SYSTEM

Introduction: System Modeling

The first step in the control design process is to develop appropriate mathematical models of the system derived either from physical

laws or experimental data. In this section, we introduce the state-space and transfer function representations of dynamic systems. We

then review some basic approaches to modeling mechanical and electrical systems and show how to enter these models into MATLAB

for further analysis.

Key MATLAB commands used in this tutorial are: ss , tf

Contents

Dynamic Systems

State-Space Representation

Transfer Function Representation

Mechanical Systems

Example: Mass-Spring-Damper System

Entering State-Space Models into MATLAB

Entering Transfer Function Models into MATLAB

Electrical Systems

Example: RLC Circuit

System Identification

System Conversions

Dynamic Systems

Dynamic systems are systems that change or evolve in time according to a fixed rule. For many physical systems, this rule can be

stated as a set of first-order differential equations:

(1)

In the above equation, is the state vector, a set of variables representing the configuration of the system at time . For instance in a

simple mechanical mass-spring-damper system, the two state variables could be the position and velocity of the mass. is the vector

of control inputs at time , representing the externally applied "forces" on the system, and is a possibly nonlinear function giving the

time derivative (rate of change) of the state vector, for a particular state, input, and time.

The state at any future time, , may be determined exactly given knowledge of the initial state, , and the time history of the

inputs, , between and by integrating Eq.(1). Though the state variables themselves are not unique, there is a minimum number of

state variables, , required in a given system for the above to hold true. is referred to as the system order and determines the

dimensionality of the state-space. The system order usually corresponds to the number of independent energy storage elements in the

system.

The relationship given in Eq.(1) is very general and can be used to describe a wide variety of different systems; unfortunately, it may

be very difficult to analyze. There are two common simplifications which make the problem more tractable. First, if the function, ,

does not depend explicitly on time, i.e. , then the system is said to be time invariant. This is often a very reasonable

assumption, since the underlying physical laws themselves do not typically depend on time. For time invariant systems, the parameters

or coefficients of the function, , are constant. The control input, however, may still be time dependent, .

The second common assumption concerns the linearity of the system. In reality, nearly every physical system is nonlinear. In other

words, is typically some complicated function of the state and inputs. These nonlinearities arise in many different ways, one of the

most common in control systems being "saturation" in which an element of the system reaches a hard physical limit to its operation.

Fortunately, over a sufficiently small operating range (think tangent line near a curve), the dynamics of most systems are

approximately linear, that is .

Until the advent of digital computers (and to a large extent thereafter), it was only practical to analyze linear time invariant (LTI)

systems. Consequently, most of the results of control theory are based on these assumptions. Fortunately, as we shall see, these results

have proven to be remarkably effective and many significant engineering challenges have been solved using LTI techniques. In fact,

the true power of feedback control systems are that they work (are robust) in the presence of the unavoidable modeling uncertainty.

State-Space Representation

For continuous linear time invariant (LTI) systems, the standard state-space representation is given below:

(2)

(3)

where is the vector of state variables (nx1), is the time derivative of state vector (nx1), is the input or control vector (px1), is the

output vector (qx1), is the system matrix (nxn), is the input matrix (nxp), is the output matrix (qxn), is the feedforward matrix

(qxp).

The output equation, Eq.(3), is necessary because often there are state variables which are not directly observed or are otherwise not of

interest. The output matrix, , is used to specify which state variables (or combinations thereof) are available for use by the controller.

Also often there is no direct feedforward in which case is the zero matrix.

The state-space representation, also referred to as the time-domain representation, can easily handle multi-input/multi-output

(MIMO) systems, systems with non-zero initial conditions, and nonlinear systems via Eq.(1). Consequently, the state-space

representation is used extensively in "modern" control theory.

Transfer Function Representation

LTI systems have the extremely important property that if the input to the system is sinusoidal, then the output will also be sinusoidal

at the same frequency but in general with different magnitude and phase. These magnitude and phase differences as a function of

frequency are known as the frequency response of the system.

Using the Laplace transform, it is possible to convert a system's time-domain representation into a frequency-domain output/input

representation, known as the transfer function. In so doing, it also transforms the governing differential equation into an algebraic

equation which is often easier to analyze.

The Laplace transform of a time domain function, , is defined below:

(4)

where the parameter is a complex frequency variable. It is very rare in practice that you will have to directly evaluate a

Laplace transform (though you should certainly know how). It is much more common to look up the transform of the function you are

interested in in a table such as the one found here: Laplace Transform Table

The Laplace transform of the nth derivative of a function is particularly important:

(5)

Frequency-domain methods are most often used for analyzing LTI single-input/single-output (SISO) systems, e.g. those governed by

a constant coefficient differential equation as follows:

(6)

The Laplace transform of this equation is given below:

(7)

where and are the Laplace Transforms of and respectively. Note that when finding transfer functions, we always

assume that the each of the initial conditions, , , , etc. is zero. The transfer function from input to output is

therefore:

(8)

It is useful to factor the numerator and denominator of the transfer function into the so called zero-pole-gain form:

(9)

The zeros of the transfer function, , are the roots of the numerator polynomial, i.e. the values of s such that . The poles

of the transfer function, , are the roots of the denominator polynomial, i.e. the values of s such that . Both the zeros and

poles may be complex valued (have both real and imaginary parts). The system Gain is .

Note that we can also determine the transfer function directly form the state-space representation as follows:

(10)

Mechanical Systems

Newton's laws of motion form the basis for analyzing mechanical systems. Newtons second law, Eq. (11), states that the sum of the forces acting on a body equals its mass times acceleration. Newton's third law, for our purposes, states that if two bodies are

connected, then they experience the same magnitude force acting in opposite directions.

(11)

When applying this equation, it is best to construct a free body diagram (FBD) of the sysetm showing all applied forces.

Example: Mass-Spring-Damper System

The free body diagram for this system is shown below. The spring force is proportional to the displacement of the mass, , and the

viscous damping force is proportional to the velocity of the mass, . Both forces oppose the motion of the mass and are therefore

shown in the negative -direction. Note also, that corresponds to the position of the mass when the spring is unstretched.

Now we proceed by summing the forces and applying Newtons second law, Eq. (11), in each direction of the problem. In this case, there are no forces acting in the -direction; however, in the -direction we have:

(12)

This equation, known as the governing equation, completely characterizes the dynamic state of the system. Later, we will see how to

use this to calculate the response of the system to any external input, , as well as analyze system properties such as stability and

performance.

To determine the state-space representation of the mass-spring-damper system, we must reduce the second order governing equation to

a set of two first order differential equations. To this end, we choose the position and velocity as our state variables.

(13)

Note also that these state variables correspond to the potential energy in the spring and the kinetic energy of the mass respectively.

Often when choosing state variables it is helpful to consider the independent energy storage elements in the system.

The state equation in this case is as follows:

(14)

If, for instance, we are interested in controlling the position of the mass, then the output equation is as follows:

(15)

Entering State-Space Models into MATLAB

Now we will show you how to enter the equations derived above into a m-file for MATLAB. Let's assign numerical values to each of

the variables.

m mass 1.0 kg

k spring constant 1.0 N/m

b damping constant 0.2 Ns/m

F input force 1.0 N

Create a new m-file and enter the following commands.

m = 1;

k = 1;

b = 0.2;

F = 1;

A = [0 1; -k/m -b/m];

B = [0 1/m]';

C = [1 0];

D = [0];

sys = ss(A,B,C,D)

sys =

a =

x1 x2

x1 0 1

x2 -1 -0.2

b =

u1

x1 0

x2 1

c =

x1 x2

y1 1 0

d =

u1

y1 0

Continuous-time state-space model.

The Laplace transform for this system assuming zero initial conditions is

(16)

and therefore the transfer function from force input to displacement output is

(17)

Entering Transfer Function Models into MATLAB

Now we will show how to enter the transfer function derived above into MATLAB. Enter the following commands into the m-file in

which you defined the system parameters.

s = tf('s');

sys = 1/(m*s^2+b*s+k)

sys =

1

---------------

s^2 + 0.2 s + 1

Continuous-time transfer function.

Note that we have used the symbolic s variable here to define our transfer function model. We recommend using this method most of

the time; however, in some circumstances, for instance in older versions of MATLAB or when interfacing with SIMULINK, you may

need to define the transfer function model using the numerator and denominator polynomial coefficients directly. In these cases, use

the following commands:

num = [1];

den = [m b k];

sys = tf(num,den)

sys =

1

---------------

s^2 + 0.2 s + 1

Continuous-time transfer function.

