Bài giảng Control system design - Chapter VII: The Root Locus Method - Nguyễn Công Phương
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
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Nguyễn Công Phương
CONTROL SYSTEM DESIGN
The Root Locus Method
Contents
I. Introduction
II. Mathematical Models of Systems
III. State Variable Models
IV. Feedback Control System Characteristics
V. The Performance of Feedback Control Systems
VI. The Stability of Linear Feedback Systems
VII.The Root Locus Method
VIII.Frequency Response Methods
IX. Stability in the Frequency Domain
X. The Design of Feedback Control Systems
XI. The Design of State Variable Feedback Systems
XII. Robust Control Systems
XIII.Digital Control Systems
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 3
The Root Locus Concept (1)
R( s ) Y( s )
Ys() KGs () K G( s )
T( s ) = =
R() s 1+ KG () s (− )
1+KG ()0 s =
→KG( s ) = − 1
→KG() s ∠ [ KG ()] s =∠ 1180 o
KG( s )= 1
→
∠[KG ()] s =+ 180o k 360; o k =±± 0, 1, 2,...
The root locus is the path of the roots of the characteristic equation traced out
in the s-plane as a system parameter varies from zero to infinity.
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The Root Locus Concept (2)
Ex. 1
R( s ) 1 Y( s )
K +
K − s( s 2)
1+KG ()1 s =+ = 0 ( )
s( s + 2)
→∆=++=+2 2ζω + ω 2 =
()2ss sKs 2n s n 0
→ =−ζωωζ ±2 −=−± ωζ 2 −
s1,2 n n1 1 n 1
Ex. 2
R( s ) 1 Y( s )
10
s( s+ a )
(− )
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The Root Locus Concept (3)
N
∆=−+ − +
(s ) 1 ∑ Ln ∑ LL nm ∑ LLL nmp ...
n=1 nm , nmp ,,
nontouching nontouching
=1 + F ( s )
∆()0s =→ F () s =− 1
Kszszsz(+ )( + )( + )...( sz + )
F( s ) = 1 2 3 M
+ + + +
(spspsp1 )( 2 )( 3 )...( sp n )
Ks+ z s + z ...
F( s )=1 2 = 1
+ +
→ s ps1 p 2 ...
∠ =∠+ +∠+ +−∠+ +∠+ + =o + o
Fs()[( sz12 ) ( sz )...][( sp 1 ) ( sp 2 )...]180 k 360
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 7
The Root Locus Procedure (1)
1. Prepare the root locus sketch.
2. Locate the open – loop poles and zeros of P(s) in the s
– plane with selected symbols.
3. The loci proceed to the zeros at infinity along
asymptotes centered at σA and with angle ϕA.
4. Determine the points at which the locus crosses the
imaginary axis (if it does so).
5. Determine the breakaway point on the real axis (if
any).
6. Determine the angle of locus departure from complex
poles and the angle of locus arrival at complex zeros,
using the phase criterion.
7. Complete the root locus sketch.
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The Root Locus Procedure (2)
Step 1
1+F () s = 0
→+1KPs ( ) = 0, 0 ≤≤∞ K
M
+
∏ (s z i )
→ +i=1 =
1K n 0
+
∏ (s p j )
j=1
n M nM
→ ++ +=↔1 ++ +=
∏(spKszj ) ∏ ()0 i ∏ ()()0 sp ji ∏ sz
j=1 i = 1K ji == 11
n
= → + =
K0∏ ( s p j )0
j=1
M
→∞→ + =
K∏( s z j ) 0
j=1
The locus of the roots of the characteristic equation 1 + KP (s) = 0 begins at
the poles of P(s) and ends at the zeros of P(s) as K increases from zero to infinity.
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The Root Locus Procedure (3)
1. Prepare the root locus sketch.
2. Locate the open – loop poles and zeros of P(s) in the s –
plane with selected symbols: the root locus on the real
axis always lies in a section of the real axis to the left of
an odd number of poles and zeros
3. The loci proceed to the zeros at infinity along asymptotes
centered at σA and with angle ϕA.
4. Determine the points at which the locus crosses the
imaginary axis (if it does so).
