1234567898
4.1
a) b) c) d)
iii iii v v
a)
5
b)
10
4.2
c)
d) e)
10 s (10 s + 1) From the Final Value Theorem, y(t) = 10 when t→∞ Y (s) =
y(t) = 10(1−e−t/10) , then y(10) = 6.32 = 63.2% of the final value.
5 (1 − e − s ) (10s + 1) s From the Final Value Theorem, y(t) = 0 when t→∞ Y (s) =
5 1 (10 s + 1) From the Final Value Theorem, y(t)= 0 when t→∞ Y (s) =
f)
g)
Y (s) =
5 6 2 (10 s + 1) ( s + 9)
then
y(t) = 0.33e-0.1t − 0.33cos(3t) + 0.011sin(3t) The sinusoidal input produces a sinusoidal output and y(t) does not have a limit when t→∞. Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
4-1
By using Simulink-MATLAB, above solutions can be verified: 10
0.5
9
0.45
8
0.4
7
0.35
0.3
y(t)
y(t)
6
5
0.25
4
0.2
3
0.15
2
0.1
1
0.05
0
0
5
10
15
20
25
30
35
40
45
0
50
0
5
10
15
20
25
30
35
40
45
50
time
time
Fig S4.2a. Output for part c) and d)
Fig S4.2b. Output for part e) 0.7
5
0.6 4.5
0.5 4
0.4 3.5
0.3
y(t)
y(t)
3
2.5
0.2
0.1 2
0
1.5
-0.1
1
-0.2
0.5
0
-0.3 0
5
10
15
20
25
30
35
40
45
50
0
2
4
6
8
10
12
14
16
18
20
time
time
Fig S4.2c. Output for part f)
Fig S4.2d. Output for part g)
4.3
a)
The dynamic model of the system is given by dV 1 = ( wi − w) dt ρ dT wi Q = (Ti − T ) + dt Vρ VρC
(2-45) (2-46)
Let the right-hand side of Eq. 2-46 be f(wi,V,T), ∂f dT = f (wi ,V , T ) = dt ∂wi
∂f ∂f ′ wi′ + V′+ T ∂V s ∂T s s
4-2
(1)
∂f ∂wi
1 = (Ti − T ) s Vρ
w Q 1 dT ∂f =− = − 2i (Ti − T ) − 2 =0 V dt s V ρ V ρC ∂V s w ∂f =− i Vρ ∂T s w dT 1 = (Ti − T ) wi′ − i T ′ , dt V ρ Vρ
dT dT ′ = dt dt
Taking Laplace transform and rearranging T ′( s ) (Ti − T ) / wi = Wi′( s ) Vρ s + 1 wi
(2)
Laplace transform of Eq. 2-45 gives V ′( s ) =
Wi′( s ) ρs
(3)
∂f If were not zero, then using (3) ∂V s (Ti − T ) V ∂f 1 + wi ∂V s s T ′( s ) wi = Wi′( s ) Vρ s + 1 wi
(4)
Appelpolscher guessed the incorrect form (4) instead of the correct form ∂f (2) because he forgot that would vanish. ∂V s b)
From Eq. 3, V ′( s ) 1 = Wi′( s ) ρs
4-3
4.4 Y( s ) = G( s )X ( s ) =
G(s)
Interpretation of G(s)
K s( τs + 1 )
U(s)
Interpretation of u(t)
K s (τs + 1)
2nd order process *
1
δ(0)
[ Delta function]
K τs + 1
1st order process
1 s
S(0)
[Unit step function]
K s
Integrator
K τs + 1
K
Simple gain
1 s (τs + 1)
(i.e no dynamics)
1 −t / τ e [Exponential input] τ
1 − e −t / τ
[Step + exponential input]
* nd
2 order or combination of integrator and 1st order process
4.5 a)
dy 1 = -2y1 – 3y2 + 2u1 dt dy 2 = 4y1 – 6y2 + 2u1 + 4u2 dt
2
(1) (2)
Taking Laplace transform of the above equations and rearranging, (2s+2)Y1(s) + 3Y2(s) = 2U1(s)
(3)
-4 Y1(s) + (s+6)Y2(s)=2U1(s) + 4U2(s)
(4)
Solving Eqs. 3 and 4 simultaneously for Y1(s) and Y2(s), 4-4
Y1(s) =
(2 s + 6) U 1 ( s ) − 12 U 2 ( s ) 2( s + 3) U1 ( s ) − 12 U 2 ( s ) = 2( s + 3)( s + 4) 2 s 2 + 14 s + 24
Y2(s) =
(4 s + 12) U 1 ( s ) − (8s + 8) U 2 ( s ) 4( s + 3) U 1 ( s ) + 8( s + 1) U 2 ( s ) = 2( s + 3)( s + 4) 2 s 2 + 14 s + 24
Therefore,
Y1 ( s ) 1 = U1 (s) s + 4
,
Y1 ( s ) −6 = U 2 ( s ) ( s + 3)( s + 4)
Y2 ( s ) 2 = U1 (s) s + 4
,
Y2 ( s ) 4( s + 1) = U 2 ( s ) ( s + 3)( s + 4)
4.6
The physical model of the CSTR is (Section 2.4.6) V
dc A = q (c Ai − c A ) − Vkc A dt
VρC
(2-66)
dT = wC (Ti − T ) + (− ∆H )Vkc A + UA(Tc − T ) dt
(2-68)
k = ko e-E/RT
(2-63)
where:
These equations can be written as, dc A = f1 (c A , T ) dt
(1)
dT = f 2 (c A , T , Tc ) dt
(2)
Because both equations are nonlinear, linearization is required. After linearization and introduction of deviation variables, we could get an expression for c ′A (s ) / T ′(s ) .
