533571874
The simplest second-order system is one for which the L’s can be combined, the C’s can be
combined, and the R’s can be combined into one L, one C, and one R. Let us consider situations
where there is no source, or a DC source.
For an RLC Series Circuit
If the circuit is a series circuit, then there is one current.
i
+ vR –
+ vL –
R
L
C
V
+
vC
–
Note that in the above circuit, the V and R could be converted to a Norton equivalent, permitting
the incorporation of a current source into the problem instead of a voltage source.
Using KVL,
 V  v R  v L  vC  V  iR  L
di 1

idt  0 .
dt C 
Taking the derivative,
R
di
d 2i i
L
  0.
dt
dt 2 C
Note that the constant source term V is gone, leaving us with the characteristic equation (i.e.,
source free), which yields the natural response of the circuit. Rewriting,
d 2i
dt 2

R di
i

 0.
L dt LC
(1)
Guess the natural response solution i  Ae st and try it,
As 2 e st 
R
1
R
1 

Ase st 
Ae st  Ae st  s 2  s 
  0,
L
LC
L LC 

And we reason that the only possibility for a general solution is when
s2  s
R
1

0.
L LC
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This leads us to

s
2
R
4
R
   
2
L
LC
R
1
L
 R 
.

   
2
2L
LC
 2L 
Defining

R
as the damping coefficient (nepers/sec) for the series case
2L
1
 o
LC
as the natural resonant frequency (radians/sec)
(2)
(3)
then we have
s     2   o2 .
(4)
Substituting (2) and (3) into (1) yields a useful form of (2), which is
d 2i
dt 2
 2
di
 o2i  0
dt
(5)
For an RLC Parallel Circuit
If the circuit is a parallel circuit, then there is one voltage. KCL at the top node yields
R
L
C
I
iR
 I  i R  i L  iC   I 
iL
v 1
dv
  vdt  C
 0.
R L
dt
Taking the derivative
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iC
+
v
–
533571874
1 dv v
d 2v
 C
 0.
R dt L
dt 2
Rewriting,
d 2v

dt 2
1 dv
v

 0.
RC dt LC
(6)
Equation (6) has the same form as (5), except that

1
for the parallel case.
2 RC
(7)
Thus, it is clear that the solutions for series and parallel circuits have the same form, except that
the damping coefficient  is defined differently.
A nice property of exponential solutions such as Ae st is that derivatives and integrals of
Ae st also have the e st term. Thus, one can conclude that all voltages and currents throughout
the series and parallel circuits have the same e st terms.
There are Three Cases
Equation (4)
s     2   o2
has three distinct cases:
1. Case 1 is where    o , so that the term inside the radical is positive. Overdamped.
2. Case 2 is where    o , so that the term inside the radical is negative. Underdamped.
3. Case 3 is where    o , so that the term inside the radical is zero. Critically Damped.
Case 3 rarely happens in practice because the terms must be exactly equal. Case 3 should be
thought of as a transition from Case 1 to Case 2, and the transition is very gradual.
Case 1: Overdamped
When    o , then the solutions for (4) are both negative real and are
s1     2   o2 ,
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s 2     2   o2 .
The natural response for any current (or voltage) in the circuit is then
i(t )  A1e s1t  A2 e s 2 t .
Coefficients A1 and A2 are found from the boundary conditions as follows:
i(t  0)  A1e s1t
t 0
 A2 e s 2t
t 0
 A1  A2 ,
di
 A1s1e s1t
 A2 s 2 e s 2 t
 A1s1  A2 s 2 .
dt t  0
t 0
t 0
Thus, we have two equations and two unknowns. To solve, one must know i (t  0) and
di
.
dt t  0
Solving, we get
 di

 A1s1 
 dt
,
A1  i (t  0)  A2  i (t  0)   t  0


s2




so


di


i (t  0)  dt t  0 

s2 



 (Note – see modification needed to include final response) (8)
A1 
s1
1
s2
Then, find A2 from
A2  i(t  0)  A1 (Note – see modification needed to include final response)
(9)
The key to finding A1 and A2 is always to know the inductor current and capacitor voltage
at t = 0+. Remember that

