Control Automatico de procesos solucionario, Ejercicios de Control de Procesos

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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE

MUCHOS DE ESTOS LIBROS GRATIS EN

DESCARGA DIRECTA

SIGUENOS EN:

VISITANOS PARA DESARGALOS GRATIS.

(d) Automatic sprinkler system for fires.

The controller is On/Off and the control i feedback with respect to temperature.,

M, D, A: temperature element/controller TE/C. Usually a bi-metallic strip that pushes the latch on the mechanism holding the slice of bread. The bread is released and the heating element de-energized when the temperature reaches the value set by the set point. (^) SP TE/C

(c) Toaster.

The controller is On/Off and the control is feedback on the temperature variable.

M: Temperature element TE , usually a gas-filled bulb

D: Temperature controller TC.

A: Solenoid S that operates the heating element in the oven

(b) Cooking oven.

TE

TC

S

SP

The controller is On/Off and the control is feedback.

M: Temperature element TE in thermostat TE/C

D: Mercury switch in thermostat TE/C

A: Solenoid S that turns unit (AC/H) on and off. TE/C^

S

AC /H

SP

(a) House air-conditioning/heating.

Identificaton of the M-D-A components, controller type, instrumentation diagram, and type of control.

Problem 1-1. Automation in daily life.

Smith & Corripio, 3rd edition

Smith & Corripio, 3rd edition

Problem 1-2. Automatic shower temperature diagram.

TC TE

Hot water

Cold water

S

SP

M: temperature sensor TE , a gas-filled bulb

D: temperature controller TC , mechanilly integrated to the sensor, but with a signa output

A: solenoid operated control valve on the hot water line.

The cold water valve is operated manually.

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F s( ) s s^2 +ω^2

s −i ⋅ω

s +i ⋅ω

= s^ −^ i^ ⋅ω+^ s+i^ ⋅ω

2 ⋅( s^ −i ⋅ω)(^ s +i ⋅ω)

= 2 s⋅

2 ⋅(^ s^2 +ω^2 )

= s s^2 +ω^2

s − i ⋅ω

e −^ (^ s^ −i^ ⋅ω)t ∞ 0

⋅ −^1

s + i ⋅ω

e −^ (^ s^ +i^ ⋅ω)t ∞ 0

∞ e −(^ s^ −i^ ⋅ω)t t

d 0

∞ e −(^ s^ +i⋅^ ω)t t

• d

F s( ) 0

∞ cos ⋅ ωt⋅e −st t

= d 0

∞ e (^) t i ⋅ ωt−e − i⋅ωt 2

e −st

(c) f t( ) = cos⋅ ωt = d

F s( ) 1 s +a

F s( ) 0

∞ e −^ ate −st t

= d 0

∞ e −(^ s^ +a)t t

= d −^1 s + a

e −^ (^ s^ +a)t ∞ 0

= ⋅^1

s +a

(b) f t( ) = e −at^ where a is constant

F s( ) 1 s^2

F s( ) −t s

e −^ st ∞ 0

s (^0)

∞ e −st t

= + ⋅ d 0 − 0 1 s^2

e −^ st ∞ 0

= − ⋅^1

s^2

v −^1 s

du = dt = e −st

F s( ) By parts: u = t dv = e −^ stdt 0

∞ t e⋅ −st t

(a) f t( ) = t = d

F s( ) 0

∞ f t( ) e −st t

= d

Problem 2-1. Derivation of Laplace transforms from its definition

Smith & Corripio, 3rd. edition

s

s + 2

s + 1

= − ⋅^1

s

s + 2

s + 1

F s( ) 1 s

s + 2

s + 1 Used the linearity property. = −

(d) f t( ) = u t( ) − e −t+t e⋅ −t F s( ) = L u t( ( )) − L e(^ −t)+L t e(^ ⋅ −t)^1

s

s + 1

(s + 1 ) 2

F s( ) 1 s

s + 1

( s + 1 ) 2

Used the linearity property. =^ +

(e) f t( ) = u t( − 2 ) 1 −e −^2 (^ t 2−)^ sin t( − 2 ) Let^ g t( ) = u t( ) 1(^ −e −^2 tsin t⋅) Then^ f t( ) = g t( − 2 )

F s( ) = e −^2 s^ G s( ) e −^2 s^1 s

(s + 2 ) 2 + 1

Used the real translation theorem and linearity. (^) F s( ) e −^2 s^1 s

(s + 2 ) 2 + 1

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Smith & Corripio, 3rd edition

Problem 2-2. Derive Laplace transforms from the properties and Table 2-1.

