PROBLEM ME3015 SET #3 (25 points). Solutions
1. (6 points) Assume that the system is initially in equilibrium. The gravitational force of mass
causes a static deflection of the spring. The external force u applies as input and the
displacement y is the output of the system.
Find the state-space representation of the spring-mass-pulley shown below. Then convert the
state-space representation to the corresponding transfer function from u to y.
Moment of Inertia
J
R
Neutral length
m
k
without MASS
u
mg
(Origin: Equilibrium position)
y
Solution
Define the tension acting in the wire as T=T + k where k=mg. Since the displacement y is defined
from the equilibrium position, the effect of gravity is not included in the dynamic equations.
T
m
=
&&
y
m
u
&
J&
TR
=
kR 2
The equations of motion for the mass and the pulley are given as
m&& = T + u
y
&
J& = TR kR 2
Since y = R , the equation of motion for the pulley can be written as
J&& = TR 2 kR 2 y
y
By eliminating T from the equations, we have
J&& = (u m&&) R 2 kR 2 y .
y
y
Thus,
&& +
y
kR 2
R2
u
y=
J + mR 2
J + mR 2
Define state variables as
x1 = y
&
x2 = y
The output y is given by y = x1 . A state-space representation is given by
0
&
x1
kR 2
=
x
&2 J + mR 2
x
y = [1 0] 1
x2
1 x
0
1 + R 2 u
0 x2
J + mR 2
ANS
Transfer function
G(s) =
Y (s)
R2
=
(ANS)
U ( s) ( + J mR 2 ) s 2 + kR 2
2. (6 points)
1
0 1 0
G ( s ) = C ( sI A) B = [1 0] sI
a b 1
1
1
s
1 0
= [1 0]
a s b 1
s b 1 0
1
[1 0]
=
a
s 1
s ( s b) a
1
1
[1 0]
s
s bs a
1
=2
s bs a
=
2
ANS
3. (6 points) Obtain the mathematical model (differential equation) of the circuit shown below:
L
R3
R1
C
R2
e(t)
i1
From the circuit diagram we obtain
di1
+ R1i1 + R2 (i1 i2 ) = e
dt
ANS
1
R3i2 +
i2 dt + R2 (i2 i1 ) = 0
C2
L
i2
4. (7 points) Find the transfer function from Ei(s) to EO(s) for the electrical system shown below:
C
R1
R3
ei(t)
R4
R2
eo(t)
Solution
i4
i1
R1
i3
A
i2
ei(t)
C
i5
R3
R2
R4
eo(t)
e A : voltage at A from the GND.
i1 =
ei e A
R1
i2 =
eA
R2
i3 =
e A eO
R3
i4 = C
i5 =
d
(e A eO )
dt
eO
R4
Since i1 = i2 + i5 we have
ei e A e A eO
.
=
+
R1
R2 R4
Also since i4 + i3 = i5 we have
(1)
C
d
e e
e
(e A eO ) + A O = O
R3
R4
dt
(2)
From (1),
ei 1
e
1
= + e A + O
R R
R1 1
R4
2
Taking Laplace transformation of this equation, we obtain
Ei 1
EO
1
= +
R R EA + R
R1 1
2
4
(3)
Also taking Laplace transformation of (2) we obtain
Cs ( E A EO ) +
E A EO EO
=
R3
R4
From the above,
EA =
R3 R4Cs + R3 + R4
EO
R4 ( R3Cs + 1)
By substituting (4) into (3),
Ei R1 + R2 R3 R4Cs + R3 + R4
EO
=
R R R ( R Cs + 1) EO + R
R1 1 2
4
3
4
Therefore,
EO
R2 R4 ( R3Cs + 1)
ANS
=
Ei [R1 R2 R3 + ( R1 + R2 ) R3 R4 ]Cs + R1 R2 + ( R1 + R2 )( R3 + R4 )