1234567898
3.1
a)
[
1e
− bt
∞
]
sin ωt = ∫ e
− bt
0
∞
sin ωt e dt = ∫ sin ωt e − ( s + b )t dt
− st
0

[− (s + b) sin ωt − ω cos ωt ]
= e − ( s + b ) t

( s + b ) 2 + ω2

0
ω
=
( s + b) 2 + ω2
∞
b)
[
1 e
− bt
∞
]
cos ωt = ∫ e
0
− bt
∞
cos ωt e dt = ∫ cos ωt e − ( s + b )t dt
− st
0

[− (s + b) cos ωt + ω sin ωt ]
= e − ( s + b ) t

( s + b) 2 + ω2

0
s+b
=
( s + b) 2 + ω2
∞
3.2
a)
The Laplace transform provided is
Y ( s) =
4
s + 3s + 4 s 2 + 6 s + 4
4
3
We also know that only sin ωt is an input, where ω =
X ( s) =
ω
2
=
2
s +ω
s2 + 2
2
( )
2
=
2 . Then
2
s +2
2
Since Y(s) = D-1(s) X(s) where D(s) is the characteristic polynomial (when
all initial conditions are zero),
Solution Manual for Process Dynamics and Control, 2nd edition,
Copyright © 2004 by Dale E. Seborg, Thomas F. Edgar and Duncan A. Mellichamp.
3-1
Y (s) =
2 2
2
2
( s + 3s + 2) ( s + 2)
2
and the original ode was
d2y
dy
+ 3 + 2 y = 2 2 sin 2t
2
dt
dt
with y ′(0) = y (0) = 0
b)
This is a unique result.
c)
The solution arguments can be found from
Y (s) =
2 2 2
( s + 1)( s + 2) + ( s 2 + 2)
which in partial fraction form is
Y (s) =
α1
α
a s + a2
+ 2 + 12
s +1 s + 2
s +2
Thus the solution will contain four functions of time
e-t
,
e-2t ,
sin 2 t , cos 2 t
3.3
a)
Pulse width is obtained when x(t) = 0
Since x(t) = h – at
tω : h − atω = 0
or
tω = h/a
b)
h
slope = -a
slope = a
x(t)
x(t)
slope = -a
x(t) = hS(t) – atS(t) + a(t -tω) S(t-tω)
3-2
c)
h a ae − stω h e − stω − 1
X ( s) = − 2 + 2 = +
s s
s
s
s2
d)
Area under pulse = h tω/2
a)
f(t) = 5 S(t) – 4 S(t-2) – S(t- 6)
1
F (s ) = = ( 5 - 4e -2s - e -6s )
s
3.4
b)
x(t)
x1
a
a
x4
tr
2tr
3tr
-a
-a
x2
x3
x(t) = x1(t) + x2(t) + x3(t) + x4(t)
= at – a(t − tr)S(t − tr) − a(t −2tr)S(t − 2tr) + a(t − 3tr)S(t − 3tr)
following Eq. 3-101. Thus
X(s) =
[
a
1 − e −tr s − e − 2tr s + e −3tr s
2
s
]
by utilizing the Real Translation Theorem Eq. 3-104.
3-3
3.5
55
55
t S(t) –
(t-30) S(t-30)
30
30
20 55 1
55 1 −30 s 20 55 1
T (s) =
+
−
e
=
+
1 − e −30 s
2
2
2
s 30 s
30 s
s 30 s
T(t) = 20 S(t) +
(
)
3.6
a)
X ( s) =
α1 =
α2 =
α3 =
b)
α
α
α
s ( s + 1)
= 1 + 2 + 3
( s + 2)( s + 3)( s + 4) s + 2 s + 3 s + 4
s ( s + 1)
( s + 3)( s + 4)
s ( s + 1)
( s + 2)( s + 4)
s ( s + 1)
( s + 2)( s + 3)
=1
s = −2
= −6
s = −3
=6
s = −4
X (s) =
1
6
6
−
+
s+2 s+3 s+4
X (s) =
s +1
s +1
=
2
( s + 2)( s + 3)( s + 4) ( s + 2)( s + 3)( s + 2 j )( s − 2 j )
X ( s) =
α + jβ 3 α 3 − j β 3
α1
α
+ 2 + 3
+
s+2 s+3
s+2j
s+2j
α1 =
α2 =
s +1
( s + 3)( s 2 + 4)
s +1
( s + 2)( s 2 + 4)
α 3 + jβ 3 =
=−
s = −2
=
s = −3
x(t ) = e−2t − 6e −3t + 6e −4t
and
1
8
2
13
s +1
( s + 2)( s + 3)( s − 2 j )
3-4
=
s = −2 j
1− 2 j
− 3 + 11 j
=
− 40 − 8 j
208
1
2
 −3 
 11 
x(t ) = − e − 2t + e −3t + 2
 cos 2t + 2
 sin 2t
8
13
 208 
 208 
1
2
3
11
= − e − 2t + e −3t −
cos 2t +
sin 2t
8
13
104
104
c)
X ( s) =
α
α2
s+4
= 1 +
2
s + 1 ( s + 1) 2
( s + 1)
(1)
α 2 = ( s + 4) s =−1 = 3
In Eq. 1, substitute any s≠-1 to determine α1. Arbitrarily using s=0, Eq. 1
gives
4 α1 3
=
+
1 12
12
or
α1 = 1
1
3
+
and
x( t ) = e− t + 3te − t
2
s + 1 ( s + 1)
1
1
1
X (s) = 2
=
=
2
2
2
s + s +1 
1  3 (s + b ) + ω
s +  +
2 4

