I am seeking for simple Identification method for the second-order systems plus time delay.

The system is QNET vertical take-off and landing (VTOL) trainer. Using the tutorial I found the system parameters.

enter image description here

The only thing I need to know is how to measure is the time delay ( the Transport Delay in the figure )

Simulation step response without time delay

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Real system Time domain response

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the question is : How I could find the dead time from the experiment data (process transient) ?

  • \$\begingroup\$ Are you asking for group delay? This the negative slope of the phase response (vs. frequency). \$\endgroup\$ – LvW Dec 18 '15 at 15:23
  • \$\begingroup\$ It its the first time I hear of group delay . the time delay(Td) I am asking about have the form (exp(-s*Td)) in the Laplace domain. \$\endgroup\$ – John Alawi Dec 18 '15 at 16:01
  • \$\begingroup\$ A bit much to explain here, but you can easily find a number of papers on system identification of second order plus dead time (aka delay) systems. \$\endgroup\$ – Spehro Pefhany Dec 18 '15 at 16:02
  • \$\begingroup\$ @SpehroPefhany Is their a simple graphical method to identify the dead time ? \$\endgroup\$ – John Alawi Dec 18 '15 at 16:25
  • \$\begingroup\$ If you hit the input with a step and extrapolate the output from the time you see significant signal back to zero with a least squares fit to a polynomial of modest order you should get the dead time. Whether that's accurate enough for you will depend on a lot of factors such as noise. \$\endgroup\$ – Spehro Pefhany Dec 18 '15 at 16:37

Delay time is very ambiguous - it kind of implies that a filter will delay all signals (at any frequency) the same amount and this isn't true. Consider the following picture: -

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The two graphs top and bottom to the left show the (frequency response) bode plots of a 2nd order low pass filter with various damping ratios. They do not directly give an indication (to the untrained eye) of how the filter (or system) might respond to a step input.

The step-input-response is shown on the right. Clearly, for different values of zeta (damping ratio) the "response" is very different. Take the example when \$\zeta\$ = 2. The output gradually rises towards 1 - clearly there is some form of delay going on but how much is it or, put another way, what critera do we impose that gives us a time delay value?

Maybe we say the delay is the time taken to reach the 90% level but, this can't apply to the scenarios when \$\zeta\$ is significantly less than 1 because of overshoot and undershoot.

I go along with @LvW and think you should consider talking about group delay.

However, this isn't a simple matter either. Consider a bessel filter and a butterworth filter and how group delay looks: -

enter image description here

The bessel filter is well-known for having a really flat group delay but its filtering characteristics are not as good as a butterworth filter which, as you can see has quite a poor group delay. Basically, what I'm labouring to say is that any system or filter does not have a fixed delay.

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  • \$\begingroup\$ I will rewrite my question to make it clear. \$\endgroup\$ – John Alawi Dec 18 '15 at 18:26

You should be more specific because the properties of a system in the time domain are given either as

  • Group delay (negative slope of the phase vs. frequency function),

or in form of the step response with different specifications:

  • Delay time

  • Rise time

  • peak time

  • Settling time

EDIT: In your comment, the mentioned expression exp(-sTd) is not a delay term but a phase angle as a function of a fixed delay time. The corresponding phase is phi(w)=-w*Td (given in rad).

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    \$\begingroup\$ \$e^{-sT_d}\$ is the Laplace transform for a pure time delay of \$T_d\$ sec \$\endgroup\$ – Chu Dec 18 '15 at 17:23
  • \$\begingroup\$ @Chu , yep that what i mean a pure time delay \$\endgroup\$ – John Alawi Dec 18 '15 at 18:24
  • \$\begingroup\$ @Yusuf, have you looked at Pade approximant? If you can extract the dominant 2nd order TF, the remaining roots can be lumped into one pure time delay term. \$\endgroup\$ – Chu Dec 18 '15 at 18:34
  • \$\begingroup\$ @Yusuf, see, e.g., home.hit.no/~hansha/documents/control/theory/… \$\endgroup\$ – Chu Dec 18 '15 at 18:41
  • \$\begingroup\$ @Chu thanks, but i don't know how to calculate the delay time, all the tutorials in the web is about the FOSDT . \$\endgroup\$ – John Alawi Dec 18 '15 at 18:50

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