# Electric Circuit Theory

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1 Electric Circuit Theory Nam Ki Min

2 Chapter 8 Natural and Step Responses of RLC Circuits Nam Ki Min

3 8.1 Introduction to the Natural Response of 3 First-Order Circuits (Chapter 7) Circuits with a single storage element (a capacitor or an inductor). The differential equations describing them are first-order. v C dv C dt + v C V s RC = 0 i(t) di dt + Ri V s L = 0 Second-Order Circuits (Chapter 8) Circuits containing two storage elements Their responses are described by differential equations that contain second derivatives. Typical examples of second-order circuits are RLC circuits. A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements. There are two basic types of RLC circuits: parallel connected and series connected.

4 Finding Initial Values Chapter 8 Natural and Step Responses of RLC Circuits 8.1 Introduction to the Natural Response of There are two key points to keep in mind in determining the initial conditions. We must carefully handle the polarity of voltage v C (t) across the capacitor and the direction of the current i L (t) through the inductor. Keep in mind that v C (t) and i L (t) are defined strictly according to the passive sign convention. One should carefully observe how these are defined and apply them accordingly. Keep in mind that the capacitor voltage is always continuous so that v c 0 + = v C (0 ) and the inductor current is always continuous so that i L 0 + = i L (0 ) where t = 0 denotes the time just before a switching event and t = 0 + is the time just after the switching event, assuming that the switching event takes place at t = 0. 4 Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current.

5 8.1 Introduction to the Natural Response of Obtaining of the Differential Equation for Parallel RLC circuits find many practical applications, notably in communications networks and filter designs. Assume initial inductor current I 0 and initial capacitor voltage V 0, i L 0 = 1 L 0 v t dt = I o 5 v c 0 = V o Since the three elements are in parallel, they have the same voltage v across them. Applying KCL at the top node gives i R + i L + i C = 0 v R + 1 L t v t dt + C dv dt = 0 v R + I o + 1 L t v t dt 0 + C dv dt = 0 (8.1) v R + 1 L 0 v t dt + 1 L t v t dt 0 + C dv dt = 0

6 8.1 Introduction to the Natural Response of 6 Obtaining of the Differential Equation for Taking the derivative with respect to t results in 1 dv R dt + v L + C d2 v dt 2 = 0 (8.2) v R + I o + 1 L t v t dt 0 + C dv dt = 0 (8.1) Dividing by C Arranging the derivatives in the descending order, we get d 2 v dt dv RC dt + v LC = 0 (8.3) This is a second-order differential equation and is the reason for calling the RLC circuits in this chapter second-order circuits.

7 8.1 Introduction to the Natural Response of Solution of the Differential Equation Our experience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let v t = Ae st (8.4) where A and s are unknown constants. d 2 v dt dv RC dt + v LC = 0 (8.3) Substituting Eq. (8.4) into Eq. (8.3) and carrying out the necessary differentiations, we obtain or As 2 e st + As RC est + Aest LC = 0 Ae st s RC s + 1 LC = 0 (8.5) 7 Since e st 0 s RC s + 1 LC = 0 (8.6) We cannot A=0 as a general solution because the voltage is not zero for all time and Ae st is the assumed solution we are trying to find.

8 8.1 Introduction to the Natural Response of 8 Solution of the Differential Equation s RC s + 1 LC = 0 (8.6) d 2 v dt dv RC dt + v LC = 0 (8.3) Eq.(8.6) is known as the characteristic equation of the differential Eq. (8.3), since the roots of the equation determine the mathematical character of v(t). The two roots of Eq. (8.8) are s 1 = 1 2RC + 1 2RC 2 1 LC (8.7) s 2 = 1 2RC 1 2RC 2 1 LC (8.8)

9 8.1 Introduction to the Natural Response of 9 Solution of the Differential Equation The two values of s in Eqs.(8.7) and (8.8) indicate that there are two possible solutions for i, each of which is of the form of the assumed solution in Eq. (8.4); that is, v 1 = A 1 e s 1t v 2 = A 2 e s 2t d 2 v dt dv RC dt + v LC = 0 (8.3) Since Eq. (8.3) is a linear equation, any linear combination of the two distinct solutions v 1 and v 2 is also a solution of Eq. (8.3). A complete or total solution of Eq. (8.3) would therefore require a linear combination of v 1 and v 2. Thus, the natural response of the parallel RLC circuit is v t = v 1 + v 2 = A 1 e s 1t +A 2 e s 2t (8.9) where the constants A 1 and A 2 are determined from the initial values v(0) and dv(0)/dt.

