Lecture 11
This lecture covers the following topics
1) Introduction to the non zero-input circuit.
2) Derivation of the state matrix equation for circuits with nonzero
inputs.
3) State equation analysis for nonzero input circuits
After finishing this lecture, you should be able to
1) Understand the basics of the non zero-input circuit.
2) Derive state matrix equation for nonzero circuits
3) Perform the State equation analysis for nonzero input circuits
Circuits with Non zero Inputs
L iL
R
_
+ vL _ + vR
+
vs
C
vC
_
Figure 13-1
Consider the series RLC circuit shown in figure 13-1.
Consider the inductor current iL and capacitor voltage vc as the
state variables.
Since for the inductor voltage
vL =
LdiL
dt
(13-1)
Then by KVL,
L
diL
= −vR − vC + vs
dt
(13-2)
L
diL
= − RiL − vc + vs
dt
(13-3)
From which, we will find our state equation as
diL
= − R iL − 1 vc + 1 vs
L
L
L
dt
(13-4)
Also
dvc = 1 i
dt C L
(13-4)
We have already seen when vs = 0, we can write the the state
equation in the form
dX (t ) = AX (t )
dt
(13-5)
however, when vs ≠ 0, For state equation in the matrix form, we
must include another term in the matrix equation. We can write
⎡ i ⎤ ⎡⎢ − R
⎢ L⎥
d ⎢ ⎥ = ⎢⎢ L
dt ⎢⎢ ⎥⎥ ⎢
1
⎢vc ⎥ ⎢⎢
⎣ ⎦ ⎣ C
⎡1⎤
⎢ ⎥
⎢L⎥
⎥ ⎢ ⎥ + ⎢ ⎥ vs
⎥⎢ ⎥ ⎢ ⎥
⎥⎢ ⎥ ⎢ ⎥
⎥ ⎣vc ⎦ ⎢ 0 ⎥
⎢⎣ ⎥⎦
⎦
⎤
- 1 ⎥ ⎡⎢ iL ⎤⎥
L⎥ ⎢ ⎥
0
(13-6)
In general form, for a circuit with nonzero input we can write
dX (t ) = AX (t ) + Bw(t )
dt
(13-7)
Where X(t) is the state vector and w(t) results from the input.
Read Examples 7.7, 7.8, 7.9
Example 13-1
Write the state Equation for the circuit of Figure 13-2
1H
4
+
2Ω
_
_
1F
6
v
_
vC
_
Figure 13-2
Solution:
By KVL
−vs + vR + vL + vC = 0
or
di
2iL + 1 L + vc = vs
4 dt
Rearranging
diL
= −8iL − 4vc + 4vs
dt
By KCL at node x,
iL = ic + iR
iL = cdvc + vc ⇒ dvc = 1 (+iL − vc )
R
dt C
dt R
(13-8)
+
+
+
i1 + vR
X
3Ω
_
dvc = 6i − 6 v ⇒ dvc = 6i − 2v
c
L 3 c
L
dt
dt
(13-9)
Arranging in matrix form,
⎡i ⎤
⎢ L ⎥ ⎡ −8
d ⎢ ⎥=⎢
⎢
dt ⎢⎢ ⎥⎥ ⎢
6
⎢vc ⎥ ⎢⎣
⎣ ⎦
⎡i ⎤
−4⎤⎥ ⎢ L ⎥
−2
⎡ 4⎤
⎢ ⎥
⎢ ⎥ ⎢ ⎥
⎥ ⎢ ⎥ + ⎢ ⎥ vs
⎥⎢ ⎥ ⎢ ⎥
⎥⎦ ⎢v ⎥ ⎢0⎥
⎣ s⎦ ⎣ ⎦
Example 13-2
Write the state equation matrix for the circuit shown in Figure 13-3
below
2Ω
1H
3
2Ω
_
+
i1 + vR
1H
2
X
_
i2
+
+
1F
4
vs
A
3Ω
vC
_
_
B
Figure 13-3
Solution:
By applying KVL in loop (A)
−vs + vR + vL + vc = 0
2
(13-10)
1/3
Similarly by KVL in loop (B)
0 − vc + vL + vR = 0
1/ 2
3
(13-11)
and by KCL at Node (X)
i1 = ic + i2 + vc − vs = 0
2
dv
⇒ i1 = C C + i2 + vc − vs
2 2
dt
(13-12)
From (13-10) we get
di1
= 3(2i1 + 0i2 − vc + vs )
dt
Therefore
di1
= −6i1 + 0i2 − 3vc + 3vs
dt
(13-13)
from (13-11)
di2
= 2(−vR + vC ) = −6i2 + 2vC + 0vs
dt
di2
= 0i1 − 6i2 + 2vc + 0vs
dt
(13-14)
From (13-12) we get
dvC
v
= (i1 − i2 − C + vs ) × 4
2 2
dt
dvC
= 4i1 − 4i2 − 2vC + 2vs
dt
(13-15)
In Matrix form
⎡ i ⎤ ⎡ −6
d ⎢⎢ i1 ⎥⎥ = ⎢⎢ 0
dt ⎢ 2 ⎥ ⎢
⎢v ⎥ ⎢ 4
⎣ C⎦ ⎣
0 −3⎤⎥ ⎡⎢ i1 ⎤⎥ ⎡⎢3⎤⎥
−6 2 ⎥ ⎢ i2 ⎥ + ⎢0⎥ vs
⎥⎢ ⎥ ⎢ ⎥
−4 −2 ⎥⎦ ⎢vC ⎥ ⎢⎣2⎥⎦
⎣
⎦
(13-16)