14.2 Transfer Function

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14.2 Transfer Function

The transfer function H(ω) (also called the network function) is a useful analytical tool for finding the frequency response of a circuit. In fact, the frequency response of a circuit is the plot of the circuit’ s transfer function H(ω) versus ω, with ω varying from ω = 0 to ω = ∞.

A transfer function is the frequenc y-dependent ratio of a forced function to a forcing function (or of an output to an input). The idea of a transfer function was implicit when we used the concepts of impedance

The frequency response of a circuit may also be considered as the variation of the gain and phase with frequency.

and admittance to relate v oltage and current. In general, a linear net work can be represented by the block diagram shown in Fig. 14.1.

The transfer function H(ω) of a circuit is the frequency-dependent ratio of a phasor output Y(ω) (an element voltage or current) to a phasor input X(ω) (source voltage or current).

Thus,

$$\mathbf{H}(\omega) = \frac{\mathbf{Y}(\omega)}{\mathbf{X}(\omega)}\tag{14.1}$$

assuming zero initial conditions. Since the input and output can be ei ther voltage or current at any place in the circuit, there are four possible transfer functions:

$$\mathbf{H}(\omega) = \text{Voltage gain} = \frac{\mathbf{V}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2a}$$ $$\mathbf{H}(\omega) = \text{Current gain} = \frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)}\tag{14.2b}$$ $$\mathbf{H}(\omega) = \text{Transfer impedance} = \frac{\mathbf{V}_o(\omega)}{\mathbf{I}_i(\omega)} \tag{14.2c}$$ $$\mathbf{H}(\omega) = \text{Transfer admittance} = \frac{\mathbf{I}_o(\omega)}{\mathbf{V}_i(\omega)}\tag{14.2d}$$

where subscripts i and o denote input and output values. Being a complex quantity, H(ω) has a magnitude H(ω) and a phase ϕ; that is, H(ω) = H(ω)⧸ϕ.

To obtain the transfer function using Eq. (14.2), we first obtain the frequenc y-domain equi valent of the circuit by replacing resistors, inductors, and capacitors with their impedances R, jωL, and 1∕jωC. We then use an y circuit technique(s) to obtain the appropriate quantity in Eq. (14.2). We can obtain the frequency response of the circuit by plot ting the magnitude and phase of the transfer function as the frequenc y varies. A computer is a real time-saver for plotting the transfer function.

The transfer function H(ω) can be expressed in terms of its numerator polynomial N(ω) and denominator polynomial D(ω) as

$$\mathbf{H}(\omega) = \frac{\mathbf{N}(\omega)}{\mathbf{D}(\omega)}$$

(14.3)

where N(ω) and D(ω) are not necessarily the same e xpressions for the input and output functions, respecti vely. The representation of H(ω) in Eq. (14.3) assumes that common numerator and denominator f actors in H(ω) have canceled, reducing the ratio to lo west terms. The roots of N(ω) = 0 are called the zeros of H(ω) and are usually represented as jω = z1, z2, …. Similarly, the roots of D(ω) = 0 are the poles of H(ω) and are represented as jω = p1, p2,….

A zero, as a root of the numerator polynomial, is a value that results in a zero value of the function. A pole, as a root of the denominator polynomial, is a value for which the function is infinite.

To avoid complex algebra, it is expedient to replace jω temporarily with s when working with H(ω) and replace s with jω at the end.

Figure 14.1

A block diagram representation of a linear network.

In this context, X(ω) and Y(ω) denote the input and output phasors of a network; they should not be confused with the same symbolism used for reactance and admittance. The multiple usage of symbols is conventionally permissible due to lack of enough letters in the English language to express all circuit variables distinctly.

Some authors use H( jω) for transfer instead of H(ω), since ω and j are an inseparable pair.

A zero may also be regarded as the value of s = jω that makes H(s) zero, and a pole as the value of s = jω that makes H(s) infinite.

Solution:

The frequency-domain equivalent of the circuit is in Fig. 14.2(b). By voltage division, the transfer function is given by

Figure 14.2 For Example 14.1: (a) time-domain RC circuit,

Comparing this with Eq. (9.18e), we obtain the magnitude and phase of H(ω) as

$$H = \frac{1}{\sqrt{1 + (\omega/\omega_0)^2}}, \qquad \phi = -\tan^{-1}\frac{\omega}{\omega_0}$$

where ω0 = 1∕RC. To plot H and ϕ for 0 < ω< ∞, we obtain their values at some critical points and then sketch.

At ω = 0, H = 1 and ϕ = 0. At ω = ∞, H = 0 and ϕ = −90°. Also, at ω = ω0, H = 1∕ √ __ 2 and ϕ = −45°. With these and a few more points as shown in Table 14.1, we find that the frequency response is as shown in Fig. 14.3. Additional features of the frequency response in Fig. 14.3 will be explained in Section 14.6.1 on low-pass filters.

TABLE 14.1 For Example 14.1. ω∕ω**⁰** H ϕ ω∕ω**⁰** H ϕ 0 1 0 10 0.1 −84° 1 0.71 −45° 20 0.05 −87° 2 0.45 −63° 100 0.01 −89°

Practice Problem 14.1 Obtain the transfer function Vo∕Vs of the RL circuit in Fig. 14.4, assuming vs = Vm cos ωt. Sketch its frequency response.

3 0.32 −72° ∞ 0 −90°

Answer: jωL∕(R + jωL); see Fig. 14.5 for the response.

H 1

0.707

Frequency response of the RC circuit: (a) amplitude response, (b) phase response.

Figure 14.4 RL circuit for Practice Prob. 14.1.

⁽b) frequency-domain RC circuit.

For the circuit in Fig. 14.6, calculate the g ain Io(ω)∕Ii(ω) and its poles Example 14.2 and zeros.

Solution:

By current division,

$$\mathbf{I}_o(\omega) = \frac{4 + j2\omega}{4 + j2\omega + 1/j0.5\omega} \mathbf{I}_i(\omega)$$

or

$$\frac{\mathbf{I}_o(\omega)}{\mathbf{I}_i(\omega)} = \frac{j0.5\omega(4+j2\omega)}{1+j2\omega + (j\omega)^2} = \frac{s(s+2)}{s^2 + 2s + 1}, \qquad s = j\omega$$

The zeros are at

s(s + 2) =0 ⇒ z1 = 0, z2 = −2

The poles are at

$$s^2 + 2s + 1 = (s + 1)^2 = 0$$

Thus, there is a repeated pole (or double pole) at p = −1.

Find the transfer function Vo(ω)∕Ii(ω) for the circuit in Fig. 14.7. Obtain Practice Problem 14.2 its zeros and poles.

its zeros and poles.
\nAnswer:

$$\frac{10(s + 2)(s + 5)}{s^2 + 10s + 10}$$

, $s = j\omega$ ; zeros: -2, -5; poles: -1.127, -8.873.