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The Laplace Transform
M. J. Roberts - All Rights Reserved.
Edited by Dr. Robert Akl
1
Generalizing the Fourier Transform
The CTFT expresses a time-domain signal as a linear
combination of complex sinusoidsof the form ej!t
. In thegeneralization of the CTFT to the Laplace transform, the
complex sinusoids become complex exponentialsof the
form est
where scan have any complex value. Replacing
the complex sinusoids with complex exponentials leads to
this definition of the Laplace transform.
L x t( )( ) = X s( ) = x t( )e" stdt"#
#
$
x t( ) L% &' X s( )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 2
Generalizing the Fourier Transform
The variable sis viewed as a generalization of the variable !of
the form s = "+ j!. Then, when ",the real part of s, is zero, the
Laplace transform reduces to the CTFT. Using s = "+ j!the
Laplace transform is
X s( ) = x t( )e# " +j!( )t dt#$
$
%
= F x t( )e#"t&' ()which is the Fourier
transform of x t( )e#"t
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 3
Generalizing the Fourier TransformThe extra factor e!
"t is sometimes called a convergence factor
because, when chosen properly, it makes the integral converge
for some signals for which it would not otherwise converge.
For example, strictly speaking, the signal A u t( ) does not have
a CTFT because the integral does not converge. But if it is
multiplied by the convergence factor, and the real part of s
is chosen appropriately, the CTFT integral will converge.
A u t( )e!j#t dt!$
$
% = A e!j#t dt
0
$
% &Does not converge
Ae!"t u t( )e!j#t dt
!$
$
% =A e! " +j#( )t
dt0
$
% &Converges (if "> 0)
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 4
Complex Exponential Excitation
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 5
Complex Exponential Excitation
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 6
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Pierre-Simon Laplace
3/23/1749 - 3/2/1827M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 7
The Transfer Function
Let x t( ) be the excitation and let y t( ) be the response of a
system with impulse response h t( ). The Laplace transform ofy t( ) is
Y s( ) = y t( )e!stdt!"
"
# = h t( )$x t( )%& '(e!stdt!"
"
#
Y s( ) = h )( )x t!)( )d)!"
"
#*
+,
-
./e
!stdt
!"
"
#
Y s( ) = h )( )d)!"
"
# x t!)( )e!stdt!"
"
#
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 8
The Transfer Function
Let x t( ) = u t( ) and let h t( ) =e!4tu t( ). Find yt( ).
yt( ) = x t( )"h t( ) = x #( )h t! #( )d#!$
$
% = u #( )e!4t!#( )
u t! #( )d#!$
$
%
yt( ) =e!4t!#( )
d#
0
t
% =e!4t
e4#
d#
0
t
% =e!4te
4t!1
4=
1!e!4t
4 , t> 0
0 , t< 0
&
'(
)(
yt( ) = 1/ 4( ) 1!e!4t( )u t( )
X s( ) = 1/ s, H s( ) =1
s+ 4*Y s( ) =
1
s+
1
s+ 4=
1/ 4
s!
1/ 4
s + 4
x t( ) = 1/ 4( ) 1!e!4t( )u t( )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 9
Cascade-Connected Systems
If two systems are cascade connected the transfer function of
the overall system is the product of the transfer functions of the
two individual systems.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 10
Direct Form II Realization
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 11
Direct Form II Realization
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 12
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Direct Form II Realization
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 13
Direct Form II Realization
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 14
Direct Form II Realization
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 15
Direct Form II Realization
A system is defined by !!y t( )+ 3 !y t( )+ 7y t( ) = !x t( )" 5 x t( ).
H s( ) =s " 5
s2+ 3s + 7
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 16
Inverse Laplace Transform
There is an inversion integral
y t( ) =1
j2!Y s( )estds
"#j$
"+j$
% , s = "+ j&
for finding y t( ) from Y s( ), but it is rarely used in practice.Usually inverse Laplace transforms are found by using tables
of standard functions and the properties of the Laplace transform.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 17
Existence of the Laplace
TransformTime Limited Signals
If x t( ) = 0 for t< t0 and t> t1it is a time limitedsignal. If
x t( ) is also bounded for all t, the Laplace transform integralconverges and the Laplace transform exists for alls.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 18
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Existence of the Laplace
Transform
Let x t( ) = rect t( ) = u t+ 1/ 2( )! u t!1/ 2( ).
