MIT OpenCourseWare
http://ocw.mit.edu
2.004 Dynamics and Control II
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Massachusetts Institute of Technology
Department of Mechanical Engineering
2.004 Dynamics and Control II
Spring Term 2008
Lecture 161
Reading:
• Class Handout - Modeling Part 3: Two-Port Energy Transducing Elements
1 Arrow Conventions on Ideal Sources
To this point we have simply told you to draw the arrows on source elements
(a) In the direction of the assumed across-variable drop for across-variable sources (voltage
and velocities),
v
V
a
in d ic a te s th a t w e
a s s u m e va > v b
s
v
b
(b) in the direction of the assumed through-variable direction for through-variable sources
(currents and forces)
a
F
s
in d ic
flo w
o r
th a t
in th
m e c
a te s th a t w e a s s u m e th e c u rre n t
in th e d ir e c tio n o f th e a r r o w
th e s o u rc e a c ts to m o v e n o d e a
e p o s itiv e r e fe r e n c e d ir e c tio n in a
h a n ic a l s y s te m
b
When we draw a branch on a graph we make the assumption that P > 0, that is that
power is flowing into the element. There are in fact two arrows implicit on each branch:
one representing the assumed across-variable drop, and a second representing the assumed
through-variable direction. If P > 0 (power is flowing into the element), the two arrows are
in the same direction. For example, consider a capacitor
1
c D.Rowell 2008
copyright �
16–1
v
a
a
a
iC
d ro p
C
v
v
C
v o lta g e d r o p
a n d c u rre n t
in th e s a m e
d ir e c tio n
v
b
b
iC
v o lta g e d r o p
a n d c u rre n t
in th e o p p o s ite
d ir e c tio n
C
b
P = vc ic > 0 when either
P = vc ic < 0 when either
1. vc > 0 and ic > 0,or
1. vc > 0 and ic < 0,or
2. vc < 0 and ic < 0.
2. vc < 0 and ic > 0.
On
passive elements, with the assumption P > 0 we can combine the two arrows into one because
they point in the same direction.
For sources, the assumption is the opposite - that is we assume that the source is supplying
energy/power to the system, P < 0.
There are two arrows (pointing in opposite directions) associated with any source –
one for the across-variable, the second for the through-variable.
th r o u g h - v a r ia b le
d ir e c tio n
i
Is
+
S y s te m
iR
d ro p
R
d ro p
-
a c r o s s - v a r ia b le
d ro p
In modeling sources, we choose to show the arrow that will normally be used to solve the
system;
(a) An across-variable source will usually be included in a loop-equation, therefore the
convention is to show the arrow associated with the across-variable drop.
Z
1
L o o p 1 : v
Fs
V
1
Z
2
L o o p 2 : v
2
Z
s
Z
+ v
Z
1
3
- v
Z
Z
2
2
- V s = 0
= 0
3
v = 0
(b) A through-variable source will be included in a node-equation, therefore the convention
is to show the arrow representing the direction of the assumed through variable.
16–2
(a )
a c r o s s - v a r ia b le
d ro p
F
o p p o s ite to
d ir e c tio n o f F
2
Z
s
s
Z
1
a t n o d e ( a ) : Fs - FZ
1
- FZ
2
= 0
2
v = 0
Energy Transduction – Two-Port Elements
Reading: Class Handout - Modeling Part 3: Two-Port Energy Transducing Ele­
ments
Many systems involve two or more energy domains, for example a system containing a dc
motor
W
T
d r iv e
m o to r
s
J
a d c s e
a tra n s
e le c tr ic
e n e rg y
i
+
v
-
e le c tr ic a l
rv o m
d u c e
a l a n
d o m
o to r is
r b e tw e e n th e e
d r o ta tio n a l
a in s
r o ta tio n a l
or there may be a scaling of the across- and through-variables within a single domain, for
example a mechanical lever
v1
a le v e r is a tr a n s d u c e r b e tw e e n
tw o tr a n s la tio n a l d o m a in s ( fo r
s m a ll d is p la c e m e n ts )
l1
F1
l2
F
v
2
2
The following two pages show some examples of two-port elements.
