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2.1 a) Nonlinear because of the y yฬ term. b) Nonlinear because of the sin y term. c)
โ
Nonlinear because of the y term. d) Variable coefficient, but Linear. e) Nonlinear because
of the sin y term. f) Variable coefficient, but linear.
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2.2 a)
Z x
4
Z t
dx = 3
t dt
2
0
3
x(t) = 2 + t2
8
b)
Z x
5
Z t
eโ4t dt
dx = 2
3
0
x(t) = 3.1 โ 0.1eโ4t
c) Let v = xฬ.
Z v
3
Z t
dv = 5
7
t dt
0
5
dx
= 7 + t2
dt
6
Z t
Z x
5 2
dx =
7 + t dt
6
2
0
5
x(t) = 2 + 7t + t3
18
v(t) =
d) Let v = xฬ.
Z v
4
Z t
dv = 7
2
eโ2t dt
0
23 7 โ2t
โ e
v(t) =
8
8
Z x
Z t
23 7 โ2t
dx =
โ e
dt
8
8
4
0
7
57 23
+ t + eโ2t
x(t) =
16
8
16
e) xฬ = C1 , but xฬ(0) = 5, so C1 = 5. x = 5t + C2 , but x(0) = 2, so C2 = 2. Thus
x = 5t + 2.
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.3 a)
t
dx
=
dt = t
2
0
3 25 โ 5x
โ ”
โ !
โ !#
Z x
5
5x
dx
3 5
=
arctanh
โ arctanh
=t
2
25
5
5
3 25 โ 5x
Z x
Let
Z
โ !
3 5
C = arctanh
5
Solve for x to obtain
x=
โ
โ
5 tanh(5 5t + C)
b)
Z x
dx
=
2
10 36 + 4x
Z t
dt = t
0
1
x x
tanโ1
=t
12
3 10
x(t) = 3 tan(12t + C)
C = tanโ1
10
3
c)
Z x
4
x
x dx
=
5x + 25
Z t
dt
0
4
x
x
โ ln(x + 5) = โ ln(x + 5) โ + ln 9 = t
5
5
5
4
x โ 5 ln(x + 5) = 5t + 4 โ 5 ln 9
So a closed form solution does not exist.
(continued on the next page)
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Problem 2.3 continued:
d)
t
dx
= โ2
eโ4t dt
x
0
5
1 โ4t
ln x|x5 =
e
โ1
2
x
1 โ4t
ln =
e
โ1
5
2
5 1 โ4t
x(t) = โ e 2 e
e
Z x
Z
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.4 From the transform definition, we have
“Z
L[mt] = lim
#
T
โst
mte
T โโ
“Z
dt = m lim
T โโ
0
#
T
โst
te
dt
0
The method of integration by parts states that
Z T
0
u dv = uv|T0 โ
Z T
v du
0
Choosing u = t and dv = eโst dt, we have du = dt, v = โeโst /s, and
L[mt] = m lim
“Z
T โโ
๏ฃฎ
= m lim ๏ฃฐ t
T โโ
T
eโst
โs 0
โ
0
T
๏ฃฎ
#
T
eโst
โ
teโst dt = m lim ๏ฃฐ t
T โโ
โs 0
eโst
T
(โs)2 0
๏ฃน
”
๏ฃป = m lim
T โโ
Z T โst
e
0
โs
๏ฃน
dt๏ฃป
T eโsT
eโsT
e0
+
โ0โ
โs
(โs)2 (โs)2
#
m
s2
because, if we choose the real part of s to be positive, then
=
lim T eโsT = 0
T โโ
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2.5 From the transform definition, we have
“Z
2
L[t ] = lim
#
T
2 โst
t e
T โโ
dt
0
The method of integration by parts states that
Z T
0
u dv = uv|T0 โ
Z T
v du
0
Choosing u = t2 and dv = eโst dt, we have du = 2t dt, v = โeโst /s, and
L[t2 ] = lim
T โโ
”
= lim
T โโ
โT
“Z
T
0
โst
2e
๏ฃฎ
#
T
eโst
โ
t2 eโst dt = lim ๏ฃฐ t2
T โโ
โs 0
Z T โst
e
#
#
2
+
s
s
Z T
โst
te
”
dt = lim
T โโ
0
โT
0
โst
2e
s
โs
๏ฃน
2t dt๏ฃป
2
+
s
1
s2
2
s3
because, if we choose the real part of s to be positive, then,
=
lim T 2 eโsT = 0
T โโ
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2.6 a)
X(s) =
2
10
+ 3
s
s
b)
X(s) =
6
1
+
2
(s + 5)
s+3
c) From Property 8,
X(s) = โ
dY (s)
ds
where y(t) = eโ3t sin 5t. Thus
Y (s) =
5
5
= 2
(s + 3)2 + 52
s + 6s + 34
10s + 30
dY (s)
=โ 2
ds
(s + 6s + 34)2
Thus
X(s) =
10s + 30
(s2 + 6s + 34)2
d) X(s) = eโ5s G(s), where g(t) = t. Thus G(s) = 1/s2 and
X(s) =
eโ5s
s2
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2.7
f (t) = 5us (t) โ 7us (t โ 6) + 2us (t โ 14)
Thus
F (s) =
eโ6s
eโ14s
5
โ7
+2
s
s
s
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2.8 a)
2 sin 3t
b)
4 cos 2t +
5
sin 2t
2
c)
2eโ2t sin 3t
d)
5 5 eโ3 t
โ
3
3
e)
5 eโ3 t 5 eโ7 t
โ
2
2
f)
eโ3 t 3 eโ7 t
+
2
2
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2.9 a)
5 cos(3 t)
b)
e3 t โ eโ3 t
c)
5 โ 15 t eโ3 t โ 5 eโ3 t
d)
2
โ
13
2 eโ2 t cos 3t + 2 sin3 3 t
13
e)
5 โ 5 cos 2t
f)
5 t sin 2t
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2.10 a)
5
5
=
3s + 7
3
5
x(โ) = lim s
=0
sโ0 3s + 7
x(0+) = lim s
sโโ
b)
x(0+) = lim s
sโโ
x(โ) = lim s
sโ0
10
3s2 + 7s + 4
=0
10
=0
3s2 + 7s + 4
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.11 a)
3 1
1
X(s) =
โ
2 s s+4
3
x(t) =
1 โ eโ4t
2
b)
5 1 31 1
+
3s
3 s+3
5 31
x(t) = + eโ3t
3
3
X(s) =
c)
13 1
1 1
+
3s+2
3 s+5
1 โ2t 13 โ5t
x(t) = โ e
+ e
3
3
X(s) = โ
d)
X(s) =
5/2
s2 (s + 4)
=
5 1
5 1
5 1
โ
+
2
8s
32 s 32 s + 4
5
5
5
x(t) = t โ
+ eโ4t
8
32 32
(continued on the next page)
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
Problem 2.11 continued:
