4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK

← Back to LINEAR SYSTEMS AND SIGNALS Overview

4.4 ANALYSIS OF ELECTRICAL NETWORKS: THE TRANSFORMED NETWORK

Example 4.12 shows how electrical networks may be analyzed by writing the integro-differential equation(s) of the system and then solving these equations by the Laplace transform. We now show that it is also possible to analyze electrical networks directly without having to write the

374 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

integro-differential equations. This procedure is considerably simpler because it permits us to treat an electrical network as if it were a resistive network. For this purpose, we need to represent a network in the “frequency domain” where all the voltages and currents are represented by their Laplace transforms.

For the sake of simplicity, let us first discuss the case with zero initial conditions. If v(t) and i(t) are the voltage across and the current through an inductor of L henries, then

$$v(t) = L \frac{di(t)}{dt}$$

The Laplace transform of this equation (assuming zero initial current) is

$$V(s) = LsI(s)$$

Similarly, for a capacitor of C farads, the voltage-current relationship is i(t) = C(dv/dt) and its Laplace transform, assuming zero initial capacitor voltage, yields I(s) = CsV(s); that is,

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

For a resistor of R ohms, the voltage-current relationship is v(t) = Ri(t), and its Laplace transform is

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

Thus, in the “frequency domain,” the voltage-current relationships of an inductor and a capacitor are algebraic; these elements behave like resistors of “resistance” Ls and 1/Cs, respectively. The generalized “resistance” of an element is called its impedance and is given by the ratio V(s)/I(s) for the element (under zero initial conditions). The impedances of a resistor of R ohms, an inductor of L henries, and a capacitance of C farads are R, Ls, and 1/Cs, respectively.

Also, the interconnection constraints (Kirchhoff’s laws) remain valid for voltages and currents in the frequency domain. To demonstrate this point, let vj(t) (j = 1, 2,…, k) be the voltages across k elements in a loop and let ij(t)(j = 1, 2,…,m) be the j currents entering a node. Then

$$\sum_{j=1}^{k} v_j(t) = 0 \quad \text{and} \quad \sum_{j=1}^{m} i_j(t) = 0$$

Now if

$$v_j(t) \Longleftrightarrow V_j(s)$$

and $i_j(t) \Longleftrightarrow I_j(s)$

then

$$\sum_{j=1}^{k} V_j(s) = 0 \quad \text{and} \quad \sum_{j=1}^{m} I_j(s) = 0$$

This result shows that if we represent all the voltages and currents in an electrical network by their Laplace transforms, we can treat the network as if it consisted of the “resistances” R, Ls, and 1/Cs corresponding to a resistor R, an inductor L, and a capacitor C, respectively. The system equations (loop or node) are now algebraic. Moreover, the simplification techniques that have been developed for resistive circuits—equivalent series and parallel impedances, voltage and current divider rules, Thévenin and Norton theorems—can be applied to general electrical networks. The following examples demonstrate these concepts.

Figure 4.10 (a) A circuit and (b) its transformed version.

In the first step, we represent the circuit in the frequency domain, as illustrated in Fig. 4.10b. All the voltages and currents are represented by their Laplace transforms. The voltage 10u(t) is represented by 10/s and the (unknown) current i(t) is represented by its Laplace transform I(s). All the circuit elements are represented by their respective impedances. The inductor of 1 henry is represented by s, the capacitor of 1/2 farad is represented by 2/s, and the resistor of 3 ohms is represented by 3. We now consider the frequency-domain representation of voltages and currents. The voltage across any element is I(s) times its impedance. Therefore, the total voltage drop in the loop is I(s) times the total loop impedance, and it must be equal to V(s), (transform of) the input voltage. The total impedance in the loop is

$$Z(s) = s + 3 + \frac{2}{s} = \frac{s^2 + 3s + 2}{s}$$

The input”voltage” is V(s) = 10/s. Therefore, the “loop current” I(s) is

$$I(s) = \frac{V(s)}{Z(s)} = \frac{10/s}{(s^2 + 3s + 2)/s} = \frac{10}{s^2 + 3s + 2} = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2}$$

