# To find the step response of an RC circuit

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1 To find the step response of an RC circuit v( t) v( ) [ v( t) v( )] e tt The time constant = RC The final capacitor voltage v() The initial capacitor voltage v(t ) To find the step response of an RL circuit i( t) i( ) [ i( t) i( )] e tt The final capacitor voltage v() The initial capacitor voltage v(t ) The time constant = L/R

2 Lecture 6 DC Circuits Transient Circuits nd Order circuits

3 Contents Examples of nd order RLC circuit The source-free series RLC circuit The source-free parallel RLC circuit Step response of a series RLC circuit Step response of a parallel RLC 3

4 Examples of Second Order RLC circuits What is a nd order circuit? A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements of different type or the same type. 4

5 Source-Free Series RLC Circuits The solution of the source-free series RLC circuit is called as the natural response of the circuit. The circuit is excited by the energy initially stored in the capacitor and inductor. The nd order of expression d i R L di i LC How to derive and how to solve? 5

6 Source-Free Series RLC Circuits: to derive the equation v R =Ri v L L di v C C t i KVL: Ri L di C t i To eliminate the integral R di L d i i C

7 Source-Free Series RLC Circuits: to solve the equation d Ae st s i R L di i LC Assume: the solution has exponential form, s R L s LC s Neper frequency/ damping factor Define: Characteristic equation R L st i Ae To be determined LC To be determined Resonant frequency/ undamped natural factor Solution: s, Natural Frequencies Linear equation: st st i( t) A e Ae

8 i( t) Still solving st s t A e Ae How to get A? Initial conditions: The inductor: i( ) A A I A and A i( ) I The capacitor: v( ) i V C di( t) st st A se A se di Ri( ) L V RI t A s A s V L RI V A A I and V Ls not independent d i di d di di i i i di t i A e What if s = s (when ) To get A i i( t) t A A te

9 Source-Free Series: RLC Circuits There are three possible solutions for the following nd order differential equation: d i di i. If > o, over-damped case st st i( t) A e Ae where s,. If = o, critical damped case i t ( t) ( A At ) e where s, 3. If < o, under-damped case Damped natural frequency t i( t) e ( B cos B sin) where d 9

10 Finding initial and final values Initial values: Passive convention The polarity of voltage The direction of current Variables that cannot change abruptly The capacitor: The inductor: v( i( v(), ) v( ) i( ) ) i(), dv( t), di( t) t t Focus on these variables Final values: i( ), v( ) DC steady state The capacitor: Open circuit The inductor: short circuit The time just before a switching event which takes place at t =

11 Example If R = Ω, L = 5 H, and C = mf in 8.8, find α, ω, s and s. What type of natural response will the circuit have? Answer: underdamped R L LC s 9. 95, j

12 Example The circuit shown below has reached steady state at t = -. If the make-before-break switch moves to position b at t =, calculate i(t) for t >. Answer: i(t) = e.5t [5cos.6583t 7.538sin.6583t] A

13 Step-Response Series: RLC Circuits The step response is obtained by the sudden application of a dc source. The nd order of expression d v R L dv v LC v s LC The above equation has the same form as the equation for source-free series RLC circuit. The same coefficients (important in determining the frequency parameters). Different circuit variable in the equation. 3

14 Step-Response Series: RLC Circuits The solution of the equation should have two components: the transient response v t (t) & the steady-state response v ss (t): v( t) v ( t) v ( t) t The transient response v t is the same as that for source-free case st st vt ( t) A e Ae v t ( t) ( A v t The steady-state response is the final value of v(t). v ss (t) = v( ) The values of A and A are obtained from the initial conditions: v() and dv()/. ss (over-damped) t A t) e (critically damped) t ( t) e ( A cos A sin) (under-damped) 4

15 Example Having been in position for a long time, the switch in the circuit below is moved to position b at t =. Find v(t) and v R (t) for t >. Answer: v(t) = { + [( cos3.464t.547sin3.464t)e t]} V v R (t)= [.3sin3.464t]e t V 5

16 Lecture 7 AC Circuits Sinusoids and Phasors

17 Appendix

18 Source-Free Parallel: RLC Circuits Let v( t) L i( ) I v() = V Apply KCL to the top node: v R L t v C dv Taking the derivative with respect to t and dividing by C The nd order of expression d v RC dv LC v 8

19 Source-Free Parallel: RLC Circuits There are three possible solutions for the following nd order differential equation: d v dv v where and RC LC. If > o, over-damped case s t st v( t) A e A e where s,. If = o, critical damped case v t ( t) ( A A t) e where s, 3. If < o, under-damped case t v( t) e ( B cos B sin) where d 9

20 Example Refer to the circuit shown below. Find v(t) for t >. Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = 66.67(e t e.5t ) V

21 Step-Response Parallel: RLC Circuits The step response is obtained by the sudden application of a dc source. The nd order of expression d i RC di i LC I s LC It has the same form as the equation for source-free parallel RLC circuit. The same coefficients (important in determining the frequency parameters). Different circuit variable in the equation.

22 Step-Response Parallel : RLC Circuits The solution of the equation should have two components: the transient response v t (t) & the steady-state response v ss (t): i( t) i ( t) i ( t) The transient response i t is the same as that for source-free case st st it ( t) A e Ae i t ss (over-damped) t t ( t) ( A At ) e (critical damped) t it ( t) e ( A cos A sin) (under-damped) The steady-state response is the final value of i(t). i ss (t) = i( ) = I s The values of A and A are obtained from the initial conditions: i() and di()/.

23 Example Find i(t) and v(t) for t > in the circuit shown in circuit shown below: Please refer to lecture or textbook for more detail elaboration. Answer: v(t) = Ldi/ = 5xsint = sint V 3

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