1. Chapter 4 Second Order Circuit (11th & 12th week)
SEE 1023 Circuit Theory
Chapter 4
Second Order Circuit
(11th & 12th week)
Prepared by : jaafar shafie

2. 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.

3. Second Order Circuit Types of second order circuit that may exist:-
Series RLC circuit,
Parallel RLC circuit,
RLL circuit,
RCC circuit. V R L C I R L C V R2 L2 L1 R1 C2 R2 C1 I

4. Second Order Circuit
As usual, the circuit will be analyzed in two parts:-
Source–free Circuit (natural response) Energy is initially stored in the element – thus no effect of current or voltage sources.
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().

5. Second Order Circuit
Consider the circuit shown below where the switch has been closed for a long time. Find:-
v(0+), i(0+),
dv(0+)/dt, di(0+)/dt,
v(), i().
24V 4Ω
0.4F
t=0s 1H +
v(t) —
i(t)

6. Second Order Circuit
When t=0+s=0–s, the inductor is shorted and the capacitor is opened. The equivalent circuit is shown below,
When the switch is opened, the equivalent circuit is
thus,
24V 4Ω +
v(0+) -
i(0+)
v(0+)=(4/8)24=12V
i(0+)= 24/8 =3A
i(0+)
24V 4Ω 1H
0.4F
i(0+)= iC (0+) =3A
and it is know that

7. Second Order Circuit
Applying KVL to the circuit when switch is opened,
thus,
i(0+)
24V 4Ω 1H
0.4F

8. Second Order Circuit
The steady state value,
i()
24V 4Ω +
v() -

9. Natural Response - Series RLC Circuit
Natural response is obtained with Series RLC circuit without source.
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, IO R L C + VO -

10. Natural Response - Series RLC Circuit
To eliminate the integral, differentiate with respect to t and rearrange the terms such that
Now, we have second order differential equation.
It is known that in first-order differential equation, the current is
where A and s are constants.
Substitute Aest to the second order equation.

11. Natural Response - Series RLC Circuit
Thus, we may write as or
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
CHARACTERISTIC
EQUATION

12. Natural Response - Series RLC Circuit
Solve the characteristic equation, one might find the two roots, which are
and

13. Natural Response - Series RLC Circuit
On the other hand, we may represent the two roots as
and
where
The two solutions for s (i.e. s1 and s2) shows that there are two values of the current, which are
The total response of the current would be

14. Natural Response - Series RLC Circuit
Thus, we may found three type of response which are
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

15. 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
which decays to zero when t increased.
A1 and A2 are determined from the initial inductor current and the rate of change of current.
Solve for
A1 and A2

16. 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

17. Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial values.
apply KVL to the loop
thus,

18. Natural Response - Series RLC Circuit
Step3:
It is found that,
A1 = , A2 = 78.36m
Time
-2ms
0ms
2ms
4ms
6ms
8ms
10ms
12ms
14ms
16ms
18ms
I(L1) 0A
0.5A
1.0A

19. 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
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.

20. Natural Response - Series RLC Circuit
In this type,  = ω0 = R/2L = 1/LC. Thus, the second order differential equation become
Let , thus
In first order differential equation, it is found that
, thus
, or ,or

21. Natural Response - Series RLC Circuit
By integrating the equation, one will found the response of critically damped response as
Thus, the derivative of the current is,
when t=0s,

22. 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

23. Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial values.
apply KVL to the loop
thus,

24. Natural Response - Series RLC Circuit
Step3:
It is found that,
A1 = 8k, A2 = 1
Time
-0.5ms
0ms
0.5ms
1.0ms
1.5ms
2.0ms
I(L1) 0A
0.5A
1.0A

25. Natural Response - Series RLC Circuit
3rd Type :  < ω0 , UNDERDAMPED
In this type,  < ω0, or C < 4L/R2.
The roots can be written as
where
and
The response can be further written as

26. Natural Response - Series RLC Circuit
Rearranging the response such that,
Applying Euler’s identities to the above equation, where
Thus,
Let B1=A1+A2 and B2=j(A1–A2), thus

27. Natural Response - Series RLC Circuit
Differentiate the i(t), we have

28. 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

29. Natural Response - Series RLC Circuit
Step2: Find the value of A1 and A2 from the initial values.
apply KVL to the loop
thus,

30. Natural Response - Series RLC Circuit
Step3:
It is found that,
B1 = 1, B2 = 0.4,
Time
-0.4ms
-0.2ms
0ms
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
I(L1)
-0.5A 0A
0.5A
1.0A

31. Natural Response - Series RLC Circuit
In summary, the response for series RCL are
1st Type :  > ω0 , (OVERDAMPED)
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
3rd Type :  < ω0 , (UNDERDAMPED)

32. Natural Response - Series RLC Circuit
The comparison of the three responses are shown below
Time 0s
0.5ms
1.0ms
1.5ms
2.0ms
2.5ms
3.0ms
3.5ms
4.0ms
4.5ms
5.0ms
I(L1)
-0.5A 0A
0.5A
1.0A
C=100F
C=10F
C=2F

33. Natural Response - Parallel RLC Circuit
Natural response for Series RLC circuit is obtained without any connection to source.
Applying KCL, thus we have
Differentiate with t, IO R L C + VO - v

34. 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,
The roots are characterized by these equations

35. Natural Response - Parallel RLC Circuit
On the other hand, we may represent the two roots as
and
where

36. 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
A1 and A2 are determined from the initial inductor current and the rate of change of current.
Solve for
A1 and A2

37. 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
*It is clear that  > ωO , and the response is the 1st type which is ‘overdamped’.