The inverse transform of this equation yields the desired result:

$$i(t) = 10(e^{-t} - e^{-2t})u(t)$$

376 CHAPTER 4 CONTINUOUS-TIME SYSTEM ANALYSIS

INITIAL CONDITION GENERATORS

The discussion in which we assumed zero initial conditions can be readily extended to the case of nonzero initial conditions because the initial condition in a capacitor or an inductor can be represented by an equivalent source. We now show that a capacitor C with an initial voltage v(0−) (Fig. 4.11a) can be represented in the frequency domain by an uncharged capacitor of impedance 1/Cs in series with a voltage source of value v(0−)/s (Fig. 4.11b) or as the same uncharged capacitor in parallel with a current source of value Cv(0−) (Fig. 4.11c). Similarly, an inductor L with an initial current i(0−) (Fig. 4.11d) can be represented in the frequency domain by an inductor of impedance Ls in series with a voltage source of value Li(0−) (Fig. 4.11e) or by the same inductor in parallel with a current source of value i(0−)/s (Fig. 4.11f).

Figure 4.11 Initial condition generators for a capacitor and an inductor.

To prove this point, consider the terminal relationship of the capacitor in Fig. 4.11a:

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

The Laplace transform of this equation yields

$$I(s) = C[sV(s) - v(0^-)]$$

4.4 Analysis of Electrical Networks: The Transformed Network 377

This equation can be rearranged as

$$V(s) = \frac{1}{Cs}I(s) + \frac{v(0^{-})}{s}$$

\n(4.33)

Observe that V(s) is the voltage (in the frequency domain) across the charged capacitor and I(s)/Cs is the voltage across the same capacitor without any charge. Therefore, the charged capacitor can be represented by the uncharged capacitor in series with a voltage source of value v(0−)/s, as depicted in Fig. 4.11b. Equation (4.33) can also be rearranged as

$$V(s) = \frac{1}{Cs}[I(s) + Cv(0^{-})]$$

This equation shows that the charged capacitor voltage V(s) is equal to the uncharged capacitor voltage caused by a current I(s) + Cv(0−). This result is reflected precisely in Fig. 4.11c, where the current through the uncharged capacitor is I(s)+Cv(0−). †

For the inductor in Fig. 4.11d, the terminal equation is

$$v(t) = L \frac{di(t)}{dt}$$

and

$$V(s) = L[sI(s) - i(0^{-})] = LsI(s) - Li(0^{-})$$

\n(4.34)

This expression is consistent with Fig. 4.11e. We can rearrange Eq. (4.34) as

$$V(s) = Ls \left[ I(s) - \frac{i(0^{-})}{s} \right]$$

This expression is consistent with Fig. 4.11f.

Let us rework Ex. 4.13 using these concepts. Figure 4.12a shows the circuit in Fig. 4.7b with the initial conditions y(0−) = 2 and vC(0−) = 10. Figure 4.12b shows the frequency-domain representation (transformed circuit) of the circuit in Fig. 4.12a. The resistor is represented by its impedance 2; the inductor with initial current of 2 amperes is represented according to the arrangement in Fig. 4.11e with a series voltage source Ly(0−) = 2. The capacitor with initial voltage of 10 volts is represented according to the arrangement in Fig. 4.11b with a series voltage source v(0−)/s = 10/s. Note that the impedance of the inductor is s and that of the capacitor is 5/s. The input of 10u(t) is represented by its Laplace transform 10/s.

The total voltage in the loop is (10/s) + 2 − (10/s) = 2, and the loop impedance is (s+2+(5/s)). Therefore,

$$Y(s) = \frac{2}{s + 2 + 5/s} = \frac{2s}{s^2 + 2s + 5}$$

which confirms our earlier result in Ex. 4.13.

