Part 3 - Modelling in Frequency Domain

Introduction

We will deal with SISO (Single-Input Single-Output) systems and introduce the Laplace transform.

MIMO (Multiple-Input Multiple-Output) usually use a different set of mathematical machinery (matrix representation of systems, called state space form). We will uncover this later on, but the primary focus is on SISO.

As we saw last time, differential equations can be used to represent relationship between input and output of a system.

a_2 \frac{d^2y}{dt^2} + a_1 \frac{dy}{dt} + y = b_2 \frac{d^2x}{dt^2} + b_1 \frac{dx}{dt} + b_0 x.

However, there is a problem, system parameters and input and output appear throughout the equation.

We would like to represent the system where the input and output are separated from the system parameters.

Block Diagrams

As we have seen in Lecture 1, we want to be able to represent the system as series of cascading subsystems, which can easily be combined together.

However, this can not be achieved (easily) with differential equations.

Laplace Transform

The Laplace transform is a mathematical tool that allows us to convert a differential equation into an algebraic equation.

With this, we can represent the mathematical description of a given system with its block diagram representation.

Definition 1 (Laplace transform)

\mathcal{L}\{f(t)\} = \int_0^{\infty} f(t) e^{-st} \, dt = F(s),

where $F$ is a complex-valued function of complex numbers.

$s$ is called the (complex) frequency variable, with units $\mathrm{s}^{-1}$ .

$t$ is called the time variable, with units $\mathrm{s}$ .

Later on we will use $s = \sigma + j\omega$ where $\sigma$ is the real part and $\omega$ is the imaginary part.

Notice that $F(s)$ contains no information about $f(t)$ for $t < 0$ , this is called the one-sided Laplace transform.

In controls, the mathematical definition of the Laplace transform is rarely used, instead we use tables of Laplace transforms.

Table 1: Common one-sided Laplace transforms.

$f(t)$	$F(s)$
$\delta(t)$	1
$u(t)$	$\frac{1}{s}$
$tu(t)$	$\frac{1}{s^2}$
$t^n u(t)$	$\frac{n!}{s^{n+1}}$
$e^{-at} u(t)$	$\frac{1}{s+a}$
$\sin(\omega t) u(t)$	$\frac{\omega}{s^2 + \omega^2}$
$\cos(\omega t) u(t)$	$\frac{s}{s^2 + \omega^2}$

Important Properties of Laplace Transform

Linearity $\mathcal{L}\{k_1 f_1(t) \pm k_2 f_2(t)\} = k_1 F_1(s) \pm k_2 F_2(s).$
Differentiation $\begin{align*} \mathcal{L}\left\{\frac{d f(t)}{dt}\right\} & = s F(s) - f(0), \newline \mathcal{L}\left\{\frac{d^2 f(t)}{dt^2}\right\} & = s^2 F(s) - s f(0) - f'(0), \newline & \vdots \end{align*}$
Frequency Shifting $\mathcal{L}\{e^{-at} f(t)\} = F(s+a).$

There are some other ones, but these are the most important ones.

Inverse Laplace Transform

The inverse Laplace transform is the operation that allows us to go from the Laplace domain to the time domain.

\mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi j} \int_{\sigma-j\infty}^{\sigma+j\infty} F(s) e^{st} \, ds,

where $j$ is the imaginary unit, $j = \sqrt{-1}$ .

We will not use the definition of the inverse Laplace transform, instead we will use tables of inverse Laplace transforms.

Transfer Functions

A transfer function is a relationship between the Laplace Transform of 2 signals (input and output) of a system.

Say that we have a block diagram of a system, with input $U(s)$ and output $Y(s)$ . Where $U(s)$ is the Laplace transform of the input signal $u(t)$ and $Y(s)$ is the Laplace transform of the output signal $y(t)$ .

Definition 2 (Transfer function)

The transfer function $G(s)$ is defined as

G(s) = \frac{Y(s)}{U(s)},

when the initial conditions are zero.

Example 1

Consider the following system

a_0 \frac{d^n y}{dt^n} + a_1 \frac{d^{n-1} y}{dt^{n-1}} + \cdots + a_{n-1} \frac{dy}{dt} + a_n y = b_0 \frac{d^m u}{dt^m} + b_1 \frac{d^{m-1} u}{dt^{m-1}} + \cdots + b_{m-1} \frac{du}{dt} + b_m u.

