NATURAL AND
STEP RESPONSES
OF RLC CIRCUITS
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CONTENTS
8.1 Linear Second Order Circuits
8.2 Solution Steps
8.3 Finding Initial Values
8.4 The Natural Response of a Series/Parallel
RLC Circuit
8.5 The Step Response of a Series/Parallel
RLC Circuit
8.6 State Equations
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8.1 Linear Second Order Circuits
n
Circuits containing two energy storage elements.
n
Described by differential equations that contain
second order derivatives.
n
Need two initial conditions to get the unique
solution.
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8.1 Linear Second Order Circuits
n Examples
(a) RLC parallel circuit
+
v(t )
−
is (t )
(b) RLC series circuit
vs (t )
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i (t )
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8.1 Linear Second Order Circuits
R1
(c) 2L+R , RL circuit
vs (t )
R2
L2
L1
(d) 2C+R , RC circuit
R
is (t )
C1
C2
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8.2 Solution Steps
Step1 : Choose nodal analysis or mesh
analysis approach
Step2 : Differentiate the equation as many
times as required to get the standard
form of a second order differential
equation .
d 2x
dx
a 2 + b + x = y (t )
dt
dt
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8.2 Solution Steps
Step 3 : Solving the differential equation
(1) homogeneous solution xh (t )
(2) particular solution x p (t )
x (t ) = xh (t ) + x p (t )
Step 4 : Find the initial conditions
d
+
x(0+ ) and dt x(0 ) and then get the
unique solution
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8.3 Finding Initial Values
Under DC steady state, L is like a short circuit
and C is like an open circuit.
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8.3 Finding Initial Values
Under transient condition, L is like an open circuit
and C is like a short circuit because iL(t) and vC(t)
are continuous functions if the input is bounded.
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8.3 Finding Initial Values
Under transient condition, L is like an open circuit
and C is like a short circuit because iL(t) and vC(t)
are continuous functions if the input is bounded.
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8.3 Finding Initial Values
dvC (0+ )
diL (0+ )
, use the
To find
and
dt
dt
following relations
diL (0+ )
d
v (0+ )
= vL (0+ ) => iL (0+ ) = L
dt
dt
L
+
+
dv (0 )
dv (0 ) iC (0+ )
C C
= ic (0+ ) => C
=
dt
dt
C
+
+
One can find vL (0 ) and iC (0 ) using either
L
nodal or mesh analysis .
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8.3 Finding Initial Values
n Example 1
The circuit is under steady state.
The switch is opened at t = 0
d
Find : (a ) i (0+ ) ,
i (0+ )
dt
d
(b) v(0+ ) ,
v(0+ )
dt
(c) i (∞) , v(∞)
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8.3 Finding Initial Values
n Example 1 (cont.)
t<0
i(0-) = 2 A
v(0-) = 4 V
∴ i(0+) = i(0-) = 2 A
v(0+) = v(0-) = 4 V
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8.3 Finding Initial Values
n Example 1 (cont.)
t = 0+
V (0+ )
di
d
= VL , ∴ i (0+ ) = L
