Canete J.F. System Engineering and Automation: An Interactive Educational Approach

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128 4 System Response Analysis

As you can observe, when studying system stability, sometimes it is not needed

to know where the poles are exactly placed, but to know whether all the poles are

on the left part of the s-plane. Moreover, if you do not have a computer program

(like Matlab) to solve high-order equations, it can be difficult to decide about the

stability of a given system. For that, there are some simple rules, given by the

Routh-Hürwitz theorem, which yields the number of poles that are placed in the

“wrong” place, that is, on the right part of the s-plane. The steps to easily check

the stability of a system are:

Obtain the characteristic equation of the system.

(4.64)

Assuming that a

≠0, if there is one (or more) term a

<0 when another

term a

>0 then one of the roots of the equation, at least, are positive (the

system is unstable) or complex conjugated in the form s=0±j

ω (critical

stability). This condition is necessary but not sufficient.

If a characteristic equation passes step 2, there is a possibility of being

unstable yet. To completely check the stability of the system we proceed

as follows. We construct a table like the one shown in fig. 4.33 where the

first two rows are filled up with the coefficients of the characteristic

equation: the first row lists the odd coefficients starting from a

while the

second row lists the even coefficients starting from a

n-1

. The rest of cells

of this table are calculated as:

(4.65)

…

t(3,1) t(3,2) t(3,3) t(3,4) …

… … … … …

t(n+1,1) t(n+1,2) t(n+1,3) t(n+1,4) …

Fig. 4.33 Routh-Hürwitz table. First two rows are filled up with the coefficients of the

characteristic equation, while the rest are computed through the expression (4.65).

The Routh-Hürwitz theorem assures that the number of poles located at the

right part of the s-plane coincides with the changes of signs of the elements of the

first column of the table. Thus, a system is stable if and only such elements are all

positive or all negative.

...

as as a++ +

()()

( 1,1) * ( 2, 1) ( 1, 1) * ( 2,1)

(, )

(1,1)

ti ti j ti j ti

ti j

−−+−−+−

−

− 4n

− 6n

−

− 3n

− 5n

− 7n

−

4.5 System Stability 129

Example 4.15

Check the stability of a system given by:

(4.66)

Solution

The characteristic equation for the system (4.66) is

(4.67)

All the coefficients of (4.67) are positive so we cannot determine the stability of

the system yet. We proceed to construct the table (note that empty cells are

considered as zero):

1 3 5

2 4

t(3,1) t(3,2)

t(4,1)

t(5,1)

being

(4.68)

In this example, notice that the first column of the table (in grey) is 1,2,1,-6, and 5.

So there are two changes of sign (from 1 to -6 and from -6 to 5), and therefore the

system at hand has 2 poles at the right part of the s-plane: it is unstable.

As expected, solving the characteristic equation we have that:

>> roots([1 2 3 4 5])

ans =

0.2878 + 1.4161i

0.2878 - 1.4161i

-1.2878 + 0.8579i

-1.2878 - 0.8579i

432

10 1

()

2345

ssss

++++

432

23450ssss++++=

(2*3) (1*4)

(3,1) 1

−

(2*5) (1*0)

(3,2) 10

−

(1* 4) (2 * 5)

(4,1) 6

−

==−

(6*5) (1*0)

(5,1) 5

−−

−

130 4 System Response Analysis

Example 4.16

Sometimes, system stability depends on a particular parameter which has to be

tuned up to make the system stable. For example given the system:

Check its stability in function of the parameter k.

Solution

We can proceed in a similar way to calculate the range of values that the

parameter can take to assure the system stability. The characteristic equation is

From the second rule of the method we can restrict the value of k that makes all

the coefficient of the equation to have the same sign as:

As commented this is only a necessity but not a sufficient condition for stability

and we have to complete the process:

1 k+2

3k 4

t(3,1)

t(4,1)

Now we have to calculate the range of the value of k such as:

Certainly, when k=0.5275 the roots of the characteristic equation are:

()

3(2)4

sksks

++++

3(2)40sksks++++=

30 0

20 2

⇒ >

⎫

⇒

⎬

+>⇒ >−

⎭

(3 *( 2)) (1*4) 3 6 4

(3,1)

kk k k

+− +−

(3,1)* 4 (0 *3

(4,1) 4

(3,1)

−

30 0

0.5275

364

0 2.5 0.5275

⇒ >

⎫

⎪

⇒

⎬

+−

⇒ <− ∧ >

⎪

⎭

4.6

System Stationary R

>> roots([1 3*k k+2 +

ans =

-1.5825

0.0000 + 1.5898i

0.0000 - 1.5898i

and the system is at the li

>> roots([1 3*k k+2 +

ans =

-1.5510

0.0255 + 1.6057i

0.0255 - 1.6057i

the system becomes unst

>> roots([1 3*k k+2 +

ans =

-1.6741

-0.0629 + 1.5445i

-0.0629 - 1.5445i

the system is stable.

