smith & corripio, 3rd edition solution

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DaladierTypewritten textLIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS

LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRORESUELTOS Y EXPLICADOSDE FORMA CLARA

VISITANOS PARADESARGALOS 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. TE/CSP

(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.

TETC

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

Watermain

TE/C

M, D, A: temperature element/controller TE/C, a rod that gives to the water pressure at a set temperature, allowing the water to spray over the fire.

The controller is On/Off with single action, and the control is feedback.

(e) Automatic cruise speed control.

ST

SCSP

S

Air

EngineTransmission

M: Speed sensor and transmitter ST on the transmission

D: Speed controller SC

A: Damper on air intake to the engine throttles the air varying the power delivered by the engine

Controller is regulating and control is feedback.

(f) Refrigerator.

TE/C SSP

M: temperature sensor TE, usually a gas-filled bulb

D: Temperature controller C, mechanically linked to the sensor

A: Solenoid S that turnsd the refrigeration compressor on and off

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

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Problem 1-2. Automatic shower temperature diagram.

TETC

Hotwater

Coldwater

S

SPM: 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.


F s( )s

s2 2+=

12

1s i

1s i ++

=

s i s+ i +2 s i ( ) s i +( )=

2 s2 s2 2+( )=

s

s2 2+=

12

1s i e

s i ( )t 0

1s i + e

s i +( )t 0

+

=

12 0

te s i ( )t

d 0

te s i +( )t

d+

=

F s( )0

tcos t e st d=

0

t

ei t e i t2

e st

d=f t( ) cos t=(c)

F s( )1

s a+=

F s( )0

te at e st

d= 0

te s a+( )t

d=1

s a+ es a+( )t

0

= 1s a+=

where a is constantf t( ) e at=(b)

F s( )1

s2=

F s( )t

se st

0

1s 0

te st

d+= 0 01

s2e st

0

= 1s2

=

v1s

e st=du dt=

dv e st dt=u t=By parts:F s( )0

tt e st d=f t( ) t=(a)

F s( )0

tf t( ) e st

d=

Problem 2-1. Derivation of Laplace transforms from its definitionSmith & Corripio, 3rd. edition

(d) f t( ) e at coss t=F s( )

0

te at cos t e st d=

0

te at e

i t e i t+2

e st

d=

12 0

te s a+ i +( )t

d 0

te s a+ i ( ) t

d+

=

12

1s a+ i + e

s a+ i +( )t 0

1s a+ i e

s a+ i ( )t 0

+

=

12

1s a+ i +

1s a+ i +

=

s a+ i s+ a+ i +2 s a+ i +( ) s a+ i ( )=

2 s a+( )2 s a+( )2 2+

= s a+s a+( )2 2+

= F s( ) s a+s a+( )2 2+

=

All the results match results in Table 2-1.1


1s

1s 2++ 2

1s 1+=

1s

1s 2++

2s 1+=

F s( )1s

1s 2++

2s 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( )+= 1s

1s 1+

1

s 1+( )2+=

F s( )1s

1s 1+

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

s1

s 2+( )2 1+

=

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

1

s 2+( )2 1+

=


Smith & Corripio, 3rd editionProblem 2-2. Derive Laplace transforms from the properties and Table 2-1.1

(a) f t( ) u t( ) 2 t+ 3 t2+= F s( ) L u t( ) 2 t+ 3 t2+( )= L u t( )( ) 2 L t( )+ 3 L t2( )+=1s

21

s2+ 3 2!

s3+= F s( ) 1

s2

s2+ 6

s3+=

Used the linearity property.