X s( ) = rect t( )e!stdt!"
"
# = e!st
dt
!1/2
1/2
# =e!s/2 ! es/2
!s=
es/2 ! e!s/2
s, Alls
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 19
Existence of the Laplace
TransformRight- and Left-Sided Signals
Right-Sided Left-Sided
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 20
Existence of the Laplace
TransformRight- and Left-Sided Exponentials
Right-Sided Left-Sidedx t( ) = e!t u t" t0( ) , !#! x t( ) = e!t u t0 " t( ) , !#!
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 21
Existence of the Laplace
Transform
Right-Sided Exponentialx t( ) =e!t u t" t0( ) , !#!
X s( ) = e!te" stdtt0
$
% = e! "&( )t
e" j'tdtt0
$
%
If Re s( ) = & >!the asymptotic
behavior of e! "&( )t
e" j't
as t($
is to approach zero and the Laplace
transform integral converges.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 22
Existence of the Laplace
Transform
Left-Sided Exponentialx t( ) =e!t u t0 " t( ) , !#!
X s( ) = e!te"stdt"$
t0
% = e!"&( )t
e" j't
dt"$
t0
%
If &< !the asymptotic behavior of
e!"&( )t
e" j't as t("$is to approach
zero and the Laplace transform
integral converges.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 23
Existence of the Laplace
TransformThe two conditions !> "and !< #define the region of
convergence(ROC) for the Laplace transform of right- and
left-sided signals.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 24
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Existence of the Laplace
Transform
Any right-sided signal that grows no faster than an exponentialin positive time and any left-sided signal that grows no faster
than an exponential in negative time has a Laplace transform.
If x t( ) = xr t( ) + x
l t( ) where x
r t( ) is the right-sided part and
xl t( ) is the left-sided part and if x
r t( ) < K
re!t and x
l t( ) < K
le"t
and !and "are as small as possible, then the Laplace-transform
integral converges and the Laplace transform exists for !< # &' & e
&'tsin (0t( )u &t( ) L" #$
(0
s+'( )2
+(0
2, %< &'
e&'tcos (
0t( )u t( ) L" #$
s+'
s+'( )2
+(0
2
, %> &' & e&'tcos (
0t( )u &t( ) L" #$
s+'
s+'( )2
+(0
2
, %< &'
Some of the most common Laplace transform pairs(There is more extensive table in the book.)
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 28
Laplace Transform Example
Find the Laplace transform of
x t( ) = e!t u t( )+ e2t u !t( )
e!t u t( ) L" #$ 1
s +1, %> !1
e2t u !t( ) L" #$ !
1
s ! 2 , %< 2
e!t u t( )+ e2t u !t( ) L" #$
1
s +1!
1
s ! 2 , ! 1
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Laplace Transform Example
Find the inverse Laplace transform of
X s( ) =4
s + 3!
10
s ! 6, " > 6
The ROC tells us that both terms must inverse transform into a
right-sided signal.
x t( ) = 4e!3t u t( )!10e6t u t( )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 31
Laplace Transform Example
Find the inverse Laplace transform of
X s( ) =4
s + 3!
10
s ! 6, "
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Partial-Fraction Expansion
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 37
Partial-Fraction ExpansionH s( ) =
10s
s + 4( ) s + 9( )=
K1
s + 4+
K2
s + 9 , ! >"4
K1= s + 4( )
10s
s + 4( ) s + 9( )
#
$%%
&
'((s="4
=10s
s + 9
#
$%&
'(s="4=
"40
5="8
K2= s + 9( )
10s
s + 4( ) s + 9( )
#
$%%
&
'((s="9
=10s
s + 4
#
$%&
'(s="9=
"90
"5=18
H s( ) ="8
s + 4+
18
s + 9=
"8s" 72+18s + 72
s + 4( ) s + 9( ) =
10s
s + 4( ) s + 9( ) . Check.
