system dynamics

ECM2105 - Control Engineering Dr Mustafa M Aziz (2013)

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SYSTEM DYNAMICS (REVISION)

1. Mechanical Systems

2. Electrical Systems

3. Electromechanical Systems – The DC Motor

4. Linearisation of Nonlinear Systems

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The common form of representing control systems is to use block diagrams; the system is

represented by a box and the signals of temperature, voltage, fluid flow etc. are represented by

directed lines. A system is a black box which has an input and an output. A general engineering

system is shown as:

SystemInput, u(t) Output, y(t)

Examples:

Power

station

Input

Fuel

Output

Electricity

Power station

Electric

motor

Input

Electric

power

Output

Mechanical

rotation

Electric motor

Normally the system is described by a mathematical model.

The question you may ask: Why use mathematical models to represent engineering systems?

The answer to this question is to predict the dynamical behaviour of the system. The model can be

used to predict the outcome of an experiment without actually undertaking that experiment. This

saves time and money in most cases including time for the process design, comparing process

designs and undertaking speculative process redesigns. In some industries, it also improve safety as

the model can be used to predict the outcome of unsafe operation and fault conditions that could

lead to a situation where the system is dangerous and/or is damaged. For example:

• How will the system perform if a particular form of control system is used? (Save money by

providing an answer before the control system is implemented)

• How will the heating system perform if a larger boiler/power plant is commissioned? (Is it

worthwhile purchasing the more costly piece of plant?)

• What will happen in a nuclear reactor if a pump fails and the coolant is lost to part of the reactor

core? (By predicting the peak temperature in the core, the possibility of thermal damage and

hence unsafe operation can be assessed)

• Can a crane lift and transfer a specified load quickley and smoothly without introducing

excessive oscillation (or swing)? (Can the crane actually do a useful job carrying this load?)

• What is the effect of driving a heavy lorry over a bridge? What are the dynamic forces that the

bridge must cope with? Are there any undesirable dynamic effects? What is the effect of a cross

wind? (Finding out before the bridge is built seems to be a good idea! - see pictures below of


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Tacoma Narrows Bridge -Washington- that collapsed on the 7th of Nov 1940 due to wind-

induced vibrations: wind frequency matched the natural or resonance frequency of the structure

leading to rapid periodic oscillations of the structure – aerodynamic flutter).

• The proposed aircraft design satisfies all previous design specifications for normal flight

conditions but how will it cope with extreme conditions?

• The pipework in the factory/ship/whatever is connected to the power unit which acts as a source

of vibration. How will this vibration travel around the area? What are its acoustic effects? What

is the relative effect of placing passive dampers at a number of selected point on the circuit?

Can active vibration control technology improve on passive mechanisms?

1. Mechanical Systems

The basic forms of the mechanical system building blocks are springs, dashpots (dampers) and

masses. Springs represent the stiffness of a system, dashpots the forces opposing motion, i.e.

frictional or damping effects, and masses the inertial or resistance to acceleration. The equations

governing the motion of mechanical systems are often directly or indirectly formulated from

Newton’s law of motion.

1.1. Translational Motion

In translational motion, the system building blocks can be considered to have:

- a force, u(t) in Newtons (N), as an input and

- a displacement, y(t) in metres (m), as an output.

These variables are functions of time.

Newton’s law of motion states that the algebraic sum of forces acting on a rigid body in a given

direction is equal to the product of the mass of the body and its acceleration in the same direction:

∑ forces = ma

where m denotes the mass and a = d2y(t)/dt

2 is the acceleration in the direction of motion.

1.1.1. Spring

Any mechanical element that undergoes a change in shape when

subjected to a force can be characterised by a stiffness element

(usually the spring). The stiffness of a spring is described by the

relationship between the force u(t) used to extend or compress a spring

output, y(t)

input, u(t)

k


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and the resulting extension or compression y(t), as shown in the figure.