(a) f t( ) = u t( ) + 2 t⋅ +3 t ⋅ 2 F s( ) = L u t(^ ( ) + 2 t⋅ +3 t ⋅ 2 ) = L u t( ( )) + 2 L t⋅ ( )+3 L t⋅^ (^2 )

s

s^2

s^3

= + ⋅ F s( ) 1 s

s^2

s^3

Used the linearity property.

(b) f t( ) = e −^2 ⋅t^ (u t^ ( ) + 2 t⋅ +3 t ⋅ 2 ) F s( ) L u t(^ ( ) + 2 t⋅ +3 t ⋅ 2 )^

s + 2

= ⋅^1

s

s^2

s^3

 s + 2

s + 2

(s + 2 ) 2

(s + 2 ) 3

F s( ) 1 s + 2

(s + 2 ) 2

(s + 2 ) 3

Used the complex translation theorem. =^ +

(c) f t( ) = u t( ) + e −^2 t−2e −t F s( ) = L u t(^ ( ) + e −^2 t−2 e⋅ −t) = L u t( ( )) + L e(^ −^2 t)−2 L e⋅ (^ −t)

Must apply L'Hopital's rule:

s ∞

2 s( + 2 )

3 s( + 2 ) 2

lim = 1 Final value: →

t ∞

lim e −^2 t^ (^ u t( ) + 2 t⋅ +3t 2 )^ = 0 ⋅∞

→ s 0

s 1 s + 2

(s + 2 ) 2

(s + 3 ) 2

lim = 0 L'Hopital's rule:^ →

t ∞

2e 2t

2e 2t

• 6t 2e 2t

lim = 0 → Check!

(c) f t( ) = u t( ) + e −^2 t−2e −t F s( ) 1

s

s + 2

s + 1

Initial value:

t 0

lim (^ u t( ) + e −^2 t−2e −t)^ = ( 1 + 1 − 2 ) + 0

→ s ∞

s 1 s

s + 2

s + 1

lim = → L'Hopital's rule:

s ∞

lim = 0 → Final value:

t ∞

lim (u t^ ( ) + e −^2 t−2e −t)^ = 1 + 0 + 0 = 1

→ s 0

s 1 s

s + 2

s + 1

lim = 1 + 0 + 0 = 1 →

Smith & Corripio, 3rd edition

Problem 2-3. Initial and final value check of solutions to Problem 2-

(a) f t( ) = u t( ) + 2 t⋅ +3t 2 F s( ) 1

s

s^2

s^3

Initial value:

t 0

lim (^ u t( ) + 2t +3t 2 )^ = 1

→ s ∞

s 1 s

s^2

s^3

s ∞

s

s^2

lim = 1 →

lim = → Final value:

t ∞

lim (u t^ ( ) + 2t +3t 2 )^ = ∞

→ s 0

s

s^2

lim = ∞ →

Check!

(b) f t( ) = e −^2 t^ (u t^ ( ) + 2t +3t 2 ) F s( ) 1

s + 2

(s + 2 ) 2

(s + 2 ) 3

Initial value:

t 0

lim e −^2 t^ (^ u t( ) + 2t +3t 2 )

→ s ∞

s 1 s + 2

(s + 2 ) 2

(s + 2 ) 3

lim = →

= 1 1( + 0 + 0 ) = 1

Smith & Corripio, 3rd edition

Problem 2-4. Laplace transform of a pulse by real translation theorem

f t( ) = H u t⋅ ( )−H u t⋅ ( −T)

F s( ) H^1 s

⋅ H e⋅ −sT^1 s

= − ⋅ H^1 e^ − −sT s

= ⋅ F s( ) H s

= (^1 −e −sT)

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(^00 2 )

2 fd ( )t

t

(^00 2 )

2 f t( )

t

f t( ) e

t 0 τ (^) e

−t := ⋅ τ fd ( )t^ u t(^ −t^0 )^ e

−( t t− 0 )

:= ⋅ τ

u t( ) 0 if t < 0 1 ift ≥ 0

Sketch the functions: (^) t 0 := 1 τ := 1 := F s( )^

τ e

− t 0 ⋅s ⋅ τ ⋅s + 1

The result to part (b) agrees with the real translation theorem.

