Nguyễn Công Phương
CONTROL SYSTEM DESIGN
The Root Locus Method
Contents
Introduction
I. 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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2
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
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3
The Root Locus Concept (1)
( )R s
( )Y s
( )G s
K
=
=
T s ( )
( )-
Y s ( ) R s ( )
KG s ( ) + KG s ( )
1
+
1
KG s
= ( ) 0
= - ( ) KG s
1
o
fi
KG s ( )
= — ( )] 1 180 KG s
[
=
KG s ( )
1
fi —
=
KG s
+ o ( )] 180
[
k
o 360 ;
= 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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4
fi — – –
The Root Locus Concept (2)
Ex. 1
( )R s
( )Y s
K
1 s s + (
2)
( )-
+
= +
=
1
KG s
( ) 1
0
K +
s s (
2)
2
2
+
=
+
=
= ( ) s
s
+ 2
+ s K s
zw 2
w s
0
n
2 n
2
zw
z
fi D
1
w z 1
1
= - s 1,2
w n
- = - 2 n
n
fi – – -
Ex. 2
( )R s
( )Y s
10
)
1 s s a+ (
( )-
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5
The Root Locus Concept (3)
N
= - ( ) 1 s
...
+ L n
L L n m
+ L L L n m p
∑
∑
∑
= 1
n
,
n m , nontouching
n m p , nontouching
= + 1
( )F s
D -
= fi s ( ) 0
= - ( ) F s
1
+
3
=
F s ( )
+ s +
+ +
K s ( + s (
)( z 1 s )(
z p
+ s )( 2 + s )(
)...( s s )...(
) )
2
p 1
z p 3
z M p n
+
2
=
=
F s ( )
1
+ s +
D
K s + s
z p
... ...
2
o
o
+
+
=
fi
F s ( )
= — + ( [
s
(
s
z
)
— + ...]
[
)
(
s
)
+ ...] 180
k
360
—
z 1 p s 1 + — + z ) 1
2
+ — + ( p s 1
p 2
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6
-
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
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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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8
£ £ ¥
The Root Locus Procedure (2) Step 1 = + ( ) 0 F s 1 = ( ) KP s
0, 0
K
fi + 1 M
+
(
s
) z i =
fi + 1
K
0
= 1 i n
+
(
s
p
)
(cid:213)
j
= 1
j
M
n
+
=
(cid:213)
+ ( s
+ ( s
K
(
s
p
)
(
s
) 0
+ ) p j
= « z ) 0 i
j
z i
1 K
= 1
i
= 1
j
n + = 1
j
M + = 1
i
n
=
fi (cid:213) (cid:213) (cid:213) (cid:213)
K
0
+ ( s
p
= ) 0 j
= 1
j
M
+
fi (cid:213)
K
= s
(
z
) 0
j
= 1
j
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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9
fi ¥ fi (cid:213)
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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10
The Root Locus Procedure (4) Step 2
Ex. 1
+
=
1
K
0
+ +
2( s s s (
2) 4)
2
s + 1
1s
0
4-
2-
2s
1s
4
s + 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.
- -
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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11
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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12
M
The Root Locus Procedure (6) Step 3 +
(
s
)
z i
+
= +
=
(cid:213)
1
KP s
( ) 1
K
0, 0
K
= i 1 n
+
(
s
p
)
£ £ ¥
j
= 1
j
n
M
(cid:213)
(
(
p
)
)
j
z i
∑
∑
poles of
P s ( )
zeros of
P s ( )
∑
= 1
j
= 1
s
=
=
A
- - - -
∑ n M
i n M
+
f
=
=
- -
o 180 ,
k
0,1, 2,..., (
n M
1)
A
2 k 1 n M
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13
- - -
The Root Locus Procedure (7) Step 3
Ex. 2
+
=
+
0
1
K
2
( +
s 2)(
1) + s
4)
poles of
P s ( )
zeros of
P s ( )
( s s ∑
s
=
A
-
∑ n M
-
[( 2)
- + - + - ( 4)
( 4)]
( 1)
=
= -
3
- -
4 1
0
+
4-
2-
1-
f
=
=
-
o 180 ,
k
0,1, 2,..., (
n M
1)
A
k 1 2 n M
+
o
=
=
+
=
180
(2
k
o 1)60 ,
k
0,1, 2.