1
3
where b =
and ω =
2
2
X (s) =
d)
t
1
2 −2
3
x(t ) = e −bt sin ω t =
e sin
t
ω
2
3
e)
X(s) =
s +1
e −0.5 s
s ( s + 2)( s + 3)
To invert, we first ignore the time delay term. Using the Heaviside
expansion with the partial fraction expansion,
Xˆ ( s ) =
s +1
A
B
C
= +
+
s ( s + 2)( s + 3) s s + 2 s + 3
Multiply by s and let s → 0
3-5
1
1
=
(2)(3) 6
Multiply by (s+2) and let s→ −2
A=
B=
− 2 +1
−1
1
=
=
(−2)(−2 + 3) (−2)(1) 2
Multiply by (s+3) and let s→-3
C=
− 3 +1
−2
2
=
=−
(−3)(−3 + 2) (−3)(−1)
3
Then
1 6 1 2 −2 3
Xˆ ( s ) =
+
+
s s+2 s+3
xˆ (t ) =
1 1 − 2 t 2 −3t
+ e − e
6 2
3
Imposing shift theorem
x(t ) = xˆ (t − 0.5) =
1 1 − 2 (t −0.5) 2 −3( t − 0.5)
+ e
− e
6 2
3
for t ≥ 0.5
3.7
a)
Y (s) =
α
6( s + 1)
6 α
= 2 = 1 + 22
2
s
s ( s + 1) s
s
6
s2
6
Y (s) = 2
s
α2 = s2
b)
Y (s) =
=6
α1 = 0
s =0
12( s + 2) α1 α 2 s + α 3
=
+ 2
s
s ( s 2 + 9)
s +9
3-6
Multiplying both sides by s(s2+9)
12( s + 2) = α 1 ( s 2 + 9) + (α 2 s + α 3 )( s )
or
12 s + 24 = (α 1 + α 2 ) s + α 3 s + 9α 1
2
Equating coefficients of like powers of s,
s2: α1 + α2 = 0
s1: α3
= 12
0
s : 9α1
= 24
Solving simultaneously,
−8
3
 8