(b) f t( ) e 2 t u t( ) 2 t+ 3 t2+( )= F s( ) L u t( ) 2 t+ 3 t2+( )s 2+

= 1s

2

s2+ 6

s3+

s 2+=

1s 2+

2

s 2+( )2+ 6

s 2+( )3+=

F s( )1

s 2+2

s 2+( )2+ 6

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:

s11

22 s 2+( )+

6

3 s 2+( )2+

1=lim

Final value:

te 2 t u t( ) 2 t+ 3t2+( ) 0 =lim

0s

s1

s 2+2

s 2+( )2+ 6

s 3+( )2+

0=lim

L'Hopital's rule:

t0

2e2t2

2e2t+ 6t

2e2t+

0=lim Check!

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

1s 2++

2s 1+=

Initial value:

0tu t( ) e 2 t+ 2e t( ) 1 1+ 2( ) 0+=lim

ss

1s

1s 2++

2s 1+

=lim

L'Hopital's rule:

s1

11

+ 21

0=lim

Final value:

tu t( ) e 2 t+ 2e t( ) 1 0+ 0+= 1=lim

0ss

1s

1s 2++

2s 1+

1 0+ 0+= 1=lim


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

(a) f t( ) u t( ) 2 t+ 3t2+= F s( ) 1s

2

s2+ 6

s3+=

Initial value:

0tu t( ) 2t+ 3t2+( ) 1=lim

ss

1s

2

s2+ 6

s3+

s

12s

+ 6s2

+

1=lim

=lim

Final value:

tu t( ) 2t+ 3t2+( ) =lim

0s1

2s

+ 6s2

+

=lim

Check!

(b) f t( ) e 2 t u t( ) 2t+ 3t2+( )= F s( ) 1s 2+

2

s 2+( )2+ 6

s 2+( )3+=

Initial value:

0te 2 t u t( ) 2t+ 3t2+( )lim

ss

1s 2+

2

s 2+( )2+ 6

s 2+( )3+

=lim

1 1 0+ 0+( )= 1=


Check!

0ss

1s

1

s 1+( )2 1+

1 0+= 1=lim

t1 e 2 tsin t( ) 1=lim

Final value:

ss

1s

1

s 1+( )2 1+

1 0= 1=lim

0t1 e 2 tsin t( ) 1=lim

Initial value:

The test of the delayed fnction is not useful. Better to test the term in brackets, g(t):

F s( ) e 2 s 1s

1

s 1+( )2 1+

=f t( ) u t 2( ) 1 e 2 t 2( ) sin t 2( ) =(e)

Check!t1 0 1

1 et+

1=lim

L'Hopital's rule:

tu t( ) e t t e t+( ) 1 0 0+=lim

0s1

ss 1+

s

s 1+( )2+

1 0 0+= 1=lim

Final value: s1

11

12 s 1+( )+

1 1 0+= 0=lim

L'Hopital's rule:

ss

1s

1s 1+

1

s 1+( )2+

=lim0t

u t( ) e t t e t+( ) 1 1 0 1+= 0=lim

Initial value:

F s( )1s

1s 1+

1

s 1+( )2+=f t( ) u t( ) e t t e t+=(d)

Smith & Corripio, 3rd editionProblem 2-4. Laplace transform of a pulse by real translation theorem

f t( ) H u t( ) H u t T( )=

F s( ) H1s

H e sT 1s

= H 1 esT

s= F s( ) H

s1 e sT( )=


0 2 40

2fd t( )

t0 2 4

0

2f t( )

t

f t( ) e

t0

et

:= fd t( ) u t t0( ) et t0( ):=

u t( ) 0 t 0 0:f t( ) e

t t0( )=

Problem 2-5. Delayed versus non-delayed function

Y t( ) 2.5 e t 2.5 u t( )+= (Table 2-1.1)

(b)9

d2 y t( )dt2

18 d y t( )dt

+ 4 y t( )+ 8 x t( ) 4=

Initial steady state: 4 y 0( ) 8 x 0( ) 4=Subtract:

