Download SEE 1003 – Basic Electrical Engineering

SEE 1023
Circuit Theory
Chapter 4
Second Order Circuit
(11th & 12th week)
Prepared by : jaafar shafie
Second Order Circuit
•
•
•
•
•
In first order circuit, the RC and RL circuits are
represented in first order differential equation.
This is due to the existence of only one storage
element at any particular circuit.
In this chapter, two storage elements will exist in
a particular circuit.
Thus, this circuit are characterized by second order
differential equation.
A circuit with second order differential equation is
called SECOND ORDER CIRCUIT.
Second Order Circuit
•
Types of second order circuit that may exist:1. Series RLC circuit,
2. Parallel RLC circuit,
3. RLL circuit,
4. RCC circuit.
R
L
C
V
V
R1
L1
R2
I
L2
R
I
C1
C
R2
L
C2
Second Order Circuit
•
As usual, the circuit will be analyzed in two parts:1. Source–free Circuit (natural response) Energy
is initially stored in the element – thus no effect of
current or voltage sources.
2. Circuit with source (forced response) Current
or voltage sources is directly connected to the
first order circuits.
• Before the circuits are being analyzed, one should
find the initial and steady state value of the
capacitor voltage and inductor current and it’s
derivative; i.e.:v(0), i(0), dv(0)/dt, di(0)/dt, v(), i().
Second Order Circuit
•
Consider the circuit shown below where the switch
has been closed for a long time. Find:1. v(0+), i(0+),
2. dv(0+)/dt, di(0+)/dt,
3. v(), i().
4Ω
24V
1H
i(t)
4Ω
0.4F
+
v(t)
—
t=0s
Second Order Circuit
1. When t=0+s=0–s, the inductor is shorted and the capacitor
is opened. The equivalent circuit is shown below,
i(0+)
24V
+
v(0+)
-
4Ω
4Ω
v(0+)=(4/8)24=12V
i(0+)= 24/8 =3A
2. When the switch is opened, the equivalent circuit is
i(0+)
i(0+)= iC (0+) =3A
4Ω
24V
thus,
1H
0.4F
and it is know that

dv
(
0
)

i (0 )  C
dt
dv(0  ) i(0 )
3


 7.5 V/s
dt
C
0.4
Second Order Circuit
Applying KVL to the circuit when switch is opened,
i(0+)
1H
0.4F
4Ω
24V


 24  4 i(0 )  vL (0 )  v(0 )  0
 24  43  vL (0 )  12  0
vL (0 )  0

di
(
0
)

thus, vL (0 )  L
dt
di (0  ) vL (0  )

0
dt
L
Second Order Circuit
3. The steady state value,
i()
24V
4Ω
i ()  0
+
v()
-
v()  24V
Natural Response - Series RLC Circuit
•
Natural response is obtained with Series RLC circuit
without source.
R
L
IO
•
•
C
+
VO
-
Energy is initially stored in the L and C, where the
initial voltage at capacitor is VO and initial current at
inductor is IO.
Applying KVL to the loop,
di 1 t
iR  L   idt  0
dt C 
Natural Response - Series RLC Circuit
•
To eliminate the integral, differentiate with respect to
t and rearrange the terms such that
2
•
•
d i R di
i


0
2
dt
L dt LC
Now, we have second order differential equation.
It is known that in first-order differential equation, the
current is
i(t)  Ae
•
st
where A and s are constants.
Substitute Aest to the second order equation.
Natural Response - Series RLC Circuit
•
•
Thus, we may write as
st
AR st Ae
2 st
As e 
se 
0
L
LC
or
R
1 
st  2
Ae  s  s 
0
L
LC 

And we should find the value of A, thus Aest must
not equal to zero. The only part that should equal to
zero is
known as
R
1
2
CHARACTERISTIC
s  s
0
L
LC
EQUATION
Natural Response - Series RLC Circuit
•
Solve the characteristic equation, one might find the
two roots, which are
2
R
1
R
s1  
   
2L
 2L  LC
and
2
R
1
R 
s2  
   
2L
 2L  LC
Natural Response - Series RLC Circuit
•
On the other hand, we may represent the two roots
as
2
2
2
2
and
s      
s      
1
where
•
•
2
O
R