 − s + 12 
8 1  3

Y (s) =
+
3s
s2 + 9
α1 =
c)
α2 =
α3 =
α2 =
,
,
( s + 2)( s + 3)
( s + 5)( s + 6)
( s + 2)( s + 3)
( s + 4)( s + 6)
( s + 2)( s + 3)
( s + 4)( s + 5)
=1
s = −4
= −6
s = −5
=6
s = −6
Y (s) =
1
6
6
−
+
s+4 s+5 s+6
Y (s) =
[(s + 1)
=
α 3 = 12
α
α
α
( s + 2)( s + 3)
= 1 + 2 + 3
( s + 4)( s + 5)( s + 6) s + 4 s + 5 s + 6
Y (s) =
α1 =
d)
8
3
1
2
]
2
+ 1 ( s + 2)
=
1
( s + 2 s + 2) 2 ( s + 2)
2
α s + α4
α
α1 s + α 2
+ 2 3
+ 5
2
2
s+2
s + 2s + 2 ( s + 2s + 2)
3-7
Multiplying both sides by ( s 2 + 2s + 2) 2 ( s + 2) gives
1 = α1s4 + 4α1s3 + 6α1s2 +4α1s + α2s3 +4α2s2 +6α2s +4α2 + α3s2 +2α3s +
α4s + 2α4 + α5s4 + 4α5s3 + 8α5s2 + 8α5s + 4α5
Equating coefficients of like power of s,
s4 : α1 + α5 = 0
s3 : 4α1 + α2 + 4α5 = 0
s2 : 6α1 + 4α2 + α3 + 8α5 = 0
s1 : 4α1 + 6α2 + 2α3 + α4 + 8α5 = 0
s0 : 4α2 + 2α4 + 4α5 = 1
Solving simultaneously:
α1 = -1/4
Y (s) =
α2 = 0
α3=-1/2
α4=0
− 1 / 4s
− 1 / 2s
1/ 4
+ 2
+
2
s+2
s + 2 s + 2 ( s + 2 s + 2)
2
3.8
a)
From Eq. 3-100
t
 1
1  ∫ f (t * )dt *  = F ( s )
0
 s
 t −τ  1
1
we know that 1  ∫ e dτ = 21 e − τ =
s ( s + 1)
0
 s
[ ]
∴
Laplace transforming yields
2
s ( s + 1)
2
or (s2 + 3s + 1) X(s) =
s ( s + 1)
s2X(s) + 3X(s) + 2X(s) =
3-8
α5 = ¼
X(s) =
x(t) = 1 − 2te-t − e-2t
and
b)
2
s ( s + 1) 2 ( s + 2)
Applying the final Value Theorem
lim x(t ) = lim sX ( s ) = lim
t →∞
s →0
s →0
2
=2
( s + 1) ( s + 2)
2
[ Note that Final Value Theorem is applicable here]
3.9
a)
X (s) =
6( s + 2)
6( s + 2)
=
( s + 9s + 20)( s + 4) ( s + 4)( s + 5)( s + 4)
2
 6s ( s + 2) 
=0
x(0) = lim 
s →∞ ( s + 5)( s + 4) 2 


 6s ( s + 2) 
x(∞) = lim
=0
s →0 ( s + 5)( s + 4) 2 


x(t) is converging (or bounded) because [sX(s)] does not have a limit at
s = −4, and s = −5 only, i.e., it has a limit for all real values of s ≥ 0.
x(t) is smooth because the denominator of [sX(s)] is a product of real
factors only. See Fig. S3.9a.
b)
10 s 2 − 3
10 s 2 − 3
X (s) = 2
=
( s − 6 s + 10)( s + 2) ( s − 3 + 2 j ) ( s − 3 − 2 j )( s + 2)


10s 3 − 3s
x(0) = lim  2
 = 10
s →∞ ( s − 6 s + 10)( s + 2)