- - -
k 2 1 4 1
o
k
60
o
k
= fi 0 = fi 1
180
o
k
= fi 2
300
f = A = f A = f A
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14
-
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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15
D
The Root Locus Procedure (9) Step 5 = = fi ( ) ( ) 0 p s s
K
=
breakaway point :
0
dp s ( ) ds
Ex. 3
+
+
=
1
K
0
+
s ( + 2)(
1) s
3)
s s (
+ s s (
+ s
3) =
= K
p s ( )
2)( + 1
s
+ 3
- fi
+ s s (
+ s
2
s
+ s
6
3) =
10 2
+ 2 +
dp = ds
d dt
2)( + 1
s
s 8 s (
1)
= -
- fi
= fi 0
+ 3 s
2
+ 2 s
8
10
+ = fi s
6 0
= - 2.46 ,
0.77
j
0.79
s 1
s 2,3
dp ds
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16
–
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
q
1
q
3
0
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).
q
2
o
+ q
q
+
=
q o
q
180
q 180
+ (
)
1
q 2
3
= 1
2
3
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18
fi -
The Root Locus Procedure (12)
5
1p
=
4
Ex. 2 +
0
1
4
3
2
+
+
s
12
s
128
s
K + 64
3
0
fi + 1
+
s K + + 4
s
j
4)(
s
+ - 4
= 4)
j
2
( s s = - + 4
4)( 4
j
= -
1
j
4
p 1 p
4
2
3p
4p
= -
p
4
3
- 0
0
-1
= p 4 (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
1. 2.
-2
3.
-3
2p
4.
-4
5.
-5 -7 -6 -5 -2 -1 0 1 -4 -3
poles of
P s ( )
zeros of
P s ( )
∑
s
=
6.
A
-
∑ n M
-
4 4
4 4
j
4
=
= -
3
7.
Prepare the root locus sketch. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. Determine the points at which the locus crosses the imaginary axis (if it does so). Determine the breakaway point on the real axis (if any). Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. Complete the root locus sketch.
- - -
- + j 4 0
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19
-
The Root Locus Procedure (13)
5
1p
=
4
Ex. 2 +
0
1
4
3
2
+
+
s
12
s
s
128
s
K + 64
3
+
=
f
=
o 180 ,
k
0,1, 2,..., (
n M
1)
A
2 k 1 n M
2 - - - 1
+
=
=
o 180 ,
k
0,1, 2, 3
3p
4p
0
k 2 1 4 0
- -1
1. 2.
-2
(cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
3.
-3
2p
4.
-4
5.
o
45
k
-5 -7 -3 -2 -1 0 1 -6
6.
o
135
k
o 225
k
= fi 1 = fi 2
7.
Prepare the root locus sketch. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. Determine the points at which the locus crosses the imaginary axis (if it does so). Determine the breakaway point on the real axis (if any). Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. Complete the root locus sketch.
o
315
k
= fi 3
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20
-5 = fi 0
-4 f = A = f A = f A = f A
The Root Locus Procedure (14)
5
1p
4
Ex. 2 +
=
1
0
4
3
2
+
+
s
12
s
s
128
s
K + 64
3
2
1
3p
4p
64 128 K 0
K 0 0 0
0
1 12 1b 1c K
-1
1. 2.
-2
4s 3s 2s 1s 0s (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
3.
-3
2p
(cid:1)(cid:1)(cid:1)(cid:1)
4.
-4
5.
-5 -7 -6 -5 -4 -3 -2 -1 0 1
K
=
=
=
53.33;
c 1
b 1
6.
53.33 128 12 53.33
568.89
> fi 0
12 64 128 12 < K
c 1
2
7.
