Continuous Transfer Functions

\(
\require{AMSsymbols}\def\lfz#1{\overset{\Large#1}{\,\circ\kern-6mu-\kern-7mu-\kern-7mu-\kern-6mu\bullet\,}}
\def\laplace{\lfz{\mathscr{L}}}
\def\fourier{\lfz{\mathcal{F}}}
\def\ztransform{\lfz{\mathcal{Z}}}
\require{cancel}
\newcommand\ccancel[2][black]
{\color{#1}{\cancel{\color{black}{#2}}}}
\)Consider a black box with input signal \(x(t)\) and output \(y(t)\). This black box processes the input signal and it produces the output signal.

Black box image
Blackbox model , source: wikipedia

The black box

When we model the transfer function of this black box as \(h(t)\), the output signal \(y(t)\) is a convolution ‘\(\ast\)’ of the input \(x(t)\) and the transfer function \(h(t)\).
$$y(t)=h(t)*x(t)$$

Likewise, in the \(s\)-domain, the transfer function describes how the output signal \(Y(s)\) responds to an arbitrary input signal \(X(s)\). This convolution in the time-domain becomes a multiplication in the \(s\)-domain. Working in this \(s\)-domain makes the convolution, into a multiplication and is thereby easier to solve.
$$Y(s)=H(s)X(s)$$

It allows one to determine the system response characteristics without having to solve the convolution.

Albeit for the discrete case, Discrete Transfer Functions describes why convolution is used in the time domain, and multiplication in the \(z\) domain.

Poles and zeroes

The generic form of the transfer function is
$$H(s)=\frac{Y(s)}{X(s)}=\frac{b_ms^m+b_{m-1}s^{m-1}+\dots+b_1s+b_0}{a_ns^n+a_{n-1}s^{n-1}+\dots+a_1s+a_0}\label{eq:tf_polynominal}$$

where \(s=\sigma+j\omega\). \(X(s)\)  and \(Y(s)\) are the Laplace transform of the time representation of the input and output voltages \(x(t)\) and \(y(t)\). The highest power of the variable \(s\) determines the order of the system, usually corresponding to total number of capacitors and inductors in the circuit.

It can be convenient to factor the polynomials in the numerator and denominator of the transfer function, and to write the function in terms of those factors [MIT]

$$\begin{array}{cr}
H(s)=K\frac{N(s)}{D(s)}=K\frac{(s-z_1)(s-z_2)\dots(s-z_m)}{(s-p_1)(s-p_2)\dots(s-p_n)},&K=\frac{b_m}{a_n}
\end{array}\label{eq:tf_factors}$$

The \(z_i\)’s are the roots of the equation \(N(s)=0\) and are defined as the system zeros.  The the \(p_i\)’s are the roots of the equation \(D(s)=0\) and are defined as the system poles.

The (complex) poles and zeros are properties of the transfer function, and therefore of the differential equation describing the input-output system dynamics. Together with the gain constant \(K\) they completely characterize the differential equation, and provide a complete description of the system.

Pole-Zero plot

The system dynamics may be represented graphically by plotting the pole and zero locations on the complex \(s\)-plane, whose axes represent the real and imaginary parts of the complex variable \(s\). Such plots are known as pole-zero plots.

It is usual to mark a zero location by a circle (\(\circ\)) and a pole location a cross (\(\times\)). The location of the poles and zeros provide qualitative insights into the response characteristics of a system. [MIT]

s-plane complex poles and zero
Poles and zeroes in \(s\)-plane

Transfer function

The transfer function may be evaluated for any value of \(s=\sigma+j\omega\). It is common to express the complex value of the transfer function in polar form.

$$H(s)=\left|H(s)\right|e^{j\angle H(s)}\label{eq:tf}$$

where magnitude \(|H(s)|\) and phase \(\angle{H(s)}\) are given by

$$|H(s)| \equiv \sqrt{\Re\left\{H(s)\right\}^2 + \Im\left\{H(s)\right\}^2}$$

$$\angle{H(s)} \equiv \mathrm{atan2}\left( \Im\left\{H(s)\right\}, \Re\left\{H(s)\right\} \right)$$

where \(\Re\) is the real operator, and \(\Im\) is the imaginary operator, and \(\mathrm{atan2}\) returns a value between \(-\pi\) and \(\pi\) [wiki], as defined in

$$\mathrm{atan2}(y,x)=\begin{cases}
\arctan\left(\frac{y}{x}\right) & x > 0 \\
\arctan\left(\frac{y}{x}\right)+\pi & x < 0 \land y \geq 0 \\
\arctan\left(\frac{y}{x}\right)-\pi & x < 0 \land y < 0 \\
\frac{\pi}{2} & x= 0 \land y > 0 \\
-\frac{\pi}{2} & x= 0 \land y < 0 \\
\text{undefined} & x= 0 \land y = 0
\end{cases}$$

