BASIC PENGENALAN SISTEM KONTROL

BASIC PENGENALAN SISTEM KONTROL PENGENALAN SISTEM-SISTEM KONTROL Sistem Kontrol Terbuka/Open-Loop INPUT

OUTPUT CONTROLLER ER

PLANT / PROCESS

-

output tidak diukur maupun di –feedback-kan

-

bergantung pada kalibrasi

-

hubungan antara output dan input diketahui

-

tidak ada ‘internal disturbance’ maupun ‘eksternal disturbance’

Contoh : - kontrol traffic (lalu lintas) - mesin cuci Keuntungan : mudah terjadi kestabilan Kekurangan : komponen-komponen relatif mahal dan memiliki akurasi tinggi

Sistem Kontrol Tertutup / Close-Loop

CONTROLLER ER INPUT

ELEMEN PENGUKUR

PLANT / PROCESS OUTPUT

→Terdapat ‘feedback’ untuk mengurangi ‘error’ A. Manual Feedback Control / Manual Close-Loop Control System

Blok Diagram : ‘Manual Feedback Control’ dari sebuah sistem thermal

B. Automatic Feedback Control / Automatic Close-Loop Control System

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Modul : Pengenalan Sistem Kontrol Oleh : Ir..Hasanuddin Sirait, MT

BASIC PENGENALAN SISTEM KONTROL Blok Diagram :

Kelebihan : komponen-komponen relatif lebih murah dan cukup akurat Kekurangan : stabilitas menjadi persoalan utama

TRANSFORMASI LAPLACE

∞

α[f(t)] = F(s) = ∫ f(t).e −st .dt 0

A. Fungsi Step F(t) = 0 untuk t < 0 = A untuk t > 0

A

0

t

∞ A F(s) = ∫ A.e − st .dt = − e − st |∞ 0 s 0 =

−

A (0 − 1) s

=

A s

B. Fungsi Pulse F(t) = 0 untuk t < 0 & t >T

2


BASIC PENGENALAN SISTEM KONTROL = A untuk 0 ≤ t ≤ T

A

0

T

t

∞ F(s) = ∫ f(t).e − st .dt 0 T

= ∫ A.e − st .dt 0 A = − .e − st | T 0 s A = − (e − st − 1) s A = (1 − e − st ) s Fungsi Unit Step : f(t) = 1 (t) →F(s) = 1/s C. Fungsi Impulse

f(t) =

A lim to → 0 to

=0

F(s) =

3

untuk 0 < t < to untuk t < 0 & to < t

− st A o) lim (1 − e t s t →∞ 0 0


BASIC PENGENALAN SISTEM KONTROL d [A(1 − e − sto )] dto = lim d to → 0 (tos) dto A = s s =A Fungsi Unit-Impulse : f(t) = δ(t) F(s) = 1 D. Fungsi Ramp F(t) = 0 untuk t < 0 = At untuk t ≥ 0 A

0

1

t

∞ F(s) = ∫ A.t.e− st .dt 0

∞ = A ∫ t.e- st .dt 0 e − st ∞ ∞ A.e− st | −∫ dt = A.t. −s 0 0 −s A ∞ − st = .dt ∫e s 0 A = s2 E. Fungsi Eksponensiil F(t) = o untuk t < 0 = Ae

F(s)

4

− αt untuk t ≥ 0

∞ = ∫ A.e− αt .e− st .dt 0


BASIC PENGENALAN SISTEM KONTROL ∞ − (α + s)t =A∫e .dt 0 A − (s + α)t ∞ =− e | 0 s+α A =− (0 − 1) s+α A = s+α F. Fungsi Sinus f(t) = A sin ωt

∞ F(s) = ∫ A.sinω.si − st .dt 0 ∞ e jωt − e − jωt = ∫ A. .e− st dt 2j 0 A ∞ jωt − st − jωt − st = −e .e )dt ∫ (e .e 2j 0 A ∞ (jω − s)t − (jω + s)t = −e )dt ∫ (e 2j 0 A 1 1 (jω − s)t − (jω + s)t ∞ = [ e + e ] 0 2j jω − s jω + s A 1 1 = [ − ] 2j s − jω s + jω A s + jω − s + jω = . 2j s 2 + ω2 ω = A. 2 s + ω2 G. Fungsi Cosinus f(t) = A cos ωt F(s) = A.

s s + ω2 2

TEOREMA-TEOREMA TRANSFORMASI LAPLACE 1. Teorema Translasi Bila F(s) = L [ f(t) ], Maka L [f (t - α)] =

