lti system analysis with the laplace...
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LTI System Analysis withthe Laplace Transform
Laplace transforms
The diagram commutesSame answer whichever way you go
Linearsystem
Differentialequation
Classicaltechniques
Responsesignal
Laplacetransform L
Inverse Laplacetransform L-1
Algebraicequation
Algebraictechniques
Responsetransform
Tim
e do
mai
n (t
dom
ain) Complex frequency domain
(s domain)
Laplace Transform - definition
Function f(t) of timePiecewise continuous and exponential order
0- limit is used to capture transients and discontinuities at t=0s is a complex variable (σ+jω)
There is a need to worry about regions of convergence ofthe integral
Units of s are sec-1=Hz
A frequency
btKetf <)(
∫∞
−=0-
)()( dtetfsF st
Laplace transform examplesStep function – unit Heavyside Function
After Oliver Heavyside (1850-1925)
Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)
Delta (impulse) function δ(t)
0if1
)()(
0
)(
000
>=+
!=!===
"+!"!"
!
!"
!
!## $
%$
%$
sj
e
s
edtedtetusF
tjststst
≥
<=
0for,10for,0
)(tt
tu
! !" "
"+#
+###>
+=
+#===
0 0 0
)()( if
1)( $%
$$
$$$
ss
edtedteesF
tstsstt
sdtetsFst allfor1)()(
0
== !"
#
#$
Laplace Transform Pair Tables
damped cosine
damped sine
cosine
sine
damped ramp
exponential
ramp
step
impulse
TransformWaveformSignal)(t!
22)( !"
"
++
+
s
s
22)( !"
!
++s
22 !
!
+s
22 !+s
s
1
s
1
2
1
s
!+s
1
2)(
1
!+s
)(tu
)(ttu
)(tuet!"
)(tut
te!"
( ) )(sin tut!
( ) )(cos tut!
( ) )(sin tutt
e !"#
( ) )(cos tutt
e !"#
Laplace Transform Properties
Linearity – absolutely critical propertyFollows from the integral definition
Example
!
L Af1(t) + Bf
2(t){ } = AL f1 (t){ } + BL f2 (t){ } = AF
1(s) + BF
2(s)
!
L Acos("t){ } = LA
2ej"t + e# j"t[ ]
$ % &
' ( )
=A
2L e
j"t{ } +A
2L e
# j"t{ }
=A
2
1
s# j"+A
2
1
s+ j"
=As
s2 + " 2
Laplace Transform Properties
Integration property
Proof
Denote
so
Integrate by parts
( )s
sFdf
t
=!"#
$%&'0
)( ((L
dtstet
dft
df !"#
"=" $%
&'(
)
*+,
-./
0 0
)(
0
)( 0000L
)(and,
0
)(and,
tfdt
dye
dt
dx
tdfy
s
stex
st==
!=
""=
"
##
!!!"
#"
#+
$$%
&
''(
)#=
$$%
&
''(
)
0000
)(1
)()( dtetfs
dfs
edf st
tstt
****L
Laplace Transform Properties
Differentiation Property
Proof via integration by parts again
Second derivative
)0()()(
!!="#$
%&'
fssFdt
tdfL
)0()(
0
)(0
)(
0
)()(
!!=
"#
!
!+
#
!$%
&'(
) !="
#
!
!=
*+,
-./
fssF
dtstetfsstedt
tdfdtste
dt
tdf
dt
tdfL
)0()0()(2
)0()()(
2
)(2
!"!!!=
!!#$%
&'(
=#$%
&'(
)*
+,-
.=
/#
/$%
/&
/'(
fsfsFs
dt
df
dt
tdfs
dt
tdf
dt
d
dt
tfdLLL
Laplace Transform PropertiesGeneral derivative formula
Translation propertiess-domain translation
t-domain translation
)0()0()0()()( )(21 !!!!"!!!=#$
#%&
#'
#() !! mmm
m
m
ffsfssFdt
tfdL
msL
)()}({ !!+=
" sFtfe tL
{ } 0for)()()( >=!!! asFeatuatf as
L
Laplace Transform Properties
Initial Value Property
Final Value Property
Caveats:Laplace transform pairs do not always handle
discontinuities properlyOften get the average value
Initial value property no good with impulsesFinal value property no good with cos, sin etc
)(lim)(lim0
ssFtfst !"+"
=
)(lim)(lim0
ssFtfst !"!