10 Solution of the Differential Equation Chapter 8 Natural and Step Responses of RLC Circuits 8.1 Introduction to the Natural Response of We can show that Eq.(8.9) also is a solution. v t = A 1 e s 1t +A 2 e s 2t dv dt = A 1s 1 e s 1t + A 2 s 2 e s 2t (8.9) d 2 v dt 2 = A 1s 1 2 e s 1t + A 2 s 2 2 e s 2t 10 d 2 v dt dv RC dt + v LC = 0 (8.3) A 1 e s 1t s RC s LC + A 2e s 2t s RC s LC = 0 (8.12) Each parenthetical term is zero because by definition s 1 and s 2 are roots of the characteristic equation. s RC s LC = 0 s RC s LC = 0 s RC s + 1 LC = 0 (8.6)

11 Definition of Frequency Terms Chapter 8 Natural and Step Responses of RLC Circuits 8.1 Introduction to the Natural Response of We have shown that v 1 is a solution, v 2 is a solution, and v 1 +v 2 is a solution. Therefore, the general solution of Eq.(8.3) has the form given in Eq.(8.13). 11 v t = A 1 e s 1t +A 2 e s 2t (8.13), (8.9) The behavior of v(t) depends on the values of s 1 and s 2. s 1 = 1 2RC + 1 2RC 2 1 LC = α + α2 ω o 2 (8.14) s 2 = 1 2RC 1 2RC 2 1 LC = α α2 ω o 2 (8.15) α = 1 2RC ω o = 1 LC (8.16) (8.17) Since the exponents s 1 t and s 2 t must be dimensionless, s 1 and s 2 must have the unit of per second.

12 Definition of Frequency Terms Chapter 8 Natural and Step Responses of RLC Circuits 8.1 Introduction to the Natural Response of Therefore, the units of α and ω o must also be per second or s 1. A unit of this type is called frequency. Let us define new terms: Neper frequency (exponential damping coefficient) α = 1 2RC (8.16) Resonant frequency 12 ω o = 1 LC (8.17) Complex frequency s 1, s 2

13 8.2 The Forms of the Natural Response of 13 Types of Responses(solutions) From Eqs. (8.14),(8.15), we can infer that there are three types of solutions depending on the relative sizes of α and ω o : s 1 = α + α 2 ω o 2 s 2 = α α 2 ω o 2 (8.14) (8.15) α = 1 2RC ω o = 1 LC Overdamped response : If α 2 > ω o 2 Critically damped response If, both roots s 1 and s 2 are negative and real. α 2 2 = ω o, s 1 = s 2 = α L > 4R 2 C L = 4R 2 C α > ω o α = ω o Underdamped response If α 2 < ω o 2, both roots s 1 and s 2 are complex and, in addition, are conjugates of each other. s 1 = α + α 2 ω o 2 = α + ω o 2 α 2 = α + jω d L < 4R 2 C α < ω o s 2 = α α 2 ω o 2 = α ω o 2 α 2 = α jω d

14 8.2 The Forms of the Natural Response of 14 What is the damping? Undamped System The following physical systems are some examples of simple harmonic oscillator. An undamped spring mass system undergoes simple harmonic motion The motion of an undamped pendulum approximates to simple harmonic motion if the angle of oscillation is small

15 Damped System Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of Damping is a dissipation of energy from a vibrating structure. The term dissipate is used to mean the transformation of mechanical energy into other form of energy and, therefore, a removal of mechanical energy from the vibrating system. 15 Overdamped Response The system returns (exponentially decays) to equilibrium without oscillating. Underdamped spring mass system Critically Damped Response The system returns to equilibrium as quickly as possible without oscillating. This is often desired for the damping of systems such as doors. x o m x = 0 x m Elastic potential energy A release from rest at a position x 0. Underdamped Response The mass will overshoot the zero point and oscillate about x=0. x o Damped harmonic oscillator

16 8.2 The Forms of the Natural Response of Electrical System The capacitor stores energy in its electric field E and the inductor stores energy in its magnetic field B (green). The charge flows back and forth between the plates of the capacitor, through the inductor. The energy oscillates back and forth between the capacitor and the inductor until (if not replenished from an external circuit) internal resistance makes the oscillations die out. V o 16 Energy oscillations in the LC Circuit and the mass-spring system V o x = 0 x o

17 Electrical System Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of We now consider a RLC circuit which contains a resistor, an inductor and a capacitor. Unlike the LC circuit energy will be dissipated through the resistor. 17 Charging system.

