8.5 Step Response of a Series RLC Circuit

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8.5 Step Response of a Series RLC Circuit

As we learned in the preceding chapter, the step response is obtained by the sudden application of a dc source. Consider the series RLC circuit shown in Fig. 8.18. Applying KVL around the loop for t > 0,

$$L\frac{di}{dt} + Ri + v = V_s \tag{8.39}$$

Figure 8.17 For Practice Prob. 8.6.

Vs R L C i t = 0 v + ‒ + ‒

Figure 8.18 Step voltage applied to a series RLC circuit.

But

$$i = C \frac{dv}{dt}$$

Substituting for i in Eq. (8.39) and rearranging terms,

$$\frac{d^2v}{dt^2} + \frac{R}{L}\frac{dv}{dt} + \frac{v}{LC} = \frac{V_s}{LC}$$

(8.40)

which has the same form as Eq. (8.4). More specifically, the coefficients are the same (and that is important in determining the frequency parameters) but the variable is different. (Likewise, see Eq. (8.47).) Hence, the characteristic equation for the series RLC circuit is not af fected by the presence of the dc source.

The solution to Eq. (8.40) has two components: the transient re sponse vt(t) and the steady-state response vss(t); that is,

$$v(t) = v_t(t) + v_{ss}(t)$$

(8.41)

The transient response vt(t) is the component of the total response that dies out with time. The form of the transient response is the same as the form of the solution obtained in Section 8.3 for the source-free circuit, given by Eqs. (8.14), (8.21), and (8.26). Therefore, the transient response vt(t) for the overdamped, underdamped, and critically damped cases are:

$$v_t(t) = A_1 e^{s_1 t} + A_2 e^{s_2 t} \qquad \text{(Overdamped)} \tag{8.42a}$$ $$v_t(t) = (A_1 + A_2 t)e^{-\alpha t}$$

(Critically damped) (8.42b)

$$v_t(t) = (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t}$$

(Underdamped) (8.42c)

The steady-state response is the final value of v(t). In the circuit in Fig. 8.18, the final value of the capacitor voltage is the same as the source voltage Vs. Hence,

$$v_{ss}(t) = v(\infty) = V_s \tag{8.43}$$

Thus, the complete solutions for the o verdamped, underdamped, and critically damped cases are:

v(t) = Vs + A1es1t + A2es2t (Overdamped)	(8.44a)
v(t) = Vs + (A1 + A2t)e−αt (Critically damped)	(8.44b)
v(t) = Vs + (A1 cos ωdt + A2 sin ωdt)e−αt (Underdamped)	(8.44c)

The values of the constants A1 and A2 are obtained from the initial conditions: v(0) and dv(0)∕dt. Keep in mind that v and i are, respectively, the voltage across the capacitor and the current through the inductor. Therefore, Eq. (8.44) only applies for finding v. But once the capacitor voltage vC = v is known, we can determine i = C dv∕dt, which is the same current through the capacitor , inductor, and resistor. Hence, the v oltage across the resistor is vR = iR, while the inductor voltage is vL = L di∕dt.

Alternatively, the complete response for any v ariable x(t) can be found directly, because it has the general form

$$x(t) = x_{ss}(t) + x_t(t)$$

(8.45)

where the xss = x(∞) is the final value and xt(t) is the transient response. The final value is found as in Section 8.2. The transient response has the same form as in Eq. (8.42), and the associated constants are determined from Eq. (8.44) based on the values of x(0) and dx(0)∕dt.

Solution:

■ CASE 1 When R = 5 Ω. For t < 0, the switch is closed for a long time. The capacitor behaves like an open circuit while the inductor acts like a short circuit. The initial current through the inductor is

$$i(0) = \frac{24}{5+1} = 4 \text{ A}$$

and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor; that is,

$$v(0) = 1i(0) = 4 \text{ V}$$

For t > 0, the switch is opened, so that we ha ve the 1- Ω resistor disconnected. What remains is the series RLC circuit with the v oltage source. The characteristic roots are determined as follows:

$$\alpha = \frac{R}{2L} = \frac{5}{2 \times 1} = 2.5, \qquad \omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 0.25}} = 2$$ $$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -1, -4$$

Since α > ω0, we have the overdamped natural response. The total response is therefore

$$v(t) = v_{ss} + (A_1 e^{-t} + A_2 e^{-4t})$$

where vss is the steady-state response. It is the final value of the capacitor voltage. In Fig. 8.19, vf = 24 V. Thus,

$$v(t) = 24 + (A_1 e^{-t} + A_2 e^{-4t})$$

(8.7.1)

We now need to find A1 and A2 using the initial conditions.

$$v(0) = 4 = 24 + A_1 + A_2$$

or

$$-20 = A_1 + A_2 \tag{8.7.2}$$

The current through the inductor cannot change abruptly and is the same current through the capacitor at t = 0⁺ because the inductor and capacitor are now in series. Hence,

$$i(0) = C \frac{dv(0)}{dt} = 4 \qquad \Rightarrow \qquad \frac{dv(0)}{dt} = \frac{4}{C} = \frac{4}{0.25} = 16$$

Before we use this condition, we need to take the derivative of v in Eq. (8.7.1).

$$\frac{dv}{dt} = -A_1 e^{-t} - 4A_2 e^{-4t}$$

(8.7.3)

At t = 0,

$$\frac{dv(0)}{dt} = 16 = -A_1 - 4A_2 \tag{8.7.4}$$

R 1 H

i

24 V

• ‒

Example 8.7

t = 0

0.25 F 1 Ω

v + ‒

From Eqs. (8.7.2) and (8.7.4), A1 = −64∕3 and A2 = 4∕3. Substituting A¹ and A2 in Eq. (8.7.1), we get

$$v(t) = 24 + \frac{4}{3}(-16e^{-t} + e^{-4t}) \text{ V}$$

(8.7.5)

Since the inductor and capacitor are in series for t > 0, the inductor current is the same as the capacitor current. Hence,

$$i(t) = C \frac{dv}{dt}$$

Multiplying Eq. (8.7.3) by C = 0.25 and substituting the values of A¹ and A2 gives

$$i(t) = \frac{4}{3}(4e^{-t} - e^{-4t}) \text{ A}$$

(8.7.6)

Note that i(0) = 4 A, as expected.

