stepinfo

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ahmed salama
ahmed salama on 14 Jun 2011
why stepinfo function don't give me any numerical results for
>> q=tf([1 2 ],[ 1 2 3 4 5 7]); >> stepinfo(q)
ans =
RiseTime: NaN
SettlingTime: NaN
SettlingMin: NaN
SettlingMax: NaN
Overshoot: NaN
Undershoot: NaN
Peak: Inf
PeakTime: Inf

Accepted Answer

Walter Roberson
Walter Roberson on 14 Jun 2011
1 and 2 are roots of the numerator and are also roots of the denominator. Therefore at 1 and 2, the value of the function is not simply one of the infinities but is instead 0/0 = NaN . Transfer functions that have point singularities like this do not have defined values for the parameters you are looking for.
  2 Comments
ahmed salama
ahmed salama on 14 Jun 2011
i don't understand u ... see this
s=tf([1000],[1 3 2 6592 36354 119894 230060 241392 120960])
S = stepinfo(s,'RiseTimeLimits',[0.05,0.95])
S =
RiseTime: NaN
SettlingTime: NaN
SettlingMin: NaN
SettlingMax: NaN
Overshoot: NaN
Undershoot: NaN
Peak: Inf
PeakTime: Inf
it is obvious that this results appears when the denominator is bigger than the forth order
please answer....
Walter Roberson
Walter Roberson on 14 Jun 2011
Sorry, I do not have that toolbox available to test.
The documentation shows a 5th order example,
http://www.mathworks.com/help/toolbox/control/ref/stepinfo.html

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