4-5
But it is not possible to get an expression for T ′(s ) / Tc′(s ) from (2) due to the presence of cA in (2). Thus the proposed approach is not feasible because the CSTR is an interacting system. Better approach: After linearization etc., solve for T ′(s ) from (1) and substitute into the linearized version of (2). Then rearrange to obtain the desired, C A′ ( s ) / Tc′(s ) (See Section 4.3)
4.7
a)
The assumption that H is constant is redundant. For equimolal overflow, L0 = L1 = L
, V1 = V2 = V
dH = L0 + V2 − L1 − V1 = 0 dt
, i.e., H is constant.
The simplified stage concentration model becomes dx1 = L( x0 − x1 ) + V ( y 2 − y1 ) dt y1 = a0 + a1x1 + a2x12 +a3x13
(1)
H
b)
(2)
Let the right-hand side of Eq. 1 be f(L, x0, x1, V, y1, y2)
H
∂f ∂f dx1 ∂f x 0′ + x1′ = f ( L, x0 , x1 , V , y1 , y 2 ) = L ′ + dt ∂L s ∂x1 s ∂x 0 s
∂f ′ ∂f ′ ∂f ′ + V + y1 + ∂y y 2 ∂V s ∂y1 s 2 s Substituting for the partial derivatives and noting that H
dx1 dx1′ = dt dt
dx1′ = ( x0 − x1 ) L ′ + L x0′ − L x1′ + ( y 2 − y1 )V ′ + V y 2′ − V y1′ dt
4-6
(3)
Similarly, ∂g x1′ = (a1 + 2a 2 x1 + 3a 3 x1 2 ) x1′ y1′ = g ( x1 ) = ∂x1 s c)
(4)
For constant liquid and vapor flow rates, L ′ = V ′ = 0 Taking Laplace transform of Eqs. 3 and 4, HsX 1′ ( s ) = L X 0′ ( s ) − L X 1′ ( s ) + V Y2′ ( s ) − V Y1′( s )
(5)
Y1′( s ) = (a1 + 2a 2 x1 + 3a3 x1 ) X 1′ ( s )
(6)
2
From Eqs. 5 and 6, the desired transfer functions are L τ X 1′ ( s ) = H X 0′ ( s ) τs + 1 Y1′( s ) = X 0′ ( s ) Y1′( s ) = Y2′( s )
V τ X 1′ ( s ) = H Y2′ ( s ) τs + 1
,
(a1 + 2a 2 x1 + 3a 3 x1 ) 2
τs + 1
(a1 + 2a 2 x1 + 3a 3 x1 ) 2
τs + 1
L τ H
V τ H
where
τ=
H L + V (a1 + 2a 2 x1 + 3a3 x1 ) 2
4.8
From material balance, d (ρAh) = wi − Rh1.5 dt
dh 1 R 1.5 = wi − h dt ρA ρA 4-7
We need to use a Taylor series expansion to linearize dh 1 R 1.5 1 1.5Rh 0.5 = wi − h + ( wi − wi ) − (h − h ) dt ρA ρA ρA ρA Since the bracketed term is identically zero at steady state,
dh ′ 1 1.5Rh 0.5 = wi′ − h′ dt ρA ρA Rearranging ρ A dh ′ 1 + h′ = wi′ 0.5 dt 1.5 Rh 1.5 Rh 0.5
Hence where
H ′( s ) K = Wi′( s ) τs + 1
1 h h [height ] = = = 0.5 1.5 1.5w [ flowrate] 1.5 Rh 1.5 Rh [mass] = [time] ρA ρAh ρV τ= = = = 0.5 1.5 1.5w [mass / time] 1.5 Rh 1.5 Rh K=
4.9
a)
The model for the system is given by mC
dT = wC (Ti − T ) + h p A p (Tw − T ) dt
mw C w
dTw = hs As (Ts − Tw ) − h p A p (Tw − T ) dt
(2-51)
(2-52)
Assume that m, mw, C, Cw, hp, hs, Ap, As, and w are constant. Rewriting the above equations in terms of deviation variables, and noting that
4-8
dT dT ′ = dt dt
dTw dTw′ = dt dt
dT ′ = wC (Ti′ − T ′) + h p A p (Tw′ − T ′) dt dT ′ m w C w w = hs As (0 − Tw′ ) − h p A p (Tw′ − T ′) dt mC
Taking Laplace transforms and rearranging, (mCs + wC + h p A p )T ′( s ) = wCTi′( s ) + h p A pTw′ ( s )
(1)
(mw C w s + hs As + h p A p )Tw′ ( s ) = h p A p T ′( s )
