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Mechanical Engineering Science 8

Dr. Daniil Yurchenko

First Order Systems:

Response to external excitation


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Introduction Mechanical System

uMF

kuFucFsd

,

0 extsd FFFF

extFkuuc

0M

aT

extF


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Introduction Thermal System

Cthermal capacity

of the water

R

TTqqqdt

dTC

aww 00;

C

q

RC

T

RC

T

dt

dT aww

aT


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Introduction Fluid System

;; 012

11 qq

dt

dVqq

dt

dVi


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Introduction

Fluid System

We got a set of two first order ODEs.

;; 012

1

1

qqdt

dV

qqdt

dVi

;;,,2

20

1

211222111

R

hq

R

hhqhCVhCV

2111

11

1

2111 hqRh

dt

dhRC

R

hhq

dt

dhC ii

2

2

1

212201

22

Rh

Rhh

dtdhCqq

dtdhC


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Introduction

In this section we will consider the response of thestandard form first order differential equation tofour different input functions

i.e. we will solve this differential equation for 4specified input functions xi(t)

T

y

T

xx Equ (1)


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Introduction

The specified inputs will be:

1) Step

2) Ramp 3) Impulse

4) Harmonic (sine or cosine wave input)

These are chosen because they are oftenthe signals used to test physical systemsand thus we can determine an expectedresponse


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8

Introduction

The solution to this differential equation can befound in a number of ways: e.g. LaplaceTransform, D-operator, integrating factor,

complementary function + particular integral(CF+PI)

In this course we will use the Laplace transform

and CF + PI technique.


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9

Introduction

The advantage of the CF + PI technique is

that the complementary function part iscommon to the solution for all possibleinput functions and needs only to bederived once.


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10

1stOrder System: Complementary function

This solution is found by setting the right handside of equation (1) equal to zero andconsidering forms of xO(t) which will satisfy

the expression on the right hand side. This isessentially an informed guessing game. Atypical function which achieves this is:

Where A and s are constants to be found. st

Aetx


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Because this part of the solution does not dependon any input (we set xi= 0) and it (usually)disappears with time we call it the transientsolution.

Starting with equation (1) andy= 0 we have

0Txx

Equ (2)


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We now guess a solution x based on therequirement that x and its first derivative must beof a similar form to enable a realistic chance ofsolving the equation.

As we said above such a solution is:

Substituting into equation (2)

stAetx whence stsAetx

0

steT

AAs


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Assuming estisnt zero, we get

Which gives:

0 T

A

As

Ts 1


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1stOrder System: Particular Integral

This solution determines the steady stateresponse of the system and depends on the

complete differential equation. A solution forx(t) is developed which is

based on the form of the input functiony(t).

If the input is a rampy(t) = t, the expectedresponse will be of the form x(t) = At+B


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1stOrder System: Particular Integral

The constants A and B are determined bysubstituting the knowny(t) and theassumedform ofx

(t) into the differentialequation.

The general solution is then the sum of theparticular integral and the complementary

function. Any remaining unknown constants are

determined from the initial conditions.


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1stOrder System: 1) Step response

A step input is defined as:

And

Where H is the magnitude of the step.

0,0 tty

0, tHty


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If we plot this function against t we see thereason for calling it a step:

x

t

H

0 +ve

-ve

y


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With this input, equation (1) becomes (droppingthe t in brackets):

For the particular integral we will try a constant

as the solution (reasonable since the input is aconstant). i.e.:

HTxTx 11

0, txBtx

Equ (4)


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Substitution into equation (4) gives:

Hence

Therefore the particular solution is

HT

BT

110

HB

HtxPI Equ (5)


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The general solution is the sum of thecomplementary function and particular

integral. From equations (3) & (5)

The constant of integration A is foundfrom the initial conditions.

Tt

CFPI AeHtxtxtx


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If we assume that at t = 0 the system was atrest and that before the step arrived

Then

Since

Then

00 x

10 AH

1

0

Te

HA


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From the above we can now see that theresponse of a first order linear system to astep input is given by:

This function is often referred to as asimple lag or an exponential lag

T

t

eHtx 1


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1stOrder System: 1) Step response For a unit stepH = 1 the response is:


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For this function, as t , the outputx H, the magnitude of the input.Therefore there is no difference betweenthe steady state response and the input. Wesay that there is no steady state error. Theoutput approaches the input via a negativeexponential such that:

At t = T seconds, x = 63% ofy

At t = 4T seconds, x = 98% ofy


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The significance of the time constant is that itcontrols the speed of response of the system.