5. Determine the breakaway point on the real axis (if any).
6. Determine the angle of locus departure from complex
poles and the angle of locus arrival at complex zeros,
using the phase criterion.
7. Complete the root locus sketch.
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The Root Locus Procedure (4)
Ex. 1 Step 2
2(s + 2)
1+K = 0
s( s + 4) +
s1 2 s1
− −
s2 4 2 s1 0
+
s1 4
The locus of the roots of the characteristic equation 1 + KP (s) = 0 begins at
the poles of P(s) and ends at the zeros of P(s) as K increases from zero to infinity.
- The number of separate loci is equal to the number of poles.
- The root loci must be symmetrical with respect to the horizontal real axis.
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The Root Locus Procedure (5)
1. Prepare the root locus sketch.
2. Locate the open – loop poles and zeros of P(s) in the s
– plane with selected symbols.
3. The loci proceed to the zeros at infinity along
asymptotes centered at σA and with angle ϕA.
4. Determine the points at which the locus crosses the
imaginary axis (if it does so).
5. Determine the breakaway point on the real axis (if
any).
6. Determine the angle of locus departure from complex
poles and the angle of locus arrival at complex zeros,
using the phase criterion.
7. Complete the root locus sketch.
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The Root Locus Procedure (6)
Step 3
M
+
∏ (s z i )
+ =+i=1 = ≤≤∞
1KPsK ()1n 0,0 K
+
∏ (s p j )
j=1
n M
− − −
∑(pj ) ∑ () z i
polesofPs ( )− zerosof Ps ( )
σ =∑ ∑ = j=1 i = 1
A nM− nM −
2k + 1
φ =180o ,k = 0,1,2,...,( n −− M 1)
A n− M
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The Root Locus Procedure (7)
Ex. 2 Step 3
(s + 1)
1+K = 0
s( s+ 2)( s + 4) 2
polesofPs ()− zerosof Ps ( )
σ = ∑ ∑
A n− M
[(− 2) +− ( 4) +− ( 4)] −− ( 1)
= = − 3
4− 1
2k + 1 − − − 0
φ =180o ,k = 0,1,2,...,( n −− M 1) 4 2 1
A n− M
2k + 1
=180o =+ (2k 1)60, o k = 0,1,2.
4− 1
= →φ = o
k 0A 60
= →φ = o
k 1A 180
= →φ = o
k 2A 300
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The Root Locus Procedure (8)
1. Prepare the root locus sketch.
2. Locate the open – loop poles and zeros of P(s) in the s
– plane with selected symbols.
3. The loci proceed to the zeros at infinity along
asymptotes centered at σA and with angle ϕA.
4. Determine the points at which the locus crosses the
imaginary axis (if it does so).
5. Determine the breakaway point on the real axis (if
any).
6. Determine the angle of locus departure from complex
poles and the angle of locus arrival at complex zeros,
using the phase criterion.
7. Complete the root locus sketch.
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The Root Locus Procedure (9)
Step 5
∆()s =→ 0 psK () =
dp( s )
breakaway point := 0
ds
Ex. 3
(s + 1)
1+K = 0
s( s+ 2)( s + 3)
−s( s + 2)( s + 3)
→K = = p( s )
s +1
−++3 +++ 2
→dpdss =( 2)( s 3) = 2 s 8 s 10 s 6
ds dt s +1 (s + 1) 2
dp
=→0 2sss3 + 8 2 + 10 +=→=− 60 s 2.46, s =− 0.77 ± j 0.79
ds 1 2,3
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The Root Locus Procedure (10)
1. Prepare the root locus sketch.
2. Locate the open – loop poles and zeros of P(s) in the s – plane with
selected symbols.
3. The loci proceed to the zeros at infinity along asymptotes centered
at σA and with angle ϕA.
4. Determine the points at which the locus crosses the imaginary axis
(if it does so).
5. Determine the breakaway point on the real axis (if any).
6. Determine the angle of locus departure from complex poles and
the angle of locus arrival at complex zeros, using the phase
criterion: The angle of locus departure from a pole is the
difference between the net angle due to all other poles and
zeros and the criterion angle of ±180 °(2k + 1).
