Part I L6 - Poles, Zeros and System Response

Lecture Video

Part A

Part B

Next: Part I L7 - The General Second-Order System

Objectives

Introduce the concepts of Poles and Zeros of a Transfer Function
Review the response of a First-Order System
Introduce Second-Order Systems

In this lecture, we will start to classify systems based on their response characteristics. So far, we introduced the transfer function as a way to represent a system model, now we will introduce a different way to represent a system. We will introduce a graphical way to represent a system. We will also introduce standard performance specifications, that will help us in analyzing the performance of systems.

The Control System Design Process

Poles, Zeros and System Response

The output response of a system is the sum of the forced response and natural response

c_t(t) = c_{forced}(t)+c_{natural}(t)

Given the equation of motion for a system, we can mathematically obtain its output response But there are qualitative ways of studying the output response of the system. One technique is to look at the poles and zeros of a system and their relationship to the output response.

Poles $\times$ and zeros $o$ can be derived from a system’s transfer function.

Poles of a transfer function - definition

Poles (denoted by the symbol $\times$ ) of a transfer function $G(s)$ can either be:

The values of $s$ that cause the transfer function $G(s)$ to become infinite, or
Any roots of the denominator of the transfer function $G(s)$ that are common to the numerator’s.

Example: The two poles of $G(s)=\dfrac{1}{s(s+4)}$ , are $s=0,s=−4$

Example: The three poles of $G(s)=\dfrac{(s+3)}{(s^2+6s+18)(s+3)}$ , are $s=−3, s=−3±3i$

Note that mathematically $G(s)=\dfrac{(s+3)}{(s^2+6s+18)(s+3)}=\dfrac{1}{(s^2+6s+18)}$ where the latter form has only two poles; however, $s=−3$ is still is a pole of the original system, it is only that the effect of this pole is cancelled in this case.

Zeros of a transfer function - definition

Zeros (denoted by the symbol $o$ ) of a transfer function $G(s)$ can either be

The values of $s$ that cause the transfer function $G(s)$ to become zero, or
Any roots of the numerator of the transfer function 𝐺(𝑠) that are common with the denominator’s

Example: The zero of the transfer function $G(s)=\dfrac{(3s+1)}{s}$ , is $s=-\dfrac{1}{3}$

Example: The two zeros of the transfer function $G(s)=\dfrac{s(2s+1)}{(2s+1)}$ , are $s=-\dfrac{1}{2}, s= 0$

In later sections, we will learn why its important to keep in mind the pole-zero cancellation behavior.

Poles, Zeros and the S-Plane

In the Laplace domain, we defined $s=\sigma+\omega j$

The s-plane is where we plot the values of $s$ . Poles and Zeros are graphically placed on the s-plane. The figure shows a graphical representation of the transfer function:

G(s)=\dfrac{5s}{(s+5)}

Note that the gain 5 is not captured on the graph (we will deal with expressing the gain value under the Root-Locus section)

Pole Location and Time Response

With respect to the pole location on the s-plane:

A pole on the real negative axis produces an exponentially decaying response.
- $G(s)=\dfrac{1}{(s+3)}$
A pole pair on the imaginary axis produce a sinusoidal response.
- $G(s)=\dfrac{1}{s^2+18}$
A pole on the real positive axis produces an exponentially growing response. (unstable response)
- $G(s)=\dfrac{1}{(s-4)}$
A pole at the origin produces a step response.
- $G(s)=\dfrac{1}{s}$

Poles and Zeros and Response Function to a Step Input

Example 1

For the system with the transfer function $G(s)=\dfrac{C(s)}{R(s)}=20\dfrac{(s+4)(s+3)(s-8)}{(s+1)(s+2)^2}$ , write, by inspection, the output $c(t)$ in general terms, to a ramp input $R(s)=\dfrac{1}{s^2}$ and specify the natural and forced response parts.

First-Order Systems

A first-order system is a system whose highest derivative order is 1, or who’s characteristic equation is of degree 1.

Example: The system (internal system dynamics) with $G(s)=\dfrac{(s+2)}{(s+3)(s+3)}=\dfrac{1}{(s+3)}$ , is a not first order system per se, but the system response is a first-order response, due to the pole-zero cancellation.
Example: The system with $G(s)=10\dfrac{(s+3)}{(s+5)}$ is a first-order system with a zero

Examples of First Order Systems:

Heat Transfer (Thermometer)
Interest Rate Growth
RC Circuit

System Performance Specifications

In the context of control system design, there are well defined performance specifications that we evaluate, such as: Rise Time $T_r$ , Settling Time $T_s$ , Time Constant $\tau$ , Percentage Overshoot, Peak Time $T_p$ . These are common characteristics, but performance specifications are not limited to them in the design of real control systems. We will define some performance specifications for First-Order systems.

Performance specifications for first-order systems are defined for $G(s)=\dfrac{a}{(s+a)}$ , a first-order system with no zero.

Moreover, the specifications are defined for a response to a unit-step input.

$R(s)=\dfrac{1}{s}$
$C(s)=\dfrac{a}{s(s+a)}$
$c(t)=c_{forced}(t)+c_{natural}(t)=1-e^{-at}$

Time Constant

$\tau$ : $\tau=\dfrac{1}{a}$ , $c(t=\dfrac{1}{a})=1-e^{-1}=0.63$

Time constant is the time required for $e^{-at}$ to decay to $37\%$ of its initial value. The reciprocal of the time constant, is called the exponential frequency

Rise Time $T_r$

The time for the response to go from 0.1 to 0.9 of its final value

For the $1^{st}$ Order System $G(s)=\dfrac{a}{(s+a)}$ , $T_r=\dfrac{2.2}{a}$

Settling Time $T_s$

The time required for the response to reach and stay within $2\%$ of its final value.

For the 1st Order System $G(s)=\dfrac{a}{(s+a)}$ , $T_s=\dfrac{4}{a}$

Other times, the settling time is defined with a different than $2\%$ target, $5\%$ is common as well. In this course we will use the $2\%$ target when defining $T_s$

Remember that the above defined equations are strictly for a system with the transfer function of the form $G(s)=\dfrac{a}{(s+a)}$ , in response to a unit step input.

Example 2

The figure shows the response of a system measured experimentally, identify the type of system response and find the time constant, rise time and settling time of the system.

Second-Order Systems