) ) ) ) )
h t( ) = "8e"4 t +18e"9t( )u t( )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 38
Partial-Fraction Expansion
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 39
Partial-Fraction Expansion
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 40
Partial-Fraction Expansion
H s( ) =10s
s + 4( )2s + 9( )
=K12
s + 4( )2 +
K11
s + 4+
K2
s + 9 , ! > 4
"
Repeated Pole
K12 = s + 4( )2 10s
s +4( )
2s +
9( )
#
$
%%
&
'
((s=)4
=)40
5=)8
Using
Kqk=1
m )k( )!d
m)k
dsm)k
s ) pq( )m
H s( )#$
&'s*pq
, k= 1,2,!,m
K11 =1
2 )1( )!
d2)1
ds2)1 s + 4( )
2H s( )#$
&'s*)4
=d
ds
10s
s + 9
#
$%&
'(s*)4
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 41
Partial-Fraction Expansion
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 42
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Partial-Fraction Expansion
H s( ) =10s
2
s + 4( ) s + 9( ) , !> "4# Improper in s
H s( ) =10s
2
s2+13s + 36
, ! >"4
Synthetic Division$ s 2 +13s + 36 10s2
10s2+130s + 360
"130s" 360
10
H s( ) =10 "130s + 360
s + 4( ) s + 9( )= 10 "
"32
s + 4+
162
s + 9
%
&'(
)* , !> "4
h t( ) = 10+ t( )" 162e"9t " 32e"4 t%& ()u t( )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 43
Method 1
G s( ) =s
s! 3( ) s2 ! 4s + 5( ) , " < 2
G s( ) =s
s! 3( ) s!2 + j( ) s!2 ! j( ) , "< 2
G s( ) =3 / 2
s! 3!
3+ j( ) / 4
s!2 + j!
3! j( ) / 4s!2 ! j
, " < 2
g t( ) = !3
2e
3t+
3+ j
4e
2!j( )t+
3! j4
e2+j( )t#
$% &
'(u !t( )
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 44
g t( ) = !3
2e
3t+
3+ j
4e
2!j( )t+
3! j4
e2+j( )t"
#$ %
&'u ! t( )
This looks like a function of time that is complex-valued. But,
with the use of some trigonometric identities it can be put into
the form
g t( ) = 3 / 2( ) e2t cos t( )+ 1 / 3( )sin t( )() *+!e3t{ }u ! t( )
which has only real values.
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 45
Method 2
G s( ) =s
s ! 3( ) s2 ! 4s + 5( ) , " < 2
G s( ) =s
s ! 3( ) s ! 2 + j( ) s ! 2 ! j( ) , " < 2
G s( ) =3 / 2
s ! 3!
3+ j( ) / 4
s ! 2 + j!
3! j( ) / 4
s ! 2 ! j , " < 2
Getting a common denominator and simplifying
G s( ) =3 / 2
s ! 3!
1
4
6s !10
s2! 4s + 5
=3 / 2
s ! 3!
6
4
s ! 5 / 3
s ! 2( )2+1
, " < 2
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 46
Method 2
G s( ) =3 / 2
s! 3!
6
4
s! 5 / 3
s! 2( )2+1
, "< 2
The denominator of the second term has the form of the Laplace
transform of a damped cosine or damped sine but the numerator
is not yet in the correct form. But by adding and subtracting thecorrect expression from that term and factoring we can put it into
the form
G s( ) =3 / 2
s! 3!
3
2
s! 2
s! 2( )2+1
+1 / 3
s! 2( )2+1
#
$%
&
'( , " < 2
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 47
Method 2
G s( ) =3 / 2
s! 3!
3
2
s! 2
s! 2( )2+1
+1 / 3
s! 2( )2+1
"
#$
%
&' , (< 2
This can now be directly inverse Laplace transformed into
g t( ) = 3 / 2( ) e2t cos t( )+ 1 / 3( )sin t( )"# %&!e3t{ }u !t( )
which is the same as the previous result.
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 48
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Method 3
When we have a pair of poles p2 and p3 that are complex conjugates
we can convert the form G s( ) =A
s ! 3+
K2
s ! p2+
K3
s ! p3into the
form G s( ) =A
s ! 3+
s K2 +
K3( )! K3p2 ! K2p3
s2! p1 + p2( ) s + p1p2
=A
s ! 3+
Bs + C
s2! p1 + p2( )s +p1p2
In this example we can find the constants A, Band Cby realizing that
G s( ) =s
s ! 3( ) s2 ! 4s + 5( )"
A
s ! 3+
Bs + C
s2! 4s + 5
, #< 2
is not just an equation, it is an identity. That means it must be an
equality for any value of s.
