chapter 3fivedots.coe.psu.ac.th/software.coe/computer control/comctrl3... · 241-482 root locus...
TRANSCRIPT
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Chapter 3Design of Discrete-Time control
systemsRoot Locus
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Outline
1 Complex variables2 Characteristic equation3 example
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Complex variables
s=+j
M=22
s=M∡
=tan−1
When
s=M e j
e j =cos j sin
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Vector representation of complex numbers : s = + j
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Vector representation of complex numbers : (s + a)
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Alternate representation of : (s + a)
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Vector representation of complex numbers : (s + 7)|s5 + j2
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Characteristic equation
T s =KG s
1KG s H s
R(s) E(s) C(s)KG(s)
H(s)
+-
1KG s H s=0Char. Eq.
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1+KG(s)H(s) = 0
KG(s)H(s) = -1
Characteristic Equation
G(s)H(s) is a complex function. We have :
KG s H s =M∡=12k1
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-a-b-c123
s1
js-plane
123=180°
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∡KG s H s =2k1
∣KG sH s∣=1
K=1
∣G s∣∣H s ∣
Value of K :
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G s =K∏i=1
m
sz i
∏j=1
n
sp i
=K∣sz1∣∣sz2∣...∣szm∣
∣sp1∣∣sp2∣...∣spn∣e j 1...m
e j 1...n
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Example (Nise)
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Vector representationof G(s) at -2+ j3
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12−3−4=56.31°71.57°−90°−108.43°
=−70.55°
Point -2+j3 is not on root locus.
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Point -2+j2/2 is on the Root Locus. We have 180 degree of angle of Ch. Eq.
For the gain
K=1
∣G sH s∣=
1M= pole lengths zero lengths
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G s=K s3s4s1s2
Example
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Complete root locus for the system example
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Example (Nise)
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Root locus and asymptotes for the system example
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Example (Nise) Sketch the root locus for the system below.
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T s =K s3
s 47s314s28K s3K
Closed-loop transfer function is
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Complex number
s47s314s28K s3K=0
Characteristic Equation
Let s= j
4− j73−142 j 8K 3K=0
4−1423K j −738K =0
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4−1423K=0
−738K =0
K=9.65 = ±1.59
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s4 1 14 3Ks3 7 8K
s2 90−K 21K
s1 −K 2−65K72090−K
s0 21K
Routh table
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From s1 row
−K 2−65K720=0
K = 9.65
Form s2 row with K=9.65
90−K s221K=80.35s2202.7=0
S = ±j1.59
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Example (Nise)
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Root locus for example showing angle of departure
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Plotting and Calibrating the Root Locus
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Transient Response Design via Gain Adjustment
The conditions justifying a second-order approximation are :
1 High order poles are much farther into the left half of the s-plane than the dominant second- order pair of poles. The response that results from a higher-order pole does not appreciably change the transient response expected from the dominant second-order poles.
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2 Closed-loop zeros near the closed-loop second-order pole pair are nearly canceled by the close proximity of higher-order closed- loop poles.3 Closed-loop zeros not canceled by the close proximity of higher-order closed-loop poles are far removed from the closed-loop second- order pole pair
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Making second-order approximations
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Design procedure for higher-order system
1 Sketch the root locus for the given system2 Assume the system is a second-order system without any zeros and then find the gain to meet the meet the transient response specification.3 Justify the second-order assumption by finding the location of all higher-order poles and evaluating the fact that they are much farther from the j-axis than the dominant second-order pair.
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4 If the assumptions cannot be justified, you solution will have to be simulated in order to be sure it meets the transient response specification.
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Example (Nise) Design the value of gain K, to yield 1.52% overshoot for the system
below. Also estimate the settling time, peak time, and steady-state error.
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Root locus of the system in slide 54
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Second- and third-order responses for case 2
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Second- and third-order responses for case 3