Optimization Of The Pi Controller According To The Symmetrical Optimum; Setpoint Smoothing - Siemens SINAMICS DCM 6RA80 Faq

Optimization of the PI controller according to the symmetrical optimum

If a controlled system contains not only 1st-order lags, proportional-action elements and
dead-time elements, but also elements with an integral character, optimization must be
performed by a method other than the absolute value optimum to avoid controller oscillation
due to series connection of the integral controller and the integral of the controlled system. A
differing optimization rule must therefore be found for the reset time Tn. This is achieved by
adjustment according to the symmetrical optimum.
If the controlled system contains an integral-action element and also a number of first-order
lag element whose time constants can be grouped together into a total time constant , you
can provide the controller with a PI action.
The frequency response of the open loop F
The control parameters are set as follows:
Kp = T / (2*Vs* ); Tn = 4* ;
0
Vs is the controlled-system gain and
T is the integral-action time of the integral-action element.
0
The rise time t is 3.1* ; the settling time t
rise
2%. This also applies of the controlled system contains an integral.
If, among the small lags, there is one with the constant T2 that exceeds the other, this must
be compensated for with a derivative-action element, i.e. a PID controller must be used.
The frequency response of the open loop is therefore:
F (p) = Kp *Vs * (1+pTn) *(1+ pTv) / [(pTn)*(1+ pT
0
If T = Tv, the following results:
2
F (p) = Kp * Vs * (1 + pTn) / [(pTn) *( pT
0
resulting in the following for the optimization: Kp = T
If the controlled system consists of first-order lags among which one T
times greater than the sum of the others, the effect of this variable is approximately integral,
which should result in optimization according to the symmetrical optimum in this case, too.
The set values are therefore: Kp = T
If the controlled system contains two especially large lags, a PID controller must be used with
an integral-action time corresponding to the smaller of the two.

Setpoint smoothing

Overshoots of 43.4% are only permitted in very rare cases. Measures must be taken to
reduce these.
Setpoint smoothing with t
the suitable results, if Ti >=1
The rise time is then 7.6 , the settling time with reference to a tolerance band of
+/- 2% is 13.6 , and the overshoot is 8.1%, with reference to a change to the settled state of
100%.
If the overshoots of the controlled variable are to disappear completely on a step change in
the setpoint channel, stronger setpoint smoothing with t
the symmetrically optimized control loop, that only contains first-order lags in the controlled
system.
If you want to avoid the disadvantage of a slower control due to smoothing of the setpoint
and still achieve a steeper rise, the smoothing element must be bypassed in the setpoint
channel by a derivate-action element t
(p) = Kp * Vs * (1 + Tn) / [(pTn) (pT
0
the sum of the small time constants.
settle
) * (1 + p )], the same expression as above,
0
/ (2*Vs* ); Tn = 4 .
1
= 4 , i.e. with the magnitude of the Tn of the controller provides
ss
= 4 .
g1
65
is 16.5* ; the overshoot is 43.3% for a x of
)* (1+p )].
2
/ (2*Vs* ) and Tn = 4* .
0
= 6 can be connected in front of
ss
) (1+p )]
0
is more than four
1