e)
13 1 13 1
2 1
+
โ
2
5s
25 s 25 s + 5
2
13 13 โ5t
x(t) = t +
โ e
5
25 25
X(s) =
f)
X(s) = โ
1
79 1
79 1
31
+
โ
2
4 (s + 3)
16 s + 3 16 s + 7
x(t) = โ
31 โ3t 79 โ3t 79 โ7t
te
+ e
โ e
4
16
16
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.12 a)
X(s) =
5
s+3
7s + 2
= C1
+ C2
2
2
2
2
(s + 3) + 5
(s + 3) + 5
(s + 3)2 + 52
or
X(s) = โ
19
5
s+3
+7
2
2
5 (s + 3) + 5
(s + 3)2 + 52
x(t) = โ
19 โ3t
e sin 5t + 7eโ3t cos 5t
5
b)
X(s) =
C1
5
s+3
4s + 3
=
+ C3
+ C2
s[(s + 3)2 + 52 ]
s
(s + 3)2 + 52
(s + 3)2 + 52
or
X(s) =
5
s+3
3 1 127
3
โ
+
2
2
34 s 170 (s + 3) + 5
34 (s + 3)2 + 52
x(t) =
3
127 โ3t
3
+
e sin 5t โ eโ3t cos 5t
34 170
34
(continued on the next page)
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
Problem 2.12 continued:
c)
X(s) =
4s + 9
[(s + 3)2 + 52 ][(s + 2)2 + 42 ]
= C1
5
s+3
4
s+2
+ C2
+ C3
+ C4
2
2
2
2
2
2
(s + 3) + 5
(s + 3) + 5
(s + 2) + 4
(s + 2)2 + 42
or
44
5
19
s+3
โ
2
2
205 (s + 3) + 5
82 (s + 3)2 + 52
69
4
19
s+2
+
2
2
328 (s + 2) + 4
82 (s + 2)2 + 42
X(s) = โ
+
x(t) = โ
44 โ3t
19
69 โ2t
19
e sin 5t โ eโ3t cos 5t +
e sin 4t + eโ2t cos 4t
205
82
328
82
d)
X(s) = 2.625
1
1
1
โ 18.75
+ 21.125
s+2
s+4
s+6
x(t) = 2.625eโ2t โ 18.75eโ4t + 21.125eโ6t
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.13 a) xฬ = 7t/5
Z x
7 t
t dt
5 0
3
7
x(t) = t2 + 3
10
Z
dx =
b) xฬ = 3eโ5t /4
3 t โ5t
e dt
dx =
4 0
4
3
x(t) =
1 โ eโ5t + 4
20
Z x
Z
c) xฬ = 4t/7
4 t
t dt
xฬ(t) โ xฬ(0) =
7 0
4
xฬ(t) = t2 + 5
14
Z x
Z t
4 2
t + 5 dt
dx =
14
3
0
4
x(t) = t3 + 5t + 3
42
Z
d) xฬ = 8eโ4t /3
8 t โ4t
xฬ(t) โ xฬ(0) =
e dt
3 0
8
17
xฬ(t) =
โ eโ4t
3
12
Z x
Z t
8 โ4t
17
dx =
โ e
dt
3
12
3
0
17
1
17
x(t) = t + eโ4t +
3
6
6
Z
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.14 a) The root is โ7/5 and the form is x(t) = Ceโ7t/5 . With x(0) = 4, C = 4 and
x(t) = 4eโ7t/5
b) The root is โ7/5 and the form is x(t) = C1 eโ7t/5 + C2 . At steady state, x = 15/7 =
C2 . With x(0) = 0, C1 = โ15/7. Thus
x(t) =
15
1 โ eโ7t/5
7
c) The root is โ7/5 and the form is x(t) = C1 eโ7t/5 + C2 . At steady state, x = 15/7 =
C2 . With x(0) = 4, C1 = 13/7. Thus
x(t) =
13
1 + eโ7t/5
7
d)
sX(s) โ x(0) + 7X(s) =
X(s) =
4
s2
5s2 + 4
4
249 โ7t
4
= 2โ
+
e
2
s (s + 7)
7s
49
49
4
249 โ7t
4
+
e
x(t) = t โ
7
49
49
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.15 a) The roots are โ7 and โ3. The form is
x(t) = C1 eโ7t + C2 eโ3t
Evaluating C1 and C2 for the initial conditions gives
25
9
x(t) = โ eโ7t + eโ3t
4
4
b) The roots are โ7 and โ7. The form is
x(t) = C1 eโ7t + C2 teโ7t
Evaluating C1 and C2 for the initial conditions gives
x(t) = eโ7t + 10teโ7t
c) The roots are โ7 ยฑ 3j. The form is
x(t) = C1 eโ7t sin 3t + C2 eโ7t cos 3t
Evaluating C1 and C2 for the initial conditions gives
x(t) =
20 โ7t
e
sin 3t + 4eโ7t cos 3t
3
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.16 a)
x = 6 eโ2 t โ 3 eโ5 t + 2
b)
x=
18 eโ2 t 76 t eโ2 t 7
+
+
5
5
5
c)
x = 3 sin 4t โ 4 cos 4t + 9
d)
x = 3 cos 5t eโ3 t +
16 sin 5t eโ3t
+2
5
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.17 a) The roots are โ3 and โ7. The form is
x(t) = C1 eโ3t + C2 eโ7t + C3
At steady state, x = 5/63 so C3 = 5/63. Evaluating C1 and C2 for the initial conditions
gives
5
5
5
x(t) = โ eโ3t + eโ7t +
36
84
63
b) The roots are โ7 and โ7. The form is
x(t) = C1 eโ7t + C2 teโ7t + C3
At steady state, x = 98/49 = 2 so C3 = 2. Evaluating C1 and C2 for the initial conditions
gives
x(t) = โ2eโ7t โ 14teโ7t + 2
c) The roots are โ7 ยฑ 3j. The form is
x(t) = C1 eโ7t sin 3t + C2 eโ7t cos 3t + C3
At steady state, x = 174/58 = 3 so C3 = 3. Evaluating C1 and C2 for the initial conditions
gives
x(t) = โ7eโ7t sin 3t โ 3eโ7t cos 3t + 3
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2.18 a)
X(s) =
60
s2 + 8s + 12
x = 15 eโ2t โ 15eโ6t
b)
X(s) =
288
s2 + 12s + 144
โ
โ
x = 16 3eโ6t sin 6 3t
c)
147
s2 + 49
x = 21 sin 7t
X(s) =
d)
X(s) =
x=
170
s2 + 14s + 85
85 eโ7t sin 6t
3
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.19 a)
6
6 1
6
=
โ
s(s + 5)
5s 5 s + 5
x(t) =
6
1 โ eโ5t
5
b)
4
4 1
4 1
=
โ
s + 3)(s + 8)
5s+3 5s+8
x(t) =
4 โ3t
e
โ eโ8t
5
c)
1
8s + 5
27 1
43 1
8s + 5
=
=โ
+
2s2 + 20s + 48
2 (s + 4)(s + 6)
4 s+4
4 s+6
x(t) = โ
27 โ4t 43 โ6t
e
+ e
4
4
d) The roots are s = โ4 ยฑ 10j.
