Principios e Pratica do Controle de Processos 3ed Soluções

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(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
Water
main
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
SC
SP
S
Air
Engine
Transmission
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 S
SP
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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Smith & Corripio, 3rd edition
Problem 1-2. Automatic shower temperature diagram.
TETC
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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is unlawful. 
F s( )
s
s2 ω2+
=
1
2
1
s i ω⋅−
1
s i ω⋅++


=
s i ω⋅− s+ i ω⋅+
2 s i ω⋅−( )⋅ s i ω⋅+( )=
2 s⋅
2 s2 ω2+( )⋅=
s
s2 ω2+
=
1
2
1−
s i ω⋅− e
s i ω⋅−( )t− ∞
0
⋅ 1−
s i ω⋅+ e
s i ω⋅+( )t− ∞
0
⋅+

=
1
2 0
∞
te s i ω⋅−( )t−
⌠⌡ d 0
∞
te s i ω⋅+( )t−
⌠⌡ d+


=
F s( )
0
∞
tcos ωt⋅ e st−⋅⌠⌡ d=
0
∞
t
ei ωt⋅ e i− ωt⋅−
2
e st−
⌠

⌡
d=f t( ) cos ωt⋅=(c)
F s( )
1
s a+=
F s( )
0
∞
te at− e st−
⌠⌡ d= 0
∞
te s a+( )t−
⌠⌡ d=
1−
s a+ e
s a+( )t− ∞
0
⋅= 1
s a+=
where a is constantf t( ) e at−=(b)
F s( )
1
s2
=
F s( )
t−
s
e st− ∞
0
⋅ 1
s 0
∞
te st−
⌠⌡ d⋅+= 0 0−
1
s2
e st− ∞
0
⋅−= 1
s2
=
v
1−
s
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 definition
Smith & Corripio, 3rd. edition
(d) f t( ) e at− coss ωt⋅=
F s( )
0
∞
te at− cos ωt⋅ e st−⋅⌠⌡ d=
0
∞
te at− e
i ωt⋅ e i− ωt⋅+
2
⋅ e st−
⌠

⌡
d=
1
2 0
∞
te s a+ i ω⋅+( )t−
⌠⌡ d 0
∞
te s a+ i ω⋅−( )− t
⌠⌡ d+


=
1
2
1−
s a+ i ω⋅+ e
s a+ i ω⋅+( )t− ∞
0
⋅ 1−
s a+ i ω⋅− e
s a+ i ω⋅−( )t− ∞
0
⋅+

=
1
2
1
s a+ i ω⋅+
1
s a+ i ω⋅−+


=
s a+ i ω⋅− s+ a+ i ω⋅+
2 s a+ i ω⋅+( ) s a+ i ω⋅−( )=
2 s a+( )
2 s a+( )2 ω2+ ⋅
= s a+
s a+( )2 ω2+
= F s( ) s a+
s a+( )2 ω2+
=
All the results match results in Table 2-1.1
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is unlawful. 
1
s
1
s 2++ 2
1
s 1+⋅−=
1
s
1
s 2++
2
s 1+−=
F s( )
1
s
1
s 2++
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
1
s 1+−
1
s 1+( )2
+=
F s( )
1
s
1
s 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
s
1
s 2+( )2 1+
−

=
Used the real translation theorem and linearity. F s( ) e 2− s 1
s
1
s 2+( )2 1+
−

=
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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 
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Smith & Corripio, 3rd edition
Problem 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( )⋅+=
1
s
2
1
s2
⋅+ 3 2!
s3
⋅+= F s( ) 1
s
2
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+
⋅= 1
s
2
s2
+ 6
s3
+

 s 2+
⋅=
1
s 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:
∞s
1
1
2
2 s 2+( )+
6
3 s 2+( )2
+

1=lim
→Final value:
∞t
e 2− t u t( ) 2 t⋅+ 3t2+( ) 0 ∞⋅=lim
→
0s
s
1
s 2+
2
s 2+( )2
+ 6
s 3+( )2
+

0=lim
→L'Hopital's rule:
∞t
0
2e2t
2
2e2t
+ 6t
2e2t
+


0=lim
→ Check!
(c) f t( ) u t( ) e 2− t+ 2e t−−= F s( ) 1
s
1
s 2++
2
s 1+−=
Initial value:
0t
u t( ) e 2− t+ 2e t−−( ) 1 1+ 2−( ) 0+=lim
→ ∞s
s
1
s
1
s 2++
2
s 1+−



∞
∞=lim→
L'Hopital's rule:
∞s
1
1
1
+ 2
1
−

 0=lim→
Final value:
∞t
u t( ) e 2− t+ 2e t−−( ) 1 0+ 0+= 1=lim
→ 0s
s
1
s
1
s 2++
2
s 1+−


 1 0+ 0+= 1=lim→
Smith & Corripio, 3rd edition
Problem 2-3. Initial and final value check of solutions to Problem 2-2
(a) f t( ) u t( ) 2 t⋅+ 3t2+= F s( )1
s
2
s2
+ 6
s3
+=
Initial value:
0t
u t( ) 2t+ 3t2+( ) 1=lim
→ ∞s
s
1
s
2
s2
+ 6
s3
+


⋅
∞s
1
2
s
+ 6
s2
+


1=lim
→
=lim
→
Final value:
∞t
u t( ) 2t+ 3t2+( ) ∞=lim
→ 0s
1
2
s
+ 6
s2
+


∞=lim
→
Check!
(b) f t( ) e 2− t u t( ) 2t+ 3t2+( )= F s( ) 1
s 2+
2
s 2+( )2
+ 6
s 2+( )3
+=
Initial value:
0t
e 2− t u t( ) 2t+ 3t2+( )lim
→ ∞s
s
1
s 2+
2
s 2+( )2
+ 6
s 2+( )3
+

∞
∞=lim→
1 1 0+ 0+( )= 1=
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
Check!
0s
s
1
s
1
s 1+( )2 1+
−

1 0+= 1=lim
→∞t
1 e 2− tsin t( )⋅−  1=lim
→
Final value:
∞s
s
1
s
1
s 1+( )2 1+
−

1 0−= 1=lim
→0t
1 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 1
s
1
s 1+( )2 1+
−

=f t( ) u t 2−( ) 1 e 2− t 2−( ) sin t 2−( )− =(e)
Check!∞t
1 0− 1
1 et⋅
+


1=lim
→
L'Hopital's rule:
∞t
u t( ) e t−− t e t−⋅+( ) 1 0− ∞ 0⋅+=lim
→ 0s
1
s
s 1+−
s
s 1+( )2
+

1 0− 0+= 1=lim
→
Final value: ∞s
1
1
1
− 1
2 s 1+( )+

 1 1− 0+= 0=lim→
L'Hopital's rule:
∞s
s
1
s
1
s 1+−
1
s 1+( )2
+

∞
∞=lim→0t
u t( ) e t−− t e t−⋅+( ) 1 1− 0 1⋅+= 0=lim
→
Initial value:
F s( )
1
s
1
s 1+−
1
s 1+( )2
+=f t( ) u t( ) e t−− t e t−⋅+=(d)
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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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. 
0 2 4
0
2
fd t( )
t
0 2 4
0
2
f t( )
t
f t( ) e
t0
τ e
t−
τ⋅:= fd t( ) u t t0−( ) e
t t0−( )−
τ⋅:=
u t( ) 0 t 0<if
1 t 0≥if
:=τ 1:=t0 1:=Sketch the functions: F s( )
τ e t0− s⋅⋅
τ s⋅ 1+=
The result to part (b) agrees with the real translation theorem.
e
t0− s⋅ 1−
s
1
τ+
⋅ e
s
1
τ+


− λ⋅⋅ ∞
0
⋅= e
t0− s⋅
s
1
τ+
= τ e
t0− s⋅⋅
τ s⋅ 1+=
F s( )
t0−
∞
λu λ( ) e
λ−
τ e
s λ t0+( )−
⌠

⌡ d= e
t0− s⋅
0
∞
λe
s
1
τ+


λ−
⌠

⌡ d⋅=
λ t t0−=Let
F s( )
0
∞
tu t t0−( ) e
t t0−( )−
τ e st−
⌠

⌡ d=f t( ) u t t0−( ) e
t t0−( )−
τ=
(b) Function is delayed and zero from t = 0 to t = t0:
F s( )
τ e
t0
τ⋅
τ s⋅ 1+=F s( ) e
t0
τ 1
s
1
τ+
= τ e
t0
τ⋅
τ s⋅ 1+=f t( ) e
t0
τ e
t−
τ=
(from Table 2-1.1)
(a) Function is non-zero for all values of t > 0:
f t( ) e
t t0−( )−
τ=
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:
9
d2 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+
1
s
= r1
18− 182 4 9⋅ 4⋅−+
2 9⋅:= r1 0.255−=
r2
18− 182 4 9⋅ 4⋅−−
2 9⋅:= r2 1.745−=
Expand in partial fractions:
Y s( )
8
9 s 0.255+( ) s 1.745+( )s=
A1
s 0.255+
A2
s 1.745++
A3
s
+=
A1
0.255−s
8
9 s 1.745+( )s
8
9 0.255− 1.745+( )⋅ 0.255−( )⋅= 2.342−=lim→
=
Smith & Corripio, 3rd edition
Problem 2-6. Solution of differential equations by Laplace transforms
Input 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 1
s
⋅= Y 0( ) y 0( ) y 0( )−= 0=
Solve for Y(s):
Y s( )
5
s 2+
1
s
=
A1
s 2+
A2
s
+=
Partial fractions:
A1
2−s
5
s
2.5−=lim
→
= A2
0s
5
s 2+ 2.5=lim→
=
Y s( )
5−
s 1+
5
s
+= Invert:
Y 0( ) 0=9
d2 Y t( )⋅
dt2
⋅ 12 d Y t( )⋅
dt
⋅+ 4 Y t( )+ 8 X t( )=Subtract initial steady state:
9
d2 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+ 0.441i−
1− 1.134i−
s 0.5+ 0.441i++
2
s
+=
A3
0s
8
9s2 9s+ 4+
2=lim
→
=A2 1− 1.134i−=
8
9 2 0.441i⋅( ) 0.5− 0.441i+( ) 1− 1.134i+=A1 0.5− 0.441i+s
8
9 s 0.5+ 0.441i+( ) slim→
=
A1
s 0.5+ 0.441i−
A2
s 0.5+ 0.441i++
A3
s
+=
Y s( )
8
9 s 0.5+ 0.441i−( ) s 0.5+ 0.441+( )s=Solve for Y(s), expand:
A2
1.745−s
8
9 s 0.255+( )s
8
9 1.745− 0.255+( ) 1.745−( )= 0.342=lim→
=
A3
0s
8
9 s 0.255+( ) s 1.745+( )
8
9 0.255( ) 1.745( )
= 2.0=lim
→
=
Y s( )
2.342−
s 0.255+
0.342
s 1.745++
2
s
+=
Invert with Table 2-1.1:
Y t( ) 2.342− e 0.255− t 0.342e 1.745− t+ 2 u t( )+=
(c) 9
d2 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
⋅=
r1
9− 92 4 9⋅ 4⋅−+
2 9⋅:= r2
9− 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+=
A1
1.5− 2.598i+s
3
2 s 1.5+ 2.598i+( ) s 0.5+( )s 0.027 0.022i+=lim→
=
A1
s 1.5+ 2.598i−
A2
s 1.5+ 2.598i++
A3
s 0.5++
A4
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
9
21
7
2