4.6 System Station

In previous sections we

of systems, paying atten

Although such features ar

temperature control syste

is still another relevant p

can be defined the stea

reference signal and the

controlled systems that

The steady-state error of

to follow an input refere

paramount importance f

example, in the altitude

accurate and changing alt

error in the system on foll

Fig. 4.34 Close-loop config

output. The error is defined

one.

sponse

])

it of the stability. For example for k=0.5

])

ble, while for greater values, for example k=0.6

])

ry Response

ere dealing with characteristics of the transient respon

ion to issues like maximum overshoot, peak times, e

highly relevant for real systems, e.g. it is desirable tha

responds quickly after a reference signal is given, the

rameter which has to do with the stationary response.

-state error of a system as the difference between t

stationary system output. This parameter has sense f

xhibit a close-loop configuration as shown in fig. 4.3

ystems gives information about the ability of the syste

ce signal. The study and analysis of this parameter is

r designing reliable control systems. Let’s think, f

control system of an airplane, in charge of keeping

tude reference given by the pilot; obviously any inhere

owing the pilot instructions may lead to a crash.

uration considering a sensorial system that

easures the pl

s the difference between the reference signal and the measur

132 4 System Response Analysis

In the temporal domain, let r(t) be the reference signal and b(t) be system

output, probably measured by a sensorial system, the steady-state error can be

defined as:

(4.69)

Whereas in the complex domain, the error can be expanded and expressed as

shown in equation (4.70).

(4.70)

Using the equation (4.70) and the final value theorem, we can mathematically

define the steady-state error as:

(4.71)

Note that the error, obviously depends on the involved systems’ nature and also on

the characteristics of the system output. That is, regarding the stationary response,

a system will behave differently according to the given reference input.

In the following we will particularize the study of the steaty-state error for

different input references and types of systems. For that the concept of type of a

system is considered and defined as the number of integrators of the open-loop

system G(s)H(s). That is, rewritting G(s)H(s) as:

(4.72)

It is said that G(s)H(s) is a N-type system. Next, let’s check the dependence of the

type of a system on the steady-state error when a particular input reference is

considered.

lim ( ) lim ( ) ( )

eetrtbt

∞

→∞ →∞

==−

() () () () () ()

()

() () () ()

1()()

() ()

() ()1

1()()

1()() ()()

() ()

1()()

() ()

1()()

Es Rs Bs Rs HsCs

Es Rs Hs Rs

GsHs

HsGs

Es Rs

GsHs

GsHs HsGs

Es Rs

GsHs

Es Rs

GsHs

=−=−

=−

⎛⎞

=−

⎜⎟

⎝⎠

⎛⎞

+−

⎜⎟

⎝⎠

lim ( ) lim ( ) lim ( )

1()()

ts s

eetsEs Rs

GsHs

∞

→∞ → →

== =

(1)(1)

() ()

(1)(1)

Ks s

GsHs

ss s

λλ

ττ

+ ⋅⋅⋅ +

4.6 System Stationary Response 133

4.6.1 Steady-State Error for Step Reference

Equation (4.71), considering R(s) as an input reference of amplitude A, is rewritten

as:

(4.73)

Considering , known as the position constant, the steady-state

error for step references is defined as:

(4.74)

Notice that as long as K

tends to infinite, the steady-state error tends to zero, and

this is only achieved when the system, G(s)H(s), is of type greater than one. That

is, from expression (4.72) it can be noted that for N

≥1, the limit will tend to

infinite.

Example 4.17

For the open-loop system

Indicate its type and its steady-state error for unitary step input.

