Analisis Kestabilan Sistem Pengendalian Umpan Balik

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

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

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 C

₁

e

pt

_{which grows continuously}

with time Æ unstable

)

(

)

(

)

(

s

G

s

m

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

)

(

d

s

s

s

m

s

s

y

−

+

−

=

Pole has positive root

s – 1 = 0 Æ s = +1

y(t) = C

₁

e

t

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:

(

)

(

1

10

)

(

)

5

)

(

10

1

10

)

(

d

s

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

)

(

−

−

=

Example 5.1.1 (Continued)

and

The transfer function:

The closed loop will have negative pole if

10

1

>

c

Gambar 5.1.5. Penerapan FBC

Note: P only controller with Kc = 1

Example 5.1.1 (Continued)

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

Example 5.1.2:

Destabilization of a stable process

with PI Control

• process with 2

nd

order TF:

2

2

1

)

(

₂

+

+

=

s

s

s

Gp

• System has two complex poles:

• Therefore, System is stable

p

₁

= -1 + j

p

₂

= -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

Example 5.1.2 (Continued)

• Implementation of PI Control

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

s

K

s

G

I

C

C

τ

1

1

)

(

• Assume that G

_m

= G

_f

= 1

• The closed loop

response for servo

problem

(

)

1

G

G

y

(

s

)

G

(

s

)

y

(

s

)

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

Example 5.1.2 (Continued)

• Let K

_c

= 100 and τ

_I

= 0.1

• Poles of G

_sp

: s

3

_{+ 2s}

2

_{+ (2+100)s + 100/0.1}

• P

₁

= -7.18 ; p

₂

= 2.59 + 11.5j ; P

₃

= 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

Example 5.1.2 (Continued)

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

y

Gambar 5.1.9. Tanggapan stabil loop tertutup

5.2 The Characteristic Equation

)

(

1

)

(

1

)

(

d

s

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

G

s

y

s

G

s

d

s

y

=

_sp

_sp

+

_load

The closed loop response of FBC :

The characteristic equation :

1 + G

_p

G

_f

G

_c

G

_m

= 0

Let p

₁

, p

₂

, …, p

_n

be the n roots of the

Characteristic Equation

1 + G

_p

G

_f

G

_c

G

_m

= (s – p

₁

) (s – p

₂

) … (s – p

_n

)

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)

Example 5.2.1: Stability analysis of FBC based

on characteristic equation

c

c

m

f

p

G

G

G

K

s

G

=

=

=

−

=

1

1

1

10

( )( )( )

1

1

0

1

10

1

1

⎟

=

⎠

⎞

⎜

⎝

⎛

−

+

=

+

_p

_f

_c

_m

K

_c

s

G

G

G

G

Again, consider Ex. 5.1.1

characteristic equation:

Which has the root: p = 1 – 10K

_c

∴ The system is stable IF p < 0

10

1

>

c

K

5.2 Persamaan Karakteristik

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

=

=

+

+

=

s

K

G

G

G

s

s

G

I

c

c

m

f

p

τ

1

1

;

1

;

1

;

2

2

1

2

( )( )

1

1

0

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

characteristic equation:

Which has three roots,

Æ

K

_c

and

τ

_I

affect the value of roots

(i.e. positive or negative)

(

2

)

0

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, K

_cu

: gain pengendali yang mana dapat menghasilkan

Gambar 5.3.1 respon loop tertutup dengan K

_c

= K

_cu

( )

t

=

(

ω

t

+

θ

)

C

sin

_u

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 -0.10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.91 1.1 0 5 10 15 20 25 30

c(t)

T

_u

t

Ultimate period

ω

_u

= ultimate frequency

( ) ( ) ( ) ( )

0

1

+

G

_p

s

G

_f

s

G

_c

s

G

_m

s

=

Persamaan Karakteristik loop tertutup: denominator dari FT loop tertutup

... (5.3.1)