18 8.2 The Forms of the Natural Response of 18 The Overdamped Voltage Response If L > 4R 2 C, α > ω o. The roots of the characteristic equation are negative real numbers. s 1 = α + α 2 ω o 2 s 2 = α α 2 ω o 2 α = 1 2RC ω o = 1 LC V o The response is v t = A 1 e s 1t +A 2 e s 2t (8.18) The process for finding the overdamped response, v(t): (1) Find s 1 and s 2, using the values of R, L, and C. (2) Find v(0 + ) and dv(0 + ) dt. (3) Find the values of A 1 and A 2 using v(0 + ) and dv(0 + ) dt. (4) Substitute the values for s 1, s 2, A 1 and A 2 into Eq.(8.18).

19 8.2 The Forms of the Natural Response of 19 The Overdamped Voltage Response The process for finding the overdamped response, v(t): (2) Find v(0 + ) and dv(0 + ) dt. Two initial conditions v 0 + = V o V o i C = C dv dt dv(0 + ) dt = i C(0 + ) C (8.21) = i L i R 0 + C i C + i L + i R = 0 i C (0 + ) + i L (0 + ) + i R (0 + ) = 0 = I o C V o CR i C 0 + = i L 0 + i R 0 + = I o V 0 R (8.22)

20 8.2 The Forms of the Natural Response of 20 The Overdamped Voltage Response The process for finding the overdamped response, v(t): (3) Find the values of A 1 and A 2 using v(0 + ) and dv(0 + ) dt. v t = A 1 e s 1t +A 2 e s 2t (8.18) v 0 + = A 1 + A 2 = V o (8.23) dv(t) dt dv(0 + ) dt = s 1 A 1 e s 1t + s 2 A 2 e s 2t = s 1 A 1 + s 2 A 2 (8.24)

21 The Underdamped Voltage Response Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of 21 If L < 4R 2 C, α < ω o. The roots of the characteristic equation are complex, and the response is underdamped. s 1 = α + α 2 ω o 2 = α + ω o 2 α 2 = α + j ω o 2 α 2 = α + jω d (8.25) s 2 = α α 2 ω o 2 = α jω d (8.26) ω d = ω o 2 α 2 α = 1 2RC ω o = 1 LC The response is v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) (8.28)

22 The Underdamped Voltage Response Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of Finding The Underdamped Voltage Response v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) (8.28) 22 The natural response is v t = A 1 e α jω d t + A 2 e α+jω d t = e αt (A 1 e jωdt + A 2 e jωdt ) Using Euler s identities, e jθ = cos θ + j sin θ e jθ = cos θ j sin θ v t (8.29) = A 1 e s 1t +A 2 e s 2t s 1 = α + jω d s 1 = α jω d We get v(t) = e αt [A 1 (cos ω d t + j sin ω d t) + A 2 (cos ω d t j sin ω d t)] = e αt [(A 1 + A 2 ) cos ω d t + j(a 1 A 2 ) sin ω d t] v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) (8.28) B 1 = A 1 + A 2 B 2 = j(a 1 A 2 )

23 8.2 The Forms of the Natural Response of 23 The Underdamped Voltage Response v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) (8.28) With the presence of sine and cosine functions, it is clear that the natural response for this case is exponentially damped and oscillatory in nature. The response has a time constant of 1/α and a period of T = 2π/ω d. Figure depicts a typical underdamped response. [Figure assumes for each case that i(0) = 0]. e αt V o

24 The Underdamped Voltage Response Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of 24 Find the values of B 1 and B 2 using v(0 + ) and dv(0 + ) dt. v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) (8.28) v 0 + = B = V o (8.30) dv(t) dt = ( αe αt )(B 1 cos ω d t ) + e αt ( ω d B 1 sin ω d t ) +( αe αt )(B 2 sin ω d t ) + e αt (ω d B 2 cos ω d t ) dv(0 + ) dt = αb 1 + ω d B 2 (8.31)

25 8.2 The Forms of the Natural Response of 25 The Critically Damped Voltage Response If L = 4R 2 C, α = ω o. The two roots of the characteristic equation are real and equal. s 1 = α + s 2 = α α 2 ω o 2 = α α 2 ω o 2 = α s 1 = s 2 = α = 1 2RC (8.32) α = 1 2RC ω o = 1 LC The response is v t = (D 1 t + D 2 )e αt (8.34)

26 8.2 The Forms of the Natural Response of 26 The Critically Damped Voltage Response Finding the natural response (Eq.8.34) d 2 v dt dv RC dt + v LC = 0 (1) (8.3) d 2 v dv dt2 + 2α dt + α2 v = 0 (2) α = ω o = 1 2RC = 1 LC d dt dv dt + αv + α dv dt + αv = 0 (3) Eq.(3) is a first-order differential equation with solution f = A 1 e αt (6) If we let f = dv dt + αv then Eq.(3) becomes df dt + αf = 0 (4) (5) Eq.(4) then becomes dv dv dt + αv = D 1e αt or eαt dt + eαt αv = D 1 This can be written as d(e αt v) dt = D 1 (8) (7)