For a linear spring, we have:

u(t) = ky(t)

where k is the spring constant or stiffness (N/m). The bigger the value of k the bigger the forces

have to be to stretch or compress the spring and so the greater the stiffness.

Laplace transform is: U(s) = kY(s)

1.1.2. Dashpot

The dashpot (damper) represents the type of forces experienced when

we endeavour to push an object through a fluid or move an object

against frictional forces. In the ideal case, the damping or resistive force

u(t) is proportional to the velocity dy/dt of the piston, i.e.

u(t) = bdy(t)/dt

where b is the frictional coefficient (N-s/m). The larger the value of b

the greater is the damping force at a particular speed.

Laplace transform is: U(s) = bsY(s)-by(0)

1.1.3. Mass

The mass block exhibits the property that the bigger the mass the

greater is the force required to give it a specific acceleration. By

Newton's law of motion:

2

2

dt

y(t)dmu(t) =

where the constant of proportionality between the force and the acceleration is the mass m (in kg).

Laplace transform is:

−−=

dt

dy(0)sy(0)Y(s)smU(s) 2

Example:

Develop a model for the mass, spring and dashpot

system shown in the figure.

m input, u(t)

output, y(t)

output, y(t)

input, u(t)

b

m

k

b

y(t)

u(t)

Viscous fluid

Piston


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Solution: The net force applied to the mass m can be found from the free-body diagram:

m u(t)ky(t)

bdy(t)/dt

y(t)

and according to Newton's law ∑ forces = ma:

2

2

dt

)t(ydm

dt

)t(dyb)t(ky)t(u =−−

Rearranging:

)t(u)t(kydt

)t(dyb

dt

y(t)dm

2

2

=++

Laplace transform is:

)s(Udt

)0(dym)0(y)kms()s(Y)kbsms( 2 =−+−++

1.2. Rotational Motion

Rotational mechanical systems are handled in the same way as translational mechanical systems,

except that:

- torque, T(t) (Newton-metres), replaces force and

- angular displacement, θ(t) (radians), replaces the translational displacement.

The equivalent three system blocks for rotational motion are a torsional spring, a rotary damper and

the moment of inertia, i.e. the inertia of a rotating mass.

The extension of Newton’s law of motion for rotational motion states that the algebraic sum of

moments or torques about a fixed axis is equal to the product of the inertia and the angular

acceleration about the axis:

∑ torques = Jα

where J denotes the moment of inertia (kg.m2) and α = d2θ(t)/dt2 is the angular acceleration (rad/s2).

1.2.1. Torsional spring

Rotational stiffness is usually associated with a torsional spring (such as the

mainspring of a clock, or with a relatively thin shaft). With a torsional

spring the angle rotated θ(t) is proportional to the torque T(t), i.e.

T(t) = Kθθθθ(t)

where K is the spring constant (N-m/rad).

Laplace transform is: T(s) = Kθ(s)

K

T(t) θ(t)

T(t) θ(t)

K

T(t) θ(t)

K


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1.2.2. Rotary damper

With the rotary damper a cylinder is rotated in a fluid and the resistive torque T(t) is proportional to

the angular velocity, i.e.

T(t) = Bdθθθθ(t)/dt

where B is the friction constant (N-m-s/rad).

Laplace transform is: T(s) = bsθ(s) - bθ(0)

1.2.3. Moment of inertia

Is the resistance of an object to changes in its angular velocity. It

is related to the torque through:

2

2

dt

θ(t)dJT(t) =

where J is the inertia (kg-m2 = N-m-s

2/rad).

Laplace transform is: )0(J)0(Js)s(Js)s(T 2 θ−θ−θ= &

Example:

Derive the differential equation describing the rotational system

shown in the following figure.