dt
dt
L
i (0+ )
dv
d
QC
= iC , ∴ v(0+ ) = C
dt
dt
C
QL
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8.3 Finding Initial Values
n Example 1 (cont.)
KVL : 2A × 4 + vL (0+ ) + 4V = 12V
∴ vL (0+ ) = 0
d
i (0+ ) = 0
dt
KCL : iC (0+ ) = 2A
∴
d
v(0+ ) = 0
dt
i (0+ ) 2
∴C
=
= 20 V/S
C
0.1
∴
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8.3 Finding Initial Values
n Example 1 (cont.)
t →∞
0.25H
4Ω
i
12V
2Ω
0.1F
t=0
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+
v
-
L is short circuitd
C is open
∴ i (∞ ) = 0
v(∞) = 12V
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8.3 Finding Initial Values
n Example 2
4Ω
2Ω
3u(t)A
+
vR
-
__
1
2F
iL
+
vc
-
0.6H
20V
d
iL (0+ ) , iL (∞)
dt
d
(b) vC (0+ ) ,
vC (0+ ) , vC (∞)
dt
d
(c) vR (0+ ) ,
vR (0+ ) , vR (∞)
dt
Find : (a ) iL (0+ ) ,
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8.3 Finding Initial Values
n Example 2 (cont.)
4Ω
4Ω
3u(t)A
2Ω
+
vR
-
__
1
2F
20V
+
vc
-
iL
0.6H
t<0
+
2Ω vR
-
iL
+
vc
20V
∴ iL (0− ) = 0
vc (0− ) = −20V
vR (0− ) = 0
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8.3 Finding Initial Values
n Example 2 (cont.)
4Ω
3u(t)A
2Ω
+
vR
-
+
vc
-
__
1
2F
iL
t = 0+
0.6H
20V
∴ iL (0+ ) = iL (0− ) = 0
vc (0+ ) = vc (0− ) = −20V
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8.3 Finding Initial Values
n Example 2 (cont.)
3u(t)A
2Ω
+
vR
-
__
1
2F
+
vc
-
20V
t = 0+
+ vo (0+) 4Ω
3A
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= 2A
2+4
vR (0+ ) = 2 A × 2 = 4V
2
ic (0+ ) = 3 A ×
= 1A
2+4
d
v (0+ ) 0
∴ iL (0+ ) = L
= =0
dt
L
L
+
i (0 ) 1
d
vc (0+ ) = c
=
=2 V
S
1
dt
c
2
d
vR (0+ ) = ?
dt
i2 Ω (0+ ) = 3 A ×
4Ω
2Ω v +(0+)
R
-
ic(0+)
iL
0.6H
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8.3 Finding Initial Values
n Example 2 (cont.)
+ vo
4Ω
3u(t)A
2Ω
+
vR
-
__
1
2F
-
4Ω
iL
+
vc
-
t > 0+
3A
0.6H
2Ω
20V
__
1
2 F
+
vR
-
+
vc
-
iL
0.6H
20V
From KVL : − vR + vo + vc + 20 = 0
Take derivative
d
d
d
vR (t ) = vo (t ) + vc (t )
dt
dt
dt
d
d
d
∴ vR (0+ ) = vo (0+ ) + vc (0+ )LL ( A)
dt
dt
dt
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8.3 Finding Initial Values
n Example 2 (cont.)
+ vo
4Ω
3u(t)A
2Ω
+
vR
-
__
1
2F
iL
+
vc
-
t > 0+
0.6H
3A
2Ω
20V
+
vR
-
__
1
2 F
+
vc
-
iL
0.6H
20V
Also , from KCL : 3 A =
Take derivative
0=
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4Ω
vR (t ) vo (t )
+
2
4
1 d
1 d
vR (t ) +
vo (t )LL ( B)
2 dt
4 dt
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8.3 Finding Initial Values
n Example 2 (cont.)
d
d
vo (0+ ) = −2 vR (0+ )LL (C )
dt
dt
From ( A) and (C )
d
d
vR (0+ ) = −2 vR (0+ ) + 2
dt
dt
d
2
∴ vR (0+ ) = V
dt
3 S
From ( B)
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8.3 Finding Initial Values
n Example 2 (cont.)
4Ω
3u(t)A
2Ω
+
vR
-
4Ω
__
1
2F
+
vc
-
iL
t →∞
0.6H
3A
2Ω
20V
+
vR( )
-
iL
+
vc( )
20V
2
= 1A
2+4
vc (∞) = −20V
∴ iL (∞) = 3 A ×
vR (∞) = 3 A × (2Ω P 4Ω) = 4V
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
(a) The source-free series RLC circuit
This section is an important background for studying
filter design and communication networks .
initial conditions
i (0) = I 0

