BAB 7 analisis-kestabilan (Materi tambahan) (1).pdf

5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 1
Analisis Kestabilan Sistem
Pengendalian Umpan Balik
05
05
Tujuan
Tujuan: Mhs mampu menganalisis kestabilan sistem
pengendalian umpan balik
Materi
Materi:
1. Konsep Kestabilan Sistem Loop Tertutup (Stability Concept
of The Closed Loop System)
2. Persamaan Karakteristik (Characteristic Equation)
3. Kestabilan Berdasarkan Ultimate Response
4. Kriteria Kestabilan Routh-Hurwitz
5. Analisis Kestabilan dengan Root Locus

5.1 Konsep Kestabilan Sistem
Loop Tertutup
• Designing a FBC (i.e. selecting its
components and tuning its controller)
seriously affects its stability characteristics.
• The notion of stability and the stability
characteristics of closed loop systems is
therefore considered to be useful.
Notion of Stability Ξ Pemahaman Kestabilan
Stability Characteristic Ξ Karakteristik Kestabilan

The Notion of Stability
(Pemahaman Kestabilan)
• How to define about stable and unstable?
Æ A dynamic system is considered to be
stable IF every bounded input produces
bounded output
• Bounded : input that always remains
between an upper and a lower limit
Æ sinusoidal, step, but not the rump
5.1 Konsep Kestabilan Sistem Loop Tertutup

Gambar 5.1.1. Bounded Input
Input
m(t)
time
Upper Limit
Lower Limit
sinusoidal
step

Gambar 5.1.2. Tanggapan (response) stabil dan tidak stabil
y
time
Desired value
Stable
Unstable
Unstable

Transfer
Function
• m = input, and y = output
• If G(s) has a pole with positive real part, it gives
rise to a term C1ept which grows continuously
with time Æ unstable
)
(
)
(
)
( s
m
s
G
s
y =
If TF of dynamic system has even one pole with
positive real part, the system is UNSTABLE
∴ all poles of TF must be in the left-hand part
of a complex plane, for the system to be
STABLE

Gambar 5.1.3. Complex Plane
Imaginary
Real
Stable Unstable

Example 5.1.1: Stabilization of an unstable process
with P Control
Let’s consider a process with the following response:
)
(
1
5
)
(
1
10
)
( s
d
s
s
m
s
s
y
−
+
−
=
Pole has positive root s – 1 = 0 Æ s = +1
y(t) = C1 et
Open loop unstable response

Gambar 5.1.4. Tanggapan sistem tak-terkendali terhadap
perubahan satu unit gangguan dengan fungsi tahap
0 0.5 1 1.5 2
0
10
20
30
40
Time
y
unstable
Example 5.1.1 (Continued)

The closed loop response:
( ) ( )
)
(
10
1
5
)
(
10
1
10
)
( s
d
K
s
s
y
K
s
K
s
y
c
sp
c
c
−
−
+
−
−
=
( )
c
c
sp
K
s
K
s
G
10
1
10
)
(
−
−
=
( )
c
load
K
s
s
G
10
1
5
)
(
−
−
=
and
The transfer function:
The closed loop will have negative pole if
10
1
>
c
K STABLE

Gambar 5.1.5. Penerapan FBC
Note: P only controller with Kc = 1

Gambar 5.1.6. Tanggapan dinamik sistem terkendali terhadap
perubahan satu unit gangguan dengan fungsi tahap
0 0.5 1 1.5 2
0
0.2
0.4
0.6
0.8
1
Time
y
Note: Regulatory problem; with Kc = 1
Stable
offset

Example 5.1.2: Destabilization of a stable process
with PI Control
• process with 2nd order TF:
2
2
1
)
( 2
+
+
=
s
s
s
Gp
• System has two complex poles:
• Therefore, System is stable
p1 = -1 + j
p2 = -1 − j
0 2 4 6 8 10
0
0.1
0.2
0.3
0.4
0.5
0.6
Time (second)
y
Gambar 5.1.7. Tanggapan stabil
loop terbuka terhadap perubahan
satu unit gangguan dengan fungsi
tahap

