8.3 The Source-Free Series RLC Circuit

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8.3 The Source-Free Series RLC Circuit

An understanding of the natural response of the series RLC circuit is a necessary background for future studies in filter design and communications networks.

Consider the series RLC circuit sho wn in Fig. 8.8. The circuit is being excited by the energy initially stored in the capacitor and inductor. The energy is represented by the initial capacitor v oltage V0 and initial inductor current I0. Thus, at t = 0,

$$v(0) = \frac{1}{C} \int_{-\infty}^{0} i \, dt = V_0 \tag{8.2a}$$ $$i(0) = I_0 \tag{8.2b}$$

Applying KVL around the loop in Fig. 8.8,

$$Ri + L\frac{di}{dt} + \frac{1}{C} \int_{-\infty}^{t} i(\tau) d\tau = 0$$

(8.3)

A source-free series RLC circuit.

To eliminate the integral, we differentiate with respect to t and rearrange terms. We get

$$\frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0$$

\n(8.4)

This is a second-order differential equation and is the reason for calling the RLC circuits in this chapter second-order circuits. Our goal is to solve Eq. (8.4). To solv e such a second-order dif ferential equation requires that we have two initial conditions, such as the initial v alue of i and its first derivative or initial values of some i and v. The initial value of i is given in Eq. (8.2b). We get the initial v alue of the deri vative of i from Eqs. (8.2a) and (8.3); that is,

$$Ri(0) + L\frac{di(0)}{dt} + V_0 = 0$$

or

$$\frac{di(0)}{dt} = -\frac{1}{L} (RI_0 + V_0)$$

\n(8.5)

With the two initial conditions in Eqs. (8.2b) and (8.5), we can no w solve Eq. (8.4). Our e xperience in the preceding chapter on first-order circuits suggests that the solution is of exponential form. So we let

$$i = Ae^{st} \tag{8.6}$$

where A and s are constants to be determined. Substituting Eq. (8.6) into Eq. (8.4) and carrying out the necessary differentiations, we obtain

$$As2est + \frac{AR}{L}sest + \frac{A}{LC}est = 0$$

or

$$Ae^{st}(s^2 + \frac{R}{L}s + \frac{1}{LC}) = 0$$

(8.7)

Since i = Aest is the assumed solution we are trying to find, only the expression in parentheses can be zero:

$$s^2 + \frac{R}{L}s + \frac{1}{LC} = 0$$

(8.8)

This quadratic equation is kno wn as the characteristic equation of the differential Eq. (8.4), since the roots of the equation dictate the character of i. The two roots of Eq. (8.8) are

$$s_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}}$$

(8.9a)

$$s_2 = -\frac{R}{2L} - \sqrt{\left(\frac{R}{2L}\right)^2 - \frac{1}{LC}}$$

(8.9b)

A more compact way of expressing the roots is

$$s_1 = -\alpha + \sqrt{\alpha^2 - \omega_0^2}$$

, $s_2 = -\alpha - \sqrt{\alpha^2 - \omega_0^2}$ (8.10)

See Appendix C.1 for the formula to find the roots of a quadratic equation. where

$$\alpha = \frac{R}{2L}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.11}$$

The roots s1 and s2 are called natural frequencies, measured in nepers per second (Np/s), because they are associated with the natural response of the circuit; ω0 is kno wn as the resonant frequency or strictly as the undamped natural frequency, expressed in radians per second (rad/s); and α is the neper frequency expressed in nepers per second. In terms of α and ω0, Eq. (8.8) can be written as

$$x^2 + 2\alpha s + \omega_0^2 = 0$$

(8.8a)

The variables s and ω0 are important quantities we will be discussing throughout the rest of the text.

s

The two values of s in Eq. (8.10) indicate that there are two possible solutions for i, each of which is of the form of the assumed solution in Eq. (8.6); that is,

$$i_1 = A_1 e^{s_1 t}, \qquad i_2 = A_2 e^{s_2 t} \tag{8.12}$$

Since Eq. (8.4) is a linear equation, an y linear combination of the tw o distinct solutions i1 and i2 is also a solution of Eq. (8.4). A complete or total solution of Eq. (8.4) would therefore require a linear combination of i1 and i2. Thus, the natural response of the series RLC circuit is

$$i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}$$

(8.13)

where the constants A1 and A2 are determined from the initial values i(0) and di(0)∕dt in Eqs. (8.2b) and (8.5).

