I'm trying to determine the open-loop transfer function of the system below (in order to determine PID gains).

enter image description here

If I account for the disturbance \$(9.81*6700)\$ in the system and use equation 2, I get very different results when I compare a simulation of the simplified function to a simulation of the full network.

If I disregard the disturbances as zero and use equation 3, and I also disconnect the disturbances in the simulation, the results match.

How should I account for the disturbances when determining the transfer function? And why is the method I'm using incorrect?

Thanks in advance.


It is the purpose of such a (complicated) feedback arrangement to process the input signal in a certain manner ("Follower" control, mostly specified in the time domain) as well as to reject any disturbances as much as possible.

This leads to TWO different transfer functions which govern the design of the whole system: (a) Signal Transfer Function (disturbances Vd=0) Hs(s)=Vout(s)/Vin(s) and (b) Disturbance Transfer Function (Vin=0) Vout(s)/Vd(s) (corrected).

That means: You cannot define a transfer function that considers both inputs (signal and disturbance) at the same time.

Comment 1: In your first sentence you speak about the "open-loop" transfer function. Don`t forget that - finally - it is the closed-loop transfer functions that matters for both cases.

Comment 2: Of course, you can determine the combined output signal in presence of both sigbnal and disturbance.

  • \$\begingroup\$ I'm a bit surprised and curious at the fact the disturbance transfer function is not Vout/Vd: isn't it supposed to characterise the effect of disturbance on output? \$\endgroup\$ Dec 13 '14 at 17:03
  • \$\begingroup\$ Thank you for pointing to the typo: Of course it must be Vout/Vd. \$\endgroup\$
    – LvW
    Dec 14 '14 at 10:17

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