9d2 Y t( )

dt2 18 d Y t( )

dt+ 4 Y t( )+ 8 X t( )=

Y t( ) y t( ) y 0( )= Y 0( ) 0=X t( ) x t( ) x 0( )=

Laplace transform:9s2 Y s( ) 18s Y s( )+ 4 Y s( )+ 8 X s( )= 8 1

s=

Solve for Y(s): Y s( )8

9s2 18s+ 4+1s

= r118 182 4 9 4+

2 9:= r1 0.255=

r218 182 4 9 4

2 9:= r2 1.745=

Expand in partial fractions:Y s( )

89 s 0.255+( ) s 1.745+( )s=

A1s 0.255+

A2s 1.745++

A3s

+=

A10.255s

89 s 1.745+( )s

89 0.255 1.745+( ) 0.255( )= 2.342=lim

=

Smith & Corripio, 3rd editionProblem 2-6. Solution of differential equations by Laplace transformsInput function: X t( ) u t( )= X s( ) 1

s= (Table 2-1.1)

(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: sY s( ) Y 0( ) 2 Y s( )+ 5 X s( )= 5 1s

= Y 0( ) y 0( ) y 0( )= 0=

Solve for Y(s):Y s( )

5s 2+

1s

=A1

s 2+A2s

+=Partial fractions:

A12s

5s

2.5=lim

= A20s

5s 2+ 2.5=lim

=

Y s( )5

s 1+5s

+= Invert:

Y 0( ) 0=9d2 Y t( )

dt2 12 d Y t( )

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

9d2 y t( )

dt2 12 d y t( )

dt+ 4 y t( )+ 8 x t( ) 4=(d)

Y t( ) 1 1.134i+( )e 0.5 0.441i+( )t 1 1.134i( )e 0.5 0.441i( )t+ 2 u t( )+=Invert using Table 2-1.1:

Y s( )1 1.134i+

s 0.5+ 0.441i1 1.134i

s 0.5+ 0.441i++2s

+=

A30s

8

9s2 9s+ 4+2=lim

=A2 1 1.134i=

89 2 0.441i( ) 0.5 0.441i+( ) 1 1.134i+=A1 0.5 0.441i+s

89 s 0.5+ 0.441i+( ) slim

=

A1s 0.5+ 0.441i

A2s 0.5+ 0.441i++

A3s

+=

Y s( )8

9 s 0.5+ 0.441i( ) s 0.5+ 0.441+( )s=Solve for Y(s), expand:

A21.745s

89 s 0.255+( )s

89 1.745 0.255+( ) 1.745( )= 0.342=lim

=

A30s

89 s 0.255+( ) s 1.745+( )

89 0.255( ) 1.745( )

= 2.0=lim

=

Y s( )2.342

s 0.255+0.342

s 1.745++2s

+=

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

(c) 9d2 y t( )

dt2 9 d y t( )

dt+ 4 y t( )+ 8 x t( ) 4=

Subtract initial steady state:9

d2 Y t( )dt2

9 d Y t( )dt

+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=

Laplace transform:9s2 9s+ 4+( )Y s( ) 8 X s( )= 8 1

s=

r19 92 4 9 4+

2 9:= r29 92 4 9 4

2 9:= r1 0.5 0.441i+=Find roots:r2 0.5 0.441i=

A2 0.027 0.022i=3

2 2 2.598i( ) 1 2.598i+( ) 1.5 2.598i+( ) 0.027 0.022i+=

A11.5 2.598i+s

32 s 1.5+ 2.598i+( ) s 0.5+( )s 0.027 0.022i+=lim

=

A1s 1.5+ 2.598i

A2s 1.5+ 2.598i++

A3s 0.5++

A4s

+=

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

9

21

7

2

1.5 2.598i1.5 2.598i+

0.5

=Find roots:

2s3 7s2+ 21s+ 9+( )Y s( ) 3 X s( )= 3 1s

=Laplace transform:

Y 0( ) 0=

2d3 Y t( )

dt3 7 d

2 Y t( )dt2

+ 21 d Y t( )dt

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

2d3 y t( )

dt3 7 d

2 y t( )dt2

+ 21 d y t( )dt

+ 9 y t( )+ 3 x t( )=(e)

