Fundamentals of Electric Circuit - Chapter 8: Second-order Circuits



 
 
  

R
L
LC
1
0
2,5
2
1
2
  
 
      
 
s
s
s
12 2
1,2 0
2
1
4
 R1 = 5Ω 
1Ω 
 For t > 0: 
        t t t t
f
v t v A e A e A e A e
4 4
1 2 1 2
( ) 24
 At t = 0:        v A A A A1 2 1 2(0) 24 4 20
            t t
dv t
i t C C A e A e i C A A
dt
4
1 2 1 2
( )
( ) 4 (0) 4 4
      t tv t e e V4
4
( ) 24 16
3
     t t
dv
i t C e e A
dt
44( ) 4
3
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
R 1
2 4 V
1 H
L
0,5F
C
R 2
21 
Chapter 8: Second-order circuits 
IV. Step response of a series/parallel RLC circuit 
IV.1. Step response of a series RLC circuit 
Ex 8.5: Find v(t) and i(t) for t > 0 in the case of the 
different values of R1 = 5Ω, 4Ω, 1Ω 
   

i A v i V
R R
1 2
24
(0) 4,5 ; (0) 1. (0) 4,5
i t = 0 
- 
+ 
v 
 For t = 0: 



 
 
  

R
L
LC
1
0
2
2
1
2
 R1 = 4Ω 
1Ω 
 For t > 0: 
      t t
f
v t v A A t e A A t e
2 2
1 2 1 2
( ) ( ) 24 ( )
 At t = 0:      v A A1 1(0) 24 4,5 19,5
              t
dv t
i t C C A tA A e i C A A A
dt
2
1 2 2 1 2 2
( )
( ) 2 2 (0) 2 4,5 57
       tv t t e V2( ) 24 19,5 57       t
dv
i t C t e A
dt
2( ) 4,5 28,5
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
R 1
2 4 V
1 H
L
0,5F
C
R 2
22 
Chapter 8: Second-order circuits 
IV. Step response of a series/parallel RLC circuit 
IV.1. Step response of a series RLC circuit 
Ex 8.5: Find v(t) and i(t) for t > 0 in the case of the 
different values of R1 = 5Ω, 4Ω, 1Ω 
   

i A v i V
R R
1 2
24
(0) 12 ; (0) 1. (0) 12
i t = 0 
- 
+ 
v 
 For t = 0: 



 
 
  

R
L
LC
1
0
0,5
2
1
2
 R1 = 1Ω 
1Ω 
 For t > 0: 
   tv t A t A t e 0,5
1 2
( ) 24 ( cos1,936 sin1,936 )
 At t = 0:      v A A1 1(0) 24 12 12
      t t
dv t
e A t A t e A t A t
dt
0,5 0,5
1 2 1 2
( )
( 1,936 sin1,936 1,936 cos1,936 ) 0,5 cos1,936 sin1,936
     t
t
v t t e V
0,5
0 0
( ) 24 (21,694sin 12cos )     ti t t t e A0,50 0( ) (3,1sin 12cos )
         s j2 2
1,2 0
0,5 1,936
      
dv i
A A A
dt C
2 1 2
(0) (0)
1,936 0,5 48 21,694
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
R
L
C
23 
Chapter 8: Second-order circuits 
IV. Step response of a series/parallel RLC circuit 
IV.2. Step response of a parallel RLC circuit 
i 
t = 0 
- 
+ 
V 
 Applying KCL at the top node for t > 0 
IS 
  
n
n f
f
i t natural response
i t i t i t 
i t forced response 
( ) :
( ) ( ) ( )
( ) :

      
di
v L
Sdt
S
Iv dv d i di i
i C I
R dt dt RC dt RC LC
2
2
1
  s t s t
S
i t I A e A e1 2
1 2
( ) Over damped: 
     tSi t I A A t e1 2( ) Critically damped: 
       tS d di t I A t A t e1 2( ) cos sin Under damped: 
A1, 2: determine from 
the initial conditions: 
i(0), di(0)/dt 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
R 2
R 1
2 0 H
L
8mF
C
30u(-t)V
24 
Chapter 8: Second-order circuits 
IV. Step response of a series/parallel RLC circuit 
IV.2. Step response of a parallel RLC circuit i 
t = 0 
- 
+ 
V Ex 8.6: Find i(t), iR(t) for t > 0 
4A 
i A 
R di v
v V
R R dt L
1
1 2
(0) 4
(0) (0)
(0) 30 15 0,75



