Analisis Kestabilan Sistem Pengendalian Umpan Balik

05

Analisis Kestabilan Sistem Pengendalian Umpan Balik

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 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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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 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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5.1 Konsep Kestabilan Sistem Loop Tertutup

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


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Gambar 5.1.1. Bounded Input Input

Upper Limit

m(t) step

sinusoidal

Lower Limit

time


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Gambar 5.1.2. Tanggapan (response) stabil dan tidak stabil y

Unstable

Stable Desired value Unstable

time 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Transfer Function

y ( s ) = G ( s ) m( s )

• 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 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 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Gambar 5.1.3. Complex Plane Imaginary

Stable

Unstable Real


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Example 5.1.1: Stabilization of an unstable process with P Control Let’s consider a process with the following response:

10 5 y(s) = m( s ) + d (s) s −1 s −1 Pole has positive root

s – 1 = 0 Æ s = +1

y(t) = C1 et Open loop unstable response 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.1.1 (Continued) Gambar 5.1.4. Tanggapan sistem tak-terkendali terhadap perubahan satu unit gangguan dengan fungsi tahap 40

unstable

y

30 20 10 0

0

0.5

1 Time

1.5


2

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Example 5.1.1 (Continued)

The closed loop response: 10 K c 5 y( s) = y sp ( s ) + d ( s) s − (1 − 10 K c ) s − (1 − 10 K c )

The transfer function: 5 Gload ( s ) = s − (1 − 10 K c )

and

10 K c Gsp ( s ) = s − (1 − 10 K c )

The closed loop will have negative pole if

1 Kc > 10

STABLE


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Gambar 5.1.5. Penerapan FBC

Note: P only controller with Kc = 1 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Gambar 5.1.6. Tanggapan dinamik sistem terkendali terhadap perubahan satu unit gangguan dengan fungsi tahap 1 0.8

Stable

y

0.6 0.4

offset 0.2 0

0

0.5

1 Time

1.5

2

Note: Regulatory problem; with Kc = 1 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.1.2: Destabilization of a stable process

with PI Control

Gambar 5.1.7. Tanggapan stabil loop terbuka terhadap perubahan satu unit gangguan dengan fungsi tahap

• process with 2nd order TF:

1 Gp ( s ) = 2 s + 2s + 2 0.6

• System has two complex poles:

0.4

y

p1 = -1 + j p2 = -1 − j

0.5

0.3 0.2 0.1

• Therefore, System is stable

0 0

2


4

6

8

10

Time (second)

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Example 5.1.2 (Continued) • Implementation of PI Control • Assume that Gm = Gf = 1 • The closed loop response for servo problem

y( s) =

⎛ 1 ⎞ ⎟⎟ GC ( s ) = K C ⎜⎜1 + ⎝ τIs ⎠

G p Gc 1 + G p Gc

y sp ( s ) = Gsp ( s ) ysp ( s )

K c (τ I s + 1) 1 τ I s +1 Kc 2 τIs τI s + 2s + 2 = Gsp ( s ) = 1 τ I s +1 Kc 3 2 1+ 2 Kc s + 2 s + (2 + K c )s + τIs s + 2s + 2 τI


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• 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 Gambar 5.1.8. Tanggapan tidak stabil dari loop tertutup 10

y

5

0

-5

-10 0

0.2

0.4

0.6

0.8

1

Time (second) 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.1.2 (Continued) Gambar 5.1.9. Tanggapan stabil loop tertutup 1.8 1.6

Kc = 10 and τI = 0.5

1.4 1.2

y

1 0.8 0.6 0.4 0.2 0

0

5

10

15

20 25 Time (second)


30

35

40

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5.2 The Characteristic Equation The closed loop response of FBC : G p G f Gc

Gd y sp ( s ) + d ( s) y ( s) = 1 + G p G f Gc Gm 1 + G p G f Gc Gm

y ( s ) = Gsp ( s ) ysp ( s ) + Gload ( s )d ( s )

denominator

The characteristic equation : 1 + GpGfGcGm = 0 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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5.2 Persamaan Karakteristik

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 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.2.1: Stability analysis of FBC based on characteristic equation Again, consider Ex. 5.1.1

10 Gp = s −1

Gf =1

Gm = 1

Gc = K c

characteristic equation:

⎛ 10 ⎞ 1 + G p G f Gc Gm = 1 + ⎜ ⎟(1)(K c )(1) = 0 ⎝ s −1 ⎠ Which has the root: p = 1 – 10Kc ∴ The system is stable IF p < 0 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

1 Kc > 10 INDALPRO / 19


Example 5.2.1 (Continued) Again, consider Ex. 5.1.2

1 Gp = 2 s + 2s + 2

⎛ 1 ⎞ ⎟⎟ ; G f = 1 ; Gm = 1 ; Gc = K c ⎜⎜1 + ⎝ τIs ⎠

characteristic equation:

⎛ K cτ I s + K c ⎞ 1 ⎛ ⎞ ⎟⎟ = 0 1 + G p G f Gc Gm = 1 + ⎜ 2 ⎟ (1) (1)⎜⎜ τIs ⎝ s + 2s + 2 ⎠ ⎝ ⎠

s + 2 s + (K c + 2)s + 3

2

Kc

τI

=0

Which has three roots, Æ Kc and τI affect the value of roots (i.e. positive or negative) 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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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 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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5.3 Kestabilan Berdasarkan Ultimate Response

Gambar 5.3.1 respon loop tertutup dengan Kc = Kcu Tu

c(t)

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 -0.3 -0.4 -0.5 -0.6 -0.7 -0.8 -0.9 -1 -1.1

5

10

15

20

C (t ) = sin (ωu t + θ ) Tu =

2π

ωu

25

30

t

Ultimate period

ωu = ultimate frequency 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Persamaan Karakteristik loop tertutup: denominator dari FT loop tertutup

1 + G p (s )G f (s )Gc (s )Gm (s ) = 0

... (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 To set (t), oC

Steam M(t), %CO Ws(t), kg/s

TC

C(t), %TO TT

Process fluid Ti(t), oC

To(t), oC

W(t), kg/s Condensate


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Gambar 5.3.2. Diagram blok pengendali suhu pada HE W(s) Gw(s)

kg/s set (t), oC

To

Ksp

R(s) +

E(s)

%TO

%TO

M(s) Gc(s)

%CO

Gv(s)

+

Ws (s)

GS(s)

kg/s

+

To(t), oC

− C(s), %TO

Gm(s)

dimana: o 50 C ( ) GS s = 30 s + 1 kg / s

0.016 kg / s Gv (s ) = 3s + 1 %CO

1 %TO Gm (s ) = 10s + 1 o C

%CO Gc (s ) = K c %TO


P Control

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Persamaan Karakteristik loop tertutup:

1 + Gm (s )GS (s )Gv (s )Gc (s ) = 0 ⎛ 1 ⎞⎛ 50 ⎞⎛ 0.016 ⎞ 1+ ⎜ ⎟⎜ ⎟⎜ ⎟Kc = 0 ⎝ 10 s + 1 ⎠⎝ 30 s + 1 ⎠⎝ 3s + 1 ⎠ Penyusunan kembali persamaan di atas diperoleh:

(10s + 1)(30s + 1)(3s + 1) + 0.80 K c = 0 900 s 3 + 420 s 2 + 43s + 1 + 0.80 K c = 0 Substitusi s = i ωu pada Kc = Kcu:

900i 3ωu3 + 420i 2ωu2 + 43iωu + 1 + 0.80 K c = 0 i2 = –1

(− 420ω

2 u

) (

)

+ 1 + 0.80 K c + i − 900ωu3 + 43ωu = 0 + i 0


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Bagian real dan imaginer harus sama dengan nol:

− 420ωu2 + 1 + 0.80 K c = 0

− 900ωu3 + 43ωu = 0 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:

2π Tu = = 28.7 sec 0.2186


Jika diinginkan respon STABIL, maka Kc < Kcu

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Gambar 5.3.3. Tanggapan kritis pengendali suhu terhadap perubahan satu unit set point dengan fungsi tahap (Contoh 5.3.1)

Temperature [degC]

Kc = Kcu = 23.8 1 0.5 0 -0.5

0

50

100 Time [sec]

150

200

0

50

100 Time [sec]

150

200

Steam [kg/sec]

0

-0.1

-0.2


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Gambar 5.3.4. Tanggapan stabil pengendali suhu terhadap perubahan satu unit set point dengan fungsi tahap (Contoh 5.3.1)

Temperature [degC]

Kc = 10 2 1 0 -1

0

50

100

150

200

150

200

Time [sec]

Steam [kg/sec]

0

-0.1

-0.2

0

50

100 Time [sec]


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Gambar 5.3.5. Tanggapan tidak stabil pengendali suhu terhadap perubahan satu unit set point dengan fungsi tahap (Contoh 5.3.1)

Temperature [degC]

Kc = 30 5

0

-5

0

50

100 Time [sec]

150

200

0

50

100 Time [sec]

150

200

Steam [kg/sec]

0.5 0 -0.5 -1


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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 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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5.4 Kriteria Kestabilan Routh-Harwitz

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 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Routh Array Row

1 2 3 4 5 . n+1

a0 a1

a2 a3

a4 a5

a6 a7

A1 B1 C1 . W1

A2 B2 C2 . W2

A3 B3 A3 . .

