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Lecture 14
- 1. Lecture 14
Second Order Transient Response
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 2. Damped Harmonic Motion
Mechanical oscillator
2
d x dx
m 2 + b + kx = f m (t )
dt dt
Electrical
El t i l oscillator
ill t
d 2q
dq 1
L 2 +R + q = v s (t )
dt dt C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 3. The Series RLC Circuit and its
Mechanical Analog
• Drive voltage vS(t) Applied force f(t)
• Charge q Displacement x
• Current i Velocity v
• Inductance L Mass m
• Inductive Energy U=(1/2)Li2 Inertial Energy (1/2)mv2
• Capacitance C Spring constant 1/k
• Capac. Energy U=(1/2)(1/C)q2 Spring energy U=(1/2)kx2
• Resistance R Dampening constant b
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 4. Second –Order Circuits
di (t )
t
+ Ri (t ) + ∫ i(t )dt + v (0) = v (t )
1
L C s
dt C 0
Differentiating with respect to time:
2
d i (t ) di (t ) 1 dv s (t )
L +R + i (t ) =
dt 2
dt C dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 5. Second –Order Circuits
d 2 i (t ) R di (t ) 1 1 dv s (t )
+ + i (t ) =
dt 2
L dt LC L dt
R Dampening
Define: α=
2L coefficient
1
ω0 = Undamped resonant
LC frequency
1 dv s (t )
f (t ) = Forcing function
L dt
2
d i (t ) di (t )
+ 2α + ω 0 i (t ) = f (t )
2
dt 2
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 6. Solution of the Second-Order Equation
Particular solution
d i (t )
2
di (t )
+ 2α + ω 0 i (t ) = f (t )
2
dt 2
dt
Complementary solution
d i (t )
2
di (t )
+ 2α + ω 0 i (t ) = 0
2
dt 2
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 7. Solution of the Complementary
Equation
d i (t )
2
di (t )
+ 2α + ω 2 i (t ) = 0 0
dt 2
dt
Try iC (t ) = Ke st :
s 2 Ke st + 2αsKe st + ω0 Ke st = 0
2
Factoring :
( s 2 + 2αs + ω 0 ) Ke stt = 0
2
Characteristic equation :
s 2 + 2αs + ω 0 = 0
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 8. Solution of the Complementary
Equation
Qaudratic Equation : ax 2 + bx + c = 0
− b ± b − 4ac 2
x=
2a
Characteristic Equation : s 2 + 2αs + ω0
2
− 2α ± (2α ) 2
− 4ω0
2
s= = −α ± α 2 − ω0
2
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 9. Solution of the Complementary
Equation
Roots of the characteristic equation:
s1 = −α + α − ω 2 2
0
s2 = −α − α − ω 2 2
0
α
ζ = Dampening ratio
ω0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 10. 1. Overdamped case (ζ > 1). If ζ > 1 (or
equivalently, if α > ω0), the roots of the
characteristic equation are real and distinct.
Then the complementary solution is:
ic (t ) = K1e s1t
+ K 2e s2t
In this case, we say that the circuit is
overdamped.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 11. 2. Critically damped case (ζ = 1). If ζ = 1 (or
equivalently, if α = ω0 ), the roots are real
and equal. Then the complementary solution
is
ic (t ) = K1e s1t
+ K 2te s1t
In this case, we say that the circuit is
critically damped.