v(0) = V0
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
initial conditions
i (0) = I 0

v(0) = V0
Step 1 : Mesh analysis
di 1 t
Ri + L + ∫ i dt = 0
dt C −∞
To eliminate the integral , take derivative
L
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d 2i
di 1
+R + i=0
2
dt
dt C
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
initial conditions
i (0) = I 0

v(0) = V0
Step 2 : Homogeneous solution , characteristic equation
R
1
S+
=0
L
LC
S 2 + 2αS + ω0 2 = 0
S2 +
R
1
, ω0 2 @
2L
LC
ω0 : undamped resonant frequency (rad/s)
⇒α@
α : damping factor or neper frequency
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
initial conditions
i (0) = I 0

v(0) = V0
characteristic roots (natural frequencies)
S1 = -α + α 2 - ω0 2
S2 = -α - α 2 - ω0 2
∴ i ( t ) = A1e S1t + A2 e S2t
Need two initial conditions , i.e. , i (0+ ) and
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d
i (0+ ) .
di
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
Case 1 Overdamped Case (α > ω0 )
Two real roots
i ( t ) = A1e S1t + A2 e S2t
Case 2 Critically Damping Case (α =ω0 )
Equal real roots
R
S1 = S 2 = −
= −α
2L
i ( t ) = ( A2 + A1t ) e −α t
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
Case 3 Underdamped Case (α < ω0 )
Complex conjugate roots
S1 = −α + jωd
S 2 = −α − jωd
ωd @ ω0 2 − α 2 damping frequency
i ( t ) = e −α t ( B1 cos ωd t + B2 sin ωd t )
Once i(t) is obtained ,solutions of other variables can be
obtained from this mesh current.
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
n
The damping effect is due to the presence of resistance R .
n
The damping factor α determines the rate at which the
response is damped .
n
If R=0 , the circuit is said to be lossless and the
oscillatory response will continue .
n
The damped oscillation exhibited by the underdamped
response is known as ringing . It stems from the ability of
the L and C to transfer energy back and forth between
them .
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
Step 3 : Initial Condition
t = 0+ , i (0+ ) = i (0− ) = I 0
From mesh equation , let t = 0+
d
1 0+
Ri (0+ ) + L i (0+ ) + ∫ idt = 0
dt
C −∞
V
V
d
R
R
∴
i (0+ ) = − i (0+ ) − 0 = − I 0 − 0
L
dt
L
L
L
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
or from equivalent circuit at t = 0+
d
i (0+ ) = vL = −( I 0 R + V0 )
dt
I R V
d
∴ i (0+ ) = − 0 − 0
dt
L
L
I0
R
L
vL
V0
i
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
(b) The source-free parallel RLC circuit
initial inductor current Io
initial capacitor voltage Vo
Step 1 : Nodal Equation
v 1 t
dv
+ ∫ vdt + C
=0
R L
dt
Taking derivative to eliminate the integral
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d 2 v 1 dv 1
+
+
v=0
dt 2 RC dt LC
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
Step 2 : Homogeneous solution
Characteristic equation
1
1
=0
S2 +
S+
RC
LC
⇒α @
S 2 + 2α S + ω0 2 = 0
1
1
, ω0 2 @
LC
2 RC
characteristic roots (natural frequencies)
S1 = −α + α 2 − ω0 2
S 2 = −α − α 2 − ω0 2
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
CASE 1. Overdamped Case (α>w0 )
v ( t ) = A1e S1t + A2 e S2t
CASE 2. Critically Damped Case (α=w0 )
v ( t ) = ( A1 + A2t ) e −α t
CASE 3. Underdamped Case (α<w0 )
v ( t ) = e −α t ( A1 cos wd t + A2 sin wd t )
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
Step 3 : Initial Condition
v(0+ ) = i (0− ) = V0
From nodal equation
v ( 0+ )
1 0+
d
vdt + C v ( 0+ ) = 0
∫
−∞
R
L
dt
+
v ( 0 ) 1 0+
d
∴ C v ( 0+ ) = −
− ∫ vdt
dt
R
L −∞
V
= − 0 − I0
R
V
I
d
∴ v ( 0+ ) = − 0 − 0
dt
RC C
+
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8.4 The Natural Response of a Series/Parallel
RLC Circuit
or from equivalent circuit at t = 0+
d
C v(0+ ) = iC ( 0+ )
dt
i (0+ )
d
∴ v(0+ ) = C
dt
C
V
From KCL, iC (0+ ) = − I 0 − 0
R
Once the nodal voltage is obtained, any other unknown
of the circuit can be found .
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8.5 The Step Response of a Series/Parallel RLC
Circuit
(a) Step response of a series
RLC circuit
Step 1. Mesh analysis (i=iL )
L
1
di
+ Ri + ∫ idt = Vs , t > 0
dt
C
Case(i) take derivative
L
d 2i
di i
+R + =0
2
dt
dt C
Same as natural response but with i(0+)=I0
V − I R − V0
d
i ( 0+ ) = s 0
dt
L
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8.5 The Step Response of a Series/Parallel RLC
Circuit
Case(ii) use v as unknown
i=C
dv
dt
V
d 2 v R dv v
+
+
= s
2
dt
L dt LC LC
Step 2. Complete solution = vh+vp
v p ( t ) = Vs
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 A1e S1t + A2 e S2t

