To find the step response of an RC circuit
v(t )  v()  [v(t0 )  v()]e

t t 0

The time constant  = RC
The final capacitor
voltage v()
The initial capacitor
voltage v(t0)
To find the step response of an RL circuit
i(t )  i()  [i(t0 )  i()]e

t t 0

The time constant
 = L/R
The final capacitor
voltage v()
The initial capacitor
voltage v(t0)
Lecture 6 DC Circuits
Transient Circuits – 2nd Order circuits
2
Contents
•
•
•
•
•
Examples of 2nd order RLC circuit
The source-free series RLC circuit
The source-free parallel RLC circuit
Step response of a series RLC circuit
Step response of a parallel RLC
3
Examples of Second Order RLC circuits
What is a 2nd order circuit?
A second-order circuit is characterized by a second-order
differential equation. It consists of resistors and the equivalent of
two energy storage elements of different type or the same type.
4
Source-Free Series RLC Circuits
• The solution of the source-free
series RLC circuit is called as the
natural response of the circuit.
• The circuit is excited by the
energy initially stored in the
capacitor and inductor.
2nd
The
order of
expression
d 2 i R di
i


0
2
L dt LC
dt
How to derive and how to solve?
5
Source-Free Series RLC Circuits:
to derive the equation
vL  L
vR=Ri
di
dt
t
1
vC   idt
C 
t
KVL:
di 1
Ri  L   idt  0
dt C 
di
d 2i i
R L 2  0
dt
dt
C
To eliminate the
integral
Source-Free Series RLC Circuits:
to solve the equation
d 2 i R di
i


0
2
L dt LC
dt
To be determined
Assume: the solution has exponential form,
 2 R 1 
Ae  s  s    0
 L LC 
st
Neper frequency/
damping factor
R
Define:  
2L
Characteristic equation
s 2  2s  02  0
Solution:
s1, 2     2  02
Linear equation:
i  Ae
st
To be determined
0 
Resonant frequency/
undamped natural factor
Natural Frequencies
i (t )  A1e s1t  A2 e s2t
1
LC
Still solving…
i (t )  A1e s1t  A2 e s2t
How to get A?
Initial conditions:
The inductor:
i ( 0)  I 0
i (0)  A1  A2  I 0
0
1
The capacitor: v(0) 
idt  V0

C 
Ri (0)  L
di
 V0  0
dt t 0
di (t )
 A1s1e s1t  A2 s2 e s2t
dt
RI 0  L A1s1  A2 s2   V0  0
A1 and A2
What if s1 = s2 (when  0 )
d 2i
di

2

 2 i 0
2
dt
dt
To get Ai
A1  A2 
 RI 0  V0
Ls
d  di

 di

   i      i   0
dt  dt

 dt

i (t )   A2  A1t e t
I0 and V0
not independent
di
  i  A1e t
dt
Source-Free Series: RLC Circuits
There are three possible solutions for the following
2nd order differential equation:
d 2i
di
2

2



i 0
0
2
dt
dt
1. If  > o, over-damped case
i (t )  A1e s1t  A2 e s2t
2
where s1, 2       0
2
2. If  = o, critical damped case
i (t )  ( A2  A1t )e t
where
s1, 2   
3. If  < o, under-damped case
Damped natural
frequency
i (t )  e t ( B1 cos  d t  B2 sin  d t ) where d  02   2
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Finding initial and final values
• Initial values:
v(0), i(0),
– Passive convention
dv(t )
,
dt t 0
di (t )
dt t 0
• The polarity of voltage
• The direction of current
– Variables that cannot change abruptly
• The capacitor: v(0  )  v(0  )
• The inductor: i (0  )  i (0  )
• Final values:
Focus on these variables
i (), v()
– DC steady state
• The capacitor: Open circuit
• The inductor: short circuit
The time just before a
switching event which
takes place at t = 0
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Example
If R = 10 Ω, L = 5 H, and C = 2 mF
in 8.8, find α, ω0, s1 and s2.
What type of natural response
will the circuit have?
Answer: underdamped
R

 1 0 
2L
1
 10
LC
s1, 2     2  02  1  j9.95
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Example
The circuit shown below
has reached steady state
at t = 0-.
If the make-before-break
switch moves to position
b at t = 0, calculate i(t) for
t > 0.
Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A
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Step-Response Series: RLC Circuits
• The step response is
obtained by the sudden
application of a dc source.
The 2nd order of
expression
vs
d 2 v R dv v



2
L dt LC LC
dt
The above equation has the same form as the equation for source-free series
RLC circuit.
• The same coefficients (important in determining the frequency
parameters).
• Different circuit variable in the equation.
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Step-Response Series: RLC Circuits
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
v (t )  vt (t )  v ss (t )
 The transient response vt is the same as that for source-free case
vt (t )  A1e s1t  A2 e s2t
(over-damped)
vt (t )  ( A1  A2t )e t
(critically damped)
vt (t )  e t ( A1 cos d t  A2 sin d t ) (under-damped)
 The steady-state response is the final value of v(t).
• vss(t) = v(∞)
 The values of A1 and A2 are obtained from the initial conditions:
• v(0) and dv(0)/dt.
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Example
Having been in position for a long time, the switch in the circuit
below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0.
Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V
vR(t)= [2.31sin3.464t]e–2t V
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Lecture 7 AC Circuits
Sinusoids and Phasors
Appendix
Source-Free Parallel: RLC Circuits
0
1
i (0)  I 0   v (t )dt
L
Let
v(0) = V0
Apply KCL to the top node:
t
v 1
dv
  vdt  C  0
R L 
dt
Taking the derivative with
respect to t and dividing by C
2nd
The
order of
expression
d 2 v 1 dv 1


v0
2
RC dt LC
dt
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Source-Free Parallel: RLC Circuits
There are three possible solutions for the following
2nd order differential equation:
d 2v
dv
 2
 02v  0
2
dt
dt
where  
1
2 RC
and 0 
1
LC
1. If  > o, over-damped case
v(t )  A1 e s1t  A2 e s2t where
s1, 2      2  0
2
2. If  = o, critical damped case
v(t )  ( A2  A1t ) e t where s1, 2   
3. If  < o, under-damped case
v(t )  e t ( B1 cos d t  B2 sin d t ) where
 d  02   2
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Example
Refer to the circuit shown below.
Find v(t) for t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer:
v(t) = 66.67(e–10t – e–2.5t) V
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Step-Response Parallel: RLC Circuits
• The step response is
obtained by the sudden
application of a dc source.
The 2nd order of
expression
d 2i 1 di
i
Is



2
dt
RC dt LC LC
It has the same form as the equation for source-free parallel RLC circuit.
• The same coefficients (important in determining the frequency
parameters).
• Different circuit variable in the equation.
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Step-Response Parallel : RLC Circuits
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
i (t )  it (t )  iss (t )
 The transient response it is the same as that for source-free case
it (t )  A1e s1t  A2 e s2t
(over-damped)
it (t )  ( A1  A2t )e t
(critical damped)
it (t )  e t ( A1 cos d t  A2 sin d t )
(under-damped)
 The steady-state response is the final value of i(t).
 iss(t) = i(∞) = Is
 The values of A1 and A2 are obtained from the initial conditions:
 i(0) and di(0)/dt.
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Example
Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown below:
• Please refer to lecture or textbook for more detail elaboration.
Answer:
v(t) = Ldi/dt = 5x20sint = 100sint V
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