Solution:

T(t) θ(t)

m

J

B

T(t) θ(t)

T

θ

K

J

B

Kθ(t) Bdθ(t)/dt

T


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2. Electrical Systems

The basic elements of passive electrical systems are resistors, capacitors and inductors. Kirchhoff's

two laws are used to develop mathematical models of electrical circuits. They are:

Kirchhoff's current law: The total current flowing towards a junction is equal to the total current

flowing out from that junction, i.e. the algebraic sum of the currents at the junction is zero:

∑ currents = 0

Kirchhoff's voltage law: The algebraic sum of all voltages taken around a closed path in a circuit is

zero:

∑ voltages = 0

2.1. Resistor

Using Ohm's law, the potential difference v(t) across a resistor at

any instant is proportional to the current i(t) through it, i.e.

v(t) = Ri(t)

where R is the resistance in Ohms (Ω).

Laplace transform is: V(s) = RI(s)

2.2. Capacitor

For a linear capacitor we have,

dt

dv(t)Ci(t) =

where i(t) is the current through the capacitor, v(t) is the potential

difference across it and C is the capacitance in Farads (F).

Laplace transform is: I(s) = CsV(s)-Cv(0)

The voltage across the capacitor can be obtained by integrating the current through it over time

from 0 to t:

constd)(iC

1)t(v

t

0

+ττ= ∫

Laplace transform is: sC

)s(I)s(V =

2.3. Inductor

For a fixed linear inductor, the potential difference v(t) across it is

proportional to the rate of change of current through it, i.e.

dt

di(t)Lv(t) =

Ri(t)

v(t)

Ci(t)

v(t)

Li(t)

v(t)


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where L is the inductance in Henrys (H).

Laplace transform is: V(s) = LsI(s)-Li(0)

Example: A resistor-inductor-capacitor network is shown below. Find a mathematical model that

describes the relationship between the input voltage vi(t) and the output voltage vo(t).

vi(t) vo(t)

R L

C

i(t)

Solution:

3. Electromechanical Systems – The DC Motor

There are many electromechanical devices, such as potentiometers, motors, and generators. Here

we present the mathematical model of a DC motor which is one of the most widely used prime

movers in industry today and is used in variable speed applications (e.g. domestic appliances,

robotics, automation, toys, computer peripherals). A motor is an electromechanical component that

yields a displacement output for a voltage input, that is, a mechanical output generated by an

electrical input. The mechanical output is used to drive an external load.

A DC motor consists of a rotating cylinder called the armature with a current carrying conductor

wrapped around it. The armature is placed in a magnetic field between two stationary poles as

shown below.


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Magnetic flux

Force

Current

Left-hand rule

The principles of operation are as follows:

- The brushes (spring-loaded carbon contacts) contact the rotating commutator causing current ia

to flow in the armature conductors. The current carrying armature conductor, in the presence of

the magnetic field Bf produced by a stationary permanent or electro-magnet, experiances a force

F given by:

F = Bfial

whose direction is given by the left-hand rule, and where l is the length of

the armature conductor cutting the magnetic field. The commutator

insures that the current changes direction in the armature coil every half

cycle to maintain the contineous motion of the rotor.

- The force F develops a turning torque proportional to the armature current,

i.e. Tm α ia

that rotates the armature and shaft where the load is normally connected.

- The movement of the armature conductor in the magnetic field induces a voltage at the

conductor terminals according to Faraday's law. This voltage is called the back electromotive

force or emf and is proportional to the angular velocity of the armature,

i.e. vb α dθ/dt.

There are two methods of controlling the speed of a DC motor:

1. Field control: hold the armature current constant and control the torque via the field voltage

(using electromagnets).

2. Armature control: hold the field constant and control the torque via the armature voltage (using

permanent magnets).

We will develop the dynamic equations for the armature-controlled, permanent magnet DC motor

system shown schematically below.