4s + 13
4s + 13
+
s2 + 8s + 116 (s + 4)2 + 102
x(t) = โ
10
s+4
+ C2
(s + 4)2 + 102
(s + 4)2 + 102
3
10
s+4
+4
= โ
10 (s + 4)2 + 102
(s + 4)2 + 102
= C1
3 โ4t
e sin 10t + 4eโ4t cos 10t
10
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.20 a)
3s + 2
s2 (s + 10)
=
1 1
7 1
7
1
+
โ
2
5s
25 s 25 s + 10
1
7
x(t) = t +
1 โ eโ10t
5
25
b)
5
15
5 1
1
5 1
=โ
โ
+
2
2
(s + 4) (s + 1)
9 (s + 4)
9s+4 9s+1
x(t) = โ
15 โ4t 5 โ4t 5 โt
te
โ e
+ e
9
9
9
c)
s2 + 3s + 5
5 1
1 1
31 3 1
=
+
+
โ
3
3
2
s (s + 2)
2s
4s
8s 8s+2
5
1
3 3
x(t) = t2 + t + โ eโ2t
4
4
8 8
d)
1
1
1 1
11 1 1
s3 + s + 6
=3 4 โ 3 +
+
โ
4
2
s (s + 2)
s
s
2s
4s 4s+2
1
1
1
1 1
x(t) = t3 โ t2 + t + โ eโ2t
2
2
2
4 4
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.21 a)
5[sX(s) โ 2] + 3X(s) =
X(s) =
2
10
+ 3
s
s
10s3 + 10s2 + 2
2s3 + 2s2 + 2/5
2 1
10 1
140 1 86
1
=
=
โ
+
โ
5s3 (s + 3)
s3 (s + 3/5)
3 s3
9 s2
9 s 27 s + 3/5
10
140 86 โ3t/5
1
โ e
x(t) = t2 โ t +
3
9
27
27
b)
4[sX(s) โ 5] + 7X(s) =
X(s) =
=
1
6
+
2
(s + 5)
s+3
1 20s3 + 261s2 + 1116s + 1543
4 (s + 5)2 (s + 7/4)(s + 3)
1
24
1
96 1
18056
1
4 1
โ
+
โ
โ
4
13 (s + 5)2 169 s + 5
845 s + 7/4 5 s + 3
x(t) = โ
24 โ5t 4514 โ7t/4 1 โ3t
6 โ5t
te
โ
e
+
e
โ e
13
169
845
5
(continued on the next page)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.
Problem 2.21 continued:
c) This simple-looking problem actually requires quite a lot of algebra to find the solution, and thus it serves as a good motivating example of the convenience of using MATLAB.
The algebraic complexity is due to a pair of repeated complex roots.
First obtain the transform of the forcing function. Let f (t) = teโ3t sin 5t. From Property 8,
dY (s)
F (s) = โ
ds
where y(t) = eโ3t sin 5t. Thus
Y (s) =
5
5
= 2
2
2
(s + 3) + 5
s + 6s + 34
10s + 30
dY (s)
=โ 2
ds
(s + 6s + 34)2
Thus
F (s) =
10s + 30
(s2 + 6s + 34)2
(1)
(continued on the next page)
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Problem 2.21 continued:
Using the same technique, we find that the transform of teโ3t cos 5t is
2s2 + 12s + 18
1
โ 2
2
2
(s + 6s + 34)
s + 6s + 34
(2)
This fact will be useful in finding the forced response.
From the differential equation,
4[s2 X(s) โ 10s + 2] + 3X(s) = F (s) =
10s + 30
(s2 + 6s + 34)2
Solve for X(s).
X(s) =
40s โ 8
10s + 30
+
2
2
4s + 3 [(s + 3) + 25]2 (4s2 + 3)
The free response is given by the first fraction, and is
โ
โ
4
3
3
xfree (t) = โ โ sin
t + 10 cos
t = โ2.3094 sin 0.866t + 10 cos 0.866t
2
2
3
(3)
The forced response is given by the second fraction, which can be expressed as
2.5s + 7.5
[(s + 3)2 + 25]2 (s2 + 3/4)
(4)
(continued on the next page)
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Problem 2.21 continued:
โ
The roots of this are s = ยฑj 3/2 and the repeated pair s = โ3 ยฑ 5j. Thus, referring
to (1), (2), and (3), we see that the form of the forced response will be
xforced (t) = C1 teโ3t sin 5t + C2 teโ3t cos 5t
+ C3 eโ3t sin 5t + C4 eโ3t cos 5t
โ
โ
3
3
t + C6 cos
t
(5)
+ C5 sin
2
2
The forced response can be obtained several ways. 1) You can substitute the form (5)
into the differential equation and use the initial conditions to obtain equations for the Ci
coefficients. 2) You can use (1) and (2) to create a partial fraction expansion of (4) in terms
of the complex factors. 3) You can perform an expansion in terms of the six roots, of the
form
A1
(s + 3 + 5j)2
+
+
A2
A3
A4
+
+
2
s + 3 + 5j (s + 3 โ 5j)
s + 3 โ 5j
โ
A6 s
3A5 /2
+ 2
2
s + 3/4 s + 3/4
4) You can use the MATLAB residue function.
The solution for the forced response is
xforced (t) = โ0.0034teโ3t sin 5t + 0.0066teโ3t cos 5t
โ 0.0026eโ3t sin 5t + 2.308 ร 10โ4 eโ3t cos 5t
+ 0.00796 sin 0.866t โ 2.308 ร 10โ4 cos 0.866t
The initial condition xฬ(0) = 0 is not exactly satisfied by this expression because of the
limited number of digits used to display it.
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2.22 The denominator roots are s = โ3 and s = โ5, which are distinct. Factor the
denominator so that the highest coefficients of s in each factor are unity:
7s + 4
1
7s + 4
=
X(s) = 2
2s + 16s + 30
2 (s + 3)(s + 5)
The partial-fraction expansion has the form
1
C2
7s + 4
C1
=
+
2 (s + 3)(s + 5)
s+3 s+5
X(s) =
Using the coefficient formula, we obtain
7s + 4
17
7s + 4
= lim
=โ
C1 = lim (s + 3)
sโโ3 2(s + 5)
sโโ3
2(s + 3)(s + 5)
4
C2 = lim
sโโ5
(s + 5)
7s + 4
31
7s + 4
= lim
=
sโโ5 2(s + 3)
2(s + 3)(s + 5)
4
(continued on the next page)
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Problem 2.22 continued:
Using the LCD method we have
1
7s + 4
C1
C2
C1 (s + 5) + C2 (s + 3)
=
+
=
2 (s + 3)(s + 5)
s+3 s+5
(s + 3)(s + 5)
=
(C1 + C2 )s + 5C1 + 3C2
(s + 3)(s + 5)
Comparing numerators, we see that C1 + C2 = 7/2 and 5C1 + 3C2 = 4/2 = 2, which give
C1 = โ17/4 and C2 = 31/4.
