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TRANSCRIPT
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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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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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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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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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1stOrder System: Complementary function
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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1stOrder System: Complementary function
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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1stOrder System: Complementary function
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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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
Substitution into equation (4) gives:
Hence
Therefore the particular solution is
HT
BT
110
HB
HtxPI Equ (5)
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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
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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1stOrder System: 1) Step response
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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1stOrder System: 2) Ramp response
Unit ramp function Ramp function
x
t0 +ve
-ve
slope = 1
yx
t0 +ve
-ve
slope = 1slope=M
y
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1stOrder System: 2) Ramp response
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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1stOrder System: 2) Ramp response
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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1stOrder System: 2) Ramp response
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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1stOrder System: 2) Ramp response
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1stOrder System: 2) Ramp response
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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1stOrder System: 2) Ramp response
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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1stOrder System: 3) Impulse response
x
t
0 +ve-ve
t infinitesimal width
The impulse function is shown schematically below
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1stOrder System: 3) Impulse response
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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1stOrder System: 3) Impulse response
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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1stOrder System: 3) Impulse response
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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1stOrder System: 3) Impulse response
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1stOrder System: 3) Impulse response
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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1stOrder System: 3) Impulse response
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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1stOrder System: 3) Impulse response
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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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
Thus the response becomes
From this we see
titi eBBetx
tiBeitx
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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
211
T
21 TT
R
imaginary
real
22 111
T
Ti
TA
B
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1stOrder System: 4) Harmonic response
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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1stOrder System: 4) Harmonic response
Amplitude attenuation |B|/|A|