7. Complete the root locus sketch.
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The Root Locus Procedure (11)
Step 6
θ
1
The angle of locus departure from a
θ
pole is the difference between the 3
net angle due to all other poles and 0
zeros and the criterion angle of
±180 °(2k + 1).
θ
2
θθθ++=o →= θ o −+ θθ
123180 1 180 ( 23 )
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The Root Locus Procedure (12)
Ex. 2 5
4 p
+K = 1
14 3 2 0
s+12 s + 64 s + 128 s 3
K
→1 + = 0 2
ss(+ 4)( s ++ 4 js 4)( +− 4 j 4)
= − + 1
p1 4 j 4
= − −
p2 4 j 4 0
= − p3 p4
p3 4
-1
=
p4 0
-2
1.VVV Prepare the root locus sketch.
2.VVV Locate the open – loop poles and zeros of
P(s) in the s – plane with selected symbols. -3
3. The loci proceed to the zeros at infinity
along asymptotes centered at σ and with -4
A p2
angle ϕA.
4. Determine the points at which the locus
-5
crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1
5. Determine the breakaway point on the real
poles ofPs ( )− zeros of Ps ( )
axis (if any). σ = ∑ ∑
6. Determine the angle of locus departure A
from complex poles and the angle of locus n− M
arrival at complex zeros, using the phase
criterion. −−−44j 44 −+ j 4
= = − 3
7. Complete the root locus sketch. 4− 0
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The Root Locus Procedure (13)
Ex. 2 5
4 p
+K = 1
14 3 2 0
s+12 s + 64 s + 128 s 3
2k + 1 2
φ =180o ,k = 0,1,2,...,( n −− M 1)
A −
n M 1
2k + 1 0
=180o ,k = 0,1,2,3 p p
− 3 4
4 0 -1
-2
1.VVV Prepare the root locus sketch.
2.VVV Locate the open – loop poles and zeros of
P(s) in the s – plane with selected symbols. -3
3.VVV The loci proceed to the zeros at infinity
along asymptotes centered at σ and with -4
A p2
angle ϕA.
4. Determine the points at which the locus
-5
crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1
5. Determine the breakaway point on the real = →φ = o
axis (if any). k 0A 45
6. Determine the angle of locus departure = →φ = o
from complex poles and the angle of locus k 1A 135
arrival at complex zeros, using the phase
criterion. = →φ = o
k 2A 225
7. Complete the root locus sketch.
= →φ = o
sites.google.com/site/ncpdhbkhn k 3A 315 20
The Root Locus Procedure (14)
Ex. 2 5
4 p
+K = 1
14 3 2 0
s+12 s + 64 s + 128 s 3
s4 1 64 K 2
3
s 12 128 0 1
s2 b 0
1 K 0
1
p3 p4
s c1 0 0
-1
s0 K
-2
1.VVV Prepare the root locus sketch.
2.VVV Locate the open – loop poles and zeros of
P(s) in the s – plane with selected symbols. -3
3.VVV The loci proceed to the zeros at infinity
along asymptotes centered at σ and with -4
A p2
angle ϕA.
4. Determine the points at which the locus
VVV -5
crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1
5. Determine the breakaway point on the real ×− ×−
axis (if any). =12 64 128 = = 53.33 128 12 K
6. Determine the angle of locus departure b1 53.33; c 1
from complex poles and the angle of locus 12 53.33
arrival at complex zeros, using the phase c>0 → K < 568.89
criterion. 1
7. Complete the root locus sketch. 2 + =→ =±
53.33s 568.89 0 s1,2 j 3.27
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The Root Locus Procedure (15)
5
θ
Ex. 2 1
K 4
1+ = 0
4 3 2 p1
s+12 s + 64 s + 128 s 3
2
→=−+K( s4 12 s 3 + 64 s 2 + 128 sps ) = ( )
dp 1
→ =−(4s3 + 36 s 2 + 128 s + 128) θ θ
4 3
ds 0
p p
dp 3 4
=→=−0s 1.58; s =− 3.71 ± 2.55 -1
ds 1 2,3
-2
1.VVV Prepare the root locus sketch.