Unless there is an infinite impulse of current through a capacitor, the voltage across a
capacitor (and the stored energy in the capacitor) remains constant during a switching
transition from t = 0- to t = 0+.
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
Unless there is an infinite impulse of voltage across an inductor, the current through an
inductor (and the stored energy in the inductor) remains constant during a switching
transition from t = 0- to t = 0+.
For example, consider the series circuit at t = 0+. Current i(t  0)  I L0  .
IL0+
+ RIL0+ –
–
+ VL0+ –
R
L
+
C
V
Once IL0+ and VC0+ are known, then we get
VC0+
–
di
as follows. From KVL
dt t  0
 V  RI L0   VL0   VC 0   0
thus
VL0   V  RI L0   VC 0  .
(10)
Since
VL0   L
di
dt t  0
then we have
V
di
 L0  .
dt t  0
L
(11)
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Similarly, for the parallel circuit at t = 0+, the voltage across all elements v(t  0)  VC 0  .
R
L
C
I
IR0+
We get
IL0+
IC0+
+
VC0+
–
dv
as follows. From KCL
dt t  0
 I  I R0   I L0   I C 0   0
thus
I C 0   I  I R 0   I L0  .
(12)
Since
IC0  C
dv
dt t  0
then we have
I
dv
 C0 .
dt t  0
C
(13)
Case 2: Underdamped
When    o , then the solutions for (4) are both complex and are
s1     2   o2    (1)( o2   2 )    j  o2   2 ,
s 2    j  o2   2 .
Define damped resonant frequency d as
 d   o2   2 ,
(14)
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so that
s1    jd ,
s2    jd
The natural response for any current (or voltage) in the circuit is then
i(t )  A1e s1t  A2 e s 2 t  A1e (  j d )t  A2 e (  j d )t ,
 A1e t e j d t  A2 e t e  j d t .
Expanding the e jd t and e  j d t terms using Euler’s rule, e j  cos( )  j sin(  ) ,
i (t )  A1e t cos( d t )  j sin(  d t )  A2 e t cos(  d t )  j sin(  d t ) ,
 A1e t cos( d t )  j sin(  d t )  A2 e t cos( d t )  j sin(  d t ) ,
 e t ( A1  A2 ) cos( d t )  j ( A1  A2 ) sin(  d t ) .
(15)
Since i(t) is a real value, then (15) cannot have an imaginary component. This means that
( A1  A2 ) is real, and j ( A1  A2 ) is real (which means that ( A1  A2 ) is imaginary). These
conditions are met if A2  A1* . Thus, we can write (15) in the form of
i (t )  e t B1 cos( d t )  B2 sin(  d t )
(16)
where B1 and B2 are real numbers.
To evaluate B1 and B2, it follows from (16) that
i(t  0)  e 0 B1 cos(0)  B2 sin( 0)  B1 ,
so
B1  i(t  0) (Note – see modification needed to include final response)
(17)
To find B2 , take the derivative of (16) and evaluate it at t = 0,
di
 e t B1 cos( d t )  B2 sin(  d t )  e t  B1 d sin(  d t )  B2 d cos( d t ) ,
dt
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di
  B1   B2 d   B1  B2 d ,
dt t  0
so we can find B2 using
di
 B1
dt t  0
B2 
(18)
d
Case 2 Solution in Polar Form
The form in (16), i (t )  e t B1 cos( d t )  B2 sin(  d t ) is most useful in evaluating coefficients
B1 and B2. But in practice, the answer is usually converted to polar form. Proceeding, write
(16) as
i(t ) 

B12  B22  e t 
B1
 B2  B2
2
 1
cos( d t ) 

sin(  d t )