^† In the time domain, a charged capacitor C with initial voltage v(0−) can be represented as the same capacitor uncharged in series with a voltage source v(0−)u(t), or in parallel with a current source Cv(0−)δ(t). Similarly, an inductor L with initial current i(0−) can be represented by the same inductor with zero initial current in series with a voltage source Li(0−)δ(t) or with a parallel current source i(0−)u(t).

Figure 4.12 A circuit and its transformed version with initial-condition generators.

Figure 4.13 Using initial condition generators and Thévenin equivalent representation.

Inspection of this circuit shows that when the switch is closed and the steady-state conditions are reached, the capacitor voltage vC = 16 volts, and the inductor current y² = 4 amperes. Therefore, when the switch is opened (at t = 0), the initial conditions are vC(0−) = 16 and y2(0−) = 4. Figure 4.13b shows the transformed version of the circuit in Fig. 4.13a. We have used equivalent sources to account for the initial conditions. The initial capacitor voltage of 16 volts is represented by a series voltage of 16/s and the initial inductor current of 4 amperes is represented by a source of value Ly2(0−) = 2.

From Fig. 4.13b, the loop equations can be written directly in the frequency domain as

$$\frac{Y_1(s)}{s} + \frac{1}{5}[Y_1(s) - Y_2(s)] = \frac{4}{s}$$ $$-\frac{1}{5}Y_1(s) + \frac{6}{5}Y_2(s) + \frac{s}{2}Y_2(s) = 2$$ $$\begin{bmatrix} \frac{1}{s} + \frac{1}{5} & -\frac{1}{5} \\ -\frac{1}{5} & \frac{6}{5} + \frac{s}{2} \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} = \begin{bmatrix} \frac{4}{5} \\ \frac{5}{2} \end{bmatrix}$$

Application of Cramer’s rule to this equation yields

$$Y_1(s) = \frac{24(s+2)}{s^2 + 7s + 12} = \frac{24(s+2)}{(s+3)(s+4)} = \frac{-24}{s+3} + \frac{48}{s+4}$$

and

$$y_1(t) = (-24e^{-3t} + 48e^{-4t})u(t)$$

Similarly, we obtain

$$Y_2(s) = \frac{4(s+7)}{s^2 + 7s + 12} = \frac{16}{s+3} - \frac{12}{s+4}$$

and

$$y_2(t) = (16e^{-3t} - 12e^{-4t})u(t)$$

We also could have used Thévenin’s theorem to compute Y1(s) and Y2(s) by replacing the circuit to the right of the capacitor (right of terminals ab) with its Thévenin equivalent, as shown in Fig. 4.13c. Figure 4.13b shows that the Thévenin impedance Z(s) and the Thévenin source V(s) are

$$Z(s) = \frac{\frac{1}{5} \left(\frac{s}{2} + 1\right)}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{s+2}{5s+12}$$ $$V(s) = \frac{-\frac{1}{5}}{\frac{1}{5} + \frac{s}{2} + 1} = \frac{-4}{5s+12}$$

According to Fig. 4.13c, the current Y1(s) is given by

$$Y_1(s) = \frac{\frac{4}{s} - V(s)}{\frac{1}{s} + Z(s)} = \frac{24(s+2)}{s^2 + 7s + 12}$$

which confirms the earlier result. We may determine Y2(s) in a similar manner.

EXAMPLE 4.19 Transformed Analysis of a Coupled Inductive Network

The switch in the circuit in Fig. 4.14a is at position a for a long time before t = 0, when it is moved instantaneously to position b. Determine the current y1(t) and the output voltage v0(t) for t ≥ 0.

Just before switching, the values of the loop currents are 2 and 1, respectively, that is, y1(0−) = 2 and y2(0−) = 1.