,
2L
O
1
O 
LC
The two solutions for s (i.e. s1 and s2) shows that
there are two values of the current, which are
s1t
s2t
i1  A1e and i2  A 2e
The total response of the current would be
i(t)  i1  i2  A1e  A 2e
s1t
s2t
Natural Response - Series RLC Circuit
•
Thus, we may found three type of response which
are
1. 1st Type :  > ω0 ,
the response is called OVERDAMPED
2. 2nd Type :  = ω0 ,
the response is called CRITICALLY DAMPED
3. 3rd Type :  < ω0 ,
the response is called UNDERDAMPED
Natural Response - Series RLC Circuit
• 1st Type :  > ω0 , OVERDAMPED
• In this type,  > ω0, or C > 4L/R2. It is found that
both roots, (i.e. s1 and s2) are negative and real.
Thus, the current response is
i(t)  i1  i2  A1e  A 2e
s1t
s2t
which decays to zero when t increased.
• A1 and A2 are determined from the initial inductor
current and the rate of change of current.
i(0)  A1  A 2

vL (0 ) di(0)

 A1s1  A 2s2
L
dt
Solve for
A1 and A2
Natural Response - Series RLC Circuit
•
For example
• A series RLC circuit has R=20Ω, L=1mH and
C=100F. If i(0+)=1A and vC(0+)=18V, find the
current response.
*It is clear that C > 4L/R2, and the response is the 1st
type which is ‘overdamped’.
Step1: Find the value of s1 and s2
2
R
1
R
s1  
   
 513.167 rad / s
2L
 2L  LC
2
R
1
R 
s2  
   
 19.487k rad / s
2L
 2L  LC
Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial
values.
i(0) 1  A1  A2
apply KVL to the loop


20 i(0 )  vL (0 )  v(0 )  0
20  vL (0 )  18  0
vL (0 )  2V
thus,


di(0 ) vL (0 )  2


 2k A / s  A1s1  A2 s2
dt
L
1m
Natural Response - Series RLC Circuit
Step3:
It is found that,
A1 = 0.9216, A2 = 78.36m
i(t)  0.9216 e
513.17 t
 78.36me
19486.8 t
A
1.0A
0.5A
0A
-2ms
0ms
2ms
4ms
6ms
8ms
10ms
12ms
14ms
16ms
18ms
Natural Response - Series RLC Circuit
• 2nd Type :  = ω0 , CRITICALLY DAMPED
• In this type,  = ω0, or C = 4L/R2. It is found that
both roots, (i.e. s1 and s2) are equal to – or
–R/2L. Thus, the current response is
i(t)  A1e t  A 2e t  ( A1  A 2 )e t  A 3 e t
• It can be seen that solution for A3 could not be
obtained with two initial condition [i(0) & di(0)/dt)].
• Thus, there might be another method to find the
response of critically damped.
Natural Response - Series RLC Circuit
•
In this type,  = ω0 = R/2L = 1/LC. Thus, the second
order differential equation become
d 2i
di
2

2



i0
2
dt
dt
d  di

 di

  i       i   0
dt  dt

 dt

•
di
Let f   i , thus
dt
d
f  f  0
dt
•
In first order differential equation, it is found that
f  A1e t , thus
 
di
d t
t
t di
t
 i  A 1e , or e
 e i  A 1 ,or
e i  A1
dt
dt
dt
Natural Response - Series RLC Circuit
•
By integrating the equation, one will found the
response of critically damped response as
i(t )  A1t  A 2 e
•
t
Thus, the derivative of the current is,


di (t )
t
t
 A 1t e 1  t   A 2e
dt
when t=0s,
di (0)
 A 1  A 2
dt
Natural Response - Series RLC Circuit
•
For example
• A series RLC circuit has R=20Ω, L=1mH and
C=10F. If i(0+)=1A and vC(0+)=18V, find the
current response.
*It is clear that C = 4L/R2, and the response is the 2nd
type which is ‘critically damped’.
Step1: Find the value of s1 and s2
2
R
1
R
s1  
   
   10k rad / s
2L
 2L  LC
2
R
1
R 
s2  
   
   10k rad / s
2L
 2L  LC
Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial
values.
i(0) 1  A 2
apply KVL to the loop