(2)
Substituting in Eq. 1 for Tw′ (s ) from Eq. 2, (mCs + wC + h p A p )T ′( s ) = wCTi′( s ) + h p A p
h p Ap (mw C w s + hs As + h p A p )
T ′( s )
Therefore,
wC (mw C w s + hs As + h p A p ) T ′( s ) = Ti′( s ) (mCs + wC + h p A p )(mw C w s + hs As + h p Ap ) − (h p Ap ) 2
b)
c)
wC (hs As + h p A p ) T ′( s ) The gain is = Ti′( s ) s =0 wC (hs As + h p A p ) + hs As h p A p
No, the gain would be expected to be 1 only if the tank were insulated so that hpAp= 0. For heated tank the gain is not 1 because heat input changes as T changes.
4.10
Additional assumptions 1) 2) 3)
perfect mixing in the tank constant density, ρ , and specific heat, C. Ti is constant.
4-9
Energy balance for the tank, ρVC
dT = wC (Ti − T ) + Q − (U + bv) A(T − Ta ) dt
Let the right-hand side be f(T,v), ρVC
dT ∂f ′ ∂f ′ = f (T , v) = T + v dt ∂T s ∂v s
(1)
∂f = − wC (U + bv ) A ∂T s ∂f = − bA(T − Ta ) ∂v s Substituting for the partial derivatives in Eq. 1 and noting that ρVC
dT ′ = − [wC + (U + bv ) A]T ′ − bA(T − Ta )v ′ dt
dT dT ′ = dt dt
Taking the Laplace transform and rearranging
[ρVCs + wC + (U + bv ) A] T ′(s) = − bA(T − T )v′ (s) a
− bA(T − Ta ) wC + (U + bv ) A T ′( s ) = v ′( s ) ρVC wC + (U + bv ) A s + 1
4.11
a)
Mass balances on surge tanks dm1 = w1 − w2 dt
(1)
dm2 = w2 − w3 dt
(2)
4-10
Ideal gas law m1 RT M m P2V2 = 2 RT M P1V1 =
Flows
(3) (4)
(Ohm's law is I =
E Driving Force = ) R Resistance
1 ( Pc − P1 ) R1 1 w2 = ( P1 − P2 ) R2 1 w3 = ( P2 − Ph ) R3 w1 =
(5) (6) (7)
Degrees of freedom: number of parameters : 8 (V1, V2, M, R, T, R1, R2, R3) number of variables : 9 (m1, m2, w1, w2, w3, P1, P2, Pc, Ph) number of equations : 7 ∴
number of degrees of freedom that must be eliminated = 9 − 7 = 2
Since Pc and Ph are known functions of time (i.e., inputs), NF = 0. b)
Development of model Substitute (3) into (1) :
MV1 dP1 = w1 − w2 RT dt
(8)
Substitute (4) into (2) :
MV2 dP2 = w2 − w3 RT dt
(9)
Substitute (5) and (6) into (8) :
MV1 dP1 1 1 = ( Pc − P1 ) − ( P1 − P2 ) RT dt R1 R2 MV1 dP1 1 1 1 1 = Pc (t ) − ( + ) P1 + P2 RT dt R1 R1 R2 R2
4-11
(10)
Substitute (6) and (7) into (9):
MV2 dP2 1 1 = ( P1 − P2 ) − ( P2 − Ph ) RT dt R2 R3 MV2 dP2 1 1 1 1 = P1 − ( + ) P2 + Ph (t ) RT dt R2 R2 R3 R3 Note that
dP1 = f 1 ( P1 , P2 ) dt
from Eq. 10
dP2 = f 2 ( P1 , P2 ) dt
from Eq. 11
(11)
This is exactly the same situation depicted in Figure 6.13, therefore the two tanks interact. This system has the following characteristics: i) ii) iii) iv) v)
Interacting (Eqs. 10 and 11 interact with each other ) 2nd-order denominator (2 differential equations) Zero-order numerator (See example 4.4 in text) No integrating elements W ′( s ) is not equal to unity. (Cannot be because Gain of 3 Pc′( s ) the units on the two variables are different).