1.5− 2.598i−
1.5− 2.598i+
0.5−



=Find roots:
2s3 7s2+ 21s+ 9+( )Y s( ) 3 X s( )= 3 1
s
⋅=Laplace transform:
Y 0( ) 0=
2
d3 Y t( )⋅
dt3
⋅ 7 d
2 Y t( )⋅
dt2
⋅+ 21 d Y t( )⋅
dt
⋅+ 9 Y t( )+ 3 X t( )=Subtract initial steady state:
2
d3 y t( )⋅
dt3
⋅ 7 d
2 y t( )⋅
dt2
⋅+ 21 d y t( )⋅
dt
⋅+ 9 y t( )+ 3 x t( )=(e)
Y t( )
4−
3
t 2−

e
0.667− t 2 u t( )+=Invert using Table 2-1.1:
A3
0s
8
9 s 0.667+( )2
2=lim
→
=
A2
0.667−s
d
ds
8
9s


 0.667−s
8−
9s2
2−=lim
→
=lim
→
=A1
0.667−s
8
9s
4−
3
=lim
→
=
Y s( )
8
9 s 0.667+( )2s
=
A1
s 0.667+( )2
A2
s 0.667++
A3
s
+=Solve for Y(s) and expand:
r2 0.667−=
r1 0.667−=r2
12− 122 4 9⋅ 4⋅−−
2 9⋅:=r1
12− 122 4 9⋅ 4⋅−+
2 9⋅:=
Find roots:
9s2 12s+ 4+( )Y s( ) 8 X s( )= 8 1
s
⋅=Laplace transform:
A3
0.5−s
3
2 s 1.5+ 2.598i−( ) s 1.5+ 2.598i+( )s 0.387−=lim→
=
3
2 1 2.598i−( ) 1 2.598i+( ) 0.5−( ) 0.387−= A4 0s
3
2s3 7s2+ 21s+ 9+
1
3
=lim
→
=
Y s( )
0.027 0.022i+
s 1.5+ 2.598i−
0.027 0.022i−
s 1.5+2.598i++
0.387−
s 0.5++
1
3
1
s
+=
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− 1
3
u t( )+=
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is unlawful. 
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
1
s 0.667+( )2
8
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−
s
1
3
+
=X t( ) u t 1−( ) e
t 1−( )−
3=(b) Forcing function:
Y t( )
8−
3
t 8−

e
0.667− t 8e 0.333− t+=Invert using Table 2-1.1:
Y s( )
8−
3
1
s 0.667+( )2
8−
s 0.667++
8
s 0.333++=
A2
0.667−s
d
ds
8
9 s 0.333+( )

 0.667−s
8−
9 s 0.333+( )2
8−=lim
→
=lim
→
=
A3
0.333−s
8
9 s 0.667+( )2
8=lim
→
=A1
0.667−s
8
9 s 0.333+( )
8−
3
=lim
→
=
8
9 s 0.667+( )2 s 0.333+( )
=
A1
s 0.667+( )2
A2
s 0.667++
A3
s 0.333++=
Y s( )
8
9s2 12s+ 4+( ) s 1
3
+


=
X s( )
1
s
1
3
+
=From Table 2-1.1:X t( ) e
t−
3=(a) Forcing function:
Y 0( ) 0=9 d
2 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
Smith & Corripio, 3rd edition
(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: 9
d2 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: r1
18− 182 4 9⋅ 4⋅−+
2 9⋅ min:= r2
18− 182 4 9⋅ 4⋅−−
2 9⋅ min:= r1 0.255− min
1−=
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−
r1
19.6 min=
Final steady-state value for unit step input:
0s
s
8
9s2 18s+ 4+
⋅ 1
s
lim
→
2→
(Final value theorem)
Smith & Corripio, 3rd edition
Problem 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( )⋅
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( )=
A1
s 2+= + terms of X(s)
Invert using Table 2-1.1: Y t( ) A1 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:
0s
s
5
s 2+⋅
1
s
lim
→
5
2
→ 2.5=
Time for oscillations to die: 5−
0.5− min 1−
10 min=
Final steady state value for a unit step imput:
0s
s
8
9s2 9s+ 4+
⋅ 1
s
lim
→
2→
(Final value theorem)
(d) 9 d
2 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=
Laplace transform and solve for Y(s): Y s( )
8
9s2 12s+ 4+
X s( )=
Find roots: r1
12− 122 4 9⋅ 4⋅−+
2 9⋅ min:= r2
12− 122 4 9⋅ 4⋅−−
2 9⋅ min:= r1 0.667− min
1−=
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 d
2 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 and solve for Y(s): Y s( )
8
9s2 9s+ 4+
X s( )=
Find the roots: r1
9− 92 4 9⋅ 4⋅−+
2 9⋅ min:= r2
9− 92 4 9⋅ 4⋅−−
2 9⋅ min:= r1 0.5− 0.441i+ min
1−=
r2 0.5− 0.441i− min 1−=
Invert using Table 2-3.1: Y t( ) D e 0.5− t⋅ sin 0.441t θ+( )= + terms of X(t)
The response is stable and oscillatory. The dominant roots are r1 and r2.
Period of the oscillations: T
2π
0.441min 1−
:= T 14.25 min=
Decay ratio: e 0.5− min
1− T 0.00081=
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(Final value theorem) 0s
s
3
2s3 7s2+ 21s+ 9+
⋅ 1
s
lim
→
1
3
→Final steady state value for a unit step input:
5−
r2
10 min=Time for response to die out:e 1.5− min
1− T 0.027=Decay ratio:
T 2.42 min=T 2π
2.598min 1−
:=The period of the oscillations is:
r2 0.5− min
1−=The response is stable and oscillatory. The dominant root is
r
1.5− 2.598i−
1.5− 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):
2
d3 Y t( )⋅
dt3
⋅ 7 d
2 Y t( )⋅
dt2
⋅+ 21 d Y t( )⋅
dt
⋅+ 9 Y t( )+ 3 X t( )=Subtract initial steady state:
2
d3 y t( )⋅
dt3
⋅ 7 d
2 y t( )⋅
dt2
⋅+ 21 d y t( )⋅
dt
⋅+ 9 y t( )+ 3 x t( )=(e)
(Final value theorem) 0s
s
8
9s2 12s+ 4+
⋅ 1
s
lim
→
2→Final steady state value for a unit step input:
5−
r1
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: k
M− g⋅
y0
:= k 1.816 N
m
=
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 i
k
M
⋅−
A2
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 2π 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:= y0 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 y0⋅−=
Subtract and substitute:
M
d2 Y t( )⋅
dt2
⋅ k− Y t( )⋅ f t( )+=
Y 0( ) 0=
0 2 4
1
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:
M
d2 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− b24M k⋅−+
2M
= b−
2M
i
k
M
b2
4M2
−⋅+=
Invert:
b2 4M k⋅<
Y t( ) D e
b−
2M
t⋅
⋅ sin k
M
b2
4M2
− t θ+



= + terms of f(t)
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
H 1:=T 1:=τ 1:=KH 1:=Invert using Table 2-1.1, and the real translation theorem:
Y s( ) K H
1
s
1
s
1
τ+
−


⋅ 1 e sT−−( )=
A2
0s
K H⋅
τ s⋅ 1+ K H⋅=lim→
=A1
1−
τs
K H⋅
τ s⋅ K− H⋅=lim→
=
Y s( )
K
τ s⋅ 1+ H⋅
1 e sT−−
s
⋅=
A1
s
1
τ+
A2
s
+


1 e sT−−( )=Substitute:
X s( ) H
1 e sT−−
s
⋅=
From Example 2-1.1b:
(b) Pulse of Fig. 2-1.1b
0 2 4
0
0.5
1
Y t( )
t
Y t( )
K
τ e
t−
τ:=
Invert using Table 2-1.1:
Y s( )
K
τ s⋅ 1+=
X s( ) 1=From Table 2-1.1:X t( ) δ t( )=(a) Unit impulse:
Y s( )
K
τ s⋅ 1+ X s( )=
Laplace transform and solve for Y(s):
Y 0( ) 0=τ d Y t( )⋅
dt
⋅ Y t( )+ K X t( )⋅=
Problem 2-10. Responses of general first-order differential equation
Smith & Corripio, 3rd edition
Y t( ) KH u t( ) e
t−
τ− u t T−( ) 1 e
t T−( )−
τ−

