Bai Tap Ly Thuyet Dieu Khien Tu Dong 01

Bµi tËp Lý thuyÕt ®iÒu khiÓn tù ®éng

§µo V¨n S¬n T§H K47

Bµi tËp lý thuyÕt ®iÒu khiÓn tù ®éng

Hä tªn: §µo V¨n S¬n Líp: Tù ®éng ho¸47

§Ò bµi: HÖ thèng cã hµm truyÒn ®¹t W(p) =

5 p + 3p2 + 3p +1 3

-Thµnh lËp ph−¬ng tr×nh vi ph©n tæng qu¸t cña hÖ -Thµnh lËp hÖ ph−¬ng tr×nh tr¹ng th¸i -Gi¶i hÖ ph−¬ng tr×nh tr¹ng th¸i kh«ng thuÇn nhÊt,dõng víi ®iÒu kiÖn ®Çu: x1(0) = 0, x2(0) = 1, x3(0) = 1, u(t) = 1.

Bµi lµm W(p) =

-Tõ hµm truyÒn ®¹t cña hÖ thèng

5 p + 3p + 3p +1 3

2

=

Y ( p) U ( p)

⇒ ( p + 3 p + 3 p + 1)Y ( p) = 5U ( p ) ⇒ &y&&+ 3 & y&+ 3 y&+ y = 5u 3

2

§ã chÝnh lµ ph−¬ng tr×nh vi ph©n cña hÖ thèng. -

§Æt: x

1

x

2

x 3 x& 3

=

y = x&1 = x& 2 = & x&1 = & x&2 = &x&&1 = &y&&

x& 3 = − x

1

− 3 x

2

− 3 x

3

+ 5 u

VËy ta cã hÖ ph−¬ng tr×nh tr¹ng th¸i:

⎧ x&1 = x 2 ⎪ ⎨ x&2 = x 3 ⎪ x& = − x − 3 x − 3 x + 5 u 1 2 3 ⎩ 3

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1

Bµi tËp Lý thuyÕt ®iÒu khiÓn tù ®éng

§µo V¨n S¬n T§H K47

Hay: ⎡ x&1 ⎢ x& ⎢ 2 ⎢⎣ x&3

⎤ ⎡ 0 ⎥ = ⎢ 0 ⎥ ⎢ ⎥⎦ ⎢⎣ − 1

1 0 − 3

0 ⎤ ⎡ x1 1 ⎥⎥ ⎢⎢ x 2 − 3 ⎥⎦ ⎢⎣ x 3

⎤ ⎡0 ⎤ ⎥ + ⎢0 ⎥u ⎥ ⎢ ⎥ ⎥⎦ ⎢⎣ 5 ⎥⎦

-Gi¶i ph−¬ng tr×nh tr¹ng th¸i kh«ng thuÇn nhÊt, dõng : 1 0⎤ ⎡0 ⎢ A=⎢0 0 1 ⎥⎥ ; B = ⎢⎣− 1 − 3 − 3⎥⎦ At X¸c ®Þnh Φ(t ) = e theo biÕn ®æi Laplace:

⎡0 ⎤ ⎢0 ⎥ ⎢ ⎥ ⎢⎣ 5 ⎥⎦

Φ( p) = [ pI − A] ⎡1 0 [ pI − A ] = p ⎢⎢ 0 1 ⎢⎣ 0 0

−1

0⎤ ⎡ 0 1 0 ⎤ ⎡p −1 ⎥ ⎢ ⎥ 0⎥ − ⎢ 0 0 1 ⎥ = ⎢⎢ 0 p ⎢⎣ 1 1 ⎥⎦ ⎢⎣ − 1 − 3 − 3 ⎥⎦ 3 2 3 2 3 det( pI − A) = p( p + 3 p + 3) + 1 = p + 3 p + 3 p + 1 = ( p + 1) −1

⎤ − 1 ⎥⎥ p + 3 ⎥⎦ 0

⎡ p2 + 3 p + 3 ⎡ p2 + 3p + 3 p+3 −1 −p ⎤ ⎢ ⎥ ⎢ 2 −1 p2 + 3 p adj ( pI − A) = ⎢ p + 3 p + 3 p − 3 p − 1⎥ = ⎢ ⎢ ⎢ 1 −p − 3 p −1 p p 2 ⎥⎦ ⎣ ⎣ ⎡ p2 + 3p + 3 1 ⎤ p+3 ⎢ ⎥ 3 3 ( p + 1) ( p + 1) 3 ⎥ ⎢ ( p + 1) −1 p2 + 3p p ⎥ adj ( pI − A) ⎢ ⇒ Φ( p ) = =⎢ 3 3 det( pI − A) ( p + 1) ( p + 1) ( p + 1) 3 ⎥ ⎢ ⎥ −p − 3p −1 p2 ⎥ ⎢ ⎢⎣ ( p + 1) 3 ( p + 1) 3 ( p + 1) 3 ⎥⎦

