1st
Slide
STABILITY OF CLOSED-LOOP SYSTEM
2nd
Slide
Open Loop Control System
Input
H(S)
G(S)
Control Process
Plant
Reference
Output
Observed State
Closed Loop Control System
Input
Reference
H(S)
G(S)
Control Process
Plant
Output
Observed State
I
3rd
Slide
Introduction
Oscillatory Response - consequence of feedback control
If oscillation has….
small amplitude and damps out quickly = STABLE
amplitude that increases with time until limit is reached and undamped = UNSTABLE
Feedback control system must be stable as prerequisite for satisfactory control
4th
Slide
:
Illustration:
≈_ ☐
Block flow diagram:
Disturbance Force
> i
✗
Reference
Force
Force
Ms
2
Position
(x)
5th Slide
Illustration:
Block flow diagram:
IT:O
Disturbance Force
"=
✗
Force
Reference
Force
Ms
2
Position
(x)
k
6ᵗʰ Slide
General Stability Criterion
7th
slide
Definition of Stability
An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs. Otherwise it is
said to be unstable
unconstrained - ideal situation where there are no physical limits on the input and output variables
bounded input - input variable that stays within upper and lower limits for all values of time
8th
slide
DETERMINING STABILITY OF AN OPEN LOOP SYSTEM:
N(s)
G(s) =
D(s)
characteristic
equation
proving:
example:
L (G(s))
'
1
G(s) =
roots: -1, -2
(s+1)(s+2)
1
Imaginary part
A
=
(s+1)(s+2)
B
(s+2)
(s+1)
St
Ae
-
t
Ae + Be
-
2T
^
Unstable region
Stable region
exponentially decays
co
STABLE
0
Real part
9ᵗʰ Slide
DETERMINING THE STABILITY OF CLOSED-LOOP SYSTEMS
General Stability Criterion:
Feedback control system is stable if and only if all roots of the characteristic equation are negative or have negative real parts.
Otherwise, the system is unstable
pic graph
Assumption:
1. Rewrite as open loop (reduce block diagrams)
2. Solve for roots of characteristic equation
3. Determine stability
10th
11th
Slide
HG
1+HG
H
G
controller
plant
slide
Routh Stability Criterion
12th
Slide
Routh Technique - determining whether any roots of a polynomial have positive real parts
Routh Stability Criterion: a necessary and sufficient condition for all roots of the characteristics equation to have negative real
parts is that all of the elements in the left column of the Routh array are positive
13th
slide
1
G(s) =
4
3
2
s + 3s - 5s + s + 2
same sign coefficient = stable
different sign coefficient = unstable
17
4
3
unstable
4
3
2
s+s+s+s+1
2
s + 3s - 5s + s + 2
used to determine stability
4
3
2
- s - s - s - s - 1 = -1 s + s + s + s + 1
☆
affect response not stability
*same sign (even if all negative) will be referred to as POSITIVE
14ᵗʰ
slide
1
s+1
G(s) =
1
•
s+3
1
G(s) =
roots:
0.5 j 2.75
the negative must come from a root in the
RHP, hence unstable
s-2
•
s-s+4
☒
1
•
1
1
1
•
4
2
3
s+2
s+1
s + 2s + 3s + 10s + 8
-2
-1
all positive
unstable
15th
slide
second row
example:
Routh Array
4
C,
Cz
Cz
3
2
s + 2s + 3s + 10s + 8
top row
R,
powers descending
Note:
"
ˢ
s + 2s + 5s + 3
Rz
g4
g
Rz
}
g2
÷
I
3
2
10
8
sˢ
52
S
'
50
/
2
5
slide
16th
6
Polynomial =
6
s
s
5
s
17th
4
5
2
A
C
E
G
B
D
F
◦
-
B*C-A*D
B
4
s
3
s
2
s
I
s
0
Pattern of 8
- (2)
* (3)
* (1)
/ (4)
slide
6
s
s
5
s
18th
A
C
E
G
B
D
F
0
4
s
3
s
2
s
I
s
0
determine the number of roots in RHP by
counting the number of sign changes
slide
Is
I
S4
g
3
52
s
2
10
2
8
_
0
1
+
2
+
s + 2s + 3s + 10s + 8
L
0
-2
1
l
18
0
18
t
8
+
8
RHP root
Unstable
-
'
go
19th
3
As + Bs + Cs + Ds + Es + Fs + G
RHP
root
Slide
Special Case 1: A zero in a row with at least one non-zero appearing later in that same row
1
0
2
3
4
her
1
2
2
4
0
5
to
unstable
5
Slide
20th
Special Case 2: The entire row is zero, not just a single entry
3
4
5
2
s + 2s + 6s + 10s + 8s + 12
Situation 1
5
s
s
"
s
3
s
2
s
s
21st
:
Situation 2
1
6
8
2
10
12
1
2
0
6
12
0
0
0
0
1
6
8
6s + 12s = 0
2
10
12
take the derivative: 2s
1
2
0
6
12
0
0
0
0
✗
✗
✗
✗
✗
5
s
4
s
3
s
2
s
s
I
✗
✗
slide
s
Situation 3
i
s+2=0
replace all zero with this
auxiliary polynomial
5
s
1
6
8
2
10
12
1
2
0
6
12
0
2
0
12
no sign changes
0
s
4
s
3
s
2
s
s
:
stable
auxiliary polynomial
✗