Chapter 8
Natural and Step Responses of
RLC Circuits
8.1-2 The Natural Response of a Parallel RLC
Circuit
8.3
The Step Response of a Parallel RLC
Circuit
8.4
The Natural and Step Response of a
Series RLC Circuit
1
Key points

What do the response curves of over-, under-,
and critically-damped circuits look like? How to
choose R, L, C values to achieve fast switching
or to prevent overshooting damage?

What are the initial conditions in an RLC circuit?
How to use them to determine the expansion
coefficients of the complete solution?

Comparisons between: (1) natural & step
responses, (2) parallel, series, or general RLC.
2
Section 8.1, 8.2
The Natural Response of a
Parallel RLC Circuit
1.
2.
3.
4.
ODE, ICs, general solution of parallel
voltage
Over-damped response
Under-damped response
Critically-damped response
3
The governing ordinary differential equation (ODE)
V0, I0, v(t) must
satisfy the
passive sign
convention.
dv 
v
1 t

  I 0   v (t )dt    0.
 By KCL: C
dt 
L 0
 R

Perform time derivative, we got a linear 2ndorder ODE of v(t) with constant coefficients:
d 2 v 1 dv v


 0.
2
dt
RC dt LC
4
The two initial conditions (ICs)

The capacitor voltage cannot change abruptly,
 v (0 )  V0 (1)

The inductor current cannot change abruptly,
 iL (0 )  I 0 , iC (0 )  iL (0 )  iR (0 )   I 0  V0 R ,
dvC
 iC (0 )  C
dt

I 0 V0
,  vC (0 )  v(0 )   
 ( 2)
C RC
t 0


5
General solution

Assume the solution is v(t )  Ae st, where A, s are
unknown constants to be solved.

Substitute into the ODE, we got an algebraic
(characteristic) equation of s determined by the circuit
parameters:
s
1
2
s 

RC

LC
 0.
Since the ODE is linear,  linear combination of
solutions remains a solution to the equation. The
general solution of v(t) must be of the form:
v (t )  A1e s1t  A2e s2t ,
where the expansion constants A1, A2 will be determined
by the two initial conditions.
6
Neper and resonance frequencies

In general, s has two roots, which can be (1)
distinct real, (2) degenerate real, or (3) complex
conjugate pair.
2
s1, 2
1
1
 1 

 
    2  02 ,
 
2 RC
 2 RC  LC
where
1

, …neper frequency
2 RC
0 
1
…resonance (natural) frequency
LC
7
Three types of natural response

How the circuit reaches its steady state depends
on the relative magnitudes of  and 0:
The Circuit is
Over-damped
Under-damped
Critically-damped
When
Solutions
  
real, distinct
roots s1, s2
  
complex roots
s1 = (s2)*
  
real, equal roots
s1 = s2
8
Over-damped response ( > 0)

The complete solution and its derivative are of
the form:
s1t
s2 t
v (t )  A1e  A2e ,

v(t )  A1s1e s1t  A2 s2e s2t .
where s1, 2     2  02 are distinct real.

Substitute the two ICs:
v (0 )  A1  A2  V0 (1)
 solve


I 0 V0
A1, A2.

v(0 )  s1 A1  s2 A2   C  RC ( 2)

9
Example 8.2: Discharging a parallel RLC circuit (1)

Q: v(t), iC(t), iL(t), iR(t) = ?
12 V
30 mA
1
1

 >0,
  2 RC  2( 200)( 2  107 )  12.5 kHz,

over
1
1
0 

 10 kHz. damped
LC

(5  102 )( 2  107 )
10
Example 8.2: Solving the parameters (2)

The 2 distinct real roots of s are:
 s     2   2  5 kHz, …|s1| < (slow)
0
1

2
2




s



 2
0  20 kHz. …|s2| > (fast)

The 2 expansion coefficients are:
 A1  A2  V0
 A1  A2  12


I 0 V0  
 5 A1  20 A2  450
 s1 A1  s2 A2   C  RC
 A1  14 V, A2  26 V.
11
Example 8.2: The parallel voltage evolution (3)
v (t )  A1e s1t  A2e s2t   14e 5000t  26e 20000t  V.
|s2| > (fast)
dominates
“Over”
damp
Converge
to zero
|s1| < (slow)
dominates
12
Example 8.2: The branch currents evolution (4)

The branch current through R is:


v(t )
iR (t ) 
  70e 5000t  130e  20000t mA.
200 

The branch current through L is:


t
1
5000 t
 20000 t


iL (t )  30 mA 
v
(
t
)
d
t

56
e

26
e
mA.