vh ( t ) = ( A1 + A2t ) e −α t
 −α t
e ( A1 cos wd t + A2 sin wd t )
overdamped
critically damped
underdamped
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8.5 The Step Response of a Series/Parallel RLC
Circuit
Step 3. Initial conditions
v ( 0+ ) = v ( 0− ) = V0
d
v ( 0+ ) = ?
dt
dv
C
= iC
dt
∴
iC ( 0
d
v ( 0+ ) =
dt
C
+
)=I
0
C
Then the unique solution can be determined
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8.5 The Step Response of a Series/Parallel RLC
Circuit
(b) Step response of a parallel RLC circuit
I0 , V0 Given
Step 1. Nodal equation
v 1 t
dv
+ ∫ vdt + C
= Is , t > 0
R L
dt
Case (i) Take derivative
C
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d 2 v 1 dv 1
+
+ v=0
dt 2 R dt L
Same as natural response
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8.5 The Step Response of a Series/Parallel RLC
Circuit
Case (ii) Let
v=L
I
di
d 2i
1 di 1
⇒
+
+
i = s ,t > 0
2
dt
dt
RC dt LC
LC
Step 2. Complete solution = ih(t)+ip(t)
ip (t ) = Is
 A1e S1t + A2 e S2t

ih ( t ) = ( A1 + A2t ) e −α t
 −α t
e ( A1 cos wd t + A2 sin wd t )
overdamped
critically damped
undererdamped
Step 3. Initial Condition
i ( 0+ ) = i ( 0− ) = I 0
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L
v (0+ ) V0
di
d
= vL ⇒ i ( 0 + ) = L
=
dt
dt
L
L
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8.5 The Step Response of a Series/Parallel RLC
Circuit
n Example
1
F
2
t < 0 , i (0- ) = 0 , v(0- ) = 12V
t>0
1
F
2
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8.5 The Step Response of a Series/Parallel RLC
Circuit
Method (a)
v 1 dv
+
2 2 dt
di
From KVL ⇒ 4i + 1 + v = 12
dt
From KCL ⇒ i =
(A)
(B)
Substitute (A) into (B)
dv
1 dv 1 d 2 v
)+(
+
)+ v = 12
dt
2 dt 2 dt 2
d 2v
dv
⇒ 2 +5
+ 6 v = 24V
dt
dt
(2 v + 2
Characteristic equation
( s + 2)( s + 3) = 0
s1 = −2, s2 = −3
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8.5 The Step Response of a Series/Parallel RLC
Circuit
vh (t ) = A1e−2t + A 2 e −3t
24
= 4V
6
∴ v(t ) = 4 + A1e−2t + A 2 e −3t
v p (t ) =
Initial Condition v(0+ )=v(0- )=12V
t=0+
1
F
2
dv
= ic
dt
dv(0+ ) ic (0+ ) −6
∴
=
=
= −12 V / S
dt
C
1/ 2
C
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8.5 The Step Response of a Series/Parallel RLC
Circuit
∴ 4 + A1 + A 2 = 12
−2A1 − 3A 2 = −12
∴ A1 = 12, A 2 = −4
∴ v(t ) = 4 + 12e −2t − 4e −3t , t ≥ 0
Method (b) Using Mesh Analysis
t>0
i
4Ω
1H
2Ω
12V
i1
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i2
v
1
F
2
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8.5 The Step Response of a Series/Parallel RLC
Circuit
di
+ 4i + (i1 − i2 )2 = 12
......(A)
dt
1 t
(i2 − i1 ) × 2Ω +
i2 dt = 0 ......(B)
1/ 2 ∫
di
di
From (B) , 2 2 − 2 1 + 2i2 = 0
dt
dt
di
+ 6i1 − 2i2 = 12
dt
di
di
−2 1 + 2 2 + 2i2 = 0
dt
dt
1
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1
F
2
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8.5 The Step Response of a Series/Parallel RLC
Circuit
In matrix form with operator D @
d
dt
−2   i1  12 
D + 6
 −2 D 2 D + 2  i  =  0 