⋅−

⋅:=
X t( ) H u t( ) u t T−( )−( )⋅:=
0 2 4
0
0.5
1
Y t( )
X t( )
t
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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. 
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. 
The tank is an integrating process because its ouput, the level, is the time integral of its input, the 
inlet flow.
0 5 10
0
5
10
h t( )
t
f(t)
h(t)
A 1:=
h t( )
1
A
t:=Invert using Table 2-1.1:H s( ) 1
A
1
s2
=Substitute:
(Table 2-1.1)F s( )
1
s
=f t( ) u t( )=Response to a unit step in flow:
H s( )
F s( )
1
A s⋅=Transfer function of the tank:
H s( )
1
A s⋅ F s( )=Laplace transform and solve for H(s):
h 0( ) 0=A d h t( )⋅
dt
⋅ f t( )=
Problem 2-11. Response of an integrating process
Smith & Corripio, 3rd edition
r2 1.745− min 1−= τe2
1−
r2
:= τe2 0.573 min=
5 τe1⋅ 19.64 min=Time for response to decay within 0.67% of its initial value:
(b) 9 d
2 y t( )⋅
dt2
⋅ 9 d y t( )⋅
dt
⋅+ 4 y t( )+ 8 x t( ) 4−=
Subtract initial steady state
and divide by the Y(t) coefficient:
9
4
d2 Y t( )⋅
dt2
⋅ 9
4
d Y t( )⋅
dt
⋅+ Y t( )+ 2 X t( )= Y 0( ) 0=
Compare coefficients to standard form: τ 9
4
min:= τ 1.5 min= ζ 9min
4 2⋅ τ⋅:= ζ 0.75=
K 2:=
Underdamped.Find roots: r1
9− 92 4 9⋅ 4⋅−+
2 9⋅ min:= r1 0.5− 0.441i+ min
1−=
Frequency of oscillations: ω 0.441 rad
min
:= Period of oscillations: T 2πω:= T 14.25 min=
Smith & Corripio, 3rd edition
Problem 2-12. Second-order differeential equations of Problem 2-6.
Standard form of the second-order equation: τ2 d
2 Y t( )⋅
dt2
⋅ 2 ζ⋅ τ⋅ d Y t( )⋅
dt
⋅+ Y t( )+ K X t( )⋅=
(b) 9 d
2 y t( )⋅
dt2
⋅ 18 d y t( )⋅
dt
⋅+ 4 y t( )+ 8 x t( ) 4−=
Subtract the initial steady state:
9
d2 Y t( )⋅
dt2
⋅ 18 d Y t( )⋅
dt
⋅+ 4 Y t( )+ 8 X t( )= Y 0( ) 0=
Divide by Y(t) coefficient: 9
4
d2 Y t( )⋅
dt2
⋅ 18
4
d Y t( )⋅
dt
⋅+ Y t( )+ 2 X t( )=
Match coeffients to standard form: τ 9
4
min:= τ 1.5 min= ζ 18min
4 2⋅ τ⋅:= ζ 1.5=Equivalent time constants:
K 2:= Overdamped.
Find roots: r1
18− 182 4 9⋅ 4⋅−+
2 9⋅ min:= r1 0.255− min 1−= τe1
1−
r1
:= τe1 3.927 min=
r2
18− 182 4 9⋅ 4⋅−−
2 9⋅ min:=
ζ 1=
K 2:= Critically damped.Equivalent time constants:
Find roots: r1
12− 122 4 9⋅ 4⋅−+
2 9⋅ min:= r1 0.667− min
1−= τe1
1−
r1
:= τe1 1.5 min=
r2
12− 122 4 9⋅ 4⋅−−
2 9⋅ min:= r2 0.667− min 1−= τe2
1−
r2
:= τe2 1.5 min=
Time for response to decay to within 0.67% of its initial value: 5 τe1⋅ 7.5 min=
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. 
Decay ratio: e 0.5− min
1− T 0.00081= Percent overshoot: e
0.5− min 1− T
2 2.8 %=
Rise time: T
4
3.56 min= Settling time: 5−
0.5− min 1−
10 min=
(c) 9 d
2 y t( )⋅
dt2
⋅ 12 d y t( )⋅
dt
⋅+ 4 y t( )+ 8 x t( ) 4−=
Subtract initial steady state and
divide by the coefficient of Y(t): 9
4
d2 Y t( )⋅
dt2
⋅ 3 d Y t( )⋅
dt
⋅+ Y t( )+ 2 X t( )=
Y 0( ) 0=
Compare coefficients to standard form: τ 9
4
min:= τ 1.5 min= ζ 3min
2 τ⋅:=
Y s( ) K ∆x 1−τ
1
s
1
τ+



2
1
s
1
τ+



− 1
s
+

⋅=
A2
1−
τs
d
ds
K ∆x⋅
τ2s


 1−
τs
K− ∆x⋅
τ2 s2
K− ∆x⋅=lim
→
=lim
→
=
A3
0s
K ∆x⋅
τ s⋅ 1+( )2
K ∆x⋅=lim
→
=A1
1−
τs
K ∆x⋅
τ2s
K− ∆x⋅
τ=lim→
=
Y s( )
K
τ s⋅ 1+( )2
∆x
s
=
A1
s
1
τ+



2
A2
s
1
τ+
+
A3
s
+=
Step response for the critically damped case:
Y t( ) K ∆x u t( )
τe1
τe1 τe2−
e
t−
τe1−
τe2
τe2 τe1−
e
t−
τe2−




⋅=
(2-5.10)Invert using Table 2-1.1:
Y s( ) K ∆x
τe1−
τe1 τe2−
1
s
1
τe1
+
τe2
τe2 τe1−
1
s
1
τe2
+
− 1
s
+



⋅=
A3
0s
K ∆x⋅
τe1 s⋅ 1+( ) τe2 s⋅ 1+( ) K ∆x⋅=lim→=
A2
K− ∆x⋅ τe2⋅
τe2 τe1−
=A1
1−
τe1
s
K ∆x⋅
τe1 τe2⋅ s
1
τe2
+


⋅ s
K− ∆x⋅ τe1⋅
τe1 τe2−
=lim
→
=
Y s( )
K
τe1 s⋅ 1+( ) τe2 s⋅ 1+( )
∆x
s
=
A1
s
1
τe1
+
A2
s
1
τe2
+
+
A3
s
+=
X s( )
∆x
s
=Step response, over-damped second-order differential equation:
Problem 2-13. Partial fraction expansion coefficients for Eqs. 2-5.10 to 2-5.13
Smith & Corripio, 3rd edition
Y s( )
K
τ s⋅ 1+( )2
r
s2
=
A1
s
1
τ+



2
A2
s
1
τ+
+
A3
s2
+
A4
s
+=
Ramp response for critically damped case:
Y t( ) K r
τe1
2
τe1 τe2−
e
t−
τe1 τe2
2
τe2 τe1−
e
t−
τe2+ t+ τe1 τe2+( )−


⋅= (2-5.12)
Invert using Table 2-1.1:
Y s( ) K r
τe1
2
τe1 τe2−
1
s
1
τe1
+
τe2
2
τe2 τe1−
1
s
1
τe2
+
+ 1
s2
+
τe1 τe2+
s
−




⋅=
K r τe1− τe2−( )⋅=
A4
0s
d
ds
K r⋅
τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅

⋅
0s
K r⋅
τe1− τe2 s⋅ 1+( )⋅ τe2 τe1 s⋅ 1+( )⋅−
τe1 s⋅ 1+( )2 τe2 s⋅ 1+( )2
⋅lim
→
=lim
→
=
A3
0s
K r⋅
τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅ K r⋅=lim→=
A2
K r⋅ τe2
2⋅
τe2 τe1−
=A1
1−
τe1
s
K r⋅
τe1 τe2⋅ s
1
τe2
+


⋅ s2⋅
K r⋅ τe1
2⋅
τe1 τe2−
=lim
→
=
Y s( )
K
τe1 s⋅ 1+( ) τe2 s⋅ 1+( )⋅
r
s2
=
A1
s
1
τe1+
A2
s
1
τe2
+
+
A3
s2
+
A4
s
+=
X s( )
r
s2
=Ramp response for the over-damped case:
Y t( ) K ∆x u t( ) tτ 1+


e
t−
τ−

⋅=
(2-5.11)
Invert using Table 2-1.1:
A1
1−
τs
K r⋅
τ2 s2
K r⋅=lim
→
= A3
0s
K r⋅
τ s⋅ 1+( )2
K r⋅=lim
→
=
A2
1−
τs
d
ds
K r⋅
τ2 s2


 1−
τs
2− K r⋅
τ2 s3
⋅ 2 K⋅ r⋅ τ⋅=lim
→
=lim
→
=
A4
0s
d
ds
K r⋅
τ s⋅ 1+( )2

 0s
2− K r⋅ τ⋅
τ s⋅ 1+( )3
⋅ 2− K⋅ r⋅ τ⋅=lim
→
=lim
→
=
Y s( ) K r
1
s
1
τ+



2
2 τ⋅
s
1
τ+
+ 1
s2
+ 2 τ⋅
s
−

⋅=
Invert using Table 2-1.1:
Y t( ) K r⋅ t 2 τ⋅+( )e
t−
τ t+ 2 τ⋅−

⋅= (2-5.13)
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Smith & Corripio, 3rd edition
X s( )
∆x
s
=Problem 2-14. Derive step reponse of n lags in series
Y s( )
K
1
n
k
τk s⋅ 1+( )∏
=
∆x
s
=
A0
s
1
n
k
Ak
s
1
τk
+∑=+=
A0
0s
K ∆x⋅
1
n
k
τk s⋅ 1+( )∏
=
K ∆x⋅=lim
→
=
Invert using Table 2-1.1:
Y t( ) K ∆x⋅ u t( )⋅
1
n
k
Ak e
t−
τk⋅∑
=
+=
Ak
1−
τk
s
K ∆x⋅
s
1 j k≠( )⋅
n
j
s
1
τ j
+

∏=⋅ 1
n
j
τ j∏
=
⋅
K ∆x⋅
1−
τk 1 j k≠( )
n
j
1−
τk
1
τ j
+

 1
n
j
τ j∏
=
⋅∏
=
⋅
=lim
→
=
K− ∆x⋅
1
τk
1
τk
n 1−⋅ τk⋅
1 j k≠( )⋅
n
j
τk τ j−( )∏
=
⋅
=
K− ∆x⋅ τk
n 1−⋅
1 j k≠( )
n
j
τk τ j−( )∏
=
=
Substitute:
Y t( ) K ∆x u t( )
1
n
k
τk
n 1−
1 j k≠( )
n
j
τk τ j−( )∏
=
e
t−
τk∑
=
−


⋅= (2-5.23)
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is unlawful. 
r1
τ1 τ2+( )− τ1 τ2+( )2 4τ1 τ2 1 k2−( )⋅−+
2 τ1⋅ τ2⋅
=
(b) The response is stable if both roots are negative if 0 < k2 < 1.
This term is positive as long as τ1, τ2, and k2 are positive, so the response is overdamped.
τ1 τ2−( )2 4τ1 τ2⋅ k2⋅+=
τ1
2
2τ1 τ2⋅− τ2
2+ 4τ1 τ2⋅ k2⋅+=
τ1 τ2+( )2 4τ1 τ2⋅ 1 k2−( )⋅− τ12 2τ1 τ2⋅+ τ22+ 4τ1 τ2⋅− 4τ1 τ2⋅ k2⋅+=
(a) The response is overdamped if the term in the radical is positive:
r1
τ1 τ2+( )− τ1 τ2+( )2 4τ1 τ2 1 k2−( )⋅−+
2 τ1⋅ τ2⋅
=
τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ k2− 0=Find the roots of the denominator:
ζ
τ1 τ2+
2 τ⋅ 1 k2−( )⋅=
τ1 τ2+
2 τ1 τ2⋅ 1 k2−( )⋅⋅=Damping ratio:
τ
τ1 τ2⋅
1 k2−
=Time constant:K
k1
1 k2−
=Gain:Comparing coefficients:
Y s( )
k1
1 k2−
τ1 τ2⋅
1 k2−