2.VVV Locate the open – loop poles and zeros of
P(s) in the s – plane with selected symbols. -3
θ
3.VVV The loci proceed to the zeros at infinity 2
along asymptotes centered at σA and with -4
angle ϕA. p2
4. Determine the points at which the locus
VVV -5
crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1
5. Determine the breakaway point on the real
VVV θ+ θ + θ + θ = o
axis (if any). 1 2 3 4 180
6.VVV Determine the angle of locus departure
from complex poles and the angle of locus →=θo − θθθ ++
arrival at complex zeros, using the phase 1180( 234 )
criterion.
7. Complete the root locus sketch. =180oooo − (90 + 135 + 90) =− 135 o = 225 o
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The Root Locus Procedure (16)
Ex. 2 5
K 4
1+ = 0
4 3 2 p1
s+12 s + 64 s + 128 s 3
2
1
0
p3 p4
-1
-2
1.VVV Prepare the root locus sketch.
2.VVV Locate the open – loop poles and zeros of
P(s) in the s – plane with selected symbols. -3
3.VVV The loci proceed to the zeros at infinity
along asymptotes centered at σA and with -4
angle ϕA. p2
4. Determine the points at which the locus
VVV -5
crosses the imaginary axis (if it does so). -7 -6 -5 -4 -3 -2 -1 0 1
5.VVV Determine the breakaway point on the real
axis (if any).
6.VVV Determine the angle of locus departure
from complex poles and the angle of locus
arrival at complex zeros, using the phase
criterion.
7.VVV Complete the root locus sketch.
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The Root Locus Procedure (17)
Ex. 2
K K
1+ = 0 →1 + = 0
s4+12 s 3 + 64 s 2 + 128 s ss(+ 4)( s ++ 4 js 4)( +− 4 j 4)
5
4
p1
3
2
1 s
1
s
0 2
s
3 p3 p4
s
-1 4
-2
-3
-4
p2
-5
-7 -6 sites.google.com/site/ncpdhbkhn-5 -4 -3 -2 -1 0 1 24
The Root Locus Procedure (18)
Ex. 3
K
Given the characteristic 1 + = 0. Find K so that s = –1 + j2?
s4+12 s 3 + 64 s 2 + 128 s
K 5 + +
→ = − 1 s1 4 j 4
ss(+ 4)( s ++ 4 js 4)( +− 4 j 4)
4 p1
→=Kss +4 s ++ 44 js +− 44 j 3
2
+=2 + 2 =
s1 4 2 3 3.61
s1
1 s +
1 s1 4
2 2 s
s++4 j 4 = 3 + 6 = 6.71 0 2
1 s
3 p3 p4
s s+4 − j 4
-1 4 1
+− =2 + 2 =
s1 4 j 4 2 3 3.61
-2
=2 + 2 =
s1 1 2 2.24 -3
→K =×××3.61 6.71 3.61 2.24 -4
s=−1 + j 2
p2
-5
= 195 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4
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The Root Locus Procedure (19)
Ex. 4
K( s2 + 16 s + 113)
Given the characteristic equation 1 + = 0 . Find K so that:
s2 ( s + 16)
1. The damping ratio ζ = 0.5.
2. The settling time is less than 2 seconds.
1 −ζω
s2+2ζωω s += 2 0;()1 rt =→=− yt ()1 ent sin( ωβθ t + )
n n β n
1.8
ζ = 0.1
1.6 ζ = 0.2
ζ = 0.5
1.4 ζ = 0.7
ζ = 1.1
1.2 ζ = 2.0
1
0.8
0.6
0.4
0.2
0
-0.2
0 0.5 1 1.5 2 2.5 3
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The Root Locus Procedure (20)
Ex. 4
K( s2 + 16 s + 113)
Given the characteristic equation 1 + = 0 . Find K so that:
s2 ( s + 16)
1. The damping ratio ζ = 0.5.
2. The settling time is less than 2 seconds.
Ks(++ 8 js 7)( +− 8 j 7)
1+ = 0
s2 ( s + 16) 8
poles ofPs ( )− zeros of Ps ( ) 6
σ = ∑ ∑
A n− M 4
− −−− +−+
= (0 16) [( 8j 7) ( 8 j 7)] 2
3− 2
0
= 0
-2
2k + 1
φ =180o ,k = 0,1,...,( n −− M 1)
A n− M -4
2k + 1 -6
φ =180,o k = 0
A 1
-8
o
= 180 -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0
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The Root Locus Procedure (21)
Ex. 4
K( s2 + 16 s + 113)
Given the characteristic equation 1 + = 0 . Find K so that:
s2 ( s + 16)