Visualization

The Laplace transform’s \(s\)-domain uses a rectangular coordinate system by defining \(s\triangleq\sigma+j\omega\), where \(\sigma\) on the horizontal axis represents the exponential decay, and \(\omega\) on the vertical axis represents the frequency.

The factorized transfer function \(\eqref{eq:tf_factors}\) can be written as
$$
H(z)=K \frac{\prod_{i=1}^m(z-q_i)}{\prod_{i=1}^n(z-p_i)}
$$

In the complex plane, the difference between two number \(s_1\) and \(s_2\) can be visualized by an vector from \(s_2\) to \(s_1\)
$$
\begin{align}
s_1-s_2&=(\sigma_1+j\omega_1)-(\sigma_2+j\omega_2)\nonumber\\
&=(\sigma_1-\sigma_2)+j(\omega_1-\omega_2)
\end{align}
$$

This can be visualized with an vector drawn from the tip of \(s_2\) to the tip of \(s_1\). Note that the length of the vector is unaffected by translation away from the origin. But the angle of the vector must be measured relative to a translated copy of the real axis.

Therefore, each of the factors in the numerator and denominator may be interpreted as a vector in the s-plane, originating from the zero \(z_i\) or pole \(p_i\) and directed to the point \(s\) at which the function is to be evaluated.

s-plane complex poles and zero evaluated for s
Pole ‘\(p\)’ evaluated at point ‘\(s\)’

Each of these vectors may be written in polar form, for example for a pole \(p_i=\sigma_i+j\omega_i\), the magnitude and angle of the vector to the point \(s=\sigma+j\omega\) are
$$\begin{aligned}
|s-p_i| &= \sqrt{(\sigma-\sigma_i)^2+(\omega-\omega_i)^2} \\
\angle(s-p_i) &= \mathrm{atan2}\left(\omega-\omega_i,\sigma-\sigma_i\right)
\end{aligned}$$

Multiplication and division

While we’re on the subject, a quick note: Multiplication and division of complex numbers is most easily done in polar form

$$\begin{eqnarray}
Z_1Z_2&
=|Z_1|e^{j\angle{Z_1}}\ |Z_2|e^{j\angle{Z_2}}&
=|Z_1||Z_2|e^{j(\angle{Z_1}+\angle{Z_2{)}}}\\
\frac{Z_1}{Z_2}&
=\frac{|Z_1|e^{j\angle{Z_1}}}{|Z_2|e^{j\angle{Z_2}}}&
=\frac{|Z_1|}{|Z_2|}e^{j(\angle{Z_1}-\angle{Z_2{)}}}
\end{eqnarray}$$

Applying \(|K|=K\) and \(\angle{K}=\mathrm{atan2}(0,K)=0\), the magnitude and angle of the complete transfer function \(H(s)\) may be written as

$$\left\{
\begin{align}
H(s) &= |H(s)|\ e^{j\angle{H(s)}}\nonumber\\
|H(s)| &= K \frac{\prod_{i=1}^m\left|(s-z_i)\right|}{\prod_{i=1}^n\left|(s-p_i)\right|}\nonumber\\
\angle{H(s)}&=\sum_{i=1}^m\angle(s-z_i)-\sum_{i=1}^n\angle(s-p_i)\nonumber
\end{align}
\right.$$

Some other tidbits:

  • A time-continuous system is unstable when poles are in the right half of the \(s\)-plane.
  • In the \(s\)-plane, the values along the vertical axis are equal to the frequency response of the system. That is, the Fourier transform is the Laplace transform evaluated at \(\sigma=0\).

Next

Suggested next reading is Evaluating Transfer Functions.

Embedded software developer
Passionately curious and stubbornly persistent. Enjoys to inspire and consult with others to exchange the poetry of logical ideas.

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