Y t( )4

3t 2

e

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

A30s

8

9 s 0.667+( )22=lim

=

A20.667s

dds

89s

0.667s

89s2

2=lim

=lim

=A10.667s

89s

43

=lim

=

Y s( )8

9 s 0.667+( )2s=

A1

s 0.667+( )2A2

s 0.667++A3s

+=Solve for Y(s) and expand:

r2 0.667=r1 0.667=r2

12 122 4 9 42 9:=r1

12 122 4 9 4+2 9:=

Find roots:

9s2 12s+ 4+( )Y s( ) 8 X s( )= 8 1s

=Laplace transform:

A30.5s

32 s 1.5+ 2.598i( ) s 1.5+ 2.598i+( )s 0.387=lim

=

32 1 2.598i( ) 1 2.598i+( ) 0.5( ) 0.387= A4 0s

3

2s3 7s2+ 21s+ 9+13

=lim

=

Y s( )0.027 0.022i+s 1.5+ 2.598i

0.027 0.022is 1.5+ 2.598i++

0.387s 0.5++

13

1s

+=

Invert using Table 2-1.1:

Y t( ) 0.027 0.022i+( )e 1.5 2.598i+( )t 0.027 0.022i( )e 1.5 2.598i( )t+ 0.387e 0.5 t 13

u t( )+=



Y t( ) u t 1( ) 83

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

31

s 0.667+( )28

s 0.667+8

s 0.333++

e s=

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

(Real translation theorem)

X s( )e s

s13

+=X t( ) u t 1( ) e

t 1( )3=(b) Forcing function:

Y t( )8

3t 8

e

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

Y s( )8

31

s 0.667+( )28

s 0.667++8

s 0.333++=

A20.667s

dds

89 s 0.333+( )

0.667s

89 s 0.333+( )2

8=lim

=lim

=

A30.333s

8

9 s 0.667+( )28=lim

=A1

0.667s8

9 s 0.333+( )8

3=lim

=

8

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

A1

s 0.667+( )2A2

s 0.667++A3

s 0.333++=

Y s( )8

9s2 12s+ 4+( ) s 13

+

=

X s( )1

s13

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

t3=(a) Forcing function:

Y 0( ) 0=9 d2 Y t( )dt2

12 d Y t( )dt

+ 4 Y t( )+ 8 X t( )=

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


(Final value theorem)

(b)9

d2 y t( )dt2

18 d y t( )dt

+ 4 y t( )+ 8 x t( ) 4=

Subtract initial steady state: 9d2 Y t( )

dt2 18 d Y t( )

dt+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=

Laplace transform and solve for Y(s): Y s( )8

9s2 18s+ 4+X s( )=

Find roots: r118 182 4 9 4+

2 9 min:= r218 182 4 9 4

2 9 min:= r1 0.255 min1=

r2 1.745 min 1=Invert using Table 2-1.1: Y t( ) A1 e

0.255 t A2 e 1.745 t+=+ terms of X(s)

The response is stable and monotonic. The domnant root is: r1 0.255 min 1=Time for the response to decay to 0.67% of its initial value: 5

r119.6 min=

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

s8

9s2 18s+ 4+ 1

slim

2(Final value theorem)

Smith & Corripio, 3rd editionProblem 2-8. Response characteristics of the equations of Problem 2-6

(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( )

dt2 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( )

5s 2+ X s( )=

A1s 2+= + terms of X(s)

Invert using Table 2-1.1: Y t( ) A1 e2 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

r2.5 min=

Final steady-state value for unit step input:

0ss

5s 2+

1s

lim

52

2.5=

Time for oscillations to die: 50.5 min 1

10 min=

Final steady state value for a unit step imput:0s

s8

9s2 9s+ 4+ 1

slim

2(Final value theorem)