0
50 mH

The branch current through C is:


dv
iC (t )  (0.2 μF)  14e 5000t  104e  20000t mA.
dt
13
Example 8.2: The branch currents evolution (5)
Converge
to zero
14
General solution to under-damped response ( < 0)

The two roots of s are complex conjugate pair:
s1, 2     2  02    jd ,
where d  02   2 is the damped frequency.

The general solution is reformulated as:
v (t )  A1e(   jd ) t  A2e(   jd ) t
 e t A1 cos d t  j sin d t   A2 cos d t  j sin d t 
 e t  A1  A2  cos d t  j  A1  A2 sin d t 
 e t B1 cos d t  B2 sin d t .
15
Solving the expansion coefficients B1, B2 by ICs

The derivative of v(t) is:
v(t )  B1  e t cos d t  d e t sin d t 
 B2  e t sin d t  d e t cos d t 
 e t  B1  d B2  cos d t  B2  d B1 sin d t .

Substitute the two ICs:
v (0 )  B1  V0 (1)


I 0 V0

v(0 )  B1  d B2   C  RC (2)

 solve
B1, B2.
16
Example 8.4: Discharging a parallel RLC circuit (1)

Q: v(t), iC(t), iL(t), iR(t) = ?
0V
-12.25 mA
1
1

  2 RC  2( 2  104 )(1.25  107 )  0.2 kHz,  < 0,

under
1
1
0 

 1 kHz.
damped
7
LC

(8)(1.25  10 )
17
Example 8.4: Solving the parameters (2)

The damped frequency is:
d      1  0.2  0.98 kHz.
2
0

2
2
2
The 2 expansion coefficients are:
 B1  V0  0(1)
 B1  0,



I 0 V0
 B1  d B2   C  RC ( 2)  B2  100 V
18
Example 8.4: The parallel voltage evolution (3)
v (t )  B1e t cos d t  B2e t sin d t  100e 200t sin 980t  V.

The voltage
oscillates (~d) and
approaches the final
value (~), different
from the overdamped case (no
oscillation, 2 decay
constants).
19
Example 8.4: The branch currents evolutions (4)

The three branch currents are:
20
Rules for circuit designers

If one desires the circuit reaches the final value
as fast as possible while the minor oscillation is
of less concern, choosing R, L, C values to
satisfy under-damped condition.

If one concerns that the response not exceed its
final value to prevent potential damage,
designing the system to be over-damped at the
cost of slower response.
21
General solution to critically-damped response ( = 0)

Two identical real roots of s make
v (t )  A1e st  A2e st  ( A1  A2 )e st  A0e st ,
not possible to satisfy 2 independent ICs (V0, I0)
with a single expansion constant A0.

The general solution is reformulated as:
v (t )  e t D1t  D2 .

You can prove the validity of this form by
substituting it into the ODE:
v(t )  ( RC ) 1 v(t )  ( LC ) 1 v (t )  0.
22
Solving the expansion coefficients D1, D2 by ICs

The derivative of v(t) is:
v(t )  D1 e t  te t   D2e t  D1  D2   D1t e t .

Substitute the two ICs:
v (0 )  D2  V0  (1)

 solve D1, D2.

I 0 V0

v(0 )  D1  D2   C  RC ( 2)

23
Example 8.5: Discharging a parallel RLC circuit (1)

Q: What is R such that the circuit is criticallydamped? Plot the corresponding v(t).
0V
1
  0 ,

2 RC

R
-12.25 mA
1
1 L 1
8
, R

 4 k.
7
2 C 2 1.25  10
LC
Increasing R tends to bring the circuit from overto critically- and even under-damped.
24
Example 8.5: Solving the parameters (2)

The neper frequency is:
1
1


 1 kHz,
3
7
2 RC 2(4  10 )(1.25  10 )
1
    1 ms.