4.12
a)
A
dh = q i − C v h1 / 2 dt
Let f = qi − C v h1 / 2 1 Then f ≈ q i − C v h 1 / 2 + qi − qi − C v h −1 / 2 (h − h ) 2
so A
C dh ′ = q i′ − 1v/ 2 h ′ dt 2h
because
4-12
qi − C v h 1 / 2 ≡ 0
Cv sA + 2h 1 / 2 H ′( s ) = Qi′ ( s )
H ' ( s) = Qi′ ( s )
1
Note: Not a standard form
C sA + 1v/ 2 2h
2h 1 / 2 / C v H ' (s) = Qi′ ( s ) 2 Ah 1 / 2 s +1 Cv where K =
b)
and τ =
2 Ah 1 / 2 Cv
Because q = C v h1 / 2 q′ = Cv
C 1 −1 / 2 1 h h ′ = 1v/ 2 h ′ = h ′ 2 K 2h
Q ′( s ) 1 = , H ′( s ) K
∴
and c)
2h 1 / 2 Cv
Q ′( s ) H ′( s ) 1 K = H ′( s ) Qi′( s ) K τs + 1
Q ′( s ) 1 = Qi′ ( s ) τs + 1
For a linear outlflow relation A
dh * = qi − C v h dt
A
dh ′ * = q i′ − C v h ′ dt
A
dh ′ * + C v h ′ = q i′ dt
Note that C v ≠ C v *
A dh′ 1 + h′ = * qi′ * C v dt Cv
or
Multiplying numerator and denominator by h on each side yields Ah dh ′ h + h ′ = * qi′ * C v h dt Cv h
4-13
or
V dh′ h + h′ = qi′ q i dt qi
τ∗ =
V qi
K∗ =
h qi
q.e.d
To put τ and K in comparable terms for the square root outflow form of the transfer function, multiply numerator and denominator of each by h 1/ 2 . K=
2h 1 / 2 h 1 / 2 2h 2h = = = 2K * 1/ 2 1/ 2 Cv h qi Cv h
2 Ah 1 / 2 h 1 / 2 2 Ah 2V τ= = = = 2τ * 1/ 2 1/ 2 Cv h qi Cv h Thus level in the square root outflow TF is twice as sensitive to changes in qi and reacts only ½ as fast (two times more slowly) since τ = 2 τ∗ .
4.13 a)
The nonlinear dynamic model for the tank is:
(
dh 1 = qi − Cv h dt π( D − h)h
)
(1)
(corrected nonlinear ODE; model in first printing of book is incorrect) To linearize Eq. 1 about the operating point (h = h , qi = qi ) , let f =
qi − Cv h π( D − h ) h
Then, ∂f ∂f f (h, qi ) ≈ h′ + qi′ ∂h s ∂qi s where
4-14
∂f 1 = ∂qi s π( D − h )h 1 Cv 1 −πD + 2πh ∂f = − + q − C h i v ( π( D − h ) h ) 2 2 h π( D − h ) h ∂h s
)
(
Notice that the second term of last partial derivative is zero from the steady-state relation, and the term π( D − h )h is finite for all 0 < h < D . Consequently, the linearized model of the process, after substitution of deviation variables is,
dh′ 1 Cv 1 1 = − h′ + qi′ dt 2 h π( D − h )h π( D − h ) h Since
qi = Cv h
dh′ 1 qi 1 1 = − h′ + qi′ dt 2 h π( D − h )h π ( D − h ) h or
dh′ = ah′ + bqi′ dt
where 1q q 1 a = − i =− i Vo 2 h π( D − h ) h
,
1 b= π( D − h ) h
Vo = volume at the initial steady state
b)
Taking Laplace transform and rearranging s h′( s ) = ah′( s ) + bqi′( s ) Therefore h′( s ) b = qi′( s ) ( s − a )
or
h′( s ) (−b / a ) = qi′( s ) (−1/ a ) s + 1
Notice that the time constant is equal to the residence time at the initial steady state.