=
Laplace Transform Properties
Multiplication-Convolution Property
Critical property for notion of transfer function
More in a bit…
!
L{ f (t) * g(t)} = F(s)G(s)
Rational FunctionsWe shall mostly be dealing with LTs which are
rational functions – ratios of polynomials in s
pi are the poles and zi are the zeros of the function
K is the scale factor or (sometimes) gain
A proper rational function has n≥mA strictly proper rational function has n>mAn improper rational function has n<m
)())((
)())((
)(
21
21
011
1
011
1
n
m
nn
nn
mm
mm
pspsps
zszszsK
asasasa
bsbsbsbsF
!!!
!!!=
++++
++++=
!!
!!
L
L
L
L
How Laplace transforms are born…in MAE 143A
From linear ODEs:
Under zero initial conditions, Laplace tranforms produce
For any given X(s) we can compute Y(s)!
This is the same as convolution in the time domain:
!
y(n )(t) + a
1y(n"1)(t) + ...+ any(t) = b
0x(m )(t) + b
1x(m"1)
(t) + ...+ bmx(t)
!
sn + a
1sn"1 + ...+ a
n( )Y (s) = b0sm + b
1sm"1 + ...+ b
m( )X(s)
!
Y (s) = H(s)X(s), H(s) :=b0sm
+ b1sm"1
+ ...+ bm
sn
+ a1sn"1
+ ...+ an
!
H(s) = L h(t){ }, Y (s) = H(s)X(s), y(t) = h(" )x(t # " )d"#$
$
%
Example
Compute the impulse response of the RC circuitdescribed by the ODE:
Apply Laplace tranform:
The impulse response is then
!
" y (t) +1
RCy(t) =
1
RCx(t)
!
Y (s) = H(s)X(s), H(s) ="
c
s+"c
, "c:=
1
RC
!
h(t) = L"1H(s){ } =
1
RCe"1
RCt
u(t)
Inverting Laplace Transforms
We have a table of inverse LTsWrite F(s) as a partial fraction expansion
Now appeal to linearity to invert via the tableSurprise!Nastiness: computing the partial fraction expansion is best
done by calculating the residues
!
F(s) =bms
m + bm"1sm"1 +L+ b1s+ b0
ansn + an"1s
n"1 +L+ a1s+ a0
= K(s" z1)(s" z2)L(s" zm)
(s" p1)(s" p2)L(s" pn )
=#1
s" p1( )+
#2
s" p2( )+
#31
(s" p3)+
#32
s" p3( )2
+#33
s" p3( )3
+ ...+#q
s" pq( )
Inverting Laplace Transforms
Compute residues at the poles
Bundle complex conjugate pole pairs into second-order terms ifyou want
but you will need to be careful
Inverse Laplace Transform is a sum of complex exponentials
In Matlab, check out [r,p,k]=residue(b,a), where b = coefficients of numerator; a = coefficients of denominator r = residues; p = poles; k = result of long division
!
(s"# " j$)(s"# + j$) = s2 " 2#s+ # 2 + $ 2( )[ ]
)()(lim sFas
as
!"
!
1
( j "1)!lims#a
dj"1
dsj"1
(s" a)mF(s)[ ]
Strictly Proper Laplace Transforms
Find the inverse LT of)52)(1(
)3(20)( 2 +++
+=
sssssF
!