If the time constant Tis 10 sec the system willtake 10 sec to reach 63% of its final value and 40sec to reach 98% of its final value following astep input

Similarly if T = 1 sec the system requires 1 sec toreach 63% of its final value and 4 sec to reach98%.

Class to demonstrate this to themselves


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1stOrder System: 2) Ramp response

A unit rampinput function is defined by

A non-unit ramp of slope M is defined by

A graph of the function illustrates theorigin of the name

tty

Mtty


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Unit ramp function Ramp function

x

t0 +ve

-ve

slope = 1

yx

t0 +ve

-ve

slope = 1slope=M

y


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With a unit rampinput equation (1) maybe written

We will guess that the response is also a

ramp of the form:

tTxTx 11

BtxCBttx

Equ (6)


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Substituting into equation (6) gives

Comparing coefficients of t

Comparing constants

tT

CT

BtTB

111

111

BT

BT

01

CTB Which gives TC

T

C 01


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Hence the particular solution is

As before the general solution is the sumof the particular integral (equ(7)) and the

complementary function (equ(3))

TttxPI Equ (7)

Tt

AeTttx

Equ (8)


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From eq. (9) and the figure it can be seenthat the output increasingly lags behind theinput until t >> T when the lag becomesconstant at T.

This means that if we minimise Ttheresponse becomes quicker and the steadystate error (difference between input and

input as t

) is also minimised.

TTtteTtttxty Tt

tt

1lim)(lim


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Steady-state error


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1stOrder System: 3) Impulse response

A unit impulse or Dirac delta function isan infinitesimally short pulse of arbitrary

magnitude but which has an area under theimpulse curve of unity. It has the symbol(t) and the unit area property may beexpressed:

1

dtt


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x

t

0 +ve-ve

t infinitesimal width

The impulse function is shown schematically below


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Thus the impulse input to our first ordersystem is

Then the standard first order equation maybe written:

tty

tTxTx 11


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Laplace Transform of

Impulse Function

f(t)

t

otherwise,0

0 t,1

)()( ttf

)0()()(0

fdttft

1)]([ tL

t

1)()(0

dtetsF st


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Thus

t

TLx

TxL

11

T

tLT

sXT

xssX 11)(

1)0()(

T

TxsX

T

sT

xT

sXTs

)0(1)(

1

)0(1

)(1

sT

TxsX

1

)0(1)(


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Applying inverse L-transform

Assuming x(0)=0 we get

sTL

T

Tx

sT

TxLsXL

/1

1)0(1

1

)0(1)( 111

TteTtx /

1)(


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The initial value of the response is 1/T andthis response then decays exponentially to

zero as time increases. At time t = T seconds the response is 37%

of its initial value.

The figure also shows that the initialtangent to the curve intercepts the timeaxis at t = T.


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Equation of the tangent is:

Substituting t = T yields y = 0 thus

confirming that this tangent intercepts thetime axis at t = T.

TtTy 112


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Example: the mass/damper system of thenext slide. Note that there is no spring. Toproduce impulsive force we may use a

hammer

m

C

v(t)F(t)


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1stOrder System: 4) Harmonic response

In this case the input is either

Therefore we can expect the steady stateresponse to be of the form

tAty sin tAty cosor

tBtx sin tBtx cosor


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For harmonic excitation we are interestedin two aspects of the response

The magnification or attenuation of theoutput relative to the input

And the phase lag

y

x


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The manipulation of the equationsbecomes easier if we use complex notation

The response is then

In general B is a complex number whichmay be written

titAAety ti sincos

tiBetx

ieBB


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Thus the response becomes

From this we see

titi eBBetx

tiBeitx


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With the given input the standard form ofthe equation becomes

If we now substitute the assumed input

from the previous slide

tiAeTxTtx 11

tititiAeT

BeT

Bei

11


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Thus

And the amplitude ratio is

We wish to get this expression into a moreunderstandable form. Therefore we mustrationalise it by multiplying numerator anddenominator by the complex conjugate of thedenominator.

AT

BT

i 11

TiA

B

1

1


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Hence

i.e.

We can plot this expression on an argand

diagram to obtain a more manageablerelationship between B/A and frequency and between phase lag and

211

1

1

1

1

T

Ti

Ti

Ti

TiA

B

22 111

T

Ti

TA

B


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211

T

21 TT

R

imaginary

real

22 111

T

Ti

TA

B


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From the diagram we can see:

and the phase lag is

These two parameters are plotted againstfrequency in the next slide

22222

222

22

2

22

1

1

1

1

11

1

TT

T

T

T

T

R

A

B

TTTT

T

11

2222

1

1

1

1

tantantan


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Amplitude attenuation |B|/|A|

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