Lecture 10
State Equations and Computer aided circuit Analysis
This lecture will cover the following topics.
9 Introduction to the zero-input circuit.
9 Definition of state variables.
9 The matrix state equation.
After finishing this lecture, you should be able to
9 Understand the concept of the zero-input circuit.
9 Understand the definition of state variables.
9 Derive matrix state equation of a zero-input circuit
Zero-Input Circuit:
L
R
+
IL
vC
_
Figure 11-1 Series RLC zero-input circuit.
Let us consider the series RLC circuit of Figure 11-1.
¾ At t = 0, Let iL(0) ≠0 and Vc (0) ≠ 0
¾ We can say that the “Condition” or “state” of the circuit at
time t = 0 is specified by the inductor current and capacitor
voltage. For this reason we call the pair of numbers
[i L (0), V c (0)] the initial state of the circuit.
¾ Extending this concept, we can refer to the pair ⎡⎣iL (t ),Vc (t )⎤⎦
as the state of the circuit at time t.
¾ The variables iL and vc as the are called state variables of the
circuit.
Applying KVL to this circuit, we obtain
L
diL
+ Vc + RiL = 0
dt
(11.1)
iL = C dvc
dt
(11-2)
With
Substituting equation (11-2) into equation (11-1) after division by
LC, we get
d 2vc + vc + CR dvc = 0
dt 2 LC LC dt
(11-3)
d 2vc + ( R ) dvc + ( 1 )v = 0
L dt
LC c
dt 2
(11-4)
Since α
= R,
2L
ωn = 1
LC
d 2vc + 2α dvc + ω 2v = 0
n c
dt
dt 2
(11-5)
Let us rewrite equations (11-1) and (11-2), such that the
coefficients of the derivatives are unity and the derivatives alone
are on one side of the equation.
diL
= − R iL − 1 vc
L
L
dt
(11-6)
and
dvc = 1 i
dt C L
(11-7)
We call equation of this form, where all the variables present are
state variables, state equations.
⎡ di ⎤ ⎡
⎢ L ⎥ ⎢− R
⎢ dt ⎥ ⎢ L
⎢
⎥=⎢
⎢
⎥ ⎢
⎢ dv ⎥ ⎢ 1
⎢ c⎥ ⎢
⎢⎣ dt ⎥⎦ ⎣ C
⎤
− 1 ⎥ ⎡⎢ iL ⎤⎥
L⎥ ⎢ ⎥
⎥⎢ ⎥
⎥⎢ ⎥
⎥⎢ ⎥
⎥ ⎣vc ⎦
⎦
0
⎡ i ⎤ ⎡⎢ − R
⎢ L⎥
d ⎢ ⎥ = ⎢⎢ L
dt ⎢⎢ ⎥⎥ ⎢
⎢ 1
⎢vc ⎥ ⎢
⎣ ⎦ ⎣ C
(11-8)
⎤
− 1 ⎥ ⎡⎢ iL ⎤⎥
L⎥ ⎢ ⎥
0
⎥⎢ ⎥
⎥⎢ ⎥
⎥
⎥ ⎢⎣vc ⎥⎦
⎦
(11-9)
If we define X(t) and A by:
⎡i ⎤
X (t ) = ⎢ L ⎥
⎢vc ⎥
⎣ ⎦
⎡ R
⎢−
⎢ L
and A = ⎢
⎢
⎢ 1
⎢
⎣ C
⎤
−1⎥
L⎥
0
⎥
⎥
⎥
⎥
⎦
Thus the matrix state equation can be written.
dX (t ) = AX (t )
dt
(11-10)
X(t) : State vector
X(0): Will be called the initial-state vector
state-variable analysis
We have from state equation of the form of equation (11-10).
The process of finding the solution of the state equation (11-10) is
known as state-variable analysis
dx(t ) = ax(t )
dt
has a solution x(t ) = x(0)eat
(11-11)
The equation
for
t ≥0
(11-12)
for t ≥ 0
(11-13)
Similarly X (t ) = e At X (0)
At
e : is a matrix and also known as the state transition matrix
One technique for evaluating for eAt is by using a power series
expansion. We know from Taylor series expansion of eat is
2 2
3 3
eat = 1+ at + a t + a t + ......
2!
3!
(11-4)
The analogous results for eAt is
2 2
3 3
e At = I + At + a t + a t + .......
2!
3!
(11-15)
where I is the identity matrix.
