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by Eq. (5.168) as
Gc(s) ~ =p
We have to find the locations of the zero Zc and the pole pc on the real axis so that
the design objectives are met.
From the analysis of PD compensator as iri 1) above, we know that the net
angle contribution due to the zero at s - -zc and pole at s = - pc must be equal
to 86.2498 deg. Let us assume that the angle contribution due to zero at s - -Zc
is Ozc = 90 deg so that zc - 2.5435. Then, the angle contribution due to the pole at
s = -Pc is Op = 3.7502 deg so that Pc = 88.9. Then, the transfer function of the
lead-compensated system is given by
' k(s + 2.5435)
Gc(s) : -
= s~s+3~s+5~s+88.90)
Now we draw the root-locus of the lead-lag compensated system as shown in
Fig. 5.36d (the root-locus around the origin is shown in tfus figure) and select the
operating point for < -.0.4.We get k = 3451.4 and closed-loop poles at -89.3603,
-2.54051- j5.7879, and -2.45
Let us verify the designs bjAB;foThing the simulation, i.e., we determine the
unit-step response using MATLJ rhc results are shownin Fig.5.37.We observe
that the peak amplitudes (hence the percent overshoot Os) for all three cases are
nearly equal. For the basic system, Tp = 1.80 s and, for PD- and lead-compensated
systems, Tp ~ 0.6 s. Thus, the design objectives have been realized.
Example 5-11
Design a PID controller for a unity feedback system with
{- 15)
G(s) = ~+_j~ :;) +g~
to operate at a peak time, which is 50qo of the basic system and has a zero steady-
state error for a unit-step input while continuing to operate at a damping ratio of0.3.
LINEAR SYSTEMS, THEORY, AND DESIGN: A BRIEF REVIEW 503
:Fig. 5.37 Urut-step responses for Example 5.10.
So/ution. The approach we take here is to design the PD controller first and
then add a PI controller.
Draw the root-locus for the basic system using MATLAB4 as shown in Fig.
5.38a and pick the point on the root-locus corresponding to C =0.3. We get
k-14.2869 and p=-10.089,-1.4555+j4.6689. Using this information,
we get Tp = 0.6729 s. Then, for the compensated system, Tpc = Tpl2 = 0.3365 s,
which corresponds to O>dc=9.3378, COnc:tOdc/,~~-9.7887. and CJdc-
< COnc - 2.9366.
Now we calculate the angle contributions. Proceeding as before in Example
5.10, we get 0i = 101.1851, 02 - 89.6355, 03 - 39.0601, and 04 = 58.2290 so that
Ozc -29.9885 giving zc =19.9014. With this, the transfer function of the PD-
compensated system is given by
k(s + 15)(s + 19.9014)
G(s) = ~s +1)(s + 3) s + 9)
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PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL3(70)