1⎤ ⎥ p⎥ p 2 ⎥⎦

Tra b¶ng biÕn ®æi Laplace ta cã: ⎧ 1 ⎫ t 2 −t = e L−1 ⎨ 3⎬ ⎩ ( p + 1) ⎭ 2 ⎧ p ⎫ t 2 −t L−1 ⎨ = t − ( )e ⎬ 3 2 ⎩ ( p + 1) ⎭

⎧ p2 ⎫ t 2 −t L−1 ⎨ = ( 1 − 2 t + )e ⎬ 3 2 ⎩ ( p + 1) ⎭

Suy ra: ⎧ p 2 + 3 p + 3⎫ t 2 −t Φ 11 (t ) = L−1 ⎨ = ( 1 + t + )e ⎬ 3 2 ⎩ ( p + 1) ⎭ ⎧ −1 ⎫ t 2 −t Φ 21 (t ) = L−1 ⎨ = − e 3⎬ 2 ⎩ ( p + 1) ⎭ ⎧ −p ⎫ t 2 −t Φ 31 (t ) = L−1 ⎨ = − ( t − )e ⎬ 3 2 ⎩ ( p + 1) ⎭

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Bµi tËp Lý thuyÕt ®iÒu khiÓn tù ®éng

§µo V¨n S¬n T§H K47

⎧ p+3 ⎫ = (t 2 + t )e −t Φ 12 (t ) = L−1 ⎨ 3⎬ ⎩ ( p + 1) ⎭ ⎧ p2 + 3p ⎫ Φ 22 (t ) = L−1 ⎨ = (1 + t − t 2 )e −t 3 ⎬ ⎩ ( p + 1) ⎭ ⎧ − 3 p − 1⎫ = (−3t + t 2 )e −t Φ 32 (t ) = L−1 ⎨ 3 ⎬ p ( + 1 ) ⎭ ⎩ ⎧ 1 ⎫ t 2 −t = e Φ 13 (t ) = L−1 ⎨ 3⎬ ⎩ ( p + 1) ⎭ 2 ⎧ p ⎫ t 2 −t = t − Φ 23 (t ) = L−1 ⎨ ( )e 3⎬ 2 ⎩ ( p + 1) ⎭ ⎧ p2 ⎫ t 2 −t Φ 33 (t ) = L−1 ⎨ ( 1 2 )e = − t + 3⎬ 2 ⎩ ( p + 1) ⎭

⎡ t 2 −t ( 1 t )e + + ⎢ 2 ⎢ t2 Φ(t ) = ⎢ − e −t ⎢ 2 ⎢ t 2 −t ⎢ − (t − )e 2 ⎣

(t 2 + t )e −t (1 + t − t 2 )e −t (−3t + t 2 )e −t

⎤ t 2 −t e ⎥ 2 ⎥ 2 t −t ⎥ (t − )e ⎥ 2 t 2 −t ⎥ (1 − 2t + )e ⎥ 2 ⎦

⎤ ⎡ ⎤ ⎡ 5t 2 −t 5(t − τ ) 2 −(t −τ ) e e ⎥ ⎢ ⎥ ⎢ 2 2 ⎥ ⎢ ⎥ ⎢ 2 2 t (t − τ ) −(t −τ ) ⎥ ]e Φ(t ) × B × U = ⎢ 5(t − )e −t ⎥ ⇒ Φ (t − τ ) × B × U = ⎢ 5[(t − τ ) − ⎥ ⎢ ⎥ ⎢ 2 2 ⎢ ⎢ t 2 −t ⎥ (t − τ ) 2 −(t −τ ) ⎥ ]e ⎥ ⎢5[1 − 2(t − τ ) + ⎢5(1 − 2t + )e ⎥ 2 2 ⎦ ⎣ ⎦ ⎣