The equivalent circuits for two types of inductive coupling are illustrated in Figs. 4.14b and 4.14c. For our situation, the circuit in Fig. 4.14c applies. Figure 4.14d shows the transformed version of the circuit in Fig. 4.14a after switching. Note that the inductors L¹ + M, L² + M, and −M are 3, 4, and −1 henries with impedances 3s, 4s, and −s respectively. The initial condition voltages in the three branches are (L¹ + M)y1(0−) = 6, (L² + M)y2(0−) = 4, and −M[y1(0−)−y2(0−)]=−1, respectively. The two loop equations of the circuit are

$$(2s+3)Y1(s) + (s-1)Y2(s) = \frac{10}{s} + 5$$ $$(s-1)Y1(s) + (3s+2)Y2(s) = 5$$

or

$$\begin{bmatrix} 2s+3 & s-1 \ s-1 & 3s+2 \end{bmatrix} \begin{bmatrix} Y_1(s) \\ Y_2(s) \end{bmatrix} \begin{bmatrix} \frac{5s+10}{s} \\ 5 \end{bmatrix}$$

Solving for Y1(s), we obtain

$$Y_1(s) = \frac{2s^2 + 9s + 4}{s(s^2 + 3s + 1)} = \frac{4}{s} - \frac{1}{s + 0.382} - \frac{1}{s + 2.618}$$

Therefore,

$$y_1(t) = (4 - e^{-0.382t} - e^{-2.618t})u(t)$$

Similarly,

$$Y_2(s) = \frac{s^2 + 2s + 2}{s(s^2 + 3s + 1)} = \frac{2}{s} - \frac{1.618}{s + 0.382} + \frac{0.618}{s + 2.618}$$

and

$$y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t)$$

The output voltage is therefore

$$v_0(t) = y_2(t) = (2 - 1.618e^{-0.382t} + 0.618e^{-2.618t})u(t)$$

DR ILL 4.10 Transformed Analysis of an RLC Circuit with a Switch

For the RLC circuit in Fig. 4.15, the input is switched on at t = 0. The initial conditions are y(0−) = 2 amperes and vC(0−) = 50 volts. Find the loop current y(t) and the capacitor voltage vC(t) for t ≥ 0.

ANSWERS

4.4-1 Analysis of Active Circuits

Although we have considered examples of only passive networks so far, the circuit analysis procedure using the Laplace transform is also applicable to active circuits. All that is needed is to replace the active elements with their mathematical models (or equivalent circuits) and proceed as before.

The operational amplifier (depicted by the triangular symbol in Fig. 4.16a) is a well-known element in modern electronic circuits. The terminals with the positive and the negative signs correspond to noninverting and inverting terminals, respectively. This means that the polarity of the output voltage v² is the same as that of the input voltage at the terminal marked by the positive sign (noninverting). The opposite is true for the inverting terminal, marked by the negative sign.

Figure 4.16b shows the model (equivalent circuit) of the operational amplifier (op amp) in Fig. 4.16a. A typical op amp has a very large gain. The output voltage v² = −Av1, where A is typically 105 to 106. The input impedance is very high, of the order of 1012 , and the output impedance is very low (50–100 ). For most applications, we are justified in assuming the gain A and the input impedance to be infinite and the output impedance to be zero. For this reason we see an ideal voltage source at the output.

Consider now the operational amplifier with resistors Ra and Rb connected, as shown in Fig. 4.16c. This configuration is known as the noninverting amplifier. Observe that the input polarities in this configuration are inverted in comparison to those in Fig. 4.16a. We now show that the output voltage v² and the input voltage v¹ in this case are related by

$$v_2 = Kv_1, \qquad \text{where } K = 1 + \frac{R_b}{R_a}$$

First, we recognize that because the input impedance and the gain of the operational amplifier approach infinity, the input current ix and the input voltage vx in Fig. 4.16c are infinitesimal and may be taken as zero. The dependent source in this case is Avx instead of −Avx because of the input polarity inversion. The dependent source Avx (see Fig. 4.16b) at the output will generate current io, as illustrated in Fig. 4.16c. Now

$$v_2 = (R_b + R_a)i_o$$

Figure 4.16 Operational amplifier and its equivalent circuit.

and also

$$v_1 = v_x + R_a i_o = R_a i_o$$

Therefore,

$$\frac{v_2}{v_1} = \frac{R_b + R_a}{R_a} = 1 + \frac{R_b}{R_a} = K$$

or

$$v_2(t) = Kv_1(t)$$

The equivalent circuit of the noninverting amplifier is depicted in Fig. 4.16d.