F(s) =k1
s+1+
k2
s+1" 2 j+
k2
*
s+1+ 2 j
!4
5
2555
21)21)(1(
)3(20)()21(
21lim2
10
1522
)3(20)()1(
1lim1
jej
jsjss
ssFjs
jsk
sss
ssFs
sk
=""=
+"=+++
+="+
+"#=
=
"=++
+=+
"#=
!
f (t) = 10e" t
+ 5 2e("1+2 j )t+ j
5
4#
+ 5 2e("1"2 j )t" j
5
4#$
% &
'
( ) u(t)
= 10e"t
+10 2e"tcos(2t +
5#
4)
$
% & '
( ) u(t)
Laplace Transforms with Multiple Poles
Compute residues at the poles
Example!
1
( j "1)!lims# a
dj "1
dsj "1
(s" a)mF(s)$ % &
' ( )
( ) ( ) ( ) ( )31
3
21
1
1
2
31
3)1(2)1(2
31
522
+!
++
+=
+
!+++=
+
+
ssss
ss
s
ss
3)1(
)52()1(lim 3
23
1−=
+
++
−→ ssss
s1
)1()52()1(lim 3
23
1=
+
++
−→ ssss
dsd
s
2)1(
)52()1(lim!2
13
23
2
2
1=
+
++
−→ ssss
dsd
s
( ) )(32)1(52 23
21 tutte
sss t −+=
+
+ −−L
Not Strictly Proper Laplace Transforms
Find the inverse LT of
Convert to polynomial plus strictly proper rational functionUse polynomial division
Invert as normal
348126)( 2
23
++
+++=
ssssssF
3
5.0
1
5.02
34
22)(
2
++
+++=
++
+++=
sss
ss
sssF
!
f (t) = " # (t) + 2#(t) + 0.5e$t + 0.5e$3t[ ]u(t)
System stability
Easy to determine using the transfer function
System is BIBO stable ifall the poles of the transfer function lie in theopen left half of the s plane
The system is called marginally stable if there aresimple poles on the imaginary axis and no poles inthe right half-plane.
Marginally stable is a particular case of BIBOunstable.
!
H(s)
System interconnections
Cascade interconnection Parallel interconnection!
+
!
H1(s)
!
H2(s)
!
H2(s)
!
H1(s)
!
X(s)
!
X(s)
!
Y (s)
!
Y (s)
!
Y (s)
!
Y (s)
!
X(s)
!
X(s)
!
H1(s)H
2(s)
!
H1(s) +H
2(s)
Feedback interconnection
is the plant or original system we want to control is the sensor/controller selected to make closed-loopsystem have certain property (e.g., stability)
Example:
Much more about this in MAE143B…
!
+
!
+
!
"
!
H1(s)
!
H2(s)
!
X(s)
!
Y (s)
!
H1(s)
1+H1(s)H
2(s)
!
X(s)
!
Y (s)
!
E(s)
!
H1(s)
!
H2(s)
!
H1(s) =
1
s,
!
H2(s) = K,
!
H1(s)
1+H1(s)H
2(s)
=1
s+K
Block diagrams - revisited
Multiple ways of drawing a system block diagram for strictlyproper (n≥m) transfer functions
(if n>m, then some of the are zero)• as a product,
!
H(s) =Y (s)
X(s)=bnsn
+ bn"1s
n"1+L+ b
1s+ b
0
ansn
+ an"1s
n"1+L+ a
1s+ a
0
!
bi
!
H(s)
!
H2(s) =
Y (s)
Y1(s)
= bnsn
+ bn"1s
n"1+L+ b
1s+ b
0
!
H1(s) =
Y1(s)
X(s)=
1
ansn
+ an"1s
n"1+L+ a
1s+ a
0
!
H1(s)H
2(s)
!
H1(s)
!
H2(s)
!
X(s)
!
Y (s)
!
Y1(s)
Block diagrams - revisited
sNY1s( ) =
1
aN
X s( )! aN!1s
N!1Y1s( ) +L+ a
1sY
1s( ) + a
0Y1s( )"# $%{ }
!
H(s)
!
H1(s)
Block diagrams - revisited• in cascade form
• in parallel form
!
H(s) = A(s " z
1)
(s " p1)L(s " zm )
(s " pm )
1
(s " pm+1)L
1
(s " pn )
!
H(s) =K1
(s " p1)
+L+Kn
(s " pn )