Figure 16.3

← Back to Fundamentals of Electric Circuits Overview i(t) + ‒ v(t) i(0) L (a) I(s) + ‒ V(s) sL (b) Li(0‒) (c) V(s) I(s) + ‒ sL ⁱ(0‒) s + ‒ Figure 16.1

Representation of an inductor: (a) timedomain, (b,c) s-domain equivalents.

As one can infer from step 2, all the circuit analysis techniques applied for dc circuits are applicable to the ^s-domain.

The elegance of using the Laplace transform in circuit analysis lies in the automatic inclusion of the initial conditions in the transformation process, thus providing a complete (transient and steady-state) solution.

Figure 16.3

Time-domain and s-domain representations of passive elements under zero initial conditions.

The s-domain equi valents are sho wn in Fig. 16.2. With the s-domain equivalents, the Laplace transform can be used readily to solv e first- and second-order circuits such as those we considered in Chapters 7 and 8. We should observe from Eqs. (16.3) to (16.8) that the initial conditions are part of the transformation. This is one advantage of using the Laplace transform in circuit analysis. Another advantage is that a complete re sponse—transient and steady state—of a netw ork is obtained. We will illustrate this with Examples 16.2 and 16.3. Also, observe the duality of Eqs. (16.5) and (16.8), confirming what we already know from Chapter 8 (see Table 8.1), namely, that L and C, I(s) and V(s), and v(0) and i(0) are dual pairs.

If we assume zero initial conditions for the inductor and the capacitor, the above equations reduce to:

Resistor:

$$V(s) = RI(s)$$

Inductor: $V(s) = sLI(s)$ (16.9)
Capacitor: $V(s) = \frac{1}{sC}I(s)$

The s-domain equivalents are shown in Fig. 16.3.

We define the impedance in the s-domain as the ratio of the voltage transform to the current transform under zero initial conditions; that is,

$$Z(s) = \frac{V(s)}{I(s)}$$

(16.10)

Thus, the impedances of the three circuit elements are

Resistor:

$$Z(s) = R$$

Inductor: $Z(s) = sL$ (16.11)
Capacitor: $Z(s) = \frac{1}{sC}$

Table 16.1 summarizes these. The admittance in the s-domain is the reciprocal of the impedance, or

$$Y(s) = \frac{1}{Z(s)} = \frac{I(s)}{V(s)}$$

(16.12)

The use of the Laplace transform in circuit analysis f acilitates the use of various signal sources such as impulse, step, ramp, e xponential, and sinusoidal.

The models for dependent sources and op amps are easy to develop drawing from the simple fact that if the Laplace transform of f(t) is F(s),

TABLE 16.1

Impedance of an element in the s-domain.*

Element	Z(s) = V(s)∕I(s)
Resistor	R
Inductor	sL
Capacitor	1∕sC

* Assuming zero initial conditions

then the Laplace transform of af(t) is aF(s)—the linearity property. The dependent source model is a little easier in that we deal with a single value. The dependent source can have only two controlling values, a constant times either a voltage or a current. Thus,

$$\mathcal{L}[av(t)] = aV(s) \tag{16.13}$$ $$\mathcal{L}[ai(t)] = aI(s) \tag{16.14}$$

The ideal op amp can be treated just lik e a resistor. Nothing within an op amp, either real or ideal, does an ything more than multiply a voltage by a constant. Thus, we only need to write the equations as we always do using the constraint that the input voltage to the op amp has to be zero and the input current has to be zero.

Find vo(t) in the circuit of Fig. 16.4, assuming zero initial conditions. Example 16.1

Solution:

We first transform the circuit from the time domain to the s-domain.

^u(t) ⇒ __1 s 1 H ⇒ sL = s __1 3 ^F⇒ ___1 sC = __3 s

The resulting s-domain circuit is in Fig. 16.5. We now apply mesh analysis. For mesh 1,

$$\frac{1}{s} = \left(1 + \frac{3}{s}\right)I_1 - \frac{3}{s}I_2\tag{16.1.1}$$

For mesh 2,

$$0 = -\frac{3}{s}I_1 + \left(s + 5 + \frac{3}{s}\right)I_2$$

or

$$I_1 = \frac{1}{3}(s^2 + 5s + 3)I_2$$

(16.1.2)

