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SECOND ORDER CIRCUIT
SECOND ORDER CIRCUIT
• Revision of 1st order circuit
• Second order circuit
• Natural response (source-free)
• Forced response
Revision of 1st order circuit
NATURAL RESPONSE (SOURCE-FREE)
+
vR

iR
iC
+
vC

R
C
Solution:
KCL 

- initial energy in capacitor
- i.e. vC(0) = Vo
iC  iR  0
dv C v C

0
dt RC
Solving this first order differential equation gives:
v C ( t )  Vo e  t RC
Revision of 1st order circuit
R
Vsu(t)
+

Solution:
FORCED RESPONSE
+
vC

C
KCL 

- no initial energy in capacitor
- i.e. vC(0) = 0
iC  iR  0
dv C v C  Vs
C

0
dt
R
Solving this first order differential equation gives:

v C ( t )  Vs 1  e  t RC


dv C v C
Vs


dt RC RC
Revision of 1st order circuit
COMPLETE RESPONSE
Complete response = natural response + forced response
v(t) = vn(t) + vf(t)

v C ( t )  Vo e  t RC  Vs 1  e  t RC

Complete response = Steady state response + transient response
v(t) = vss(t) + vt(t)
v C ( t )  Vs  Vo  Vs e  t RC
Revision of 1st order circuit
COMPLETE RESPONSE
In general, this can be written as:
x(t )  x()  x(0)  x()et 
- can be applied to voltage or current
- x(0) : initial value
v C ( t )  Vs  Vo  Vs e  t RC
- x() : steady state value
For the 2nd order circuit, we are going to adopt the same approach
Before we begin …..
To successfully solve 2nd order equation, need to know how to
get the initial condition and final values CORRECTLY
INCORRECT initial conditions /final values will result in a wrong
solution
In 1st order circuit
•need to find initial value of inductor current (RL circuit) OR capacitor
voltage (RC circuit): iL(0) or vC(0)
•Need to find final value of inductor current OR capacitor voltage: iL(∞)
or vC(∞)
In 2nd order circuit
•need to find initial values of iL and/or vC : iL(0) or vC(0)
•Need to find final values of inductor current and/or capacitor voltage: iL(∞)
, vC(∞)
•Need to find the initial values of first derivative of iL or vC : diL(0)/dt
dvC(0)/dt
Section 8.2 of Alexander/Sadiku
Finding initial and final values
Example 8.1
Switch closed for a long time and open at t=0. Find:
i(0+), v(0+),
di(0+)/dt, dv(0+)/dt,
i(∞), v(∞)
Finding initial and final values
PP 8.2
Find:
iL(0+), vC(0+), vR(0+)
diL(0+)/dt, dvC(0+)/dt, dvR(0+)/dt,
iL(∞), vC(∞), vR(∞)
Second order circuit
Natural Response of Series RLC Circuit
(Source-Free Series RLC Circuit)
R
L
We want to solve for i(t).
C
i
Applying KVL,

di 1 t
Ri  L 
i dt  0
dt C 
Differentiate once,
di
d2i i
R L 2   0
dt
dt
C
d2i R di
i


0
2
dt
L dt LC
 This is a second order differential equation
with constant coefficients
Second order circuit
Assuming
i( t )  Aest
AR st A st
As e 
se 
e 0
L
LC
2
st
1 
 2 R
Ae  s  s 
0
L
LC 

st
Since
Aest cannot become zero,
1 
 2 R
s  s 
0
L
LC 
d2i R di
i


0
2
L dtas the
LCCHARACTERISTIC EQUATION of the diff. equation
Thisdtis known
Second order circuit
Solving for s,
2
R
1
R
s1  
   
2L
 2L  LC
2
R
1
R
s2  
   
2L
 2L  LC
Which can also be written as
s1    2  o2
where  
R
,
2L
s2    2  o2
o 
1
LC
1 
 2 R
s

s

0

 (nepers/s)
s1, s2 – known as natural frequencies
L
LC 

 – known as neper frequency, o – known as resonant frequency
Second order circuit
i( t )  A1es1t  A 2es2t
A1 and A2 are determined
from initial conditions
Case 1
  o
2
s1    Case
2  
2o
  o
Overdamped solution
s2    2  o2
Critically damped solution
Case 3
  o
Underdamped solution
Second order circuit
Case 1
Overdamped response
  o
Roots to the characteristic equation are real and negative
i( t )  A1es1t  A 2es2t
A1 and A2 are determined from initial conditions:
(i) At t = 0,
i(0)  A1  A 2
(ii) At t = 0,
di(0)
 s1A 1  s2 A 2
dt
Second order circuit
Case 1
Overdamped response
  o
100
0.05H
R
100

 1000,
2L 2(0.05)
1
1
o 

 200
LC
0.05(0.0005)

+
vc

0.5mF
Initial condition vc(0) =100V
Second order circuit
Case 2
Critically damped response
  o
i( t )  A1e  t  A 2e  t  A 3 e  t
A3 is determined from 2 initial conditions: NOT POSSIBLE
 solution should be in different form:
i(t )  A1te t  A 2et
A1 and A2 are determined from initial conditions:
(i) At t = 0,
i(0)  A 2
(ii) At t = 0,
di(0)
 A 1  A 2
dt
Second order circuit
Case 2
Critically damped response
  o
20
0.05H
R
20

 200,
2L 2(0.05)
1
1
o 

 200
LC
0.05(0.0005)

+
vc

0.5mF
Initial condition vc(0) =100V
Second order circuit
Case 3
Underdamped response
  o
Roots to the characteristic equation are complex
i( t )  A1es1t  A 2es2t
s2    o2  2
s1    j o2  2
   jd
   jd
d  o2  2
i(t) = A1e- (a-jw
[
- known as damped natural frequency
d )t
+ A 2e-( a+jw
d )t
i(t) = e- at A1ejw t + A 2e-jw t
d
d
]
Second order circuit
Case 3
Underdamped response
  o
Using Euler’s identity: ej = cos  + jsin 
i( t )  e  t A1cos d t  j sin d t   A 2 cos d t  j sin d t 
 e  t A 1  A 2 cos d t  jA1  A 2  sin d t 
i( t )  e  t B1 cos d t  B2 sin d t 
where
B1  A1  A2 
[
and B2  jA1  A2 
i(t) = e- at A1ejw t + A 2e-jw t
d
d
]
Second order circuit
Case 3
Underdamped response
  o
(i) At t = 0,
di(0)
 B1  dB2
B1 cos dtdt  B2 sin dt
(ii) At tt = 0,
i( t )  e
i(0)  B1
Second order circuit
Case 3
Underdamped response
  o
10
0.05H
R
10

 100,
2L 2(0.05)
1
1
o 

 200
LC
0.05(0.0005)

+
vc

0.5mF
Initial condition vc(0) =100V
Second order circuit
Underdamped, overdamped and critically damped responses