20 i(0 )  vL (0 )  v(0 )  0
20  vL (0 )  18  0
vL (0 )  2V
thus,


di(0 ) vL (0 )  2


 2k  A1  A2
dt
L
1m
Natural Response - Series RLC Circuit
Step3:
It is found that,
A1 = 8k, A2 = 1
i(t )  8000t  1e
10000t
1.0A
0.5A
0A
-0.5ms
I(L1)
0ms
0.5ms
1.0ms
Time
1.5ms
2.0ms
Natural Response - Series RLC Circuit
• 3rd Type :  < ω0 , UNDERDAMPED
• In this type,  < ω0, or C < 4L/R2.
• The roots can be written as
s1     (O   2 )    jd
2
s2     (O   2 )    jd
2
where
j  1
and
d  O 2   2
• The response can be further written as
i(t)  A1e
 (  jd ) t
 A 2e
 (  jd ) t
Natural Response - Series RLC Circuit
•
Rearranging the response such that,
i(t)  e
•
•
A e
1
jd t
 A 2e
 jd t

Applying Euler’s identities to the above equation,
where
e j  cos   j sin  , and e j  cos  j sin 
Thus,
i(t)  e
•
t
t
A1  A 2 cos d t  j A1  A 2 sin d t 
Let B1=A1+A2 and B2=j(A1–A2), thus
i(t)  e
t
B1 cos d t  B2 sin d t 
Natural Response - Series RLC Circuit
•
Differentiate the i(t), we have


di (t )
 B1 e t  sin d t d  cos d t (e t )
dt
 B 2 e t cos d t d  sin d t (e t )


Natural Response - Series RLC Circuit
•
For example
• A series RLC circuit has R=20Ω, L=1mH and
C=2F. If i(0+)=1A and vC(0+)=18V, find the
current response.
*It is clear that C < 4L/R2, and the response is the 3rd
type which is ‘underdamped’.
Step1: Find the value of s1 and s2
2
R
1
R 
s1  
   
 10k  j 20k rad / s
2L
 2L  LC
  10k
and
d  20k
Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial
values.
i(0) 1  B1
apply KVL to the loop


20 i(0 )  vL (0 )  v(0 )  0
20  vL (0 )  18  0
vL (0 )  2V
thus,
di(0 ) vL (0  )  2


 2k A / s  B1  B2d
dt
L
1m
Natural Response - Series RLC Circuit
Step3:
It is found that,
B1 = 1, B2 = 0.4,
  10k
and
d  20k
i(t)  e 10000t cos(20000t )  (0.4) sin( 20000t )
1.0A
0.5A
0A
-0.5A -0.4ms
I(L1)
-0.2ms
0ms
0.2ms
0.4ms
Time
0.6ms
0.8ms
1.0ms
Natural Response - Series RLC Circuit
•
In summary, the response for series RCL are
1. 1st Type :  > ω0 , (OVERDAMPED)
s1t
s2t
1
2
i(t)  A e  A e
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
i(t )  A1t  A 2 e
t
3. 3rd Type :  < ω0 , (UNDERDAMPED)
i(t)  e
t
B1 cos d t  B2 sin d t 
Natural Response - Series RLC Circuit
• The comparison of the three responses are
shown below
1.0A
C=100F
0.5A
C=10F
0A
C=2F
-0.5A 0s
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
I(L1)
Time
3.0ms
3.5ms
4.0ms
4.5ms
5.0ms
Natural Response - Parallel RLC Circuit
•
Natural response for Series RLC circuit is obtained
without any connection to source.
R
+
v
-
L
IO
C
•
Applying KCL, thus we have
•
v 1 t
dv
  vdt  C  0
R L 
dt
Differentiate with t,
2
d v 1 dv 1


v0
2
dt
RC dt LC
+
VO
-
Natural Response - Parallel RLC Circuit
•
Replace the first derivative as s and the second
derivative as s2. Thus, the characteristic equation is
obtained as follows,
1
1
2
s 
s
0
RC
LC
•
The roots are characterized by these equations
2
1
1
 1 
s1  
 
 
2RC
 2RC  LC
2
1
1
 1 
s2  
 
 
2RC
 2RC  LC
Natural Response - Parallel RLC Circuit
•
On the other hand, we may represent the two roots
as
2
2
2
2
and
s      
s      
1
where
O
1

,
2RC
2
1
O 
LC
O
Natural Response - Parallel RLC Circuit
• 1st Type :  > ω0 , OVERDAMPED
• In this type,  > ω0, or L > 4R2C. The roots, (i.e. s1
and s2) are negative and real. Thus, the voltage
response is
v(t)  A1e  A 2e
s1t
s2t
• A1 and A2 are determined from the initial inductor
current and the rate of change of current.
v(0)  A1  A 2
 v(0)  RiL (0)  dv(0)