The inverse transform is
x(t) = C1 eโ3t + C2 eโ5t = โ
17 โ3t 31 โ5t
e
+ e
4
4
In this example the LCD method requires more algebra, including the solution of two
equations for the two unknowns C1 and C2 .
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2.23 a) The roots are โ3 and โ5. The form of the free response is
x(t) = A1 eโ3t + A2 eโ5t
Evaluating this with the given initial conditions gives
x(t) = 27eโ3t โ 17eโ5t
The steady-state solution is xss = 30/15 = 2. Thus the form of the forced response is
x(t) = 2 + B1 eโ3t + B2 eโ5t
Evaluating this with zero initial conditions gives
x(t) = 2 โ 5eโ3t + 3eโ5t
The total response is the sum of the free and the forced response. It is
x(t) = 2 + 22eโ3t โ 14eโ5t
The transient response consists of the two exponential terms.
(continued on the next page)
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Problem 2.23 continued:
b) The roots are โ5 and โ5. The form of the free response is
x(t) = A1 eโ5t + A2 teโ5t
Evaluating this with the given initial conditions gives
x(t) = eโ5t + 9teโ5t
The steady-state solution is xss = 75/25 = 3. Thus the form of the forced response is
x(t) = 3 + B1 eโ5t + B2 teโ5t
Evaluating this with zero initial conditions gives
x(t) = 3 โ 3eโ5t โ 15teโ5t
The total response is the sum of the free and the forced response. It is
x(t) = 3 โ 2eโ5t โ 6teโ5t
The transient response consists of the two exponential terms.
(continued on the next page)
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Problem 2.23 continued:
c) The roots are ยฑ5j. The form of the free response is
x(t) = A1 sin 5t + A2 cos 5t
Evaluating this with the given initial conditions gives
x(t) =
4
sin 5t + 10 cos 5t
5
The form of the forced response is
x(t) = B1 + B2 sin 5t + B3 cos 5t
Thus the entire forced response is the steady-state forced response. There is no transient
forced response. Evaluating this function with zero initial conditions shows that B2 = 0
and B3 = โB1 . Thus
x(t) = B1 โ B1 cos 5t
Substituting this into the differential equation shows that B1 = 4 and the forced response
is
x(t) = 4 โ 4 cos 5t
The total response is the sum of the free and the forced response. It is
x(t) = 4 + 6 cos 5t +
4
sin 5t
5
The entire response is the steady-state response. There is no transient response.
(continued on the next page)
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Problem 2.23 continued:
d) The roots are โ4 ยฑ 7j. The form of the free response is
x(t) = A1 eโ4t sin 7t + A2 eโ4t cos 7t
Evaluating this with the given initial conditions gives
x(t) =
44 โ4t
e
sin 7t + 10eโ4t cos 7t
7
The form of the forced response is
x(t) = B1 + B2 eโ4t sin 7t + B3 eโ4t cos 7t
The steady-state solution is xss = 130/65 = 2. Thus B1 = 2. Evaluating this function with
zero initial conditions shows that B2 = โ8/7 and B3 = โ2. Thus the forced response is
8
x(t) = 2 โ eโ4t sin 7t โ 2eโ4t cos 7t
7
The total response is the sum of the free and the forced response. It is
x(t) = 2 +
36 โ4t
e
sin 7t + 8eโ4t cos 7t
7
The transient response consists of the two exponential terms.
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2.24 a) The root is s = 5/3, which is positive. So the model is unstable.
b) The roots are s = 5 and โ2, one of which is positive. So the model is unstable.
c) The roots are s = 3 ยฑ 5j, whose real part is positive. So the model is unstable.
d) The root is s = 0, so the model is neutrally stable.
e) The roots are s = ยฑ2j, whose real part is zero. So the model is neutrally stable.
f) The roots are s = 0 and โ5, one of which is zero and the other is negative. So the
model is neutrally stable.
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2.25 a) The system is stable if both of its roots are real and negative or if the roots are
complex with negative real parts. Assuming that m 6= 0, we can divide the characteristic
equation by m to obtain
k
c
= s2 + as + b = 0
s2 + s +
m
m
where a = c/m and b = k/m. The roots are given by the quadratic formula:
โ
โa ยฑ a2 โ 4b
s=
2
(continued on the next page)
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Problem 2.25 continued:
Thus the condition that m, c, and k have the same sign is equivalent to a > 0 and b > 0.
There are three cases to be considered:
1. Complex roots (a2 โ 4b 0.
2. Repeated, real roots (a2 โ 4b = 0). In this case both roots are โa/2 and are negative
if a > 0.
3. Distinct, real roots (a2 โ 4b > 0). Let the two roots be denoted r1 and r2 . We can
factor the characteristic equation as s2 + as + b = (s โ r1 )(s โ r2 ) = 0. Expanding
this gives
(s โ r1 )(s โ r2 ) = s2 โ (r1 + r2 )s + r1 r2 = 0
Comparing the two forms shows that
r1 r2 = b
(1)
and
r1 + r1 = โa
(2)
If b > 0, condition (1) shows that both roots have the same sign. If a 0 and b > 0.
b) Neutral stability occurs if either 1) both roots are imaginary or 2) one root is zero
โ
while the other root is negative. Imaginary roots occur when a = 0 (the roots are s = ยฑ b)
In this case the free response is a constant-amplitude oscillation. Case 2 occurs when b = 0
and a > 0 (the roots are s = 0 and s = โa). In this case the free response decays to a
non-zero constant.
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2.26 a) ฯ = 5
b) ฯ = 4
c) ฯ = 3
d) The roots is s = 3/8, so the model is unstable, so no time constant is defined.
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2.27 a) The root is s = โ4/13, so the model is stable, and xss = 16/4 = 4. Since ฯ = 13/4,
it takes about 4ฯ = 13 to reach steady state.
b) The root is s = โ4/13, so the model is stable, and xss = 16/4 = 4. Since ฯ = 13/4,
it takes about 4ฯ = 13 to reach steady state.
c) The root is s = 7/15, so the model is unstable, and no steady state exists.
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2.28 1)
X(s) =
s+1 5
1 s+1 5
C1
C2
=
=
+
4s + 1 s
4 s + 1/4 s
s
s + 1/4
C1 = 5, C2 = โ15/4, so
x(t) = 5 โ
15 โt/4
e
4
2)
X(s) =
C1 = 5, C2 = โ5, so
1
C1
C2
1 5
1
5
=
=
+
4s + 1 s
4 s + 1/4 s
s
s + 1/4
x(t) = 5 โ 5eโt/4
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2.29
3[sX(s) โ 4] + X(s) = 6
X(s) =
6
s + 1/3
x(t) = 6eโt/3
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2.30 a)
โ
r
โ
4
10
40
ฮถ= โ =
ฯn =
= 2 10
10
1
2 40
s = โ2 ยฑ 6j
so ฯ = 1/2 and ฯd = 6.
b)
s = 1 ยฑ 4.7958j
So the model is oscillatory but unstable, and thus ฮถ and ฯ are not defined.
r
ฯn =
โ
24
=2 6
1
ฯd = 4.7958
c)
20
ฮถ= โ
=1
2 100
s = โ10, โ10
so ฯ = 1/10. Since the roots are real, the response is not oscillatory, and ฯn and ฯd have
no meaning.
d) The root is s = โ10, so ฯ = 1/10. Since the model is first order, ฮถ, ฯn and ฯd have
no meaning.