s2
τ1 τ2+
1 k2−
s+ 1+
X s( )=
Rerrange interacting equation:
Y s( )
K
τ2 s2 2ζ τ⋅ s⋅+ 1+
X s( )=
Standard form of the second-order differential equaton, Eq. 2-5.4:
Y s( )
k1
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ k2− X s( )=
k1
τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ k2− X s( )=
Problem 2-15. Transfer function of second-order interacting systems.
Smith & Corripio, 3rd edition
If τ1, τ2, and k2 are positive, and if k2 < 1, then the positive term in the numerator is always less in 
magnitude than the negative term, and the root is negative. The other root has to be negative 
because both terms in the numerator are negative. So, the response is stable. 
(c) Effective time constants
As the response is overdamped, we can derive the formulas for the two effective time constants. 
These are the negative reciprocals of the two real roots:
τe1
2 τ1⋅ τ2⋅
τ1 τ2+ τ1 τ2−( )2 4τ1 τ2⋅ k2⋅+−
= τe1
2 τ1⋅ τ2⋅
τ1 τ2+ τ1 τ2−( )2 4τ1 τ2⋅ k2⋅++
=
The first of these is the dominant time constant.
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The response canot be unstable for positive Kc. The time constant and damping ratio are always 
real and positive for positive gain.
Cannot be undamped for finite Kc.
ζ 0=(iii) Undamped:
ζ cannot be negative for positive Kc13 Kc< ∞<0 ζ< 1<(ii) Underdamped:
Kc
1
3
<4
3
1 Kc+>
2
3 1 Kc+( ) 1>ζ 1>(i) Overdamped:
Ranges of the controller gain for which the response is:
ζ 4
2 τ⋅ 1 Kc+( )⋅=
2
3 1 Kc+( )⋅=Damping ratio:
τ 3
1 Kc+
=Time constant:K
Kc
1 Kc+
=Gain:
C s( )
Kc
1 Kc+
3
1 Kc+
s2
4
1 Kc+
s+ 1+
R s( )=
Rearrange feedback loop transfer function and compare coefficients:
C s( )
K
τ2 2ζ τ⋅ s⋅+ 1+
R s( )=Standard second-order transfer function, Eq. 2-5.4:
This is a second-order process with a proportional controller.
C s( )
Kc
3s 1+( ) s 1+( )⋅ Kc+
R s( )=
Kc
3s2 4s+ 1+ Kc+
=
Problem 2-16. Transfer function of a second-order feedback control loop
Smith & Corripio, 3rd edition
Y X t( )( )
α
1 α 1−( )xb+ 2
X t( )=
Y X t( )( ) y x t( )( ) y xb( )−=X t( ) x t( ) xb−=Let 
y x t( ) y xb( ) 1 α 1−( ) xb⋅+  α⋅ α xb⋅ α 1−( )⋅−
1 α 1−( )xb+ 2
x t( ) xb−( )+=
y x t( )( )
α x t( )⋅
1 α 1−( )x t( )+=
(c) Eqilibrium mole fraction by relative volatility, Eq. 2-6.3:
Po Γ t( )( ) B p
o⋅ Tb( )
Tb C+( )2
Γ t( )=
Po Γ t( )( ) po T t( )( ) po Tb( )−=Γ t( ) T t( ) Tb−=Let
po T t( )( ) po Tb( ) B
Tb C+( )2
e
A
B
Tb C+
−
T t( ) Tb−( )+=
po T t( )( ) e
A
B
T t( ) C+−=
(b) Antoine equation for vapor pressure, Eq. 2-6.2:
Hd Γ t( )( ) a1 2a2 Tb⋅+ 3a3 Tb2⋅+ 4a4 Tb3⋅+ Γ t( )=
Hd Γ t( )( ) H T t( )( ) H Tb( )−=Γ t( ) T t( ) Tb−=Let
H T t( )( ) H Tb( ) a1 2a2 Tb⋅+ 3a3 Tb2⋅+ 4a4 Tb3⋅+  T t( ) Tb−( )+=
H T t( )( ) H0 a1 T t( )⋅+ a2 T2⋅ t( )⋅+ a3 T3⋅ t( )+ a4 T4⋅ t( )+=
(use subscript b for base value)(a) Enthalpy as a function of temperature, Eq. 2-6.1:
Problem 2-17. Linearization of common process model functions.
Smith & Corripio, 3rd edition
(d) Flow as a function of pressure drop, Eq. 2-6.4:
f ∆p t( )( ) k ∆p t( )⋅=
f ∆p t( )( ) f ∆pb( ) k
2 ∆pb⋅
∆p t( ) ∆pb−( )+=
Let ∆P t( ) ∆p t( ) ∆pb−= F ∆P t( )( ) f ∆p t( )( ) f ∆pb( )−=
F ∆P t( )( ) k
2 ∆pb⋅
∆P t( )=
(e) Radiation heat transfer rate as a function of temperature, Eq. 2-6.5:
q T t( )( ) ε σ⋅ A⋅ T4⋅ t( )=
q T t( )( ) q Tb( ) 4 ε⋅ σ⋅ A⋅ Tb3⋅ T t( ) Tb−( )+=
Let Γ t( ) T t( ) Tb−= Q Γ t( )( ) q T t( )( ) q Tb( )−=
Q Γ t( )( ) 4 ε⋅ σ⋅ A⋅ Tb3⋅ Γ t( )⋅=
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is unlawful. 
Tmax 610 K= Tmin 590 K=
Temperature range for which the heat transfer rate is within 5% of the linear 
approximation:
error ε σ⋅ A⋅ T4⋅ ε σ⋅ A⋅ Tb4⋅ 4ε σ⋅ A⋅ Tb3⋅ T Tb−( )+ −= 0.05 ε σ⋅ A T4⋅⋅( )=
Simplify and rearrange: T4 4 Tb
3⋅ T⋅− 3Tb4+ 0.05T4=
As the error is always positive, the absolute value brackets can be dropped. Rearrange into a 
polynomial and find its roots:
0.95
T
Tb



4
4
T
Tb
− 3+ 0=
polyroots
3
4−
0
0
0.95








1.014− 1.438i−
1.014− 1.438i+
0.921
1.108




=
Ignore the complex roots. The other two roots are the lower and upper limits of the range:
0.921
T
Tb
≤ 1.108≤
For Tb 400K:= Tmin 0.921 Tb⋅:= Tmax 1.108Tb:= Tmin 368 K= Tmax 443 K=
Smith & Corripio, 3rd edition
Problem 2-18. Linearization of radiation heat transfer--range of accuracy.
q T( ) 4ε σ⋅ A⋅ T4⋅= Use subscript "b" for base value for linearization.
From the solution to Problem 2-17(e), the slope is: d q T( )⋅
dT
4 ε⋅ σ⋅ A⋅ T3⋅=
Temperature range for which the slope is within 5% of the slope at the base value
K 1.8R:=
error 4 ε⋅ σ⋅ A⋅ T3⋅ 4 ε⋅ σ⋅ A⋅ Tb3⋅−= 0.05 4 ε⋅ σ⋅ A⋅ Tb3⋅ ⋅=
Tmax
3 1.05 Tb= 1.0164Tb=T
Tb



3
1− 0.05=Simplify and rearrange:
Tmin
3 0.95 Tb= 0.983Tb=
For Tb 400K:= Tmax 3 1.05 Tb:= Tmin 3 0.95 Tb:= Tmax 407 K= Tmin 393 K=
Tb 600K:= Tmax 3 1.05 Tb:= Tmin 3 0.95 Tb:=
Tb 600K:= Tmin 0.921 Tb⋅:= Tmax 1.108Tb:= Tmin 553 K= Tmax 665 K=
So the range for which the linear approximation is within 5% of the heat rate is much wider than the 
range for which the value of the slope is within 5% of the actual slope. We must keep in mind that 
the parameters of the dynamic model are a function of the slope, not the heat rate. 
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0 x≤ 0.362≤
(b) xmin 1.1 0.9,( ) 0.637= xmax 1.1 0.9,( ) 1.183= (one) 0.637 x≤ 1≤
(c) xmin 5 0.1,( ) 0.092= xmax 5 0.1,( ) 0.109= 0.092 x≤ 0.109≤
(d) xmin 5 0.9,( ) 0.872= xmax 5 0.9,( ) 0.93= 0.872 x≤ 0.93≤
The range of accuracy is narrower the higher α and the higher xb.
For the vapor composition: y x( )
α x⋅
1 α 1−( )x+=
error
α x⋅
1 α 1−( )x+
α xb⋅
1 α 1−( )xb+
α
1 α 1−( )xb+ 2
x xb−( )+
1−= 0.05=
α x⋅
1 α 1−( )x+
1 α 1−( )xb+ 2
α xb 1 α 1−( )xb+ ⋅ α x⋅+ α xb⋅−
1− 0.05=
The error is always negative, so we can change signs and drop the absolute value bars:
Smith & Corripio, 3rd edition
Problem 2-19. Equilibrium vapor composition--range of accuracy
y x( )
α x⋅
1 α 1−( )x+= Use subscript "b" for base value for linearization.
From the solution to Problem 2-17(c): d y x( )⋅
dx
α
1 α 1−( )x+ 2
=
For the slope:
error
α
1 α 1−( )x+ 2
α
1 α 1−( )xb+ 2
−= 0.05 α
1 α 1−( )xb+ 2
=
Simplify and rearrange: 1 α 1−( )xb+
1 α 1−( )x+


2
1− 0.05=
Lower limit:
1 α 1−( )xb+
1 α 1−( )xmin+ 1.05= xmin α xb,( )
1 α 1−( )xb+ 1.05−
1.05 α 1−( ):=
Upper limit: 1 α 1−( )xb+
1 α 1−( )xmax+ 0.95= xmax α xb,( ) 1 α 1−
( )xb+ 0.95−
0.95 α 1−( ):=
(a) xmin 1.1 0.1,( ) 0.143−= (zero) xmax 1.1 0.1,( ) 0.362=
0.40 x≤ 1≤
(c) α 5:= xb 0.1:=
polyroots
0.95 α 1−( )⋅
0.05− α 1−( )2 xb 0.05xb− 2 α 1−( )−
0.95 α 1−( )




0.605
1.653


=
xmin 0.605xb:= xmax 1.653xb:= xmin 0.061= xmax 0.165= 0.061 x≤ 0.165≤
(d) α 5:= xb 0.9:=
polyroots
0.95 α 1−( )⋅
0.05− α 1−( )2 xb 0.05xb− 2 α 1−( )−
0.95 α 1−( )




0.577
1.732


=
xmin 0.577xb:= xmax 1.732xb:= xmin 0.519= xmax 1.559= 0.519 x≤ 1≤
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1 α 1−( )xb+ 2α x⋅ 0.95 1 α 1−( )x+  α α 1−( ) xb2 α x⋅+ =
0.95 α 1−( )⋅ x2⋅ 0.95 α 1−( )2⋅ xb2⋅ 0.95+ 1− 2 α 1−( )⋅ xb⋅− α 1−( )2 xb2⋅−  x⋅+ 0.95 α 1−( )⋅ xb⋅+
0.95 α 1−( ) x
xb



2
0.05− α 1−( )2⋅ xb 0.05xb− 2 α 1−( )−


x
xb
⋅+ 0.95 α 1−( )+ 0=
Find the roots, one is the lower limit and the other one the upper limit:
(a) α 1.1:= xb 0.1:=
polyroots
0.95 α 1−( )⋅
0.05− α 1−( )2 xb 0.05xb− 2 α 1−( )−
0.95 α 1−( )