 A 1s1  A 2s2

RC
dt


Solve for
A1 and A2
Natural Response - Parallel RLC Circuit
•
For example
• A parallel RLC circuit has R=20Ω, L=10mH and
C=1F. If iL(0+)=0.5A and v(0+)=10V, find the
current response.
Step1: Find the value of  and ω0
1
1


 25000
2RC 2(20)(1 )
1
1
O 

 10000
LC
(10m)(1 )
*It is clear that  > ωO , and the response is the 1st
type which is ‘overdamped’.
Natural Response - Parallel RLC Circuit
• Find s1 and s2.
s1      O  25k  (25k ) 2  (10k ) 2  2.087k
2
2
s2      O  25k  (25k ) 2  (10k ) 2  47.9k
2
2
• The initial condition are
v(0)  A1  A 2  10
dv(0)
v(0)  RiL (0)
10  ( 20)(0.5)


 1000k
dt
RC
( 20)(1 )
dv(0)
 A1s1  A 2s2  A1( 2.087k )  A 2 ( 47.9k )
dt
Natural Response - Parallel RLC Circuit
Step3:
It is found that,
A1 = -18.09, A2 = 28.09
v(t)  - 11.367 e
2.087k t
 21.367 e
47.9 k t
V
10V
5V
0V
-5V
-10V0s
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
V(R5:2)
Time
1.2ms
1.4ms
1.6ms
1.8ms
2.0ms
Natural Response - Parallel RLC Circuit
• 2nd Type :  = ω0 , CRITICALLY DAMPED
• In this type,  = ω0, or L = 4R2C. From the series
RLC circuit, the response found in parallel RLC
circuit is
v(t )  A1t  A 2 e
t
• The derivative is represented as


dv(t )
t
t
 A 1t e 1  t   A 2e
dt
and at t=0s,
dv(0)
 A 1  A 2
dt
Natural Response - Parallel RLC Circuit
•
For example
• A parallel RLC circuit has R=50Ω, L=10mH and
C=1F. If iL(0+)=0.5A and v(0+)=10V, find the
current response.
Step1: Find the value of  and ω0
1
1


 10000
2RC 2(50)(1 )
1
1
O 

 10000
LC
(10m)(1 )
*It is clear that  = ωO , and the response is the 2nd
type which is ‘critically damped’.
Natural Response - Parallel RLC Circuit
Step2: Find the value of A1 and A2 from the initial
values.
v(0) 10  A2
apply KCL to the loop
dv(0)
v(0)  RiL (0)
10  (50)(0.5)


 700k
dt
RC
(50)(1 )
dv(0)
 A1  A 2  A 1  A 2 (10k )
dt
Natural Response - Parallel RLC Circuit
Step3:
It is found that,
A1 = -600k, A2 = 10
v(t )  - 600kt  10e
10000t
10V
0V
-10V
-20V 0s
0.1ms
0.2ms
0.3ms
0.4ms
0.5ms
V(R5:2)
Time
0.6ms
0.7ms
0.8ms
0.9ms
1.0ms
Natural Response - Parallel RLC Circuit
• 3rd Type :  < ω0 , UNDERDAMPED
• In this type,  < ω0, or L < 4R2C.
• The roots can be written as
s1     (O   2 )    jd
2
s2     (O   2 )    jd
2
where
j  1
and
d  O 2   2
• The response can be further written as
v(t)  A1e
 (  jd ) t
 A 2e
 (  jd ) t
Natural Response - Parallel RLC Circuit
•
Applying the same method as in series, the
response of the parallel is
v(t)  e t A1  A 2  cos d t  j A1  A 2 sin d t 
•
Let B1=A1+A2 and B2=j(A1–A2), thus
v(t)  e
 t
B1 cos d t  B2 sin d t 
Natural Response - Parallel RLC Circuit
•
Differentiate the v(t), we have


dv(t )
 B1 e t  sin d t d  cos d t (e t )
dt
 B 2 e t cos d t d  sin d t (e t )


Natural Response - Parallel RLC Circuit
•
For example
• A parallel RLC circuit has R=80Ω, L=10mH and
C=1F. If iL(0+)=0.5A and v(0+)=10V, find the
current response.
Natural Response - Parallel RLC Circuit
•
In summary, the responses for parallel RCL are
1. 1st Type :  > ω0 , (OVERDAMPED)
v(t)  A1e  A 2e
s1t
s2t
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
v(t )  A1t  A 2 e
t
3. 3rd Type :  < ω0 , (UNDERDAMPED)
v(t)  e
 t
B1 cos d t  B2 sin d t 
Step Response - Series RLC Circuit
•
Step response is obtained in Series RLC circuit with
source.
R
L
VS
i
C
+
v
-
•
Applying KVL to the loop,
dv
di
iR  L  v  VS ,where i  C
dt
dt
•
Substitute i to the derivative terms,
2
VS
d v R dv v



