The transfer function H(ω) (also called the network function) is a useful analytical tool for finding the frequency response of a circuit. You will find this function quite often when you try to analyze a closed-loop electrical circuit.
In fact, the frequency transfer function of a circuit is the plot of the circuit’s transfer function H(ω) versus ω, with ω varying from ω = 0 to ω = ∞.
What is Transfer Function
The transfer function definition is the frequency-dependent ratio of a forced function to a forcing function (or of output to 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 can be represented by the block diagram shown below.
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,
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 function formulas:
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.
They are easy to remember since the voltage transfer function is the comparison of the output voltage and input voltage. The same thing applies to the current transfer function where it is a comparison of the output current and input current.
To obtain the transfer function using the equation above, 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 the equation above. We can obtain the frequency response of the circuit by plotting the magnitude and phase of the transfer function as the frequency varies.
A computer is a real time-saver for plotting the transfer function.
Some 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
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 the equation above 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. H(s) transfer function will make solving the analysis easier since it gets rid of the complex algebra without changing the result.
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.
Transfer Function of Circuit Examples
Let’s review the transfer function examples below:
1. For the RC circuit below, obtain the transfer function Vo/Vs and its frequency response. Let vs = Vm cos ωt. The left is the time-domain RC circuit and the right is the frequency-domain RC circuit.
Solution:
The frequency-domain equivalent of the circuit is on the right. By voltage division, the transfer function is given by
We obtain the magnitude and phase of H(ω) as
where ω0 = 1/RC. To plot H and 𝝓 for 0 < ω < ∞, we obtain their values at some critical points and then sketch. The left is the amplitude response and the right is the phase response
At ω = 0, H = 1 and 𝝓 = 0. At ω = ∞, H = 0 and 𝝓 = −90o. Also, at ω = ω0, H = 1/√2 and 𝝓 = −45o. With these and a few more points as shown in the table below, we find that the frequency response is as shown above.
2. For the circuit below, calculate the gain Io(ω)/Ii(ω) and its poles and zeros.
Solution:
By current division,
or
The zeros are at
The poles are at
Thus, there is a repeated pole (or double pole) at p = −1