Application of final value theorem is not valid because [sX(s)] does not
have a limit for some real s ≥ 0, i.e., at s = 3±2j. For the same reason, x(t)
is diverging (unbounded).
x(t) is oscillatory because the denominator of [sX(s)] includes complex
factors. See Fig. S3.9b.
3-9
16 s + 5
16s + 5
=
2
( s + 9) ( s + 3 j ) ( s − 3 j )
X (s) =
16s 2 + 5s 
x(0) = lim  2
 = 16
s →∞
 ( s + 9) 
Application of final value theorem is not valid because [sX(s)] does not
have a limit for real s = 0. This implies that x(t) is not diverging, since
divergence occurs only if [sX(s)] does not have a limit for some real value
of s>0.
x(t) is oscillatory because the denominator of [sX(s)] is a product of
complex factors. Since x(t) is oscillatory, it is not converging either. See
Fig. S3.9c
0.4
0.35
0.3
0.25
x(t)
0.2
0.15
0.1
0.05
0
-0.05
0
0.5
1
1.5
2
2.5
Time
3
3.5
4
4.5
Figure S3.9a. Simulation of X(s) for case a)
8000
6000
4000
2000
x(t)
c)
0
-2000
-4000
-6000
-8000
-10000
-12000
0
0.5
1
1.5
2
2.5
Time
Figure S3.9b. Simulation of X(s) for case b)
3-10
5
20
15
10
x(t)
5
0
-5
-10
-15
-20
0
0.5
1
1.5
2
2.5
Time
3
3.5
4
4.5
5
Figure S3.9c. Simulation of X(s) for case c)
The Simulink block diagram is shown below. An impulse input should be
used to obtain the function’s behavior. In this case note that the impulse
input is simulated by a rectangular pulse input of very short duration. (At
time t = 0 and t =0.001 with changes of magnitude 1000 and –1000
respectively). The MATLAB command impulse might also be used.
Figure S3.9d. Simulink block diagram for cases a), b) and c).
3-11
3.10
a)
i)
ii)
iii)
iv)
Y(s) =
2
2
A B
C
= 2
= 2+ +
s s+4
s ( s + 4 s ) s ( s + 4) s
∴
y(t) will contain terms of form: constant, t, e-4t
Y(s) =
2
2
A
B
C
=
= +
+
s ( s + 4 s + 3) s ( s + 1)( s + 3) s s + 1 s + 3
∴
y(t) will contain terms of form: constant, e-t, e-3t
Y (s) =
2
2
A
B
C
=
= +
+
2
2
s ( s + 2)
s+2
s ( s + 4 s + 4) s ( s + 2)
∴
y(t) will contain terms of form: constant, e-2t , te-2t
Y (s) =
2
s ( s + 4 s + 8)
2
2
2
2
s 2 + 4 s + 8 = ( s 2 + 4 s + 4) + (8 − 4) = ( s + 2) 2 + 2 2
2
Y (s) =
s[(s + 2) 2 + 2 2 ]
∴
b)
y(t) will contain terms of form: constant, e-2t sin2t, e-2tcos2t