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2.31 a) The roots are
s=
โ10d ยฑ
p
100d2 โ 4(29)d2
= (โ5 ยฑ 2j) d
2
So if d > 0, the real part is negative, and the system is stable.
b)
10
10d
= โ 4ฯ = 8/3.
d) With F (s) = 1/s,
2
ฯ=
3
X(s) =
or
1
1
1
1
=
2
4 s(s + 3s + 3)
4 s[(s + 23 )2 + 34 ]
โ
C1 (s + 23 ) + C2 23
C3
+
X(s) =
3
3 2
s
(s + 2 ) + 4
โ
where C1 = โC3 = โ1/12 and C2 = โ 3/12. Thus
โ
โ
โ !
3
3
3
1
1
โ3t/2
x(t) = e
โ
cos
tโ
sin
t +
12
2
12
2
12
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.36 a) Transform both equations to obtain
4sX(s) = โ4X(s) + 2Y (s) + F (s)
sY (s) = โ9Y (s) โ 5X(s) + G(s)
These can be solved using Cramerโs rule to obtan
s+9
X(s)
= 2
F (s)
4s + 40s + 46
X(s)
2
= 2
G(s)
4s + 40s + 46
b) The roots are s = โ1.3258 and s = โ8.6742. The time constants are ฯ = 0.7543 and
ฯ = 0.1153. The response does not oscillate.
c) The free response is governed by the dominant time constant, which is ฯ = 0.7543.
The response is essentially zero for t > 4ฯ = 3.0172.
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2.37 a)
7[sX(s) โ 3] + 5X(s) = 4
X(s) =
25/7
25
=
7s + 5
s + 5/7
25 โ5t/7
e
7
Note that this gives x(0+) = 25/7. From the initial value theorem
x(t) =
x(0+) = lim s
sโโ
25/7
25
=
s + 5/7
7
which is not the same as x(0โ).
b)
(3s2 + 30s + 63)X(s) = 5
X(s) =
5
3s2 + 30s + 63
=
x(t) =
5/3
s2 + 10s + 21
=
5 1
5 1
โ
12 s + 3 12 s + 7
5 โ3t
e
โ eโ7t
12
From the initial value theorem
5/3
x(0+) = lim s 2
=0
sโโ s + 10s + 21
which is the same as x(0โ). Also
5/3
5
xฬ(0+) = lim s2 2
=
sโโ
s + 10s + 21
3
which is not the same as xฬ(0โ).
(continued on the next page)
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
Problem 2.37 continued:
c)
s2 X(s) โ 2s โ 3 + 14[sX(s) โ 2] + 49X(s) = 3
X(s) =
2s + 34
s2 + 14s + 49
= 20
1
1
+2
2
(s + 7)
s+7
x(t) = 20teโ7t + 2eโ7t
From the initial value theorem
2s + 35
x(0+) = lim s 2
=2
sโโ s + 14s + 49
which is the same as x(0โ). However, the initial value theorem is invalid for computing
xฬ(0+) and gives an undefined result because the orders of the numerator and denominator
of sX(s) are equal.
d)
s2 X(s) โ 4s โ 7 + 14[sX(s) โ 4] + 58X(s) = 4
X(s) =
4s + 67
s2 + 14s + 58
=
4s + 67
3
s+7
= 13
+4
2
2
2
2
(s + 7) + 3
(s + 7) + 3
(s + 7)2 + 32
x(t) = 13eโ7t sin 3t + 4eโ7t cos 3t
From the initial value theorem
4s + 67
=4
x(0+) = lim s 2
sโโ s + 14s + 58
which is the same as x(0โ). However, the initial value theorem is invalid for computing
xฬ(0+) and gives an undefined result because the order of the numerator of sX(s) is greater
than the denominator.
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2.38 a)
1
7[sX(s) โ 3] + 5X(s) = 4s = 4
s
25
25/7
X(s) =
=
7s + 5
s + 5/7
x(t) =
25 โ5t/7
e
7
From the initial value theorem
x(0+) = lim s
sโโ
25/7
25
=
s + 5/7
7
which is not the same as x(0โ).
b)
1 6
7[sX(s) โ 3] + 5X(s) = 4s +
s s
25s + 6
1 25s + 6
6 1 83
1
X(s) =
=
=
+
s(7s + 5)
7 s(s + 5/7)
5 s 35 s + 5/7
6 83 โ5t/7
+ e
5 35
which gives x(0+) = 25/7, which is not the same as x(0โ). However, the initial value
theorem is invalid for computing x(0+) and gives an undefined result because the orders of
the numerator and denominator of X(s) are equal.
x(t) =
(continued on the next page)
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Problem 2.38 continued:
c)
1
3[s2 X(s) โ 2s โ 3] + 30[sX(s) โ 2] + 63X(s) = 4s = 4
s
1
55 1
31 1
6s + 73
X(s) =
=
โ
3 (s + 3)(s + 7)
12 s + 3 12 s + 7
55 โ3t 31 โ7t
e
โ e
12
12
This gives x(0) = 2, which is the same as x(0โ), and xฬ(0) = 13/2, which is not the same
as xฬ(0โ).
From the initial value theorem
x(t) =
x(0+) = lim s
sโโ
1
6s + 73
=2
3 (s + 3)(s + 7)
which is the same as x(0โ). However, the initial value theorem is invalid for computing
xฬ(0+) and gives an undefined result because the order of the numerator of sX(s) is greater
than the denominator.
(continued on the next page)
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
Problem 2.38 continued:
d)
1 6
3[s2 X(s) โ 4s โ 7] + 30[sX(s) โ 4] + 63X(s) = 4s +
s s
X(s) =
1
1 12s2 + 145s + 6
1
1
= 0.0952 + 8.9167
โ 5.0119
2
3 s(s + 10s + 21)
s
s+3
s+7
x(t) = 0.0952 + 8.9167eโ3t โ 5.0119eโ7t
This gives x(0) = 4, which is the same as x(0โ), and xฬ(0) = 8.3332, which is not the same
as xฬ(0โ).
The initial value theorem gives x(0+) = 4 but is invalid for computing xฬ(0+) because
the orders of the numerator and denominator of sX(s) are equal.
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2.39 Transform each equation.
3[sX(s) โ 5] = Y (s)
sY (s) โ 10 =
4
โ 3Y (s) โ 15X(s)
s
Solve for X(s) and Y (s).
X(s) =
1 15s2 + 55s + 4
15s2 + 55s + 4
=
3s3 + 9s2 + 15s
3 s(s2 + 3s + 5)
30s โ 213
1 30s โ 213
=
2
3s + 9s + 15
3 s2 + 3s + 5
The denominator roots are s = โ1.5 ยฑ 1.658j. Thus
Y (s) =
s + 1.5
C1 1
1.658
C1
+
+ C2
s
3
(s + 1.5)2 + 2.75
(s + 1.5)2 + 2.75
X(s) =
and
โ
โ
11
t + 313 11 sin
2
”
1
1 โ3t/2
x(t) = +
e
781 cos
4 165
!