. . .

… … … … …

.

…


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Where a1a2 − a0 a3 A1 = a1

a1a4 − a0 a5 A2 = a1

A1a3 − a1 A2 B1 = A1

A1a5 − a1 A3 B2 = A1

B1 A2 − A1 B2 C1 = B1

B1 A3 − A1 B3 C2 = B1

a1a6 − a0 a7 A3 = a1

. . .

. . .

. . .

etc.


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Examine the elements of the first column of the array above

a0 a1

A1

B1

C1

...

W1

• 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 system is STABLE IF all the elements in the first column of the Routh Array are POSITIVE 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.4.1: Stability Analysis with the RouthHurwitz Criterion Consider FBC system in Ex. 5.1.2, Its Characteristic eq. is

s + 2 s + (K c + 2 )s + 3

2

Routh array is: Row

Kc

τI

=0 1st Column

1

1

2 + Kc

2

2

Kc

3 4

2(2 + K c ) − 2

Kc

τI

τI 0

Kc

τI


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Example 5.4.1 (Continued) The elements of the first column are:

⎡ ⎢ ⎢1, 2, ⎢ ⎣

2(2 + K c ) − 2

Kc

τI

,

⎤ Kc ⎥ τ I ⎥⎥ ⎦

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:

2(2 + K c ) >


Kc

τI

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Example 5.4.2: Critical stability Conditions for FBC Return to Ex.5.4.1, and Let τI= 0.1 3rd

element becomes:

2(2 + K c ) − 10 K c 2

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 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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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 2

CV

1.5 1 0.5 0

0

5

10

15

10

15

Time [sec] 6

MV

4 2 0 -2

0

5

Time [sec]


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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 1.5

CV

1 0.5 0

0

5

10

15

10

15

Time [sec] 3

MV

2 1 0

0

5 Time [sec]


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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 20

CV

10 0 -10 -20

0

5

10

15

10

15

Time [sec] 100

MV

0 -100 -200

0

5 Time [sec]


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5.5 Analisis Kestabilan dengan Root-Locus

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


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Example 5.5.1: Root locus for a Reactor with P Control R Flow rate = F – m

A Flow rate = m

Reactions: A+RÆB B+RÆC C+RÆD

Flow rate = F

D+RÆE

Concentration in C = y

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 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Example 5.5.1: (Continued) 2.98(s + 2.25) y( s) = TF of the process: G p ( s ) = m( s ) (s + 1.45)(s + 2.85)2 (s + 4.35)

Implementation of P Control with Gc(s) = Kc , and Assume that Gm = Gf = 1, we have the following characteristic equation: 1 +

2.98(s + 2.25) Kc = 0 2 (s + 1.45)(s + 2.85) (s + 4.35)

S4 + 11.5 s3 + 47.49 s2 + (2.98Kc + 83.0633)s + (6.705Kc + 51.2327) = 0 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Roots of the characteristic equation (for ex. 5.5.1)

Kc

p1

p2

p3

p4

0

− 1.45

−2.85

1

−1.92

−2.34 + 0.82 j −2.34 − 0.82 j −4.89

5

−2.20

−1.76 + 1.88 j −1.76 − 1.88 j −5.77

20

−2.23

−1.06 + 3.23 j −1.06 − 3.23 j −7.14

50

−2.24

−0.38 + 4.48 j −0.38 − 4.48 j −8.49

100

−2.25

+0.30 + 5.70 j +0.30 − 5.70 j −9.85

−2.85


−2.85

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Gambar 5.5.1. Root-Locus of the reactor in Ex. 5.5.1 Root Locus 20

15

10

Imaginary Axis

5

0

-5

-10

-15

-20 -20

-15

-10

-5

0

5

10

15

Real Axis 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN

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Gambar 5.5.2. Tanggapan stabil loop terkendali terhadap perubahan satu unit set-point (Ex. 5.5.1) 1.5

New set-point

1

y

Kc = 50 Kc = 20 Offset

0.5

Kc = 5 Kc = 1 0

0

2

4

6

8

10 12 Time (Sec)

14


16

18

20

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Gambar 5.5.3. Tanggapan kritis dari loop terkendali terhadap perubahan satu unit set-point (Ex. 5.5.1) 2

Critical condition

Kc = 75

y

1.5

1

0.5

0

0

2

4

6

8

10 12 Time (Sec)

14


16

18

20

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Gambar 5.5.4. Tanggapan tak-stabil dari loop terkendali terhadap perubahan satu unit set-point (Ex. 5.5.1) 300

Kc = 100

200

y

100

0

-100

-200

-300

0

2

4

6

8

10 12 Time (Sec)

14


16

18

20

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Analisis Kestabilan Sistem Pengendalian Umpan Balik

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