16–3
v
F
T
r
F
W
W
v
T
W
P in io n
F
R a c k
v
F
r
V
v =
W
=
rW
1_
F = - r T
( a ) R a c k a n d p in io n
1_
F =
v
-rW
r T
( b ) S lid e r - c r a n k
Q
W
P
i
+
T
E
Q
W
P
-
W = - 1_ Q
D
T = D P
_1 _
v =
F =
- 1_ Q
A
A P
F
( d ) M o v in g - c o il lo u d s p e a k e r
A re a A
i
+
Q
m a g n e t
c o il
i = - K
( c ) R o ta r y p o s itiv e d is p la c e m e n t p u m p
F
F
E = K v
R e s e r v o ir
v
v
v
T
-
v = K v W
1
i = - _ _ T
K v
P
( d ) F lu id p is to n - c y lin d e r
W
( b ) E le c tr ic a l m o to r /g e n e r a to r
16–4
N 1 te e th
1_ F
2
v1 = -2 v2
F
F 2
=
1
2
_1
W 1 = - N W 2
W 1
v 2
T2
( b ) G e a r tr a in
F 2
L = l2 / l 1
1
T1 = N T 2
W 2
T1
( a ) B lo c k a n d ta c k le
F
N 2 te e th
N = N 2 /N 1
F 1
v1
W 2
W 1
l2
l1
v
v
v 1
1
W
2
T 1
W
1
= - _ v2
L
1
1
W
W 1 = R W 2
T 1 = - _1 T
2
T 2
R
W
F1 = L F2
R
r1
N
= N 2 / N
1
+
i1
-
N
2
tu rn s
i2
v 2
v1
F o r a .c . in p u ts
= r /r
2
2
r
1
2
( d ) B e lt d r iv e
( c ) M e c h a n ic a l le v e r
N 1 tu rn s
2
v1 =
P1
+
A re a : A 1
A re a : A 2
Q 1
1_ v
2
N
Q 2
A = A 2 /A 1
Po = 0
i1 = - N i 2
( f) F lu id tr a n s fo r m e r
( e ) E le c tr ic a l tr a n s fo r m e r
16–5
P2
P1 = A P
2
1
_
Q 1 = - Q 2
A
In all of these examples the energy transduction is lossless and static, that is there is no
energy storage.
P1 = f1 v
f1
1
+
-
f2
T W O -P O R T
v1
v2
T R A N S D U C E R
+
-
P2 = f2 v
2
P1 + P2 = f1 v1 + f2 v2 = 0
where the power flow at each port is defined to be positive into the port.
There are two possibilities:
(a) There is an proportional relationship between the across-variables on the two sides of
the two-port element, and an proportional relationship between the through-variables.
v1 = kv2
across-variable1 ∝ across-variable2
f1 = (1/k)f2 through-variable1 ∝ through-variable2 ,
where clearly P1 = P2 . This relationship defines a transformer.
tra n s fo rm e r
v1
f2
f1
v2
(b) There is an proportional relationship between the across-variables on one side and the
through-variable on the other side of the two-port element.
v1 = kf2
across-variable1 ∝ through-variable2
f1 = (1/k)v2 through-variable1 ∝ across-variable2 ,
where clearly P1 = P2 . This relationship defines a gyrator.
g y ra to r
v1
f2
f1
16–6
v2
Examples:
(1) Rack and Pinion:
r
W
V
P in io n
R a c k
v
=
rW
1_
F = - r T
}
a tra n s fo rm e r
F
It can be seen that the linear velocity of the rack is proportional to the angular velocity
of the pinion. Similarly the force needed to balance the torque applied to the pinion
is proportional to the torque. The rack and pinion is therefore a lossless transformer.
(2) DC motor:
d r iv e
m o to r
T
W
T = -K i
1
W =
v
K
}
a tra n s fo rm e r
i
+
v
-
In the dc motor the torque produced is proportional to the current flowing, while the
back emf produced is proportional to the angular velocity of the shaft. The motor is a
lossless transformer.
Example 1
A DC motor with an inertial load
in h e r e n t r e s is ta n c e a n d in d u c ta n c e o f
m o to r w in d in g
+
L
V
s
(t)
f
R
f
ro to r
+
e
b
id e a l tr a n s d u c e r
J
W
-
L
B
1
V s (t)
e le c tr ic a l
R
v = 0
r o ta tio n a l
16–7
2
B
W
J
= 0
Impedance Relationships across Two-Port Elements
(1) The Transformer:
b
a
1
Z = ?