e

− t 0 ⋅ s − 1

s 1 τ

⋅ e

s 1 τ

− ⋅λ ⋅ ∞ 0

= ⋅ e

− t 0 ⋅s

s 1 τ

= τ^ e

− t 0 ⋅s ⋅ τ ⋅s + 1

F s( ) −t 0

∞

u (^ λ)^ e λ

− λ τ (^) e −s^ (^ λ^ +t 0 )

= (^) ⌡ d e

− t 0 ⋅s

0

∞

e λ

s 1 τ

− λ

= ⋅⌡ d Let λ = t −t 0

F s( ) 0

∞

u t( −t 0 ) e t

−( t t− 0 ) τ (^) e −st

f t( ) u t( −t 0 ) e = d

−( t t− 0 )

= τ

(b) Function is delayed and zero from t = 0 to t = t 0 :

F s( ) τ^ e

t 0

⋅^ τ τ ⋅ s+ 1

F s( ) e =

t 0 τ 1 s 1 τ

= τ^ e

t 0

⋅^ τ τ ⋅ s+ 1

f t( ) e =

t 0 τ (^) e

−t = τ^ (from Table 2-1.1)

(a) Function is non-zero for all values of t > 0:

f t( ) e

−( t t− 0 )

= τ

Problem 2-5. Delayed versus non-delayed function

9 d^ Y 0( ) = 0

(^2) ⋅Y t( )

dt^2

⋅ 12 d Y t⋅ ( ) dt

Subtract initial steady state: + ⋅ + 4 Y t( ) = 8 X t( )

9 d^

(^2) ⋅y t( )

dt^2

⋅ 12 d y t⋅ ( ) dt

(d) + ⋅ + 4 y t( ) = 8 x t( ) − 4

Invert usingTable 2-1.1: Y t( ) = ( − 1 +1.134i)e ( − 0.5+0.441i)t (^) + (− 1 −1.134i)e ( − 0.5−0.441i)t+2 u t( )

Y s( ) −^1 +1.134i s + 0.5−0.441i

− 1 −1.134i s + 0.5+0.441i

s

A 3

s 0

9s^2 + 9s+ 4

lim = 2 →

A 2 = − 1 −1.134i =

9 2 0.441i( ⋅ ) ( −0.5 +0.441i)

A 1 =− 1 +1.134i s − 0.5+0.441i

9 s( + 0.5+0.441i) s

lim →

A 1

s + 0.5−0.441i

A 2

s + 0.5+0.441i

A 3

s

Y s( ) 8 9 s( + 0.5−0.441i) s( + 0.5+0.441)s

Solve for Y(s), expand: =

A 2

s −1.

9 s( +0.255)s

lim = = 0. →

A 3

s 0

9 s( +0.255) s( +1.745)

lim = = 2. →

Y s( ) −2. s +0.

s +1.

s

Invert with Table 2-1.1: (^) Y t( ) = −2.342 e − 0.255t (^) + 0.342e − 1.745t+2 u t( )

(c) 9 d^

(^2) ⋅y t( )

dt^2

⋅ 9 d y t⋅ ( ) dt

• ⋅ + 4 y t( ) = 8 x t( ) − 4

Subtract initial steady state: 9 d 2 ⋅Y t( ) dt^2

⋅ 9 d Y t⋅ ( ) dt

• ⋅ + 4 Y t( ) = 8 X t( ) Y 0( ) = 0

Laplace transform: ( 9s 2 + 9s+ 4 )Y s( ) = 8 X s( ) 8 1

s

r 1 −^9

+^2 −4 9⋅ ⋅ 4

:= r 2 −^9

−^2 −4 9⋅ ⋅ 4

Find roots: :=^ r^1 =−0.5^ +0.441i r 2 =−0.5 −0.441i

(^3) A 2 = 0.027 −0.022i 2 2 2.598i( ⋅ ) ( − 1 +2.598i) (− 1.5 +2.598i)

=0.027 +0.022i

A 1

s − 1.5+2.598i

2 s( + 1.5+2.598i) s( +0.5)s

lim = 0.027 +0.022i →

A 1

s + 1.5−2.598i

A 2

s + 1.5+2.598i

A 3

s +0.