0.138
7.231


=
xmin 0.138xb:= xmax 7.231xb:= xmin 0.014= xmax 0.723= 0.014 x≤ 0.723≤
(b) α 1.1:= xb 0.9:=
polyroots
0.95 α 1−( )⋅
0.05− α 1−( )2 xb 0.05xb− 2 α 1−( )−
0.95 α 1−( )




0.444
2.25


=
xmin 0.444xb:= xmax 2.25xb:= xmin 0.4= xmax 2.025=
2 k⋅ cAb⋅ cBb⋅ 2 hr 1−= k cAb2⋅ 2 hr 1−=
R CA t( ) CB t( ),( ) 2hr 1− CA t( ) 2hr 1− CB t( )+=
For cA 3
kmole
m3
:= 2 k⋅ cA⋅ cBb⋅ 2 k⋅ cAb⋅ cBb⋅− 1 hr 1−=
(off by 50%)
k cA
2⋅ k cAb2⋅− 2.5 hr 1−= (off by 125%)
For cB 2
kmole
m3
:= 2 k⋅ cAb⋅ cB⋅ 2 k⋅ cAb⋅ cBb⋅− 2 hr 1−=
(off by 100%)
k cAb
2⋅ k cAb2⋅− 0 hr 1−= (same as the base value)
These errors on the parameters of the linear approximation are significant, meaning that it is only 
valid for very small deviations of the reactant concentrations from their base values.
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Smith & Corripio, 3rd edition
Problem 2-20. Linearization of chemical reaction rate. kmole 1000mole:=
r cA t( ) cB t( ),( ) k cA t( )2⋅ cB t( )= Use subscript "b" for base value for linearization.
Problem parameters: k 0.5
m6
kmole2hr
:= cAb 2
kmole
m3
:= cBb 1
kmole
m3
:=
Linearize: r cA t( ) cB t( ),( ) r cAb cBb,( ) 2k cAb⋅ cBb cA t( ) cAb−( )⋅+ k cAb2⋅ cB t( ) cBb−( )+=
Let R CA t( ) CB t( ),( ) r cA t( ) cB t( ),( ) r cAb cBb,( )−= CAb t( ) cA t( ) cAb−=
CB t( ) cB t( ) cBb−=
R CA t( ) CB t( ),( ) 2k cAb⋅ cBb⋅ CA t( )⋅ k cAb2⋅ CB t( )⋅+=
At the given base conditions:
degC K:= mmHg atm
760
:= mole% %:=
Numerical values for benzene at: pb 760mmHg:= Tb 95degC:= xb 50mole%:=
A 15.9008:= B 2788.51degC:= C 220.80degC:=
Let pob p
o Tb( )=
pob e
A
B
Tb C+
−
mmHg:= pob 1177 mmHg=
xb B⋅ pob⋅
pb Tb C+( )2⋅
0.022
1
degC
=
pob
pb
1.549=
pob xb⋅
pb
2
0.00102
1
mmHg
=
Smith & Corripio, 3rd edition
Problem 2-21. Linearization of Raoult's Law for equilibrium vapor 
composition.
Raoult's Law: y T t( ) p t( ), x t( ),( ) p
o T t( )( )
p t( )
x t( )= po T t( )( ) e
A
B
T t( ) C+−=
Linearize: Use subscript "b" for base value for linearization.
y T t( ) p t( ), x t( ),( ) y Tb pb, xb,( ) xbpb
δ
δT⋅ p
o T t( )( )⋅ ⋅ T t( ) Tb−( )⋅+ p
o Tb( )
pb
x t( ) xb−( )+=
po− Tb( )xb
pb
2
p t( ) pb−( )+δ
δT e
A
B
T t( ) C+−


⋅ B
Tb C+( )2
e
A
B
Tb C+
−
⋅=
B po⋅ Tb( )⋅
Tb C+( )2
=
Let Y Γ t( ) P t( ), X t( ),( ) y T t( ) p t( ), x t( ),( ) y Tb pb, xb,( )−= Γ t( ) T t( ) Tb−= P t( ) p t( ) pb−=
X t( ) x t( ) xb−=
Y Γ t( ) P t( ), X t( ),( ) xb B⋅ p
o⋅ Tb( )⋅
pb Tb C+( )2⋅
Γ t( )
po Tb( )
pb
X t( )+
po Tb( ) xb⋅
pb
2
P t( )−=
Y Γ t( ) P t( ), X t( ),( ) 0.022
degC
Γ t( ) 1.549 X t( )+ 0.00102
mmHg
P t( )−=
pob xb⋅
pb
77.441 %= y Tb pb, xb,( ) 77.44mole%=
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From the initial steady state: 0 fb cA.b cAb−( )⋅ k Tb( ) V⋅ cAb⋅−=
cAb
fb cAib⋅
fb kb V⋅+
:= cAb 9.231 10 5−×
kmole
m3
=
Calculate parameters: τ V
fb kb V⋅+
:= K1
cAib cAb−
fb V kb⋅+
:= K2
fb
fb V kb⋅+
:= τ 0.01 s=
K1 0.046
s kmole⋅
m6
=
K3
V− kb⋅ E⋅ cAb⋅
1.987
kcal
kmole K⋅ Tb
2⋅ fb V kb⋅+( )⋅
:=
K2 7.692 10
6−×=
fb V kb⋅+ 260.002
m3
s
= K3 3.113− 10 6−×
kmol
m3K
=
Linearized equation:
0.01 sec⋅
d CA t( )⋅
dt
⋅ CA t( )+ 0.046
kmole
m3
s
m3
F t( ) 7.692 10 6−⋅ CAi t( )+ 3.113
kmole
m3K
Γ t( )−=
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
Smith & Corripio, 3rd edition
Problem 2-22. Linearization of reactor of Examples 2-6.4 and 2-6.1.
From the results of Example 2-6.4: τ
d CA t( )⋅
dt
⋅ CA t( )+ K1 F t( )⋅ K2 CAi t( )⋅+ K3 Γ t( )⋅+=
Use subscript "b" for base value for linearization.
τ V
fb V k Tb( )⋅+= K1 cAib cAb−fb V k Tb( )⋅+= K2
fb
fb V k Tb( )⋅+= K3
V− k Tb( )⋅ E cAb⋅
R Tb
2⋅ fb V k Tb( )⋅+( )
=
Problem parameters: V 2.6m3:= fb 0.002
m3
s
:= cAib 12
kmole
m3
:=
Let kb k Tb( )=
Tb 573K:= kb 100s 1−:= E 22000
kcal
kmole
:=
p t( ) ρ t( ) v
2 t( )
2
⋅ po+= v t( ) 2
p t( ) po−( )
ρ t( )⋅=
Flow through the orifice caused by the bullet: wo t( ) ρ t( ) Ao⋅ v t( )⋅= Ao 2 ρ t( )⋅ p t( ) po−( )⋅⋅=
Ideal gas law: ρ t( ) M p t( )⋅
Rg T 273K+( )⋅
=
Substitute into mass balance:
V M⋅
Rg T 273 K⋅+( )⋅
d p t( )⋅
dt
⋅ wi t( ) Ao
2 M⋅
Rg T 273K+( )⋅
p t( ) p t( ) po−( )⋅−=
Solve for the derivative:
d p t( )⋅
dt
g wi t( ) p t( ),( )= Rg T 273K+( )⋅ V M⋅ wi t( ) Ao 2 M⋅Rg T 273K+( )⋅ p t( ) p t( ) po−( )⋅⋅−


=
Linearize: d p t( )⋅
dt
δ g⋅
δ wi⋅ b
⋅ wi t( ) wb−( ) δ g⋅δ p⋅
b
⋅ p t( ) pb−( )+=
Let P t( ) p t( ) pb−= Wi t( ) wi t( ) wb−=
a1
δ g⋅
δ wi⋅ b
⋅= a1
Rg T 273K+( )⋅
V M⋅:= a1 65.56
kPa
kg
=
Smith & Corripio, 3rd edition
Problem 2-23. Pressure in a compressed air tank when punctured.
V
p(t)
wi(t)
wo(t)
po
Assumptions:
Air obeys ideal gas law•
Constant temperature•
Design conditions: kPa 1000Pa:=
pb 500 101.3+( )kPa:= M 29
kg
kmole
:=
Ao 0.785cm
2:= T 70degC:= V 1.5m3:=
Rg 8.314
kPa m3⋅
kmole K⋅⋅:= po 101.3kPa:= Use subscript "b" for base value for linearization.
Solution:
Mass balance on the tank: V
d ρ t( )⋅
dt
⋅ wi t( ) wo t( )−=
Bernoulli's equation:
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is unlawful. 
K 1.8R:=
If the compressor shuts down it will take approximately 5(42.8) = 214 sec (3.5 min) for the 
pressure transient to die out, according to the linear approximation. (See the results of the 
simulation, Problem 13-3, to see how long it actually takes.) 
P s( )
Wi s( )
K
τ s⋅ 1+=
Transfer function:
K 2.8 103× kPa sec⋅
kg
=τ 42.9 sec=K
a1
a2−
:=τ 1
a2−
:=Then 
τ d P t( )⋅
dt
⋅ P t( )+ K Wi t( )⋅=Compare to standard form of first-order equation:
P 0( ) 0=1
a2−
d P t( )⋅
dt
⋅ P t( )+
a1
a2−
Wi t( )=
d P t( )⋅
dt
a1 Wi t( )⋅ a2 P t( )⋅+=Substitute:
a2 0.023− sec 1−=a2
Ao−
2 V⋅
2 Rg⋅ T 273 K⋅+( )⋅
M pb⋅ pb po−( )⋅
kPa
1000Pa
⋅
2 pb⋅ po−( )1000Pa
kPa
⋅ m
100cm



2
:=
a2
δ g⋅
δ p⋅
b
⋅=
Ao−
V
2 Rg⋅ T 273K+( )⋅
M
⋅ 1
2
pb pb p0−( ) 
1−
2⋅ 2pb po−( )=
Γ t( ) T t( ) Tb−=
Substitute: d Γ t( )⋅
dt
a1 Γs t( )⋅ a2 Γ t( )⋅+= Γ 0( ) 0= (base is initial steady state)
Standard form of the first-order differential equation: τ d Γ t( )⋅
dt
⋅ Γ t( )+ K Γs t( )⋅=
Divide by -a2 and rearrange: 1
a2−
d Γ t( )⋅
dt
⋅ Γ t( )+
a1
a2−
Γs t( )=
M cv⋅
4 ε⋅ σ⋅ A⋅ Tb3⋅
d Γ t( )⋅
dt
⋅ Γ t( )+
Tsb
Tb