1. The damping ratio ζ = 0.5.
2. The settling time is less than 2 seconds.
3
s 1 16 8
s2 0 0
θ
1 6 1
s b1
0 4
s c1
2 θ θ
−1 1 16 3, 4
b = θ
1 5
0 0 0 0
3 2
s+16 s -2
K= − = p( s )
s2 +16 s + 113
-4
dp =
1 -6 θ
ds 2
θ+ θ + θ + θ + θ = o -8
1 2 3 4 5 180
→θ = o -20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0
3 90 sites.google.com/site/ncpdhbkhn 28
The Root Locus Procedure (22)
Ex. 4
K( s2 + 16 s + 113)
Given the characteristic equation 1 + = 0 . Find K so that:
s2 ( s + 16)
1. The damping ratio ζ = 0.5.
2. The settling time is less than 2 seconds.
2 2
+ζω + ω =
s2n s n 0
s
8 1
2
→=−ζω ± ω ζ − s
s1,2 n j n 1 2 s *
6 s
3
4
Im{s } ω ζ 2 −1 1
k =1 =n =−1
ζω ζ 2 2
Re{s1 } n
0
ζ =0.5 →k = 1.73
-2
→s* =− 4.5 + j 8.0 -4
-6
4 4
→T = == 0.89 second -8
settling ζω
n 4.5
-20 -18 -16 -14 -12 -10 -8 -6 -4 -2 0
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus
Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 30
Parameter Design by the Root
Locus Method (1)
1+KP ( s ) = 0, 0 ≤≤∞ K
n+ n −1 +++=
asn as n −1... asa 1 0 0
a s
→1 +1 = 0
n+ n −1 ++ 2 +
asn as n −1... asa 2 0
s3+ s 2 +β s + α = 0
α
β=0 →s3 + s 2 + α = 01 → + = 0 → α *
2 +
s( s 1)
β
α*32→ + + βα + * → +s = → β
s s s 1 0
s3+ s 2 + α *
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Parameter Design by the Root
Ex. Locus Method (2)
Design the system to satisfy the following
R( s ) K1 Y( s )
specifications: G( s ) =
s( s + 2)
1. Steady – state error for a ramp input is less (− ) (− )
than 35% of input slope.
=
2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2
2. The settling time is less than 3 seconds.
R/ s 2 R/ s2 R s+( K K + 2)
E( s ) = = = 1 2
G( s ) K s 2 + + +
1+ 1 s( KK1 2 2) sK 1
1+ G () s H () s s( s + 2)
1+
K
1+ 1 K s
s( s + 2) 2
R + + +
= = =sKK(12 2) = KK 12 2
esteady state lim e () t lim sE () s lim s R
t→∞ s → 0 s→0 2 + + +
ss( KK1 2 2) sK 1 K 1
e K K + 2
ss ≤ 0.35 →1 2 ≤ 0.35
R K1
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Parameter Design by the Root
Ex. Locus Method (3)
Design the system to satisfy the following
R( s ) K1 Y( s )
specifications: G( s ) =
s( s + 2)
1. Steady – state error for a ramp input is less (− ) (− )
than 35% of input slope.
=
2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2
2. The settling time is less than 3 seconds.
2+ζωω +=→=−± 2 ζωωζ 2 −
ss2nn 0 s1,2 nn j 1
ζ = 0.707
Im{s } ω ζ 2 −1 1
k =1 =n =−1
ζω ζ 2
Re{s1 } n
1.33
ζ ≥0.707 →k ≤ 1
4
T = ≤→=3σ ζω ≥ 1.33
settlingζω n
n
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Parameter Design by the Root
Ex. Locus Method (4)
Design the system to satisfy the following
R( s ) K1 Y( s )
specifications: G( s ) =
s( s + 2)
1. Steady – state error for a ramp input is less (− ) (− )
than 35% of input slope.
