3. Apply KCL to the top node ,we have

4. We normalize the highest derivative by dividing by C , we get
Since the highest derivative in the equation is 2 , then this equation is

5. We can not solve this equation by separating variables and integrating as we did in
The first order equation in Chapter 7
The classical approach is to assume a solution
Question : what is a solution we might assume that satisfies the above equation
what is a solution that when differentiated twice and added to its first derivative
multiplied by a constant and then added to it the solution itself divided by a constant
will give zero
The only candidate ( مرشح ) that satisfies the above equation will be
an exponential function similar to chapter 7

6. A ≠ 0 else the whole proposed solution will be zero
Substituting the proposed or assumed solution into the differential equation we get
A ≠ 0 else the whole proposed solution
will be zero
That leave
The equation is called the characteristic equation of the differential equation because
the roots of this quadratic equation determine the mathematical character of v(t)

7. characteristic equation of the differential equation
The two roots of characteristic equation are
The solutions
either one satisfies the differential equation
Denoting these two solutions as v1(t) and v2(t)

8. characteristic equation
are solutions regardless of A1 and A2

9. are solutions
Also a combinations of the two solution is also a solution
is a solutions as can be shown:
Substituting the above in the differential equation , we have
is a solutions

10. are solutions
however
Is the most general solution because it contain
all possible solutions
( Recall how the time constant on RL and RC circuits depend on R,L and C)

11. where characteristic equation of the differential equation
Roots of the characteristic equation
Rewriting the roots s1,2 as follows:
where

12. where characteristic equation of the differential equation
So to distinguish between the two frequencies we name them as

13. where characteristic equation of the differential equation
s1,2 are also frequencies since they are summation of frequencies
To distinguish them from the Neper and Resonant frequencies and because s1,2 can be complex we call them complex frequencies
All these frequencies have the dimension of angular frequency per timer (rad/s)

14. complex frequencies

15. We will discuss each case separately

18. did not change

20. So far we have seen that the behavior of a second-order RLC circuit depends on the values of s1 and s2, which in turn depend on the circuit parameters R,L, and C.
Therefore, the first step in finding that natural response is to calculate these values
and determine whether the response is overdamped, underdamped or critically damped
Completing the description of the natural response requires finding two unknown coefficients, such as A1 and A2 .The method used to do this is based on matching the solution for the natural response to the initial conditions imposed by the circuit
In this section, we analyze the natural response form for each of the three types of damping, beginning with the overdamped response as will be shown next

21. where A1 , A2 are determined from Initial conditions as follows:
(2)
But what is

22. Summarizing
(1) From the circuit elements R,L,C we can find
(2) A1 , A2 are determined from Initial conditions as follows:
But what is

25. Because the roots are real and distinct, we know that the response is overdamped

27. 3- Form the general solution
4- solve (1) and (2)

28. were
Damped Radian Frequency
Using Euler Identities

29. Using Euler Identities

30. It is clear that the natural response for this case is exponentially damped and oscillatory in nature

31. The constants B1 and B2 are real because v(t) real (the left hand side)
Therefore the right hand side is also real
The constants A1 and A2 are complex conjugate ( can be shown )
Therefore

32. Summary
were
The constants B1 and B2 can be determined from the initial conditions
This was shown previously using KCL and
Solving (1) and (2) , w obtain B1 and B2

33. (a) Calculate the roots of the characteristic equation
underdamped
For the underdamped case, we do not ordinarily solve for s1 and s2 because we do not use them explicitly.

35. (c) Calculate the voltage response for t ≥ 0

36. (d) Plot v(t) versus t for the time interval 0 ≤ t ≤ 11 ms

37. The Critically Damped Voltage Response
The second-order circuit is critically damped when or
where A0 is an arbitrary constant
The solution above can not satisfies the two initial conditions (V0, I0) with only on constant A0
It seems that there is a problem ?
We can trace this dilemma back to the assumption that the solution takes the form of
When the roots of the characteristic equation are equal, the solution for the differential equation takes a different form, namely

38. The Critically Damped Voltage Response
Is not possible assumption because it leads to
which can not satisfies (V0, I0 )
Another solution for the differential equation takes a different form, namely
The justification of is left for an introductory course in differential equations.
Finding the solution involves obtaining D1 and D2 by following the same pattern set in the overdamped and underdamped cases:
We use the initial values of the voltage v(0+)and the derivative of the voltage with respect to time dv(0+)/dt to write two equations containing D1 and D2

39. (a) For the circuit above ,find the value of R that results in a critically damped voltage response.

40. (b) Calculate v(t) for t > 0
critically damped

41. (c) Plot v(t) versus t for 0 ≤ t ≤ 7 ms

42. Initial voltage on the Capacitor
Initial current through the Indictor

46. where A1 , A2 are determined from Initial conditions as follows:
(2)

47. where

49. Summary

50. follows
We will show first the indirect approach solution , then the direct approach

51. KCL
Differentiating once we obtain
(Natural Response Solutions)

52. Indirect Continue
the equations into

54. The initial energy stored in the RLC circuit above is zero
At t = 0 , a dc current source of 24 mA is applied to the circuit.
The value of R is 400 W

57. Direct

58. Indirect

65. KVL
Differentiating once we obtain

66. 2ed order differential equation
The 2ed order differential equation has the same form as the parallel RLC
The characteristic equation for the series RLC circuit is
where

69. Underdammped

70. From the circuit , we note the followings:

72. Whichever expression is used ( the second is recommended ) , we get

73. The step response for a series RLC circuit
KVL
Substitute in the KVL equation
The differential equation has the same form as the step parallel RLC
Therefore the procedure of finding in the series RLC similar the one used
to find for the parallel RLC

77. Solution
First we have to decided what type of RLC connection , Parallel or Series also natural or step
We look at the circuit after the switch moved , t > 0
Series RLC
Step Response

78. To find i(t) directly