EXAMPLE 4.20 Transform Analysis of a Sallen–Key Circuit

The circuit in Fig. 4.17a is called the Sallen–Key circuit, which is frequently used in filter design. Find the transfer function H(s) relating the output voltage vo(t) to the input voltage vi(t).

Figure 4.17 (a) Sallen–Key circuit and (b) its equivalent.

We are required to find

$$H(s) = \frac{V_o(s)}{V_i(s)}$$

assuming all initial conditions to be zero.

Figure 4.17b shows the transformed version of the circuit in Fig. 4.17a. The noninverting amplifier is replaced by its equivalent circuit. All the voltages are replaced by their Laplace transforms, and all the circuit elements are shown by their impedances. All the initial conditions are assumed to be zero, as required for determining H(s).

We shall use node analysis to derive the result. There are two unknown node voltages, Va(s) and Vb(s), requiring two node equations.

At node a, IR¹ (s), the current in R¹ (leaving the node a), is [Va(s) − Vi(s)]/R1. Similarly, IR² (s), the current in R² (leaving the node a), is [Va(s) − Vb(s)]/R2, and IC¹ (s), the current in capacitor C¹ (leaving the node a), is [Va(s)−Vo(s)]C1s = [Va(s)−KVb(s)]C1s.

The sum of all the three currents is zero. Therefore,

$$\frac{V_a(s) - V_i(s)}{R_1} + \frac{V_a(s) - V_b(s)}{R_2} + [V_a(s) - KV_b(s)]C_1s = 0$$

or

$$\left(\frac{1}{R_1} + \frac{1}{R_2} + C_1 s\right) V_a(s) - \left(\frac{1}{R_2} + KC_1 s\right) V_b(s) = \frac{1}{R_1} V_i(s)$$

Similarly, the node equation at node b yields

$$\frac{V_b(s) - V_a(s)}{R_2} + C_2 s V_b(s) = 0$$

or

$$-\frac{1}{R_2}V_a(s) + \left(\frac{1}{R_2} + C_2s\right)V_b(s) = 0$$

The two node equations in two unknown node voltages Va(s) and Vb(s) can be expressed in matrix form as

$$\begin{bmatrix} G_1 + G_2 + C_1 s & -(G_2 + KC_1 s) \ -G_2 & (G_2 + C_2 s) \end{bmatrix} \begin{bmatrix} V_a(s) \ V_b(s) \end{bmatrix} = \begin{bmatrix} G_1 V_i(s) \ 0 \end{bmatrix}$$

where

$$G_1 = \frac{1}{R_1} \quad \text{and} \quad G_2 = \frac{1}{R_2}$$

Application of Cramer’s rule yields

$$\frac{V_b(s)}{V_i(s)} = \frac{G_1 G_2}{C_1 C_2 s^2 + [G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)]s + G_1 G_2}$$

=

$$\frac{\omega_0^2}{s^2 + 2\alpha s + \omega_0^2}$$

where

$$K = 1 + \frac{R_b}{R_a} \quad \text{and} \quad \omega_0^2 = \frac{G_1 G_2}{C_1 C_2} = \frac{1}{R_1 R_2 C_1 C_2}$$ $$2\alpha = \frac{G_1 C_2 + G_2 C_2 + G_2 C_1 (1 - K)}{C_1 C_2} = \frac{1}{R_1 C_1} + \frac{1}{R_2 C_1} + \frac{1}{R_2 C_2} (1 - K)$$

Now

$$V_o(s) = KV_b(s)$$

Therefore,

$$H(s) = \frac{V_o(s)}{V_i(s)} = K \frac{V_b(s)}{V_i(s)} = \frac{K\omega_0^2}{s^2 + 2\alpha s + \omega_0}$$

2