A
Bs
C
2( s + 1)
2( s + 1)
=
= + 2
+ 2
2
2
2
2
s ( s + 4) s ( s + 2 ) s s + 2
s + 22
2( s + 1) 1
A = lim 2
=
s → 0 ( s + 4)
2
Y (s) =
2(s+1) = A(s2+4) + Bs(s) + Cs
2s+2 = As2 + 4A + Bs2 + Cs
Equating coefficients on like powers of s
s2 :
0=A+B
→
B = −A = −
s1 :
2= C
→
s0 :
2 = 4A
→
C=2
1
A=
2
3-12
1
2
∴
Y(s) =
1 2 − (1 2) s
2
+ 2
+ 2
2
s
s +2
s + 22
y(t) =
1 1
2
− cos 2t + sin 2t
2 2
2
y(t) =
1
(1 − cos 2t ) + sin 2t
2
3.11
a)
Since convergent and oscillatory behavior does not depend on initial
dx 2 (0) dx(0)
conditions, assume
=
= x ( 0) = 0
dt
dt 2
Laplace transform of the equation gives
s 3 X ( s ) + 2 s 2 X ( s ) + 2 sX ( s ) + X ( s ) =
3
1
3
1
3
s ( s + 1)( s + +
j )( s + −
j)
2 2
2 2
Denominator of [sX(s)] contains complex factors so that x(t) is oscillatory,
and denominator vanishes at real values of s= −1 and -½ which are all <0
so that x(t) is convergent. See Fig. S3.11a.
2
s 2 X ( s) − X (s) =
s −1
2
2
X (s) =
=
2
( s − 1)( s − 1) ( s − 1) 2 ( s + 1)
X (s) =
b)
3
s
3
=
s ( s + 2 s 2 + 2 s + 1)
3
The denominator contains no complex factors; x(t) is not oscillatory.
The denominator vanishes at s=1 ≥0; x(t) is divergent. See Fig. S3.11b.
c)
s 3 X ( s) + X (s) =
X (s) =
1
s +1
2
1
=
( s + 1)( s 3 + 1)
2
1
1
3
1
3
( s + j )( s − j )( s + 1)( s − +
j )( s − −
j)
2 2
2 2
The denominator contains complex factors; x(t) is oscillatory.
The denominator vanishes at real s = 0, ½; x(t) is not convergent. See Fig.
S3.11c.
3-13
s 2 X ( s ) + sX ( s ) =
4
s
4
4
= 2
s ( s + s ) s ( s + 1)
X (s) =
2
The denominator of [sX(s)] contains no complex factors; x(t) is not
oscillatory.
The denominator of [sX(s)] vanishes at s = 0; x(t) is not convergent. See
Fig. S3.11d.
3.5
3
2.5
x(t)
2
1.5
1
0.5
0
-0.5
0
1
2
3
4
5
time
6
7
8
9
10
Figure S3.11a. Simulation of X(s) for case a)
700
600
500
400
x(t)
d)
300
200
100
0
0
0.5
1
1.5
2
2.5
time
3
3.5
4
4.5
5
Figure S3.11b. Simulation of X(s) for case b)
3-14
80
60
x(t)
40
20
0
-20
-40
0
1
2
3
4
5
time
6
7
8
9
10
Figure S3.11c. Simulation of X(s) for case c)
18
16
14
12
x(t)
10
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
time
Figure S3.11d. Simulation of X(s) for case d)