2
2
B1  B2

B2
i(t )  B12  B22  e t cos( ) cos(d t )  sin(  ) sin( d t )
The
B1
B12  B22
and
B2
B12  B22
terms are the cosine and sine, respectively, of the angle shown
in the right triangle. Unless B1 is zero, you can find θ using
B2
θ
 B2 
.
 B1 
  tan 1 
B1
But be careful because your calculator will give the wrong answer 50% of the time. The reason
is that tan( )  tan(  180) . So check the quadrant of your calculator answer with the
quadrant consistent with the right triangle. If the calculator quadrant does not agree with the
figure, then add or subtract 180° from your calculator angle, and re-check the quadrant.
The polar form for i(t) comes from a trigonometric identity. The expression
i(t )  B12  B22  e t cos( ) cos(d t )  sin(  ) sin( d t )
becomes the damped sinusoid
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i(t )  B12  B22  e t cos( d t   ) .
Case 3: Critically Damped
When    o , then there is only one solution for (4), and it is s1  s2   . Thus, we have
repeated real roots. Proceeding as before, then there appears to be only one term in the natural
response of i(t),
i(t )  Ae t .
But this term does not permit the natural response to be zero at t = 0, which is a problem. Thus,
let us propose that the solution for the natural response of i(t) has a second term of the form
te t , so that
i(t )  A1tet  A2 e t   A1t  A2 e t .
(19)
At this point, let us consider the general form of the differential equation given in (5), which
shown again here is
d 2i
dt
2
 2
di
 o2i  0 .
dt
For the case with    o , the equation becomes
d 2i
dt
2
 2
di
  2i  0 .
dt
We already know that Ae t satisfies the equation, but let us now test the new term Ate  at .
Substituting in,




A  e t  e t   2tet  2A e t  tet   2 Atet  0 ?
Factoring out the Ae t term common to all yields
      2t  2 1  t    2t  0 ? (Yes!)
So, we confirm the solution of the natural response.
i(t )  A1tet  A2 e t   A1t  A2 e t
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(20)
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To find coefficient A2, use
i(t  0)   A1  0  A2 e 0  A2 .
Thus,
A2  i(t  0) . (Note – see modification needed to include final response)
(21)
To find A1 , take the derivative of (20),
di
 A1e t  A1te t  A2 e t   A1  A1t  A2 e t .
dt
Evaluating at t = 0 yields
di
 A1  A2 .
dt t  0
Thus,
A1 
di
 A2 .
dt t  0
(22)
The question that now begs to be asked is “what happens to the te t term as t → ∞? Determine
this using the series expansion for an exponential.
We know that
et  1 
t   t 2  t 3   .
te t 
t
1!
2!
3!
Then,
et

t
t   t 2  t 3  
1
1!
2!
.
3!
Dividing numerator and denominator by t yields,
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te t 
As t → ∞, the
t
e t

1
1 t  t 2 t 3




t t  1! t  2! t  3!

1
1   2 t  3t 2
 


t 1!
2!
3!
.
1
in the denominator goes to zero, and the other “t” terms are huge. Since the
t
numerator stays 1, and the denominator becomes huge, then tet → 0.
The Total Response = The Natural Response Plus The Final Response
If the circuit has DC sources, then the steady-state (i.e., “final”) values of voltage and current
may not be zero. “Final” is the value after all the exponential terms have decayed to zero. And
yet, for all the cases examined here so far, the exponential terms all decayed to zero after a long
time. So, how do we account for “final” values of voltages or currents?
Simply think of the circuit as having a total response that equals the sum of its natural response
and final response. Add the final term, e.g.
i(t )  I final  A1e s1t  A2 e s 2 t ,
so
i(t )  I final   A1e s1t  A2e s2t (for overdamped).
(23)
Likewise,
i(t )  I final  e t B1 cos( d t )  B2 sin(  d t ),
i(t )  I final   e t B1 cos(d t )  B2 sin( d t ) (for underdamped).
(24)
Likewise,
i(t )  I final   A1t  A2 e t ,
i(t )  I final    A1t  A2 e t (for critically damped).
(25)
You can see that the presence of the final term I final will affect the A and B coefficients because
the initial value of i(t) now contains the I final term. Take I final into account when you
evaluate the A’s and B’s by replacing i (t  0) in (8), (9), (17), and (21) with
i(t  0)  I final .


