Solution:

← Back to Fundamentals of Electric Circuits Overview Hence,

$$\frac{dv(0^{+})}{dt} = \frac{-6}{0.5} = -12 \text{ V/s}$$

(8.9.2)

The final values are obtained when the inductor is replaced by a short circuit and the capacitor by an open circuit in Fig. 8.26(b), giving

$$i(\infty) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(\infty) = 2i(\infty) = 4 \text{ V}$$

(8.9.3)

Next, we obtain the form of the transient response for t > 0. By turning off the 12-V voltage source, we have the circuit in Fig. 8.27. Applying KCL at node a in Fig. 8.27 gives

$$i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt}$$

(8.9.4)

Applying KVL to the left mesh results in

$$4i + 1\frac{di}{dt} + v = 0$$

(8.9.5)

Since we are interested in v for the moment, we substitute i from Eq. (8.9.4) into Eq. (8.9.5). We obtain

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

or

$$\frac{d^2v}{dt^2} + 5\frac{dv}{dt} + 6v = 0$$

From this, we obtain the characteristic equation as

$$s^2 + 5s + 6 = 0$$

with roots s = −2 and s = −3. Thus, the natural response is

$$v_n(t) = Ae^{-2t} + Be^{-3t}$$

(8.9.6)

where A and B are unknown constants to be determined later. The steadystate response is

$$v_{ss}(t) = v(\infty) = 4 \tag{8.9.7}$$

The complete response is

$$v(t) = v_t + v_{ss} = 4 + Ae^{-2t} + Be^{-3t}$$

(8.9.8)

We now determine A and B using the initial values. From Eq. (8.9.1), v(0) = 12. Substituting this into Eq. (8.9.8) at t = 0 gives

$$12 = 4 + A + B \qquad \Rightarrow \qquad A + B = 8 \tag{8.9.9}$$

Taking the derivative of v in Eq. (8.9.8),

$$\frac{dv}{dt} = -2Ae^{-2t} - 3Be^{-3t}$$

(8.9.10)

Substituting Eq. (8.9.2) into Eq. (8.9.10) at t = 0 gives

$$-12 = -2A - 3B \qquad \Rightarrow \qquad 2A + 3B = 12 \tag{8.9.11}$$

From Eqs. (8.9.9) and (8.9.11), we obtain

$$A = 12, \qquad B = -4$$

so that Eq. (8.9.8) becomes

$$v(t) = 4 + 12e^{-2t} - 4e^{-3t} \text{ V}, \qquad t > 0 \tag{8.9.12}$$

From v, we can obtain other quantities of interest by referring to Fig. 8.26(b). To obtain i, for example,

$$i = \frac{v}{2} + \frac{1}{2} \frac{dv}{dt} = 2 + 6e^{-2t} - 2e^{-3t} - 12e^{-2t} + 6e^{-3t}$$

= 2 - 6e^{-2t} + 4e^{-3t} A, \t t > 0 (8.9.13)

Notice that i(0) = 0, in agreement with Eq. (8.9.1).

Determine v and i for t > 0 in the circuit of Fig. 8.28. (See comments Practice Problem 8.9 about current sources in Practice Prob. 7.5.)

Answer: 20(1 − e⁻5*^t* ) V, 5(1 − e⁻5*^t* ) A.

Figure 8.28 For Practice Prob. 8.9.

Find vo(t) for t > 0 in the circuit of Fig. 8.29.

Solution:

This is an example of a second-order circuit with two inductors. We first obtain the mesh currents i1 and i2, which happen to be the currents through the inductors. We need to obtain the initial and final values of these currents.

For t < 0, 7u(t) = 0, so that i1(0⁻ ) = 0 = i2(0⁻ ). For t > 0, 7u(t) = 7, so that the equivalent circuit is as shown in Fig. 8.30(a). Due to the continuity of inductor current,

$$i_1(0^+) = i_1(0^-) = 0,$$

$i_2(0^+) = i_2(0^-) = 0$ (8.10.1)

$$v_{L_2}(0^+) = v_o(0^+) = 1[(i_1(0^+) - i_2(0^+)] = 0 \tag{8.10.2}$$

Applying KVL to the left loop in Fig. 8.30(a) at t = 0⁺,

$$7 = 3i_1(0^+) + v_{L_1}(0^+) + v_o(0^+)$$

Figure 8.29 For Example 8.10.