Electrical Systems

Like Newtons laws in mechanical systems, Kirchoffs circuit laws are the basic analytical tool in electrical systems. Kirchoffs current law (KCL) states that the sum of the electrical currents entering and exiting a node in a circuit must be equal. Kirchoffs voltage law (KVL) states that the sum of voltage differences around any closed loop in the circuit is zero. When applying KVL, the

source voltages are typically taken as positive and the load voltages taken as negative.

Example: RLC Circuit

We will now consider a simple series combination of three passive electrical elements: a resistor, an inductor, and a capacitor, known

as an RLC Circuit.

Since this circuit is a single loop, each node only has one input and output; therefore, application of KCL simply shows that the current

is the same throughout the circuit at any given time, . Now applying KVL around the loop and using the sign conventions indicated

in the diagram, we arrive at the following governing equation.

(18)

We note that that the governing equation for the RLC circuit has an analogous form to the mass-spring-damper mechanical system. In

particular, they are both second order systems where the charge (integral of current) corresponds to displacement, the inductance to

mass, the resistance to viscous damping, and the inverse capacitance to the spring stiffness. These analogies and others like them turn

out to be quite useful conceptually in understanding the behavior of dynamical systems.

The state-space representation is found by choosing the charge and current as the state variables.

(19)

where,

(20)

The state equation is therefore:

(21)

We choose the current as ouput as follows:

(22)

The transfer function representation may be found by taking the Laplace transform as we did for the mass-spring-damper or from the

state-space equation as follows:

(23)

(24)

The RLC state-space and transfer fcuntion models can be entered into MATLAB using the same procedure as discussed for the mass-

spring-damper system above.

System Identification

In this section, we have seen how to model systems using basic physical principles; however, often this is not possible either because

the parameters of the system are uncertain, or the underlying processes are simply not known. In these cases, we must rely on

experimental measurements and statistical techniques to develop a system model, a process known as system identification.

System identification may be performed using either time-domain or frequency-domain data, see the Introduction: System

Identification page.

Also refer to MATLABs System Identification Toolbox for more information on this subject.

Introduction: System Analysis

Once appropriate mathematical models of a system have been obtained, either in state-space or transfer function form, we may then

analyze these models to predict how the system will respond in both the time and frequency domains. To put this in context, control

systems are often designed to improve stability, speed of response, steady-state error, or prevent oscillations. In this section, we will

show how to determine these dynamic properties from the system models.

Key MATLAB commands used in this tutorial are: tf , ssdata , pole , eig , step , pzmap , bode , ltiview

Contents

Time Response Overview

Frequency Response Overview

Stability

System Order

First Order Systems

Second Order Systems

Time Response Overview

The time response represents how the state of a dynamic system changes in time when subjected to a particular input. Since the

models we have derived consist of differential equations, some integration must be performed in order to determine the time response

of the system. For some simple systems, a closed-form analytical solution may be available. However, for most systems, especially

nonlinear systems or those subject to complicated input forces, this integration must be carried out numerically. Fortunately,

MATLAB provides many useful resources for calculating time responses for many types of inputs, as we shall see in the following

sections.

The time response of a linear dynamic system consists of the sum of the transient response which depends on the initial conditions

and the steady-state response which depends on the system input. These correspond to the free (homogeneous or zero input) and the

forced (inhomogeneous or non-zero input) solutions of the governing differential equations respectively.

Frequency Response Overview

All the examples presented in this tutorial are modeled by linear constant coefficient differential equations and are thus linear time-

invariant (LTI). LTI systems have the extremely important property that if the input to the system is sinusoidal, then the steady-state

output will also be sinusoidal at the same frequency but in general with different magnitude and phase. These magnitude and phase

differences as a function of frequency comprise the frequency response of the system.

The frequency response of a system can be found from the transfer function in the following way: create a vector of frequencies

(varying between zero or "DC" to infinity) and compute the value of the plant transfer function at those frequencies. If is the open-

loop transfer function of a system and is the frequency vector, we then plot versus . Since is a complex number, we can

plot both its magnitude and phase (the Bode Plot) or its position in the complex plane (the Nyquist Diagram). Both methods display

the same information in different ways.

Stability

For our purposes, we will use the Bounded Input Bounded Output (BIBO) definition of stability which states that a system is stable

if the output remains bounded for all bounded (finite) inputs. Practically, this means that the system will not blow up while in operation.

The transfer function representation is especially useful when analyzing system stability. If all poles of the transfer function (values of

s at which the denominator equals zero) have negative real parts, then the system is stable. If any pole has a positive real part, then the

system is unstable. If we view the poles on the complex s-plane, then all poles must be in the left half plane (LHP) to ensure stability.

If any pair of poles is on the imaginary axis, then the system is marginally stable and the system will oscillate. The poles of a LTI

system model can easily be found in MATLAB using the pole command, an example if which is shown below:

s = tf('s');

G = 1/(s^2+2*s+5)

pole(G)

G =

1

-------------

s^2 + 2 s + 5

Continuous-time transfer function.

ans =

-1.0000 + 2.0000i

-1.0000 - 2.0000i

Thus this system is stable since the real parts of the poles are both negative. The stability of a system may also be found from the state-

space representation. In fact, the poles of the transfer function are the eigenvalues of the system matrix A. We can use the eig

command to calculate the eigenvalues using either the LTI system model directly, eig(G) or the system matrix as shown below.

[A,B,C,D] = ssdata(G);

eig(A)

ans =

-1.0000 + 2.0000i

-1.0000 - 2.0000i

System Order

The order of a dynamic system is the order of the highest derivative of its governing differential equation. Equivalently, it is the

highest power of s in the denominator of its transfer function. The important properties of first, second, and higher order systems will

be reviewed in this section.

First Order Systems

First order systems are the simplest dynamic systems to analyze. Some common examples include cruise control systems and RC

circuits.

The general form of the first order differential equation is as follows

(1)

The first order transfer function is

(2)

DC Gain

The DC gain, , is the ratio of the magnitude of the steady-state step response to the magnitude of the step input. From the Final

Value Theorem, for stable transfer functions the DC gain is the value of the transfer function when s=0. For first order systems equal

to .

Time Constant

The time constant is the time it takes for the system to reach 63% of the steady-state value for a step response or to

decrease to 37% of the inital value for an impulse response. More generally, it represents the time scale for which the dynamics of the

system are significant.

Poles/Zeros

There is a single real pole at . Therefore, the system is stable if is positive and unstable if is negative. There are no zeros.

Step Response

We can calculate the system time response to a step input of magnitude u using the MATLAB following commands:

k_dc = 5;

Tc = 10;

u = 2;

s = tf('s');

G = k_dc/(Tc*s+1)

step(u*G)

G =

5

--------

10 s + 1

Continuous-time transfer function.

Note: MATLAB also provides a powerful GUI (LTI Viewer) for analyzing LTI systems which can be accessed using,

ltiview('step',G).

If you right click on the step response graph and select Characteristics, you can choose to have several system metrics overlaid on the

response: peak response, settling time, rise time, and steady-state.

Settling Time

The settling time, , is the time required for the system ouput to fall within a certain percentage (i.e. 2%) of the steady state value for a

step input or equivalently to decrease to a certain percentage of the initial value for an impulse input. The settling times for first order

system for the most common tolerances are provided in the table below. Note that the tighter the tolerance, the longer the system

response takes to settle to within this tolerance, as expected.

10% 5% 2% 1%

Ts=2.3/a=2.3Tc Ts=3/a=3Tc Ts=3.9/a=3.9Tc Ts=4.6/a=4.6Tc

Rise Time

The rise time, , is the time required for the system output to rise from some lower level x% to some higher level y% of the final

steady-state value. For first order systems, the typical range is 10% - 90%.

Bode Plots

The Bode Plots show the magnitude and phase of the system frequency response, . We can generate the Bode plots in MATLAB

using the bode(G) command.

bode(G)

Again the same results could be obtained using the LTI viewer GUI, ltiview('bode',G)

The Bode plots use a logarithmic frequency scale, so that a larger range of frequencies are visible. Also, the magnitude is represented

using the logarithmic decibel unit (dB) defined as:

(3)

Like frequency, the decibel scale allows us to view a much larger range of magnitudes on a single plot. Also, as we shall see in

subsequent tutorials, when systems are combined or controllers are added, transfer functions are often multiplied together. Using the

dB scale, we may simply add the magnitudes of the transfer functions. Note, we may also add the phase angles though these are not

shown on a log scale.