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How do you get the final values? If the problem has a DC source, then remember that after a
long time when the time derivatives are zero, capacitors are “open circuits,” and inductors are
“short circuits.” Compute the “final values” of voltages and currents according to the “open
circuit” and short circuit” principles.
The General Second Order Case
Second order circuits are not necessarily simple series or parallel RLC circuits. Any two noncombinable storage elements (e.g., an L and a C, two L’s, or two C’s) yields a second order
circuit and can be solved as before, except that the  and  o are different from the simple series
and parallel RLC cases. An example follows.
R2
C
R1
V
IL
+ VC –
L
General Case Circuit #1 (Prob. 8.56)
Work problems like Circuit #1 by defining capacitor voltages and inductor currents as the N state
variables. For Circuit #1, N = 2. You will write N circuit equations in terms of the state
variables, and strive to get an equation that contains only one of them. Use variables instead of
numbers when writing the equations.
For convenience, use the simple notation VC and I L to represent time varying capacitor voltage


and inductor current, and VC and I L to represent their derivatives, and so on.
To find the natural response, turn off the sources. Write your N equations by using KVL and
KCL, making sure to include each circuit element in your set of equations.
For Circuit #1, start with KVL around the outer loop,

R1 I L  VC  L I L  0 .
(26)
Now, write KCL at the node just to the left of the capacitor,

V
 I L  C  C VC  0 ,
R2
which yields
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
V
I L  C  C VC .
R2
(27)
Taking the derivative of (27) yields

IL


V
 C  C VC .
R2
(28)

We can now eliminate I L and I L in (26) by substituting (27) and (28) into (26), yielding
 

 

VC
VC

R1 
 C VC   VC  L 
 C VC   0 .
R
 R2

 2

Gathering terms,


LC VC   R1C 

L    R1 
 1VC  0 ,
 VC  
R2 
 R2 
and putting into standard form yields
  R
1  
1  R1 
VC   1 
 1VC  0
 VC 

LC  R2 
 L R2C 
(29)
Comparing (29) to the standard form in (5), we see that the circuit is second-order and
o2 
R
1 
1  R1
1 
1  R1 
 1 , 2   1 
 , so     
.

2  L R2C 
LC  R2 
 L R2C 
The solution procedure for the natural response and total response of either vC (t ) or i L (t ) can
then proceed in the same way as for series and parallel RLC circuits, using the  and  o values
shown above.
Notice in the above two equations that when R2   ,
 o2 
R
1
, and   1 ,
LC
2L
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533571874
which correspond to a series RLC circuit. Does this make sense for this circuit?
Note that when R1  0 , then
 o2 
1
1
, and  
,
LC
2 R2 C
which correspond to a parallel RLC circuit. Does this make sense, too?
Now, consider Circuit #2.
R2
I
R1
IL1
L1
IL2
L2
General Case Circuit #2 (Prob. 8.60)
Turning off the independent source and writing KVL for the right-most mesh yields


 L1 I L1  R2 I L 2  L2 I L 2  0 .
(30)
KVL for the center mesh is

R1 I L1  I L 2   L1 I L1  0 ,
yielding
L 
I L 2   I L1  1 I L1 .
R1
(31)
Taking the derivative of (31) yields


L 
I L 2   I L1  1 I L1 .
R1
(32)
Substituting into (31) and (32) into (30) yields


 L1 I L1  R2  I L1 

 
L1  
I L1   L2  I L1 
R1


L1   
I L1   0 .
R1

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Gathering terms,
 L1L2   
 
R2 L1

I


L


L
2  I L1  R2 I L1  0 ,

 L1  1
R
R
1
1




and putting into standard form yields
  R
R
R   R R 
I L1   1  2  1  I L1   1 2  I L1  0 .
 L2 L2 L1 
 L1L2 
(33)
Comparing (33) to (5),
o2 
R
R1R2
R
R 
R
R 
1 R
, 2   1  2  1  , so     1  2  1  .
L1L2
2  L2 L2 L1 
 L2 L2 L1 
(34)
The solution procedure for the natural response and total response of either i L1 (t ) or i L2 (t ) can
then proceed in the same way as for series and parallel RLC circuits, using the  and  o values
shown above.
Normalized Damping Ratio
Our previous expression for solving for s was
s 2  2s   o2  0 .
(35)
Equation (35) is sometimes written in terms of a “normalized damping ratio”  as follows:
s 2  2 o s   o2  0 .
(36)
Thus, the relationship between damping coefficient  and normalized damping ratio  is
 

