Page 822 - The Mechatronics Handbook

066_Frame_C26 Page 7 Wednesday, January 9, 2002 1:58 PM

imaginary axis dominate the transient response of y(t). To generalize this observation, let r 1 ,…,r n be the
poles of T(s), such that Re(r k )<<Re(r 2 ) = Re(r 1 ) < 0, for all k ≥ 3 . Then, the pair of complex conjugate
poles r 1,2 are called the dominant poles. We have seen that the desired transient response properties, e.g.,
PO and t s , can be translated into requirements on the location of the dominant poles.

26.3 Root Locus Construction

As mentioned above, the root locus primarily deals with finding the roots of a characteristic polynomial
that is an affine function of a single parameter, K,

c s() = Ds() + KN s() (26.5)

where D(s) and N(s) are fixed monic polynomials (i.e., coefficient of the highest power is normalized to 1).
If N and/or D are not monic, the highest coefficient(s) can be absorbed into K.

Root Locus Rules
Recall that the usual root locus shows the locations of the closed-loop system poles as K varies from 0
to +∞. The roots of D(s), p 1 ,…, p n , are the poles, and the roots of N(s), z 1 ,…, z m , are the zeros, of the
open-loop system, G(s) = KF(s). Since P(s) and C(s) are proper, G(s) is proper, and hence n ≥ m . So the
degree of the polynomial χ(s) is n and it has exactly n roots.
Let the closed-loop system poles, i.e., roots of χ(s), be denoted by r 1 (K),…, r n (K). Note that these are
functions of K; whenever the dependence on K is clear, they are simply written as r 1 ,…, r n . The points
in that satisfy (26.5) for some K > 0 are on the root locus. Clearly, a point r ∈ is on the root locus

if and only if

1
K = – ---------- (26.6)
Fr()

The condition (26.6) can be separated into two parts:

1
K = – ------------- (26.7)
Fr()

±
±
∠ K = 0° = ( – 2 + 1) × 180° ∠ Fr(), = 0, 1, 2,… . (26.8)
–

The phase rule (26.8) determines the points in that are on the root locus. The magnitude rule (26.7)
determines the gain K > 0 for which the root locus is at a given point r. By using the definition of F(s),
(26.8) can be rewritten as
n m
( 2 + 1) × 180° = ∑ ∠ ( rp i ) – ∑ ∠ ( rz j ) (26.9)
–
–
i=1 j=1
Similarly, (26.7) is equivalent to

n
–
K = ∏ i=1 rp i (26.10)
--------------------------
m
–
∏ j=1 rz j
©2002 CRC Press LLC