The low frequency magnitude of the first order bode plot is . The magnitude plot has a bend at the frequency equal to the

absolute value of the pole (ie. ), and then decreases at 20dB for every factor of ten increase in frequency (-20dB/decade). The

phase plot is asymptotic to 0 degrees at low frequency, and asymptotic to -90 degrees at high frequency. Between frequency 0.1a and

10a, the phase changes by approximately -45 degrees for every factor of ten increase in frequency (-45 degrees/decade).

We will see in the Frequency Methods for Controller Design Section how to use Bode Plots to calculate closed loop stability and

performance of feedback systems.

Second Order Systems

Second order systems are commonly encountered in practice, and are the simplest type of dynamic system to exhibit oscillations. In

fact many real higher order systems are modeled as second order to facilitate analysis. Examples include mass-spring-damper systems

and RLC circuits.

The general form of the first order differential equation is as follows

(4)

The first order transfer function is

(5)

DC Gain

The DC gain, , again is the ratio of the magnitude of the steady-state step response to the magnitude of the step input, and for stable

systems it is the value of the transfer function when . For second order systems,

(6)

Damping Ratio

The damping ratio is a dimensionless quantity charaterizing the energy losses in the system due to such effects as viscous friction or

electrical resistance. From the above definitions,

(7)

Natural Frequency

The natural frequency is the frequency (in rad/s) that the system will oscillate at when there is no damping, .

(8)

Poles/Zeros

The second order transfer function has two poles at:

(9)

Under Damped System

If , then the system is under damped. Both poles are complex valued with negative real parts; therefore the system is stable but

oscillates while approaching the steady-state value.

k_dc = 1;

w_n = 10;

zeta = 0.2;

s = tf('s');

G1 = k_dc*w_n^2/(s^2 + 2*zeta*w_n*s + w_n^2);

pzmap(G1)

axis([-3 1 -15 15])

step(G1)

axis([0 3 0 2])

Settling Time

The settling time, , is the time required for the system ouput to fall within a certain percentage of the steady state value for a step

input or equivalently to decrease to a certain percentage of the initial value for an impulse input. For a second order, underdamped

system, the settling time can be approximated by the following equation:

(10)

The settling times for the most common tolerances are presented in the following table:

10% 5% 2% 1%

Ts=2.3/(zeta*wn) Ts=3/(zeta*w_n) Ts=3.9/(zeta*w_n) Ts=4.6/(zeta*w_n)

Percent Overshoot

The percent overshoot is the percent by which a system exceeds its final steady-state value. For a second order under damped system,

the percent overshoot is diretly related to the damping ratio by the following equation:

(11)

For second order under damped systems, the 2% settling time, , rise time, , and percent overshoot, %OS, are related to the damping

and natural frequency as shown below.

(12)

(13)

(14)

Over Damped Systems

If , then the system is over damped. Both poles are real and negative; therefore the system is stable and does not oscillate. The

step response and a pole-zero map of an over damped system are calculated below:

zeta = 1.2;

G2 = k_dc*w_n^2/(s^2 + 2*zeta*w_n*s + w_n^2);

pzmap(G2)

axis([-20 1 -1 1])

step(G2)

axis([0 1.5 0 1.5])

Critically Damped Systems

If , then the system is critically damped. Both poles are real and have the same magnitude, . Critically damped

systems approach steady-state quickest without oscillating. Now change the value of the damping to 1, and replot the step response and

pole-zero map.

zeta = 1;

G3 = k_dc*w_n^2/(s^2 + 2*zeta*w_n*s + w_n^2);

pzmap(G3)

axis([-11 1 -1 1])

step(G3)

axis([0 1.5 0 1.5])

Undamped Systems

If , then the system is undamped. In this case, the poles are purely imaginary; therefore the system is marginally stable and

oscillates indefinitely.

zeta = 0;

G4 = k_dc*w_n^2/(s^2 + 2*zeta*w_n*s + w_n^2);

pzmap(G4)

axis([-1 1 -15 15])

step(G4)

axis([0 5 -0.5 2.5])

Bode Plot

We show the Bode Magnitude and Phase Plots for all damping conditions of a second order system below:

bode(G1,G2,G3,G4)

legend('under damped: zeta < 1','over damped: zeta > 1','critically damped: zeta = 1','undamped: zeta = 0')

The magnitude of the bode plot of a second order system drops off at -40dB per decade, while the relative phase changes from 0 to -

180 degrees at -90 degrees per decade. For the under damped systems, we also see a resonance peak near the natural frequency, = 10

rad/s. The sharpness of the peak depends on the damping in the system, and is charaterized by the quality factor, or Q-Factor, defined

below. The Q-factor is an important property in signal processing.

(15)

CONTROL

Introduction: PID Controller Design

In this tutorial we will introduce a simple yet versatile feedback compensator structure, the Proportional-Integral-Derivative (PID)

controller. We will discuss the effect of each of the pid parameters on the closed-loop dynamics and demonstrate how to use a PID

controller to improve the system performance.

Key MATLAB commands used in this tutorial are: tf , step , pid , feedback , pidtool , pidtune

Contents

PID Overview The Characteristics of P, I, and D Controllers Example Problem Open-Loop Step Response

Proportional Control Proportional-Derivative Control Proportional-Integral Control Proportional-Integral-Derivative Control General Tips for Designing a PID Controller Automatic PID Tuning

PID Overview

In this tutorial, we will consider the following unity feedback system:

The output of a PID controller, equal to the control input to the plant, in the time-domain is as follows:

(1)

First, let's take a look at how the PID controller works in a closed-loop system using the schematic shown above. The variable ( )

represents the tracking error, the difference between the desired input value ( ) and the actual output ( ). This error signal ( ) will be

sent to the PID controller, and the controller computes both the derivative and the integral of this error signal. The control signal ( ) to

the plant is equal to the proportional gain ( ) times the magnitude of the error plus the integral gain ( ) times the integral of the error

plus the derivative gain ( ) times the derivative of the error.

This control signal ( ) is sent to the plant, and the new output ( ) is obtained. The new output ( ) is then fed back and compared to the

reference to find the new error signal ( ). The controller takes this new error signal and computes its derivative and its integral again,

ad infinitum.

The transfer function of a PID controller is found by taking the Laplace transform of Eq.(1).

(2)

= Proportional gain = Integral gain = Derivative gain

We can define a PID controller in MATLAB using the transfer function directly, for example:

Kp = 1;

Ki = 1;

Kd = 1;

s = tf('s');

C = Kp + Ki/s + Kd*s

C =

s^2 + s + 1

-----------

s

Continuous-time transfer function.

Alternatively, we may use MATLAB's pid controller object to generate an equivalent continuous-time controller as follows:

C = pid(Kp,Ki,Kd)

C =

1

Kp + Ki * --- + Kd * s

s

with Kp = 1, Ki = 1, Kd = 1

Continuous-time PID controller in parallel form.

Let's convert the pid object to a transfer function to see that it yields the same result as above:

tf(C)

ans =

s^2 + s + 1

-----------

s

Continuous-time transfer function.

The Characteristics of P, I, and D Controllers

A proportional controller ( ) will have the effect of reducing the rise time and will reduce but never eliminate the steady-state error.

An integral control ( ) will have the effect of eliminating the steady-state error for a constant or step input, but it may make the

transient response slower. A derivative control ( ) will have the effect of increasing the stability of the system, reducing the

overshoot, and improving the transient response.

The effects of each of controller parameters, , , and on a closed-loop system are summarized in the table below.

CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S-S ERROR

Kp Decrease Increase Small Change Decrease

Ki Decrease Increase Increase Eliminate

Kd Small Change Decrease Decrease No Change

Note that these correlations may not be exactly accurate, because , , and are dependent on each other. In fact, changing one of

these variables can change the effect of the other two. For this reason, the table should only be used as a reference when you are

determining the values for , and .

Example Problem

Suppose we have a simple mass, spring, and damper problem.

The modeling equation of this system is

(3)

Taking the Laplace transform of the modeling equation, we get

(4)

The transfer function between the displacement and the input then becomes

(5)

Let

M = 1 kg

b = 10 N s/m

k = 20 N/m

F = 1 N

Plug these values into the above transfer function

(6)

The goal of this problem is to show you how each of , and contributes to obtain

Fast rise time

Minimum overshoot

No steady-state error

Open-Loop Step Response

Let's first view the open-loop step response. Create a new m-file and run the following code:

s = tf('s');

P = 1/(s^2 + 10*s + 20);

step(P)

The DC gain of the plant transfer function is 1/20, so 0.05 is the final value of the output to an unit step input. This corresponds to the

steady-state error of 0.95, quite large indeed. Furthermore, the rise time is about one second, and the settling time is about 1.5 seconds.