.
o
(37)
Normalized damping ratio  has the convenient feature of being 1.0 at the point of critical
damping. When  < 1.0, the response is underdamped. When  < 1.0, the response is
overdamped. Examples for a unit step input are shown in the following figure.
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533571874
Response of Second Order System
(zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)
0.1
1.8
0.2
1.6
1.4
0.4
1.2
1
0.8
0.6
0.99
0.4
0.2
0
0
2
4
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533571874
Notes on Solving the Three Cases
s     2   o2 has three distinct cases:
 Case 1 is where    o , so that the term inside the radical is positive. Overdamped.
 Case 2 is where    o , so that the term inside the radical is negative. Underdamped.
 Case 3 is where    o , so that the term inside the radical is zero. Critically Damped.
Case 1: Overdamped (    o , so s1 and s2 are both negative real)
The natural (i.e., source free) response for any current or voltage in the network has the form

A1e s1t  A2 e s 2 t

s1     2   o2 ,

s 2     2   o2 .
The key to finding A1 and A2 is always to know the inductor current and capacitor voltage at t =
0+. Remember that

Unless there is an infinite impulse of current through a capacitor, the voltage across a
capacitor (and the stored energy in the capacitor) remains constant during a switching
transition from t = 0- to t = 0+.

Unless there is an infinite impulse of voltage across an inductor, the current through an
inductor (and the stored energy in the inductor) remains constant during a switching
transition from t = 0- to t = 0+.
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Case 2: Underdamped (    o ,so s1 and s2 are complex conjugates)
The natural (i.e., source free) response for any current or voltage in the network has the
rectangular form

i (t )  e t B1 cos( d t )  B2 sin(  d t )

s1    j  o2   2 ,

s 2    j  o2   2 .

Damped resonant frequency  d   o2   2 ,
so that

s1    jd ,

s2    jd

B1  i(t  0) (Note – see modification needed to include final response)
(17)

di
 B1
dt t  0
B2 
(18)
(16)
(14)
d
In polar form,

i(t )  B12  B22  e t cos( d t   )
B2
θ
1  B2 

  tan

Regarding the arctangent - be careful because your calculator will give the wrong
answer 50% of the time. The reason is that tan( )  tan(  180) . So check the
quadrant of your calculator answer with the quadrant consistent with the right
triangle. If the calculator quadrant does not agree with the figure, then add or subtract
180° from your calculator angle, and re-check the quadrant.
 .
 B1 
B1
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Case 3: Critically Damped (    o , so s1  s2   )

i(t )  A1tet  A2 e t   A1t  A2 e t .
(19)

A2  i(t  0) . (Note – see modification needed to include final response)
(21)

A1 
di
 A2 .
dt t  0
(22)
Total Response = Natural Response Plus Final Response
If the circuit has DC sources, then the final response is also DC. The total response is

i(t )  I final  A1e s1t  A2 e s 2 t ,

i(t )  I final   A1e s1t  A2e s2t
(23)
The effect of Ifinal on determining A1 and A2 simply means that you use (23) at t = 0 instead of
i(t = 0) by itself.
If the circuit has AC sources, the final response is the phasor solution.
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Normalized Damping Ratio
Our previous expression for solving for s was
s 2  2s   o2  0 .
(35)
Equation (35) is sometimes written in terms of a “normalized damping ratio”  as follows:
s 2  2 o s   o2  0 .
(36)
Thus, the relationship between damping coefficient  and normalized damping ratio  is
 

.
o
(37)
Normalized damping ratio  has the convenient feature of being 1.0 at the point of critical
damping. When  < 1.0, the response is underdamped. When  < 1.0, the response is
overdamped. Examples for a unit step input are shown in the following figure.
Response of Second Order System
(zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)
0.1
1.8
0.2
1.6
1.4
0.4
1.2
1
0.8
0.6
0.99
0.4
0.2
0
0
2
4
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