The transfer function H(w) (sometimes called the "network function") is an analytical tool used to find the frequency response of a circuit. The frequency response of the circuit is the plot of H(w) versus frequency (w) as "w" is varied from 0 to infinity. In essence, the transfer function is the frequency-dependent ratio of a forced function (output) to a forcing function (input).
Formal definition
When examining the transfer function in terms of phasors, it is defined as the frequency-dependent ratio of a phasor output Y(w) to a phasor input X(w). $$ \mathbb{H}(\omega) = \frac{\mathbb{Y}(\omega)}{\mathbb{X}(\omega)} $$ ...where: $$ \mathbb{Y}(\omega) = output \; voltage \; or \; current $$ $$ \mathbb{X}(\omega) = source (input) \; voltage \; or \; current $$
Since inputs and outputs can be either voltage or current we have four possible transfer functions: $$ 1) \; \mathbb{H}(\omega) = \frac{\mathbb{V}_o(\omega)}{\mathbb{V}_i(\omega)} = voltage \; gain $$ $$ 2) \; \mathbb{H}(\omega) = \frac{\mathbb{I}_o(\omega)}{\mathbb{I}_i(\omega)} = current \; gain $$ $$ 3) \; \mathbb{H}(\omega) = \frac{\mathbb{V}_o(\omega)}{\mathbb{I}_i(\omega)} = transfer \; impedance $$ $$ 4) \; \mathbb{H}(\omega) = \frac{\mathbb{I}_o(\omega)}{\mathbb{V}_i(\omega)} = transfer \; admittance $$
The resulting transfer function H(w) is a complex quantity containing both a magnitude and a phase angle: $$ \mathbb{H}(\omega) = H(\omega) \angle \phi $$
Determining the transfer function
The following three steps are used to determine the transfer function:
- Convert the circuit from the time-domain to the frequency-domain. (Recall the following impedance definitions for inductors and capacitors): $$ \mathbb{Z}_L = j\omega L $$ $$ \mathbb{Z}_C = \frac{1}{j \omega C} = \frac{-j}{\omega C} $$
- Use circuit techniques to determine H(w)
- Determine the frequency response by plotting magnitude and phase as frequency varies. (Best left to a computer.)
General expression for the transfer function
More generally, the transfer function can be written/defined as: $$ \mathbb{H}(\omega) = \frac{\mathbb{N}(\omega)}{\mathbb{D}(\omega)} \qquad,Expression \; 1 $$ ...where: $$ \mathbb{N}(\omega) = numerator \; polynomial $$ $$ \mathbb{D}(\omega) = denominator \; polynomial $$
Expression #1 assumes that common factors in the numerator and denominator have been canceled and the ratio has been reduced to its lowest terms.
Zeroes of H(w):
The zeroes of the transfer function are obtained by: $$ letting \; \mathbb{N}(\omega) = 0 $$ The zeroes of the transfer function are values that result in a zero value for the function and are represented as: $$ j\omega = z_1, z_2, ... $$
Poles of H(w):
The poles of the transfer function are obtained by: $$ letting \; \mathbb{D}(\omega) = 0 $$ The poles of the transfer function are values for which the function is infinite and are represented as: $$ j\omega = p_1, p_2, ... $$
Often, the "jw" term is replaced with "s" in order to avoid complex algebra. At the end of the calculations, the "jw" term is placed back in.
In the next page we will look at a transfer function example problem...