5

e− α.s .F(s)


BASIC PENGENALAN SISTEM KONTROL Bukti :

L[f(t-α)]

∞ = ∫ f(t − α)e − st .dt 0 ∞ − s(t − α) = ∫ f(t − α)e − αs .e 0 ∞ = ∫ f(τ(τ − sτdτ −α ∞ = e − αs ∫ f(τ(τ)− sτ dτ −α ∞ = e − αs ∫ f(t).e− st dt 0 = e − αs .F(s)

2. Teorema Perkalian Dengan

e− αt

Bila F(s) = L [ f(t) ], Maka : L [

e− αt .f(t) ] = F ( s + α )

Bukti : L[

∞ e− αt .f(t) ] = ∫ e − αt f(t)e− st .dt 0 ∞ − (s + α)t = ∫ f(t).e .dt 0 = F(s + α)

3. Teorema Diferensiasi Bila F(s) = L [ f(t) ], Maka : L [

df(t) ] = sF(s) – f(0) dt

Dimana f(0) adalah harga f(t) untuk t=0

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BASIC PENGENALAN SISTEM KONTROL L[

d 2f(t) 2 I ] = s F(s) - sf(0) – f (0) 2 dt

L[

d3f(t) ] = s3F(s) – s2f(0) – sfI(0) – fii(0) 3 dt

L[

∞ df(t) df(t) ]= ∫ ( )e− st .dt dt dt 0

Bukti :

∞ = ∫ e− st df(t) 0 ∞ = e − st .f(t) |∞ − ∫ f(t)de− st 0 0 ∞ = 0 − f(0) + s ∫ f(t).e− st dt 0 = −f(0) + sF(s) 4. Teorema Integrasi Bila F(s) = L [ f(t) ], Maka : L [

∫ f(t)dt ] =

F(s) f − 1(0) + s s

Dimana f-1(0) adalah

∫ f(t)dt untuk t = 0

Bukti : L[

∫ f(t)dt ] =

∞ − st ∫ [ ∫ f(t)dt]e dt 0

1∞ = − ∫ [ ∫ f(t)dt]de− st s0 ∞ 1 = − [e− st ∫ f(t)dt |∞ − ∫ e − st f(t)dt ] 0 s 0 1 = − [0 − f − 1(0) − F(s)] s F(s) f − 1(0) = + s s

7


BASIC PENGENALAN SISTEM KONTROL ∴L[

∫ ( ∫ f(t)dt)dt ] =

F(s) f − i (0) f − ii (0) + + s s2 s2

5. Teorema Harga Awal Dan Harga Akhir A.

lim f(t) = lim sF(s) t→0 s→∞

B.

lim f(t) = lim sF(s) t→∞ s→0

Bukti : A.

∞ df(t) lim ∫ [ ]e− st dt = 0 dt s→∞0 lim sF(s) − f(0) = 0 s→∞

f(0) = lim f(t) = lim sF(s) t→0 s→∞ B.

df(t) lim L[ ] = lim [sF(s) − f(0)] dt s→0 s→0

= lim sF(s) − f(0) s→0 karena

lim e − st = 1 s→0

∞ df(t) ]dt = f(t) |∞ ∫[ 0 0 dt

= f(∞) − f(0) = lim sF(s) - f(0) s→0 ∴

f(∞) = lim f(t) = lim sF(s) t→∞ s→0

INVERSI TRANSFORMASI LAPLACE Untuk mencari fungsi waktu f(t) dari transformasi laplacenya -I

L [ F(s) ] = f(t) Metode Ekspansi Pembagian Parsial (Partial Fraction Expantion) F(s) = F1(s) + F2(s) + ….. + Fn(s)

8


BASIC PENGENALAN SISTEM KONTROL -I

-I

-I

-I

L [ F(s) ] = L [ F1(s) ] + L [ F2(s) ] + ….. + L [ Fn(s) ] F(t) = f1(t) + f2(t) + ….. + fn(t) Contoh :

s+3 (s + 1)(s + 2)

1. F(s) =

s+3 a1 a = + 2 (s + 1)(s + 2) s + 1 s + 2

F(s) =

=

[

s+3 (s + 1)] =2 s = −1 (s + 1)(s + 2)

a2 =

[

s+3 (s + 2)] = −1 s = −2 (s + 1(s + 2)

a1

-1

f(t) = L [ F(s) ] -1

=L [ = 2. e 2. G(s) =

2 −1 -1 ]+L [ ] s +1 s+2

− t − e − 2t

s3 + 5s 2 + 9s + 7 (s + 1)(s + 2)