2
dt
L dt LC LC
2nd order diff.
equ for capacitor
voltage
Step Response - Series RLC Circuit
•
•
•
•
To obtain the total response, consider the KVL of the
loop
di 1 t
iR  L   idt  VS
dt C 
Differentiate the equation, we get
di
d 2i i
R L 2   0
dt
dt
C
It can be seen that the characteristic equation is the
same with source-free series RLC.
Thus, by looking at the roots of the characteristic
equation, one may find the type of the response
(either overdamped, critically damped or
underdamped).
Step Response - Series RLC Circuit
•
Furthermore, it can be concluded that the total
response can be represented in terms of ‘transient’
and ‘steady-state’ value,
v(t )  vt (t )  vSS (t )
•
where vt(t) is the ‘transient response’ and vss(t) is the
‘steady-state’ response.
The vt(t) is the part which will determine the type of
the response, and it reflects to the response for
source-free series RLC circuit.
Step Response - Series RLC Circuit
•
The transient responses are:
vt(t)  A 1e s1t  A 2e s2t
vt (t )  A 1t  A 2 e
overdamped
 t
critically damped
vt(t)  e t B1 cos d t  B2 sin d t 
•
underdamped
While vss(t) is the final voltage value (i.e. when t=)
across the capacitor. Normally, it will equal to Vs.
vSS(t)  v()  VS
Step Response - Series RLC Circuit
•
Thus, the complete responses for series RLC with
source are:
1. 1st Type :  > ω0 , (OVERDAMPED)
v(t)  VS  A1e  A 2e
s1t
s2t
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
v(t )  VS  A 1t  A 2 e
 t
3. 3rd Type :  < ω0 , (UNDERDAMPED)
v(t)  VS  B1 cos d t  B2 sin d t e
 t
Step Response - Series RLC Circuit
Example
• A series RLC circuit is shown below. If i(0+)=1A and
vC(0+)=18V, find i(t) and v(t).
20Ω 1mH
i(t) 100F
20V
+
v(t)
-
Step 1: Find  and ω0. Determine type of response.
1
R
 3.162k rad / s

 10k rad / s 0 
LC
2L
It is found that  > ω0, type of response  overdamped.
s1,2      0  513.167,  19.487k rad / s
2
2
Step Response - Series RLC Circuit
Step 3: The steady state voltage across capacitor is
vSS(t)  v()  VS  20V
Step2: Find the value of A1 and A2 from the initial
values.
v(0)  18  20  A1  A 2
A1  A 2  2
the derivative of the response at initial is
dv(0 ) i(0 )
1


 10k V / s  A1s1  A2 s2
dt
C
100
thus,
A1  1.527,
A 2  0.473
Step Response - Series RLC Circuit
Step3:
It is found that the complete response is
v(t)  20  1.527 e513.17t  0.473 e19486.8 tV
dv(t )
i(t)  C
dt
513.17 t
19486.8 t
 78.361m e
 921.68me
A
•
Next, find the voltage response if R is changed to
6.324Ω and then to 2Ω
Step Response - Series RLC Circuit
•
Comparison of the three responses with different R
value.
2Ω
22V
6.324Ω
21V
20V
20Ω
19V
18V
0s
1ms
2ms
3ms
4ms
5ms
V(L2:2)
Time
6ms
7ms
8ms
9ms
10ms
Step Response - Parallel RLC Circuit
•
Step response is obtained in Parallel RLC circuit
with source.
IS
R
C
L
i
•
Applying KCL to the loop,
di
v
dv
 C  i  I S ,where v  L
dt
R
dt
•
Substitute i to the derivative terms,
2
IS
d i
1 di
i