• Implementation of PI Control ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
s
K
s
G
I
C
C
τ
1
1
)
(
• Assume that Gm = Gf = 1
• The closed loop
response for servo
problem
)
(
)
(
)
(
1
)
( s
y
s
G
s
y
G
G
G
G
s
y sp
sp
sp
c
p
c
p
=
+
=
( )
( )
I
c
c
I
I
c
I
I
c
I
I
c
sp
K
s
K
s
s
s
K
s
s
K
s
s
s
s
K
s
s
s
G
τ
τ
τ
τ
τ
τ
τ
+
+
+
+
+
=
+
+
+
+
+
+
+
=
2
2
1
1
2
2
1
1
1
2
2
1
)
(
2
3
2
2

• Let Kc = 100 and τI = 0.1
• Poles of Gsp: s3 + 2s2 + (2+100)s + 100/0.1
• P1 = -7.18 ; p2 = 2.59 + 11.5j ; P3 = 2.59 – 11.5j
0 0.2 0.4 0.6 0.8 1
-10
-5
0
5
10
Time (second)
y
Gambar 5.1.8. Tanggapan tidak stabil dari loop tertutup

0 5 10 15 20 25 30 35 40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time (second)
y
Gambar 5.1.9. Tanggapan stabil loop tertutup
Kc = 10 and τI = 0.5

5.2 The Characteristic Equation
)
(
1
)
(
1
)
( s
d
G
G
G
G
G
s
y
G
G
G
G
G
G
G
s
y
m
c
f
p
d
sp
m
c
f
p
c
f
p
+
+
+
=
)
(
)
(
)
(
)
(
)
( s
d
s
G
s
y
s
G
s
y load
sp
sp +
=
The closed loop response of FBC :
The characteristic equation :
1 + GpGfGcGm = 0
denominator

Let p1, p2, …, pn be the n roots of the
Characteristic Equation
1 + GpGfGcGm = (s – p1) (s – p2) … (s – pn)
Stability criterion of a closed loop system
A FBC system is stable if all the roots of its
characteristic equation have negative real
part (i.e. are to the left of the imaginary axis)
5.2 Persamaan Karakteristik

Example 5.2.1: Stability analysis of FBC based
on characteristic equation
c
c
m
f
p K
G
G
G
s
G =
=
=
−
= 1
1
1
10
( )( )( ) 0
1
1
1
10
1
1 =
⎟
⎠
⎞
⎜
⎝
⎛
−
+
=
+ c
m
c
f
p K
s
G
G
G
G
Again, consider Ex. 5.1.1
characteristic equation:
Which has the root: p = 1 – 10Kc
∴ The system is stable IF p < 0
10
1
>
c
K

⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
=
=
+
+
=
s
K
G
G
G
s
s
G
I
c
c
m
f
p
τ
1
1
;
1
;
1
;
2
2
1
2
( )( ) 0
1
1
2
2
1
1
1 2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
⎟
⎠
⎞
⎜
⎝
⎛
+
+
+
=
+
s
K
s
K
s
s
G
G
G
G
I
c
I
c
m
c
f
p
τ
τ
Again, consider Ex. 5.1.2
Which has three roots,
Æ Kc and τI affect the value of roots
(i.e. positive or negative)
( ) 0
2
2 2
3
=
+
+
+
+
I
c
c
K
s
K
s
s
τ

5.3 Kestabilan Berdasarkan Ultimate
Response
Pada bagian ini kita akan membahas batasan kestabilan berdasarkan respon
kritis (ultimate response).
Ingat kembali! Respon stabil jika akar-akar denominator pada FT adalah
negatif (di sebelah kiri sumbu imaginer pada bidang kompleks).
Metode Substitusi Langsung
Metode ini sangat mudah untuk mencari parameter-parameter
pengendalian yang mana dapat menghasilkan respon stabil.
Jika pers. karakteristik mempunyai satu atau dua akar yang terletak tepat
pada sumbu imaginer, maka menghasilkan respon kritis (osilasi terjaga).
Ultimate Gain, Kcu: gain pengendali yang mana dapat menghasilkan
respon kritis.