2
Z 1
v
1 = k v 2
1
f1 = - f2
k
Let an element (or system) with impedance Z1 be connected across a transformer as
shown, then the impedance seen from side 1 of the transformer is
V1 (s)
.
F1 (s)
Z=
At node (b)
f2 = −fZ1
v2 = vZ1
so that
Z=
V1 (s)
kV2 (s)
VZ (s)
=
= k2 1
= k 2 Z1
F1 (s)
(−1/k)F2 (s)
FZ1 (s)
Example 2
Find the apparent inertia of an inertia J with a step-gear box with a 10:1
ratio.
W
g e a r tr a in
N :1
T
d r iv e
m o to r
s
W
1
J
W
2
Z = ?
1
W
1
2
2
w ith th e r o ta tio n a l d ir e c tio n s a s s h o w n : W
1
Z 1 =
s J
2
= -1 0 W
1
With the above definition k = Ω1 /Ω2 = −0.1, so that
Zin = k 2 Z1 (s) =
0.01
1
=
Js
(100J)s
or in other words the inertia J is seen at the input shaft as an equivalent
inertia 100J.
16–8
(2) The Gyrator:
b
a
1
Z = ?
2
Z 1
v
= k f2
1
f1 = - v
k
1
2
At node (b)
f2 = −fZ1
v2 = vZ1
so that
Z=
V1 (s)
kF2 (s)
FZ (s)
1
=
= k2 1
= k2
= k 2 Y1 .
F1 (s)
(−1/k)V2 (s)
vZ1 (s)
Z1
The nature of the apparent impedance as seen at the input has been changed to its
reciprocal. The result is that an A-type element appears to be a T-type element when
connected behind a gyrator, and vice-versa.
No example is given here because there are no naturally occurring gyrators
in the energy domains covered this term.
Impedance Based Modeling with Two-Port Elements
Consider a Thévenin source coupled to a load Z2 through a transformer.
Z
1
o u tp u t is v
1
2
4
V s
3
The transformer provides the constraints
v4 = kv3
f4 = −(1/k)f3
16–9
Z
2
Z
2
(a) If Z2 is reflected to the l.h. side of the transformer:
Z
1
V s
V4 (s) =
k
2
k 2 Z2
Vs (s)
Z1 + k 2 Z2
Z
but
2
VZ2 = v3 =
1
v4
k
so that the transfer function H(s) is
H(s) =
(b)
VZ2
kZ2
=
.
Vs (s)
Z1 + k 2 Z2
Alternatively, we can transfer all elements to the r.h. side of the transformer
vZ1 = v3 =
1
v4
k
From loop (1)
vZ1 + v4 − Vs = 0
and
1
(v3 − vZ1 )
but vZ1 = Z1 fZ1
k
1
=
(Vs − Z1 f4 )
(fZ1 = f4 )
k�
�
1
Z1
=
Vs −
VZ
k
kZ2 1
v Z2 =
so that
�
VZ2
1 Z1
1+
k Z2
�
=
1
Vs
k
or
H(s) =
VZ2
kZ2
=
.
Vs (s)
Z1 + k 2 Z2
which is the same result as above.
Example 3
Use this result to find the transfer function
ΩJ (s)
H(s) =
Vin (s)
for the system shown below, which includes a dc servo motor, driven from a
voltage source Vin (s), driving an inertial load J and bearing damping B
16–10
W
T u r n ta b le
B e a r in g
J
B
R
L
i
+
V s (t)
L
M o to r
1
V s (t)
R
2
B
W
J
v = 0
= 0
Combine the series and parallel impedances and redraw the graph
Z
1
Z
V s
2
Z
= L s + R
Z
1
2
=
1
J s + B
where Z1 = R + sL, and Z2 = 1/(Js + B). Assume that for the motor vb = v1 =
Km Ωm . From the previous result
H(s) =
ΩJ (s)
kZ2
=
Vin (s)
Z1 + k 2 Z2
Km /(Js + B)
=
2 /(Js + B)
(R + sL) + Km
H(s) =
JLs2
Km
2)
+ (RJ + BL)s + (BR + Km
16–11