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

1. 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
2. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 2 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
3. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 3 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
4. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 4 Gambar 5.1.1. Bounded Input Input m(t) time Upper Limit Lower Limit sinusoidal step 5.1 Konsep Kestabilan Sistem Loop Tertutup
5. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 5 Gambar 5.1.2. Tanggapan (response) stabil dan tidak stabil y time Desired value Stable Unstable Unstable 5.1 Konsep Kestabilan Sistem Loop Tertutup
6. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 6 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
7. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 7 Gambar 5.1.3. Complex Plane Imaginary Real Stable Unstable 5.1 Konsep Kestabilan Sistem Loop Tertutup
8. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 8 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
9. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 9 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) 5.1 Konsep Kestabilan Sistem Loop Tertutup
10. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 10 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 ) ( − − = Example 5.1.1 (Continued) and The transfer function: The closed loop will have negative pole if 10 1 > c K STABLE 5.1 Konsep Kestabilan Sistem Loop Tertutup
11. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 11 Gambar 5.1.5. Penerapan FBC Note: P only controller with Kc = 1 Example 5.1.1 (Continued) 5.1 Konsep Kestabilan Sistem Loop Tertutup
12. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 12 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
13. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 13 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
14. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 14 Example 5.1.2 (Continued) • 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
15. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 15 Example 5.1.2 (Continued) • 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 5.1 Konsep Kestabilan Sistem Loop Tertutup
16. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 16 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 Time (second) y Gambar 5.1.9. Tanggapan stabil loop tertutup Kc = 10 and τI = 0.5 5.1 Konsep Kestabilan Sistem Loop Tertutup
17. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 17 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
18. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 18 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
19. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 19 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 5.2 Persamaan Karakteristik
20. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 20 Example 5.2.1 (Continued) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = = = + + = 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 characteristic equation: 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 5.2 Persamaan Karakteristik
21. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 21 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.
22. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 22 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
23. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 23 ( ) ( ) ( ) ( ) 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 5.3 Kestabilan Berdasarkan Ultimate Response
24. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 24 ( ) 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: 5.3 Kestabilan Berdasarkan Ultimate Response
25. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 25 ( ) ( ) ( ) ( ) 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 5.3 Kestabilan Berdasarkan Ultimate Response
26. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 26 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 5.3 Kestabilan Berdasarkan Ultimate Response
27. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 27 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 5.3 Kestabilan Berdasarkan Ultimate Response
28. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 28 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 5.3 Kestabilan Berdasarkan Ultimate Response
29. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 29 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] 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) Kc = 30
30. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 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
31. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 31 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
32. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 32 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 5.4 Kriteria Kestabilan Routh-Harwitz
33. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 33 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. 5.4 Kriteria Kestabilan Routh-Harwitz
34. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 34 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 5.4 Kriteria Kestabilan Routh-Harwitz
35. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 35 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 5.4 Kriteria Kestabilan Routh-Harwitz
36. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 36 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 Kc and τI The system is STABLE if Kc and τI satisfy the condition: ( ) I c c K K τ > + 2 2 5.4 Kriteria Kestabilan Routh-Harwitz
37. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 37 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 5.4 Kriteria Kestabilan Routh-Harwitz
38. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 38 5.3 Kestabilan Berdasarkan Ultimate Response 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]
39. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 39 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]
40. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 40 5.3 Kestabilan Berdasarkan Ultimate Response 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 Time [sec]
41. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 41 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
42. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 42 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 5.5 Analisis Kestabilan dengan Root-Locus
43. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 43 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 characteristic equation: S4 + 11.5 s3 + 47.49 s2 + (2.98Kc + 83.0633)s + (6.705Kc + 51.2327) = 0 5.5 Analisis Kestabilan dengan Root-Locus
44. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 44 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 5.5 Analisis Kestabilan dengan Root-Locus
45. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 45 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 5.5 Analisis Kestabilan dengan Root-Locus
46. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 46 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 5.5 Analisis Kestabilan dengan Root-Locus
47. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 47 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)
48. 5 - ANALISIS KESTABILAN SISTEM PENGENDALIAN UMPAN BALIK – DR. ENG. Y. D. HERMAWAN INDALPRO / 48 0 2 4 6 8 10 12 14 16 18 20 -300 -200 -100 0 100 200 300 Time (Sec) y Kc = 100 5.5 Analisis Kestabilan dengan Root-Locus Gambar 5.5.4. Tanggapan tak-stabil dari loop terkendali terhadap perubahan satu unit set-point (Ex. 5.5.1)
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