Part 2 - Transfer functions

Introduction

In this part we’ll define what a transfer function is for a system, and why they are important.

Let’s firstly cover feedback again.

Feedback system

As we discussed last time we’ll encounter:

$v(t)$ - Interference
$u(t)$ - Control signal/Input signal
$y(t)$ - Response signal/Output signal
$r(t)$ - Reference value
$e(t)$ - Control error

Let’s take a real world example

Example 1

Say that we have a system which controls temperature. We have an initial temperature for 19 $^\circ$ .

At the time instance, $t_0$ , we change $r(t)$ (via a knob) to be at 21 $^\circ$ .

From a time period, $t_1$ to $t_2$ , we open a window which cools down the room.

From this real world example we can describe the system as:

$t < t_0$ , the system is stable at 19 $^\circ$

$t = t_0$ , $r(t) = 21^\circ$ . We will get a control error, $e(t) > 0$ . The control device steps in and increases the temperature, which will stabilize at 21 $^\circ$ .

$t = t_1$ , The window is opened, which means we have $v(t) < 0$ , or in other words, the temperature decreases. This means again that $e(t) > 0$ . The control device needs to readjust to this change.

$t = t_2$ , The window is closed, which will yield a control error, the control device will readjust to this change as well.

$t > t_2$ , The system will remain stable at 21 $^\circ$ again.

Controllers

But how do we actually control a control error? There are many different types of controllers, but the most common are the PID variants.

P-controller $u(t) = K_p e(t).$
I-controller $u(t) = K_i \int_0^t e(\tau) \, d\tau.$
D-controller $u(t) = K_d \dfrac{de(t)}{dt}.$

We also have the combinations:

PI-controller $u(t) = K_p e(t) + K_i \int_0^t e(\tau) \, d\tau.$
PD-controller $u(t) = K_p e(t) + K_d \dfrac{de(t)}{dt}.$
PID-controller $u(t) = K_p e(t) + K_i \int_0^t e(\tau) \, d\tau + K_d \dfrac{de(t)}{dt}.$

Laplace transform

We’ve already covered the definition and the formality of the Laplace transform so let’s quickly recover it.

Recall (Laplace transform)

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

Where $s = \mu + j\omega$

Transfer function

Let’s define the transfer function in words first.

Definition 1 (Transfer function)

The transfer function for a system is equal to the Laplace transformation of the output signal divided by the Laplace transformation of the input signal.

In other words:

G(s) = \dfrac{Y(s)}{U(s)}, \quad Y(s) = G(s) U(s).

Properties

We denote the Laplace transform as $\mathcal{L}$ .

$\mathcal{L} \{\dot{y}\} = sY(s) - y(0)$
$\mathcal{L} \{\ddot{y}\} = s^2Y(s) - sy(0) - \dot{y}(0)$
$\mathcal{L} \{\int_0^t y(\tau) \, d\tau \} = \dfrac{1}{s} Y(s)$

Theorem 1 (Initial value theorem)

y(0) = \lim_{s \to \infty} s \cdot Y(s).

Theorem 2 (Final value theorem)

\lim_{t \to \infty} y(t) = \lim_{s \to 0} s \cdot Y(s).

Given that, $\lim_{t \to \infty} y(t)$ exists.

Example 2

Say that we have a car that moves, we have $u(t)$ which is the driving force forward. $y(t)$ is the speed of the car, and of course we have friction/air resistance which we’ll denote $v(t)$ .

Let’s find the transfer function of this system.

From Newton’s second law of motion we know that:

F = ma.

We know from high-school physics that:

a = \dfrac{dy(t)}{dt} = \dot{y}(t).

Now the tricky part, $v(t)$ , let’s define a simple model which is that the air resistance is proportional to the speed with some constant:

v(t) = b \cdot y(t).

Therefore:

\begin{align*} u(t) - b \cdot y(t) & = m \cdot \dot{y}(t) \newline m \cdot \dot{y}(t) + b \cdot y(t) & = u(t). \end{align*}

Let’s transform into Laplace (NB: All our initial conditions are zero!).

\begin{align*} ms \cdot Y(s) + b \cdot Y(s) & = U(s) \newline (ms + b)Y(s) & = U(s) \newline G(s) = \dfrac{Y(s)}{U(s)} & = \dfrac{1}{ms + b}. \end{align*}

Let’s introduce $T = \dfrac{m}{b}$ and $K = \dfrac{1}{b}$ . Therefore:

G(s) = \dfrac{K}{1 + sT}.

Unit step function

Definition 2 (Unit step function)

In this series we’ll denote the unit step function with $\sigma(t)$ .

The definition for the unit step is:

\sigma(t) = \begin{cases} 1 & t \geq 0 \newline 0 & t < 0. \end{cases}

Let $u(t) = \sigma(t)$ now.

$U(s) = \dfrac{1}{s}$

This means that:

Y(s) = G(s) U(s) = \dfrac{K}{s(1 + sT)}.

Using partial fraction decomposition:

Y(s) = \dfrac{K}{s} - \dfrac{KT}{1 + sT}.