3
Γs t( )=
Compare coefficients: τ
M cv⋅
4 ε⋅ σ⋅ A⋅ Tb3⋅
= K
Tsb
Tb



3
=
Laplace transform: Γ s( )
Γs s( )
K
τ s⋅ 1+=
The input variable is the temperature of the oven wall. See problem 13-4 for the simulation.
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
Smith & Corripio, 3rd edition
Problem 2-24. Temperature of a turkey in an oven.
T(t)
Ts(t)
M
Assumptions
Uniform turkey temperature•
Negligible heat of cooking•
Radiation heat transfer only•
Energy balance on the turkey:
M cv⋅
d T t( )⋅
dt
⋅ ε σ⋅ A⋅ Ts4 t( ) T4 t( )− ⋅=
Use subscript "b" for linearization base values.
Solve for the derivative: d T t( )⋅
dt
g Ts t( ) T t( ),( )= ε σ⋅ A⋅M cv⋅ Ts
4 t( ) T4 t( )− =
Linearize: d T t( )⋅
dt
a1 Ts t( ) Tsb−( )⋅ a2 T t( ) Tb−( )⋅+=
where a1
δ g⋅
δTs b
⋅= 4 ε⋅ σ⋅ A⋅
M cv⋅
Tsb
3= a2
δ g⋅
δT
b
⋅= 4− ε⋅ σ⋅ A⋅
M cv⋅
Tb
3=
Let Γs t( ) Ts t( ) Tsb−=
Q t( ) q t( ) qb−= a1
δ g⋅
δq
b
⋅= a2
δ g⋅
δT
b
⋅=
a1
1
C
:= a2
4− α⋅ Tb3⋅
C
:= a1 5.556 10 3−×
R
BTU
= a2 0.381− hr 1−=
Substitute: d Γ t( )⋅
dt
a1 Q t( )⋅ a2 Γ t( )⋅+= Γ 0( ) 0= (base is initial value)
Standard form of first-order differential equation: τ d Γ t( )⋅
dt
⋅ Γ t( )+ K Q t( )⋅=
Divide by -a2 and rearrange: 1
a2−
d Γ t( )⋅
dt
⋅ Γ t( )+
a1
a2−
Q t( )=
C
4 α⋅ Tb3⋅
d Γ t( )⋅
dt
⋅ Γ t( )+ 1
4α Tb3⋅
Q t( )=
Compare coefficients: τ C
4α Tb3⋅
:= K 1
4α Tb3⋅
:= τ 2.62 hr= K 0.01458 R hr⋅
BTU
=
Smith & Corripio, 3rd edition
Problem 2-25. Slab heated by an electric heater by radiation.
T(t)
Ts
q(t)
Assumptions:
Uniform temperature of the slab•
Heat transfer by radiation only•
Energy balance on the slab:
M cv⋅
d T t( )⋅
dt
⋅ q t( ) ε σ⋅ A⋅ T4 t( ) Ts4− ⋅−=
Let C M cv⋅= α ε σ⋅ A⋅=
Substitute C
d T t( )⋅
dt
⋅ q t( ) α T4 t( ) Ts4− −=
Problem parameters: Use subscript "b" to denote linearization base value.
C 180
BTU
R
:= α 5 10 8−⋅ BTU
hr R4⋅
:= Ts 540R:=Tb 700R:=
Solve for the derivative: d T t( )⋅
dt
g q t( ) T t( ),( )= 1
C
q t( )
α
C
T4 t( ) Ts
4− −=
Linearize: d T t( )⋅
dt
a1 q t( ) qb−( )⋅ a2 T t( ) Tb−( )⋅+=
Let Γ t( ) T t( ) Tb−=
Transfer function: Γ s( )
Q s( )
K
τ s⋅ 1+=
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
r1
1.8− 1.82 4 0.8⋅ 1 0.1Kc+( )⋅−+
2 0.8⋅=
1.8−
1.6
1.8
1.6



2 1 0.1Kc+
0.8
−+=
Roots of the characteristic equation:
(b) Values of the controller gain for which the response is over-damped, critically 
damped, and under-damped
0.8s2 1.8s+ 1+ 0.1Kc+ 0=Characteristic equation:
C s( )
R s( )
0.1Kc
0.8s2 1.8s+ 1+ 0.1Kc+
=Closed-loop transfer function:
τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ Kc K⋅+ 0=Characteristic equation:
C s( )
R s( )
Kc
K
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅
1 Kc
K
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅⋅+
=
Kc K⋅
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅ Kc K⋅+=
(a) Closed loop transfer function and characteristic equation of the loop.
τ2 0.8min:=τ1 1min:=K 0.10
%TO
%CO
:=Problem parameters:
Gc s( ) Kc=G1 s( )
K
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Problem 6-1. Second-order loop with proportional controller.
Smith & Corripio, 3rd edition %CO %:=%TO %:=
τe1
1−
r1
= τe1
2 τ1⋅ τ2⋅
τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−−
:= τe1 0.889 min=
τe2
1−
r2
= τe2
2 τ1⋅ τ2⋅
τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−+
:= τe2 0.889 min=
Kc 0.2
%CO
%TO
:= (under-damped, time constant and damping ratio)
τ2 s2 2ζ τ⋅ s⋅+ 1+
τ1 τ2⋅
1 Kc K⋅+
s2
τ1 τ2+
1 Kc K⋅+
s+ 1+=
Match coefficients: τ
τ1 τ2⋅
1 Kc K⋅+
:= ζ
τ1 τ2+
2 τ⋅ 1 Kc K⋅+( )⋅:= τ 0.886 min= ζ 0.996=
(d) Steady-state offset for a unit step change in set point.
Final value theorem:
∞t
Y t( )
0s
s Y s( )⋅lim
→
=lim
→
R s( )
1
s
= (Table 2-1.1)
The response is critically damped when the term in the radical is zero: 1.8
1.6



2 1 0.1Kc+
0.8
− 0=
Kccd
1
0.1
0.8
1.8
1.6



2
1−

:= Kccd 0.125
%CO
%TO
=Critically damped:
Over-damped (real roots): Kc 0.125
%CO
%TO
< Under-damped: Kc 0.125
%CO
%TO
>
The loop cannot be unstable for positive gain because,
for real roots the radical cannot be greater than the negative term, so both roots are negative•
for complex conjugate roots the real part is always negative, -1.8/1.6, or -(τ1+τ2)/2τ1τ2•
This is true for all positive values of the time constants and the product K. cK.
(c) Equivalent time constants for different values of the gain:
Kc 0.1
%CO
%TO
:= (over-damped, two equivalent time constans)
τe1
1−
r1
= τe1
2 τ1⋅ τ2⋅
τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−−
:= τe1 0.935 min=
τe2
1−
r2
= τe2
2 τ1⋅ τ2⋅
τ1 τ2+( ) τ1 τ2+( )2 4 τ1⋅ τ2⋅ 1 Kc K⋅+( )⋅−+
:= τe2 0.847 min=
Kc 0.125
%CO
%TO
:= (critically damped, two equal real time constants)
Kc 0.1
%CO
%TO
:=
0s
s
Kc K⋅
τ1 τ2⋅ s2⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅
1
s
lim
→
9.9009900990099009901 10-3⋅→
offset 1 0.0099−( )%TO:= offset 0.99 %TO=
Kc 0.125
%CO
%TO
:=
0s
s
Kc K⋅
τ1 τ2⋅ s2⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅
1
s
lim
→
1.2345679012345679012 10-2⋅→
offset 1 0.01235−( )%TO:= offset 0.988 %TO=
Kc 0.2
%CO
%TO
:=
0s
s
Kc K⋅
τ1 τ2⋅ s2⋅ τ1 τ2+( ) s⋅+ 1+ Kc K⋅+⋅
1
s
lim
→
1.9607843137254901961 10-2⋅→
offset 1 0.01961−( )%TO:= offset 0.98 %TO=
These are very large offsets because the loop gains are so small.
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0.00842
%CO
%TO
Kc< 0.825
%CO
%TO
<Under-damped (complex conjugate roots):
Kc 0.825
%CO
%TO
>andKc 0.00842
%CO
%TO
<Over-damped (two real roots):
Kc 0.00842
%CO
%TO
=Kc
30 302 4 0.25⋅ 36⋅−−
2 36⋅:=
Kc 0.82491
%CO
%TO
=Kc
30 302 4 0.25⋅ 36⋅−+
2 36⋅:=0.25 30Kc− 36Kc
2+ 0=
The response is critically damped when the term in the radical is zero:
r1
1.5 6Kc−( )− 1.5 6Kc−( )2 4 0.5⋅ 1 6Kc+( )⋅−+
2 0.5⋅= 1.5− 6Kc+ 0.25 30Kc− 36Kc
2++=
Roots:
(b) Values of the gain for which the response is over-, critically, and under-damped
0.5 s2⋅ 1.5 6Kc−( )s+ 1+ 6Kc+ 0=Characteristic equation:
C s( )
R s( )
Kc 6⋅ 1 s−( )
s 1+( ) 0.5s 1+( )⋅ Kc 6⋅ 1 s−( )+
=Closed-loop transfer function:
(a) Closed-loop transfer function and characteristic equation of the loop.
Gc s( ) Kc
%CO
%TO
⋅=G1 s( )
6 1 s−( )
s 1+( ) 0.5 s⋅ 1+( )⋅
%TO
%CO
=
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Problem 6-2. Inverse-response second-order system with proportional 
controller.
Smith & Corripio, 3rd edition
ζ
1.5 6Kc−( )min
2 τ⋅ 1 6Kc+( )⋅:= τ 0.423 min= ζ 0.127−=(unstable)
Try values that result in equivalent time constants:
Kc 0.005
%CO
%TO
:= τe1
1min
1.5 6 Kc⋅− 0.25 30Kc− 36Kc2+−
:= τe1 0.868 min=
τe2
1min
1.5 6 Kc⋅− 0.25 30Kc− 36Kc2++
:= τe2 0.559 min=
Kc 1
%CO
%TO
:= τe1
1min
1.5 6 Kc⋅− 0.25 30Kc− 36Kc2+−
:= τe1 0.143− min=
(unstable)
τe2
1min
1.5 6 Kc⋅− 0.25 30Kc− 36Kc2++
:= τe2 0.5− min=
(d) Offset for various values of the gain and a unit step change in set point.
Kc 0.10
%CO
%TO
:=
0s
s
Kc 6⋅ 1 s−( )⋅
0.5s2 1.5 6 Kc⋅−( )s+ 1+ 6Kc+
⋅ 1
s
lim
→
.37500000000000000000→
offset 1 0.375−:= offset 0.625 %CO
%TO
=
The response is unstable when Kc 0.25
%CO
%TO
> (one real root is positive or the real part of the 
complex roots i positive)
(c) Effective time constants or time constant and damping ratio for various values o
the gain:
0.5
1 6Kc+
s2
1.5 6Kc−
1 6Kc+
s+ 1+ τ2 s2 2ζ τ⋅ s⋅+ 1+=
Kc 0.1
%CO
%TO
:= τ 0.5min
2
1 6Kc+
:= ζ
1.5 6Kc−( )min
2 τ⋅ 1 6Kc+( )⋅:= τ 0.559 min= ζ 0.503=
Kc 0.125
%CO
%TO
:= τ 0.5min
2
1 6Kc+
:= ζ
1.5 6Kc−( )min
2 τ⋅ 1 6Kc+( )⋅:= τ 0.535 min= ζ 0.401=
Kc 0.2
%CO
%TO
:= τ 0.5min
2
1 6Kc+
:= ζ
1.5 6Kc−( )min
2 τ⋅ 1 6Kc+( )⋅:= τ 0.477 min= ζ 0.143=
Kc 0.3
%CO
%TO
:= τ 0.5min
2
1 6Kc+
:=
Kc 0.125
%CO
%TO
:=
0s
s
Kc 6⋅ 1 s−( )⋅
0.5s2 1.5 6 Kc⋅−( )s+ 1+ 6Kc+
⋅ 1
s
lim
→
.42857142857142857143→
offset 1 0.429−:= offset 0.571 %CO
%TO
=
Kc 0.20
%CO
%TO
:=
0s
s
Kc 6⋅ 1 s−( )⋅
0.5s2 1.5 6 Kc⋅−( )s+ 1+ 6Kc+
⋅ 1
s
lim
→
.54545454545454545455→
offset 1 0.545−:= offset 0.455 %CO
%TO
=
The offsets are high because the gains are small. Of course, since for gains greater than 
0.25%CO/%TO the loop is unstable, offsets can only be high with a proportional controller.
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
K 1.8R:=
The real root cannot be negative for any positive value of the loop gain KK c because the radical is 
always smaller than the negative term. Also, for complex conjugate roots, the real part is always 
r1
1 KKc+( )− τI⋅ 1 KKc+( )2 τI2⋅ 4 τI⋅ τ⋅ KKc⋅−+
2τ I τ⋅
=
No, there is no ultimate gain. This result just means that a negative loop gain will make the loop 
unstable. Another way to show it is to determine the roots of the characteristic equation:
KKcu 0:=ωu 0:=τ I− τ⋅ ωu
2⋅ KKcu+ i 1 KKcu+( ) ωu+ 0=Substitute s = iω:
(b) Is there an ultimate gain for this loop?
(no offset)
0s
KKc τ I s⋅ 1+( )⋅
τI τ⋅ s2⋅ 1 KKc+( )τ I s⋅+ KKc+
KKc
KKc
= 1=lim
→
Offset: the steady state gain is:
τI τ⋅ s2⋅ 1 KKc+( )τ I s⋅+ KKc+ 0=Characteristic equation:
C s( )
R s( )
KKc τI s⋅ 1+( )⋅
τI s⋅ τ s⋅ 1+( )⋅ KKc τ I s⋅ 1+( )⋅+=Closed-loop transfer functon:
(a) Closed-loop transfer function and characteristic equation of the loop. Offset.
τ 1:=To work in dimensionless units, t/τ, set:
Gc s( ) Kc 1
1
τ I s⋅
+