iti ll d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 12. 3. Underdamped case (ζ < 1). Finally, if ζ < 1
(or equivalently, if α < ω0) the roots are
equivalently ),
complex. (By the term complex, we mean that
the roots involve the square root of –1.) In other
1)
words, the roots are of the form:
s1 = −α + jωn a n d s2 = −α − jωn
in which j is the square root of -1 and the
natural frequency is given by:
f q y g y
ωn = ω − α 2
0
2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 13. I electrical engineering, we use j rather than i
In l t i l i i th th
to stand for square root of -1,because we use i
for
f current. F complex roots, the
t For l t th
complementary solution is of the form:
ic (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin (ω n t )
In this case, we say that the circuit is
underdamped.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 14. Analysis of a Second-Order Circuit
with a DC Source
Apply KVL around the loop:
di (t )
L + Ri (t ) + vC (t ) = VS
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 15. Analysis of a Second-Order Circuit
with a DC Source
di (t )
L + Ri (t ) + vC (t ) = VS
dt
dvC (t )
Substitute i (t ) = C
dt
d 2 vC (t ) dvC (t )
LC + RC + vC (t ) = Vs
dt 2
dt
Divide by
Di id b LC :
d 2 vC (t ) R dv(t ) C 1 Vs
+ + vC (t ) =
d2
dt L dt LC LC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 16. Analysis of a Second-Order Circuit
with a DC Source
To find the particular solution, we consider the steady
solution
state. For DC steady state, we can replace the inductors
by a short circuit and the capacitor by an open circuit:
vCp (t ) = Vs = 10V
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 17. Analysis of a Second-Order Circuit
with a DC Source
To fi d the homogeneous solution, we consider three cases;
find h h l i id h
R=300Ω, 200Ω and 100Ω. These values will correspond to
over-damped, critically damped and under-damped cases.
p , y p p
1 1
ω0 = = = 10 4 rads / sec
LC (10mH )(1μF )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 18. Analysis of a Second-Order Circuit
with a DC Source
Case I : R = 300Ω
R 300Ω α 1.5 x10 4
α= = = 1.5 x10 4 ζ = = = 1.5 → Overdamped
2 L 2(10mH ) ω0 10 4
s1 = −α + α 2 − ω0 = −1.5 x10 4 − (1.5 x10 4 ) 2 − (10 4 ) 2 = −2.618 x10 4
2
s 2 = −α − α 2 − ω0 = −0.382 x10 4
2
Adding the particular and complementary solutions :
vC (t ) = 10 + K1e s1t + K 2 e s2t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 19. Analysis of a Second-Order Circuit
with a DC Source
To find K1 and K2 we use the initial conditions:
vc (t ) = 10 + K1e s1t + K 2 e s2t
vc (0) = 0 = 10 + K1e 0 + K 2 e 0 = 10 + K1 + K 2
i (0) = 0
dvc (t ) dvc (0)
i (t ) = C i (0) = 0 → =0
dt dt
dvc (t )
= s1 K1e s1t + s 2 K 2 e s2t
dt
dvc (0)
= 0 = s1 K1e 0 + s 2 K 2 e 0 = s1 K1 + s 2 K 2
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 20. Analysis of a Second-Order Circuit
10 + K + K = 0
1
with a DC Source
2
s1 K1 + s 2 K 2 = 0
K1 + K 2 = −10
s1 K1 + s 2 K 2 = 0
− 10 1
0 s2 (−10)( s 2 ) − (0)(1) − 10 s 2 3.82 x10 4
K1 = = = = = 1.71
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.3820 x10 4 − (−2.618 x10 4 )
s1 s2
1 − 10
s1 0 (1)(0) − ( s1 )(−10) 10 s1 − 26.18 x10 4
K2 = = = = = −11.7
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.3820 x10 − (−2.618 x10 )
4 4
s1 s2
vc (t ) = 10 + 1.708e s1t − 11.708e s2t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 21. Analysis of a Second-Order Circuit
with a DC Source
Case II : R = 200Ω
R 200Ω α 10 4
α= = = 10 4 ζ = = 4 = 1 → Critically damped
2 L 2(10mH ) ω0 10
s1 = s 2 = −α + α 2 − ω 0 = −α = −10 4
2