5.3 Kestabilan Berdasarkan Ultimate Response
Gambar 5.3.1 respon loop tertutup dengan Kc = Kcu
( ) ( )
θ
ω +
= t
t
C u
sin
u
u
T
ω
π
2
=
-1.1
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
0 5 10 15 20 25 30
c(t)
Tu
t
Ultimate period
ωu = ultimate frequency

( ) ( ) ( ) ( ) 0
1 =
+ s
G
s
G
s
G
s
G m
c
f
p
Persamaan Karakteristik loop tertutup: denominator dari FT loop tertutup
... (5.3.1)
Substitusi langsung: s = i ωu
Bagian real = 0
Bagian imaginer = 0
Penyelesaian secara simultan
menghasilkan ultimate gain, Kcu
Contoh 5.3.1: mencari Kcu dan ωu untuk pengendali suhu pada HE
Steam
Process fluid
Ti(t), oC
Condensate
TC
W(t), kg/s
To(t), oC
TT
To
set (t), oC
Ws(t), kg/s
M(t), %CO
C(t), %TO

( )
TO
CO
K
s
G c
c
%
%
=
( )
CO
s
kg
s
s
Gv
%
/
1
3
016
.
0
+
=
( )
s
kg
C
s
s
G
o
S
/
1
30
50
+
=
( )
C
TO
s
s
G o
m
%
1
10
1
+
=
+
W(s)
+
C(s), %TO
M(s)
E(s)
R(s) +
−
Gc(s)
Gm(s)
Gv(s)
Ksp
%TO %CO
GS(s)
Gw(s)
Ws (s)
kg/s
To(t), oC
To
set (t), oC
%TO
kg/s
Gambar 5.3.2. Diagram blok pengendali suhu pada HE
P Control
dimana:

( ) ( ) ( ) ( ) 0
1 =
+ s
G
s
G
s
G
s
G c
v
S
m
Persamaan Karakteristik loop tertutup:
0
80
.
0
1
43
420
900 2
3
=
+
+
+
+ c
K
s
s
s
0
1
3
016
.
0
1
30
50
1
10
1
1 =
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
+
+ c
K
s
s
s
Penyusunan kembali persamaan di atas diperoleh:
( )( )( ) 0
80
.
0
1
3
1
30
1
10 =
+
+
+
+ c
K
s
s
s
Substitusi s = i ωu pada Kc = Kcu:
0
80
.
0
1
43
420
900 2
2
3
3
=
+
+
+
+ c
u
u
u K
i
i
i ω
ω
ω
( ) ( ) 0
0
43
900
80
.
0
1
420 3
2
i
i
K u
u
c
u +
=
+
−
+
+
+
− ω
ω
ω
i2 = –1

Bagian real dan imaginer harus sama dengan nol:
0
80
.
0
1
420 2
=
+
+
− c
u K
ω
0
43
900 3
=
+
− u
u ω
ω
Kedua pers. di atas diselesaikan secara simultan, menghasilkan:
ωu = 0 Kcu = –1.25 %CO/TO
ωu = 0.2186 rad/sec Kcu = 23.8 %CO/TO relevant
Jadi diperoleh ultimate period:
sec
7
.
28
2186
.
0
2
=
=
π
u
T
Jika diinginkan
respon STABIL,
maka Kc < Kcu

0 50 100 150 200
-0.5
0
0.5
1
Temperature
[degC]
Time [sec]
0 50 100 150 200
-0.2
-0.1
0
Steam
[kg/sec]
Time [sec]
Gambar 5.3.3. Tanggapan kritis pengendali suhu terhadap perubahan
satu unit set point dengan fungsi tahap (Contoh 5.3.1)
Kc = Kcu = 23.8

