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2.004 Dynamics and Control II
Spring 2008
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Massachusetts Institute of Technology
Department of Mechanical Engineering
2.004 Dynamics and Control II
Spring Term 2008
Lecture 151
Reading:
• Class Handout: Modeling Part 1: Energy and Power Flow in Linear Systems
Sec. 3.
• Class Handout: Modeling Part 2: Summary of One-Port Primitive Elements
•
1
Nise: Secs. 2.4 and 2.6.
Rotational Systems (continued)
Example 1
The diagram shows a mechanical tachometer that uses frictional drag plates to
create a torque proportional to angular velocity difference. The angular velocity
is indicated by the displacement θ of a torsional spring.
K
q
B
W
J
J
fr ic tio n a l d r a g p la te s
W
in
Find the transfer function relating the displacement of the indicator θ to the
input angular velocity Ωin
θ(s)
H(s) =
Ωin (s)
and show that for a constant input angular velocity the steady-state indicated
speed θss ∝ Ωi .
Solution: There are two distinct angular velocities, and the system graph is:
1
c D.Rowell 2008
copyright �
15–1
B
J
a c r o s s - v a r ia b le
s o u rc e
W
K
in
= 0
W
Using impedances, redraw the graph combining the inertia and the spring into a
single impedance Z2 :
Z
1
W
J
T
W
Z
in
W
Z
Z
2
1
=
1
2
B
=
1
1
J s
s
K
=
1 /J K
s /K + 1 /J s
= 0
Then
Z2
Ωin (s)
Z1 + Z 2
s/(Js2 + K)
=
Ωin (s)
1/B + s/(Js2 + K)
Bs
=
Ωin (s).
Js2 + Bs + K
ΩJ =
But the angular displacement θ(s) = ΩJ (s)/s so that
H(s) =
Js2
B
+ Bs + K
If the input velocity is a step function at t = 0, the Ωin (s) = Ω/s and the
steady-state indicated value will be
θss = lim θ(t) = lim sθ(s) = lim sH(s)
t→∞
s→0
s→0
Ω
B
= Ω
s
K
show that the indicated speed (θ) is proportional to the input angular velocity.
(This example is repeated using mesh equations below.)
15–2
2
Transfer Function Generation Using Mesh/Loop Equations
This method is useful when the system contains an across variable source.
We will develop the method using the following mechanical system as an example.
K
K
V (t)
K
1
B
m
2
1
m
B
1
v
K
o
3
B
1
B
V (t)
2
B
2
m
1
1
v
3
B
2
m
o
2
2
The input is a velocity source V (t), and the output is the velocity of the mass m2 .
Step 1: (This step is optional.) Do some simple impedance combinations to reduce the
complexity, being careful not to eliminate any nodes or branches that specify the output
variable.
Z
Z
Z
5
Z
1
Z
V (t)
v
3
Z
2
o
Z
4
K
=
=
5
B
1
m
K
2
s
1
3
B
s
1
m
1
1
=
4
Z
1
1
B
3
Z
v = 0
2
=
Z
=
s
1
1
2
s
2
Step 2: Define a set of closed loops/meshes, making sure that each branch is covered by
at least one loop.
Z
Z
5
3
1
V (t)
Step 3:
Z
1
Z
2
2
v
3
Z
4
v = 0
Write a loop equation for each loop:
vZ1 + vZ2 − V = 0
vZ3 + vZ4 − vZ2 = 0
vZ5 − vZ3 − vZ1 = 0
15–3
o
Step 4: At this point we assume that each loop has a continuous through variable with
it, that is loop n has Fn associated with it. When a branch is contacted by more
that one loop, the branch through variable is the sum of all of the contacting loop
through variables. We use impedance relationships and substitute loop through variable
expressions for the across variables:
Z1 (F1 − F3 ) + Z2 (F1 − F2 ) − V = 0
Z3 (F2 − F3 ) + Z4 F2 − Z2 (F1 − F2 ) = 0
Z5 F3 − Z3 (F2 − F3 ) − Z1 (F1 − F3 ) = 0
Step 5: Rewrite these equations collecting terms in the loop through variables to create a
set of linear equations:
(Z1 + Z2 )F1 − Z2 F2 − Z1 F3 = V
−Z2 F1 + (Z2 + Z3 + Z4 )F2 − Z3 F3 = 0
−Z1 F1 + Z3 F2 + (Z1 + Z3 + Z5 )F3 = 0
It may be useful to express these equations in matrix form:
⎡
⎤⎡
⎤ ⎡
⎤
(Z1 + Z2 )
−Z2
−Z1
F1
V
⎣
−Z2
⎦ ⎣ F2 ⎦
=
⎣
0
⎦
(Z2 + Z3 + Z4 )
−Z3
−Z1
Z3
(Z1 + Z3 + Z5 )
F3
0
Step 6: Identify how the output variable is related to the loop through variables. In this
case
vout = F2 Z4
Step 7: Solve the set of linear equations for the variable(s) identified in Step 6, in this
case F2 , using any method, and substitute in the output equation (Step 6).