Let's design a controller that will reduce the rise time, reduce the settling time, and eliminate the steady-state error.

Proportional Control

From the table shown above, we see that the proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces

the steady-state error.

The closed-loop transfer function of the above system with a proportional controller is:

(7)

Let the proportional gain ( ) equal 300 and change the m-file to the following:

Kp = 300;

C = pid(Kp)

T = feedback(C*P,1)

t = 0:0.01:2;

step(T,t)

C =

Kp = 300

P-only controller.

T =

300

----------------

s^2 + 10 s + 320

Continuous-time transfer function.

The above plot shows that the proportional controller reduced both the rise time and the steady-state error, increased the overshoot, and

decreased the settling time by small amount.

Proportional-Derivative Control

Now, let's take a look at a PD control. From the table shown above, we see that the derivative controller (Kd) reduces both the

overshoot and the settling time. The closed-loop transfer function of the given system with a PD controller is:

(8)

Let equal 300 as before and let equal 10. Enter the following commands into an m-file and run it in the MATLAB command

window.

Kp = 300;

Kd = 10;

C = pid(Kp,0,Kd)

T = feedback(C*P,1)

t = 0:0.01:2;

step(T,t)

C =

Kp + Kd * s

with Kp = 300, Kd = 10

Continuous-time PD controller in parallel form.

T =

10 s + 300

----------------

s^2 + 20 s + 320

Continuous-time transfer function.

This plot shows that the derivative controller reduced both the overshoot and the settling time, and had a small effect on the rise time

and the steady-state error.

Proportional-Integral Control

Before going into a PID control, let's take a look at a PI control. From the table, we see that an integral controller (Ki) decreases the

rise time, increases both the overshoot and the settling time, and eliminates the steady-state error. For the given system, the closed-loop

transfer function with a PI control is:

(9)

Let's reduce the to 30, and let equal 70. Create an new m-file and enter the following commands.

Kp = 30;

Ki = 70;

C = pid(Kp,Ki)

T = feedback(C*P,1)

t = 0:0.01:2;

step(T,t)

C =

1

Kp + Ki * ---

s

with Kp = 30, Ki = 70

Continuous-time PI controller in parallel form.

T =

30 s + 70

------------------------

s^3 + 10 s^2 + 50 s + 70

Continuous-time transfer function.

Run this m-file in the MATLAB command window, and you should get the following plot. We have reduced the proportional gain

(Kp) because the integral controller also reduces the rise time and increases the overshoot as the proportional controller does (double

effect). The above response shows that the integral controller eliminated the steady-state error.

Proportional-Integral-Derivative Control

Now, let's take a look at a PID controller. The closed-loop transfer function of the given system with a PID controller is:

(10)

After several trial and error runs, the gains = 350, = 300, and = 50 provided the desired response. To confirm, enter the

following commands to an m-file and run it in the command window. You should get the following step response.

Kp = 350;

Ki = 300;

Kd = 50;

C = pid(Kp,Ki,Kd)

T = feedback(C*P,1);

t = 0:0.01:2;

step(T,t)

C =

1

Kp + Ki * --- + Kd * s

s

with Kp = 350, Ki = 300, Kd = 50

Continuous-time PID controller in parallel form.

Now, we have obtained a closed-loop system with no overshoot, fast rise time, and no steady-state error.

General Tips for Designing a PID Controller

When you are designing a PID controller for a given system, follow the steps shown below to obtain a desired response.

1. Obtain an open-loop response and determine what needs to be improved

2. Add a proportional control to improve the rise time 3. Add a derivative control to improve the overshoot 4. Add an integral control to eliminate the steady-state error 5. Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. You can always refer to the table shown in this "PID Tutorial"

page to find out which controller controls what characteristics.

Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single

system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to

implement a derivative controller on the system. Keep the controller as simple as possible.

Automatic PID Tuning

MATLAB provides tools for automatically choosing optimal PID gains which makes the trial and error process described above

unnecessary. You can access the tuning algorithm directly using pidtune or through a nice graphical user interface (GUI) using

pidtool.

The MATLAB automated tuning algorithm chooses PID gains to balance performance (response time, bandwidth) and robustness

(stability margins). By default the algorthm designs for a 60 degree phase margin.

Let's explore these automated tools by first generating a proportional controller for the mass-spring-damper system by entering the

following commands:

pidtool(P,'p')

The pidtool GUI window, like that shown below, should appear.

Notice that the step response shown is slower than the proportional controller we designed by hand. Now click on the Show

Parameters button on the top right. As expected the proportional gain constant, Kp, is lower than the one we used, Kp = 94.85 < 300.

We can now interactively tune the controller parameters and immediately see the resulting response int he GUI window. Try dragging

the resposne time slider to the right to 0.14s, as shown in the figure below. The response does indeeed speed up, and we can see Kp is

now closer to the manual value. We can also see all the other performance and robustness parameters for the system. Note that the

phase margin is 60 degrees, the default for pidtool and generally a good balance of robustness and performance.

Now let's try designing a PID controller for our system. By specifying the previously designed or (baseline) controller, C, as the

second parameter, pidtool will design another PID controller (instead of P or PI) and will compare the response of the system with the

automated controller with that of the baseline.

pidtool(P,C)

We see in the output window that the automated controller responds slower and exhibits more overshoot than the baseline. Now

choose the Design Mode: Extended option at the top, which reveals more tuning parameters.

Now type in Bandwidth: 32 rad/s and Phase Margin: 90 deg to generate a controller similar in performance to the baseline. Keep in

mind that a higher bandwidth (0 dB crossover of the open-loop) results in a faster rise time, and a higher phase margin reduces the

overshoot and improves the system stability.

Finally we note that we can generate the same controller using the command line tool pidtune instead of the pidtool GUI

opts = pidtuneOptions('CrossoverFrequency',32,'PhaseMargin',90);

[C, info] = pidtune(P, 'pid', opts)

C =

1

Kp + Ki * --- + Kd * s

s

with Kp = 320, Ki = 169, Kd = 31.5

Continuous-time PID controller in parallel form.

info =

Stable: 1

CrossoverFrequency: 32

PhaseMargin: 90

Introduction: Root Locus Controller Design

In this tutorial we will introduce the root locus, show how to create it using MATlAB, and demonstrate how to design feedback

controllers using the root locus that satisfy certain performance criteria.

Key MATLAB commands used in this tutorial are: feedback , rlocus , step , sisotool

Contents

Closed-Loop Poles Plotting the Root Locus of a Transfer Function Choosing a Value of K from the Root Locus Closed-Loop Response Using SISOTOOL for Root Locus Design

Closed-Loop Poles

The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with

proportional gain K and unity feedback.

The closed-loop transfer function is:

(1)

and thus the poles of the closed-loop poles of the closed-loop system are values of such that .

If we write , then this equation has the form:

(2)

(3)

Let = order of and = order of (the order of a polynomial is the highest power of that appears in it).

We will consider all positive values of K. In the limit as , the poles of the closed-loop system are or the poles of . In

the limit as , the poles of the closed-loop system are or the zeros of .

No matter what we pick K to be, the closed-loop system must always have poles, where is the number of poles of . The root

locus must have branches, each branch starts at a pole of and goes to a zero of . If has more poles than zeros (as is

often the case), and we say that has zeros at infinity. In this case, the limit of as is zero. The number of zeros at

infinity is , the number of poles minus the number of zeros, and is the number of branches of the root locus that go to infinity

(asymptotes).

Since the root locus is actually the locations of all possible closed-loop poles, from the root locus we can select a gain such that our

closed-loop system will perform the way we want. If any of the selected poles are on the right half plane, the closed-loop system will

be unstable. The poles that are closest to the imaginary axis have the greatest influence on the closed-loop response, so even though the

system has three or four poles, it may still act like a second or even first order system depending on the location(s) of the dominant

pole(s).