27 8.2 The Forms of the Natural Response of 27 The Critically Damped Voltage Response Finding the natural response (Eq.8.34) d(e αt v) dt = D 1 (8) Integrating both sides yields d(e αt v) = D 1 dt V o or e αt v = D 1 t + D 2 (9) v(t) = (D 1 t + D 2 )e αt (8.34) v(t) Figure is a sketch of v(t) = te αt, which reaches a maximum value of e 1 /α at t = 1/α, one time constant, and then decays all the way to zero. This is a typical critically damped response. The natural response of the critically damped circuit is a sum of two terms: a negative exponential and a negative exponential multiplied by a linear term.

28 The Critically Damped Voltage Response Chapter 8 Natural and Step Responses of RLC Circuits 8.2 The Forms of the Natural Response of Finding the values of D 1 and D 2 using v(0 + ) and dv(0 + ) dt. 28 v(t) = (D 1 t + D 2 )e αt (8.34) v 0 + = 0 + D 2 = V o (8.35) dv(t) dt = D e αt + D 1 t + D 2 ( αe αt ) V o dv(0 + ) dt = D 1 αd 2 (8.36)

29 8.3 The Step Response of 29 Responses of Parallel RLC Circuit Natural Response: Summary The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor voltage V o and initial inductor current I o. Thus, at t = 0, v C 0 = 1 C 0 Overdamped response: i t dt = V o i 0 = I o α 2 > ω o 2 L > 4R 2 C Both roots s 1 and s 2 are negative and real. s 1 = α + α 2 ω o 2 s 2 = α α 2 ω o 2 v t = A 1 e s 1t +A 2 e s 2t Underdamped response: α 2 2 < ω o L < 4R 2 C The roots of the characteristic equation are complex. s 1 = α + jω d s 2 = α jω d v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) Critically damped response: α 2 2 = ω o L = 4R 2 C The two roots of the characteristic equation are real and equal. s 1 = s 2 = α v(t) = (D 1 t + D 2 )e αt

30 8.3 The Step Response of 30 Responses of Parallel RLC Circuit Step Response The step response is obtained by the sudden application of a dc source. i(t) I 0 t We want to find i due to a sudden application of a dc current. To develop a general approach to finding the step response of a second order circuit, we focus on finding the current in the inductor branch, i L. This current does not approach zero as t increases. i L = I

31 8.3 The Step Response of 31 Step Response Applying KCL at the top node for t > 0, or i L + i R + i C = I i L + v R + C dv dt = I (8.37) From the definition v = L di L dt dv dt = L d2 i L dt 2 (8.38) (8.39) Substituting Eqs.(8.38) and (8.39) into Eq.(8.37) gives i L + L R di L dt + LC d2 i L dt 2 = I d 2 i L dt di L RC dt + i L LC = I LC (8.40) (8.41)

32 8.3 The Step Response of 32 Step Response d 2 i L dt di L RC dt + i L LC = I LC (8.41) The complete solution to Eq. (8.41) consists of the natural response i n and the forced response i f ; that is,, i L t = i n (t) + i f (t) The natural response is the same as what we had in Section 8.2. v t = A 1 e s 1t +A 2 e s 2t v(t) = e αt (B 1 cos ω d t + B 2 sin ω d t) v(t) = (D 1 t + D 2 )e αt i n (t) = A 1 e s 1t +A 2 e s 2t i n (t) = e αt (B 1 cos ω d t + B 2 sin ω d t) i n (t) = (D 1 t + D 2 )e αt The forced response is the steady state or final value of i. In the circuit in Fig.8.11, the final value of the current through the inductor is the same as the source current I. Thus, i f t = I

33 8.3 The Step Response of 33 Step Response The complete solutions: i L t = i n (t) + i f (t) Overdamped response i n (t) = A 1 e s1t +A 2 e s 2t i f t = I i L t = I + A 1 e s 1t +A 2 e s 2t (8.47) Underdamped response i n (t) = e αt (B 1 cos ω d t + B 2 sin ω d t) i f t = I i L t = I + e αt (B 1 cos ω d t + B 2 sin ω d t) (8.48) Critically damped response i n (t) = (D 1 t + D 2 )e αt i f t = I i L t = I + (D 1 t + D 2 )e αt (8.49)

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