INC 341 PT & BP
INC341Root 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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Break-inpoint
Breakawaypoint
1. Real-axis breakaway andbreak-in points
point where the locus leaves the real axis
point where the locus returns to the real axis
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1)()( sHsKG)()(
1
sHsGK
set s = σ (on the real axis))()(
1
HGK
Breakawaypoint
Break-inpoint
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Example
)2)(1(
)5)(3()()(
ss
ssKsHsKG Find breakaway, break-in points
0158
612611
)158(
)23(
)23(
)158(1)()(
2
2
2
2
2
2
ss
ss
ds
dK
ss
ssK
ss
ssKsHsKG Condition of poles
then solve for s
s = -1.45, 3.82 is breakaway and break-in points
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Another approach without derivative
n
i
m
i pszs 11
11
82.3,45.1
0612611
2
1
1
1
5
1
3
1
2
s
ss
ssss
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2. Calculation of jω-axis crossing
Imaginary axis is a boundary of stabilityuse 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
KsKsss
sKsT
3)8(147
)3()(
234
From the closed-loop transfer function, find an imaginary axis crossing
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65.9
0720652
K
KK
Substitute K=9.65 in s2 to find the value of s
59.135.80
65.202
065.20235.8021)90( 22
js
sKsK
A complete row of zerosyields imag. axis roots
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3. Angles of departure and Arrival
Fact: root locus starts at open loop poles and ends at open loop zeros
180)12()()( ksHsKG
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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180)12(654321 k
180)12(654312 k
Angel of arrival (zero)
Angel of departure (pole)
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Example
sketch root locus and find angel of departureof complex poles
x
x
x
o-3 -2 -1
1
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43.10857.251
180)2
1(tan)
1
1(tan90
1
111
4321
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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.
180)12()()( ksHsKG
Gain at this point = pole length/zero length
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At r=0.747 71.1B
EDCAK
Intersection with damping ratio line
Coordinate on Damping line = (rcosθ, rsinθ)
Try r = 0.5, 1, 0.8, 0.7, 0.75, 0.725, …..
ζ=cosθ
θY=-mxM = tan(acos(damping ratio))
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Examplesketching 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 Designvia 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 by the third pole (c gives a better approx. than d)
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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=0.8Step 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 angleis added up to 180*n, n=1,2,3,…
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Step V: find the corresponding K at each pointStep 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 appearsas the forward path gain.
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1022
10
)(1
)()(
112
psspssG
sGsT
Try to get a general TF )()(1
)()(
sHsKG
sKGsT
102
)2()()(
102)2(
1
10210
)(
)2(102
10)(
21
21
2
12
ss
spsHsKG
ssspsssT
spsssT
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Using MATLAB with Root Locus
•tf
•pzmap
•rlocus
•sgrid
•sisotool