Plotting the Root Locus of a Transfer Function

Consider an open-loop system which has a transfer function of

(4)

How do we design a feedback controller for the system by using the root locus method? Say our design criteria are 5% overshoot and 1

second rise time. Make a MATLAB file called rl.m. Enter the transfer function, and the command to plot the root locus:

s = tf('s');

sys = (s + 7)/(s*(s + 5)*(s + 15)*(s + 20));

rlocus(sys)

axis([-22 3 -15 15])

Choosing a Value of K from the Root Locus

The plot above shows all possible closed-loop pole locations for a pure proportional controller. Obviously not all of those closed-loop

poles will satisfy our design criteria, To determine what part of the locus is acceptable, we can use the command sgrid(Zeta,Wn) to

plot lines of constant damping ratio and natural frequency. Its two arguments are the damping ratio ( ) and natural frequency ( )

[these may be vectors if you want to look at a range of acceptable values]. In our problem, we need an overshoot less than 5% (which

means a damping ratio of greater than 0.7) and a rise time of 1 second (which means a natural frequency greater than 1.8). Enter the

following in the MATLAB command window:

Zeta = 0.7;

Wn = 1.8;

sgrid(Zeta,Wn)

On the plot above, the two dotted lines at about a 45 degree angle indicate pole locations with = 0.7; in between these lines, the poles

will have > 0.7 and outside of the lines < 0.7. The semicircle indicates pole locations with a natural frequency = 1.8; inside the

circle, < 1.8 and outside the circle > 1.8.

Going back to our problem, to make the overshoot less than 5%, the poles have to be in between the two white dotted lines, and to

make the rise time shorter than 1 second, the poles have to be outside of the white dotted semicircle. So now we know only the part of

the locus outside of the semicircle and in betwen the two lines are acceptable. All the poles in this location are in the left-half plane, so

the closed-loop system will be stable.

From the plot above we see that there is part of the root locus inside the desired region. So in this case, we need only a proportional

controller to move the poles to the desired region. You can use the rlocfind command in MATLAB to choose the desired poles on

the locus:

[k,poles] = rlocfind(sys)

Click on the plot the point where you want the closed-loop pole to be. You may want to select the points indicated in the plot below to

satisfy the design criteria.

Note that since the root locus may have more than one branch, when you select a pole, you may want to find out where the other pole

(poles) are. Remember they will affect the response too. From the plot above, we see that all the poles selected (all the "+" signs) are at

reasonable positions. We can go ahead and use the chosen K as our proportional controller.

Closed-Loop Response

In order to find the step response, you need to know the closed-loop transfer function. You could compute this using the rules of block

diagrams, or let MATLAB do it for you (there is no need to enter a value for K if the rlocfind command was used):

K = 350;

sys_cl = feedback(K*sys,1)

sys_cl =

350 s + 2450

--------------------------------------

s^4 + 40 s^3 + 475 s^2 + 1850 s + 2450

Continuous-time transfer function.

The two arguments to the function feedback are the numerator and denominator of the open-loop system. You need to include the

proportional gain that you have chosen. Unity feedback is assumed.

If you have a non-unity feedback situation, look at the help file for the MATLAB function feedback, which can find the closed-loop

transfer function with a gain in the feedback loop.

Check out the step response of your closed-loop system:

step(sys_cl)

As we expected, this response has an overshoot less than 5% and a rise time less than 1 second.

Using SISOTOOL for Root Locus Design

Another way to complete what was done above is to use the interactive MATLAB GUI called sisotool. Using the same model as

above, first define the plant, .

s = tf('s');

plant = (s + 7)/(s*(s + 5)*(s + 15)*(s + 20));

The sisotool function can be used for analysis and design. In this case, we will focus on using the Root Locus as the design method

to improve the step response of the plant. To begin, type the following into the MATLAB command window:

sisotool(plant)

The following window should appear. To start, select the tab labeled Graphical Tuning. Within this window, turn off Plot 2 and make

sure Plot 1 is the Root Locus and verify that Open Loop 1 is selected. Finally, click the button labeled Show Design Plot to bring up

the tunable Root Locus plot.

In the same fashion, select the tab labeled Analysis Plots. Within this window, for Plot 1, select Step. In the Contents of Plots

subwindow, select Closed Loop r to y for Plot 1. If the window does not automatically pop up, click the button labeled Show

Analysis Plot.

The next thing to do is to add the design requirements to the Root Locus plot. This is done directly on the plot by right-clicking and

selecting Design Requirements, New. Design requirements can be set for the Settling Time, the Percent Overshoot, the Damping

Ratio, the Natural Frequency, or a Region Constraint. There is no direct requirement for Rise Time, but the natural frequency can be

used for this.

Here, we will set the design requirements for the damping ratio and the natural frequency just like was done with sgrid. Recall that

the requirements call for = 0.7 and = 1.8. Set these within the design requirements. On the plot, any area which is still white, is an

acceptable region for the poles.

Zoom into the Root Locus by right-clicking on the axis and select Properties, then click the label Limits. Change the real axis to -25

to 5 and the imaginary to -2.5 to 2.5.

Also, we can see the current values of some key parameters in the response. In the Step response, right-click on the plot and go to

Characteristics and select Peak Response. Do the same for the Rise Time. There should now be two large dots on the screen

indicating the location of these parameters. Click each of these dots to bring up a screen with information.

Both plots should appear as shown here:

As the characteristics show on the Step response, the overshoot is acceptable, but the rise time is incredibly off.

To fix this, we need to choose a new value for the gain K. Similarly to the rlocfind command, the gain of the controller can be

changed directly on the root locus plot. Click and drag the pink box on the origin to the acceptable area where the poles have an

imaginary component as shown below.

At the bottom of the plot, it can be seen that the loop gain has been changed to 361. Looking at the Step response, both of the values

are acceptable for our requirements.

Published with MATLAB 7.14

Introduction: Frequency Domain Methods for Controller Design

The frequency response method of controller design may be less intuitive than other methods you have studied previously. However, it

has certain advantages, especially in real-life situations such as modeling transfer functions from physical data. In this tutorial, we will

see how we can use the open-loop frequency response of a system to predict its behavior in closed-loop.

Key MATLAB commands used in this tutorial are: bode , nyquist , margin , lsim , step , feedback , sisotool

Contents

Gain and Phase Margin Nyquist Diagram The Cauchy Criterion Closed-Loop Performance from Bode Plots Closed-Loop Stability from the Nyquist Diagram

Gain and Phase Margin

Consider the following unity feedback system:

where is a variable (constant) gain and is the plant under consideration. The gain margin is defined as the change in open-loop

gain required to make the system unstable. Systems with greater gain margins can withstand greater changes in system parameters

before becoming unstable in closed-loop.

The phase margin is defined as the change in open-loop phase shift required to make a closed-loop system unstable.

The phase margin also measures the system's tolerance to time delay. If there is a time delay greater than in the loop (where

is the frequency where the phase shift is 180 deg), the system will become unstable in closed-loop. The time delay, can be thought

of as an extra block in the forward path of the block diagram that adds phase to the system but has no effect on the gain. That is, a time

delay can be represented as a block with magnitude of 1 and phase (in radians/second).

For now, we won't worry about where all this comes from and will concentrate on identifying the gain and phase margins on a Bode

plot.

The phase margin is the difference in phase between the phase curve and -180 degrees at the point corresponding to the frequency that

gives us a gain of 0 dB (the gain crossover frequency, ). Likewise, the gain margin is the difference between the magnitude curve

and 0 dB at the point corresponding to the frequency that gives us a phase of -180 degrees (the phase crossover frequency, ).

One nice thing about the phase margin is that you don't need to replot the Bode in order to find the new phase margin when changing

the gains. If you recall, adding gain only shifts the magnitude plot up. This is equivalent to changing the y-axis on the magnitude plot.

Finding the phase margin is simply a matter of finding the new cross-over frequency and reading off the phase margin. For example,

suppose you entered the command bode(sys). You will get the following bode plot:

s = tf('s');

sys = 50/(s^3 + 9*s^2 + 30*s +40);

bode(sys)

grid on

title('Bode Plot with No Gain')

You should see that the phase margin is about 100 degrees. Now suppose you added a gain of 100, by entering the command

bode(100*sys). You should get the following plot:

bode(100*sys)

grid on

title('Bode Plot with Gain = 100')

As you can see the phase plot is exactly the same as before, and the magnitude plot is shifted up by 40 dB (gain of 100). The phase

margin is now about -60 degrees. This same result could be achieved if the y-axis of the magnitude plot was shifted down 40 dB. Try

this, look at the first Bode plot, find where the curve crosses the -40 dB line, and read off the phase margin. It should be about 90

degrees, the same as the second Bode plot.

