Lecture26_25thFeb2009.ppt

Lecture: 26
SECOND ORDER SYSTEM

SECOND ORDER SYSTEM
 A linear time invariant second-order system is
described by a second-order differential equation:
Other Names:
Quadratic Lag
Second Order Diff. Equations
cx
dt
dx
b
dt
x
d
a
t
f 

 2
2
)
( …………..(A)

EXAMPLE OF
SECOND ORDER SYSTEM
 Classical example from mechanics
 A block of mass W resting on a horizontal, frictionless
table is attached to a linear spring.
 A viscous damper (dashpot) is also attached to the
block.
 Assume the system is free to oscillate horizontally
under the influence of a forcing function F(t).
 The origin of the coordinate system is taken as the
right edge of block when the spring is in the relaxed
or un-stretched condition.
 At time zero, block is assumed to be at rest at this
origin.
 Positive directions for force and displacement are
indicated by the arrows in Fig. 1.
SYSTEM EXPLANATION:

EXAMPLE OF
SECOND ORDER SYSTEM
SYSTEM EXPLANATION:
Fig. 1: Damped Vibrator

EXAMPLE OF
SECOND ORDER SYSTEM
 Consider the block at some instant when it is to the
right of Y = 0 and when it is moving toward the right
(positive direction).
 Under these conditions, the position Y and the
velocity dY/dt are both positive.
 At this particular instant, the following forces are
acting on the block:
1. The force exerted by the spring (toward the left) of
-KY where K is a positive constant, called Hooke’s
constant.
2. The viscous friction force (acting to the left) of
-C dY/dt, where C is a positive constant called the
damping coefficient.
3. The external force F(t) (acting toward the right).
SYSTEM EXPLANATION:

EXAMPLE OF
SECOND ORDER SYSTEM
 Consider the block at some instant when it is to the
right of Y = 0 and when it is moving toward the right
(positive direction).
 Under these conditions, the position Y and the velocity
dY/dt are both positive.
 At this particular instant, the following forces are acting
on the block:
1. The force exerted by the spring (toward the left) of -
KY where K is a positive constant, called Hooke’s
constant.
2. The viscous friction force (acting to the left) of -C
dY/dt, where C is a positive constant called the
damping coefficient.
3. The external force F(t) (acting toward the right).
FORCES ACTING ON THE SYSTEM:

EXAMPLE OF
SECOND ORDER SYSTEM
 Newton’s law of motion, which states that
 the sum of all forces acting on the mass is equal to the
rate of change of momentum (mass X acceleration),
 Takes the form:
 Rearrangements gives
 Where
 W=mass of block, lbm
 Gc=32.2(lbm)(ft)/(lbf)(sec2)
 C = viscous damping coefficient, lbf/(ft/sec)
 K = Hooke’s constant, lbf/ft
 F(t)= driving force, a function of time, lbf
MOMENTUM BALANCE:
)
(
2
2
t
F
dt
dY
C
KY
dt
y
d
g
W
c



 …………………(1)
)
(
2
2
t
F
KY
dt
dY
C
dt
y
d
g
W
c


 ………………(2)

EXAMPLE OF
SECOND ORDER SYSTEM
 Dividing Eq. (2) by K gives
 For convenience, this is written as
 Where
MOMENTUM BALANCE:
K
t
F
Y
dt
dY
K
C
dt
y
d
K
g
W
c
)
(
2
2


 ………………(3)
)
(
2
2
2
2
t
X
Y
dt
dY
dt
y
d


 
 ………………(4)
K
g
W
c

2
 …….……………….…………….(5)
K
C


2 ……...…………….…………….(6)
K
t
F
t
X
)
(
)
(  ……...…………….…………….(7)

EXAMPLE OF
SECOND ORDER SYSTEM
 Solving for τ and ζ from Eq. (5) and (6) gives:
K
g
W
c

 (sec) …………(8)
WK
C
gc
4
2

 (Dimension less)…………(9)

TRANSFER FUNCTION
 The second order equation is:
 The above equation is written in a standard form that
is widely used in control theory.
 If the block is motionless (dY/dt=0) and located at
its rest position (Y=0) before the forcing function is
applied, the Laplace transform of Eq(4) becomes:
TRANSFORM OF EQ(4)
)
(
2
2
2
2
t
X
Y
dt
dY
dt
y
d


 
 ………………(4)
)
(
)
(
)
(
2
)
(
2
2
s
X
s
Y
s
sY
s
Y
s 

 
 …………(10)
1
2
1
)
(
)
(
2
2



s
s
s
X
s
Y


…………(11)

Lecture26_25thFeb2009.ppt

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Lecture26_25thFeb2009.ppt