Adding the particular and complementary solutions :
vC (t ) = 10 + K1e s1t + K 2 te s1t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 22. Analysis of a Second-Order Circuit
with a DC Source
To find K1 and K2 we use the initial conditions:
vc (t ) = 10 + K1e s1t + K 2 te s1t
vc (0) = 0 = 10 + K1e 0 + K 2 (0)e 0 = 10 + K1
i (0) = 0
dvc (t ) dvc (0)
i (t ) = C i (0) = 0 → =0
dt dt
dvc (t )
= s1 K1e s1t + s1 K 2 te s1t + K 2 e s1t
dt
dvc (0)
= 0 = s1 K1e 0 + s1 K 2 (0)e 0 + K 2 e 0 = s1 K1 + K 2
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 23. Analysis of a Second-Order Circuit
with a DC Source
10 + K1 = 0 → K1 = −10
s1 K1 + K 2 = 0 → K 2 = − s1 K1 = 10s1 = 10(−10 4 ) = −10 5
vc (t ) = 10 − 10e s1t − 10 5 te s2t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 24. Analysis of a Second-Order Circuit
with a DC Source
Case III : R = 100Ω
R 100Ω α 0.5 x10 4
α= = = 0.5 x10 4 ζ = = = 0.5 → Underdamped
2 L 2(10mH ) ω0 10 4
ω n = ω0 − α 2 = (10 4 ) 2 − (0.5 x10 4 ) 2 = 8660
2
Adding the particular and complementary solutions :
vC (t ) = 10 + K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
dvC (t )
= −αK1e −αt cos(ω n t ) − K1e −αt ω n sin(ω n t ) − αK 2 e −αt sin(ω n t ) + K 2 e −αt ω n cos(ω n t )
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 25. Analysis of a Second-Order Circuit
with a DC Source
To find K1 and K2 we use the initial conditions:
v c ( 0) = 0
dvc (0)
=0
dt
10 + K1 = 0 → K1 = −10
α 5000
− α K1 + ω n K 2 = 0 → K 2 = K1 = (−10) = −5.774
ωn 8660
vc (t ) = 10 − 10e −αt cos(ω n t ) − 5.774e −αt sin(ω n t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 26. Overdamped
O d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 27. Critically d
C iti ll damped
d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 28. Underdamped
U d d d
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 30. Normalized Step Response of a Second
Order System
u (t ) = 0 t < 0
u (t ) = 1 t > 0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 31. Normalized Step Response of a Second
Order System
Example: Closing a switch at t=0 to apply a DC voltage A
t 0
v(t ) = Au (t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 32. Normalized Step Response of a Second
Order System
What value of ζ
should be used to get
to the steady state
position quickly
without overshooting?
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 33. Circuits with Parallel L and C
t
V dq dv
∫
1
iR = iL = vdt + iL (0) ic = =C
R L dt dt
0
We can replace the circuit with it’s Norton equivalent
and then analyze the circuit by writing KCL at the top
y y g p
node:
t
dv(t ) 1
∫
1
C + v(t ) + v(t )dt + i L (0) = in (t )
dt R L
0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 34. Circuits with Parallel L and C
t
d (t ) 1
dv
∫
1
C + v(t ) + v(t )dt + i L (0) = in (t )
dt R L
0
differentiating :
d 2 v(t ) 1 dv(t ) 1 din (t )
C + + v(t ) =
dt R dt L dt
d 2 v(t ) 1 dv(t ) 1 1 din (t )
+ + v(t ) =
dt RC dt LC C dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 35. Circuits with Parallel L and C
d 2 v(t ) 1 dv(t ) 1 1 din (t )
+ + v(t ) =
dt RC dt LC C dt
1
Dampening coefficient α =
2 RC
1
Undamped resonant frequency ω 0 =
LC
1 din (t )
Forcing function f (t ) =
C dt
d 2 v(t ) dv(t )
+ 2α + ω0 v(t ) = f (t )
2
dt dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 36. Circuits with Parallel L and C
d 2 v(t ) dv(t )
+ 2α + ω0 v(t ) = f (t )
2
dt dt
This is the same equation as we found for the series LC
circuit with the following changes for α:
1
Parallel circuit α =
2 RC
R
Series circuit α =
2L