Example 8.10

Figure 8.30 Equivalent circuit of that in Fig. 8.29 for: (a) t > 0, (b) t → ∞.

$$v_{L_1}(0^+) = 7 \, \text{V}$$

Since L¹ di1∕dt = *v^L*1,

$$\frac{di_1(0^+)}{dt} = \frac{v_{L1}}{L_1} = \frac{7}{\frac{1}{2}} = 14 \text{ A/s}$$

(8.10.3)

Similarly, since L² di2∕dt = *v^L*2,

$$\frac{di_2(0^+)}{dt} = \frac{v_{L2}}{L_2} = 0$$

\n(8.10.4)

As t → ∞, the circuit reaches steady state, and the inductors can be replaced by short circuits, as shown in Fig. 8.30(b). From this figure,

$$i_1(\infty) = i_2(\infty) = \frac{7}{3} A$$

(8.10.5)

Next, we obtain the form of the transient responses by removing the voltage source, as shown in Fig. 8.31. Applying KVL to the two meshes yields

$$4i_1 - i_2 + \frac{1}{2}\frac{di_1}{dt} = 0$$

(8.10.6)

and

Obtaining the form of the transient response for Example 8.10.

$$i_2 + \frac{1}{5} \frac{di_2}{dt} - i_1 = 0 \tag{8.10.7}$$

From Eq. (8.10.6),

$$i_2 = 4i_1 + \frac{1}{2} \frac{di_1}{dt}$$

(8.10.8)

Substituting Eq. (8.10.8) into Eq. (8.10.7) gives

$$4i_1 + \frac{1}{2}\frac{di_1}{dt} + \frac{4}{5}\frac{di_1}{dt} + \frac{1}{10}\frac{d^2i_1}{dt^2} - i_1 = 0$$ $$\frac{d^2i_1}{dt^2} + 13\frac{di_1}{dt} + 30i_1 = 0$$

From this we obtain the characteristic equation as

$$s^2 + 13s + 30 = 0$$

which has roots s = −3 and s = −10. Hence, the form of the transient response is

$$i_{1n} = Ae^{-3t} + Be^{-10t}$$

(8.10.9)

Figure 8.31

where A and B are constants. The steady-state response is

$$i_{1ss} = i_1(\infty) = \frac{7}{3} \text{ A}$$

(8.10.10)

From Eqs. (8.10.9) and (8.10.10), we obtain the complete response as

$$i_1(t) = \frac{7}{3} + Ae^{-3t} + Be^{-10t}$$

(8.10.11)

We finally obtain A and B from the initial values. From Eqs. (8.10.1) and (8.10.11),

$$0 = \frac{7}{3} + A + B \tag{8.10.12}$$

Taking the derivative of Eq. (8.10.11), setting t = 0 in the derivative, and enforcing Eq. (8.10.3), we obtain

$$14 = -3A - 10B \tag{8.10.13}$$

From Eqs. (8.10.12) and (8.10.13), A = −4∕3 and B = −1. Thus,

$$i_1(t) = \frac{7}{3} - \frac{4}{3} e^{-3t} - e^{-10t}$$

(8.10.14)

We no w obtain i2 from i1. Applying KVL to the left loop in Fig. 8.30(a) gives

$$7 = 4i_1 - i_2 + \frac{1}{2} \frac{di_1}{dt} \qquad \Rightarrow \qquad i_2 = -7 + 4i_1 + \frac{1}{2} \frac{di_1}{dt}$$

Substituting for i1 in Eq. (8.10.14) gives

$$i_2(t) = -7 + \frac{28}{3} - \frac{16}{3}e^{-3t} - 4e^{-10t} + 2e^{-3t} + 5e^{-10t}$$