     
  

R R
R
R R
1 2
1 2
10

  

 
 
 
         
  
t t
f
sRC
s i t I A e A e
s
LC
12 2 11,978 0,5218
1,2 0 1 2
2
0
1
6,25
11,9782
( )
1 0,5218
2,5
iR 
 For t < 0: 
 For t > 0: We have a parallel RCL circuit with current source 
 At t = 0: 
      
 
 
    
i A A A A
A
di
AA A
dt
1 2 1 2
1
21 2
(0) 4 4
0.0655
(0)
0.065511,978 0,5218 0,75
   t ti t e e A0.5218 11.978( ) 4 0.0655( )
t t
R
di
i t L e e A
dt
11,978 0,52181( ) 0,785 0,0342
20
   
20Ω 
20Ω 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
25 
Chapter 8: Second-order circuits 
V. General second order circuits 
 Give a second order circuit, the step response x(t) (current or voltage) can be 
determined by taking the following 5 steps: 
 Determine the initial conditions x(0) and dx(0)/dt and the final value x(∞) 
 Find the natural response xn(t) (with 2 unknown constants) by turning off 
independent sources and applying KCL and KVL. 
 Obtain the forced response as: xf(t) = x(∞) 
 The total response is the sum of the natural response and forced response 
x(t) = xn(t) + xf(t) 
 Determine the 2 unknown constants by imposing the initial conditions x(0) 
and dx(0)/dt. 
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
26 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.7: Find the complete response v and i for t > 0. 
Applying KCL: 
1 2 V
0,5F
C
R 2
1 H
L
R 1
2Ω 
4Ω i 
t = 0 
 Find the initial and final values: 
v V v v V
i i i 
( 0) 12 ( 0) ( 0) 12
( 0) 0 ( 0) ( 0) 0
      
 
      
C
C C
iv dv
i i i A V s
R dt C
2
( 0)( 0) ( 0)
( 0) ( 0) ( 0) 6 12 /
 
           
f
i A v i V v t
R R
1 2
12
( ) 2 , ( ) 2. ( ) 4 ( )       

 Find the natural response: Turn off the voltage source, and apply KCL, KVL 
t t
n
v dv
i C d v dv
R dt v
v t Ae Bedt dt
di
s si L v
dt
2
2 32 2
2
5 6 0
( )
5 6 04 0
 

    
   
     

Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
27 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.7: Find the complete response v and i for t > 0. 
 Imposing the initial condition gives: 
1 2 V
0,5F
C
R 2
1 H
L
R 1
2Ω 
4Ω i 
t = 0 
 The complete response is: 
t t
n f
v t v t v t A e B e
2 3( ) ( ) ( ) 4 . .     
A B 
A
dv
BA B
dt
8
12
(0)
42 3 12
 

 
      
t t t t t tv dv
i C e e e e e e A t
R dt
2 3 2 3 2 3
2
2 6 2 12 6 2 6 4 , 0
               
t t
v t e e V t
2 3( ) 4 12 4 , 0     
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
7u(t)V
R 2
0,5H
L 1
R 1
0,2H
L 2
28 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.8: Find v0(t) for t > 0 
 Applying KVL to the left loop at t = +0 
1Ω 
3Ω 
 Obtain the initial and final values of 2 currents 
For t u t i i
1 2
0 : 7 ( ) 0 ( 0) 0 ( 0)      

 
         
 
L
L L
L
di v
V s
dt L
R i v v v V
di t v
dt L
1 1
1
1 1 1 0 1
2 2
2
( 0)
14 /
7 ( 0) ( 0) ( 0) ( 0) 7
( )
0
 L
i i i i
v v R i i
1 1 2 2
2 0 2 1 2
( 0) ( 0) 0, ( 0) ( 0) 0
( 0) ( 0) ( 0) ( 0) 0
       

