Solution:

← Back to Fundamentals of Electric Circuits Overview (s

• 3s + 2)V1 = 30 ^V1 = ____________ 30 (^s + 1)(s + 2) = _____ ³⁰ s + 1 − _____ 30 s + 2

or

$$v_1(t) = (30e^{-t} - 30e^{-2t})u(t) \text{ V} \tag{16.5.1}$$

For Fig. 16.13(b) we get,

$$\frac{V_2 - 0}{10/3} + \frac{V_2 - 0}{5s} - \frac{1}{s} + \frac{V_2 - 0}{1/(0.1s)} = 0$$

or

$$0.1\left(s+3+\frac{2}{s}\right)V_2 = \frac{1}{s}$$

This leads to

$$V_2 = \frac{10}{(s+1)(s+2)} = \frac{10}{s+1} - \frac{10}{s+2}$$

Taking the inverse Laplace transform, we get

$$v_2(t) = (10e^{-t} - 10e^{-2t})u(t) \text{ V}$$

(16.5.2)

For Fig. 16.13(c),

$$\frac{V_3 - 0}{10/3} + \frac{V_3 - 0}{5s} - 0 + \frac{V_3 - 5/s}{1/(0.1s)} = 0$$ $$0.1\left(s+3+\frac{2}{s}\right)V_3 = 0.5$$ $$V_3 = \frac{5s}{(s+1)(s+2)} = \frac{-5}{s+1} + \frac{10}{s+2}$$

This leads to

$$v_3(t) = (-5e^{-t} + 10e^{-2t})u(t) \text{ V}$$

(16.5.3)

Now all we need to do is to add Eqs. (16.5.1), (16.5.2), and (16.5.3):

$$v(t) = v_1(t) + v_2(t) + v_3(t)$$

= { (30 + 10 - 5)e^-t + (-30 + 10 - 10)e^-2t}u(t) V

or

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

which agrees with our answer in Example 16.4.

Practice Problem 16.5 For the circuit shown in Fig. 16.12, and the same initial conditions in Example 16.4, find the current through the inductor for all time t > 0 using superposition.

Answer:

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

A.

Figure 16.14 For Example 16.6.

Example 16.6 Assume that there is no initial ener gy stored in the circuit of Fig. 16.14 at t = 0 and that is = 10u(t) A. (a) Find Vo(s) using Thevenin’s theorem. (b) Apply the initial- and final-value theorems to find vo(0⁺) and vo(∞). (c) Determine vo(t).

Solution:

Because there is no initial energy stored in the circuit, we assume that the initial inductor current and initial capacitor voltage are zero at t = 0.

(a) To find the Thevenin equivalent circuit, we remove the 5-Ω resistor and then find Voc (VTh) and Isc. To find VTh, we use the Laplacetransformed circuit in Fig. 16.15(a). Since Ix = 0, the dependent voltage source contributes nothing, so

$$V_{\text{oc}} = V_{\text{Th}} = 5\left(\frac{10}{s}\right) = \frac{50}{s}$$

To find ZTh, we consider the circuit in Fig. 16.15(b), where we first find Isc. We can use nodal analysis to solve for V1 which then leads to Isc(Isc = Ix = V1∕2s).

s).

$$-\frac{10}{s} + \frac{(V_1 - 2I_x) - 0}{5} + \frac{V_1 - 0}{2s} = 0$$

along with

$$I_x = \frac{V_1}{2s}$$

leads to

$$V_1 = \frac{100}{2s + 3}$$

Hence,

$$I_{\rm sc} = \frac{V_1}{2s} = \frac{100/(2s+3)}{2s} = \frac{50}{s(2s+3)}$$

and

$$Z_{\text{Th}} = \frac{V_{\text{oc}}}{I_{\text{sc}}} = \frac{50/s}{50/[s(2s+3)]} = 2s + 3$$

The given circuit is replaced by its Thevenin equivalent at terminals a-b as shown in Fig. 16.16. From Fig. 16.16,

$$V_o = \frac{5}{5 + Z_{\text{Th}}} V_{\text{Th}} = \frac{5}{5 + 2s + 3} \left(\frac{50}{s}\right) = \frac{250}{s(2s + 8)} = \frac{125}{s(s + 4)}$$

(b) Using the initial-value theorem we find

$$v_o(0) = \lim_{s \to \infty} sV_o(s) = \lim_{s \to \infty} \frac{125}{s+4} = \lim_{s \to \infty} \frac{125/s}{1+4/s} = \frac{0}{1} = 0$$

Using the final-value theorem we find

$$v_o(\infty) = \lim_{s \to 0} sV_o(s) = \lim_{s \to 0} \frac{125}{s+4} = \frac{125}{4} = 31.25 \text{ V}$$

(c) By partial fraction,

$$V_o = \frac{125}{s(s+4)} = \frac{A}{s} + \frac{B}{s+4}$$

\n

$$A = sV_o(s) \Big|_{s=0} = \frac{125}{s+4} \Big|_{s=0} = 31.25$$

\n

$$B = (s+4)V_o(s) \Big|_{s=-4} = \frac{125}{s} \Big|_{s=-4} = -31.25$$

\n

$$V_o = \frac{31.25}{s} - \frac{31.25}{s+4}$$

Taking the inverse Laplace transform gives

$$v_o(t) = 31.25(1 - e^{-4t})u(t)$$

V

Notice that the values of vo(0) and vo(∞) obtained in part (b) are confirmed.