(we will not do this step here).
Example 2
Repeat the mechanical tachometer problem of Example 1, this time using mesh
equations.
B
K
q
1
B
W
J
2
J
W
fr ic tio n a l d r a g p la te s
W
J
in
15–4
K
in
W
= 0
With the system graph and two loops defined as above, the loop equations are:
ΩB + ΩJ − Ωin = 0
ΩK − ΩJ = 0
Substitute for the across variables (Ω):
ZB T1 + ZK (T1 − T2 ) = Ωin
ZJ T2 − ZK (T1 − T2 ) = 0
Rearrange:
(ZB + ZK )T1 − ZK T2 = Ωin
−ZK (T1 + (ZJ + ZK )T2 = 0
and in matrix form
�
ZB + Z K
−ZK
−ZK
ZJ + Z K
��
T1
T2
�
�
=
Ωin
0
�
The output variable is ΩJ = ZJ T2 , therefore use Cramer’s rule to solve for T2 .
�
�
� ZB + ZK Ωin �
�
�
� −ZK
0 �
ZK Ωin
�=
T2 = ��
ZB ZJ + Z B ZK + Z K ZJ
−ZK ��
� ZB + Z K
� −ZK
ZJ + Z K �
and since ΩJ = ZJ T2 , ZJ = 1/(Js), ZK = s/K, and ZB = 1/B
ΩJ =
ZJ ZK Ωin
Bs
= 2
Ωin
ZB ZJ + Z B ZK + Z K ZJ
Js + Bs + K
and as before, θ(s) = ΩJ /s giving
H(s) =
3
θ(s)
B
= 2
Ωin (s)
Js + Bs + K
Transfer Function Generation Using Node Equations
This method is useful when the system contains a through variable source.
We will demonstrate the method using the following graph:
15–5
Z
2
(a )
(b )
Z
F
1
Z
4
s
Z
3
(c )
Step 1: Write node equations for each node (except the reference node) - let f be a
generalized through variable.
Fs − fZ1 − fZ2 = 0
f Z2 − f Z4 − f Z3 = 0
node(a)
node(b)
Step 2: Express the branch through variables in terms of admittances (Y = 1/Z) and
across variables
(va − vb )Y2 + (va − vc )Y1 = Fs
(va − vb )Y2 + (vb − vc )Y4 − (vb − vc )Y3 = 0
Step 3: Collect terms in the across variables, and form a set of linear equations (note that
node (c) is the reference node, therefore vc = 0).
(Y1 + Y2 )va − Y2 vb = Fs
Y2 va − (Y2 + Y3 + Y4 )vb = 0
Step 4: Identify the output variable in terms of nodal across variables and solve the linear
equations.
Example 3
Find H(s) = Xm2 (s)/F (s) for the following system:
F
v
v
1
m
B
1
K
2
m
(a )
s
(b )
B
K
2
m
v = 0
m
1
(c )
The graph has been simplified by combining parallel elements:
15–6
2
F
s
(a )
=
Y
2
=
Y
3
=
(b )
1
Z
Z
3
Z
1
2
1
Y
= m
Z
1
1
= m
Z
2
Z
1
3
s
1
2
= B +
s
K
s
(c )
At the nodes (a) and (b):
−FB − Fm1 − Fs = 0
FB − Fm2 + Fs = 0
Substitute for through variables
−(va − vb )Y3 − va Y1 = Fs
(va − vb )Y3 − vb Y1 = −Fs
Reorganize and write in matrix form:
�
��
� �
�
va
Fs
−(Y3 + Y1 )
Y3
=
Y3
−(Y2 + Y3 )
vb
−F (s)
Recognize that the output xm2 (s) = vb (s)/s, use Cramer’s rule to solve for vb :
�
�
� −(Y3 + Y1 ) F (s) �
�
�
�
Y3
−Fs ) �
Y1 F s
�=
vb = ��
�
Y1 Y2 + Y3 (Y1 + Y2 )
Y3
� −(Y3 + Y1 )
�
�
�
−(Y2 + Y3 )
Y3
Y1 = m1 s, Y2 = m2 s, and Y3 = B + K/s, giving
vm2 (s) = vb (s) =
m1 m2
s2
m1 s
Fs (s)
+ B(m1 + m2 )s + K(m1 + m2 )
and
H(s) =
xm2 (s)
1 vm2 (s)
m1
=
=
Fs (s)
s Fs (s)
m1 m2 s2 + B(m1 + m2 )s + K(m1 + m2 )
15–7