โ
11
t
2
!#
Also,
Y (s) = C1
and
s + 1.5
1.658
+ C2
2
(s + 1.5) + 2.75
(s + 1.5)2 + 2.75
”
2
y(t) = eโ3t/2 55 cos
11
โ
โ
11
t โ 86 11 sin
2
!
โ
11
t
2
!#
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the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.40 Transform each equation.
sX(s) โ 5 = โ2X(s) + 5Y (s)
sY (s) โ 2 = โ6Y (s) โ 4X(s) +
10
s
Solve for X(s) and Y (s).
X(s) =
5s2 + 40s + 50
s3 + 8s2 + 32s
2s2 โ 6s + 20
s3 + 8s2 + 32s
The denominator roots are s = 0 and s = โ4 ยฑ 4j. Thus
Y (s) =
X(s) =
=
C1
4
s+4
+ C2
+ C3
s
(s + 4)2 + 42
(s + 4)2 + 42
55
4
55
s+4
25
+
+
2
2
16s 16 (s + 4) + 4
16 (s + 4)2 + 42
x(t) =
25 55 โ4t
55
+ e sin 4t + eโ4t cos 4t
16 16
16
Also,
Y (s) =
=
C1
4
s+4
+ C3
+ C2
2
2
s
(s + 4) + 4
(s + 4)2 + 42
5
33
4
11
s+4
โ
+
2
2
8s
8 (s + 4) + 4
8 (s + 4)2 + 42
y(t) =
5 33 โ4t
11
โ e sin 4t + eโ4t cos 4t
8
8
8
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2.41 Transforming both sides of the equation we obtain
s2 Y (s) โ sy(0) โ yฬ(0) + Y (s) =
1
s+1
which gives
Y (s) =
s2 y(0) + [y(0) + yฬ(0)] + yฬ(0) + 1
(s + 1) [sy(0) + yฬ(0)] + 1
=
(s + 1)(s2 + 1)
(s + 1)(s2 + 1)
This can be expanded as follows.
Y (s) = C1
1
s
1
+ C2 2
+ C3 2
s+1
s +1
s +1
We find the coefficients following the usual procedure and obtain C1 = 1/2, C2 = yฬ(0)+1/2,
and C3 = y(0) โ 1/2. Thus the solution is
1
1
1
y(t) = eโt + yฬ(0) +
sin t + y(0) โ
cos t
2
2
2
(continued on the next page)
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Problem 2.41 continued:
Because the initial values can be arbitrary, the general form of the solution is
1
y(t) = eโt + A1 sin t + A2 cos t
2
(1)
This form can be used to obtain a solution for cases where y(t) or yฬ(t) are specified at points
other than t = 0. For example, suppose we are given that y(0) = 5/2 and y(ฯ/2) = 3. Then
evaluation of equation (1) at t = 0 and at t = ฯ/2 gives
1
5
y(0) = + A2 =
2
2
ฯ
2
y
1
= eโฯ/2 + A1 = 3
2
The solution of these two equations is A1 = 3 โ eโฯ/2 /2 = 2.896 and A2 = 2, and the
solution of the differential equation is
1
y(t) = eโt + 2.896 sin t + 2 cos t
2
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2.42 (a) For nonzero initial conditions, the transform gives
s2 X(s) โ sx(0) + xฬ(0) + 4X(s) =
or
X(s) =
3
s2
s3 x(0) + s2 xฬ(0) + 3
C1 C2
2
s
= 2 +
+ C3 2
+ C4 2
2
2
s (s + 4)
s
s
s +4
s +4
The solution form is thus
x(t) = C1 t + C2 + C3 sin 2t + C4 cos 2t
which can be used even if the boundary conditions are not specified at t = 0.
(b) The form from part (a) satisfies the differential equation if C1 = 3/4 and C2 = 0.
From x(0) = 10, we obtain C4 = 10. From x(5) = 30, we obtain C3 = โ63.675. Thus
3
x(t) = t โ 63.675 sin 2t + 10 cos 2t
4
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2.43 The denominator roots are s = โ3 ยฑ 5j and s = ยฑ6j. Thus we can express X(s) as
follows.
30
X(s) =
[(s + 3)2 + 52 ] (s2 + 62 )
which can be expressed as the sum of terms that are proportional to entries 8 through 11
in Table 2.2.1.
X(s) = C1
s+3
6
s
5
+ C2
+ C3 2
+ C4 2
2
2
2
2
2
(s + 3) + 5
(s + 3) + 5
s +6
s + 62
(1)
We can obtain the coefficients by noting that X(s) can be written as
5C1 (s2 + 62 ) + C2 (s + 3)(s2 + 62 ) + 6C3 (s + 3)2 + 52 + C4 s (s + 3)2 + 52
X(s) =
[(s + 3)2 + 52 ] (s2 + 62 )
(2)
Comparing the numerators of equations (1) and (2), and collecting powers of s, we see that
(C2 + C4 )s3 + (5C1 + 3C2 + 6C3 + 6C4 )s2 + (36C2 + 36C3 + 34C4 )s
+180C1 + 108C2 + 204C3 = 30
or
C2 + C4 = 0
36C2 + 36C3 + 34C4 = 0
5C1 + 3C2 + 6C3 + 6C4 = 0
180C1 + 108C2 + 204C3 = 30
These are four equations in four unknowns. Note that the first equation gives C4 = โC2 .
Thus we can easily eliminate C4 from the equations and obtain a set of three equations in
three unknowns. The solution is C1 = 6/65, C2 = 9/65, and C3 = โ1/130, and C4 = โ9/65.
(continued on the next page)
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Problem 2.43 continued:
The inverse transform is
x(t) = C1 eโ3t sin 5t + C2 eโ3t cos 5t + C3 sin 6t + C2 cos 6t
=
6 โ3t
9
1
9
e sin 5t + eโ3t cos 5t โ
sin 6t โ
cos 6t
65
65
130
65
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2.44 Transform the equation.
5
(s2 + 12s + 40)X(s) = 3 2
s + 25
The characteristic roots are s = โ6 ยฑ 2j. Thus
15
(s2 + 25)(s2 + 12s + 40)
5
s
2
s+6
= C1 2
+ C2 2
+ C3
+ C4
s + 25
s + 25
(s + 6)2 + 4
(s + 6)2 + 4
X(s) =
or
X(s) =
4
19
4
1
5
s
2
s+6
โ
+
+
2
2
2
85 s + 25 85 s + 25 170 (s + 6) + 4 85 (s + 6)2 + 4
Thus
x(t) =
1
4
19 โ6t
4
sin 5t โ
cos 5t +
e sin 2t + eโ6t cos 2t
85
85
170
85
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2.45 From the text example, the form A sin(ฯt + ฯ) has the transform
A
s sin ฯ + ฯ cos ฯ
s2 + ฯ 2
For this problem, ฯ = 5. Comparing numerators gives
A (s sin ฯ + 5 cos ฯ) = 4s + 9
Thus
A sin ฯ = 4
5A cos ฯ = 9
With A > 0, ฯ is seen to be in the first quadrant.