Taking the Laplace transform of both sides, we get

[a_0 s^n + a_1 s^{n-1} + \cdots + a_{n-1} s + a_n] Y(s) = [b_0 s^m + b_1 s^{m-1} + \cdots + b_{m-1} s + b_m] U(s).

Let’s call those polynomials $P(s)$ and $Q(s)$ , respectively.

Then, the transfer function is

G(s) = \frac{Y(s)}{U(s)} = \frac{Q(s)}{P(s)}.

$P(s)$ is the characteristic function and $P(s) = 0$ is the characteristic equation.

The order of the system is determined by the highest power of $s$ in the characteristic equation.

Poles of a transfer function are values of $s$ for which $P(s) = 0$ . In other words, they are roots (solutions) of the characteristic equation. There are always $n$ poles.

Zeroes of a transfer function are values of $s$ for which $Q(s) = 0$ . In other words, they are roots (solutions) of the numerator polynomial. There are always $m$ zeroes.

A transfer function is proper if $m \leq n$ .

Example 2 (Mass-spring-damper)

Consider a simple mass-spring-damper system.

$f(t)$ is the force input. $q(t)$ is the displacement of the mass (relative to its equilibrium position when $f = 0$ ).

The parameters $m, c, k$ have units of $\mathrm{kg}$ , $\mathrm{N\,s/m}$ and $\mathrm{N/m}$ , respectively.

Firstly, Newton’s second law gives us

m \frac{d^2 q}{dt^2} + c \frac{dq}{dt} + kq = f(t).

Taking the Laplace transform of both sides, we get

m s^2 Q(s) + c s Q(s) + k Q(s) = F(s).

Then, the transfer function is

G(s) = \frac{Q(s)}{F(s)} = \frac{1}{m s^2 + c s + k}.

Example 3 (Series RLC circuit)

Consider a simple electrical circuit, with a voltage source, an inductor $L$ , a resistor $R$ and a capacitor $C$ in series, in a closed loop.

$v(t)$ is the voltage input. $q(t)$ is the charge displacement (the rate of change of $q(t)$ is the instantaneous current, $i(t)$ ).

The parameters $L, R, C$ have units of $\mathrm{H}$ , $\Omega$ and $\mathrm{F}$ , respectively.

Firstly, Kirchhoff’s voltage law gives us

L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = v(t).

Taking the Laplace transform of both sides, we get

L s^2 Q(s) + R s Q(s) + \frac{1}{C} Q(s) = V(s).

Then, the transfer function is

G(s) = \frac{Q(s)}{V(s)} = \frac{1}{L s^2 + R s + \frac{1}{C}}.

Combining Transfer Functions

Suppose you have one system which takes in input $u_1(t)$ and produces output $y_1(t)$ .

Suppose also that some other system takes in input $u_2(t)$ and produces output $y_2(t)$ .

The transfer functions for the two systems are,

G_1(s) = \frac{Y_1(s)}{U_1(s)}, \quad G_2(s) = \frac{Y_2(s)}{U_2(s)}.

These two systems might be joined by connecting the output of the first into the input of the second so that,

u_2(t) = y_1(t).

Then, the combined system has transfer function

G(s) = G_1(s) G_2(s).

Proof:

\begin{align*} Y_1(s) & = G_1(s) U_1(s), \newline U_2(s) & = Y_1(s), \newline Y_2(s) & = G_2(s) U_2(s) = G_2(s) Y_1(s) = G_1(s) G_2(s) U_1(s). \end{align*}

If, instead of being connected in series, the two systems are connected in parallel so that they share the same input signal, but their outputs add, then,

u_1(t) = u_2(t) = u(t), \quad y(t) = y_1(t) + y_2(t).

Then, the combined system has transfer function

G(s) = G_1(s) + G_2(s).

Proof:

\begin{align*} Y_1(s) & = G_1(s) U(s), \newline Y_2(s) & = G_2(s) U(s), \newline Y(s) & = Y_1(s) + Y_2(s) = G_1(s) U(s) + G_2(s) U(s) = (G_1(s) + G_2(s)) U(s). \end{align*}

Summary

Laplace transform is a powerful tool to convert differential equations into algebraic equations.
Transfer functions are the relationship between the Laplace transform of the input and output signals of a system.
Transfer functions can be combined in series or parallel to form more complex systems.