We can have MATLAB calculate and display the gain and phase margins using the margin(sys) command. This command returns

the gain and phase margins, the gain and phase cross over frequencies, and a graphical representation of these on the Bode plot. Let's

check it out:

margin(100*sys)

Bandwidth Frequency

The bandwidth frequency is defined as the frequency at which the closed-loop magnitude response is equal to -3 dB. However, when

we design via frequency response, we are interested in predicting the closed-loop behavior from the open-loop response. Therefore, we

will use a second-order system approximation and say that the bandwidth frequency equals the frequency at which the open-loop

magnitude response is between -6 and -7.5 dB, assuming the open-loop phase response is between -135 deg and -225 deg. For a

complete derivation of this approximation, consult your textbook.

In order to illustrate the importance of the bandwidth frequency, we will show how the output changes with different input frequencies.

We will find that sinusoidal inputs with frequency less than Wbw (the bandwidth frequency) are tracked "reasonably well" by the

system. Sinusoidal inputs with frequency greater than Wbw are attenuated (in magnitude) by a factor of 0.707 or greater (and are also

shifted in phase).

Let's say we have the following closed-loop transfer function representing a system:

(1)

sys = 1/(s^2 + 0.5*s + 1);

bode(sys)

Since this is the closed-loop transfer function, our bandwidth frequency will be the frequency corresponding to a gain of -3 dB.

Looking at the plot, we find that it is approximately 1.4 rad/s. We can also read off the plot that for an input frequency of 0.3 radians,

the output sinusoid should have a magnitude about one and the phase should be shifted by perhaps a few degrees (behind the input).

For an input frequency of 3 rad/sec, the output magnitude should be about -20 dB (or 1/10 as large as the input) and the phase should

be nearly -180 (almost exactly out-of-phase). We can use the lsim command to simulate the response of the system to sinusoidal

inputs.

First, consider a sinusoidal input with a frequency lower than Wbw. We must also keep in mind that we want to view the steady state

response. Therefore, we will modify the axes in order to see the steady state response clearly (ignoring the transient response).

sys = 1/(s^2 + 0.5*s + 1);

w = 0.3;

t = 0:0.1:100;

u = sin(w*t);

[y,t] = lsim(sys,u,t);

plot(t,y,t,u)

axis([50 100 -2 2])

Note that the output (blue) tracks the input (green) fairly well; it is perhaps a few degrees behind the input as expected. However, if we

set the frequency of the input higher than the bandwidth frequency for the system, we get a very distorted response (with respect to

the input):

sys = 1/(s^2 + 0.5*s + 1);

w = 3;

t = 0:0.1:100;

u = sin(w*t);

[y,t] = lsim(sys,u,t);

plot(t,y,t,u)

axis([90 100 -1 1])

Again, note that the magnitude is about 1/10 that of the input, as predicted, and that it is almost exactly out of phase (180 degrees

behind) the input. Feel free to experiment and view the response for several different frequencies , and see if they match the Bode

plot.

Nyquist Diagram

The Nyquist plot allows us to predict the stability and performance of a closed-loop system by observing its open-loop behavior. The

Nyquist criterion can be used for design purposes regardless of open-loop stability (remember that the Bode design methods assume

that the system is stable in open-loop). Therefore, we use this criterion to determine closed-loop stability when the Bode plots display

confusing information.

Note: The MATLAB nyquist command does not provide an adequate representation for systems that have open-loop poles in

the jw-axis. Therefore, we suggest that you copy the nyquist1.m file as a new m-file. This m-file creates more accurate Nyquist

plots, since it correctly deals with poles and zeros on the jw-axis.

The Nyquist diagram is basically a plot of where is the open-loop transfer function and is a vector of frequencies which

encloses the entire right-half plane. In drawing the Nyquist diagram, both positive and negative frequencies (from zero to infinity) are

taken into account. We will represent positive frequencies in red and negative frequencies in green. The frequency vector used in

plotting the Nyquist diagram usually looks like this (if you can imagine the plot stretching out to infinity):

However, if we have open-loop poles or zeros on the jw axis, will not be defined at those points, and we must loop around them

when we are plotting the contour. Such a contour would look as follows:

Please note that the contour loops around the pole on the jw axis. As we mentioned before, the MATLAB nyquist command does not

take poles or zeros on the jw axis into account and therefore produces an incorrect plot. To correct this, please download and use

nyquist1.m. If we have a pole on the jw axis, we have to use nyquist1. If there are no poles or zeros on the jw-axis, or if we have

pole-zero cancellation, we can use either the nyquist command or nyquist1.m.

The Cauchy Criterion

The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex plane, and mapping it through a

complex function , the number of times that the plot of encircles the origin is equal to the number of zeros of enclosed by

the frequency contour minus the number of poles of enclosed by the frequency contour. Encirclements of the origin are counted as

positive if they are in the same direction as the original closed contour or negative if they are in the opposite direction.

When studying feedback controls, we are not as interested in as in the closed-loop transfer function:

(2)

If encircles the origin, then will enclose the point -1. Since we are interested in the closed-loop stability, we want to know

if there are any closed-loop poles (zeros of ) in the right-half plane. More details on how to determine this will come later.

Therefore, the behavior of the Nyquist diagram around the -1 point in the real axis is very important; however, the axis on the standard

nyquist diagram might make it hard to see what's happening around this point. To correct this, you can add the lnyquist.m function to

your files. The lnyquist.m command plots the Nyquist diagram using a logarithmic scale and preserves the characteristics of the -1

point.

To view a simple Nyquist plot using MATLAB, we will define the following transfer function and view the Nyquist plot:

(3)

s = tf('s');

sys = 0.5/(s - 0.5);

nyquist(sys)

axis([-1 0 -1 1])

Now we will look at the Nyquist diagram for the following transfer function:

(4)

Note that this function has a pole at the origin. We will see the difference between using the nyquist, nyquist1, and lnyquist

commands with this particular function.

sys = (s + 2)/(s^2);

nyquist(sys)

nyquist1(sys)

lnyquist(sys)

Note that the nyquist plot is not the correct one, the nyquist1 plot is correct, but it's hard to see what happens close to the -1 point,

and the lnyquist plot is correct and has an appropriate scale.

Closed-Loop Performance from Bode Plots

In order to predict closed-loop performance from open-loop frequency response, we need to have several concepts clear:

The system must be stable in open-loop if we are going to design via Bode plots.

If the gain crossover frequency is less than the phase crossover frequency (i.e. ), then the closed-loop system will be stable. For second-order systems, the closed-loop damping ratio is approximately equal to the phase margin divided by 100 if the phase margin

is between 0 and 60 degrees. We can use this concept with caution if the phase margin is greater than 60 degrees. For second-order systems, a relationship between damping ratio, bandwidth frequency, and settling time is given by an equation

described on the Extras: Bandwidth page. A very rough estimate that you can use is that the bandwidth is approximately equal to the natural frequency.

Let's use these concepts to design a controller for the following system:

Where is the controller and is:

(5)

The design must meet the following specifications:

Zero steady state error. Maximum overshoot must be less than 40%. Settling time must be less than 2 seconds.

There are two ways of solving this problem: one is graphical and the other is numerical. Within MATLAB, the graphical approach is

best, so that is the approach we will use. First, let's look at the Bode plot. Create a m-file with the following code:

sys = 10/(1.25*s + 1);

bode(sys)

There are several characteristics of the system that can be read directly from this Bode plot. First of all, we can see that the bandwidth

frequency is around 10 rad/sec. Since the bandwidth frequency is roughly the same as the natural frequency (for a first order system of

this type), the rise time is 1.8/BW = 1.8/10 = 1.8 seconds. This is a rough estimate, so we will say the rise time is about 2 seconds.

The phase margin for this system is approximately 95 degrees. The relation damping ratio = PM/100 only holds for PM < 60. Since

the system is first-order, there should be no overshoot.

The last major point of interest is steady-state error. The steady-state error can be read directly off the Bode plot as well. The constant

( , , or ) is found from the intersection of the low frequency asymptote with the w = 1 line. Just extend the low frequency line

to the w = 1 line. The magnitude at this point is the constant. Since the Bode plot of this system is a horizontal line at low frequencies

(slope = 0), we know this system is of type zero. Therefore, the intersection is easy to find. The gain is 20 dB (magnitude 10). What

this means is that the constant for the error function is 10. The steady-state error is 1/(1+Kp) = 1/(1+10) = 0.091.

If our system was type one instead of type zero, the constant for the steady-state error would be found in a manner similar to the

following.

