Seborg

Disponível somente no TrabalhosFeitos
  • Páginas : 11 (2642 palavras )
  • Download(s) : 0
  • Publicado : 21 de novembro de 2012
Ler documento completo
Amostra do texto
1234567898

3.1 a) 1e

[

− bt

sin ωt = ∫ e
0

]



− bt

sin ωt e dt = ∫ sin ωt e − ( s + b )t dt
− st 0




 [− (s + b) sin ωt − ω cos ωt ] = e − ( s + b ) t  ( s + b ) 2 + ω2  0 ω = ( s + b) 2 + ω2

b)

1 e

[

− bt

cos ωt = ∫ e
0

]



− bt

cos ωt e dt = ∫ cos ωt e − ( s + b )t dt
− st 0




 [− (s + b) cos ωt + ω sin ωt ] = e− ( s + b ) t  ( s + b) 2 + ω2  0 s+b = ( s + b) 2 + ω2

3.2

a)

The Laplace transform provided is
Y ( s) = 4 s + 3s + 4 s 2 + 6 s + 4
4 3

We also know that only sin ωt is an input, where ω =
X ( s) = ω 2 = 2 s +ω s2 + 2
2

2 . Then

( )

2

=

2 s +2
2

Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when all initial conditions are zero),Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.

3-1

Y (s) =

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

and the original ode was

d2y dy + 3 + 2 y = 2 2 sin 2t 2 dt dt
b) c) This is a unique result. The solution arguments can be found from Y (s) = 2 2 2 ( s + 1)( s + 2) + ( s 2 + 2)

with y ′(0) = y (0) =0

which in partial fraction form is
Y (s) = α1 α a s + a2 + 2 + 12 s +1 s + 2 s +2

Thus the solution will contain four functions of time e-t , e-2t , sin 2 t , cos 2 t

3.3

a)

Pulse width is obtained when x(t) = 0 Since x(t) = h – at tω : h − atω = 0 or tω = h/a

b)
h slope = -a x(t) x(t) slope = a

slope = -a

x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω) 3-2

c) d)

h a ae− stω h e − stω − 1 X ( s) = − 2 + 2 = + s s s s s2 Area under pulse = h tω/2

3.4

a)

f(t) = 5 S(t) – 4 S(t-2) – S(t- 6) 1 F (s ) = = ( 5 - 4e -2s - e -6s ) s

b)

x(t) x1 a a tr -a 2tr -a 3tr x4

x2

x3

x(t) = x1(t) + x2(t) + x3(t) + x4(t) = at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr) following Eq. 3-101. Thus X(s) =
a 1 − e −tr s − e − 2tr s + e −3trs 2 s

[

]

by utilizing the Real Translation Theorem Eq. 3-104.

3-3

3.5

T(t) = 20 S(t) +

55 55 t S(t) – (t-30) S(t-30) 30 30 20 55 1 55 1 −30 s 20 55 1 T (s) = + − e = + 1 − e −30 s 2 2 2 s 30 s 30 s s 30 s

(

)

3.6 a)
X ( s) = α α α s ( s + 1) = 1 + 2 + 3 ( s + 2)( s + 3)( s + 4) s + 2 s + 3 s + 4

α1 =

s ( s + 1) ( s + 3)( s + 4) s ( s + 1) ( s + 2)( s + 4) s( s + 1) ( s + 2)( s + 3)

=1
s = −2

α2 =

= −6
s = −3

α3 =

=6
s = −4

X (s) =

1 6 6 − + s+2 s+3 s+4

and

x(t ) = e−2t − 6e −3t + 6e −4t

b)

X (s) =

s +1 s +1 = 2 ( s + 2)( s + 3)( s + 4) ( s + 2)( s + 3)( s + 2 j )( s − 2 j ) α + jβ 3 α 3 − j β 3 α1 α + 2 + 3 + s+2 s+3 s+2j s+2j

X ( s) =

α1 = α2 =

s +1 ( s + 3)( s 2 + 4) s +1 ( s + 2)( s 2 + 4)

=−
s= −2

1 8

=
s = −3

2 13
=
s = −2 j

α 3 + jβ 3 =

s +1 ( s + 2)( s + 3)( s − 2 j )

1− 2 j − 3 + 11 j = − 40 − 8 j 208

3-4

1 2  −3   11  x(t ) = − e − 2t + e −3t + 2  cos 2t + 2  sin 2t 8 13  208   208 
1 2 3 11 = − e − 2t + e −3t − cos 2t + sin 2t 8 13 104 104

c)

X ( s) =

α α2 s+4 = 1 + 2 s + 1 ( s + 1) 2 ( s + 1)

(1)

α 2 = ( s + 4) s =−1 = 3In Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1 gives
4 α1 3 = + 1 12 12 or α1 = 1

d)

1 3 + and x( t ) = e− t + 3te − t 2 s + 1 ( s + 1) 1 1 1 X (s) = 2 = = 2 2 2 s + s +1  1  3 (s + b ) + ω s +  + 2 4  1 3 where b = and ω = 2 2 X (s) =

1 2 −2 3 x(t ) = e −bt sin ω t = e sin t ω 2 3 e) X(s) = s +1 e −0.5 s s ( s + 2)( s + 3)

t

To invert, we firstignore the time delay term. Using the Heaviside expansion with the partial fraction expansion, ˆ X ( s) = s +1 A B C = + + s ( s + 2)( s + 3) s s + 2 s + 3

Multiply by s and let s → 0

3-5

1 1 = (2)(3) 6 Multiply by (s+2) and let s→ −2 A= B= − 2 +1 −1 1 = = (−2)(−2 + 3) (−2)(1) 2

Multiply by (s+3) and let s→-3 C= − 3 +1 −2 2 = =− (−3)(−3 + 2) (−3)(−1) 3

Then
1 6 1 2 −2 3 ˆ X (s)...
tracking img