Prepare the root locus sketch. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. Determine the points at which the locus crosses the imaginary axis (if it does so). Determine the breakaway point on the real axis (if any). Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. Complete the root locus sketch.
+
568.89
= fi 0
3.27
s 53.33
j
= – s 1,2
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21
· - · -
The Root Locus Procedure (15)
q
1
5
4
Ex. 2 +
=
1
0
4
3
2
1p
+
+
s
12
s
s
128
s
K + 64
3
4
3
2
= - K
+ ( s
+ 12 s
+ 64 s
= 128 ) s
p s ( )
2 fi
3
2
+ (4 s
+ 36 s
+ 128 s
128)
q
q
4
1 fi
3p
3 4p
0
= fi 0
1.58 ;
3.71 2.55
= - s 1
= - s 2,3
– -1
1. 2.
-2
q
dp = - ds dp ds (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
3.
-3
2 2p
4.
-4
5.
o
+ q
q
+
+
=
180
1
q 2
4
-5 -6 -5 -4 -3 -2 -1 0 1
(cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
6.
o
+
-7 q 3
180
+ q (
)
= q 1
q 2
q 3
4
o
o
Prepare the root locus sketch. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. Determine the points at which the locus crosses the imaginary axis (if it does so). Determine the breakaway point on the real axis (if any). Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. Complete the root locus sketch.
7.
=
+ o
fi -
(90
+ o 135
= - o 90 )
= o 135
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225 22
-
The Root Locus Procedure (16)
5
4
Ex. 2 +
=
1
0
4
3
2
1p
+
+
s
12
s
s
128
s
K + 64
3
2
1
3p
4p
0
-1
1. 2.
-2
(cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
3.
-3
2p
4.
-4
5.
(cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1) (cid:1)(cid:1)(cid:1)(cid:1)
6.
7.
Prepare the root locus sketch. Locate the open – loop poles and zeros of P(s) in the s – plane with selected symbols. The loci proceed to the zeros at infinity along asymptotes centered at σA and with angle ϕA. Determine the points at which the locus crosses the imaginary axis (if it does so). Determine the breakaway point on the real axis (if any). Determine the angle of locus departure from complex poles and the angle of locus arrival at complex zeros, using the phase criterion. Complete the root locus sketch.
(cid:1)(cid:1)(cid:1)(cid:1)
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23
-5 -7 -6 -5 -4 -3 -2 -1 0 1
The Root Locus Procedure (17)
Ex. 2 +
=
fi + 1
0
1
0
4
3
2
+
+
s s (
4)(
s
K + + 4
= 4)
j
j
4)(
s
s
12
s
K + 64
s
128
s
+ 5
+ - 4
4
1p
3
2
1
0
3p
4p
s 1 s 2 s 3 s 4
-1
-2
-3
-4
2p
-5
-7
-6
-5
-3
-2
-4
0
1
-1 sites.google.com/site/ncpdhbkhn
24
The Root Locus Procedure (18)
Ex. 3
=
+
0.
1
2
3
4
+
+
K Given the characteristic Find K so that s = –1 + j2? + s 64
128
12
s
s
s 5
j
4
1
+ + s 1 4
+
s s (
4)(
s
j
4)(
s
+ - 4
K + + 4
= - 4)
j
fi
1p
4
+ = K s s
+ + 4 s
4
j
4
+ - s
4
j
4
3 fi
2
+
=
2
2 3
3.61
s + = 1 4
1s
2
1
4
s + 1
1 s
2
+ +
=
+
=
j
4
2 3
2 6
6.71
s 1 4
3
3p
4p
j
4
s 0 s
+ - 1 4 s
4
+ 2
= 4
j
2
= 2 3
3.61
+ - 1 4 s
s -1
+
=
2 1
2 2
2.24
-2
s = 1
-3
3.61 6.71 3.61 2.24
K =- + 1 s
-4 fi · · ·
= 2 j =
2p -3
195
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25
-5 -7 -6 -5 -4 -2 -1 0 1 2 4 3
The Root Locus Procedure (19)
Ex. 4
2
+
K s (
113)
=
+ Given the characteristic equation . Find K so that:
0
1
+ 2
s
(
s 16 + s 16)
1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds.