       
- 
+ 
V0 i1 
i2 
 As t  ∞:     i i A
R
1 2
1
7
( ) ( ) 2,33
 Applying KVL to 2 meshes 
to find natural responses: 

      

      

di
R R i R i L d i di
idt
dt dt
di
R i i L s s 
dt
21
1 2 1 2 2 1 1 1
12
22
2 2 1 2
( ) 0
13 30 0
( ) 0 13 30 0
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
7u(t)V
R 2
0,5H
L 1
R 1
0,2H
L 2
29 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.8: Find v0(t) for t > 0 
1Ω 
3Ω 
- 
+ 
V0 i1 
i2 
 Applying KVL to loop 
 
     
     
    
t t
A B A
i t e e A
A B B
3 10
1
2,33 0 1,33
( ) 2,33 1,33
3 10 14 1
 
 
      
 
t t
n
s
s s i Ae Be
s
12 3 10
1
2
3
13 30 0
10
      t t
f n
i t i i Ae Be
3 10
1 1 1
( ) 2,33
 Imposing the initial condition gives: 
           t t
di di
i i L i i L e e A
dt dt
3 101 1
1 2 1 2 1 1
7 4 7 4 2,33 3,33
   t tv t R i t i t e e A3 100 2 1 2( ) ( ) ( ) 2      
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
10u(t) mV
R 2R 1
4
3
7
C 1
C 2
30 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.9: Find v0(t) for t > 0 10kΩ 
- 
+ 
V0 
2 
VO 
 For the natural response, turning off the source: 
 
 


S
v v dv v v
At node C
R dt R
v v dv
At node C 
R dt
1 2 1 0
2
1 2
1 0 0
1
2
1:
2 :
 Applying KCL: 
     v v v S
v v dv dv dv
C C C
R dt dt dt
2 1 0 1 2 0 0
2 2 1
1
  
dv
v v R C
dt
0
1 0 2 1
10kΩ 
20μF 
100μF 
- + V2 
1 
V1 
 
     
 
S
vd v dv v
dt R C R C dt R R C C R R C C
2
0 0 0
2
1 2 2 2 1 2 1 2 1 2 1 2
1 1
   
S
d v dv
v v
dt dt
2
0 0
02
2 5 5
      s s s j2
1,2
2 5 0 1 2
   tnv t e A t B t0 ( ) cos2 sin2
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
10u(t) mV
R 2R 1
4
3
7
C 1
C 2
31 
Chapter 8: Second-order circuits 
V. General second order circuits 
Ex 8.9: Find v0(t) for t > 0 10kΩ 
- 
+ 
V0 
2 
VO 
       
S f
v v v v v mV
0 1 0 0
( ) ( ) ( ) 10 As t ∞: 
   v v
0 2
( 0) ( 0) 0
      tn fv t v v e A t B t0 0 0( ) 10 cos2 sin2
10kΩ 
20μF 
100μF 
- + V2 
1 
V1 
     
   
     
v v dv v v
v v v dt R C
0 2 0 1 0
1 2 0 2 1
( 0) ( 0) 0 ( 0)
0
( 0) ( 0) ( 0) 0
 The complete response is: 
 Find initial conditions: 
      

  
      

v A A 
dv
A B B
dt
0
0
( 0) 10 0 10
( 0)
2 0 5
 The complete response becomes: 
    tv t e t t mV0( ) 10 10cos2 5sin2
Fundamentals of Electric Circuits – Viet Son Nguyen - 2011 
32 
Chapter 8: Second-order circuits 
VI. Applications 
 Practical applications of RLC circuits are found in control and communications 
circuits, for examples: 
 Ringing circuits 
 Peaking circuits 
 Resonant circuits 
 Smoothing circuits 
 Filters 
 Automobile ignition 
 Most of the circuits cannot be covered until we treat AC sources.