Solution

The transfer function can be rewritten as:

It is a 0-type system and thus the error for a step input at the stationary will not be

equal to zero. That is, considering a unitary step input and equation (4.74), the

steady-state error is:

00 0

lim ( ) lim lim

1 () () 1 () ()

ss s

As A

esEss

GsHs GsHs

∞

→→ →

== =

lim ( ) ( )

KGsHs

→

∞

() ()

GsHs

22 2/2

() ()

(1)(2)(1)(1/21)32

GsHs

ss s sss

== =

++ + +++

()

,lim 1

(1) 1

0.5

∞

→

∞

⎧

⎫

⎪

== =

⎨

⎬

⎪

⎩⎭

134 4 System Response Analysis

4.6.2 Steady-State Error for Ramp Reference

Similarly, when the reference signal R(s) is a ramp, equation (4.71) becomes

(4.75)

Considering , called velocity constant, the steady-state error

for ramp references is defined as:

(4.76)

Systems aimed to follow a ramp reference without error at the steady-state must

be, at least of type 2 (N

≥2), in order to make K

to tend to infinite, and therefore

make the error to tend to zero.

Example 4.18

Check the type of the following system and its steady-state error for step and ramp

inputs

Solution

Rewriting the transfer function we can check that it is of type 1, i.e. it has an

integrator, and thus, it will follow a step input without errors at the steady-state,

while it will not accurately respond to a ramp reference.

For a step input of amplitude A, the error will be:

000

lim ( ) lim lim

1 ()() ()() ()()

sss

As A A

esEss

GsHs s sGsHs sGsHs

∞

→→→

== = =

lim ( ) ( )

KsGsHs

→

∞

() ()

GsHs

sss

55 5/2

() ()

(1)(2) (1)(1/21)32

GsHs

ss s ss ssss

== =

++ + +++

()

5/2

,lim

1 ( 1)(1 / 2 1)

step p

step

Ksss

∞

→

∞

⎧

⎫

== →∞

⎨

⎬

+++

⎩⎭

4.6

System Stationary R

Fig. 4.35 Simulink diagram

Fig. 4.36 System response.

is of type 1, the steady-state

In this case the error at the

zoomed area.

Step

Ramp

sponse

or step and ramp inputs.

op) System response for unitary step signal. Given the syst

error is zero. Bottom) System response for unitary ramp sign

steady state is not zero (it is actually 0.2) as illustrated by t

Transfer Fcn 1

s +3s +2s

Transfer Fcn

s +3s +2s

Scop

136 4 System Response Analysis

While for a ramp reference with slope R,

As expected the system has no errors at the steady-state for steps references but has

a certain error for ramps since it is a 1-type system (see fig. 4.36 and fig. 4.36).

The same reasoning can be done for other input references, for instance, for

parabolic references, R(s)=1/s

, the error will be calculated as:

(4.77)

And therefore the type of a system aimed to follow this input without error should

be of type 3 or higher.

()

5/2

,lim

(1)(1/21)

ramp v

ramp

eKs

Ksss

∞

→

∞

⎧⎫

== =

⎨⎬

⎩⎭

22 2

000

lim ( ) lim lim

1()()

() () () ()

sss

As A A

esEss

GsHs

ssGsHs sGsHs

∞

→→→

== = =

Chapter 5

Introduction to Control Systems

Control systems are aimed to modify the behavior of an existing system to

perform in a desired way. Several examples can be found in the real life in which

certain control actions are needed to achieve the wanted results. For example the

temperature of a room may change due to external and unexpected perturbations,

e.g. somebody opens/closes a window, sunshine heats the walls, etc. An air-

conditioned system can be installed to help in keeping a given temperature in spite

of any external change in the system. The air-conditioned system includes

elements to modify the current temperature (also called actuators, since they act

on the system), sensors to measure the current temperature, and some control

logic to determine the actuation to be carried out in order to achieve the target

temperature. In this chapter we will introduce the basis of the control systems,

their influence on real life systems, and their design to modify the behavior of a

given plant to fulfill particular specifications.

5.1 Control Systems

Control systems have been present in the society from the ancient past in an

attempt to design and construct machinery with a particular purpose. For instance,

the first records of the oldest control system date from the III century BC, when

Ktesibios designed a water clock, knows as Clepsydra, based on a mechanism in

charge of maintaining constant the water inflow of a tank. By means of a floating

valve the water volume of an auxiliary tank is maintained invariable and thus the

inflow of water of the controlled tank constant. The water level of the latter

indicates how much time is passing.

However, the real expansion and formal study of the control systems came into

scene at the end of the XVIII century with the arrival of the Industrial Revolution.

Representative examples of this period are the self-regulating wind machine

devised by E. Lee in 1745 which was aimed to automatically control the

orientation and inclination of the windmill’s sails according to the wind direction

improving the mill performance. Another example is the steam engine centrifugal

governor envisaged by Watt in 1788 to regulate the speed of steam locomotives

(see fig. 5.1).

Nowadays, we can find control systems almost everywhere, from highly

sophisticated and complex systems to maintain the speed and course of planes, to