zw
2
+
+
=
s
zw 2
w s
0;
= fi ( ) 1 r t
= - ( ) 1
y t
e
q wb nt sin(
t
)
n
2 n
+ n
1 b
1.8
1.6
1.4
1.2
z = 0.1 z = 0.2 z = 0.5 z = 0.7 z = 1.1 z = 2.0
1
0.8
0.6
0.4
0.2
0
-0.2
0
0.5
1.5
2
2.5
3
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26
-
The Root Locus Procedure (20)
Ex. 4
2
+
K s (
113)
=
+ Given the characteristic equation . Find K so that:
1
0
+ 2
s
(
s 16 + s 16)
1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds.
K s (
+ - 8
j
7)
+
=
1
0
8
+ + 8 2 s
j 7)( + s (
s 16)
6
poles of
P s ( )
zeros of
P s ( )
∑
s
=
A
4
-
-
2
(0 16)
[( 8
( 8
j
7)]
=
- - - -
∑ n M + - + j 7) 3 2
0
-2
+
f
=
=
-
o 180 ,
k
0,1,..., (
n M
1)
A
-4
0= k 1 2 n M +
-6
1
2
f
=
=
o 180 ,
k
0
A
k 1
-8
o
=
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
180
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27
- - -
The Root Locus Procedure (21)
Ex. 4
2
+
K s (
113)
=
+ Given the characteristic equation . Find K so that:
1
0
+ 2
s
(
s 16 + s 16)
1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds.
8
16 0
q
6
1
4
3s 2s 1s 0s
2
4,q q
3
1 0 1b 1c 1 16
=
q
b 1
5
0
0
1 0
0
3
2
+
-2
= -
K
p s ( )
2
s +
s
16
16 s = + 113 s
-4
=
1
-6
q
2
o
-8
dp ds q
q +
+
q +
+
1
q 2
q 3
4
= 5 180
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
-
o 90
q = 3
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28
fi
The Root Locus Procedure (22)
Ex. 4
2
+
K s (
113)
=
+ Given the characteristic equation . Find K so that:
1
0
+ 2
s
(
s 16 + s 16)
1. The damping ratio ζ = 0.5. 2. The settling time is less than 2 seconds.
2
+
+
=
s
zw 2
w s
0
n
2 n
s
8
1
2
s
zw
w
2
z j
1
*s
= - s 1,2
n
n
6
s
3
4
2
w
fi – -
1
n
=
=
=
k
1
1 2
2
z zw
z
s Im{ } 1 s Re{ } 1
n
0
z =
0.5
fi = k
1.73
-2
-4
- -
= - * s
+ 4.5
j 8.0
-6
4 =
fi
-8
T
0.89 second
= settling
zw
4 = 4.5
n
-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
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29
fi
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
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30
+
£ £ ¥
Parameter Design by the Root Locus Method (1) = 0, 0
KP s ( )
K
1
n
n
1
+
=
a
s
+ + ...
0
a s n
n
1
+ a s a 1
0
fi + 1
0
- -
n
n
2
+
+
a
s
a s 1 + + 1 ...
= a
a s n
n
1
a s 2
0
3
2
+
+
+ a
=
s
s
sb
0
a
*
a
a
+ 3 s
+ = fi + 2 s 0
1
0
2
s
(
s
*
b
fi + *
- -
+ 3 s
+ 2 s
+ a s
1
b 0
3
+
= fi + 1) b s = fi + a * 2
= fi b 0 a
s
s
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31
fi
Parameter Design by the Root Locus Method (2)
( )R s
( )Y s
=
G s ( )
K 1 + s s (
2)
( )-
( )-
=
( )H s
K s 2
Ex. Design the system to satisfy the following specifications: 1. Steady – state error for a ramp input is less than 35% of input slope. 2. The damping ratio ζ ≥ 0.707. 2. The settling time is less than 3 seconds.