38. Natural Response - Parallel RLC Circuit
Find s1 and s2.
The initial condition are

39. Natural Response - Parallel RLC Circuit
Step3:
It is found that,
A1 = , A2 = 28.09
Time 0s
0.2ms
0.4ms
0.6ms
0.8ms
1.0ms
1.2ms
1.4ms
1.6ms
1.8ms
2.0ms
V(R5:2)
-10V
-5V 0V 5V
10V

40. 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
The derivative is represented as
and at t=0s,

41. 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
*It is clear that  = ωO , and the response is the 2nd type which is ‘critically damped’.

42. Natural Response - Parallel RLC Circuit
Step2: Find the value of A1 and A2 from the initial values.
apply KCL to the loop

43. Natural Response - Parallel RLC Circuit
Step3:
It is found that,
A1 = -600k, A2 = 10
Time 0s
0.1ms
0.2ms
0.3ms
0.4ms
0.5ms
0.6ms
0.7ms
0.8ms
0.9ms
1.0ms
V(R5:2)
-20V
-10V 0V
10V

44. Natural Response - Parallel RLC Circuit
3rd Type :  < ω0 , UNDERDAMPED
In this type,  < ω0, or L < 4R2C.
The roots can be written as
where
and
The response can be further written as

45. Natural Response - Parallel RLC Circuit
Applying the same method as in series, the response of the parallel is
Let B1=A1+A2 and B2=j(A1–A2), thus

46. Natural Response - Parallel RLC Circuit
Differentiate the v(t), we have

47. 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.

48. Natural Response - Parallel RLC Circuit
In summary, the responses for parallel RCL are
1st Type :  > ω0 , (OVERDAMPED)
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
3rd Type :  < ω0 , (UNDERDAMPED)

49. Step Response - Series RLC Circuit
Step response is obtained in Series RLC circuit with source.
Applying KVL to the loop,
,where
Substitute i to the derivative terms, i R L C + v - VS
2nd order diff.
equ for capacitor
voltage

50. Step Response - Series RLC Circuit
To obtain the total response, consider the KVL of the loop
Differentiate the equation, we get
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).

51. 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,
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.

52. Step Response - Series RLC Circuit
The transient responses are:
While vss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Vs.
overdamped
critically damped
underdamped

53. Step Response - Series RLC Circuit
Thus, the complete responses for series RLC with source are:
1st Type :  > ω0 , (OVERDAMPED)
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
3rd Type :  < ω0 , (UNDERDAMPED)

54. 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).
Step 1: Find  and ω0. Determine type of response.
It is found that  > ω0, type of response  overdamped.
i(t)
20Ω
1mH
100F +
v(t) -
20V

55. Step Response - Series RLC Circuit
Step 3: The steady state voltage across capacitor is
Step2: Find the value of A1 and A2 from the initial values.
the derivative of the response at initial is
thus,

56. Step Response - Series RLC Circuit
It is found that the complete response is
Next, find the voltage response if R is changed to 6.324Ω and then to 2Ω

57. Step Response - Series RLC Circuit
Comparison of the three responses with different R value. 2Ω
6.324Ω
20Ω

58. Step Response - Parallel RLC Circuit
Step response is obtained in Parallel RLC circuit with source.
Applying KCL to the loop,
,where
Substitute i to the derivative terms, i R L + v - C IS
2nd order diff.
equ for capacitor
voltage

59. 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,
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.

60. Step Response - Parallel RLC Circuit
The transient responses are:
While iss(t) is the final voltage value (i.e. when t=) across the capacitor. Normally, it will equal to Is.
overdamped
critically damped
underdamped

61. Step Response - Parallel RLC Circuit
Thus, the complete responses for parallel RLC with source are:
1st Type :  > ω0 , (OVERDAMPED)
2. 2nd Type :  = ω0 , (CRITICALLY DAMPED)
3rd Type :  < ω0 , (UNDERDAMPED)

62. 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.

63. 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
where x is either in voltage or current.

64. 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.

65. General Second-Order Circuit - eg
Consider a circuit shown below. Find the complete response of the voltage v(t)?
Step 1: Turn off the independent source. Find the second order differential equation. i 4Ω 1H +
v(t) -
0.5F
12V 2Ω
t >0s i 4Ω 1H +
v(t) -
0.5F 2Ω v
using KCL and KVL
to find the 2nd order
differential equations,

66. General Second-Order Circuit - eg
KCL at node ‘v’ KVL at mesh ‘i’
We just concern on ‘v’. Thus, substitute ‘i’ to the right hand side equation
Thus, the characteristic equation is,

67. 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
The steady state voltage is
The transient voltage is

68. General Second-Order Circuit - eg
Step 4: The initial voltage and current is,
At just after the switch is closed, the circuit is shown as iC i iR 4Ω 1H +
v(t) -
0.5F
12V 2Ω

69. General Second-Order Circuit - eg
The overall response is
From the overall response, at initial
It is found that A1 = 12 and A2 = -4.
Thus,