(d) 9 d2 y t( )dt2

12 d y t( )dt

+ 4 y t( )+ 8 x t( ) 4=

Subtract initial steady state:9

d2 Y t( )dt2

12 d Y t( )dt

+ 4 Y t( )+ 8 X t( )=

Y 0( ) 0=


9s2 12s+ 4+X s( )=

Find roots: r112 122 4 9 4+

2 9 min:= r212 122 4 9 4

2 9 min:= r1 0.667 min1=

r2 0.667 min 1=Invert using Table 2-1.1: Y t( ) A1 t A2+( )e 0.667 t= + terms of X(t)

(c) 9 d2 y t( )dt2

9 d y t( )dt

+ 4 y t( )+ 8 x t( ) 4=

Subtract initial steady state: 9d2 Y t( )

dt2 9 d Y t( )

dt+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=


9s2 9s+ 4+X s( )=

Find the roots: r19 92 4 9 4+

2 9 min:= r29 92 4 9 4

2 9 min:= r1 0.5 0.441i+ min1=

r2 0.5 0.441i min 1=Invert using Table 2-3.1: Y t( ) D e 0.5 t sin 0.441t +( )= + terms of X(t)The response is stable and oscillatory. The dominant roots are r1 and r2.

Period of the oscillations: T2

0.441min 1:= T 14.25 min=

Decay ratio: e 0.5 min1 T 0.00081=


(Final value theorem) 0ss

3

2s3 7s2+ 21s+ 9+ 1

slim

13

Final steady state value for a unit step input:

5r2

10 min=Time for response to die out:e 1.5 min1 T 0.027=Decay ratio:

T 2.42 min=T 22.598min 1

:=The period of the oscillations is:

r2 0.5 min1=The response is stable and oscillatory. The dominant root is

r

1.5 2.598i1.5 2.598i+

0.5

min 1=r polyroots

9

21

7

2

min 1:=

Find roots:

Y s( )3

2s3 7s2+ 21s+ 9+X s( )=Laplace transform and solve for Y(s):

2d3 Y t( )

dt3 7 d

2 Y t( )dt2

+ 21 d Y t( )dt

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

2d3 y t( )

dt3 7 d

2 y t( )dt2

+ 21 d y t( )dt

+ 9 y t( )+ 3 x t( )=(e)

(Final value theorem) 0ss

8

9s2 12s+ 4+ 1

slim

2Final steady state value for a unit step input:

5r1

7.5 min=Time required for the response to decay within 0.67% of its initial value:r1 0.667 min 1=The response is stable and monotonic. The dominant root is

Value of k: kM gy0

:= k 1.816 Nm

=

Laplace transform:M s2 Y s( ) k Y s( )+ F s( )=

Solve for Y(s): Y s( )1

M s2 k+F s( )=

A1

s ikM

A2

s ikM

++=

+ terms of F(s)

0:=D 1:=

Invert using Table 2-3.1: Y t( ) D sinkM

t s +:= + terms of f(t)

The mobile will oscillate forever with a period of T 2 Mk

:= T 1.043 s=


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

-Mgf(t)

y(t)

-ky(t)

y = 0

Problem data: M 50gm:= y0 27 cm:=Solution:

Force balance:

Md v t( )

dt M g k y t( ) f t( )+=

Velocity: d y t( )dt

v t( )=

Initial steady state: 0 M g k y0=Subtract and substitute:

Md2 Y t( )

dt2 k Y t( ) f t( )+=

Y 0( ) 0=

0 2 41

0

1

Y t( )

t

To more accurately reflect the motion of the bird mobile, we must add the resistance of the air. If we assume it to be a force proportional to the velocity:

Md2 Y t( )

dt2 k Y t( ) b d Y t( )

dt f t( )+=

With this added term the roots will have a negative real part, causing the oscillations to decay, as they do in practice:

Y s( )1

M s2 b s+ k+F s( )= r1

b b2 4M k+2M

= b2M

ikM

b2

4M2+=

Invert:b2 4M k

smith & corripio, 3rd edition solution

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