= $\frac{7}{3} - \frac{10}{3}e^{-3t} + e^{-10t}$ (8.10.15)

From Fig. 8.29,

Answer: 14(e^−t

$$v_o(t) = 1[i_1(t) - i_2(t)]$$

(8.10.16)

Substituting Eqs. (8.10.14) and (8.10.15) into Eq. (8.10.16) yields

$$v_o(t) = 2(e^{-3t} - e^{-10t})$$

\n(8.10.17)

Note that vo(0) = 0, as expected from Eq. (8.10.2).

) V, t > 0.

− e⁻6*^t*

For t > 0, obtain vo(t) in the circuit of Fig. 8.32. ( Hint: First find v¹ Practice Problem 8.10 and v2.)

Figure 8.32 For Practice Prob. 8.10.

The use of op amps in second-order circuits avoids the use of inductors, which are undesirable in some applications.

8.8 Second-Order Op Amp Circuits

An op amp circuit with tw o storage elements that cannot be combined into a single equi valent element is second-order. Because inductors are bulky and hea vy, they are rarely used in practical op amp circuits. F or this reason, we will only consider RC second-order op amp circuits here. Such circuits find a wide range of applications in devices such as filters and oscillators.

The analysis of a second-order op amp circuit follows the same four steps given and demonstrated in the previous section.

Example 8.11 In the op amp circuit of Fig. 8.33, find vo(t) for t > 0 when v^{s =} 10u(t) mV. Let R1 = R2 = 10 kΩ, C1 = 20 μF, and C2 = 100 μF.

Figure 8.33 For Example 8.11.

Solution:

Although we could follow the same four steps given in the previous section to solve this problem, we will solve it a little differently. Due to the voltage follower configuration, the voltage across C1 is vo. Applying KCL at node 1,

$$\frac{v_s - v_1}{R_1} = C_2 \frac{dv_2}{dt} + \frac{v_1 - v_o}{R_2}$$

(8.11.1)

At node 2, KCL gives

$$\frac{v_1 - v_o}{R_2} = C_1 \frac{dv_o}{dt}$$

(8.11.2)

But

$$v_2 = v_1 - v_o \tag{8.11.3}$$

We now try to eliminate v1 and v2 in Eqs. (8.11.1) to (8.11.3). Substituting Eqs. (8.11.2) and (8.11.3) into Eq. (8.11.1) yields

$$\frac{v_s - v_1}{R_1} = C_2 \frac{dv_1}{dt} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt}$$

(8.11.4)

From Eq. (8.11.2),

$$v_1 = v_o + R_2 C_1 \frac{dv_o}{dt}$$

(8.11.5)

Substituting Eq. (8.11.5) into Eq. (8.11.4), we obtain

$$\frac{v_s}{R_1} = \frac{v_o}{R_1} + \frac{R_2 C_1}{R_1} \frac{dv_o}{dt} + C_2 \frac{dv_o}{dt} + R_2 C_1 C_2 \frac{d^2 v_o}{dt^2} - C_2 \frac{dv_o}{dt} + C_1 \frac{dv_o}{dt}$$

or

$$\frac{d^2v_o}{dt^2} + \left(\frac{1}{R_1C_2} + \frac{1}{R_2C_2}\right)\frac{dv_o}{dt} + \frac{v_o}{R_1R_2C_1C_2} = \frac{v_s}{R_1R_2C_1C_2}$$

(8.11.6)

With the given values of R1, R2, C1, and C2, Eq. (8.11.6) becomes

$$\frac{d^2v_o}{dt^2} + 2\frac{dv_o}{dt} + 5v_o = 5v_s$$

(8.11.7)

To obtain the form of the transient response, set vs = 0 in Eq. (8.11.7), which is the same as turning off the source. The characteristic equation is

$$s^2 + 2s + 5 = 0$$

which has complex roots s1,2 = −1 ± j2. Hence, the form of the transient response is