BrushesBrushes


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Tm = Kmia

θ

R Lia(t)

va

+

_

+

_

Armature circuitdt

dKv bb

θ=

B

J

Inertial load

L and R represent the inductance and resistance of the motor's armature circuit, and the voltage vb

represents the generated back emf which is proportional to the shaft velocity dθ/dt. The torque Tm

generated by the motor is proportional to the armature current ia. The inertia represents the

combined inertia of the motor armature and the load, and B is the total viscous friction acting on the

output shaft.

The differential equations of the motor armature circuit and the inertial load are:

dt

)t(dK

dt

)t(diL)t(Ri)t(v b

aaa

θ++= and

dt

)t(dB)t(iK

dt

)t(dJ am2

2 θ−=

θ

4. Linearisation of Nonlinear Systems

4.1. Nonlinearity

Definition 1: A system is said to be linear if the principle of superposition applies. The principle of

superposition states that if the input signal u1(t) produces an output y1(t), input u2(t) produces an

output y2(t) and a1 and a2 are constant numbers then the input a1u1(t) + a2u2(t) produces the output

a1y1(t) + a2y2(t).

Linear

systema1u1(t) + a2u2(t) a1y1(t) + a2y2(t)

Definition 2: A system is said to be nonlinear if the principle of superposition does not apply.

Example: Test whether f(x) = 5x is linear.

f(x)

x0

Output

Input

Solution:


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Adding individual outputs:

Input a1x1 produces output f(a1x1) = 5a1x1

Input a2x2 produces output f(a2x2) = 5a2x2

Adding: f(a1x1) + f(a2x2) = 5ax1 + 5ax2

Resultant output response:

Input a1x1 + a2x2 produces output:

f(a1x1 + a2x2) = 5(a1x1 + a2x2) = 5a1x1 + 5a2x2

Since the resultant output can be calculated as the sum of the separate output responses, then

superposition applies → linear system.

Example: Test whether f(x) = 2x2 is linear.

f(x)

x0

Output

Input

Solution:

Adding individual outputs:

Resultant output response:

Exercises: Use the above definitions to test which of the following is linear and which is nonlinear:

(i) f(x) = 5cos(x)

(ii) f(x) = 4ln(x)

(iii) )t(u)t(ydt

)t(dy

dt

)t(yd2

2

=++


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(iv) )t(u)t(ydt

)t(dy)t(y

dt

)t(yd2

2

=++

(v) )t(u)t(ydt

)t(dy)t(u

dt

)t(yd2

2

=++

4.2. Nonlinear systems

The following are examples of physical systems.

1) A RLC electrical circuit with capacitance voltage

y(t) and applied voltage u(t). By applying

Kirchhoff’s voltage law, the differential equation is:

)t(u)t(ydt

)t(ydLC

dt

)t(dyRC

2

2

=++

2) A diode allows current to flow only in one direction.

In the forward bias region, the current/voltage

characterestic (and hence resistance) is not linear

and can be described by:

i = io(eev(t)/2kT

– 1)

3) A vehicle system with mass m, external force u(t),

displacement y(t), and friction coefficient b. By

Newton’s second law, we have the following

differential equation:

)t(udt

)t(dyb

dt

)t(ydm

2

2

=+

4)

A spring-mass-damper system with mass m, damping constant b, non-linear spring f[y(t)], and

displacement y(t). By Newton’s second law, the dynamic equation is:

)t(u)]t(y[fdt

)t(dyb

dt

)t(ydm

2

2

=++

mu(t)

)t(yb&

y(t)

u(t) y(t)

R L

C

m

f(y(t))

b

y(t)

u(t)

Nonlinear springf(y)

y

v(t)

i(t)

v

i


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5) A pendulum, which can be viewed as a one-link robot

manipulator, is shown in the figure. The parameters of the

pendulum are the length of the link (from the rotating point to the

centre of the mass of the ball) L, the mass of the ball m, the

coefficient of friction in the rotating point B, the acceleration of

gravity g, the applied torque T(t) and joint angle θ(t). It is assumed that the link is massless and the ball is very small. The

dynamic equation is:

)t(T))t(sin(mgLdt

)t(dB

dt

)t(dmL

2

22 =θ+

θ+

θ

It is noted that the dynamic equations described in (1) and (3) are linear and the dynamic equations

described in (2), (4) and (5) are nonlinear. The systems described in (1), (3), (4) and (5) are single-

input and single-output systems (SISO).