From Eq. (8.10), we can infer that there are three types of solutions:

1. If α > ω0, we have the overdamped case.

• 2. If α = ω0, we have the critically damped case.
• 3. If α < ω0, we have the underdamped case.

We will consider each of these cases separately.

Overdamped Case (α > ω0)

From Eqs. (8.9) and (8.10), α > ω0 implies C > 4L∕R² . When this happens, both roots s1 and s2 are negative and real. The response is

$$i(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t}$$

(8.14)

which decays and approaches zero as t increases. Figure 8.9(a) illustrates a typical overdamped response.

Critically Damped Case (α = ω0)

When α = ω0, C = 4L∕R² and

$$s_1 = s_2 = -\alpha = -\frac{R}{2L}$$

(8.15)

The response is overdamped when the roots of the circuit’s characteristic equation are unequal and real, critically damped when the roots are equal and real, and underdamped when the roots are complex.

The neper (Np) is a dimensionless unit named after John Napier (1550–1617), a Scottish mathematician.

The ratio α/ω0 is known as the damping

ratio ζ.

For this case, Eq. (8.13) yields

$$i(t) = A_1 e^{-\alpha t} + A_2 e^{-\alpha t} = A_3 e^{-\alpha t}$$

where A3 = A1 + A2 . This cannot be the solution, because the two initial conditions cannot be satisfied with the single constant A3. What then could be wrong? Our assumption of an e xponential solution is incor rect for the special case of critical damping. Let us go back to Eq. (8.4). When α = ω0 = R∕2L, Eq. (8.4) becomes

$$\frac{d^2i}{dt^2} + 2\alpha \frac{di}{dt} + \alpha^2 i = 0$$ $$\frac{d}{dt}\left(\frac{di}{dt} + \alpha i\right) + \alpha \left(\frac{di}{dt} + \alpha i\right) = 0$$

\n(8.16)

If we let

or

$$f = \frac{di}{dt} + \alpha i \tag{8.17}$$

then Eq. (8.16) becomes

$$\frac{df}{dt} + \alpha f = 0$$

which is a first-order differential equation with solution f = A1e⁻α^t , where A1 is a constant. Equation (8.17) then becomes

$$\frac{di}{dt} + \alpha i = A_1 e^{-\alpha t}$$ $$e^{\alpha t} \frac{di}{dt} + e^{\alpha t} \alpha i = A_1 \tag{8.18}$$

This can be written as

$$\frac{d}{dt}(e^{at}i) = A_1 \tag{8.19}$$

Integrating both sides yields

$$e^{\alpha t}i = A_1t + A_2$$

or

or

$$i = (A_1 t + A_2)e^{-\alpha t}$$

(8.20)

where A2 is another constant. Hence, 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, or

$$i(t) = (A_2 + A_1 t)e^{-\alpha t}$$

(8.21)

A typical critically damped response is sho wn in Fig. 8.9(b). In f act, Fig. 8.9(b) is a sk etch of i(t) = te⁻α^t , which reaches a maximum v alue of e −1 ∕α at t = 1∕α, one time constant, and then decays all the way to zero.

(a) Overdamped response, (b) critically damped response, (c) underdamped response.

(c)

Underdamped Case (α < ω0)

For α < ω0, C < 4L∕R² . The roots may be written as

$$s_1 = -\alpha + \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha + j\omega_d$$

(8.22a)

$$s_2 = -\alpha - \sqrt{-(\omega_0^2 - \alpha^2)} = -\alpha - j\omega_d$$

(8.22b)

where j = √ ___ −1 and ωd = √ _______ ω 0 ²− α² ,which is called the damped frequency. Both ω0 and ωd are natural frequencies because they help determine the natural response; while ω0 is often called the undamped natur al fr equency, ωd is called the damped natural frequency. The natural response is

$$i(t) = A_1 e^{-(\alpha - j\omega_d)t} + A_2 e^{-(\alpha + j\omega_d)t}$$

= $e^{-\alpha t} (A_1 e^{-j\omega_d t} + A_2 e^{-j\omega_d t})$ (8.23)

Using Euler’s identities,

$$e^{j\theta} = \cos\theta + j\sin\theta, \qquad e^{-j\theta} = \cos\theta - j\sin\theta \qquad (8.24)$$

we get

$$i(t) = e^{-\alpha t} [A_1(\cos \omega_d t + j \sin \omega_d t) + A_2(\cos \omega_d t - j \sin \omega_d t)]$$

= $e^{-\alpha t} [(A_1 + A_2) \cos \omega_d t + j(A_1 - A_2) \sin \omega_d t]$ (8.25)

Replacing constants ( A1 + A2) and j(A¹ − A2) with constants B1 and B2, we write

$$i(t) = e^{-\alpha t} (B_1 \cos \omega_d t + B_2 \sin \omega_d t)$$

(8.26)

With the presence of sine and cosine functions, it is clear that the natural response for this case is e xponentially damped and oscillatory in nature. The response has a time constant of 1 ∕α and a period of T = 2π∕ωd. Figure 8.9(c) depicts a typical underdamped response. Part (a) and (b) of Fig. 8.9 assume for each case that i(0) = 0.