The 2 expansion coefficients are:
 D2  V0  0 (1)
 D1  98 kV s

-12.25 mA


I 0 V0
 D1  D2   C  RC  ( 2)  D2  0
0.125 F
25
Example 8.5: The parallel voltage evolution (3)


v(t )  D1te t  D2 e t  98,000te 1000t V.
98 V/ms
26
Procedures of solving nature response of parallel RLC

Calculate parameters   (2 RC ) 1 and 0  1

Write the form of v(t) by comparing  and :
LC .
 A e s1t  A e s2t , s  -   2   2 , if    ,
2
1, 2
0
0
 1
 t
v (t )  e B1 cos d t  B2 sin d t , d  02   2 , if   0 ,
 t
e D1t  D2 , if   0 .

Find the expansion constants (A1, A2), (B1, B2), or
(D1, D2) by two ICs: v(0 )  V0 (1),


I 0 V0


v (0 )   C  RC (2)

27
Section 8.3
The Step Response of
a Parallel RLC Circuit
1.
Inhomogeneous ODE, ICs, and general
solution
28
The homogeneous ODE
Is
+
V0

I0
1 t
v
dv 

 By KCL: C
  I 0   v (t )dt    I s .
dt 
L 0
 R

Perform time derivative, we got a homogeneous
ODE of v(t) independent of the source current Is:
d 2 v 1 dv v


 0.
2
dt
RC dt LC
29
The inhomogeneous ODE
Is

iL
V0
I0
Change the unknown to the inductor current iL(t):
v
 dv
C dt  iL  R  I s , d 2iL
Is
1 diL iL
 2 



dt
RC dt LC LC
v  L diL ,

dt
30
The two initial conditions (ICs)
iL
Is

V0
I0
The inductor current cannot change abruptly,
 iL (0 )  I 0 (1)

The capacitor voltage cannot change abruptly,
 vC (0 )  V0  vL (0 ),
diL
 vL (0 )  L
dt

V0
,  iL (0 )  ( 2)
L
t 0

31
General solution

The solution is the sum of final current If = Is and
the nature response iL,nature(t):
iL (t )  I f  iL,nature (t ),
where the three types of nature responses were
elucidated in Section 8.2:
 I f  A1e s1t  A2e s2t , if   0 ,

iL (t )   I f  e t B1 cos d t  B2 sin d t , if   0 ,

t
D1t  D2 , if   0 .

I
e
 f
32
Example 8.7: Charging a parallel RLC circuit (1)

Q: iL(t) = ?
I0 = 0
Is=
24
mA
V0 = 0
625 
1
1

  2 RC  2(625)( 2.5  108 )  32 kHz,
  < 0,

under
1
1
0 

 40 kHz. damped
2
8
LC

(2.5  10 )( 2.5  10 )
33
Example 8.7: Solving the parameters (2)

The complete solution is of the form:
iL ( t )  I s  e
t
B1 cos d t  B2 sin d t ,
2
2
2
2






40

32
 24 kHz.
where d
0

The 2 expansion coefficients are:
 I s  B1  I 0  0(1)
 B1  24 mA,



V0
 B1  d B2  L  0(2)  B2  32 mA
34
Example 8.7: Inductor current evolution (3)


iL (t )  24  24e 32,000t cos(24,000t )  32e 32,000t sin(24,000t ) mA.
35
Example 8.9: Charging of parallel RLC circuits (1)

Q: Compare iL(t) when the resistance R = 625 
(under-damp), 500  (critical damp), 400 
(over-damp), respectively.
I0 = 0
Is=
24
mA
V0 = 0
& final conditions remain: iL(0+)=0, i'L(0+)=0,
If = 24 mA. Different R’s give different functional
forms and expansion constants.
 Initial
36
Example 8.9: Comparison of rise times (2)