In this course we mainly study dynamics and control of single-input and single-output linear

systems. A general form of a single-input and single-output linear time-invariant system is defined

by:

)t(ubdt

)t(dub

dt

)t(udb

dt

)t(udb

)t(yadt

)t(dya

dt

)t(yda

dt

)t(yd

011m

1m

1mm

m

m

011n

1n

1nn

n

++++=

++++

−

−

−

−

−

−

L

L

where y(t) is the output of the system and u(t) is the input. The coefficients an and bn are real

constants and n ≥ m.

4.3. Linearisation Concepts

Most physical systems have generally nonlinear characteristics. However, over a small range of

input values around an operating point a system can be described by a linear model. Linear

approximations allow us to apply linear analysis techniques to design controllers and study the

behaviour of systems.

For example, the following graph shows the measured output voltage from a depth sensor that

regulates the level of a liquid in a tank. The characteristic of the sensor is clearly nonlinear, but for

small operating values around the point (us,ys), called the operating or equilibrium point, the

behaviour of the sensor can be described by a linear approximation (straight line).

L

θ

T

mgmgsin(θ)

B


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u

y

us

ys

Level in tank

Sensor voltage

Measured results

Linear approximation

Operating region

δu

δy

If we operate the system such that liquid level in the tank stays within the operating region, δu, we can use this linear approximation. However, outside this region the linear formula would no longer

be valid.

0

f(x)

xxo

f(xo)

δx

δf

To find a linear mathematical model of the nonlinear system f(x), we can expand f(x) into a Taylor

series around the operating point (xo,f(xo)):

L+−

+−

+=== !2

)xx(

dx

fd

!1

)xx(

dx

df)x(f)x(f

2

o

xx

2

2

o

xx

o

oo

If the variation around the operating point, δx = x – xo, is small we may neglect the higher-order

terms in δx and we have:

)xx(dx

df)x(f)x(f o

xx

o

o

−+≈=

(1)

or

xdx

df)x(f)x(f

oxx

o δ=−=

Depth sensor

Outflow

InflowPump


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δδδδf(x) = cδδδδx

oxxdx

dfc

=

=

The resulting approximation yields a straight line relationship between the change in δf(x) and δx with the constant of proportionality, c, being the slope of this line at xo.

Example: Linearise f(x) = 5cos(x) about x = π/2.

f(xo) = 5cos(π/2) = 0

5)xsin(5dx

dfc

o

o

xxxx

−=−===

=

The linear approximation around xo can be found

using equation (1):

)2/x(5

)xx(dx

df)x(f)x(f o

xx

o

o

π−−≈

−+≈=

Exercises: Linearise the following functions:

(i) f(x) = 3sin(2x) about x = π/4

(ii) f(x) = 2x2 + 3 about x = 0

(iii) f(x) = 8ln(x) about x = 5

π/2

x

f(x)

f(x) ≈ -5δx

δx


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4.4. Linearising differential equations

m

Nonlinear spring,

f(y(t))

b

y(t)

u(t)

Example: Consider the mechanical system shown in the above figure with a non-linear spring. The

nonlinear model is:

)t(u)]t(y[fdt

)t(dyb

dt

)t(ydm

2

2

=++

Find the linearised model of this system around the equilibrium point (us,ys).

Solution:

1. Determine the equilibrium point (us,ys). This often corresponds to the steady-state condition of

the system. Hence at u = us the equilibrium point can be found by setting all the time

derivatives to zero in the differential equation (d2y/dt

2 = 0 and dy/dt = 0) and solving:

f(ys) = us

for ys.