2
+
+
=
=
=
E s ( )
2
R s s
s + (
2)
( K K 1 2 + K K 1
2
2) + s K 1
+
1
+
2)
2/ R s ( ) G s G s H s ( ) ( )
1
+
1
+
1
K s 2
R s / K 1 + s s ( K 1 + s s (
2)
+
+
+
2
=
=
=
=
s
R
e
steady state
2
e t lim ( ) t
sE s lim ( ) s
0
lim s 0
R s
K K 2 1 K
s + (
2)
s
1
K K ( 2 1 + K K 1
2
2) + s K 1
+
2
fi ¥ fi fi
0.35
0.35
sse R
K K 1 2 K 1
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32
£ fi £
Parameter Design by the Root Locus Method (3)
( )R s
( )Y s
=
G s ( )
K 1 + s s (
2)
( )-
( )-
=
( )H s
K s 2
Ex. Design the system to satisfy the following specifications: 1. Steady – state error for a ramp input is less than 35% of input slope. 2. The damping ratio ζ ≥ 0.707. 2. The settling time is less than 3 seconds.
2
2
+
+
w
s
zw 2
w s
= fi 0
j
1
n
2 n
= - zw s 1,2
z n
n
z =
0.707
2
w
– -
1
n
=
=
=
k
1
1 2
z zw
z
s Im{ } 1 s Re{ } 1
n
1.33
z ‡
- -
0.707
1k
4
=
=
s
zw
fi £
T
3
1.33
settling
n
zw
n
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33
£ fi ‡
Parameter Design by the Root Locus Method (4)
( )R s
( )Y s
=
G s ( )
K 1 + s s (
2)
( )-
( )-
=
( )H s
K s 2
2
+
+
=
0
(
s
2
5 5 s
Ex. Design the system to satisfy the following specifications: 1. Steady – state error for a ramp input is less than 35% of input slope. 2. The damping ratio ζ ≥ 0.707. 2. The settling time is less than 3 seconds. + s K 1 a =
1
K K 1 + 2 s
2) sb +
+ 2 s
0
2
b
= fi 0
+ = a 2 s
0
*s
fi 4 4 s s 1 s 2 3 3
fi + 1
0
2 2
z =
+ 2 s a +
0.707
= 2)
s s (
1 1
a
=
=
b
0 0
20
+ 2 s
+ 2 s
+ s
= 20 0
fi -1 -1
K 1 b
fi + 1
0
2
-2 -2
+
2
-3 -3
b =
= -
= 4.3 K K
s *
s s j
= 20 3.15
+ s + 3.15
1
2
-4 -4 fi
0.215
34
= K 2
-5 -4 -2 -6 0 -4 -2 0 fi -5 -6 sites.google.com/site/ncpdhbkhn
Parameter Design by the Root Locus Method (5)
( )R s
( )Y s
=
G s ( )
K 1 + s s (
2)
( )-
( )-
=
( )H s
K s 2
Ex. Design the system to satisfy the following specifications: 1. Steady – state error for a ramp input is less than 35% of input slope. 2. The damping ratio ζ ≥ 0.707. 2. The settling time is less than 3 seconds.