0 50 100 150 200
-1
0
1
2
Temperature
[degC]
Time [sec]
0 50 100 150 200
-0.2
-0.1
0
Steam
[kg/sec]
Time [sec]
Gambar 5.3.4. Tanggapan stabil pengendali suhu terhadap perubahan
satu unit set point dengan fungsi tahap (Contoh 5.3.1)
Kc = 10

0 50 100 150 200
-5
0
5
Temperature
[degC]
Time [sec]
0 50 100 150 200
-1
-0.5
0
0.5
Steam
[kg/sec]
Time [sec]
Gambar 5.3.5. Tanggapan tidak stabil pengendali suhu terhadap
perubahan satu unit set point dengan fungsi tahap (Contoh 5.3.1)
Kc = 30

5.4 Routh-Hurwitz Stability Criterion
• Does not require calculation of actual
values of the roots of the characteristic
equation.
• But, it only requires that we know if any
root is to the right of imaginary axis
• Make a conclusion as to the stability of
closed loop system quickly without
computing the actual values of the roots

Expand the characteristic eq. into the following
polynomial form
1 + GpGfGcGm ≡ ao sn + a1 sn-1 + … + an-1 s + an = 0
Let ao be positive, if it is negative, multiply both
sides of equation by −1.
First Test: If any of coefficients a1 , a2 , … , an-1 , an is
negative; there is at least one root of the characteristic
eq. which has positive real part Æ UNSTABLE
Second Test: If all coefficients a1 , a2 , … , an-1 , an are
positive; Then, from the 1st test we cannot conclude
anything about the location of the roots. Form the
following array (known as Routh array):
5.4 Kriteria Kestabilan Routh-Harwitz

Routh Array
…
.
.
W2
W1
n+1
.
.
.
.
…
.
A3
C2
C1
5
…
.
B3
B2
B1
4
…
.
A3
A2
A1
3
…
a7
a5
a3
a1
2
…
a6
a4
a2
a0
1
Row

Where
1
3
0
2
1
1
a
a
a
a
a
A
−
=
1
5
0
4
1
2
a
a
a
a
a
A
−
=
1
7
0
6
1
3
a
a
a
a
a
A
−
=
1
2
1
3
1
1
A
A
a
a
A
B
−
=
1
3
1
5
1
2
A
A
a
a
A
B
−
=
1
2
1
2
1
1
B
B
A
A
B
C
−
=
1
3
1
3
1
2
B
B
A
A
B
C
−
=
. . .
. . .
. . .
etc.

Examine the elements of the first
column of the array above
• If any of these elements is negative, we have at
least one root to the right of the imaginary axis
and the system is UNSTABLE
• The number of sign changes in the elements of
the first column is equal to the number of roots
to the right of the imaginary axis
a0 a1 A1 B1 C1 . . . W1
∴ A system is STABLE IF all the elements in the
first column of the Routh Array are POSITIVE

Example 5.4.1: Stability Analysis with the Routh-
Hurwitz Criterion
( ) 0
2
2 2
3
=
+
+
+
+
I
c
c
K
s
K
s
s
τ
I
c
K
τ
Consider FBC system in Ex. 5.1.2, Its Characteristic eq. is
( )
2
2
2
I
c
c
K
K
τ
−
+
I
c
K
τ
Routh array is:
Row
1st Column
1
2
3
4
1
2
2 + Kc
0

( )
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡ −
+
I
c
I
c
c K
K
K
τ
τ
,
2
2
2
,
2
,
1
The elements of the first column are:
Third element can be positive or negative
depending on the values of Kc and τI
The system is STABLE if Kc and τI satisfy
the condition: ( )
I
c
c
K
K
τ
>
+
2
2

Example 5.4.2: Critical stability Conditions for FBC
Return to Ex.5.4.1, and Let τI= 0.1
( )
2
10
2
2 c
c K
K −
+
3rd element becomes:
IF Kc = 0.5 Æ third element = 0 Æ critical condition,
two roots on imaginary axis (pure
imaginary) Æ ± j(1.58)
Æ sustained sinusoidal response
IF Kc < 0.5 Æ third element is positive Æ STABLE
IF Kc > 0.5 Æ third element is negative Æ UNSTABLE