Once the inductor current i(t) is found for the RLC series circuit as shown above, other circuit quantities such as indi vidual element voltages can easily be found. F or example, the resistor voltage is vR = Ri, and the inductor voltage is vL = L di∕dt. The inductor current i(t) is selected as the key variable to be determined first in order to take advantage of Eq. (8.1b).

We conclude this section by noting the follo wing interesting, peculiar properties of an RLC network:

• 1. The behavior of such a network is captured by the idea of damping, which is the gradual loss of the initial stored ener gy, as evidenced by the continuous decrease in the amplitude of the response. The damping effect is due to the presence of resistance R. The neper frequency α determines the rate at which the response is damped. If R = 0, then α = 0, and we have an LC circuit with 1∕√ ___ LC as the undamped natural frequency. Since α < ω0 in this case, the response is not only undamped b ut also oscillatory. The circuit is said to be loss-less, because the dissipating or damping element (R) is absent. By adjusting the value of R, the response may be made undamped, overdamped, critically damped, or underdamped.
• 2. Oscillatory response is possible due to the presence of the tw o types of storage elements. Ha ving both L and C allows the flow

^R = 0 produces a perfectly sinusoidal response. This response cannot be practically accomplished with L and C because of the inherent losses in them. See Figs 6.8 and 6.26. An electronic device called an oscillator can produce a perfectly sinusoidal response.

Examples 8.5 and 8.7 demonstrate the effect of varying R.

The response of a second-order circuit with two storage elements of the same type, as in Fig. 8.1(c) and (d), cannot be oscillatory.

of energy back and forth between the tw o. The damped oscillation exhibited by the underdamped response is kno wn as ringing. It stems from the ability of the storage elements L and C to transfer energy back and forth between them.

3. Observe from Fig. 8.9 that the w aveforms of the responses dif fer. In general, it is dif ficult to tell from the waveforms the dif ference between the overdamped and critically damped responses. The critically damped case is the borderline between the underdamped and overdamped cases and it decays the fastest. With the same initial conditions, the overdamped case has the longest settling time, because it tak es the longest time to dissipate the initial stored energy. If we desire the response that approaches the final value most rapidly without oscillation or ringing, the critically damped circuit is the right choice.

circuit that is as close as possible to a critically damped circuit.

What this means in most practical circuits is that we seek an overdamped

Example 8.3

In Fig. 8.8, R = 40 Ω, L = 4 H, and C = 1∕4 F. Calculate the charac teristic roots of the circuit. Is the natural response o verdamped, underdamped, or critically damped?

Solution:

We first calculate

$$\alpha = \frac{R}{2L} = \frac{40}{2(4)} = 5$$

, $\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{4 \times \frac{1}{4}}} = 1$

The roots are

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -5 \pm \sqrt{25 - 1}$$

or

$$s_1 = -0.101, \qquad s_2 = -9.899$$

Since α > ω0, we conclude that the response is overdamped. This is also evident from the fact that the roots are real and negative.

Practice Problem 8.3	If R = 10 Ω, L = 5 H, and C = 2 mF in Fig. 8.8, find α, ω0, s1, and s2. What
	type of natural response will the circuit have?

Answer: 1, 10, −1 ± j9.95, underdamped.

Example 8.4		Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached
	steady state at t = 0−.

Solution:

For t < 0, the switch is closed. The capacitor acts like an open circuit while the inductor acts like a shunted circuit. The equivalent circuit is shown in Fig. 8.11(a). Thus, at t = 0,

$$i(0) = \frac{10}{4+6} = 1 \text{ A}, \qquad v(0) = 6i(0) = 6 \text{ V}$$

The circuit in Fig. 8.10: (a) for t < 0, (b) for t > 0.

where i(0) is the initial current through the inductor and v(0) is the initial voltage across the capacitor.