Substituting this into Eq. (16.1.1),

$$\frac{1}{s} = \left(1 + \frac{3}{s}\right)\frac{1}{3}(s^2 + 5s + 3)I_2 - \frac{3}{s}I_2$$

Multiplying through by 3s gives

ng through by 3s gives
\n

$$3 = (s^3 + 8s^2 + 18s)I_2 \implies I_2 = \frac{3}{s^3 + 8s^2 + 18s}$$

\n

$$V_o(s) = sI_2 = \frac{3}{s^2 + 8s + 18} = \frac{3}{\sqrt{2}} \frac{\sqrt{2}}{(s + 4)^2 + (\sqrt{2})^2}$$

Taking the inverse transform yields

$$v_o(t) = \frac{3}{\sqrt{2}} e^{-4t} \sin \sqrt{2t} \text{ V}, \qquad t \ge 0$$

Figure 16.4 For Example 16.1.

Figure 16.5

Mesh analysis of the frequency-domain equivalent of the same circuit.

• ‒ 4 Ω vo(t) 1 H F 2.5u(t) V ¹ 4

u(t) V (t) 2δ(t) A +

‒

0.1 F

Answer: 10(1 − e⁻2*^t* − 2te⁻2*^t* )u(t) V.

Figure 16.6

For Practice Prob. 16.1.

Example 16.2 Find vo(t) in the circuit of Fig. 16.7. Assume vo(0) = 5 V.

10 Ω

10 Ω v^o 10e‒^t

‒

Figure 16.7 For Example 16.2.

Solution:

We transform the circuit to the s-domain as shown in Fig. 16.8. The initial condition is included in the form of the current source Cvo(0) = 0.1(5) = 0.5 A. [See Fig. 16.2(c).] We apply nodal analysis. At the top node,

16.2(c).] We apply nodal analysis. At the to
\n

$$\frac{10/(s+1) - V_o}{10} + 2 + 0.5 = \frac{V_o}{10} + \frac{V_o}{10/s}$$

or

$$\frac{1}{s+1} + 2.5 = \frac{2V_o}{10} + \frac{sV_o}{10} = \frac{1}{10}V_o(s+2)$$

Multiplying through by 10,

$$\frac{10}{s+1} + 25 = V_o(s+2)$$

or

$$V_o = \frac{25s + 35}{(s + 1)(s + 2)} = \frac{A}{s + 1} + \frac{B}{s + 2}$$

where

$$A = (s+1)V_o(s) \Big|_{s=-1} = \frac{25s+35}{(s+2)} \Big|_{s=-1} = \frac{10}{1} = 10$$ $$B = (s+2)V_o(s) \Big|_{s=-2} = \frac{25s+35}{(s+1)} \Big|_{s=-2} = \frac{-15}{-1} = 15$$

Thus,

$$V_o(s) = \frac{10}{s+1} + \frac{15}{s+2}$$

Taking the inverse Laplace transform, we obtain

$$v_o(t) = (10e^{-t} + 15e^{-2t})u(t) \text{ V}$$

Find vo(t) in the circuit shown in Fig. 16.9. Note that, since the voltage input is multiplied by u(t), the voltage source is a short for all t < 0 and iL(0) = 0.

Answer:

$$(12e^{-2t} - 2e^{-t/3})u(t)$$

V.

Figure 16.9 For Practice Prob. 16.2.

Figure 16.10 For Example 16.3.

In the circuit of Fig. 16.10(a), the switch moves from position a to posi- Example 16.3 tion b at t = 0. Find i(t) for t > 0.

Solution:

The initial current through the inductor is i(0) = Io. For t > 0, Fig. 16.10(b) shows the circuit transformed to the s-domain. The initial condition is incorporated in the form of a voltage source as Li(0) = LIo. Using mesh analysis,

$$I(s)(R + sL) - L I_o - \frac{V_o}{s} = 0$$

\n(16.3.1)

or

$$I(s) = \frac{L I_o}{R + sL} + \frac{V_o}{s(R + sL)} = \frac{I_o}{s + R/L} + \frac{V_o/L}{s(s + R/L)}$$

(16.3.2)

Applying partial fraction expansion on the second term on the right-hand side of Eq. (16.3.2) yields

$$I(s) = \frac{I_o}{s + R/L} + \frac{V_o/R}{s} - \frac{V_o/R}{(s + R/L)}$$

(16.3.3)

