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S.l.dr.ing.mat. Alina BogoiDifferential Equations
POLITEHNICA University
of
Bucharest
Faculty of Aerospace
Engineering
CHAPTER 7
Laplace Transform(Cont)
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Laplace Transform Theorems
Theorem Description
Definition of Laplace Transform
Linear Property
Derivatives
Integrals
First Shifting Property
Second Shifting Property
==0
)()()}({ sFdttfetf stL
)()()}()({ sbGsaFtbgtaf +=+L
)0()0()()}({ )1(1)( = nnnn ffssFstf LL
)(1
})({0
sFs
df =
L
)()}({ asFtfeat =L
)()}()({ sFeatuatf as
=L
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3
Transfer functionBlack-box
systemInput
x(t)
Output
y(t)
1
1 01( ) ( ) ... ( ) ( ) ... ( )
n n m
n mn n m
d d dy t a y t a y t b x t b x t
dt dt dt
+ + + = + +
Assume all initial conditions are zero, we get the transfer
function(TF) of the system as
Consider an input-output system from x(t) to y(t) whose
dynamic model has the form:
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Connection between the t- and s-domains
The transfer function, G(s), is defined by:
)t(xb...)t(xdt
db)t(ya...)t(ydt
da)t(ydt
dmm
m
nn
n
n
n++=+++
01
1
1
)s(Xb...)s(Xsb)s(Ya...)s(Ysa)s(Ys mm
n
nn
++=+++
0
1
1
n
nn
m
m
a...sas
b...sb
)s(X
)s(Y)s(G
+++
++==
1
1
0
011 =+++
n
nn a...sas Characteristic Equation
Poles of G(s) Roots of the Characteristic Eq
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Transfer functionBlack-box
systemInput
x(t)
Output
y(t)
differential
equation
x(t) y(t)
transferfunction
X(s)Y(s)
G(s)
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Example 1. Find the transfer function of the RLC
1) Writing the differential equation of the system according to
physical law:
R L
C
u(t)u
c (t)
i(t)Input Output
2) Assuming all initial conditions are zero and applying Laplace
transform
3) Calculating the transfer function as( )G s
2( ) 1( )( ) 1
cU sG sU s LCs RCs
= =+ +
( ) ( ) ( ) ( )C C CLCu t RCu t u t u t+ + =&& &
2 ( ) ( ) ( ) ( )c c cLCs U s RCsU s U s U s+ + =
Solution:
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Properties of transfer function
The transfer function is defined only for alinear time-invariant system, not fornonlinear system.
All initial conditions of the system are setto zero.
The transfer function is independent of theinput of the system.
7
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Position of poles
and zeros
-a
j
i0
( )X ss a
=+
Transfer function
( ) atx t Ae=
Time-domain impulse
response
0
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1 1
2 2( )
( )
A s BX s
s a b
+=
+ +
Transfer function
( ) sin( )atx t Ae bt = +
Time-domain
impulse response
Position of poles and
zeros
-a
j
i
b
0
0
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1010
1 1
2 2( )
s BX s
s b
+=
+
Transfer function
( ) sin( )x t A bt = +
Time-domain
impulse response
Position of poles and
zerosj
i
b
00
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1111
Position of poles
and zeros
a
j
i0
( )X ss a
=
Transfer function
( ) atx t Ae=
Time-domain impulse
response
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1212
1 1
2 2( )
( )
A s BX s
s a b
+=
+
Transfer function
( ) sin( )atx t Ae bt = +
Time-domain
dynamic response
Position of poles and
zeros
-a
j
i
b
0
0
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1313
Summary of pole position & system dynamics
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Significance of poles
The nature and value of the poles determine whether
the system is stable or instable, and the type of
response
The nature and value of any pole is classified as a
function of its location in theplan defined by:
the Real part and Imaginary part of the pole
[or in other words the s-domain]
Im(s)Re(s)
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Stability Criterion vs Pole
Locations)(sIm
UnstableStable
)(sRe
The locations of poles ins-domain determinewhether the system is stable or unstable.
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Stability Criterion based on
the Pole Locations
A system is stable if a l l its poles havenegative real parts
(i.e., there are al l st r i ct l y inside the
left-side s-plane)
and unstable otherwise
Note:
This criterion is valid only if the system is
linear time-invariant
(i.e., has constant parameters).
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How to analyze and design acontrol system
18
Controller Actuator Plant
Sensor
-r
Expectedvalue
e
Error
Disturbance
Controlledvariable
n
yu
The first thing is to establish system model(mathematical model)
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Block diagramThe transfer function relationship
19
( ) ( ) ( )Y s G s U s=
can be graphically denoted through a block diagram.
G(s)U(s) Y(s)
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Equivalent transform of blockdiagram
1 Connection in series
G(s)U(s) Y(s)
( ) ?G s =
X(s)G1(s) G2(s)
U(s) Y(s)
1 2
( )
( ) ( ) ( )( )
Y s
G s G s G sU s= =
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2.Connection in parallel
G(s)U(s) Y(s)
1 2
( )( ) ( ) ( )
( )
Y sG s G s G s
U s= = +
U(s)
G2(s)
G1(s) Y1(s)
Y2(s)
+Y(S)
( ) ?G s =
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3. Negative feedback
M(s)
R(s) Y(s)
( ) ( ) ( )
( ) ( ) ( ) ( )
Y s U s G s
U s R s Y s H s
=
=
( )( )1 ( ) ( )
G sM sG s H s= +
[ ]( ) ( ) ( ) ( ) ( )Y s R s Y s H s G s=
Y(s)G(s)
H(s)
U(s)R(s)_
Transfer function of a negative feedback system:
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Laplace
transform
Fourier
transform
Three models
Differential equation Transfer function Frequency characteristic
Transfer
function
Differential
equation
Frequency
characteristic
Linear systemStudy
time-domain
responsestudy
frequency-domainresponse
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S.l.dr.ing.mat. Alina
Bogoi
Differential
Equations
POLITEHNICA University
of
Bucharest
Faculty of Aerospace
Engineering
CHAPTER 8
The Fourier SeriesThe Fourier Transform
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Content Periodic Functions Fourier Series
Complex Form of the Fourier Series Impulse
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Odd and Even Functions A function f(x) is said to be even
if
A function f(x) is said to be odd
if
)()( xfxf =
)()( xfxf =
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Odd and Even Functions
)()( xfxf = )()( xfxf =
f(x)
x
f(x)
x
Even Function Odd Function
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Odd and Even Functions Property
The product of an even and anodd function is odd.
evenisxfifdxxfdxxfLL
L)(,)(2)(
0 =
oddisxfifdxxfL
L)(,0)( =
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Periodic Functions
A function ( )f t is periodic
if it is defined for all real tand if there is some positive number
(the lowest),T such that ( ) ( )t T f t + = .
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Example:
4cos
3cos)(
tttf +=
Find its period.
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Example:
4cos
3cos)(
tttf +=
Find its period.
)()( Ttftf += )(41cos)(
31cos
4cos
3cos TtTttt +++=+
Fact: )2cos(cos += m
= mT
23
= nT 24
= mT 6
= nT 8
= 24T smallest T
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Example:
tttf 21 coscos)( +=
Find its period.
)()( Ttftf += )(cos)(coscoscos 2121 TtTttt +++=+
= mT 21
= nT 22 n
m=
2
1
2
1
must be a
rational number
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Example:
tttf )10cos(10cos)( ++=
Is this function a periodicone?
+=
1010
2
1 not a rationalnumber
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Rectangular Pulse
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Sawtooth wave
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Orthogonal Functions
Call a set of functions {k }orthogonal on an interval a < t