A 4

s

Y s( ) 3 2 s( + 1.5−2.598i) s( + 1.5+2.598i) s( +0.5)s

Solve for Y(s) and expand: =

polyroots

− 1.5 −2.598i − 1.5 +2.598i − 0.

Find roots: =

( 2s^3 + 7s^2 + 21s+ 9 )Y s( ) = 3 X s( ) 3 1

s

Laplace transform: = ⋅

Y 0( ) = 0

2 d^

(^3) ⋅Y t( )

dt^3

⋅ 7 d^

(^2) ⋅ (^) Y t( )

dt^2

• ⋅ 21 d Y t⋅ ( ) dt

Subtract initial steady state: + ⋅ + 9 Y t( ) = 3 X t( )

2 d^

(^3) ⋅y t( )

dt^3

⋅ 7 d^

(^2) ⋅y t( )

dt^2

• ⋅ 21 d y t⋅ ( ) dt

(e) + ⋅ + 9 y t( ) = 3 x t( )

Y t( ) −^4 3

 t − 2

Invert using Table 2-1.1: = e −^ 0.667t^ +2 u t( )

A 3

s 0

9 s( +0.667) 2

lim = 2 →

A 2

s −0.

d ds

9s

 s −0.

9s^2

lim = − 2 →

lim = →

A 1 =

s −0.

9s

lim = →

Y s( ) 8 9 s( +0.667) 2 s

A 1

( s +0.667) 2

A 2

s +0.

A 3

s

Solve for Y(s) and expand: = +

r 2 =−0.

r 2 −^12 12 r 1 =−0.

−^2 −4 9⋅ ⋅ 4

r 1 −^12 12 :=

+^2 −4 9⋅ ⋅ 4

Find roots: :=

( 9s^2 + 12s+ 4 )Y s( ) = 8 X s( ) 8 1

s

Laplace transform: = ⋅

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Y t( ) u t( − 1 ) −^8 3

 ⋅( t − 1 )− 8

 ⋅e −^ 0.667⋅(^ t 1−)+8 e⋅ −^ 0.333⋅(^ t 1−)

Apply the real translation theorem in reverse to this solution:

Y s( ) −^8 3

( s +0.667) 2

s +0.

s +0.

= e −s

The partial fraction expansion of the undelayed signal is the same:

(Real translation theorem)

X s( ) e^

−s

s 1 3

X t( ) u t( − 1 ) e =

−( t 1−)

(b) Forcing function: =^3

Y t( ) −^8 3

 t − 8

Invert using Table 2-1.1: = e −^ 0.667t^ +8e −^ 0.333t

Y s( ) −^8 3

(s +0.667) 2

s +0.

s +0.

A 2

s −0.

d ds

9 s( +0.333)

 s −0.

9 s( +0.333) 2

lim = − 8 →

lim = →

A 3

s −0.

9 s( +0.667) 2

lim = 8 →

A 1 =

s −0.

9 s( +0.333)

lim = →

9 s( +0.667) 2 ( s +0.333)

A 1

(s +0.667) 2

A 2

s +0.

A 3

s +0.

Y s( ) 8

(9s 2 + 12s+ 4 ) s 1

X s( ) 1 s 1 3

X t( ) e From Table 2-1.1: =

−t

(a) Forcing function: =^3

9 d^ Y 0( ) = 0

(^2) ⋅Y t( )

dt^2

⋅ 12 d Y t⋅ ( ) dt

• ⋅ + 4 Y t( ) = 8 X t( )

Problem 2-7. Solve Problem 2-6(d) with different forcing functions

Smith & Corripio, 3rd edition

(Final value theorem)

(b) 9 d 2 ⋅y t( )

dt^2

⋅ 18 d y t⋅ ( ) dt

• ⋅ + 4 y t( ) = 8 x t( ) − 4

Subtract initial steady state: 9 d^

(^2) ⋅Y t( )

dt^2

⋅ 18 d Y t⋅ ( ) dt

• ⋅ + 4 Y t( ) = 8 X t( ) Y 0( ) = 0

Laplace transform and solve for Y(s): Y s( ) 8 9s^2 + 18s+ 4

= X s( )

Find roots: r 1 −^18

+^2 −4 9⋅ ⋅ 4

2 9⋅ min

:= r 2 −^18

−^2 −4 9⋅ ⋅ 4

2 9⋅ min

:= r 1 =−0.255 min −^1

r 2 =−1.745 min −^1

Invert using Table 2-1.1: Y t( ) = A 1 ⋅ e −^ 0.255t+A 2 ⋅e −^ 1.745t

• terms of X(s)