For t > 0, the switch is opened and the v oltage source is discon nected. The equivalent circuit is shown in Fig. 8.11(b), which is a sourcefree series RLC circuit. Notice that the 3-Ω and 6-Ω resistors, which are in series in Fig. 8.10 when the switch is opened, have been combined to give R = 9 Ω in Fig. 8.11(b). The roots are calculated as follows:

$$\alpha = \frac{R}{2L} = \frac{9}{2\left(\frac{1}{2}\right)} = 9, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{2} \times \frac{1}{50}}} = 10$$ $$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -9 \pm \sqrt{81 - 100}$$

or

s1,2 = −9 ± j4.359

Hence, the response is underdamped (α < ω); that is,

$$i(t) = e^{-9t} (A_1 \cos 4.359t + A_2 \sin 4.359t)$$

(8.4.1)

We now obtain A1 and A2 using the initial conditions. At t = 0,

$$i(0) = 1 = A_1 \tag{8.4.2}$$

From Eq. (8.5),

$$\left. \frac{di}{dt} \right|_{t=0} = -\frac{1}{L} [Ri(0) + v(0)] = -2[9(1) - 6] = -6 \text{ A/s}$$

(8.4.3)

Note that v(0) = V0 = −6 V is used, because the polarity of v in Fig. 8.11(b) is opposite that in Fig. 8.8. Taking the derivative of i(t) in Eq. (8.4.1),

$$\frac{di}{dt} = -9e^{-9t}(A_1 \cos 4.359t + A_2 \sin 4.359t) \n+ e^{-9t}(4.359)(-A_1 \sin 4.359t + A_2 \cos 4.359t)$$

Imposing the condition in Eq. (8.4.3) at t = 0 gives

$$-6 = -9(A_1 + 0) + 4.359(-0 + A_2)$$

But A1 = 1 from Eq. (8.4.2). Then

$$-6 = -9 + 4.359A_2$$

$\Rightarrow$ $A_2 = 0.6882$

Substituting the values of A1 and A2 in Eq. (8.4.1) yields the com plete solution as

$$i(t) = e^{-9t}(\cos 4.359t + 0.6882 \sin 4.359t) A$$

Practice Problem 8.4

Figure 8.12 For Practice Prob. 8.4.

Figure 8.13 A source-free parallel RLC circuit.

The circuit in Fig. 8.12 has reached steady state at t = 0⁻. If the makebefore-break switch moves to position b at t = 0, calculate i(t) for t > 0.

Answer:

$$e^{-2.5t} (10 \cos 1.6583t - 15.076 \sin 1.6583t)$$

A.

8.4 The Source-Free Parallel RLC Circuit

Parallel RLC circuits find many practical applications, notably in communications networks and filter designs.

Consider the parallel RLC circuit shown in Fig. 8.13. Assume initial inductor current I0 and initial capacitor voltage V0,

$$i(0) = I_0 = \frac{1}{L} \int_{-\infty}^{0} v(t) dt$$

(8.27a)

$$v(0) = V_0 \tag{8.27b}$$

Because the three elements are in parallel, they have the same voltage v across them. According to passive sign convention, the current is entering each element; that is, the current through each element is leaving the top node. Thus, applying KCL at the top node gives

$$\frac{v}{R} + \frac{1}{L} \int_{-\infty}^{t} v(\tau) \, d\tau + C \frac{dv}{dt} = 0 \tag{8.28}$$

Taking the derivative with respect to t and dividing by C results in

$$\frac{d^2v}{dt^2} + \frac{1}{RC}\frac{dv}{dt} + \frac{1}{LC}v = 0$$

(8.29)

We obtain the characteristic equation by replacing the first derivative by s and the second derivative by s 2 . By following the same reasoning used in establishing Eqs. (8.4) through (8.8), the characteristic equation is obtained as

$$s^2 + \frac{1}{RC} s + \frac{1}{LC} = 0$$

(8.30)

The roots of the characteristic equation are

$$s_{1,2} = -\frac{1}{2RC} \pm \sqrt{\left(\frac{1}{2RC}\right)^2 - \frac{1}{LC}}$$

or

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2}$$

(8.31)

where

$$\alpha = \frac{1}{2RC}, \qquad \omega_0 = \frac{1}{\sqrt{LC}} \tag{8.32}$$

The names of these terms remain the same as in the preceding section, as they play the same role in the solution. Again, there are three possible solutions, depending on whether α > ω0, α = ω0, or α < ω0. Let us consider these cases separately.

Overdamped Case (α > ω0)

From Eq. (8.32), α > ω0 when L > 4R² C. The roots of the characteristic equation are real and negative. The response is