Substitusi langsung: s = i

ω

_u

Bagian real = 0

Bagian imaginer = 0

Penyelesaian secara simultan

menghasilkan ultimate gain, K

_cu

Contoh 5.3.1: mencari K

_cu

dan

ω

_u

untuk pengendali suhu pada HE

Steam

Process fluid

T

_i

(t),

o

_C

Condensate

TC

W(t), kg/s

T

_o

(t),

o

_C

TT

T

_o set

_(t),

o

_C

W

_s

(t), kg/s

M(t), %CO

C(t), %TO

( )

TO

CO

K

s

G

_c

_c

%

%

=

( )

CO

s

kg

s

s

G

_v

%

/

1

3

016

.

0

+

=

( )

s

kg

C

s

s

G

o

S

/

1

30

50

+

=

( )

C

TO

s

s

G

_m

%

_o

1

10

1

+

=

+

W(s)

+

C(s), %TO

M(s)

E(s)

R(s) +

−

G

_c

(s)

G

_m

(s)

G

_v

(s)

K

_sp

%TO

%CO

G

_S

(s)

G

_w

(s)

W

_s

(s)

kg/s

T

_o

(t),

o

_C

T

_o

set

_(t),

o

_C

%TO

kg/s

P Control

dimana:

( ) ( ) ( ) ( )

0

1

+

G

_m

s

G

_S

s

G

_v

s

G

_c

s

=

Persamaan Karakteristik loop tertutup:

0

80

.

0

1

43

420

900

s

3

+

s

2

+

s

+

+

K

_c

=

0

1

3

016

.

0

1

30

50

1

10

1

1

⎟

=

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

+

⎟

⎠

⎞

⎜

⎝

⎛

+

+

K

_c

s

s

s

Penyusunan kembali persamaan di atas diperoleh:

(

10

s

+

1

)(

30

s

+

1

)(

3

s

+

1

)

+

0

.

80

K

_c

=

0

Substitusi s = i

ω

_u

pada K

_c

= K

_cu

:

0

80

.

0

1

43

420

900

i

3

ω

_u

3

+

i

2

ω

_u

2

+

i

ω

_u

+

+

K

_c

=

(

−

420

ω

_u

2

+

1

+

0

.

80

K

_c

) (

+

i

−

900

ω

_u

3

+

43

ω

_u

)

=

0

+

i

0

i

2

= –1

0

80

.

0

1

420

2

+

+

=

−

ω

_u

K

_c

0

43

900

3

+

=

−

ω

_u

ω

_u

Kedua pers. di atas diselesaikan secara simultan, menghasilkan:

ω

_u

= 0

K

_cu

= –1.25 %CO/TO

ω

_u

= 0.2186 rad/sec

K

cu

= 23.8 %CO/TO

relevant

Jadi diperoleh ultimate period:

sec

7

.

28

2186

.

0

2

₌

=

π

u

T

Jika diinginkan

respon STABIL,

maka K

_c

< K

_cu

0

50

100

150

200

-0.5

0

0.5

1

T

e

m

p

er

at

u

re [

d

eg

C

]

Time [sec]

0

50

100

150

200

-0.2

-0.1

0

S

team

[

k

g

/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

T

e

mp

er

at

u

re [

d

eg

C

]

Time [sec]

0

50

100

150

200

-0.2

-0.1

0

S

team [

k

g

/sec]

Gambar 5.3.4. Tanggapan stabil pengendali suhu terhadap perubahan

satu unit set point dengan fungsi tahap (Contoh 5.3.1)

0

50

100

150

200

-5

0

5

T

e

m

p

er

at

u

re [

d

eg

C

]

Time [sec]

0

50

100

150

200

-1

-0.5

0

0.5

S

team

[

k

g

/sec]

Time [sec]

5.3 Kestabilan Berdasarkan Ultimate Response

Gambar 5.3.5. Tanggapan tidak stabil pengendali suhu terhadap

perubahan satu unit set point dengan fungsi tahap (Contoh 5.3.1)

• 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

Expand the characteristic eq. into the following

polynomial form

1 + G

_p

G

_f

G

_c

G

_m

≡

_a

_o

s

n

+

_a

₁

s

n-1

+ … +

_a

_n-1

s +

_a

_n

= 0

Let

_a

_o

be positive, if it is negative, multiply both

sides of equation by −1.