2
dt RC dt LC LC
+
v
-
2nd order diff.
equ for capacitor
voltage
Step Response - Parallel RLC Circuit
•
•
It has the same characteristic equation to that of
natural response of parallel/series RLC.
Furthermore, it can also be concluded that the total
response can be represented in terms of ‘transient’
and ‘steady-state’ value,
i(t )  it (t )  iSS (t )
•
where it(t) is the ‘transient response’ and iss(t) is the
‘steady-state’ response.
The it(t) is the part which will determine the type of
the response, and it reflects to the response for
source-free parallel RLC circuit.
Step Response - Parallel RLC Circuit
•
The transient responses are:
it(t)  A 1e s1t  A 2e s2t
it (t )  A 1t  A 2 e
overdamped
t
critically damped
it(t)  e t B1 cos d t  B2 sin d t 
•
underdamped
While iss(t) is the final voltage value (i.e. when t=)
across the capacitor. Normally, it will equal to Is.
iSS(t)  i()  IS
Step Response - Parallel RLC Circuit
•
Thus, the complete responses for parallel RLC with
source are:
1. 1st Type :  > ω0 , (OVERDAMPED)
i(t)  I S  A 1e  A 2e
s1t
s2t
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
i (t )  I S  A 1t  A 2 e
t
3. 3rd Type :  < ω0 , (UNDERDAMPED)
i(t)  IS  B1 cos d t  B2 sin d t e
t
General Second-Order Circuit
•
•
•
•
•
Previously, only the second-order series and parallel
circuits are considered.
If the circuits were neither in series nor in parallel,
what method should be used?
In this topic, we would concentrate to find the
response for a second-order circuit which is neither
a series nor parallel circuit.
The response might be in terms of voltage or
current. Thus, generally the response are
characterized as x(t).
In general 2nd-order circuit, the most important part
is to find the characteristic equation and find the two
roots (s1 and s2). From the two roots, one should
know the type of the response.
General Second-Order Circuit
•
•
•
The type of the response are the same with series
and parallel RLC circuit.
Finally, find the initial and steady state value (x(0),
dx(0)/dt and x()) to determine the constant value;
i.e. A1 and A2.
Generally, the response is the summation of the
‘transient’ and ‘steady-state’ value and can be
expressed as
x(t )  xt (t )  xSS (t )
where x is either in voltage or current.
General 2nd-Order Circuit – step by step
Turn off the independent source to find the second
order differential equation using KCL and/or KVL.
Find the roots and determine the type
of the response (i.e. O-D, C-D or U-D)
The 2nd order differential equation would determine
whether the response is voltage or current!
Find the steady-state and initial value
to determine A1 and A2.
General Second-Order Circuit - eg
•
Consider a circuit shown below. Find the complete
response of the voltage v(t)?
4Ω
1H
i
2Ω
0.5F
12V
t >0s
+
v(t)
-
Step 1: Turn off the independent source. Find the
second order differential equation.
4Ω
1H
i
v
2Ω
0.5F
+
v(t)
-
using KCL and KVL
to find the 2nd order
differential equations,
General Second-Order Circuit - eg
KCL at node ‘v’
i  iR  iC
v
dv
i  C
R
dt
v 1 dv
 
4 2 dt
KVL at mesh ‘i’
di
iR  L  v  0
dt
di
4i  (1)  v  0
dt
We just concern on ‘v’. Thus, substitute ‘i’ to the right hand
side equation
2
d v
dv
 5  6v  0
2
dt
dt
Thus, the characteristic equation is,
s 2  5s  6  0
General Second-Order Circuit - eg
Step 2: The two roots are s1 = -2, and s2 = -3
*It is obvious that the two roots are negative and real. Thus, the
transient response is the 1st type (OVERDAMPED)
Step 3: The standard response is
v(t )  vt (t )  vSS (t )
The steady state voltage is
vSS (t )  v(0)  4V
The transient voltage is
vt(t)  A1e
2 t
 A 2e
3t
General Second-Order Circuit - eg
Step 4: The initial voltage and current is,
v(0 )  v(0 )  12V
i (0  )  i ( 0  )  0 A
At just after the switch is closed, the circuit is shown as
iC
4Ω
1H
+
i
2Ω 0.5F v(t)
12V
iR
-
i (0  )  iR (0  )  iC (0  )


v(0 ) 1 dv(0 )
i (0 ) 

2
2 dt
dv(0  )
 12V / s
dt

General Second-Order Circuit - eg
The overall response is
v(t)  4  A1e
2t
 A 2e
3t
From the overall response, at initial

v(0 )  4  A1  A 2  12

dv(0 )
 2A1  3A 2  12
dt
It is found that A1 = 12 and A2 = -4.
Thus,
v(t)  4  12e
2t
 4e
3t