4-15
4.14
Assumptions 1) Perfectly mixed reactor 2) Constant fluid properties and heat of reaction. a)
Component balance for A, dc A = q (c Ai − c A ) − Vk (T )c A dt Energy balance for the tank,
(1)
V
ρVC
dT = ρqC (Ti − T ) + ( − ∆H )Vk (T )c A dt
(2)
Since a transfer function with respect to cAi is desired, assume the other inputs, namely q and Ti, to be constant. Linearize (1) and (2) and not that
V
dc A dc ′A dT dT ′ = , = , dt dt dt dt
dc ′A 20000 = qc ′Ai − (q + Vk (T ))c ′A − Vc A k (T ) T′ dt T2
ρVC
dT ′ 20000 = − ρqC + ∆HVc A k (T ) T ′ + (−∆H )Vk (T )c′A dt T2
(3)
(4)
Taking the Laplace transforms and rearranging
[Vs + q + Vk (T )]C ′ (s) = qC ′ (s) − Vc A
Ai
A
k (T )
20000 T ′( s ) T2
(5)
20000 ρVCs + ρqC − (−∆H )Vc A k (T ) T 2 T ′( s ) = (−∆H )Vk (T )C A′ ( s ) (6)
Substituting for C ′A (s ) from Eq. 5 into Eq. 6 and rearranging,
4-16
T ′( s ) = ′ ( s) C Ai
−∆HVk (T ) q 20000 20000 Vs + q + Vk (T ) ρVCs + ρqC − ( −∆H )Vc A k (T ) + ( −∆H )c AV 2 k 2 (T ) 2 T T2
(7)
c A is obtained from Eq. 1 at steady state, cA =
qc Ai = 0.0011546 mol/cu.ft. q + Vk (T )
Substituting the numerical values of T , ρ, C, ( − ∆H), q, V, c A into Eq. 7 and simplifying,
T ′( s) 11.38 = C ′Ai ( s) (0.0722s + 1)(50s + 1) b)
T ′( s) The gain K of the above transfer function is , C ′Ai ( s ) s =0
K=
0.15766 q c q c q − 3.153 × 106 A2 + 13.84 + 4.364.107 A2 T 1000 T 1000
(8)
obtained by putting s=0 in Eq. 7 and substituting numerical values for ρ, C, ( − ∆H), V. Evaluating sensitivities, dK K K2 = − dq q 0.15766q
cA q 2 + 0 . 01384 − 3153 = −6.50 × 10 −4 6 2 10 T
6 7 dK K 2 q 3.153 × 10 c A × 2 2 × 4.364 × 10 c A =− + 13 . 84 − dT 3.153 1000 T3 T3
= −2.57 × 10 −5
dK dK dc A = × dc Ai dc A dc Ai =
6 −K2 q 3.153 × 10 + 13.84 − 0.15766q 1000 T2
= 8.87 × 10 −3
4-17
4.364 × 10 7 q + 2 T q + 13840
4.15
From Example 4.4, system equations are:
dh1′ 1 = qi′ − h1′ dt R1 dh′ 1 1 A2 2 = h1′ − h2′ dt R1 R2 Using state space representation, A1
,
q1′ =
1 h1′ R1
,
q ′2 =
1 h2′ R2
x1 = Ax + Bu y = Cx + Du
where
h′ x = 1 h2′
,
u = qi
and
y = q 2′
then, dh1′ 1 dt − R A = 1 1 1 dh2′ R1 A1 dt
q ′2 = 0
0 1 − R2 A2
1 ′ A h1 + 1 0 h2′
qi′
h1′ 1 R2 h2′
Therefore, 1 − R A A= 1 1 1 R1 A1
0 1 − R2 A2
1 A 1 , B= , C=0 0
4-18
1 , E=0 R2
4.16
Applying numerical values, equations for the three-stage absorber are:
dx1 = 0.881 y f − 1.173 x1 + 0.539 x 2 dt dx 2 = 0.634 x1 − 1.173 x 2 + 0.539 x3 dt dx3 = 0.634 x 2 − 1.173 x3 + 0.539 x f dt