=
2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2
2. The settling time is less than 3 seconds.
s2 +( KK ++= 2) sK 0
1 2 1 5 5
2 s s
→s +2 s +β s += α 0 4 1 4 1
s s
2 2
β=→0s2 + 2 s + α = 0 3 3
s *
α 2 2
→1 + = 0 1 1
s( s + 2) ζ = 0.707
0 0
α= = →+++=2 β
20K1 s 2 s s 200 -1 -1
β s -2 -2
→1 + = 0
s2 +2 s + 20 -3 -3
= − + →β = = -4 -4
s* 3.15 j 3.15 4.3 K1 K 2
-5 -5
→ = -6 -4 -2 0 -6 -4 -2 0
K2 0.215 sites.google.com/site/ncpdhbkhn 34
Parameter Design by the Root
Ex. Locus Method (5)
Design the system to satisfy the following
R( s ) K1 Y( s )
specifications: G( s ) =
s( s + 2)
1. Steady – state error for a ramp input is less (− ) (− )
than 35% of input slope.
=
2. The damping ratio ζ ≥ 0.707. Hs( ) Ks2
2. The settling time is less than 3 seconds.
= = 5
K120; K 2 0.215
s
4 1
s
2
× + 3
20 0.215 2 s *
e = =0.315 ≤ 0.35 2
ss 20
1
ζ = 0.707
0
ζ =
0.707 -1
-2
4 4 -3
T == =1.27 < 3 seconds
settling σ 3.15 -4
-5
-6 -4 -2 0
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 36
Sensitivity and the Root Locus
(1)
∂lnT ∂ T / T
The logarithmic sensitivity: ST = =
K ∂lnK ∂ K / K
∂r ∂ r ∆ r
The root sensitivity: S ri =i = i ≈ i
K ∂lnK ∂ KK / ∆ KK /
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Sensitivity and the Root Locus
Ex. (2) R( s ) K Y( s )
G( s ) =
K 2 2 + β
1+ =↔++==++ 0sβ sK 0 s β s α (− ) s( s )
s( s + β )
αα= ±∆ αββ, = ±∆ β
0 0
0.8 s
β =→+K = 1
0 1 1 0 s
s( s + 1) 2
0.6 0.1 =
r1 K 0.6
= → =− ± =
K0.5 r1,2 0.5 j 0.5 r1 K 0.5
0.4 r−0.1 K = 0.4
= →=−±0.1 1
K0.6 (20%) r1,2 0.5 j 0.59
0.2
= →−0.1 =−±
K0.4(20%) r1,2 0.5 j 0.39
0
∆ −+ −−+
r1 r1 ( 0.5 j 0.59) ( 0.5 j 0.5)
S + = =
K ∆K/ K 0.1/0.5 -0.2
= = ∠ o −0.1 =
j0.45 0.45 90 -0.4 r2 K 0.4
r2 K = 0.5
∆r( −+ 0.5 j 0.39) −−+ ( 0.5 j 0.5) -0.6 0.1 =
r1 1 r K 0.6
S − = = 2
K ∆K/ K 0.1/ 0.5
-0.8
o
=−j0.55 = 0.55 ∠− ( 90 )
-1 -0.8 -0.6 -0.4 -0.2 0
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Sensitivity and the Root Locus
Ex. (3) R( s ) K Y( s )
G( s ) =
K 2 2 + β
1+ =↔++==++ 0sβ sK 0 s β s α (− ) s( s )
s( s + β )
αα= ±∆ αββ = ±∆ β
0, 0 1
2 (∆β ) s +∆β
β=→1s + (1 +∆ β ) s + α = 0 →1 + = 0 -∆β
0 s2 + s + α 0.8
2 ∆β = − 0.2
−+∆(1β ) ± (1 +∆ β ) − 4 α 0.6
r =
1,2 2 ∆β = 0
0.4 ∆β = +
∆=+β → =−± 0.2
0.2(20%)r1,2 0.6 j 0.37
∆=−β → =−± 0.2
0.2(20%)r1,2 0.4 j 0.58
∆ −+ −−+ 0
r1 r1 ( 0.6 j 0.37) ( 0.5 j 0.5)
Sβ + = =
∆β/ β 0.2/1 -0.2
o
=0.82 ∠ ( − 127.6 ) -0.4
∆r( −+ 0.4 j 0.58) −−+ ( 0.5 j 0.55) -0.6
S r1 =1 =
β − ∆β β
/ 0.2 /1 -0.8
o
=0.64 ∠ 38.7 -1
-1 -0.8 -0.6 -0.4 -0.2 0
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Sensitivity and the Root Locus
Ex. (4) R( s ) K Y( s )
G( s ) =
+ β
(− ) s( s )
r1 = = ∠ o
SK+ j 0.45 0.45 90
r1 =− = ∠− o
SK− j 0.55 0.55 ( 90 )
r1 o
Sβ + =0.82 ∠ ( − 127.6 )
r1 o
Sβ − =0.64 ∠ 38.7
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 41
PID Controllers (1)
K K s2 + K s + K
PIDcontroller:GsK ( ) = +I + Ks = DPI
c Ps D s
K K s+ K
PIcontroller:G ( s ) = K +I = P I
c P s s
= +
PDcontroller:Gsc ( ) K P Ks D
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PID Controllers (2)
Ex. 1
2 R( s ) s2 +6 s + 10 1 Y( s )
s+6 s + 10 1 K
K D (s+ 2)( s + 4)
D s( s+ 2)( s + 4) (− ) s
