MATLAB: Step response settling time

Control System Toolboxsettling time

Hi,
The settling is the time required for the response curve to reach and stay within a range of 2% of the final value.
So why in Matalb the two following systems have the same stelling time ? >> Tz = 1
>> Tp = 3
>> SO1 = tf([ 1],[Tp 1])
>> SO2 = tf([Tz 1],[Tp 1])
>> InfoSO1 = stepinfo(SO1);
>> InfoSO2 = stepinfo(SO2);
Theorically (for me) these settling times should be :
tr1 = -Tp*log(0.02)
tr2 = -Tp*log(0.02/(1+Tz/Tp))
Thank's for any answer.
V.

Best Answer

If you look at the doc page for stepinfo , you will see that settling time is defined as:
The response has settled when the error y(t) - yfinal becomes smaller than a fraction ST of its peak value. The default value is ST=0.02 (2%).
Changing the value of zero changes the initial output value for the step response, but does not change the time constant of the first order system. stepinfo computes the settling time as a time when the difference between step response and final value (1) is 2% of the difference between initial value and final value.
So that's why the zero does not change the settling time.
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