before submitting it. I wrote it to take in one system and give the results specified in http://www.mathworks.com/help/control/ref/stepinfo.html.

I was also a little concerned with the math I did. I couldn't find a better explanation of the various parameters than Matlab's website specified. That said, I wrote it and compared it to a few examples using my universities Matlab and they mostly match.

Let me know what I should do to improve it. I am very new to the open source contributing scene, let me know what I should do to help.

Enjoy,

Aidan Plenert Macdonald

4178 Decoro Street, San Diego, CA 92122

Cell: 805 418 0174

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I just did a quick read through and I see that for slope you did time over y_yalue

I am sure that it should be the other way   rise over run

I will do more checking later

Doug

--

Thanks for looking. No, I think the slope is right. I am using is to calculate the true times below given the error in the difference from the ideal 90% and 10% voltage. The slope is in units seconds/volts. With an error from the ideal in volts (the error come from discretization of the signal), I linearly approximate the true value.

--

OK you are using a different definition than calculus. 

Like I said I will do more checking.

DAS

Sorry, Does this look better? I am using Gmail

Aidan Plenert Macdonald

4178 Decoro Street, San Diego, CA 92122

Cell: 805 418 0174

Yes very good :-)

Your calculation of  Settling Time is my next concern.

There are many definitions for settling time and I was wondering what you are using?

I see the ST=.02  but then you multiplied it to ypeak.

and you have yavg of the tail etc.

What are you really doing?

Doug

Perhaps you could point me to a better definition, but I was using what Matlab said here that the response has settled when the error abs(yfinal - y(t)) has reached ST times the peak value. So I use convolution to average the y values over a span. I picked an arbitrary averaging width, perhaps you know a better one.

I then take the use average in place of y(t), which I argue in the case of oscillating signals is required. I attempt to minimize the absolute value of the error minus the ST*ypeak value. The second abs(...) is so the min function isn't taking negative error differences.

I take the minimum value of the error - ST*ypeak to be the closest point to settled, and use that corresponding time value.

I looked around, including 2 books I have for more precise definition in terms of the step functions response. Wikipedia, from what I saw, only had it is a function of variables in the system function, but I want a numerical definition.

Thanks.

Aidan Plenert Macdonald

4178 Decoro Street, San Diego, CA 92122

Cell: 805 418 0174

I think the 2% settling time is defined to be :

The last time the signal was outside the 2% range of the final value.

So I don't agree with matlab's definition of being outside of 2% of the peek.

Now as to your software even if we stay with matlab's definition, your software does not get the same results as Matlab's example.

from matlab's example:

sys = tf([1 5],[1 2 5 7 2]);
S = stepinfo(sys,'RiseTimeLimits',[0.05,0.95])

 S =

        RiseTime: 7.4454
    SettlingTime: 13.9378
     SettlingMin: 2.3737
     SettlingMax: 2.5201
       Overshoot: 0.8032
      Undershoot: 0
            Peak: 2.5201
        PeakTime: 15.1869

from your software:

>> sys = tf([1 5],[1 2 5 7 2]);
>> S = stepinfo(sys,'RiseTimeLimits',[0.05,0.95])
yfinal =  2.5002
ysettle =  0.47566
settling_ind =  79
S =

  scalar structure containing the fields:

    RiseTime =  5.1696
    SettlingTime =  39.866
    SettlingMin =  2.1908
    SettlingMax =  2.5192
    Overshoot =  0.0076091
    Undershoot =  0.012105
    Peak =  2.5192
    PeakTime =  15.381

matlab has a settling time of 13.9 sec.

and your software has 39.8 sec.

I have taught this method for second order systems:

d=poly([ -1-4j, -1+4j])

n=[17]

sys1=tf(n,d)

a=roots(d)

ar=real(a(1))

[Y, T, X] =step(sys1);

fv=Y(end)

gg=fv*(1-exp(ar .*T));

plot(T,gg)

hold on

step(sys1)

From the graph you can see that the settling time of the first order curve gives a good smooth function, that can be used as the settling time for the second order system. We then went with 4 or 5  time constants of the first order curve as or settling time.

The definitions given at the start of my discussion both give  very discontinuous results.

That is --- a small change in gain can give a jump in the settling time, because the last place that it was outside the 2% will be a different minor peak . 

I know that there are many definitions of settling time and that for this software we want a numerical approach.

So my suggestion is to do  something like Matlab but do it to 2% of the final value not the peak value .

Lukas Reichlin is the maintainer of the control pkg. so he has the final say as to which way to go.

Doug