Fig.1
c.
The characteristic equation is s3?7s2?12s?K?0 Yields the Routh array
s3s2s117K12?7K12K
s0Since we have to determine the maximum value of K consistent with stability, according to the Routh’s criterion, we will yield:
K12??0 and K?0.
7Solving them yields: 0?K?84.
When K?84, the system will be critically stable. Thereby, we can confirm the previous results.
Module10
Problem 10.2
Plot the root locus for the feedback control system specified by the open-loop transfer function
K(s2?4s?13) GH(s)?2(s?0.5)(s?2s?2)Solution :
Root #1: Here we have two complex zeros located at the root of (s2?4s?13), which are
s??2?3j. There are three open-loop poles, s??0.5and s??1?j.
Rule #2: For there is only one pole on the real axis at s??0.5, the real-axis segment
forming part of the locus is the part on the left of ?0.5.
360?360???360?, we can get the angle between Rule #3: From the formula: ??n?m3?2adjacent asymptotes. We can see that there is only one asymptote.
?p??zi?0.5?2?4??1.5 Rule #4: The asymptote intersect the real axis at ?a?in?m3?2Rule #5: Applying this rule to the zero at s??2?3j and measuring the angles from the
formula:???180?(116.6?116.6?104)?90?67.2?. This is the angle of entry into the zero s??2?3j from complex poles. And the angle of emergence from the pole
s??1?j is ???180?(116?64?90)?76?14? .
Rule #6: The characteristic equation for this problem becomes
s3?(2.5?K)s2?(3?4K)s?(1?13K)?0 Substituting s?j? yields the following equations:
K ??2?3?4?0
and ?(2.5?K)?2?1?13K?0
the solution is K??j1.274 and ?2?3?j5.10
so we can conclude that the locus has no intersection with imaginary axis.
Rule #7: Normally, there should be a real axis breakaway point on the left of the open pole
at
s??2. From the formula:
K(s2?4s?13)(s?2)(s2?2s?2) 1??0, we can get : K??22(s?2)(s?2s?2)(s?4s?13)But at the negative real axis, we cannot calculate the maximum number of K, so there is no breakaway point on the real axis.
This is the root locus drawing by matlab, and the parameters we got from matlab accord well with the numbers we counted.
Problem 10.5－NO.4 Finished Solution：
The open-loop transfer function of the system is :
GH?
There are three poles at s=-1 ,s= -2+1.5j ,s=-2-1.5j.
K?s?1??s2?4s?6.25?
Using Rules #1 and #2 ,it may be seen that the negative real-axis segment for s<-1from part of the locus. Since there are three more poles than zeros, Rule #3 indicates there are three branches of the locus going to infinity and the negative real axis is one of them. The point where the asymptotes intersect the real axis is given by Rule #4 and may be calculated from
?2?2?15?a???3?03
360。?60。And the angle between adjacent asymptotes is 3
The locus up to this point is shown in the Fig.
The angle of emergence from the complex pole located at s= -2+1.5j may be evaluated by using Rule #5 as follows:
?k?180?(90?123.7)??33.7
The final plot may be sketched as follows:
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