Let's check our predictions by looking at a step response plot. This can be done by adding the following two lines of code into the

MATLAB command window.

sys_cl = feedback(sys,1);

step(sys_cl)

title('Closed-Loop Step Response, No Controller')

As you can see, our predictions were very good. The system has a rise time of about 2 seconds, has no overshoot, and has a steady-

state error of about 9%. Now we need to choose a controller that will allow us to meet the design criteria. We choose a PI controller

because it will yield zero steady-state error for a step input. Also, the PI controller has a zero, which we can place. This gives us

additional design flexibility to help us meet our criteria. Recall that a PI controller is given by:

(6)

The first thing we need to find is the damping ratio corresponding to a percent overshoot of 40%. Plugging in this value into the

equation relating overshoot and damping ratio (or consulting a plot of this relation), we find that the damping ratio corresponding to

this overshoot is approximately 0.28. Therefore, our phase margin should be at least 30 degrees. We must have a bandwidth frequency

greater than or equal to 12 if we want our settling time to be less than 1.75 seconds which meets the design specs.

Now that we know our desired phase margin and bandwidth frequency, we can start our design. Remember that we are looking at the

open-loop Bode plots. Therefore, our bandwidth frequency will be the frequency corresponding to a gain of approximately -7 dB.

Let's see how the integrator portion of the PI or affects our response. Change your m-file to look like the following (this adds an

integral term but no proportional term):

plant = 10/(1.25*s + 1);

contr = 1/s;

bode(contr*plant, logspace(0,2))

Our phase margin and bandwidth frequency are too small. We will add gain and phase with a zero. Let's place the zero at 1 for now

and see what happens. Change your m-file to look like the following:

plant = 10/(1.25*s + 1);

contr = (s + 1)/s;

bode(contr*plant, logspace(0,2))

It turns out that the zero at 1 with a unit gain gives us a satisfactory answer. Our phase margin is greater than 60 degrees (even less

overshoot than expected) and our bandwidth frequency is approximately 11 rad/s, which will give us a satisfactory response. Although

satisfactory, the response is not quite as good as we would like. Therefore, let's try to get a higher bandwidth frequency without

changing the phase margin too much. Let's try to increase the gain to 5 and see what happens. This will make the gain shift and the

phase will remain the same.

plant = 10/(1.25*s + 1);

contr = 5 * ((s + 1)/s);

bode(contr*plant, logspace(0,2))

That looks really good. Let's look at our step response and verify our results. Add the following two lines to your m-file:

sys_cl = feedback(contr*plant,1);

step(sys_cl)

As you can see, our response is better than we had hoped for. However, we are not always quite as lucky and usually have to play

around with the gain and the position of the poles and/or zeros in order to achieve our design requirements.

Closed-Loop Stability from the Nyquist Diagram

Consider the negative feedback system:

Remember from the Cauchy criterion that the number N of times that the plot of G(s)H(s) encircles -1 is equal to the number Z of

zeros of 1 + G(s)H(s) enclosed by the frequency contour minus the number P of poles of 1 + G(s)H(s) enclosed by the frequency

contour (N = Z - P). Keeping careful track of open- and closed-loop transfer functions, as well as numerators and denominators, you

should convince yourself that:

The zeros of 1 + G(s)H(s) are the poles of the closed-loop transfer function. The poles of 1 + G(s)H(s) are the poles of the open-loop transfer function.

The Nyquist criterion then states that:

P = the number of open-loop (unstable) poles of G(s)H(s). N = the number of times the Nyquist diagram encircles -1. clockwise encirclements of -1 count as positive encirclements. counter-clockwise encirclements of -1 count as negative encirclements. Z = the number of right-half-plane (positive, real) poles of the closed-loop system.

The important equation whih relates these three quantities is:

(7)

Note: This is only one convention for the Nyquist criterion. Another convention states that a positive N counts the counter-

clockwise or anti-clockwise encirclements of -1. The P and Z variables remain the same. In this case the equation becomes Z =

P - N. Throughout these tutorials, we will use a positive sign for clockwise encirclements.

It is very important (and somewhat tricky) to learn how to count the number of times that the diagram encircles -1. Therefore, we will

go into some detail to help you visualize this. You can view this movie as an example.

Another way of looking at it is to imagine you are standing on top of the -1 point and are following the diagram from beginning to end.

Now ask yourself: How many times did I turn my head a full 360 degrees? Again, if the motion was clockwise, N is positive, and if the

motion is anti-clockwise, N is negative.

Knowing the number of right-half plane (unstable) poles in open loop (P), and the number of encirclements of -1 made by the Nyquist

diagram (N), we can determine the closed-loop stability of the system. If Z = P + N is a positive, nonzero number, the closed-loop

system is unstable.

We can also use the Nyquist diagram to find the range of gains for a closed-loop unity feedback system to be stable. The system we

will test looks like this:

where G(s) is:

(8)

This system has a gain K which can be varied in order to modify the response of the closed-loop system. However, we will see that we

can only vary this gain within certain limits, since we have to make sure that our closed-loop system will be stable. This is what we

will be looking for: the range of gains that will make this system stable in the closed-loop.

The first thing we need to do is find the number of positive real poles in our open-loop transfer function:

roots([1 -8 15])

ans =

5

3

The poles of the open-loop transfer function are both positive. Therefore, we need two anti-clockwise (N = -2) encirclements of the

Nyquist diagram in order to have a stable closed-loop system (Z = P + N). If the number of encirclements is less than two or the

encirclements are not anti-clockwise, our system will be unstable.

Let's look at our Nyquist diagram for a gain of 1:

sys = (s^2 + 10*s + 24)/(s^2 - 8*s + 15);

nyquist(sys)

There are two anti-clockwise encirclements of -1. Therefore, the system is stable for a gain of 1. Now we will see how the system

behaves if we increase the gain to 20:

nyquist(20*sys)

The diagram expanded. Therefore, we know that the system will be stable no matter how much we increase the gain. However, if we

decrease the gain, the diagram will contract and the system might become unstable. Let's see what happens for a gain of 0.5:

nyquist(0.5*sys)

The system is now unstable. By trial and error we find that this system will become unstable for gains less than 0.80. We can verify

our answers by zooming in on the Nyquist plots as well as by looking at the closed-loop steps responses for gains of 0.79, 0.80, and

0.81.

Gain Margin

We already defined the gain margin as the change in open-loop gain expressed in decibels (dB), required at 180 degrees of phase shift

to make the system unstable. Now we are going to find out where this comes from. First of all, let's say that we have a system that is

stable if there are no Nyquist encirclements of -1, such as:

(9)

Looking at the roots, we find that we have no open loop poles in the right half plane and therefore no closed-loop poles in the right-

half-plane if there are no Nyquist encirclements of -1. Now, how much can we vary the gain before this system becomes unstable in

closed-loop? Let's look at the following figure:

The open-loop system represented by this plot will become unstable in closed loop if the gain is increased past a certain boundary. The

negative real axis area between -1/a (defined as the point where the 180 degree phase shift occurs...that is, where the diagram crosses

the real axis) and -1 represents the amount of increase in gain that can be tolerated before closed-loop instability.

If we think about it, we realize that if the gain is equal to a, the diagram will touch the -1 point:

(10)

or

(11)

Therefore, we say that the gain margin is a units. However, we mentioned before that the gain margin is usually measured in decibels.

Hence, the gain margin is:

(12)

We will now find the gain margin of the stable, open-loop transfer function we viewed before. Recall that the function is:

(13)

and that the Nyquist diagram can be viewed by typing:

sys = 50/(s^3 + 9*s^2 + 30*s + 20);

nyquist(sys)

As we discussed before, all that we need to do to find the gain margin is find a, as defined in the preceding figure. To do this, we need

to find the point where there is exactly 180 degrees of phase shift. This means that the transfer function at this point is real (has no

imaginary part). The numerator is already real, so we just need to look at the denominator. When s = jw, the only terms in the

denominator that will have imaginary parts are those which are odd powers of s. Therefore, for G(jw) to be real, we must have:

(14)

which means w = 0 (this is the rightmost point in the Nyquist diagram) or w = sqrt(30). We can then find the value of G(jw) at this

point using polyval:

w = sqrt(30);

polyval(50,j*w)/polyval([1 9 30 40],j*w)

ans =

-0.2174

The answer is: -0.2174 + 0i. The imaginary part is zero, so we know that our answer is correct. We can also verify by looking at the

Nyquist plot again. The real part also makes sense. Now we can proceed to find the gain margin.