The current of an under-damped circuit rises
faster than that of its over-damped counterpart.
37
Section 8.4
The Natural and Step
Response of a Series RLC
Circuit
1.
Modifications of time constant, neper
frequency
38
ODE of nature response
V0, I0, i(t) must
satisfy the
passive sign
convention.
1 t
di 





V

i
(
t
)
d
t
 0.
Ri
L
 By KVL:
0



dt 
C 0

2
d
i
R
di
i
 By derivative:


 0.
2
dt
L dt LC
1
in parallel RLC
RC
39
The two initial conditions (ICs)

The inductor cannot change abruptly,
 i (0 )  I 0 (1)

The capacitor voltage cannot change abruptly,
 vC (0 )  V0 , vL (0 )  vC (0 )  vR (0 )  V0  I 0 R,
diL
 vL (0 )  L
dt

V0  I 0 R
,  iL (0 )  i(0 )  
 ( 2)
L
t 0


40
General solution
st
i
(
t
)

Ae
 Substitute
into the ODE, we got a
different characteristic equation of s:
1
R
2
2

s







s  s
 0.
1, 2
0.
L
LC
2

The form of s1,2 determines the form of general
solution:
 A1e s t  A2e s t , if   0
1
2
 t
i (t )  e B1 cos d t  B2 sin d t , if   0
e t D t  D , if   
1
2
0

R
, 0 
where  
2L
1
, d  02   2 .
LC
( 2 RC ) 1 in parallel RLC
41
Example 8.11: Discharging a series RLC circuit (1)

Q: i(t), vC(t) = ?
I0 = 0
vC(0-) = 100 V,
 V0 = -100 V
+
 100 V
560
R

  2 L  2(0.1)  2.8 kHz,


1
1
0 

 10 kHz.
7
LC

(0.1)(1  10 )
 <0,
underdamped
42
Example 8.11: Solving the parameters (2)

The damped frequency is:
d      10  2.8  9.6 kHz.
2
0

2
2
2
The 2 expansion coefficients are:
 B1  I 0  0(1)
 B1  0,
-100 V



V0  I 0 R
(2)  B2  104.2 mA
 B1  d B2  
L
9.6
kHz
100 mH
43
Example 8.11: Loop current evolution (3)
i (t )  e t B1 cos d t  B2 sin d t   104.2e 2,800t sin 9,600t  mA.
44
Example 8.11: Capacitor voltage evolution (4)
vc (t )  Ri (t )  Li(t ) 
e 2,800t 100 cos 9,600t  29.17 sin 9,600t  V.

When the capacitor
energy starts to
decrease, the
inductor energy starts
to increase.

Inductor energy starts
to decrease before
capacitor energy
decays to 0.
45
ODEs of step response
I0
Vs
V0
+

1 t
di 

 By KVL: Ri  L
 V0   i (t )dt   Vs .
dt 
C 0


The homogeneous and inhomogeneous ODEs
of i(t) and vC (t) are:
d 2i R di
i
d 2vC R dvC vC
Vs


 0, and



.
2
2
dt
L dt LC
dt
L dt LC LC
46
The two initial conditions (ICs)
I0
vC (t)
Vs

+

V0
The capacitor voltage cannot change abruptly,
 vC (0 )  V0 (1)

The inductor current cannot change abruptly,
 iL (0 )  I 0  iC (0 ),
dvC
 iC (0 )  C
dt

I0
,  vC (0 )  ( 2)
C
t 0

47
General solution

The solution is the sum of final voltage Vf = Vs
and the nature response vC,nature(t):
vC (t )  V f  vC ,nature (t ),
where the three types of nature responses were
elucidated in Section 8.4.
V f  A1e s1t  A2e s2t , if   0 ,

vC (t )  V f  e t B1 cos d t  B2 sin d t , if   0 ,

t
D1t  D2 , if   0 .

V
e
 f
48
Key points

What do the response curves of over-, under-,
and critically-damped circuits look like? How to
choose R, L, C values to achieve fast switching
or to prevent overshooting damage?

What are the initial conditions in an RLC circuit?
How to use them to determine the expansion
coefficients of the complete solution?

Comparisons between: (1) natural & step
responses, (2) parallel, series, or general RLC.
49