The inverse Laplace transform of this gives

$$i(t) = \left(I_o - \frac{V_o}{R}\right)e^{-t/\tau} + \frac{V_o}{R}, \qquad t \ge 0$$

(16.3.4)

where τ = R∕L. The term in parentheses is the transient response, while the second term is the steady-state response. In other words, the final value is i(∞) = Vo∕R, which we could have predicted by applying the final-value theorem on Eq. (16.3.2) or (16.3.3); that is,

$$\lim_{s \to 0} sI(s) = \lim_{s \to 0} \left( \frac{sI_o}{s + R/L} + \frac{V_o/L}{s + R/L} \right) = \frac{V_o}{R}$$

(16.3.5)

Equation (16.3.4) may also be written as

$$i(t) = I_0 e^{-t/\tau} + \frac{V_o}{R} (1 - e^{-t/\tau}), \qquad t \ge 0$$

(16.3.6)

The first term is the natural response, while the second term is the forced response. If the initial condition Io = 0, Eq. (16.3.6) becomes

$$i(t) = \frac{V_o}{R}(1 - e^{-t/\tau}), \qquad t \ge 0$$

(16.3.7)

which is the step response, since it is due to the step input Vo with no initial energy.

Figure 16.11 For Practice Prob. 16.3.

Practice Problem 16.3 The switch in Fig. 16.11 has been in position b for a long time. It is moved to position a at t = 0. Determine v(t) for t > 0.

Answer: v(t) = (Vo − IoR)e⁻t∕^τ + IoR, t > 0, where τ = RC.

16.3 Circuit Analysis

Circuit analysis is again relatively easy to do when we are in the s-domain. We merely need to transform a complicated set of mathematical relationships in the time domain into the s-domain where we convert operators (derivatives and integrals) into simple multipliers of s and 1∕s. This now allows us to use algebra to set up and solv e our circuit equations. The exciting thing about this is that all of the circuit theorems and relation ships we developed for dc circuits are perfectly valid in the s-domain.

Remember, equivalent circuits, with capacitors and inductors, only exist in the s-domain; they cannot be transformed back into the time domain.

Example 16.4 Consider the circuit in Fig. 16.12(a). Find the value of the voltage across the capacitor assuming that the value of vs(t) = 10u(t) V and assume that at t = 0, −1 A flows through the inductor and +5 V is across the capacitor.

Solution:

Figure 16.12(b) represents the entire circuit in the s-domain with the initial conditions incorporated. We now have a straightforward nodal analysis problem. Because the value of V1 is also the value of the capacitor voltage in the time domain and is the only unknown node voltage, we only need to write one equation.

$$\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + \frac{i(0)}{s} + \frac{V_1 - [v(0)/s]}{1/(0.1s)} = 0$$

(16.4.1)

or

$$0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s} + \frac{1}{s} + 0.5\tag{16.4.2}$$

where v(0) = 5 V and i(0) = −1 A. Simplifying we get

$$(s^2 + 3s + 2) V_1 = 40 + 5s$$

or

$$V_1 = \frac{40 + 5s}{(s+1)(s+2)} = \frac{35}{s+1} - \frac{30}{s+2}$$

(16.4.3)

Taking the inverse Laplace transform yields

$$v_1(t) = (35e^{-t} - 30e^{-2t})u(t) \text{ V}$$

(16.4.4)

For the circuit shown in Fig. 16.12 with the same initial conditions, find Practice Problem 16.4 the current through the inductor for all time t > 0.

Answer:

$$i(t) = (3 - 7e^{-t} + 3e^{-2t})u(t)$$

A.

For the circuit sho wn in Fig. 16.12, and the initial conditions used Example 16.5 in Example 16.4, use superposition to find the value of the capacitor voltage.

Solution:

Inasmuch as the circuit in the s-domain actually has three independent sources, we can look at the solution one source at a time. Figure 16.13 presents the circuits in the s-domain considering one source at a time. We now have three nodal analysis problems. First, let us solve for the capacitor voltage in the circuit shown in Fig. 16.13(a).

$$\frac{V_1 - 10/s}{10/3} + \frac{V_1 - 0}{5s} + 0 + \frac{V_1 - 0}{1/(0.1s)} = 0$$

or

$$0.1\left(s+3+\frac{2}{s}\right)V_1 = \frac{3}{s}$$

Simplifying we get

2