Gambar 5.4.1. Tanggapan kritis dari sistem terkendali terhadap
Kc = 0.5 ; thoi = 0.1
0 5 10 15
0
0.5
1
1.5
2
CV
Time [sec]
0 5 10 15
-2
0
2
4
6
MV
Time [sec]

Gambar 5.4.2. Tanggapan stabil dari sistem terkendali terhadap
Kc = 0.1 ; thoi = 0.1
0 5 10 15
0
0.5
1
1.5
CV
Time [sec]
0 5 10 15
0
1
2
3
MV
Time [sec]

Gambar 5.4.3. Tanggapan tidak stabil dari sistem terkendali terhadap
Kc = 1 ; thoi = 0.1
0 5 10 15
-20
-10
0
10
20
CV
Time [sec]
0 5 10 15
-200
-100
0
100
MV
Time [sec]

5.5 Root-Locus Analysis
• Root Locus is a graphical technique that
consists of graphing the roots of the
characteristic equation
• The resulting graph allows to see whether
a root crosses the imaginary axis to the
right hand side of the s-plane
5.5 Analisis Kestabilan dengan Root-Locus

Example 5.5.1: Root locus for a Reactor with P Control
R A
Flow rate = m
Flow rate = F – m
Flow rate = F
Concentration in C = y
Reactions:
A + R Æ B
B + R Æ C
C + R Æ D
D + R Æ E
The control objective is to keep the concentration of
the desired product C as close as possible to its set
point by manipulating the flow rate of A

Example 5.5.1: (Continued)
( )
( )( ) ( )
35
.
4
85
.
2
45
.
1
25
.
2
98
.
2
)
(
)
(
)
( 2
+
+
+
+
=
=
s
s
s
s
s
m
s
y
s
Gp
TF of the process:
( )
( )( ) ( )
0
35
.
4
85
.
2
45
.
1
25
.
2
98
.
2
1 2
=
+
+
+
+
+ c
K
s
s
s
s
Implementation of P Control with Gc(s) = Kc , and
Assume that Gm = Gf = 1, we have the following
S4 + 11.5 s3 + 47.49 s2 + (2.98Kc + 83.0633)s
+ (6.705Kc + 51.2327) = 0

Roots of the characteristic equation
(for ex. 5.5.1)
−9.85
+0.30 − 5.70 j
+0.30 + 5.70 j
−2.25
100
−8.49
−0.38 − 4.48 j
−0.38 + 4.48 j
−2.24
50
−7.14
−1.06 − 3.23 j
−1.06 + 3.23 j
−2.23
20
−5.77
−1.76 − 1.88 j
−1.76 + 1.88 j
−2.20
5
−4.89
−2.34 − 0.82 j
−2.34 + 0.82 j
−1.92
1
−2.85
−2.85
−2.85
− 1.45
0
p4
p3
p2
p1
Kc

Gambar 5.5.1. Root-Locus of the reactor in Ex. 5.5.1
-20 -15 -10 -5 0 5 10 15
-20
-15
-10
-5
0
5
10
15
20
Root Locus
Real Axis
Imaginary
Axis

Gambar 5.5.2. Tanggapan stabil loop terkendali terhadap
perubahan satu unit set-point (Ex. 5.5.1)
Kc = 50
Kc = 20
Kc = 5
Kc = 1
0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
Time (Sec)
y
Offset
New set-point

0 2 4 6 8 10 12 14 16 18 20
0
0.5
1
1.5
2
Time (Sec)
y
Critical condition Kc = 75
Gambar 5.5.3. Tanggapan kritis dari loop terkendali
terhadap perubahan satu unit set-point (Ex. 5.5.1)

0 2 4 6 8 10 12 14 16 18 20
-300
-200
-100
0
100
200
300
Time (Sec)
y
Kc = 100
Gambar 5.5.4. Tanggapan tak-stabil dari loop terkendali
terhadap perubahan satu unit set-point (Ex. 5.5.1)

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