The response is stable and monotonic. The domnant root is: r 1 =−0.255 min −^1

Time for the response to decay to 0.67% of its initial value: −^5 r 1

=19.6 min

Final steady-state value for unit step input: s 0

s 8 9s^2 + 18s+ 4

s

lim →

(Final value theorem)

Smith & Corripio, 3rd edition

Problem 2-8. Response characteristics of the equations of Problem 2-

(a) d y t⋅ ( )

dt

• 2 y t( ) = 5 x t( ) + 3

Initial steady state: 2 y 0( ) = 5 x 0( ) + 3

Subtract: d Y t⋅ ( ) dt

• 2 Y t( ) = 5 X t( ) Y t( ) = y t( ) −y 0( ) X t( ) = x t( ) −x 0( )

Laplace transform: s Y s⋅ ( )+ 2 Y s( ) = 5 X s( ) Y 0( ) = y 0( ) −y 0( ) = 0

Solve for Y(s): Y s( ) 5 s + 2

= X s( )

A 1

s + 2

= + terms of X(s)

Invert using Table 2-1.1: Y t( ) = A 1 ⋅e −^2 t + terms of X(t)

The response is stable and monotonic.The dominant and only root is r :=− 2 min −^1

Time for response to decay to within 0.67% of its initial value: −^5 r

=2.5 min

Final steady-state value for unit step input: s 0

s 5 s + 2

s

lim →

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

(Final value theorem) s^0

s 3 2s^3 + 7s^2 + 21s+ 9

s

lim →

Final steady state value for a unit step input: →

r 2

e −^ 1.5min^ Time for response to die out: =10 min

− (^1) T Decay ratio: =0.

T 2 π T =2.42 min 2.598min −^1

The period of the oscillations is: :=

The response is stable and oscillatory. The dominant root is r 2 =−0.5 min −^1

r

−1.5 −2.598i − 1.5 +2.598i − 0.

r polyroots = min −^1

:= min −^1

Find roots:

Y s( ) 3 2s^3 + 7s^2 + 21s+ 9

Laplace transform and solve for Y(s): = X s( )

2 d^

(^3) ⋅Y t( )

dt^3

⋅ 7 d^

(^2) ⋅ (^) Y t( )

dt^2

• ⋅ 21 d Y t⋅ ( ) dt

Subtract initial steady state: + ⋅ + 9 Y t( ) = 3 X t( )

2 d^

(^3) ⋅y t( )

dt^3

⋅ 7 d^

(^2) ⋅y t( )

dt^2

• ⋅ 21 d y t⋅ ( ) dt

(e) + ⋅ + 9 y t( ) = 3 x t( )

(Final value theorem) s^0

s 8 9s^2 + 12s+ 4

s

lim →

Final steady state value for a unit step input: → 2

r 1

Time required for the response to decay within 0.67% of its initial value: =7.5 min

The response is stable and monotonic. The dominant root is r 1 =−0.667 min −^1

Value of k: (^) k −M^ ⋅g y 0

:= k 1.816 N m

Laplace transform: (^) M s ⋅ 2 Y s( ) + k Y s⋅ ( ) = F s( )

Solve for Y(s): (^) Y s( ) 1 M s ⋅ 2 +k

= F s( )

A 1

s i k M

A 2

s i k M

• terms of F(s) θ := 0 D := 1 Invert using Table 2-3.1: (^) Y t( ) D sin k M

 t s⋅ +θ

:= ⋅ + terms of f(t)

The mobile will oscillate forever with a period of (^) T 2 π M k

:= ⋅ T =1.043 s

Smith & Corripio, 3rd edition

Problem 2-9. Second-Order Response: Bird Mobile

• Mg

f ( t )

y ( t )

-ky ( t )

y = 0

Problem data: (^) M := 50gm y 0 :=− 27 cm

Solution:

Force balance:

M d v t⋅ ( ) dt

⋅ = −M ⋅ g− k y t⋅ ( )+f t( )

Velocity: d y t⋅ ( ) dt

= v t( )

Initial steady state: (^0) = −M ⋅g −k y⋅ 0

Subtract and substitute:

M d^

(^2) ⋅Y t( )

dt^2

⋅ = −k ⋅Y t ( )+f t( )

Y 0( ) = 0