ฯ = tanโ1
sin ฯ
4/A
20
= tanโ1
= tanโ1
= 1.148 rad
cos ฯ
9/5A
9
Because sin2 ฯ + cos2 ฯ = 1,
4
A
2
+
9
5A
2
=1
which gives A = 4.386. Thus
x(t) = 4.386 sin(5t + 1.148)
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2.46 Taking the transform of both sides of the equation and noting that both initial conditions are zero, we obtain
6
s2 X(s) + 6sX(s) + 34X(s) = 5 2
s + 62
Solve for X(s).
X(s) =
30
(s2 + 6s + 34)(s2 + 62 )
The inverse transform is
x(t) =
6 โ3t
9
1
9
e sin 5t + eโ3t cos 5t โ
sin 6t โ
cos 6t
65
65
130
65
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2.47 Transform the equation.
(s2 + 12s + 40)X(s) =
10
s
or, since the characteristic roots are s = โ6 ยฑ 2j,
X(s) =
10
s[(s + 6)2 + 22 ]
(1)
From the text example, the form Aeโat sin(ฯt + ฯ) has the transform
A
s sin ฯ + a sin ฯ + ฯ cos ฯ
(s + a)2 + ฯ 2
For this problem, a = 6 and ฯ = 2. Thus
X(s) =
or
X(s) =
C1
10
s sin ฯ + 6 sin ฯ + 2 cos ฯ
=
+ C2
2
2
s[(s + 6) + 2 ]
s
(s + 6)2 + 22
C1 (s2 + 12s + 40) + C2 s2 sin ฯ + 6C2 s sin ฯ + 2C2 s cos ฯ
s[(s + 6)2 + 22 ]
(2)
(continued on the next page)
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Problem 2.47 continued:
Collecting terms and comparing the numerators of equations (1) and (2), we have
(C1 + C2 sin ฯ)s2 + (12C1 + 6C2 sin ฯ + 2C2 cos ฯ)s + 40C1 = 10
Thus comparing terms, we see that C1 = 1/4 and
1
+ C2 sin ฯ = 0
4
3 + 6C2 sin ฯ + 2C2 cos ฯ = 0
So
C2 sin ฯ = โ
1
4
C2 cos ฯ = โ
3
4
Thus ฯ is in the third quadrant and
โ1/4
= 0.322 + ฯ = 3.463 rad
โ3/4
ฯ = tanโ1
Because sin2 ฯ + cos2 ฯ = 1,
1
4C2
2
+
3
4C2
2
=1
which gives C2 = 0.791. Thus
x(t) =
1
+ 0.791eโ6t sin(2t + 3.463)
4
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2.48 Transform the equation.
X(s) =
Thus
F (s) โ X(s) = F (s) โ
F (s)
s2 + 8s + 1
F (s)
s2 + 8s
=
F (s)
s2 + 8s + 1
s2 + 8s + 1
Because F (s) = 6/s2 ,
F (s) โ X(s) =
s+8
s2 + 8s 6
6
= 2
2
2
s + 8s + 1 s
s + 8s + 1 s
From the final value theorem,
s+8
6
fss โ xss = lim s[F (s) โ X(s)] = lim s 2
=8
sโ0
sโ0 s + 8s + 1 s
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2.49 The roots are s = โ2 and โ4. Thus
X(s) =
Let
1 โ eโ3s
(s + 2)(s + 4)
1
1
F (s) =
=
(s + 2)(s + 4)
2
1
1
โ
s+2 s+4
so
1 โ2t
e
โ eโ4t
2
From Property 6 of the Laplace transform,
f (t) =
x(t) =
1h
i
1 โ2t
e
โ eโ4t โ
eโ2(tโ3) โ eโ4(tโ3) us (t โ 3)
2
2
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2.50
2C
C
C
tus (t) โ
(t โ D)us (t โ D) + (t โ 2D)us (t โ 2D)
D
D
D
From Property 6 of the Laplace transform,
f (t) =
F (s) =
C
2C โDs
C โ2Ds
C
โDs
โ2Ds
โ
e
+
e
=
1
โ
2e
+
e
Ds2 Ds2
Ds2
Ds2
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2.51
C
C
tus (t) โ (t โ D)us (t โ D) โ Cus (t โ D)
D
D
From Property 6 of the Laplace transform,
f (t) =
F (s) =
C
C โDs C โDs
โ
e
โ e
2
Ds
Ds2
s
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2.52
f (t) = M us (t) โ 2M us (t โ T ) + M us (t โ 2T )
From Property 6,
F (s) =
M
2M โT s M โ2T s
โ
e
+
e
s
s
s
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2.53
P (t) = 3us (t) โ 3us (t โ 5)
From Property 6,
P (s) =
3 3 โ5s
โ e
s s
P (s)
3 1 โ eโ5s
3 1 โ eโ5s
X(s) =
=
=
4s + 1
s(4s + 1)
4 s(s + 1/4)
Let
Then
1
3
1
1
=3
โ
F (s) =
4 s(s + 1/4)
s s + 1/4
f (t) = 3 1 โ eโt/4
Since
X(s) = F (s) 1 โ eโ5s
we have
h
i
x(t) = f (t) โ f (t โ 5)us (t โ 5) = 3 1 โ eโt/4 โ 3 1 โ eโ(tโ5)/4 us (t โ 5)
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2.54 Let
f (t) = t +
Then
F (s) =
t3 2t5
+
3
15
1
2
16
s4 + 2s2 + 16
+
+
=
s2 s4
s6
s6
From the differential equation,
X(s) =
=
F (s)
s4 + 2s2 + 16
=
s+1
s6 (s + 1)
19
16 16 18 18 19 19
โ 5 + 4 โ 3 + 2 โ
+
6
s
s
s
s
s
s
s+1
Thus
2 5 2 4
t โ t + 3t3 โ 9t2 + 19t โ 19 + 19eโt
15
3
On a plot of this and the solution obtained from the lower-order approximation, the two
solutions are practically indistinguishable.
x(t) =
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2.55 From the derivative property of the Laplace transform, we know that
L[xฬ(t)] =
Z โ
xฬ(t)eโst dt = sX(s) โ x(0)
0
Therefore
Z โ
lim [sX(s)] = lim x(0) +
sโโ
sโโ
Z
= lim x(0) + lim
sโโ
sโโ
lim
โ0+
xฬ(t)eโst dt
0
xฬ(t)eโst dt
Z
+ lim
i
0 sโโ
โ0+
0
h
lim xฬ(t)eโst dt
The limits on and s can be interchanged because s is independent of t. Within the interval
[0, 0+], eโst = 1, and so
Z
lim [sX(s)] = x(0) + lim
sโโ
sโโ
lim
โ0+
xฬ(t) dt
0
Z
+ lim
โ0+
h
โst
lim xฬ(t)e
0 sโโ
dt
i
= x(0) + x(t)|t=0+
t=0 + 0 = x(0+)
This proves the theorem.