First Test: If any of coefficients

_a

₁

,

a

₂

, … ,

a

_n-1

,

a

_n

is

negative; there is at least one root of the characteristic

eq. which has positive real part Æ UNSTABLE

Second Test: If all coefficients

_a

₁

,

a

₂

, … ,

a

_n-1

,

a

_n

are

positive; Then, from the 1

st

_{test we cannot conclude}

anything about the location of the roots. Form the

following array (known as Routh array):

Routh Array

…

.

.

W

₂

W

₁

n+1

.

.

.

.

…

.

A

₃

C

₂

C

₁

5

…

.

B

₃

B

₂

B

₁

4

…

.

A

₃

A

₂

A

₁

3

…

a

₇

a

₅

a

₃

a

₁

2

…

a

₆

a

₄

a

₂

a

₀

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

a

0

a

1

A

1

B

1

C

1

. . . W

1

∴ 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

(

2

)

0

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

1

st

_Column

1

2

3

4

1

2

2 + K

_c

0

Example 5.4.1 (Continued)

(

)

⎥

⎥

⎥

⎦

⎤

⎢

⎢

⎢

⎣

⎡

₊

₋

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 K

_c

and

τ

_I

The system is STABLE if K

_c

and

τ

_I

satisfy

the condition:

(

)

c

c

K

K

τ

>

+

2

2

Example 5.4.2: Critical stability Conditions for FBC

Return to Ex.5.4.1, and Let

τ

_I

= 0.1

(

)

2

10

2

2

+

K

_c

−

K

_c

3

rd

_{element becomes:}

IF K

_c

= 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 K

_c

< 0.5 Æ third element is positive Æ STABLE

IF K

_c

> 0.5 Æ third element is negative Æ UNSTABLE

Gambar 5.4.1. Tanggapan kritis dari sistem terkendali terhadap

perubahan satu unit set point dengan fungsi tahap (Contoh 5.4.1)

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]

5.3 Kestabilan Berdasarkan Ultimate Response

Gambar 5.4.2. Tanggapan stabil dari sistem terkendali terhadap

perubahan satu unit set point dengan fungsi tahap (Contoh 5.4.1)

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

perubahan satu unit set point dengan fungsi tahap (Contoh 5.4.1)

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

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

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)

(

)

(

1

.

45

)(

2

.

85

) (

4

.

35

)

25

.

2

98

.

2

)

(

)

(

)

(

₂

+

+

+

+

=

=

s

s

s

s

s

m

s

y

s

G

_p

TF of the process:

(

)

(

1

.

45

)(

2

.

85

) (

4

.

35

)

0

25

.

2

98

.

2

1

₂

=

+

+

+

+

+

K

_c

s

s

s

s

Implementation of P Control with G

_c

(s) = K

_{c ,}

and

Assume that G

_m

= G

_f

= 1, we have the following

characteristic equation:

S

4

_{+ 11.5 s}

3

_{+ 47.49 s}

2

_{+ (2.98K}

c

+ 83.0633)s

+ (6.705K

_c

+ 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

p

₄

p

₃

p

₂

p

₁

K

_c

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

Im

agi

nar

y A

xi

s

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}

5.5 Analisis Kestabilan dengan Root-Locus

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

y

Kc = 100

Gambar 5.5.4. Tanggapan tak-stabil dari loop terkendali

terhadap perubahan satu unit set-point (Ex. 5.5.1)