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Unformatted text preview: Formulation of the General Problem The general form of an equation whose roots can be studied by the root locus
method is 0(3) + KN (s) = 0 (8.11.1) where K is the parameter to be varied. For now, we take the functions D(s) and N (s) to
be polynomials in 5 with constant coefﬁcients, and we consider the case K 2 0, Another standard form of the problem is obtained by rewriting (8.111) as 1 + KP(s) = 0 (8.142)
where
P(s) = {1%) (8.113) Plotting Guides for the Primary Root Locus Standard Form:
l+KP(s)=O, K20 Plat) == AIM/DB)
N(5)zsl+b~13.l+ '” +b15+bo
D(s)=s*‘+a,_,s"‘ + +als+ao m g n
Terminology:
The zeros of Pb“) are the roots of M3) = 0.
The poles of Pfs) are the roots of Dis) : 0. Guide 1 The mat locus plot is symmetric about the real axis‘ This is because complex roots
occur in conjugate pairst Thus, we need deal with only the upper half plane of the plot. Guide 2 The number of loci equals the number of poles of P(s). Guide 3 The loci Start at the poles of PIS) with K =0 and terminate with K = 00 either at
the zeros of P(s) or at inﬁnity. Guide 4 The root locus can exist on the real axis only to the left of an odd number of real poles
and/0r zeros; furthermore. it must exist there. FIGURE 8.8 Location of the root locus on the real axis for varimns pole—zero conﬁgurations.
The locus olf the real axis '5 not shown. Guide 5 The locations of breakaway and break~in points are found by determining where the
parameter K attains a local maximum or minimum on the real axis. Exampic 8.1
Determine the root locus plot for the equation +§U+® s{s + 4) The poles are 3 = 0, ~4; the zero is s = ~6. Thus, there are two loci. One starts at s = 0 and the other at s = «4, with K = 0. One path terminates at s = —6, while the
other must terminate at s 2 so. This information is given by Guides 2 and 3. Guide 4 shows that the locus exists on the real axis between s = O and s = —4 and to the left of s = —6. Therefore, the two paths must break away from the real axis between s z 0 and s = —4. From the location of the termination point at s = —6, we know that the locus must return to the real axis. From Guide I, we see that both paths
must break in at the same point. z 0 (8.2—3) These points are found from Guide 5 as follows. Solve (8.23) for K and compute
dK/ds. __ ws(s+4)
K_ 3+6
dK“ (s+6)(25+4)—s(s+4)_0
ds  (5+6)2 This is satisﬁed for ﬁnite 3 if the numerator is zero. s2 + 125 + 24 = 0
The candidates are s: «6 i 2J5 = ~.254. «9.46 There is no need to check for a minimum or a maximum. because we know from
Guide4that the breakaway point mustbe 5: —2.54 and the breakin point
= —9.46. This leaves only the shape 0f the locus off the real axis tobedetermined. Elementary geometry can be used to show that the root locus of the equation
(s+ b}(s+e)+K(s+a)=0 (8.24)
is a circle centered on the zero at 5 = —a ifa, b, c > 0 and a > b > e This case is shown in Figure 8.10. The radius of the circle can be determined once the breakaway and
breakin points are found. (a) FIGURE 8.10 Roottoms plot for FIGURE 8.9 (a) Rootlocus plot for (s + bx: + c) + Kts + a) = 0 for K 2 9
3(5+4)+K(3+6)m0,f0r K20. and a>b>c>8. Guide 6 The loci that do not terminate at a zero approach inﬁnity along usymplotes. The angles
that the asymptoles make with the real axis are found from (8.25), where n is chosen successively as n= +1. «1, +3, —3, until enough angles have been found.
n180°
= ' =+l,+3,l.. 82
6 2—1) n __  ( 5)
<—————
zpx4 l < ><
Z«P=2
ZP=—3 z1>=—4 FIGURE 8.12 Asymptotic angles for commonly occurring casts.
Guide 7 The asymprozes intersect the real axis at the common point a = 351:3: (8.2~6)
P ~ 2 where 25,, and 252 are the algebraic sums of the values of the poles and zeros.
Guide 8 T he points at which the loci cross the imaginary axis and the associated values of K can
be found by the Routh—Hurwitz criterion or by substituting s = la) into the equation of
interest. The frequency w is the crossover frequency. Example 3.2 Plot the root locus for the equation 33 + 352 + 25 + K = 0 for K 2 0 (3.237)
We can factor the equation as follows:
K
1 + ————~—~ = 0 3(3 + l)(s + 2) The poles are $20, —— l, —2, and there are no zeros. All three paths approach asymptotes as K ~+ 03‘ The locus exists on the real axis between 5 = 0 and —— 1 and to
the left of s = ~2. To ﬁnd the breakaway point, compute ddes. K .—.. —(S3 + 352 + 23)
(1K ...... .=_“2 2:0
d5 (3? +6s+) The candidates are 5: —0.423, — 1.58. The breakaway point must obviously be
5 = —0.423, because the locus cannot exist between s = ~1 and s = —2. With this
value of s, (8.27) gives K =0.385. The three asymptotic angles are found from (8.25). ”urge“ 0 0—3 =n600 n=il,i3.... Thus,
6: 450“, +60“, i180“ Note that for the last angle it does not matter whether we use it = + 3 or n = w 3. The
intersection is found from {8.26}. 0+1—1)+t—2)—0
e=~————————~—~=——l 3 — 0
The two paths that start at s = 0 and —l approach the i603 asymptotes‘ The path
starting at s = ——2 follows the 180‘” asymptote and thus lies entirely on the real axis. The 60‘> asymptotes indicate that two paths will cross the imaginary axis and
generate two unstable roots. Substituting 3 = to) into (8i2~7) gives mic? — 3602 + 2iw + K : 0
or
iw(2 __. (122) = 0
K = 3012 The solution w = 0, K = 0 corresponds to the pole at s = 0. The solution of interest is a) = iV/E, K = 6. The locus can now be sketched. This is shown in Figure 8.13:1. The
system has three roots All are real and negative for 0 g K g 0.385. For 0.385 < K < 6,
the system is stable with two complex roots and one real root, For K > 6‘ the system is
unstable due to two complex roots with positive real parts; In: FIGURE 8.13 (a) Rootlocus plot for 5(5 + “(s + 2) + K —= 0, for K 20. %%%%%%%%%%%%%%536%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
MATLAB
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%9%%%%%%%%%%%%%%%%%%%%%%%%%%%? 1. we have to have our characteristic equation (the denominator of ourtransfer function) and rearrange it as den(s)+K*num(s);
5.9., separate the polynomials with K from without K; 2. REMEMBER that the "den” and "num" are NOT the denominator and numerator of our actually system. They are for drawing use
only here. 3. For example, if our C.E. is as CsA3+2hsA2~5*s)+K*{5A2+S*s+4)=0,
then we type "den=[1 2 S 0]" and "num=[1 S 4]”. 4. Type "sys=tf(num,den)" to define the sys For draw root locus
plot purpose. 5. Type “rlocusCsys3” (c) (b) (a) FIGURE 8.15 Some common rootlocus
forms not seen in the chapter examples.
(a) (s+c)(3+a+ib)(s+a~ib)+K=0, K20.
(b) (s+a)(s+b)(s+c)+K(s+d)=0, K20
(see also Figures 8.16 and 9.12:).
(c) (5+ch +a +ib)(s +aib)+K(s+tl)=0.
K 2 0. ...
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 Spring '09

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