INC341
Root Locus (Continue)
Lecture 8
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Sketching Root Locus
(review)
1. Number of branches
2. Symmetry
3. Real-axis segment
4. Starting and ending points
5. Behavior at infinity
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Refining the sketch
1.Real-axis breakaway and break-in
points
2.Calculation of jω-axis crossing
3.Angels of departure and arrival
4.Locating specific points
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1. Real-axis breakaway and
break-in points
Breakaway
point
point where
the locus
leaves the
real axis
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Break-in
point
point where
the locus
returns to
the real axis
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KG(s) H (s)  1
set s = σ (on the real axis)
1
K
G( s) H ( s)
1
K
G( ) H ( )
Breakaway
point
Break-in
point
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Example
KG ( s) H ( s) 
K ( s  3)(s  5)
( s  1)(s  2)
Find breakaway, break-in points
K ( s 2  8s  15)
KG ( s) H ( s)  1 
( s 2  3s  2)
Condition of poles
 ( s 2  3s  2)
K 2
( s  8s  15)
dK 11s 2  26s  61
 2
0
ds
s  8s  15
then solve for s
s = -1.45, 3.82 is breakaway and break-in points
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m
Another approach
without derivative
n
1
1
1 s  z  1 s  p
i
i
1
1
1
1



s  3 s  5 s 1 s  2
11s 2  26s  61  0
s  1.45, 3.82
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2. Calculation of jω-axis crossing
Imaginary axis is a boundary of stability
use Routh-Hurwitz criterion!!!
Imaginary axis crossing
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Review of Routh-Hurwitz
“the number of roots of the polynomial
that are in the right half plane is equal
to the number of sign changes in the
first column”
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Example
From the closed-loop transfer function, find
an imaginary axis crossing
T ( s) 
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K ( s  3)
s 4  7 s 3  14s 2  (8  K ) s  3K
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A complete row of zeros
yields imag. axis roots
 K 2  65K  720  0
K  9.65
Substitute K=9.65 in s2 to find the value of s
(90  K ) s 2  21K  80.35s 2  202.65  0
 202.65
s
  j1.59
80.35
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3. Angles of departure and Arrival
Fact: root locus starts at open loop poles
and ends at open loop zeros
KG(s) H (s)  (2k  1)180

Assume a point on the root locus close to a complex
Pole, the sum of angles to this point is an odd multiple
of 180.
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Angel of departure (pole)
1  2  3 4 5  6  (2k 1)180
Angel of arrival (zero)
2  1 3  4  5 6  (2k 1)180
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Example
sketch root locus and find angel of departure
of complex poles
x
x
-3
o
-2
1
-1
x
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 1   2   3   4
1
1 1
 1  90  tan ( )  tan ( )  180
1
2
1  251.57  108.43

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1
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4. Calibrating root locus
Search a given line for the point yielding a
summation of angles equal to an odd multiple
of 180.
KG(s) H (s)  (2k  1)180

Gain at this point = pole length/zero length
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Intersection with damping ratio line
ζ=cosθ
θ
Y=-mx
M = tan(acos(damping ratio))
Coordinate on Damping line = (rcosθ, rsinθ)
Try r = 0.5, 1, 0.8, 0.7, 0.75, 0.725, …..
At r=0.747 K 
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AC D E
B
 1.71
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Example
sketching root locus
What is the exact point
and gain where the locus
crosses the imag. Axis?
Where is the breakaway
point?
What range of K that keep
the system stable?
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Transient Response Design
via Gain Adjustment
Find K that gives a desired peak time, settling time,
%OS (find K at the intersection)
Use 2 order approx. and
consider only dominant
pole
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The third pole can be ignored (a gives a better approx.
than b cause the third pole is further to the left)
Zero closed to the dominant poles can be cancelled
INC third
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by the
pole (c gives a better approx. thanPTd)
Example
Find K that yields 1.52% overshoot.
Also estimate settling time, peak time,
steady-state error corresponding to the K
Step I: 1.52% overshoot  ζ=0.8
Step II: draw a root locus
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Step III: draw a straight line of 0.8 damping ratio
Step IV: find intersection points where the net angle
is added up to 180*n, n=1,2,3,…
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Step V: find the corresponding K at each point
Step VI: find peak time, settling time corresponding to
the pole locations (assume 2nd order approx.)
Step VII: calculate Kv and ss error
Note: case 1 and 2 cannot use 2nd order approx.
cause the third pole and closed loop zero are far away.
In case 3, the approx. is valid.
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Generalized Root Locus
K is fixed, vary open loop pole instead!!!
Creating an equivalent system where p1 appears
as the forward path gain.
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G( s)
10
T ( s) 
 2
1  G( s) s  p1s  2s  2 p1  10
Try to get a general TF T ( s) 
KG ( s)
1  KG ( s) H ( s)
10
T ( s)  2
s  2 s  10  p1 ( s  2)
10
2
s
 2 s  10
T ( s) 
p1 ( s  2)
1 2
s  2 s  10
p1 ( s  2)
KG ( s ) H ( s )  2
s  2 s  10
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Using MATLAB with Root Locus
•tf
•pzmap
•rlocus
•sgrid
•sisotool
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