⋅=G1 s( )
K
τ s⋅ 1+=
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Problem 6-3. First-order process and proportional-integral controller.
Smith & Corripio, 3rd edition
R
K
1.8
:=
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is unlawful. 
As KKc increases τc decreases and the response 
is faster. 
0 2 4
0
0.5
1
Y t( )
R t( )
t
Y t( ) u t( ) e
t−
τc−:=
Invert using Table 2-1.1:
R t( ) u t( ):= τc 1:=
u t( ) 0 t 0<if
1 t 0≥if
:=
Y s( )
1
τc s⋅ 1+
1
s
=
1
s
1
s
1
τc
+
−=τc
τ
KKc
=Let 
Y s( )
KKc τ s⋅ 1+( )⋅
τ s⋅ τ s⋅ 1+( )⋅ KKc τ s⋅ 1+( )⋅+
1
s
=
KKc
τ s⋅ KKc+
1
s
=
(Table 2-1.1)R s( )
1
s
=
(c) Response of the loop to a step change in set point for τI = τ as the gain varies 
from 0 to infinity.
Real
1 KKc+( )−
2 τ⋅ 0<=
negative:
Tu 1.987 min=Tu 2
π
ωu
:=
KIu 110
%CO
%TO min⋅=ωu 3.162 min
1−=KIu
τ1 τ2+( ) ωu2⋅
K
:=ωu
1
τ1 τ2⋅
:=τ2 0.1min:=
(b) Ultimate gain and period for other values of the smaller time constant:
Tu 5.62 min=Tu
2π
ωu
:=
KIu 22.5
%CO
%TO min⋅=ωu 1.118 min
1−=KIu
τ1 τ2+( ) ωu2⋅
K
:=ωu
1
τ1 τ2⋅
:=
τ1− τ2⋅ ωu
3 ωu+ 0=τ1 τ2+( )− ωu2 KKIu+ 0=
τ1− τ2⋅ i⋅ ωu
3⋅ τ1 τ2+( )ωu2− i ωu⋅+ KKIu+ 0 0i+=Substitute s = iωu:
τ1 τ2⋅ s3 τ1 τ2+( )s2+ s+ KKI+ 0=
τ1 τ2⋅ s2⋅ τ1 τ2+( )s+ 1+ K KI⋅s+ 0=Characteristic equation:
τ2 0.8min:=τ1 1min:=K 0.1
%CO
%TO
:=
(a) Ultimate gain and period with the parameters of Problem 6-1:
Gc s( )
KI
s
=G1 s( )
K
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Problem 6-4. Second-order process with pure integral controller.
Smith & Corripio, 3rd edition
τ2 2min:= ωu
1
τ1 τ2⋅
:= KIu
τ1 τ2+( ) ωu2⋅
K
:= ωu 0.707 min 1−= KIu 15
%CO
%TO min⋅=
Tu
2π
ωu
:= Tu 8.886 min=
Reducing the non-dominat time constant increases the ultimate gain and reduces the ultimate 
period, as expected. When τ2 is increased to 2 min, it becomes the dominant time constant and 
the ultmate gain should be higher than for part (a). However, in this case K I has units of rate and, 
since the loop is slower, it results in a smaller ultimate gain.
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
τI
0.8min2
1.8min
> 0.444min=or τI
τ1 τ2⋅
τ1 τ2+
>
Notice that for some values of τI the ultimate frequency and period are complex. When this 
happens there is no ultimate gain and the loop is stable for all values of the gain. So, the loop is 
always stable as long as
Tu 2π 0.8min2 1.8min τ I⋅−⋅=KKcu
1.8min τI⋅
0.8min2 1.8min τ I⋅−
=For the given numerical values:
Tu
2π
ωu
= 2π τ1 τ2⋅ τ1 τ2+( ) τI⋅−⋅=
KKcu
τ1 τ2+( ) τ I⋅
τ1 τ2⋅ τ1 τ2+( ) τ I⋅−=ωu
1
τ1 τ2⋅ τ1 τ2+( ) τI⋅−=
τ1− τ2⋅ τ I⋅ ωu
2⋅ τ I+ τ1 τ2+( )τI2 ωu2⋅+ 0=KKcu τ1 τ2+( )τ I ωu2⋅=
τ1 τ2+( )− τ I ωu2⋅ KKcu+ i τ1− τ2⋅ τI⋅ ωu3⋅ τI 1 KKcu+( )ωu⋅+ + 0 i0+=Substitute s = iωu:
τ1 τ2⋅ τI⋅ s3⋅ τ1 τ2+( )τI s2⋅+ τI 1 KKc+( )⋅ s⋅+ KKc+ 0=Characteristic equation of the loop:
(a) Ultimate gain and period as a function of integral time τI.
τ2 0.8min:=τ1 1min:=
Gc s( ) Kc 1
1
τ I s⋅
+


⋅=G1 s( )
K
τ1 s⋅ 1+( ) τ2 s⋅ 1+( )⋅=
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Problem 6-5. Second-order process with proportional-integral controller.
Smith & Corripio, 3rd edition
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is unlawful. 
0 5 10
0
0.2
0.4
DR KKc( )
KKc
5 10
0
0.5
1
1.5
ζ KKc( )
KKc
DR e
2− π ζ⋅
1 ζ2−= e
2− π
2 0.8KKc 1
1
4 0.8⋅ KKc⋅
−


⋅⋅
= e
2− π⋅
3.2 KKc⋅ 1−= DR KKc( ) e
2− π
3.2KKc 1−:=
Under these conditions the decay ratio is, from Eq. 2-5.18:
KKc
1
4 0.8⋅> 0.3125=These are complex for
r2
1− 1 4 0.8⋅ KKc⋅−−
2 0.8⋅=r1
1− 1 4 0.8⋅ KKc⋅−+
2 0.8⋅=
Roots:
ζ KKc( ) 12 0.8KKc⋅:=ζ
1min
2 τ⋅ KKc⋅
=
1
2 0.8 KKc⋅⋅
=τ 0.8min
2
KKc
=Damping ratio:
τ2 s2 2ζ τ⋅ s⋅+ 1+ 0.8
KKc
s2
1
KKc
s+ 1+=Standard second-order differential equation:
1
KKc 1
1
s
+

⋅
s 1+( ) 0.8s 1+( )⋅+ 1
KKc s 1+( )⋅
s s 1+( ) 0.8s 1+( )⋅+= 0.8s
2 s+ KKc+= 0=
For these values there is no ultimate gain. THe characteristic equation becomes:
(b) Damping ratio and decay ratio with Kc equal to one half the ultimate and τI = 1 
min
Block diagram of the control loop:
Fsset(s) E(s) M(s)
C(s)
Fs(s)Gc(s) Gv(s)
KT
Ksp +
-
Size the flow transmitter for 150% of design flow: fsmax 1.5 fs⋅:= fsmax 225
kscf
hr
=
Transmitter gain: KT
100%TO
fsmax
:= Ksp KT:= KT 0.444
%TO hr⋅
kscf
=
PI Controller: Gc s( ) Kc 1
1
τ I s⋅
+