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2.56 From the derivative property of the Laplace transform, we know that
L[xฬ(t)] =
Z โ
xฬ(t)eโst dt = sX(s) โ x(0)
0
Therefore,
Z โ
lim [sX(s)] = lim x(0) + lim
sโ0
Z โ
= x(0) +
0
sโ0
sโ0
h
i
โst
xฬ(t)e
dt
0
lim xฬ(t)eโst dt = x(0) +
sโ0
Z โ
xฬ(t) dt
0
because s is independent of t and limsโ0 eโst = 1. Thus
“Z
lim [sX(s)] = x(0) + lim
T โโ
sโ0
T
#
xฬ(t) dt = x(0) + lim
T โโ
0
h
x(t)|t=T
t=0
i
= x(0) + lim x(T ) โ x(0) = lim x(T ) = lim x(t)
T โโ
T โโ
tโโ
This proves the theorem.
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2.57 Let
Z t
g(t) =
x(t) dt
0
Then
Z t
L
x(t) dt = L[g(t)] =
Z t
0
g(t)eโst dt
0
To use integration by parts we define u = g and dv = eโst dt, which give du = dg = x(t) dt
and v = โeโst /s. Thus
Z t
t=โ
โst
g(t)e
0
g(t)eโst
โ
dt =
โs t=0
g(0) 1
+
=0+
s
s
=
1
s
Z โ
Z โ โst
e
x(t)eโst dt =
0
Z
x(t) dt
+
t=0
0
โs
x(t) dt
g(0) X(s)
+
s
s
X(s)
s
This proves the property.
If there is an impulse in x(t) at t = 0, then g(0) equals the strength of the impulse. If
there is no impulse at t = 0, then g(0) = 0.
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2.58 a)
[r,p,k] = residue([8,5],[2,20,48])
The result is r = [10.7500, -6.7500], p = [-6.0000, -4.0000], and k = [ ]. The
solution is
x(t) = 10.75eโ6t โ 6.75eโ4t
b)
[r,p,k] = residue([4,13],[1,8,116])
The result is r = [2.0000 – 0.1500i, 2.0000 + 0.1500i], p = [-4.0000 + 10.0000i,
-4.0000 – 10.0000i], and k = [ ]. The solution is
x(t) = (2 โ 0.15j)e(โ4+10j)t + (2 + 0.15j)e(โ4โ10j)t
The solution is
x(t) = 2eโ4t (2 cos 10t + 0.15 sin 10t)
c)
[r,p,k] = residue([3,2],[1,10,0,0])
The result is r = [ -0.2800, 0.2800, 0.2000], p = [-10, 0, 0], and k = [ ]. The
solution is
x(t) = โ0.28eโ10t + 0.28 + 0.2t
(continued on the next page)
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Problem 2.58 continued:
d)
[r,p,k] = residue([1,0,1,6],[1,2,0,0,0,0])
The result is r = [-0.2500, 0.2500, 0.5000, -1.0000, 3.0000], p =[ -2, 0, 0, 0,
0], and k = [ ]. The solution is
1
1
x(t) = โ0.25eโ2t + 0.25 + 0.5t โ t2 + t3
2
2
e)
[r,p,k] = residue([4,3],[1,6,34,0])
The result is r = [-0.0441 – 0.3735i, -0.0441 + 0.3735i, 0.0882], p = [-3.0000
+ 5.0000i, -3.0000 – 5.0000i, 0], and k = [ ].The solution is
x(t) = (โ0.0441 โ 0.3735j)e(โ3+5j)t + (โ0.0441 + 0.3735j)e(โ3โ5j)t + 0.0882
The solution is
x(t) = 2eโ3t (โ0.0441 cos 5t + 0.3735 sin 5t) + 0.0882
(continued on the next page)
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Problem 2.58 continued:
f)
[r,p,k] = residue([5,3,7],[1,12,44,48])
The result is r = [21.1250 -18.7500 2.6250], p = [ -6, -4, -2], and k = [ ]. The
solution is
x(t) = 21.125eโ6t โ 18.75eโ4t + 2.625eโ2
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2.59 a)
[r,p,k] = residue(5,conv([1,8,16],[1,1]))
The result is r = [-0.5556, -1.6667, 0.5556], p = [-4.0000, -4.0000, -1.0000], k
= [ ]. The solution is
x(t) = โ0.5556eโ4t โ 1.6667teโ4t + 0.5556eโt
b)
[r,p,k] = residue([4,9],conv([1,6,34],[1,4,20]))
The result is r = [-0.1159 + 0.1073i, -0.1159 – 0.1073i, 0.1159 – 0.1052i, 0.1159
+ 0.1052i], p = -3.0000 + 5.0000i, -3.0000 – 5.0000i, -2.0000 + 4.0000i, -2.0000
– 4.0000i], and k = [ ]. The solution is
x(t) = (โ0.1159 + 0.1073j)e(โ3+5j)t + (โ0.1159 โ 0.1073j)e(โ3โ5j)t
+ (0.1159 โ 0.1052j)e(โ2+4j)t + (0.1159 + 0.1052j)e(โ2โ4j)t
The solution is
x(t) = 2eโ3t (โ0.1159 cos 5t โ 0.1073 sin 5t) + 2eโ2t (0.1159 cos 4t + 0.1052 sin 4t)
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2.60 a)
sys = tf(1,[3,21,30]);
step(sys)
b)
sys = tf(1,[5,20, 65]);
step(sys)
c)
sys = tf([3,2],[4,32,60]);
step(sys)
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2.61 a)
sys = tf(1,[3,21,30]);
impulse(sys)
b)
sys = tf(1,[5,20, 65]);
impulse(sys)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.62
sys = tf(5,[3,21,30]);
impulse(sys)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.63
sys = tf(5,[3,21,30]);
step(sys)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.64 a)
sys = tf(1,[3,21,30]);
t = [0:0.001:1.5];
f = 5*t;
[x,t] = lsim(sys,f,t);
plot(t,x)
b)
sys = tf(1,[5,20,65]);
t = [0:0.001:1.5];
f = 5*t;
[x,t] = lsim(sys,f,t);
plot(t,x)
c)
sys = tf([3,2],[4,32,60]);
t = [0:0.001:1.5];
f = 5*t;
[x,t] = lsim(sys,f,t);
plot(t,x)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.
2.65 a)
sys = tf(1,[3,21,30]);
t = [0:0.001:6];
f = 6*cos(3*t);
[x,t] = lsim(sys,f,t);
plot(t,x)
b)
sys = tf(1,[5,20,65]);
t = [0:0.001:6];
f = 6*cos(3*t);
[x,t] = lsim(sys,f,t);
plot(t,x)
c)
sys = tf([3,2],[4,32,60]);
t = [0:0.001:6];
f = 6*cos(3*t);
[x,t] = lsim(sys,f,t);
plot(t,x)
c 2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisherโs consent is unlawful.

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