3.12
Since the time function in the solution is not a function of initial
conditions, we Laplace Transform with
x ( 0) =
dx(0)
=0
dt
τ1τ2s2X(s) + (τ1+τ2)sX(s) + X(s) = KU(s)
3-15
X ( s) =
K
U (s)
τ1 τ 2 s + ( τ1 + τ 2 ) s + 1
2
Factoring denominator
X ( s) =
a)
K
U (s)
(τ1 s + 1)(τ 2 s + 1)
If u(t) = a S(t) then U(s)=
X a (s) =
a
s
Ka
s (τ1 s + 1)(τ 2 s + 1)
τ1 ≠ τ 2
xa(t) = fa( S(t), e-t/τ1, e– t/τ2)
b)
If u(t) = be-t/τ then U(s) =
X b ( s) =
bτ
τs +1
Kbτ
(τs + 1)(τ1 s + 1)(τ 2 s + 1)
τ ≠ τ1 ≠ τ 2
xb(t) = fb(e-t/τ , e-t/τ1, e– t/τ2)
c)
If u(t) =ce-t/τ where τ = τ1 , then U(s) =
X c ( s) =
τc
τ1 s + 1
Kcτ
(τ1 s + 1) 2 (τ 2 s + 1)
xc(t) = fc(e– t/τ1, t e– t/τ1, e– t/τ2)
d)
If u(t) = d sin ωt then U(s) =
X d (s) =
dω
s + ω2
2
Kd
( s + ω )(τ1 s + 1)(τ 2 s + 1)
2
2
xd(t) = fd(e– t/τ1, e– t/τ2, sin ωt, cos ωt)
3-16
3.13
dx3
+ 4 x = et
3
dt
a)
d 2 x(0) dx(0)
=
= x(0) = 0
dt 2
dt
with
Laplace transform of the equation,
s 3 X(s) + 4 X(s) =
X (s) =
=
1
s −1
1
1
=
3
( s − 1)( s + 4) ( s − 1)( s + 1.59)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j )
α 3 + jβ 3
α 3 − jβ 3
α1
α2
+
+
+
s − 1 s + 1.59 s − 0.79 + 1.37 j s − 0.79 − 1.37 j
α1 =
α2 =
1
( s + 4)
=
3
s =1
1
5
1
( s − 1)( s − 0.79 + 1.37 j )( s − 0.79 − 1.37 j )
α 3 + jβ 3 =
1
( s − 1)( s + 1.59)( s − 0.79 − 1.37 j )
=−
s = −1.59
1
19.6
= −0.74 − 0.59 j
s = 0.79 −1.37 j
1
1
−
− 0.074 − 0.059 j − 0.074 + 0.059 j
X(s) = 5 + 19.6 +
+
s − 1 s + 1.59 s − 0.79 + 1.37 j
s − 0.79 − 1.37 j
1
1 −1.59t
x(t ) = e t −
e
− 2e 0.79t (0.074 cos 1.37t + 0.059 sin 1.37t )
5
19.6
dx
− 12 x = sin 3t
with
x(0) = 0
dt
b)
sX (s) − 12 X(s) =
X (s) =
=
3
s +9
2
3
3
=
( s + 9)( s − 12) ( s + 3 j )( s − 3 j )( s − 12)
2
α3
α 1 + j β1 α 1 − j β 1
+
+
s+3j
s−3j
s − 12
3-17
α 1 + j β1 =
α3 =
=
s = −3 j
3
1
4
=−
−
j
− 18 + 72 j
102 102
3
1
=
( s + 9 ) s =12 51
2
−
X (s) =
x(t ) = −
c)
3
( s − 3 j )( s − 12)
1
4
1
4
1
−
j −
+
j
102 102 + 102 102 + 51
s+3j
s −3j
s − 12
1
1
(cos 3t + 4 sin 3t ) + e12 t
51
51
d2x
dx
+ 6 + 25 x = e −t
2
dt
dt
dx(0)
= x(0) = 0
dt
with
s 2 X ( s ) + 6 sX ( s ) + 25X ( s ) + X ( s ) =
X ( s) =
α1 =
1
or
s +1
X( s ) =
1
( s + 1 )( s + 6 s + 25 )
2