G(s) = s + 2 +

G(t) =

3. F(s) =

s+3 (s + 1)(s + 2)

d δ(t) + 2δ (t) + 2e − t − e − 2t dt

s +1 s(s + s + 1) 2

F(s) =

α s+α s +1 2 +a = 1 2 s(s + s + 1) s + s + 1 s 2

Untuk mendapatkan α1 dan α2 :

α s+α s +1 2 +a = 1 2 s(s + s + 1) s + s + 1 s 2

α s+α s +1 1 2 = s(s 2 + s + 1) s 2 + s + 1 s +1 . s2 + s + 1 = α s + α | 1 2 s = −0,5 − j0,866 s(s + s + 1) 2

0,5 − j0,866 = α (−0,5 − j0,866) + α 1 2 − 0,5 − j0,866

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BASIC PENGENALAN SISTEM KONTROL 0,5 – j0.866 = α1 (0,25 + j0,866 – 0,75) + α2 (-0,5 – j0,866) Real : 0,5 = -0,5α1 – 0,5α2 → α1 + α2 = -1 Imajiner : -0,866 = 0,866α1 – 0,866α2 → α1 + α2 = -1 α1 = -1 , α2 = 0 Untuk mendapatkan a :

a =[

s(s + 1) ] =1 s=0 s(s 2 + s + 1)

F(s) =

=

−s 1 + 2 s + s +1 s 1 s + 0,5 0,5 − + s (s + 0,5)2 + o,8662 (s + 0,5)2 + 0,8662

f(t) = L-1 [ F(s) ] =1–

4. F(s) =

F(s) =

e 0,5t cos0,866t + 0,578e − 0,5t sin0,866t

s 2 + 2s + 3 (s + 1)3 b3

+

(s + 1)3

b2 (s + 1)2

+

b1 (s + 1)

s 2 + 2s + 3 .(s + 1)3 ]s= -1 3 (s + 1)

b3 = [

2

=

(s + 2s + 3)s= -1

=2 b2 =

1 d s 2 + 2s + 3 { [ .(s + 1)3 ]} s = −1 1! ds (s + 1)3

= (2s +2)s= -1 =0 b1 =

1 d 2 s 2 + 2s + 3 { [ .(s + 1)3 ]} s = −1 2! ds 2 (s + 1)3

= ½ . (2) =1 -1

f(t) = L [ F(s) ]

2

-1

=L [ 2

(s + 1)3 -t

-1

]+L [

1 ] s +1

-t

=t .e +e

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BASIC PENGENALAN SISTEM KONTROL SOAL LATIHAN 1. F(s) =

2. F(s) =

s +1

→f(t) = … ?

s 2 + 5s + 6 5(s + 2) (s + 1)3 (s + 3)(s 2 + s + 5)

→f(t) = … ?

3. f(t) = A cos (ωt + ϕ)

→F(s)= … ?

4. f(t) = 0 untuk t < 0 & t > 2T -A untuk 0 ≤ t < T

→ F(s) = … ?

A untuk T ≤ t ≤ 2T 5. → F(s) = … ?

A

T

11

2T

t


BASIC PENGENALAN SISTEM KONTROL PENYELESAIAN PERSAMAAN DIFFERENSIAL Contoh : 1. Selesaikan persamaan differensial berikut :

.. . . x + 3 x + 6x = 0, x(0) = 0, x (0) = 3 Transformasi laplace dari persamaan differential diatas menghasilkan : 2

s X(s) – sx(0) -

. x (0) + 3(sX(s) – x(0)) + 6X(s) = 0

2

s X(s) – 0 – 3 + 3 (sX(s) – 0) + 6X(s) = 0 X(s) (s2 + 3s + 6) = 3 X(s) =

3 s 2 + 3s + 6

Untuk mendapatkan x(t) : X(s) =

=

3 9 9 s 2 + 3s + − + 6 4 4

3 (s + 3/2) 2 + 15/4 3

=

(s + 3/2) 2 + ( =

6 1/2 15 15 (s + 3/4) 2 + (1/2 15 ) 2

∴ x(t) =

12

1 15 ) 2 2

2 15e− 3/2t .Sin[(1/2 15 )t] 5


BASIC PENGENALAN SISTEM KONTROL

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