Apply KCL to the top node ,we have

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

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

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)

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)

characteristic equation are solutions regardless of A1 and A2

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

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)

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

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

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)

complex frequencies

We will discuss each case separately

did not change

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

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

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

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

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

were Damped Radian Frequency Using Euler Identities

Using Euler Identities

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

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

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

(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.

(c) Calculate the voltage response for t ≥ 0

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

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

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

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

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

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

Initial voltage on the Capacitor Initial current through the Indictor

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

where

Summary

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

KCL Differentiating once we obtain (Natural Response Solutions)

Indirect Continue the equations into

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

Direct

Indirect

KVL Differentiating once we obtain

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

Underdammped

From the circuit , we note the followings:

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

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

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

To find i(t) directly