T( s ) =
s2 +6 s + 10 1
1+ K
D s( s+ 2)( s + 4)
K( s2 + 6 s + 10)
= D →+sK3( + 6) s 2 + (6 K ++ 8) sK 10 = 0
3+ + 2 + ++ D D D
sK(D 6) s (6 K D 8) sK 10 D
2
s
1
s
1.5 2
s
3
1
0.5
0
-0.5
-1
-1.5
-2
-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0
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PID Controllers (3)
1 −ζω −
=−±−→=ζωωζ2 nt ωζ −+2 1 ζ
s1,2 n j n 1 yt () e sin(1n t cos)
y( t ) 1 −ζ 2
1.6
M pt M− final value
Percent overshoot = pt 100%
1.4 final value
Overshoot 2
= −ζπ/ 1 − ζ
1.2 100 e
1.0 + δ
1
0.9 1.0 −δ
0.8
0.6 π
Peak time T =
p 2
0.4 ω1− ζ
n 4
Rise time T Settling time T =
r s ζω
0.2 n
Tr1
0.1 t
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
PID Gain Percent Overshoot Settling Time Steady – State Error
K
Increasing P Increases Minimal impact Decreases
K
Increasing I Increases Increases Zero steady – state error
K
Increasing D Decreases Decreases No impact
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PID Controllers (4)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Step response with K = 885.5, K = 0, and K = 0
P I D
2
1.5
1
Amplitude 0.5
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Root locus with K = 0, and K = 0
I D
10
=
s KP 885.5
5 1
s
2
0
s
3
-5
-10
-15 -10 -5 0
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PID Controllers (4)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Step response with K = 0, and K = 0
I D
2
K = 885.5
1.8 P
K = 442.75
P
1.6 K = 370
P
1.4
1.2
1
Amplitude 0.8
0.6
0.4
0.2
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
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PID Controllers (5)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Root locus with K = 0, and K = 370
I P
25
s
1
20
s
2
s
15 3
10
= −ζπ/ 1 − ζ 2
Percent overshoot100 e 5
4 0
Settling Time =
ζω
n -5
-10
-15
-20
-25
-14 PID-12 Gain-10 Percent-8 Overshoot-6 Settling-4 Time -2 0
K
Increasing D Decreases Decreases
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PID Controllers (6)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
60
40
20
Percent Overshoot
0
= −ζπ/ 1 − ζ 2 0 10 20 30 40 50
Percent overshoot100 e K
D
4 4
Settling Time =
ζω
n 3
2
Settling Time
1
0 10 20 30 40 50
K
D
PID Gain Percent Overshoot Settling Time
K
Increasing D Decreases Decreases
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PID Controllers (7)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Root locus with K = 370, K = 0
P D
100
80
60
)
-1 40
−ζπ − ζ 2 20
Percent overshoot=100 e / 1
0
4 -20
Settling Time =
ζω
n -40
Imaginary Axis ImaginaryAxis (seconds
-60
-80
-100
-120 -100 -80 -60 -40 -20 0 20 40 60
Real Axis (seconds -1 )
PID Gain Percent Overshoot Settling Time
K
Increasing I Increases Increases
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PID Controllers (8)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
90
80
70
60
Percent Overshoot
50
−ζπ − ζ 2
Percent overshoot=100 e / 1 0 100 200 300 400 500 600
K
I
4 20
Settling Time =
ζω
n 15
10
Settling Time 5
0
0 100 200 300 400 500 600
K
I
PID Gain Percent Overshoot Settling Time
K
Increasing I Increases Increases