=
=
20;
K
0.215
K 1
2
5
+
*s
=
=
0.315 0.35
sse
20 0.215 2 20
4 s 1 s 2 3 · £ 2
z =
0.707
1
z =
0.707
0
-1
-2
=
=
=
<
1.27 3 seconds
T
-3
settling
4 s
4 3.15
-4
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-5 -6 -4 -2 0
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
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36
Sensitivity and the Root Locus (1)
=
=
S The logarithmic sensitivity:
T K
¶ ¶
ln ln
T K
T T / K K /
¶ ¶
=
=
S The root sensitivity:
ir K
¶ ¶ D »
r i ln
K
r i K K /
r i K K /
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37
¶ ¶ D
( )R s
( )Y s
=
G s ( )
K +
b
s s (
)
Sensitivity and the Root Locus (2) a b + s
b + s K
= = 0
= « 0
+ 2 s
+ 2 s
Ex. + 1
( )-
b
K +
s s (
a
= a
a
) b
b
=
b
,
0
0
0.8
b = fi + 1 1
0
0
s 1 s 2
K +
= 1)
0.6
=
– D – D
0.5
0.5
j
0.5
K
0.1 1r 1r
s s ( = - r 1,2
0.1
0.4
K = K = K =
0.6 0.5 0.4
1r -
=
fi –
K
0.6 (20%)
0.5
j
0.59
= - 0.1 r 1,2
0.2
fi –
=
= - 0.1
K
0.4 (20%)
0.5
j
0.39
0
- fi –
r 1,2 + ( 0.5
j
0.59)
+ ( 0.5
j
0.5)
=
=
S
+
r 1 K
-0.2
D - - -
r 1 K K /
0.1/ 0.5
o
=
=
D
-0.4
j
0.45 0.45 90
2r -
0.1 2r
—
-0.6
+ ( 0.5
j
0.39)
+ ( 0.5
j
0.5)
K = K = K =
0.4 0.5 0.6
0.1 2r
=
=
S
D - - -
r 1 K
r 1 K K /
0.1/ 0.5
-0.8
= -
=
- D
o 0.55 0.55 ( 90 )
j
-1
-0.8
-0.6
-0.4
0
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-0.2 38
— -
( )R s
( )Y s
=
G s ( )
K +
b
s s (
)
Sensitivity and the Root Locus (3) a b + s
b + s K
= = 0
= « 0
+ 2 s
+ 2 s
Ex. + 1
( )-
K +
b
s s (
= a
a
a
) b
b
=
b
,
1
– D – D
b
b
+D -D
0
0 = fi 1
+ + D 2 s (1
0 + = a ) s
fi + 1
0
0.8
0
( 2
b ) s = a + + s
s
bD = -
0.2
2
b
+ D
b
a
b D b
0.6
- + D (1
)
)
4
=
r 1,2
bD =
0
bD = +
0.4
0.2
bD = +
– -
0.2 (20%)
0.6
j
0.37
0.2
bD = -
= -
fi –
0.2 (20%)
0.4
0.58
(1 2 = - r 1,2 r 1,2
0
fi –
j + ( 0.5
+ ( 0.6
j
0.37)
j
0.5)
=
=
r Sb 1
+
-0.2
D - - -
r 1 b b /
0.2 /1
-0.4
=
D
o 0.82 ( 127.6 )
-0.6
— -
+ ( 0.4
j
0.58)
+ ( 0.5
j
0.55)
=
=
r Sb 1
D - - -
r 1 b b /
0.2 /1
-0.8
o
=
- D
-1
0.64 38.7
-1
-0.8
-0.6
-0.4
-0.2
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0 39
—
Sensitivity and the Root Locus (4)
Ex.
