root_locus

Since the G(s) H (s) is a complex quantity, so
G(s) H (s) 180(2k 1),(k 0,1, 2) G( s ) H ( s ) 1

In many cases, G(s)H(s) involves a gain K, and the characteristic equation may be written as
B( s) f ( s ) B( s) A( s) 0 A( s) or B( s) A( s) B( s) A( s) 0

On the other hand , from we obtain
f (s) B(s) KA(s) 0
B( s ) K A( s)
Angle and Magnitude Conditions

Consider the system shown
The closed-loop TF is:
C (s) G(s) R( s) 1 G ( s ) H ( s )

The characteristic equation
1 G( s) H ( s) 0 G( s) H ( s) 1

Select a test point s on the real axis. a) if the test point is on the positive real axis, then s s 1 s 2 0
b) if the test point is on the negative real axis between 0 and -1 , then
3 2
So ,the breakaway point is s=-0.4226

Determine the points where the root loci cross the imaginary axis the characteristic equation of closed-loop for the present system is
K K lim G ( s) lim lim 3 s s s ( s 1)( s 2) s s
And the angle condition becomes
3s 180(2k 1)

The angles of asymptotes =
180(2k 1) (2k 1) , (k 0,1, 2) 3 nm
K G( s) ( s 1)3

Generally , the intersection of the asymptotes and the real axis is obtained by the following equation
( p1 p1 pn ) ( z1 z2 zm ) s nm
s 3s 2s K 0
3 2

Make use of the Routh’s criterion
3 2
s s s s
1 3 6K 3 K
2 . Determine the asymptotes of the root loci. the asymptotes of the root loci as s approaches infinity can be determined as follows: if a test point s is selected very far from the origin, then
dK B(s) A(s) B(s) A(s) 0 2 ds A ( s)
B(s) A(s) B(s) A(s) 0

For the present example ,the characteristic equation
K 1 0 s ( s 1)( s 2) K ( s 3s 2s ) dK 2 (3s 6 s 2) 0 ds s 0.4226, s 1.5774

The roots of characteristic equation is
s 2 2s 2 K 0

If the parameter gain K changed from zero to infinity , all the roots of the closed-loop system is drawn , which is called the root-locus method. Simultaneously, through the root loci , we can know how the value of the K take effects on the parameter jita and omiga of the second-order system.

From the design viewpoint , in some systems simple gain adjustment may move the closed-loop poles to desired locations. Then the design problem may become the selection of an appropriate gain value. If the gain adjustment alone does not yield a desired result, addition of a compensator to the system will become necessary

Let us write the characteristic equation of closed-loop TF as: KA( s) 1 G( s) H ( s) 1 0 B( s )
f (s) B(s) KA(s) 0
Note that f(s)=0 has the multiple roots at points where
Root-Locus Analysis
qiqian
The basic principle

If a closed-loop TF is given like that

The open-loop TF is :
K s (0.5s 1)

The closed-loop TF is:
2K s 2 2s 2 K
df ( s ) 0 ds
df ( s ) B( s ) KA( s) 0 ds dA( s) dB( s) A( s) , B( s) ds ds B( s ) K A( s )
If we substitute this value of K into equation, we get

Example 1
G ( s )
K s s 1 s 2 s ( s 1)( s 2) 180(2k 1), ( k 0,1, 2 ) K 1 s ( s 1)( s 2)
G ( s)

1. Determine the root loci on the real axis the first step constructing a root-locus plot is to locate the open-loop poles, 0 , -1, -2. the starting points of the root loci (corresponding to K=0) are open-loop poles. The number of individual root loci for this system is three, which is the same as the number of open-loop poles
Where n=number of poles of G(s)H(s)
m=number of zeros of G(s)H(s)
Find the intersection :
K K G(s) 3 2 s( s 1)( s 2) s 3s
For a large value of s , this last equation may be approximate by

If the test point is between -1 and -2, then
s s 1 s 2 360
If the test point is located on the negative real axis from -2 to –infinite ,the angle condition is satisfied. Thus , root loci exit on the negative real axis between 0 and -1 and between -2 and -infinite
K ( s z1 ) G( s) H ( s) 1 ( s p1 )( s p2 )( s p3 )( s p4 )
G( s) H ( s) 1 1 2 3 4 KB1 G( s) H ( s) A1 A2 A3 A4
Illustrative examples
s 180 s 1 s 2 0 s s 1 s 2 180
The angle condition is satisfied . Therefore, the portion between 0 and -1 forms the portion of the root locus.
w 3, 3, 0 1
Which can be written as