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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 61
Reading:
• Nise: Sec. 2.4 (pages 45–55)
• Class Handout: Modeling Part 1: Energy and Power Flow in Linear Systems
Sec. 1 (Introduction)
Sec. 4 (Electrical System Elements)
1 Modeling Electrical Systems (continued)
In Lecture 5 we examined the primitive electrical elements (capacitors inductors and resis­
tors), and sources (voltage source and current source). We now look at how these elements
behave when connected together in a circuit.
Interconnection Laws:
(a) Kirchoff ’s Current Law (KCL): The sum of currents flowing into(or out of) a
junction is zero. In the figure below, at the circled junction we sum the currents into
the junction to find
i1 − i2 − i3 = 0
R
V s (t)
i 1 + ( - i2 ) + ( - i3 ) = 0
i3
i1
i2
+
C
-
L
We will define a junction as a node, and if there are n circuit branches attached to a
node
n
�
in = 0
i=1
where we define the convention that positive current flow is into the node.
1
c D.Rowell 2008
copyright �
6–1
i1
i2
in
in
ik
-1
Kirchoff ’s Voltage Law (KVL): The sum of voltage drops around any closed loop in
a circuit is zero. The assumed sign convention for the voltage drop on each element
must be defined. Two clockwise loops are shown in the figure below. For loop (1)
vR + vC − Vs (t),
while for loop (2)
vL − vC (t) = 0.
R
+
V s (t)
+
-
1
+
+
-
C
L o o p 1 : v
L
2
-
L o o p 2 :
R
+ v
v
L
C
- V
- v
C
s
= 0
= 0
N o te : T h e + a n d - o n th e d ia g r a m
s h o w s th e d ir e c tio n o f th e
a s s u m e d v o lta g e d r o p .
Electrical Impedance:
i
+
N e tw o rk
V
-
Define the impedance of an element or passive circuit as a transfer function relating current
I(s) to voltage V (s) at its terminals:
Z(s) =
V (s)
I(s)
In addition we can define the admittance Y (s) as the reciprocal of the impedance:
Y (s) =
1
I(s)
=
Z(s)
V (s)
The Impedance of Passive Electrical Elements
6–2
(a) The Capacitor:
For the capacitor
dv
.
dt
Taking the Laplace Transform:
+
i=C
i
V
I(s) = CsV (s)
C
ZC (s) =
-
V (s)
1
=
I(s)
sC
or the admittance YC (s) = sC.
(b) The Inductor:
For the inductor
i
di
.
dt
Taking the Laplace Transform:
+
V
v=L
V (s) = LsI(s)
L
ZL (s) =
-
V (s)
= sL
I(s)
or the admittance YL (s) = 1/sL.
(c) The Resistor:
For the resistor
v = Ri.
+
V
i
Taking the Laplace transform
V (s) = RI(s)
R
ZR (s) =
-
V (s)
=R
I(s)
or the admittance YR = 1/R.
Impedance Nomenclature: We now introduce a graphical representation that will be
used to denote systems in many energy domains.
b ra n c h
Z
6–3
n o d e
The impedance is drawn as a graph branch between two nodes. Nodes represent junctions
between the elements in the circuit. The arrow on the branch indicates both the assumed
direction of voltage drop across the element, and the assumed current direction.
Example 1
The electrical circuit, consisting of a capacitor C, an inductor L, and a resistor
R is shown below:
R
V s (t)
Z
+
L
C
-
V s (t)
+
Z
R
C
-
Z
L
The impedance graph is shown on the right. The nodes on the graph represent
points of distinct voltage in the circuit.
Impedance Connection Rules
(a) Series connection: Two or more elements are defined to be connected in series if
they share a common current. For the two elements Z1 and Z2 in series below:
Using KCL at the junction between Z1 and Z2 :
Z
V s (t)
+
i
1
-
iZ1 = iZ2 = i
Z
Using KVL around the loop: vZ1 + vZ2 − Vs = 0
Vs = iZ1 + iZ2
2
Zeq =
V (s)
= Z 1 + Z2
I(s)
In general with n impedances Zi (i = 1, . . . , n) in series:
Zeq =
n
�
i=1
6–4
Zi
Example 2
For the tree elements in series below:
L
C
R
Z
e q
1
+ sL + R
sC
or expressing the impedance as a transfer function (a ratio of polynomials):
Zeq = ZC + ZL + ZR =
Zeq =
V (s)
LCs2 + RCs + 1
=
I(s)
Cs
(b) Parallel connection: Two or more elements are defined to be connected in parallel
if they share a common voltage. For the two elements Z1 and Z2 in parallel below:
Using KVL:
vZ1 = vZ2 = Vs
i
V
( ts )
+
-
Z
Using KCL at the node:
1
Z
is = iZ1 + iZ2
2
1
I
iZ + iZ2
= = 1
Zeq
V
V
1
1
1
=
+
Zeq
Z1 Z2
In general for n impedances Zi (i = 1, . . . , n) in parallel, the equivalent impedance is:
n
� 1
1
=
Zeq
Zi
i=1
Alternatively, using admittances Y = 1/Z
n
�
1
yeq =
=
Yi .