( )R s
( )Y s
=
G s ( )
K +
b
s s (
)
( )-
o
=
=
S
j
0.45 0.45 90
+
= -
=
—
S
o 0.55 0.55 ( 90 )
j
r 1 K r 1 K
=
— - -
S
o 0.82 ( 127.6 )
r 1 b
+
o
=
— -
S
0.64 38.7
r 1 b
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40
— -
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
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41
PID Controllers (1)
2
+
K s D
P
I
=
+
+
=
PID controller:
K
G s ( ) c
P
K s D
K I s
+ K s K s
P
I
=
+
=
PI controller:
K
G s ( ) c
P
K I s
+ K s K s
=
+
PD controller:
K
G s ( ) c
P
K s D
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42
PID Controllers (2)
Ex. 1
2
( )R s
( )Y s
+
+
2
s
10
+
+
s
10
K
D
+
+
K
1 2)(
(
s
s
4)
D
s 6 s
+
+
( )-
1 2)(
4)
=
T s ( )
2
+
s ( 10
s
+
1
K
D
+
4)
2
+ +
=
+ 3 s
(
6)
+ 2 s
(6
+ 8) s
10
0
+ K D
+ K D
= K D
3
+ +
+
+
s 6 s + s 6 s K s ( D + 2 s 6)
( s s 6 K (6
s 1 2)( s 10) + 8)
s
10
K
s
(
K
D
D
D
2
s
1
s
1.5
2
s
3
1
0.5
0
-0.5
-1
-1.5
-2
-5
-4.5
-3.5
-2.5
-1.5
-2
-3
-1
-0.5
0
-4 sites.google.com/site/ncpdhbkhn
43
fi
PID Controllers (3)
zw
1
2
2
1
= -
zw
w
z
=
w nt
z
j
1
y t ( )
e
sin(
+ 1
z t
cos
)
s 1,2
n
n
n
2
( )y t
z
1
1.6
- - – - fi - -
ptM
final value
ptM
=
Percent overshoot
100%
1.4
final value
2
Overshoot
zp
-
z / 1
=
e 100
1.2
1.0 d+
1
0.9
1.0 d-
0.8
0.6
p
=
Peak time pT
2
w
z
- -
0.4
1n
4
=
Settling time sT
zw
n
0.2
Rise time rT 1rT
0.1
t
0
0
5
0.5
1
1.5
2
2.5
3
3.5
4
4.5
-
PID Gain
Percent Overshoot
Settling Time
Steady – State Error
Increases
Minimal impact
Decreases
Increasing KP
Increases
Increases
Zero steady – state error
Decreases
Decreases
No impact
Increasing KI Increasing KD
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44
PID Controllers (4)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
Step response with K
= 0
= 885.5, K I
= 0, and K D
P
2
1.5
e d u t i l
1
p m A
0.5
0
0
0.5
1
1.5
2
3
3.5
4
4.5
5
2.5 Time (s)
= 0
Root locus with K I
= 0, and K D
10
885.5
PK =
s
1
5
s
2
0
s
3
-5
-10
-15
-5
0
-10 sites.google.com/site/ncpdhbkhn
45
PID Controllers (4)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
= 0
Step response with K I
= 0, and K D
2
= 885.5
K
P
1.8
= 442.75
K
P
1.6
= 370
K
P
1.4
1.2
e d u
t i l
1
p m A
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
3
3.5
4
4.5
5
2.5 Time (s)
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46
PID Controllers (5)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
= 370
Root locus with K I
= 0, and K P
25
20
15
s 1 s 2 s 3
10
2
zp
z / 1
=
Percent overshoot
100
e
5
0
4
=
Settling Time
zw
-5
n
-10
-15
-20
-25
-14
-10
-4
-2
0
- -
-12 PID Gain
-8 -6 Percent Overshoot
Settling Time
Decreases
Decreases
Increasing KD
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47
PID Controllers (6)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
60
t
40
o o h s r e v O
t
20
n e c r e P
0
2
zp
10
20
30
40
50
0
z / 1
=
Percent overshoot
100
e
K
D
4
4
=
Settling Time
zw
n
i
3
e m T
g n
2
i l t t e S
1
0
10
20
30
40
50
K
D
- -
PID Gain
Percent Overshoot
Settling Time
Decreases
Decreases
Increasing KD
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48
PID Controllers (7)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
= 0
Root locus with K P
= 370, K D
100
80
60
)
1 -
40
20
2
zp
z / 1
=
Percent overshoot
100
e
i
0
-20
4
s d n o c e s ( s x A y r a n
=
i
Settling Time
zw
-40
n
g a m
I
-60
-80
-100
-120
-100
-80
-60
-40
-20
0
20
40
60
Real Axis (seconds-1)
- -
Percent Overshoot
Settling Time
Increases
Increases
PID Gain Increasing KI
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49