 2  
(C)
Initial Condition , i1 (0+ ) = i(0+ ) = i(0− ) = 0A
i2 (0+ ) = ic (0+ ) = −6A
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di1 (0+ ) vL (0+ )
=
=0
dt
L
di2 (0+ )
= 0A / S
dt
49
8.5 The Step Response of a Series/Parallel RLC
Circuit
Eliminate i2 variable from (C)
d 2i1
di1
+
+ 6i1 = 12
5
dt 2
dt
∴ i1 (t ) = 2 − 6e −2t + 4e −3t , t ≥ 0
Or eliminate i1 variable from (C)
d 2i2
di2
+
5
+ 6i2 = 0
dt 2
dt
∴ i2 (t ) = −12e −2t + 6e −3t , t ≥ 0
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8.5 The Step Response of a Series/Parallel RLC
Circuit
Problem : (a) Time consuming to eliminate the other
variable to get a higher order differential
equation.
(b) It is also necessary to obtain the desired
initial conditions.
(c) As the order gets higher when the
network contains more energy storage
elements, the process gets more
complicated.
The difficulty can be overcome by using state equation
formulation.
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8.6 State Equation
When the differential equations of a circuit is written in
the following form:
d
x = f ( x, u , t )
dt
T
x = [ x1 x2 ....xn ]
state vector
u = [u1 u2 ....um ]
input vector
f T = [ f1 f 2 .. f n ]
vector function
T
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8.6 State Equation
It is said that the circuit equations are in the state
equation form.
(a) This form lends itself most easily to analog or digital
computer programming.
(b) The extension to nonlinear and/or time varying
networks is quite easy.
(c) In this form, a number of theoretic concepts of
systems are readily applicable to networks.
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8.6 State Equation
For a linear time-invarying circuit , a simpler form
d
x = A x + Bu , state equation
dt
y = C x + Du , output equation
A : n × n matrix.
B : n × m matrix.
C : l × n matrix.
D : l × m matrix.
Note that on the right hand side of the state or output
equation, only x and u are allowed.
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8.6 State Equation
Step1. Pick a tree which contains all the capacitors and
none of inductors.
Step2. Use the tree-branch capacitor voltages and the
link inductor currents as unknown (i.e. , state)
variables.
Note: (a) Nodal Analysis
Every unknown of the circuit can be
calculated from nodal voltages.
(b) Mesh Analysis
Every unknown of the circuit can be
calculated from mesh currents.
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8.6 State Equation
(c) State Equation
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l
The chosen variables include both
voltage and current unknown. It belongs
to mixed type.
l
Every unknown of the circuit can be
calculated from the state variables by
replacing each inductor with a current
source and each capacitor with a voltage
source and then solving the resulting
resistive circuit.
56
8.6 State Equation
Step3. Write a fundamental cutset equation (i.e. KCL
equation) for each capacitor.
Note that in these cutset equations, all branch
currents must be expressed in terms of x and u.
Step4. Write a fundamental loop equation (i.e. KVL
equation) for each inductor.
Note that in these loop equations, all branch
voltages must be expressed in terms of x and u.
Step5. Rearrange the above equations into standard form
and find the solution for the given initial condition.
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8.6 State Equation
Example 1
n
Step1
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tree
58
8.6 State Equation
Step2 choose i and v as state variables.
Step3 fundamental cutset about the capacitor tree branch.
C
dv
v
=idt
2
Step4 fundamental loop for the inductor link.
L
di
+ v -12V +4i = 0
dt
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8.6 State Equation
Step5
1
d v   C
= 
dt  i   -4
 L
-1 
0
2C  v   
+ 1 (12V)