⋅=
Size the control valve for 100% overcapacity. From Eq. 5-2.3: Let Cf 0.9:=
G
M lbmole⋅
29lb
:= y 1.63
Cf
p1 p2−
p1 14.7psia+
⋅:= y 1.181= fy y 0.148y3−:= fy 0.937=
Smith & Corripio, 3rd edition degF R:= psia psi:=
kscf 1000ft3:= psig psi:=Problem 6-6. Design of gas flow control loop.
FT
FC
fs(t)
m(t)
c(t)
fsset(t)
p1 p2
Design conditions: lbmole 453.59mole:=
fs 150
kscf
hr
:= p1 150psig:=
T1 60degF:= p2 80psig:=
M 29
lb
lbmole
:= α 50:=
τv 0.06min:= τI τv:=
Kc 0.9
%CO
%TO
:=
Assume the pressures and temperatures are constant and that the flowtransmitter FT has a built-in 
square-root extractor so that the signal c(t) is proportional to the flow fs(t). The valve is 
equal-percentage and the controller is PI.
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
So the closed-loop responds faster than the valve, and has no offset.
τc 0.026 min=τc
τv
KT Kc⋅ Kv⋅
:=Closed-loop time constant:
Fs s( )
Fs
set s( )
Ksp Kc⋅ Kv⋅
τv s⋅ KT Kc⋅ Kv⋅+
= 1τc s⋅ 1+
=Ksp KT=τI τv=With 
Fs s( )
Fs
set s( )⋅
Ksp Gc s( )⋅ Gv s( )⋅
1 KT Gc s( )⋅ Gv s( )⋅+
=
Ksp Kc⋅ 1
1
τ I s⋅
+


⋅
Kv
τv s⋅ 1+
⋅
1 KT Kc⋅ 1
1
τI s⋅
+


⋅
Kv
τv s⋅ 1+
+
=
Closed-loop transfer function and time constant of the loop:
Gv s( )
Kv
τv s⋅ 1+
=
Transfer function of the valve:
Kv 5.87
kscf
hr %CO⋅=Kv
ln α( )
100%CO
fs:=
Valve gain, equal-percentage, constant pressures, Eq. 5-2.24:
Cvmax 110
gal
min psi⋅:=
From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service:
Cvmax 58.91
gal
min psi⋅=
Cvmax
200 %⋅ fs⋅ G T1 460 R⋅+( )⋅⋅
Cf p1 14.7psia+( )⋅ fy⋅
gal hr⋅
0.836kscf min⋅
psi
R
⋅:=
Size the flow transmitter for :
Transmitter gain: KT
100%TO
wmax
:= Ksp KT:= KT 0.02
%TO hr⋅
lb
=
PI Controller: Gc s( ) Kc 1
1
τ I s⋅
+


⋅=
Size the control valve for 100% overcapacity. From Eq. 5-2.3:
G
M lbmole⋅
29lb
:= y 1.63
Cf
p1 p2−
p1 14.7psia+
⋅:= y 1.319= fy y 0.148y3−:= fy 0.979=
From the steam table the saturated steam pressure at: p1 14.7psia+ 59.7 psia= Tsat 292degF:=
Smith & Corripio, 3rd edition
Problem 6-7. Steam flow control loop.
FT
FC
w(t)
m(t)
c(t)
wset(t)
p1 p2
Design conditions: lbmole 453.59mole:=
w 3500
lb
hr
:= p1 45psig:=
Tsh 50degF:= p2 20psig:=
M 18
lb
lbmole
:= wmax 5000
lb
hr
:=
τI τv:=Linear valve.
Cf 0.8:= Kc 0.5
%CO
%TO
:=
Assume the pressures and temperatures are constant and that the flow transmitter FT has a built-in 
square-root extractor so that the signal c(t) is proportional to the flow w(t). The valve is linear and 
the controller is PI.
Block diagram of the control loop:
Wset(s) E(s) M(s)
C(s)
W(s)
Gc(s) Gv(s)
KT
Ksp +
-
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
So the closed-loop responds slightly slower than the valve, and has no offset. What can be 
adjusted to speed-up the response of the closed loop?
KT Kc⋅ Kv⋅ 0.913=τc
τv
KT Kc⋅ Kv⋅
:=Closed-loop time constant:
Fs s( )
Fs
set s( )
Ksp Kc⋅ Kv⋅
τv s⋅ KT Kc⋅ Kv⋅+
=
1
τc s⋅ 1+
=Ksp KT=τI τv=With 
W s( )
Wset s( )⋅
Ksp Gc s( )⋅ Gv s( )⋅
1 KT Gc s( )⋅ Gv s( )⋅+
=
Ksp Kc⋅ 1
1
τ I s⋅
+


⋅
Kv
τv s⋅ 1+
⋅
1 KT Kc⋅ 1
1
τI s⋅
+


⋅
Kv
τv s⋅ 1+
+
=
Closed-loop transfer function and time constant of the loop:
Gv s( )
Kv
τv s⋅ 1+
=
Transfer function of the valve:
Kv 91.31
lb
hr %CO⋅=Kv
wvmax
100%CO
:=
Valve gain, linear, constant pressures, Eq. 5-2.23:
wvmax 9131
lb
hr
=
wvmax
0.836kscf min⋅
gal hr⋅
R
psi
⋅ Cvmax⋅ Cf⋅
p1 14.7psia+
G T1 460R+( )⋅⋅ fy
M lbmole⋅
0.380kscf
⋅:=
Cvmax 110
gal
min psi⋅:=
From Fig. C-10.1, p. 532, a 3-in Masoneilan valve is the smallest for this service:
Cvmax 84.3
gal
min psi⋅=
Cvmax
200 %⋅ fs⋅ G T1 460 R⋅+( )⋅⋅
Cf p1 14.7psia+( )⋅ fy⋅
gal hr⋅
0.836kscf min⋅
psi
R
⋅:=
T1 342 degF=T1 Tsat Tsh+:=fs 73.889
kscf
hr
=fs
w
M
0.380⋅ kscf
lbmole
:=
Real part: ωu
4
6ωu
2− 1+ Kcu+ 0= Kcu ωu min⋅( )4− 6min2 ωu( )2⋅+ 1− := Kcu 4 %CO%TO=
(b) G1 s( )
1
s 1+( )2
= s 1+( )2 Kc+ 0= s2 2s+ 1+ Kc+ 0=
Substitute s = iωu at Kc = Kcu: ωu
2− 2ωu i⋅+ 1+ Kcu+ 0 0i+=
Imaginary part: 2ωu 0= ωu 0:= (There is no ultimate gain)
Real part: ωu
2− 1+ Kcu+ 0= Kcu 1−
%CO
%TO
:=
The loop becomes monotonically unstable when the controller gain is less than -1%CO/%TO.
Smith & Corripio, 3rd edition
Problem 6-8. Ultimate gain and period of various process transfer functions.
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Proportional controller: Gc s( ) Kc= Characteristic equation: 1 Kc G1 s( )⋅+ 0=
(a) G1 s( )
1
s 1+( )4
= s 1+( )4 Kc+ 0= s4 4s3+ 6s2+ 4s+ 1+ Kc+ 0=
Substitute s = iωu at Kc = Kcu: ωu
4
4ωu
3
i− 6ωu
2− 4ωu i⋅+ 1+ Kcu+ 0 0i+=
Imaginary part: 4− ωu
3
4ωu+ 0= ωu 1min 1−:= Tu
2π
ωu
:= Tu 6.28 min=
Imaginary part: 8− ωu
3
7 0.5Kcu+( )ωu+ 0=
8− ωu
2
7+ 7ωu
2+ 0.5− 0= ωu 6.5 min 1−:= Tu
2π
ωu
:=
Kcu 14min
2 ωu
2⋅ 1−:= Kcu 90
%CO
%TO
= Tu 2.46 min=
(e) G1 s( )
1
4s 1+( ) 0.2s 1+( )⋅ 0.1s 1+( )⋅= 0.08s
3 1.22s2+ 4.3s+ 1+ Kc+ 0=
Substitute s = iωu at Kc = Kcu: 0.08− ωu
3
i 1.22ωu
2− 4.3ωu i⋅+ 1+ Kcu+ 0 0i+=
Imaginary part: 0.08− ωu.
3
4.3ωu+ 0= ωu
4.3
0.08
min 1−:= Tu
2π
ωu
:=
Real part: 1.22− ωu
2
1+ Kcu+ 0= Tu 0.857 min=
Kcu 1.22min
2ωu
2
1−:= Kcu 64.6
%CO
%TO
=
(c) G1 s( )
1
4s 1+( ) 2 s⋅ 1+( )⋅ s 1+( )⋅= 8s
3 14s2+ 7s+ 1+ Kc+ 0=
Substitute s = iωu at Kc = Kcu: 8− ωu
3
i 14ωu
2− 7ωu i⋅+ 1+ Kcu+ 0 0i+=
Imaginary part: 8− ωu
3
7ωu+ 0= ωu
7
8
min 1−:= Tu
2π
ωu
:= Tu 6.72 min=
Real part: 14− ωu
2
1+ Kcu+ 0= Kcu 14min2ωu
2
1−:= Kcu 11.25
%CO
%TO
=
(d) G1 s( )
0.5s 1+
4s 1+( ) 2s 1+( )⋅ s 1+( )⋅= 8s
3 14s2+ 7 0.5Kc+( )s+ 1+ Kc+ 0=
Substitute s = iωu at Kc = Kcu: 8− ωu
3
i 14ωu
2− 7 0.5Kcu+( )ωu i⋅+ 1+ Kcu+ 0 0i+=
Real part: 14− ωu
2
1+ Kcu+ 0= Kcu 14ωu
2
1−=
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only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work 
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is unlawful. 
Tu 1.797 min=
Tu
2π
ωu
:=ωu
1 Kcu+
1.8
min 1−:=1.8− ωu
2
1+ Kcu+ 0=Real part:
Kcu 21
%CO
%TO
=Kcu
6.3
0.3
:=6.3 0.3Kcu−( )ωu 0=Imaginary part:
1.8− ωu
2
6.3 0.3Kcu−( )ωu i⋅+ 1+ Kcu+ 0 0i+=Substitute s = iωu at Kc = Kcu:
1.8s2 6.3 0.3Kc−( )s+ 1+ Kc+ 0=
e 0.6− s 1 0.3s−
1 0.3s+=Padé approximation:
6s 1+ Kc e 0.6− s⋅+ 0=G1 s( )
e 0.6− s
6s 1+=(f)
KIu 4− min3 ωu
4⋅ 4min ωu
2⋅+:= Tu
2π
ωu
:=
KIu 0.569
%CO
%TO min⋅= Tu 15.17 min=
Must use the smaller ultimate frequency, as the ultimate gain for the other value is negative.
(b) G1 s( )
1
s 1+( )2
= s3 2s2+ s+ KI+ 0=
Substitute s = iωu at KI = KIu: ωu
3− i 2ωu
2− ωu i⋅+ KIu+ 0 0i+=
Imaginary part: ωu
3− ωu+ 0= ωu 1min 1−:= Tu
2π
ωu
:= Tu 6.28 min=
Real part: 2− ωu
2
KIu+ 0= KIu 2minωu
2:= KIu 2
%CO
%TO min⋅=
Smith & Corripio, 3rd edition
Problem 6-9. Ultimate gain and period with integral controller.
Gc(s) G1(s)
G2(s)
R(s) E(s) M(s)
D(s)
C(s)+ +
+-
Integral controller: Gc s( )
KI
s
= Characteristic equation: 1
KI
s
G1 s( )+ 0=
(a) G1 s( )
1