PID Controllers (8)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
90
t o o h s r e v O
80
70
t n e c r e P
60
2
zp
z / 1
=
Percent overshoot
100
e
I
50 - - 100 200 400 500 600 0 300 K
4
=
Settling Time
zw
20
n
i
e m T
15
g n
i l t t
e S
10
5
I
0 0 100 200 400 500 600 300 K
Percent Overshoot
Settling Time
Increases
Increases
PID Gain Increasing KI
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50
PID Controllers (9)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
4
60
3
i
40
e m T
t o o h s r e v O
2
g n
20
i l t t e S
1
t n e c r e P
0
0
0
20
40
60
20
40
60
0
K
K
D
D
20
90
15
80
i
t o o h s r e v O
10
70
e m T g n
i l t t e S
5
60
t n e c r e P
0
50
0
0
200
400
600
200
400
600
K
K
I
I
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51
PID Controllers (9)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
= 60
Step response with K P
=370, K I
= 100, and K D
1.4
1.2
1
0.8
e d u t i l
p m A
0.6
0.4
0.2
0
0
1
2
3
4
6
7
8
9
10
5 Time (s)
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PID Controllers (10)
Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts
Controller Type
KI
KD
P
PI
PID
KP 0.5KUltimate 0.45KUltimate 0.6KUltimate
0.54KUltimate/TUltimate 1.2KUltimate/TUltimate
0.6KUltimateTUltimate/8
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53
PID Controllers (11)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
Step response with K
= 0
= 885.5, K I
= 0, and K D
P
2
T =
0.83s
1.5
e d u t i l
1
p m A
0.5
0
0
0.5
1
1.5
2
3
3.5
4
4.5
5
2.5 Time (s)
= 0
Root locus with K I
= 0, and K D
10
885.5
PK =
s
1
5
s
2
0
s
3
-5
-10
-15
-5
0
-10 sites.google.com/site/ncpdhbkhn
54
PID Controllers (12)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
=
=
885.5;
0.83
K U
T U
b
=
=
w
)n =
z 10,
0.707,
4.
n
Ziegler-Nichols PID Controller Gain Tuning Using Closed-loop Concepts
Controller Type
KI
KD
P
PI
PID
KP 0.5KUltimate 0.45KUltimate 0.6KUltimate
0.54KUltimate/TUltimate 1.2KUltimate/TUltimate
0.6KUltimateTUltimate/8
=
=
K
0.6
= 0.6 885.5 531.3
P
K U
=
=
=
K
1.2
1.2
1280.2
I
885.5 0.83
K U T U
·
=
=
=
K
0.6
0.6
55.1
D
K T U U 8
885.5 0.83 8
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55
·
PID Controllers (13)
Ex. 2
( )R s
( )Y s
+
+
K
P
K s D
+
1 + s s b s )( (
zw 2
K I s
( )-
b
=
=
w
)n =
z 10,
0.707,
4.
n
1.6
Step response with the manual tuning Step response with the Ziegler - Nichols PID tuning
1.4
1.2
1
e d u
t i l
0.8
p m A
0.6
0.4
0.2
0
0
0.5
1
1.5
2
3
3.5
4
4.5
5
2.5 Time (s)
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56
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
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57
Negative Gain Root Locus
Ex.
+
=
1
K
0
2
s +
-
s
20 s
5
50
)
Root Locus
1 -
20
10
i
0
-10
i
-20
s d n o c e s ( s x A y r a n g a m
-20
-10
0
10
20
30
40
50
I
)
Real Axis (seconds-1) Root Locus
1 -
20
K = -
5.0
K = -
2.5
10
i
0
-10
i
-20
s d n o c e s ( s x A y r a n g a m
-20
-10
0
10
20
30
40
50
I
Real Axis (seconds-1)
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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
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59
The Root Locus Using Control Design Software (1)
Ex. 1
20
+
=
1
K
0
2
- + s + 5
s
s
50
• rlocus • rlocfind
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60
-
The Root Locus Using Control Design Software (2)
Ex. 1
+
=
=
G s ( )
0
2
s +
5
20 + s
s
20
• step • impulse
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61
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