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 frequency-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 and admittance to relate voltage and current. In general, a linear network Input Output can be represented by the block diagram shown in Fig. 1.
A block diagram representation of a linear network.
Fig. 1: A block diagram representation of a linear network.
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).
$$H(ω) ={Y(ω) \over X(ω)}$$
assuming zero initial conditions. Since the input and output can be either voltage or current at any place in the circuit, there are four possible transfer functions:
$$ \bbox[10px,border:1px solid grey]{\begin{split} H(ω) &= \text{Voltage gain} ={V_o(ω) \over V_i(ω)} \\ H(ω) &= \text{ Current gain} ={I_o(ω) \over I_i(ω)}\\ H(ω) &= \text{Transfer Impedance} ={V_o(ω) \over I_i(ω)}\\ H(ω) &= \text{Transfer Admittance} ={I_o(ω) \over V_i(ω)} \end{split}} \tag{1}$$
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, $H(ω) = H (ω) \angle φ$.
To obtain the transfer function using Eq. (1), we first obtain the frequency-domain equivalent of the circuit by replacing resistors, inductors, and capacitors with their impedances $R$, $jωL$, and $1/jωC$. We then use any circuit technique(s) to obtain the appropriate quantity in Eq. (1).
We can obtain the frequency response of the circuit by plotting the magnitude and phase of the transfer function as the frequency varies.
Some authors use H( jω) for transfer instead of H(ω), since ω and j are an inseparable pair. The transfer function H(ω) can be expressed in terms of its numerator polynomial N(ω) and denominator polynomial D(ω) as
$$\bbox[10px,border:1px solid grey]{H(ω) ={N(ω) \over D(ω)}} \tag{2}$$
where N(ω) and D(ω) are not necessarily the same expressions for the input and output functions, respectively. The representation of H(ω) in Eq. (2) assumes that common numerator and denominator factors in H(ω) have canceled, reducing the ratio to lowest 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.
Example 1: For the RC circuit in Fig. 2(a), obtain the transfer function $V_o/V_s$ and its frequency response. Let $$v_s = V_m cos ωt$$
A block diagram representation of a linear network.
Fig. 2: For Example 1: (a) time-domain RC circuit, (b) frequency-domain RC circuit.
Solution: The frequency-domain equivalent of the circuit is in Fig. 2(b). By voltage division, the transfer function is given by $$\mathbf{H}(\omega)=\frac{\mathbf{V}_{o}}{\mathbf{V}_{s}}=\frac{1 / j \omega C}{R+1 / j \omega C}=\frac{1}{1+j \omega R C}$$ we obtain the magnitude and phase of $ \mathbf{H}(\omega) $ as $$H=\frac{1}{\sqrt{1+\left(\omega / \omega_{0}\right)^{2}}}$$ and $$\phi=- tan ^{-1} \frac{\omega}{\omega_{0}}$$ where $ \omega_{0}=1 / R C $.
To plot $ H $ and $\phi $ for $ 0<\omega<\infty $, we obtain their values at some critical points and then sketch.
At $ \omega=0, H=1 $ and $ \phi=0 $.
At $ \omega=\infty, H=0 $ and $ \phi=-90^{\circ} $.
Also, at $ \omega=\omega_{0}, H=1 / \sqrt{2} $ and $ \phi=-45^{\circ} $.
With these and a few more points as shown in Table 1, we find that the frequency response is as shown in Fig. 3.
Fig. 3: Frequency response of the RC circuit: (a) amplitude response, (b) phase response.
Table 1.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024
Privacy Policy