yi = 0.72 xi Transforming into a state-space representation form: dx1 dt dx 2 dt dx 3 dt
0 − 1.173 0.539 = 0.634 − 1.173 0.539 0 0 . 634 − 1 . 173
0 0 y1 0.72 y = 0 0.72 0 2 y 3 0 0 0.72
x1 0.881 x + 0 y 2 f x3 0
x1 0 x + 0y 2 f x3 0
Therefore, because xf can be neglected in obtaining the desired transfer functions,
0 − 1.173 0.539 A = 0.634 − 1.173 0.539 0 0.634 − 1.173
4-19
0.881 B = 0 0
0 0 0.72 C= 0 0.72 0 0 0 0.72
0 D = 0 0
Applying the MATLAB function ss2tf , the transfer functions are: Y1′( s ) 0.6343s 2 + 1.4881s + 0.6560 = 3 Y f′ ( s ) s + 3.5190 s 2 + 3.443s + 0.8123
Y2′( s ) 0.4022 s + 0.4717 = 3 Y f′ ( s ) s + 3.5190s 2 + 3.443s + 0.8123 Y2′( s ) 0.2550 = 3 Y f′ ( s ) s + 3.5190s 2 + 3.443s + 0.8123
4.17
Dynamic model: dX = µ( S ) X − DX dt dS = −µ ( S ) X / Y X / S − D ( S f − S ) dt
Linearization of non-linear terms: (reference point = steady state point) 1. f 1 ( S , X ) = µ( S ) X =
µm S X Ks + S
f1 (S , X ) ≈ f1 (S , X ) +
∂f1 ∂S
(S − S ) + S ,X
∂f 1 ∂X
(X − X ) S ,X
Putting into deviation form,
f 1 ( S ′, X ′) ≈
∂f 1 ∂S
S′ + S ,X
∂f 1 ∂X
S ,X
µ (K + S ) − µ m S µ m S X ′ = m s X S '+ 2 ( K + S ) s Ks + S
4-20
X '
Substituting the numerical values for µ m , K s , S and X then:
f 1 ( S ′, X ′) ≈ 0.113S ' + 0.1X ' 2. f 2 ( D, S , S f ) = D( S f − S )
f 2 ( D' , S ′, S ′f ) ≈
∂f 2 ∂f D' + 2 ∂D D , S , S f ∂S
S' + D ,S ,S f
∂f 2 ∂S f
S ′f D ,S ,S f
f 2 ( D' , S ′, S ′f ) ≈ ( S f − S ) D' − D S ' + D S ′f f 2 ( D' , S ′, S ′f ) ≈ 9 D' − 0.1S ' + 0.1S f '
3. f 3 ( D, X ) = DX f 3 ( D' , X ' ) ≈ D' X + X ' D = 2.25 D' + 0.1X '
Returning to the dynamic equation: putting them into deviation form by including the linearized terms: dX ' = 0.113S ' + 0.1X ' − 2.25 D' − 0.1X ' dt dS ' −0.113 0.1 = S'− X ' − 9 D ' + 0.1S ' − 0.1S ′f dt 0.5 0.5
Rearranging: dX ' = 0.113S ' − 2.25 D' dt dS ' = −0.126 S ' − 0.2 X ' − 9 D ' − 0.1S ′f dt
Laplace Transforming: sX ' ( s ) = 0.113S ' ( s ) − 2.25 D' ( s )
sS '( s ) = −0.126 S '( s ) − 0.2 X '( s ) − 9 D '( s ) − 0.1S ′f ( s )
4-21
Then, 0.113 2.25 S ' (s) − D' ( s) s s −0.2 9 0.1 S '( s ) = X '( s ) − D '( s ) − S ′f ( s ) s + 0.126 s + 0.126 s + 0.126 X ' (s) =
or
0.0226 X ' ( s ) 1 + = s ( s + 0.126) =−
1.017 0.0113 2.25 D '( s ) − S ′f ( s ) − D′( s ) s + 0.126 s + 0.126 s
Therefore, X ' (s) − 1.3005 − 2.25s = 2 D' ( s ) s + 0.126 s + 0.0226
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