Seborg
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12345678983.1 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