The initial energy in the circuit of Fig. 16.17 is zero at t = 0. Assume that Practice Problem 16.6 vs = 360u(t) V. (a) Find Vo(s) using the Thevenin theorem. (b) Apply the initial- and final-value theorems to find vo(0) and vo(∞). (c) Obtain vo(t).

Answer: (a)

$$V_o(s) = \frac{288(s+0.25)}{s(s+0.3)}$$

, (b) 288 V, 240 V,
(c) $(240 + 48e^{-0.3t})u(t)$ V.

(b)

Figure 16.15 For Example 16.6: (a) finding VTh, (b) determining ZTh.

Figure 16.16 The Thevenin equivalent of the circuit in Fig. 16.14 in the s-domain.

Figure 16.17 For Practice Prob. 16.6.

16.4 Transfer Functions

The transfer function is a key concept in signal processing because it indicates how a signal is processed as it passes through a network. It is a fitting tool for finding the network response, determining (or designing for) network stability, and network synthesis. The transfer function of a network describes how the output behaves with respect to the input. It specifies the transfer from the input to the output in the s-domain, assuming no initial energy.

The transfer function H(s) is the ratio of the output response Y(s) to the input excitation X(s), assuming all initial conditions are zero.

Thus,

$$H(s) = \frac{Y(s)}{X(s)}$$

\n(16.15)

The transfer function depends on what we define as input and output. Because the input and output can be either current or voltage at any place in the circuit, there are four possible transfer functions:

$$H(s) = \text{Voltage gain} = \frac{V_o(s)}{V_i(s)}\tag{16.16a}$$ $$H(s) = \text{Current gain} = \frac{I_o(s)}{I_i(s)}\tag{16.16b}$$ $$H(s) = \text{Impedance} = \frac{V(s)}{I(s)}\tag{16.16c}$$ $$H(s) = \text{Admittance} = \frac{I(s)}{V(s)}\tag{16.16d}$$

Thus, a circuit can ha ve man y transfer functions. Note that H(s) is dimensionless in Eqs. (16.16a) and (16.16b).

Each of the transfer functions in Eq. (16.16) can be found in two ways. One w ay is to assume any convenient input X(s), use any circuit analysis technique (such as current or v oltage division, nodal or mesh analysis) to find the output Y(s), and then obtain the ratio of the two. The other approach is to apply the ladder method, which involves walking our way through the circuit. By this approach, we assume that the output is 1 V or 1 A as appropriate and use the basic laws of Ohm and Kirchhoff (KCL only) to obtain the input. The transfer function becomes unity divided by the input. This approach may be more convenient to use when the circuit has many loops or nodes so that applying nodal or mesh analysis becomes cumbersome. In the first method, we assume an input and find the output; in the second method, we assume the output and find the input. In both methods, we calculate H(s) as the ratio of output to input transforms. The two methods rely on the linearity property, since we only deal with linear circuits in this book. Example 16.8 illustrates these methods.

Some authors would not consider Eqs. (16.16c) and (16.16d) transfer functions.

For electrical networks, the transfer function is also known as the network

function.

Equation (16.15) assumes that both X(s) and Y(s) are known. Sometimes, we know the input X(s) and the transfer function H(s). We find the output Y(s) as

$$Y(s) = H(s)X(s) \tag{16.17}$$

and take the inverse transform to get y(t). A special case is when the input is the unit impulse function, x(t) = δ(t), so that X(s) = 1. For this case,

$$Y(s) = H(s)$$

or $y(t) = h(t)$ (16.18)

where

$$h(t) = \mathcal{L}^{-1}[H(s)]$$

(16.19)

The term h(t) represents the unit impulse response—it is the time-domain response of the netw ork to a unit impulse. Thus, Eq. (16.19) pro vides a new interpretation for the transfer function: H(s) is the Laplace transform of the unit impulse response of the netw ork. Once we kno w the impulse response h(t) of a network, we can obtain the response of the netw ork to any input signal using Eq. (16.17) in the s-domain or using the convolution integral (section 15.5) in the time domain.

The output of a linear system is y(t) = 10*^e* Example 16.7 ^−t cos 4t u(t) when the input is x(t) = e^−t u(t). Find the transfer function of the system and its impulse response.

Solution:

If x(t) = e^−t u(t) and y(t) = 10e^−t cos 4t u(t), then

$$X(s) = \frac{1}{s+1}$$

and $Y(s) = \frac{10(s+1)}{(s+1)^2 + 4^2}$

Hence,

$$H(s) = \frac{Y(s)}{X(s)} = \frac{10(s+1)^2}{(s+1)^2 + 16} = \frac{10(s^2 + 2s + 1)}{s^2 + 2s + 17}$$

To find h(t), we write H(s) as

$$H(s) = 10 - 40 \frac{4}{(s+1)^2 + 4^2}$$

From Table 15.2, we obtain

$$h(t) = 10\delta(t) - 40e^{-t}\sin 4t \,u(t)$$