Example 11-1:
Consider the zero-input parallel RLC circuit of Fig. 11-2. Find the
expression for
dv
diL
and c in terms of state variables of this circuit,
dt
dt
iL and vc.
iR
R
+
vL L
_
iL
C
iC
Figure 11-2: Zero input parallel RLC circuit
Solution
Since,
L
diL
= vc
dt
Thus,
diL 1
= v
dt L c
(11-16)
Also, we know the fact that
C dvc = ic
dt
By applying KCL, we have
ic = −iL − iR
(11-17)
by substituting the value of ic into equation (11-7) and dividing by C, we
get
dvc = − iL − iR
C C
dt
(11-18)
Substituting in (11-18), the value of iR = vc / R , by applying ohm’s
law,
dvc = − iL − vc
C RC
dt
(11-19)
Thus we have the required state equations given in equation (1120)
diL 1
= v
dt L c
dvc = − iL − vc
c RC
dt
In matrix form
(11-20)
⎡ i ⎤ ⎡⎢
0
⎢ L⎥ ⎢
d ⎢ ⎥=⎢
dt ⎢⎢ ⎥⎥ ⎢
⎢ 1
⎢vc ⎥ ⎢ −
⎣ ⎦ ⎣ C
⎤⎡ ⎤
⎥ ⎢ iL ⎥
⎥⎢ ⎥
⎥⎢ ⎥
⎥⎢ ⎥
1
⎥
⎢v ⎥
−
RC ⎥⎦ ⎣ c ⎦
1
L
(11-21)
Example 11-2
Write the matrix state equation for the zero-input RLC circuit
shown in Figure 11-3
+
5Ω v_R
2H
+
vC
+
vL
_
_
iL
3F
4Ω
iC
iR4
Figure: 11-3
Solution
By applying KVL,
L
diL
di
+ 5iL = vc ⇒ L = vc − 5 iL
dt
dt L L
(11-22)
Substituting the values, we will get
diL
= − 5 iL + vc
2
2
dt
(11-23)
Applying KCL,
iR 4 + iC + iL = 0 ⇒ C dvc + vc + iL = 0
dt 4
(11-24)
Therefore,
dvc = − vc − iL = − iL − vc
4c
C
3 12
dt
(11-25)
In matrix form
⎡
⎤
⎡i ⎤ ⎢ − 5 1 ⎥ ⎡i ⎤
L
L
⎢ ⎥ ⎢
2
2
⎥⎢ ⎥
d ⎢ ⎥=⎢
⎥⎢ ⎥
⎢ ⎥ ⎢
⎥⎢ ⎥
dt ⎢ ⎥
⎢ 1
1 ⎥ ⎢⎢ ⎥⎥
⎢vc ⎥ ⎢ −
−
v
⎣ ⎦ ⎢ 3
12 ⎥⎦⎥ ⎣ c ⎦
⎣
(11-26)
Example11-3: Find the state equation in matrix form
1Ω +
vL
i1 _
v1 2Ω i4 v2
+ vR _
+
vC 3F
4H
_
i3
+
5F
_
i4
iL
Figure: 11-4: circuit with three state variables
Inductor voltage: vL = 4
diL 1
= v
dt 4 2
diL
dt
(11-27)
Capacitor current:
3dvc = i = −i − i
3
1 2
dt
v v
=− 1− R
1 2
v −v
= −v1 − ( 1 2 )
2
= −3 v1 + 1 v2
2
2
and therefore
dv1
= − 1 v1 + 1 v2
2
6
dt
(11-28)
Also,
5 dvc = i4 = −i2 − iL = −iL +
dt
dv2
= − 1 iL + 1 v1 − 1 v2
5
10
10
dt
v1 − v2
2
(11-29)
Writing the state equations in matrix form
⎡i ⎤ ⎡ 0
⎢ L⎥ ⎢
d ⎢⎢ v1 ⎥⎥ = ⎢⎢ 0
dt ⎢ ⎥ ⎢
⎢v ⎥ ⎢ − 1
⎢⎣ 2 ⎥⎦ ⎢⎣
5
0
−1
2
1
10
⎤⎡ ⎤
i
⎥
4⎥⎢ L⎥
1 ⎥ ⎢⎢ v1 ⎥⎥
6 ⎥⎢ ⎥
⎥
− 1 ⎥ ⎢⎣⎢vc ⎥⎦⎥
10⎦
1
(11-30)
Example 11-4
Find the state equation in matrix form for the
circuit of Figure 11-5
i1
1H
+
+
vc
_
+
vC
3Ω
_
_
1
F
5
i2 1
H
2
vc
_
+
+
+
5Ω
4Ω
_
_
Figure 11-5
By applying KVL in the LHS loop
L1
di1
di
+ R1i1 + vc = 0 ⇒ 1 = − 1(3i1 + vc )
1
dt
dt
(11-31)
By applying KVL in the RHS loop
L2 =
di1
di
+ 4i2 − vc = 0 ⇒ 2 = − 4 i2 + vc
1
1
dt
dt
2
2
(11-32)
By KCL at vc
i1 = ic + vc + i2
5
i1 = C dvc + vc + i2
dt 5
⇒ dvc = (i1 − i2 − vc )5
5
dt
dvc = 5i − 5i − v
c
1
2
dt
(11-33)
Therefore the matrix equation becomes
⎡i ⎤
⎢ L ⎥ ⎡ −3
d ⎢ i2 ⎥ = ⎢⎢ 0
dt ⎢⎢ ⎥⎥ ⎢
⎢v ⎥ ⎢⎣ 5
⎣⎢ c ⎦⎥
⎡i ⎤
0 1⎤⎥ ⎢ L ⎥
⎢ ⎥
−8 2 ⎥ ⎢ i2 ⎥
⎥
−5 −2 ⎥⎦ ⎢⎢ ⎥⎥
⎣⎢vc ⎦⎥