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 37. Example Parallel LC Circuit: RF-ID
Tag
L
C
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 38. Circuits with Parallel L and C
Find v(t) for R=25Ω, 50Ω and 250Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 39. Circuits with Parallel L and C
Find v(t) for R=25Ω, 50Ω and 250Ω
1 1
ω0 = = = 1x10 5 rad / sec
LC (1x10 −3 H )(1x10 −7 F )
1
α=
2 RC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 40. Circuits with Parallel L and C
Vp(t) = 0
To find the particular solution vP(t) (steady state
response) replace C with an open circuit and L with a
short circuit
circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 41. Circuits with Parallel L and C
Case I : R = 25Ω
1 1 5 −1 α 2 x10 5 s −1
α= = = 2 x10 s ζ = = = 2 → Overdamped
2 RC 2(25Ω)(1x10 −7 F ) ω0 10 5 rad / s
s1 = −α + α 2 − ω 0 = −2 x10 5 + (2 x10 5 ) 2 − (10 5 ) 2 = −0.268 x10 5
2
s 2 = −α − α 2 − ω 0 = −2 x10 5 − (2 x10 5 ) 2 − (10 5 ) 2 = −3.73 x10 5
2
Adding the particular and complementary solutions :
vC (t ) = K1e s1t + K 2 e s2t
v P (t ) = 0
vT (t ) = K1e s1t + K 2 e s2t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 42. Circuits with Parallel L and C
Initial conditions :
v (0 −) = 0 → v (0 + ) = 0 Since the voltage across a capacitor
and the current through an inductor
i L (0 −) = 0 → i L (0 + ) = 0 cannot change instantaneously
KCL at t = 0 +
v (0 + ) dv(0+)
0.1 = + i L (0 + ) + C
R dt
dv(0+ ) dv(0+ ) 0.1 0.1
0.1 = C → = = = 10 6 v / s
dt dt C 0.1x10 −6
v(0+ ) = K1e 0 + K 2 e 0 → K1 + K 2 = 0
d (0 + )
dv
= K1s1e 0 + K 2 s 2 e 0 = K1s1 + K 2 s 2 = 10 6
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 43. Circuits with Parallel L and C
K1 + K2 = 0
s1 K1 + s 2 K 2 = 10 6
0 1
10 6 s2 (0)( s 2 ) − (10 6 )(1) − 10 6 − 10 6
K1 = = = = = 2.89
1 1 (1)( s 2 ) − ( s1 )(1) s 2 − s1 − 0.268 x10 − (−3.73 x10 )
5 5
s1 s2
K 2 = − K1 = −2.89
− 0.268 x105 t −3.73 x105 t − 0.268 x105 t −3.73 x105 t
v(t ) = 2.89e − 2.89e = 2.89(e −e )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 44. Circuits with Parallel L and C
Case II : R = 50Ω
1 1 5 −1 α 1x10 5 s −1
α= = = 1x10 s ζ = = = 1 → Critically damped
2 RC 2(50Ω)(1x10 − 7 F ) ω0 10 5 rad / s
s1 = −α = −1x10 5
Adding the particular and complementary solutions :
vC (t ) = K1e s1t + K 2 te s1t
v P (t ) = 0
vT (t ) = K1e s1t + K 2 te s1t
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 45. Circuits with Parallel L and C
v(t ) = K1e s1t + K 2te s1t
v(0) = K1e0 + K 2 (0)e0 = K1 = 0
v(t ) = K 2te s1t
dv(t )
= K 2e s1t + s1K 2te s1t
dt
dv(0 + )
= K 2e0 = 106 → K 2 = 106
dt
6 −105 t
v(t ) = 10 te
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 46. Circuits with Parallel L and C
Case III : R = 250Ω
1 1 5 −1 α 0.2 x10 5 s −1
α= = = 0.2 x10 s ζ = = = 0.2 → Underdamped
2 RC 2(250Ω)(1x10 −7 F ) ω0 10 5 rad / s
ω n = ω 0 − α 2 = (1x10 5 ) 2 − (0.2 x10 2 ) 2 = 0.98 x10 5
2
s1 = −α + jω n = −0.2 x10 5 + j 0.98 x10 5
s 2 = −α − jω n = −0.2 x10 5 − j 0.98 x10 5
Adding the particular and complementary solutions :
vC (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
v P (t ) = 0
vT (t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
( i (
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
- 47. Circuits with Parallel L and C
v(t ) = K1e −αt cos(ω n t ) + K 2 e −αt sin(ω n t )
dv(t )
= −αK1e −αt cos(ω n t ) − ω n K1e −αt sin(ω n t ) − αK 2 e −αt sin(ω n t ) + ω n K 2 e −αt cos(ω n t )
dt
v(0) = 0 = K1e −αt → K1 = 0
dv(0) 10 6
= 10 = −αK1 + ω n K 2 = ω n K 2 → K 2 =
6
= 10.2
dt 0.98 x10 5
− 0.2 x105 t
v(t ) = (10.2)e sin(0.98 x10 5 t )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.