2. Linearise the differential equation for small excursions about the equilibrium point. Let y = ys

+ δy and u = us + δu and substitute into the differential equation:

uu)yy(fdt

)yy(db

dt

)yy(dm ss

s

2

s

2

δ+=δ++δ+

+δ+

Since ys is a constant:

2

2

2

s

2

dt

yd

dt

)yy(d δ=

δ+

dt

yd

dt

)yy(d s δ=

δ+

Expanding f(ys + δy) using equation (1):

yc)y(f)yy(f ss δ+=δ+ where

syydy

)y(dfc

=

=


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Substituting into the differential equation:

uuyc)y(fdt

ydb

dt

ydm ss2

2

δ+=δ++δ

+δ

But since f(ys) = us, then we have:

uycdt

ydb

dt

ydm

2

2

δ=δ+δ

+δ

This equation is linear.

It is normal practice at this stage to simplify the notation by replacing the symbols δy and δu by y and u (equivalent to measuring the value of the input and output relative to the equilibrium values):

ucydt

dyb

dt

ydm

2

2

=++ (2)

Example: If f(y(t)) = ky2 in the previous example, linearise the differential equation.

Solution:

Exercise: Linearise the following differential equation for small excursions about y = π/4:

0)ycos(dt

dy2

dt

yd2

2

=++


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TUTORIAL PROBLEM SHEET 2

1. Derive the differential equation describing the relationship between the input of the force u(t)

and the output displacement y(t) for the system shown in the following figure

m

b2

y(t)

u(t)

b1

where b1 and b2 are the frictional coefficients.

2. Derive the relationship between the potential difference across the inductor, y(t), and the input

u(t) potential for the circuit shown below.

u(t) y(t)

R

L

3. Show that the following first-order differential equation is linear, i.e. superposition holds:

)t(ku)t(aydt

)t(dy+−=

where a and k are constants.

4. Use a Taylor series expansion to derive the linearised model of the function f(x) = 0.5x3 around

the point xo = 2.

5. A shock absorber system is shown with wall friction coefficient B

and a nonlinear spring. Find the linearised differential equation for

this system around ys = 1. The nonlinear spring force is defined by

f(y(t)) = 1 – e-y(t)

.

6. The relationship between the force u(t) used to stretch a spring and its extension is given by u(t)

= ky(t)3, where k is a constant. Linearise this equation for an equilibrium point of (us,ys).

m

u(t)y(t)

B

f(y(t))


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7. Derive the dynamic equation for rotational angle θ (with respect to the inertial reference) of the simplified single-axis satellite shown in the

figure. This space satellite has a moment of inertia J. Fc is the jet

control force about the centre of the mass, and there is also some

disturbance moments MD on the satellite due primarily to solar pressure

acting on the solar panels.

8. The flow of traffic in a single lane can be described by the following

equation:

)t(y/k

12ekV

dt

)t(dy −−=

where y(t) is the relative distance between two cars, V is a constant velocity of the lead car, k1 and

k2 are real positive constants. Your tasks are:

a) obtain the equilibrium value ys that results in dy(t)/dt = 0.

b) linearise the equation around ys, and write the resulting linearised equation.

9. A pendulum, which can be viewed as a one-link robot

manipulator, is shown in the figure. The parameters of the

pendulum are the length of the link (from the rotating point to

the centre of the mass of the ball) L, the mass of the ball m,

the coefficient of friction in the rotating point B, the

acceleration of gravity g, the applied torque T(t) and joint

angle θ(t). It is assumed that the link is massless and the ball

is very small.

a) develop the mathematical model of the pendulum system

b) prove that this model is nonlinear

c) linearise the model around the equilibrium point.

L

θ

T

mg

θ

Fc

d

MD

Reference

Jet

system dynamics

Documents

heating system

control systems

general engineering

engineering systems

system dynamics revision

electromechanical systems

electrical systems

process design