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PID Controllers (9)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
4 60
3
40
2
20
Settling Time 1
Percent Overshoot
0 0
0 20 40 60 0 20 40 60
K K
D D
20 90
15 80
10 70
Settling Time 5 60
Percent Overshoot
0 50
0 200 400 600 0 200 400 600
K K
I sites.google.com/site/ncpdhbkhn I 51
PID Controllers (9)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Step response with K =370, K = 100, and K = 60
P I D
1.4
1.2
1
0.8
Amplitude 0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8 9 10
Time (s)
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PID Controllers (10)
Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts
Controller Type KP KI KD
P 0.5 KUltimate
PI 0.45 KUltimate 0.54 KUltimate /TUltimate
PID 0.6 KUltimate 1.2 KUltimate /TUltimate 0.6 KUltimate TUltimate /8
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PID Controllers (11)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
Step response with K = 885.5, K = 0, and K = 0
P I D
2
T = 0.83s
1.5
1
Amplitude 0.5
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
Root locus with K = 0, and K = 0
I D
10
=
s KP 885.5
5 1
s
2
0
s
3
-5
-10
-15 -10 -5 0
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PID Controllers (12)
Ex. 2
R( s ) 1 Y( s )
+KI +
KP K D s + + ζω
− s ss( b )( s 2n )
K=885.5; T = 0.83 ( )
U U β= ζ = ω =
10, 0.707,n 4.
Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts
Controller Type KP KI KD
P 0.5 KUltimate
PI 0.45 KUltimate 0.54 KUltimate /TUltimate
PID 0.6 KUltimate 1.2 KUltimate /TUltimate 0.6 KUltimate TUltimate /8
= =× =
KP0.6 K U 0.6 885.5 531.3
=KU =885.5 =
KI 1.2 1.2 1280.2
TU 0.83
K T 885.5× 0.83
K =0.6U U = 0.6 = 55.1
D 8 8
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PID Controllers (13)
Ex. 2
R( s ) K 1 Y( s )
K+I + K s
P D ss(+ b )( s + 2ζω )
(− ) s n
β= ζ = ω =
10, 0.707,n 4.
1.6
Step response with the manual tuning
1.4 Step response with the Ziegler - Nichols PID tuning
1.2
1
0.8
Amplitude
0.6
0.4
0.2
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
Time (s)
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The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 57
Negative Gain Root Locus
Ex.
s − 20
1+K = 0
s2 +5 s − 50
) Root Locus
-1
20
10
0
-10
-20
-20 -10 0 10 20 30 40 50
Imaginary Axis Axis Imaginary (seconds
Real Axis (seconds -1 )
) Root Locus
-1
20
10 K = − 5.0 K = − 2.5
0
-10
-20
-20 -10 0 10 20 30 40 50
Imaginary Axis Imaginary Axis (seconds
Real Axis (seconds -1 )
sites.google.com/site/ncpdhbkhn 58
The Root Locus Method
1. The Root Locus Concept
2. The Root Locus Procedure
3. Parameter Design by the Root Locus Method
4. Sensitivity and the Root Locus
5. PID Controllers
6. Negative Gain Root Locus
7. The Root Locus Using Control Design
Software
sites.google.com/site/ncpdhbkhn 59
The Root Locus Using Control
Ex. 1 Design Software (1)
−s + 20
1+K = 0
s2 +5 s − 50
• rlocus
• rlocfind
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The Root Locus Using Control
Ex. 1 Design Software (2)
s + 20
G( s )= = 0
s2 +5 s + 20
• step
• impulse
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