RLC Circuits
Natural Response
Ch. 8 – RLC Circuits
RLC Parallel
2
RLC Series
Natural Response
The circuit has been working for a long time and the energy of
capacitor or inductor suddenly released to the resistive network.
Step Response
The response of a circuit to the sudden application of a DC
voltage or current source
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
3
 Natural Response of RLC Parallel
Applying KCL:
0  iC  iL  iR  C
dv 1 t
v
  vd  I 0 
dt L 0
R
d 2v 1
1 dv
 C 2  v
0
dt
L
R dt
This is a second order homogeneous DE with constant coefficients.
It admits the solution of the form
As 2 e st 
v(t )  Ae
st

d 2 v 1 dv 1


v0
dt 2 RC dt LC
1
1
1
1 

Ase st 
Ae st  0  Ae st  s 2 
s
0
RC
LC
RC
LC


Characteristic Equation
1
1
s +
s+
=0
RC
LC
2
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
4
2
The roots of characteristic
equation are
Simplify the notation as

1
and 0 
2 RC
2
1
1
 1 
s2  
 
 
2 RC
 2 RC  LC
s1     2  0 2
1
LC
s2     2  0 2
 The second order differential
equation can be rewritten using α
and ω0 as
d 2v
dv
2

2



0v  0
2
dt
dt
1
1
 1 
s1  
 
 
2 RC
 2 RC  LC
 The characteristic equation can
be rewritten using α and ω0 as
s 2  2 s  0  0
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
5
Characteristic Equation Terminology
Parameter
s1, s2
The characteristic
roots can be
Terminology
Value in
Natural Response
Characteristic Roots
s1     2  0 2
s2     2  0 2

Neper Frequency

1
2RC
0
Resonant Radian
Frequency
0 
1
LC
Real numbers and (distinct)
  2  02  0
Two identical real numbers
  2  02  0
Complex numbers (complex conjugates)
ELEC 250 – Summer 2015
  2  02  0
Ch. 8 – RLC Circuits
6
Real distinct roots
 2  02  0 
The solution

Note that the constants A1 and A2 are
determined by initial conditions as

When the roots are real and distinct, the
voltage response of parallel RLC circuit
is said to be over-damped
v ( t ) = A1es1t + A2es2t
ELEC 250 – Summer 2015
dv(0 )
v(0 ) and
dt

Ch. 8 – RLC Circuits
7
Solution strategy for real distinct roots
Step (1) Find the roots of the characteristic equation.
dv(0 )
Step (2) Find v(0 ) and
using circuit analysis.
dt
s1     2  0 2
s2     2  0 2

Step (3) Find the values of A1 and A2 using the values at step 2 as
v(0 )  A1  A2
dv(0 ) iC (0 )

 s1 A1  s2 A2
dt
C
Step (4) Substitute values of s1, s2, A1 and A2 from step (1) and (3) to
obtain the expression of solution as
v ( t ) = A1es1t + A2es2t
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
8
Duplicate real roots
 2  02  0 
The solution
v  t   D1tes1t  D2es1t

Note that the constants A1 and A2 are
determined by initial conditions as

When the roots are real and duplicate,
the voltage response of parallel RLC
circuit is said to be critically-damped
ELEC 250 – Summer 2015
dv(0 )
v(0 ) and
dt

Ch. 8 – RLC Circuits
9
Solution strategy for duplicate real roots
Step (1) Find the roots of the characteristic equation.
s1  s1  
dv(0 )
Step (2) Find v(0 ) and
using circuit analysis.
dt

Step (3) Find the values of D1 and D2 using step the values at step 2 as
v(0 )  D2
dv(0 ) iC (0 )

 D1   D2
dt
C
Step (4) Substitute values of s1, D1 and D2 from step (1) and (3) to
obtain the expression of solution as
v  t   D1tes1t  D2es1t
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
10
Complex (conjugate) roots
 2  02  0  The solution
v  t   e t  B1 cos d t   B2 sin d t 
where damped radian frequency

Note that the constants B1 and B2 are
determined by initial conditions as

When the roots are complex conjugate,
the voltage response of parallel RLC
circuit is said to be under-damped
ELEC 250 – Summer 2015
d   2   2 0
dv(0 )
v(0 ) and
dt

Ch. 8 – RLC Circuits
11
Solution strategy for complex conjugate roots
Step (1) Find the roots of the characteristic equation.
Step (2) Find v(0 ) and

s1    jd
s2    jd
dv(0 )
using circuit analysis.
dt
Step (3) Find the values of A1 and A2 using step the values at step 2 as
v(0 )  B1
dv(0 ) iC (0 )

  B1  d B2
dt
C
Step (4) Substitute values of s1, s2, B1 and B2 from step (1) and (3) to
obtain the expression of solution as
v  t   e t  B1 cos d t   B2 sin d t 
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
12
Example: The two switches in the circuit operate synchronously. When
switch 1 is in position a, switch 2 is in position d. When switch 1 moves to
position b, switch 2 moves to position c. Switch 1 has been in position a for a
long time. At t = 0, the switches move to their alternate positions. Find 𝒗𝒐 (𝒕)
for 𝒕 > 𝟎 ?
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
13
Solution:
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
14
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
15
 Natural Response of RLC Series
Applying KVL:
t
di 1
VR  VL  VC  0  Ri  L   id  V0  0 
dt C 0
d 2i
di i
 L 2 R  0
dt
dt C
d 2i R di
i



0
This is a second order homogeneous DE with constant coefficients.
dt 2 L dt LC
The RLC series DE can be written as
d 2i
di
2

2



i0
0
2
dt
dt

R
2L
0 
1
LC
Neper Frequency
Resonant Frequency
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
16
 Natural Response of RLC Series


The series RLC circuit is governed by the same second order
DE with constant coefficients as the parallel RLC circuit
It admits the same solutions
Two real roots  Over-damped
i  t   A1es1t  A2es2t
Duplicate roots  Critically-damped
i  t   D1tes1t  D2es1t
Two complex roots  Under-damped
i  t   est  B1 cos d t  B2 sin d t 
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
17
Comparison Parallel/Series RLC
Parameter
s1, s2
Terminology
Characteristic
Roots

Neper
Frequency
0
Resonant
Radian
Frequency
Value in
Natural Response
PARALLEL
Value in
Natural Response
SERIES
s1     2  0 2
s1     2  0 2
s2     2  0 2
s2     2  0 2

1
2RC
0 
1
LC
ELEC 250 – Summer 2015

0 
R
2L
1
LC
Ch. 8 – RLC Circuits
18
Example: In the following circuit, the resistor is adjusted for critical damping.
The initial capacitor voltage is 15 V, and the initial inductor current is 6 mA.
a) Find the numerical value of R.
𝒅𝒊
b) Find the numerical values of 𝒊 𝒕 and immediately after the switch is closed.
𝒅𝒕
c) Find 𝒗𝒄 (𝒕) for 𝒕 > 𝟎 ?
ELEC 250 – Summer 2015
Ch. 8 – RLC Circuits
19
Solution:
ELEC 250 – Summer 2015