Zeq
i=1
Note: For N = 2 we can write
1
1
1
Z 1 + Z2
=
+
=
Zeq
Z1 Z2
Z1 Z2
6–5
which leads to the very common representation
Zeq =
Z1 Z2
Z1 + Z 2
Example 3
Find the impedance of a capacitor C, and inductor L and a resistor R connected
in parallel:
V s (t)
1
Z
+
C
R
-
L
Z=
1
1
1
+
+
1/sC sL R
1
1
= sC +
+
sL R
LCRs2 + Ls + R
=
RLs
=
V (s)
RLs
=
2
I(s)
LCRs + Ls + R
Example 4
Find the impedance of the following circuit, assuming we should include resis­
tance and inductance of the coil:
R
V s (t)
+
-
Z
in c lu d e b o th in d u c ta n c e a n d
r e s is ta n c e o f th e c o il
C
L , R
L
V s (t)
Z = Z1 + Z4 � (Z2 + Z3 )
Z4 (Z2 + Z3 )
= Z1 +
Z4 + Z2 + Z3
(1/sC)(RL + Ls)
= R+
1/sC + RL + Ls
RL + Ls
= R+
2
LCs + RL Cs + 1
6–6
+
-
Z
4
=
1
= R
1
C s
Z
Z
3
2
= R
= L s
L
V (s)
RLCs2 + (RRL C + L)s + (R + RL )
=
I(s)
LCs2 + RL Cs + 1
Z=
The Voltage Divider:
v
Z
Z
V s (t)
+
I(s) =
1
V (s)
(Z1 + Z2 )
and
1
VZ2 (s) = I(s)Z2 = VZ2 (s) =
Z
-
Consider two impedances in series with voltage V across them:
v
2
Z
Similarly
2
VZ1 (s) =
Z2
V (s).
Z1 + Z 2
Z1
V (s).
Z1 + Z 2
The voltage divider relationship may be used to find the transfer function of many simple
systems.
Example 5
Find the transfer function relating V0 to Vs in the following circuit
R
Z
V s (t)
+
v o
C
-
V s (t)
+
v o
1
Z
-
2
v = 0
Use the voltage divider relationship
V0 =
1/sC
Z2
Vs =
Vs
Z1 + Z 2
R + 1/sC
H(s) =
V0 (s)
1
=
V (s)
RCs + 1
Example 6
Find the transfer function relating V0 to Vs in the following circuit:
6–7
Z
R
V s (t)
V s
+
C
-
L , R
v o
L
V s (t)
+
a s b e fo r e , in c lu d e b o th
in d u c ta n c e a n d r e s is ta n c e
o f th e c o il
Z
4
=
1
= R
1
C s
V s (t)
Z
v o
1
+
Z
-
5
v = 0
V0 =
Z5
Vs
Z1 + Z 5
where
Z5 = Z4 � (Z2 + Z3 ) =
=
Z4 (Z2 + Z3 )
Z4 + Z 2 + Z 3
(1/sC)(RL + Ls )
1/sC + RL + Ls
Using the voltage divider relationship, the transfer function is
R +Ls
L
V0 (s)
Z5
LCs2 +RL Cs+1
H(s) =
=
=
L +Ls
Vs (s)
Z1 + Z 5
R1 + LCsR
2 +R Cs+1
L
H(s) =
R1
LCs2
RL + Ls
+ (R1 RL C + L)s + (R1 + RL )
The Current Divider: Consider two impedances in parallel:
v
I
(a )
I
i1
Z
i2
1
Z
v = 0
6–8
2
Z
-
Reduce the impedance graph to a series connection of two elements
V s
v o
v = 0
Z
3
2
= R
= L s
L
Using KCL at the top node (a),
I − i1 − i2 = 0
or
i1 + i1 = I
But i1 = V /Z1 , and i2 = V /Z2 so that
V
V
+
=I
Z1 Z2
i1 =
or
V =
1
I
1/Z1 + 1/Z2
V
1/Z1
Y1
=
I=
I
Z1
(1/Z1 + 1/Z2 )
Y 1 + Y2
Similarly
Y2
I.
Y1 + Y 2
The current divider may be used to find transfer functions for some simple circuits.
i2 =
Example 7
Find the transfer function
H(s) =
Vo (s)
I(s)
in the following circuit:
R
(a )
I
i1
(b )
i2
I(t)
C
L
Draw the system as an impedance graph:
I
I(s )
V a (s )
(a )
i1
Z
1
i2
Z
(b )
2
V o(s )
Z
v = 0
6–9
3
v o(t)
Let Z1 = 1/sC, Z2 = R, and Z3 = sL. We will use Vo (s) = I2 (s)Z3 (at node
(b)), and find I2 (s) from the current division at node (a):
1
Z2 +Z3
1
1
+ Z2 +Z
Z1
3
1
I(s)
(1/Z1 )(Z2 + Z3 ) + 1
1
1
=
I(s) =
I(s)
2
Cs(R + Ls) + 1
LCs + RCs + 1
I2 (s) =
I(s) =
Vo (s) = I2 (s)Ls =
LCs2
Ls
I(s)
+ RCs + 1
or
H(s) =
Vo (s)
Ls
=
I(s)
LCs2 + RCs + 1
6–10