α
α + β2 j α 2 − β2 j
1
= 1 + 2
+
( s + 1)( s + 3 + 4 j )( s + 3 − 4 j ) s + 1 s + 3 + 4 j s + 3 − 4 j
1
( s + 6 s + 25)
=
2
α 2 + jβ 2 =
s = −1
1
20
1
( s + 1)( s + 3 − 4 j )
=−
s = −3− 4 j
1
1
−
j
40 80
1
1
1
1
1
− −
j − −
j
X ( s ) = 20 + 40 80 + 40 80
s +1 s + 3 + 4 j
s + 3− 4 j
x(t ) =
d)
1 −t
1
1
e − e −3t ( cos 4t + sin 4t )
20
20
40
Laplace transforming (assuming initial conditions = 0, since they do not
affect results)
sY1(s) + Y2(s) = X1(s)
(1)
sY2(s) – 2Y1(s) + 3 Y2(s) = X2(s)
(2)
3-18
From (2),
(s+3) Y2(s) = X2(s) + 2Y1(s)
Y2 ( s ) =
1
2
X 2 (s) +
Y1 ( s )
s+3
s+3
Substitute in Eq.1
sY1(s) +
1
2
X 2 (s) +
Y1 ( s ) = X1(s)
s+3
s+3
We neglect X2(s) since it is equal to zero.
[s(s + 3) + 2]Y1 (s) = ( s + 3) X 1 (s)
( s 2 + 3s + 2)Y1 ( s ) = ( s + 3) X 1 ( s )
Y1 ( s ) =
s+3
s+3
X 1 (s) =
X 1 (s)
( s + 1)( s + 2)
s + 3s + 2
2
1
s +1
s+3
A
B
C
Y1 ( s ) =
=
+
+
2
2
( s + 1) ( s + 2) ( s + 1) ( s + 1)
s+2
Now if x1(t) = e-t then X1(s) =
∴
so that y1(t) will contain e-t/τ , te-t/τ, e–2t functions of time.
For Y2(s)
Y2 ( s ) =
2
A
B
C
=
+
+
2
s+2
( s + 1) ( s + 2) ( s + 1) ( s + 1)
2
so that y2(t) will contain the same functions of time as y1(t) (although
different coefficients).
3-19
3.14
d 2 y (t )
dy (t )
d ( x − 2)
+3
+ y (t ) = 4
− x(t − 2)
2
dt
dt
dt
Taking the Laplace transform and assuming zero initial conditions,
s2Y(s) + 3sY(s) + Y(s) = 4 e-2ssX(s) −e-2sX(s)
Rearranging,
Y (s)
− (1 − 4s )e −2 s
= G(s) = 2
X ( s)
s + 3s + 1
a)
(1)
The standard form of the denominator is : τ2s2 + 2ζτs + 1
From (1) , τ = 1 , ζ = 1.5
Thus the system will exhibit overdamped and non-oscillatory response.
b)
Steady-state gain
K = lim G ( s ) = −1
(from (1))
s →0
c)
For a step change in x
1.5
− (1 − 4 s )e −2 s 1.5
and Y(s) = 2
X(s) =
s
( s + 3s + 1) s
Therefore yˆ (t ) = −1.5 + 1.5e-1.5t cosh(1.11t) + 7.38e-1.5t sinh(1.11t)
Using MATLAB-Simulink, y(t)= yˆ (t − 2) is shown in Fig. S3.14
1.5
1
0.5
0
-0.5
-1
-1.5
0
5
10
15
20
25
30
Figure S3.14. Output variable for a step change in x of magnitude 1.5
3-20
3.15
f (t ) = hS (t ) − hS (t − 1 / h)
dx
+ 4 x = h[S (t ) − S (t − 1 / h)]
dt
x(0)=0
,
Take Laplace transform,
 1 e−s / h 