■ CASE 2 When R = 4 Ω. Again, the initial current through the inductor is

$$i(0) = \frac{24}{4+1} = 4.8 \text{ A}$$

and the initial capacitor voltage is

$$v(0) = 1i(0) = 4.8 \text{ V}$$

For the characteristic roots,

$$\alpha = \frac{R}{2L} = \frac{4}{2 \times 1} = 2$$

while ω0 = 2 remains the same. In this case, s1 = s2 = −α = −2, and we have the critically damped natural response. The total response is therefore

$$v(t) = v_{ss} + (A_1 + A_2 t)e^{-2t}$$

and, as before vss = 24 V,

$$v(t) = 24 + (A_1 + A_2 t)e^{-2t}$$

(8.7.7)

To find A1 and A2, we use the initial conditions. We write

$$v(0) = 4.8 = 24 + A_1$$

$\Rightarrow$ $A_1 = -19.2$ (8.7.8)

Since i(0) = C dv(0)∕dt = 4.8 or

$$\frac{dv(0)}{dt} = \frac{4.8}{C} = 19.2$$

From Eq. (8.7.7),

$$\frac{dv}{dt} = (-2A_1 - 2tA_2 + A_2)e^{-2t}$$

\n(8.7.9)

At t = 0,

$$\frac{dv(0)}{dt} = 19.2 = -2A_1 + A_2 \tag{8.7.10}$$

From Eqs. (8.7.8) and (8.7.10), A1 = −19.2 and A2 = −19.2. Thus, Eq. (8.7.7) becomes

$$v(t) = 24 - 19.2(1 + t)e^{-2t}$$

V (8.7.11)

The inductor current is the same as the capacitor current; that is,

$$i(t) = C \frac{dv}{dt}$$

Multiplying Eq. (8.7.9) by C = 0.25 and substituting the values of A¹ and A2 gives

$$i(t) = (4.8 + 9.6t)e^{-2t} A$$

(8.7.12)

Note that i(0) = 4.8 A, as expected.

■ CASE 3 When R = 1 Ω. The initial inductor current is

$$i(0) = \frac{24}{1+1} = 12 \text{ A}$$

and the initial voltage across the capacitor is the same as the voltage across the 1-Ω resistor,

$$v(0) = 1i(0) = 12 \text{ V}$$ $$\alpha = \frac{R}{2L} = \frac{1}{2 \times 1} = 0.5$$

Since α = 0.5 < ω0 = 2, we have the underdamped response _______

$$s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -0.5 \pm j1.936$$

The total response is therefore

$$v(t) = 24 + (A_1 \cos 1.936t + A_2 \sin 1.936t)e^{-0.5t}$$

(8.7.13)

We now determine A1 and A2. We write

$$v(0) = 12 = 24 + A_1 \qquad \Rightarrow \qquad A_1 = -12 \tag{8.7.14}$$

Since i(0) = C dv(0)∕dt = 12,

$$\frac{dv(0)}{dt} = \frac{12}{C} = 48\tag{8.7.15}$$

But

$$\frac{dv}{dt} = e^{-0.5t}(-1.936A_1 \sin 1.936t + 1.936A_2 \cos 1.936t)$$ $$-0.5e^{-0.5t}(A_1 \cos 1.936t + A_2 \sin 1.936t)$$

(8.7.16)

At t = 0,

$$\frac{dv(0)}{dt} = 48 = (-0 + 1.936A_2) - 0.5(A_1 + 0)$$

Substituting A1 = −12 gives A2 = 21.694, and Eq. (8.7.13) becomes

$$v(t) = 24 + (21.694 \sin 1.936t - 12 \cos 1.936t)e^{-0.5t} \text{ V}$$

(8.7.17)

The inductor current is

$$i(t) = C \frac{dv}{dt}$$

Multiplying Eq. (8.7.16) by C = 0.25 and substituting the values of A¹ and A2 gives

$$i(t) = (3.1 \sin 1.936t + 12 \cos 1.936t)e^{-0.5t} \text{ A}$$

(8.7.18)

Note that i(0) = 12 A, as expected.

Figure 8.20 plots the responses for the three cases. From this figure, we observe that the critically damped response approaches the step input of 24 V the fastest.

For Example 8.7: response for three degrees of damping.

Practice Problem 8.7 Having been in position a for a long time, the switch in Fig. 8.21 is moved to position b at t = 0. Find v(t) and vR(t) for t > 0.

Figure 8.21 For Practice Prob. 8.7.

Answer: 15 − (1.7321 sin 3.464t + 3 cos 3.464t)e⁻2*^t* V, 3.464e⁻2*^t* sin 3.464t V.

Figure 8.22 Parallel RLC circuit with an applied current.