-1   i   
L
L 
[ A]
[ B]
The desired solutions are v and i
 v  1 0   v   0 
y=   = 
   +   (12V)
 i  0 1   i  0
[C ]
[ D]
with initial condition v(0+ ) = 12V
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i (0+ ) = 0A
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8.6 State Equation
Example 2
n
d 3v
d 2v
dv
+
5
+ 4 + 3v = u(t)
3
2
dt
dt
dt

 dx1
 x1 = v(t)
 dt = x 2


dv(t)

 dx 2
⇒ 
Let  x 2 =
= x3
dt
dt



 dx 3 d 3v
d 2 v(t)
x
=
= 3
 3

dt 3
dt

 dt
d 2 v dv
= - 5 2 - 4 - 3v + u(t)
dt
dt
= - 5x 3 - 4x 2 - 3x1 + u(t)
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8.6 State Equation
State space representation
 x1   0 1 0   x1  0 
d  
∴
x 2 = 0 0 1   x 2  + 0  u(t)

 
   
dt
 x 3  -3 -4 -5  x 3  1 
 x1 
y = [1 0 0]  x 2  + [ 0] u(t)
 
 x 3 
A high order differential equation can be represented
in the form of state equation.
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8.6 State Equation
n
Example 3 : Find vR4
i3
a
iL1
iL2
L1
R3
L2
b
g
h
R2
R1
es
C1
i4
R4
c
+
vC1
d
+
C2 vC2
e
R5
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There are 8 nodes.
63
f
8.6 State Equation
Step1 Pick a tree as follows :
a
iL1 iL2
g
b
h
c
+
d
vC1
- +
vC2
e -
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There are 7 tree branches
and 3 links.
64
8.6 State Equation
Step2 Choose iL1, iL2 , vC1 , vC2 as unknowns.
Step3 Fundamental cutsets (KCL) for capacitors.
dv
C1 C1 = iL1
dt
dv
C2 C2 = iL1 + iL2
dt
Step4 Fundamental loops (KVL) for inductors.
di
L1 L1 = -vR1 - vC1 - vC2 - vR5 + vR4
dt
= -R1iL1 - vC1 - vC2 - R 5 (iL1 + iL2 ) + vR4
Note that vR4 should be expressed in terms of x and u
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8.6 State Equation
Absorb voltages vR1
and vC1 in current
source iL1 ,
and vR2 in iL2.
Absorb voltages vC2
and vR5 in (iL1+ iL2)
current source.
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8.6 State Equation
∴ vR4 = es
R R
R4
- (iL1 + iL2 ) 3 4
R 3 +R 4
R 3 +R 4
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8.6 State Equation
Step5

0

 vC1  

 0
d vC2  
=
dt  iL1   −1

 
 iL2   L1

0

0
1
C1
0
1
C2
−1
L1
−(R1 + R 2 )
L1
−1
L2
−R
L2

vR4 = 0 0

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-R 3R 4
R3 + R 4
where R @ R 5 +



1


C2


−R

L1

−(R 2 + R) 

L2

0
 0 
 vC1   0 
v   
R4
 C2  +  1 
e
 iL1   L1  R 3 +R 4 s


 iL2   1 
 
 L2 
 vC1 


-R 3R 4  vC2 
R4
+
es
 
R 3 + R 4  iL1  R 3 + R 4


 iL2 
R 3R 4
R 3 +R 4
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8.6 State Equation
Special case
(a)
i1
i3
i2
From KCL
i1+i2+i3 = 0
∴ i3 = -i1-i2
Inductor current i3 is dependent on i1 and i2 and
is no longer a state variable.
One can choose only n-1 (here 2) inductor
currents as state variables.
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8.6 State Equation
(b)
From KVL
vC1+vC2 = vC3
One can choose n-1 (here 2) capacitor
voltages as state variables.
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Summary
n
Objective 1 : Be able to find the initial values and the
initial derivative values.
n
Objective 2 : Be able to determine the natural response
and the step response of a series RLC circuit.
n
Objective 3 : Be able to determine the natural response
and the step response of a parallel RLC circuit.
n
Objective 4 : Be able to obtain the state equation and
output equation of a linear circuit.
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Summary
Chapter Problems : 8.16
8.25
8.32
8.40
8.44
Due within one week.
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