sX ( s ) + 4 X ( s ) = h −
s 
s
α 
1
α
X ( s ) = h(1 − e − s / h )
= h(1 − e − s / h )  1 + 2 
s ( s + 4)
 s s + 4
1
1
1
1
α1 =
=
, α2 =
=−
s + 4 s =0 4
s s =−4
4
h
1 
1
(1 − e − s / h )  −

4
s s + 4
X ( s) =
=
h 1 e −s / h
1
e −s / h 
−
−
+


4 s
s
s + 4 s + 4
0
h
(1 − e − 4t )
4
h − 4( t −1 / h )
e
− e −4t
4
x(t ) =
[
t <0
0 < t < 1/h
]
t > 1/h
1
h=1
h=10
h=100
0.9
0.8
0.7
x(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
time
1.2
1.4
1.6
1.8
2
Figure S3.15. Solution for values h= 1, 10 and 100
3-21
3.16
a)
Laplace transforming
[s Y (s) − sy(0) − y′(0)]+ 6[sY (s) − y(0)] + 9Y (s) = s s+ 1
2
2
(s2 + 6s + 9)Y(s) − s(1) − 2 –(6)(1)=
(s2 + 6s + 9)Y(s) =
s
s +1
2
s
+s+8
s +1
2
s + s 3 + s + 8s 2 + 8
(s + 6s + 9)Y(s) =
s2 +1
2
Y(s) =
s 3 + 8s 2 + 2 s + 8
( s + 3) 2 ( s 2 + 1)
To find y(t) we have to expand Y(s) into its partial fractions
Y (s) =
A
B
Cs
D
+
+ 2
+ 2
2
( s + 3)
s + 3 s +1 s +1
y(t) = Ate-3t + Be-3t + C cost + D sint
b)
Y(s)=
s +1
s ( s + 4 s + 8)
Since
42
< 8 we know we will have complex factors.
4
2
∴ complete square in denominator
s2 + 4s + 8 = s2 + 4s + 4 + 8−4
= s2 + 4s + 4 + 4 = (s+2)2 + (2)2
∴ Partial fraction expansion gives
3-22
{ b = 2 , ω=2}
Y (s) =
A
B ( s + 2)
C
s +1
+ 2
+ 2
=
2
s s + 4 s + 8 s + 4 s + 8 s ( s + 4 s + 8)
Multiply by s and let s→0
A=1/8
Multiply by s(s2+4s+8)
A(s2+4s+8) + B(s+2)s + Cs = s + 1
As2 + 4As + 8A + Bs2 + 2Bs + Cs = s + 1
Y (s) =
→ B = −A = −
s2 :
A+B=0
s1 :
4A + 2B + C = 1
s0 :
8A = 1
→A=
1
1 3
C = 1 + 2  − 4  =
8
8 4
→
1
8
1
8
(This checks with above result)
1 / 8 (− 1 / 8)( s + 2)
3/ 4
+
+
2
2
s
( s + 2) + 2
( s + 2) 2 + 2 2
1 1
 3
y(t) =   −   e-2t cos 2t +   e-2t sin 2t
8 8
8
3.17
V
dC
+ qC = qCi
dt
Since V and q are constant, we can Laplace Transform
sVC(s) + qC(s) = q Ci(s)
Note that c(t = 0) = 0
Also, ci(t) = 0
ci(t) = ci
,
,
t ≤0
t>0
3-23
Laplace transforming the input function, a constant,
Ci ( s ) =
ci
s
so that
sVC(s) + qC(s) = q
ci
s
or
C(s) =
qci
( sV + q ) s
Dividing numerator and denominator by q
C(s) =
ci
V

 s +1 s
q

Use Transform pair #3 in Table 3.1 to invert (τ =V/q)
V
− t 


c(t) = ci 1 − e q 




Using MATLAB, the concentration response is shown in Fig. S3.17.
(Consider V = 2 m3, Ci=50 Kg/m3 and q = 0.4 m3/min)
50
45
40
35
c(t)
30
25
20
15
10
5
0
0
5
10
15
20
25
30
Time
Figure S3.17. Concentration response of the reactor effluent stream.
3-24
3.18
a)
If Y(s) =
KAω
s( s 2 + ω2 )
and input U(s) =
Aω
= 1 {A sin ωt}
(s + ω2 )
2
then the differential equation had to be
dy
= Ku (t )
dt
b)
Y(s) =
α1 =
with
y(0) = 0
α ω
α
α s
KAω
= 1+ 2 2 2 + 2 3 2
2
2
s( s + ω ) s s + ω
s +ω
KAω
s + ω2
2
=
s →0
KA
ω
Find α2 and α3 by equating coefficients
KAω= α1(s2+ω2) + α2s2+α3ωs
KAω = α1s2 + α1ω2 + α2s2 + α3ωs
s2 :
0 = α1 + α2
s:
0 = α3 ω
→ α2 = −α1 =
→ α3 = 0
KAω
KA / ω ( KA / ω) s
=
− 2
2
2
s
s( s + ω )
s + ω2
KA
(1 − cos ωt )
y(t) =
ω
∴ Y(s) =
3-25
− KA
ω
c)
A
u(t)
-A
2KA/ω
y(t)
0
Time
i)
We see that y(t) follows behind u(t) by 1/4 cycle = 2π/4= π/2 rad.
which is constant for all ω
ii)
The amplitudes of the two sinusoidal quantities are:
y : KA/ω
u: A
Thus their ratio is K/ω, which is a function of frequency.
3-26