We found that the 180 degrees phase shift occurs at -0.2174 + 0i. This point was previously defined as -1/a. Therefore, we now

have a, which is the gain margin. However, we need to express the gain margin in decibels:

(15)

(16)

(17)

We now have our gain margin. Let's see how accurate it is by using a gain of a = 4.6 and zooming in on the Nyquist plot:

a = 4.6;

nyquist(a*sys)

The plot appears to go right through the -1 point. We will now verify the accuracy of our results by viewing the zoomed Nyquist

diagrams and step responses for gains of 4.5, 4.6, and 4.7.

Phase Margin

We have already discussed the importance of the phase margin. Therefore, we will only talk about where this concept comes from. We

have defined the phase margin as the change in open-loop phase shift required at unity gain to make a closed-loop system unstable.

Let's look at the following graphical definition of this concept to get a better idea of what we are talking about.

Let's analyze the previous plot and think about what is happening. From our previous example we know that this particular system will

be unstable in closed-loop if the Nyquist diagram encircles the -1 point. However, we must also realize that if the diagram is shifted by

theta degrees, it will then touch the -1 point at the negative real axis, making the system marginally stable in closed-loop. Therefore,

the angle required to make this system marginally stable in closed-loop is called the phase margin (measured in degrees). In order to

find the point we measure this angle from, we draw a circle with radius of 1, find the point in the Nyquist diagram with a magnitude of

1 (gain of zero dB), and measure the phase shift needed for this point to be at an angle of 180 degrees.

Published with MATLAB 7.14

Introduction: State-Space Methods for Controller Design

In this section, we will show how to design controllers and observers using state-space (or time-domain) methods.

Key MATLAB commands used in this tutorial are: eig , ss , lsim , place , acker

Contents

Modeling Stability Controllability and Observability Control Design Using Pole Placement Introducing the Reference Input Observer Design

Modeling

There are several different ways to describe a system of linear differential equations. The state-space representation was introduced

in the Introduction: System Modeling section. For a SISO LTI system, the state-space form is given below:

(1)

(2)

where x is a n by 1 vector representing the state (commonly position and velocity variable in mechanical systems), u is a scalar

representing the input (commonly a force or torque in mechanical systems), and y is a scalar representing the output. The matrices A (n

by n), B (n by 1), and C (1 by n) determine the relationships between the state and input and output variables. Note that there are n

first-order differential equations. State space representation can also be used for systems with multiple inputs and outputs (MIMO), but

we will only use single-input, single-output (SISO) systems in these tutorials.

To introduce the state space design method, we will use the magnetically suspended ball as an example. The current through the coils

induces a magnetic force which can balance the force of gravity and cause the ball (which is made of a magnetic material) to be

suspended in midair. The modeling of this system has been established in many control text books (including Automatic Control

Systems by B. C. Kuo, the seventh edition).

The equations for the system are given by:

(3)

(4)

where h is the vertical position of the ball, i is the current through the electromagnet, V is the applied voltage, M is the mass of the ball,

g is gravity, L is the inductance, R is the resistance, and K is a coefficient that determines the magnetic force exerted on the ball. For

simplicity, we will choose values M = 0.05 Kg, K = 0.0001, L = 0.01 H, R = 1 Ohm, g = 9.81 m/sec^2. The system is at

equilibrium (the ball is suspended in midair) whenever h = K i^2/Mg (at which point dh/dt = 0). We linearize the equations about

the point h = 0.01 m (where the nominal current is about 7 amp) and get the state space equations:

(5)

(6)

where:

(7)

is the set of state variables for the system (a 3x1 vector), u is the input voltage (delta V), and y (the output), is delta h. Enter the system

matricies into a m-file.

A = [ 0 1 0

980 0 -2.8

0 0 -100 ];

B = [ 0

0

100 ];

C = [ 1 0 0 ];

Stability

One of the first things we want to do is analyze whether the open-loop system (without any control) is stable. As discussed in the

Introduction: System Analysis section, the eigenvalues of the system matrix, A, (equivalent to the poles of the transfer fucntion)

determine the stability. The eigenvalues of the A matrix are the values of s where det(sI - A) = 0.

poles = eig(A)

poles =

31.3050

-31.3050

-100.0000

One of the poles is in the right-half plane, (i.e. has positive real part which means that the system is unstable in open-loop.

To check out what happens to this unstable system when there is a nonzero initial condition, add the following lines to your m-file

and it again:

t = 0:0.01:2;

u = zeros(size(t));

x0 = [0.01 0 0];

sys = ss(A,B,C,0);

[y,t,x] = lsim(sys,u,t,x0);

plot(t,y)

title('Open-Loop Response to Non-Zero Initial Condition')

xlabel('Time (sec)')

ylabel('Ball Position (m)')

It looks like the distance between the ball and the electromagnet will go to infinity, but probably the ball hits the table or the floor first

(and also probably goes out of the range where our linearization is valid).

Controllability and Observability

A system is controllable if there exists a control input, u(t), that transfers any state of the system to zero in finite time. It can be shown

that an LTI system is controllable if and only if its controllabilty matrix, CO, has full rank (i.e. if rank(CO) = n where n is the number

of states ). The rank of the controllability matrix of an LTI model can be determined in MATLAB using the commands

rank(ctrb(A,B)) or rank(ctrb(sys)).

(8)

All the state variables of a system may not be directly measurable, for instance if the component is in an inaccessible location. In these

cases it is neccesary to estimate the values of the unknown internal state variables using only the available system outputs. A system is

observable if the initial state, x(t_0), can be determined from the system output, y(t), over some finite time t_0 < t < t_f. For LTI

systems, the system is observable if and only if the observability matrix, OB, has full rank (i.e. if rank(OB) = n where n is the number

of states). The observability of an LTI model can be determined in MATLAB using the command rank(obsv(A,C)) or

rank(obsv(sys)).

(9)

Controllability and observability are dual concepts. A system (A,B) is controllable if and only if a system (A',C,B',D) is observable.

This fact will be useful when designing an observer, as we shall see below.

Control Design Using Pole Placement

Let's build a controller for this system using pole placement. The schematic of a full-state feedback system is shown below. By full-

state, we mean that all state variables are known to the controller at all times. For instance in this system, we would need a sensor

measuring the ball position, another measuring velocity, and a third measuring current in the electro-magnet.

For simplicity, let's assume the reference is zero, R=0. The input is then

(10)

The state-space equations for the closed-loop feedback system are therefore

(11)

(12)

The stability and time domain performance of the closed-loop feedback system are determined primarily by the location of the poles

(eigenvalues) of the matrix (A-BK). Since the matrices A and B*K are both 3 by 3 matrices, there will be 3 poles for the system. By

choosing an appropriate K matrix we can place these closed-loop poles anywhere we want. We can use the MATLAB function place

to find the control matrix, K, which will give the desired poles.

Before attempting this method, we have to decide where we want the closed-loop poles to be. Suppose the criteria for the controller

were settling time < 0.5 sec and overshoot < 5%, then we might try to place the two dominant poles at -10 +/- 10i (at zeta = 0.7 or 45

degrees with sigma = 10 > 4.6*2). The third pole we might place at -50 to start, and we can change it later depending on what the

closed-loop behavior is. Remove the lsim command from your m-file and everything after it, then add the following lines to your m-

file:

p1 = -10 + 10i;

p2 = -10 - 10i;

p3 = -50;

K = place(A,B,[p1 p2 p3]);

sys_cl = ss(A-B*K,B,C,0);

lsim(sys_cl,u,t,x0);

xlabel('Time (sec)')

ylabel('Ball Position (m)')

The overshoot is too large (there are also zeros in the transfer function which can increase the overshoot; you do not see the zeros in

the state-space formulation). Try placing the poles further to the left to see if the transient response improves (this should also make the

response faster).

p1 = -20 + 20i;

p2 = -20 - 20i;

p3 = -100;

K = place(A,B,[p1 p2 p3]);

sys_cl = ss(A-B*K,B,C,0);

lsim(sys_cl,u,t,x0);

xlabel('Time (sec)')

ylabel('Ball Position (m)')

This time the overshoot is smaller. Consult your textbook for further suggestions on choosing the desired closed-loop poles.

Compare the control effort required (K) in both cases. In general, the farther you move the poles, the more control effort it takes.

Note: If you want to place two or more poles at the same position, place will not work. You can use a function called acker which

works similarly to place:

K = acker(A,B,[p1