Suy ra: t ⎡ ⎤ 5(t − τ ) 2 −(t −τ ) ⎢ ⎥ ∫0 2 e dτ ⎢ ⎥ t ⎢ t ⎥ (t − τ ) 2 −(t −τ ) ∫0 Φ(t − τ ) × B × u(τ )dτ =⎢ ∫0 5[(t − τ ) − 2 ]e dτ ⎥ ⎢t ⎥ ⎢ (t − τ ) 2 −(t −τ ) ⎥ ]e dτ ⎥ ⎢ ∫ 5[1 − 2(t − τ ) + 2 ⎣0 ⎦

Tr−íc tiªn ta tÝnh c¸c tÝch ph©n sau: t

∫e

− ( t −τ )

dτ = − e − t + 1

0 t

∫ (t − τ )e

− ( t −τ )

dτ = −(t + 1)e − t + 1

0 t

∫ (t − τ )

2

e −(t −τ ) dτ = −(t 2 + 2t + 2)e −t + 2

0

Do ®ã ta cã: [email protected]

3

Bµi tËp Lý thuyÕt ®iÒu khiÓn tù ®éng

t

*∫5 0 t

(t − τ ) 2 −(t −τ ) 5 5 t2 e dτ = ∫ (t − τ ) 2 e −(t −τ ) dτ = − (t 2 + 2t + 2)e −t + 2 = −5( + t + 1)e −t + 5 2 20 2 2

[

t

* ∫ 5[(t − τ ) − 0

§µo V¨n S¬n T§H K47

]

5t 2 −t (t − τ ) 2 −(t −τ ) 5 e ]e dτ = 5 ∫ (t − τ )e −(t −τ ) dτ - ∫ (t − τ ) 2 e −(t −τ ) dτ = 2 20 2 0 t

t

* ∫ 5[1 − 2(t − τ ) + 0

t

(t − τ ) 2 −(t −τ ) 5 ]e dτ = 5∫ e −( t −τ ) dτ − 10∫ (t − τ )e −(t −τ ) dτ + ∫ (t − τ ) 2 e −(t −τ ) dτ 2 20 0 0 t

t

t

t2 − t )e −t 2 ⎡ ⎤ t2 − + t + 1) e − t + 5 ⎥ 5 ( ⎢ 2 ⎢ ⎥ t 5t 2 −t ⎢ ⎥ Φ − × × = ( t ) B u ( ) d e τ τ τ ∫0 ⎢ ⎥ 2 ⎢ ⎥ t2 − 5( − t )e −t ⎢ ⎥ 2 ⎣ ⎦ ⎡0 ⎤ Tõ gi¶ thiÕt ta cã: X ( 0 ) = ⎢⎢ 1 ⎥⎥ ⎢⎣ 1 ⎥⎦ = −5(

Suy ra:

⎡ 3t 2 − t ⎤ + ( t )e ⎢ ⎥ 2 ⎢ ⎥ 2 3t Φ ( t ) × X ( 0 ) = ⎢ (1 + 2 t − )e −t ⎥ ⎢ ⎥ 2 ⎢ 3t 2 − t ⎥ )e ⎥ ⎢ (1 − 5 t + 2 ⎣ ⎦ t

VËy: X (t ) = Φ (t ) × X (0) + ∫ Φ (t − τ ) × B × u (τ )dτ 0

⎡ ⎤ ⎡ ⎤ t2 3t −t + − + t + 1)e −t + 5⎥ 5 ( t e ( ) ⎢ ⎥ ⎢ 2 −t 2 2 ⎢ ⎥ ⎢ ⎥ ⎡− (t + 4t + 5)e + 5⎤ 2 2 5t −t 3t ⎥ = ⎢ (t 2 + 2t + 1)e −t ⎥ = ⎢(1 + 2t − )e − t ⎥ + ⎢ e ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 2 2 −t ⎢ ⎥ 2 2 − + ( t 1 ) e ⎢ ⎥ ⎢ ⎥ 3t t ⎣ ⎦ −t −t )e ⎥ ⎢ − 5( − t )e ⎢ (1 − 5t + ⎥ 2 2 ⎣ ⎦ ⎣ ⎦ 2

KÕt luËn: ⎡ − (t 2 + 4 t + 5 ) e − t + 5 ⎤ ⎢ ⎥ X (t ) = ⎢ ( t + 1) 2 e − t ⎥ 2 −t ⎢ ⎥ ( 1 t ) e − ⎣ ⎦

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