Inverse Laplace Transform Example
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 49
Inverse Laplace Transform Example Method 3
Acan be found as before to be 3/ 2. Letting s = 0, the
identity becomes 0 ! "3 / 2
3+C
5and C= 5 / 2. Then, letting
s = 1, and solving we get B = " 3/ 2. Now
G s( ) =3 / 2
s " 3+
"3 / 2( )s + 5 / 2s
2" 4s + 5
, #< 2
or
G s( ) =3 / 2
s " 3"
3
2
s " 5 / 3
s2
" 4s + 5 , #< 2
This is the same as a result in Method 2 and the rest of the solution
is also the same. The advantage of this method is that all the
numbers are real.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 50
Use of MATLAB in Partial Fraction
Expansion
MATLAB has a function residuethat can be very helpful in
partial fraction expansion. Its syntax is[r,p,k] = residue(b,a)
where bis a vector of coefficients of descending powers of s
in the numerator of the expression and ais a vector of coefficients
of descending powers of sin the denominator of the expression,
ris a vector of residues, pis a vector of finite pole locations and
kis a vector of so-called direct terms which result when the
degree of the numerator is equal to or greater than the degree
of the denominator. For our purposes, residues are simply the
numerators in the partial-fraction expansion.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 51
Laplace Transform Properties
Let g t( ) and h t( ) form the transform pairs, g t( ) L! "# G s( )
and h t( ) L! "# H s( ) with ROC's, ROCG and ROCH respectively.
Linearity $g t( )+ %h t( ) L! "# $G s( )+ %H s( )
ROC & ROCG ' ROCH
Time Shifting g t ( t0( )L
! "# G s( )e(st0
ROC = ROCG
s-Domain Shift es0t g t( ) L! "# G s ( s0( )
ROC = ROCG shifted by s0,
(sis in ROC if s ( s0 is in ROCG )
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 52
Time Scaling g at( ) L! "# 1/ a( )G s /a( ) ROC = ROCG scaled by a
(sis in ROC if s /ais in ROCG )
Time Differentiationd
dtg t( ) L! "# sG s( )
ROC $ ROCG
s-Domain Differentiation % tg t( ) L! "#d
dsG s( )
ROC= ROCG
Laplace Transform Properties
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 53
Convolution in Time g t( )!h t( ) L" #$ G s( )H s( )
ROC % ROCG & ROCH
Time Integration g '( )d'()
t
*L" #$ G s( ) /s
ROC % ROCG & + > 0( )
If g t( ) = 0 , t< 0 and there are no impulses or higher-order
singularities at t= 0 then
Initial Value Theorem: g 0+( ) = lims#)
sG s( )
Final Value Theorem: limt#)
g t( ) = lims#0
sG s( ) if limt#)
g t( ) exists
Laplace Transform Properties
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 54
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FinalValueTheorem limt!"
g t( ) = lims!
0
sG s( )
This theorem only applies if the limit limt!"
g t( ) actually exists.
It is possible for the limit lims!0
sG s( ) to exist even though the
limit limt!"
g t( ) does not exist. For example
x t( ) = cos #0t( )L
$ !% X s( ) =s
s2+#0
2
lims!0
s X s( ) = lims!0
s2
s2+#0
2 = 0
but limt!"
cos #0t( ) does not exist.
Laplace Transform Properties
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 55
Final Value TheoremLaplace Transform Properties
The final value theorem applies to a function G s( ) if all the
poles of sG s( ) lie in the open left half of the splane. Be sure
to notice that this does not say that all the poles of G s( ) must
lie in the open left half of the splane. G s( ) could have a single
pole at s = 0 and the final value theorem would still apply.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 56
Use of Laplace Transform
Properties
Find the Laplace transforms of x t( ) = u t( ) ! u t! a( ) and
x 2t( ) = u 2t( ) ! u 2t! a( ). From the table u t( ) L" #$ 1 /s, %> 0.
Then, using the time-shifting property u t! a( ) L" #$ e!as / s, % > 0.
Using the linearity property u t( ) ! u t! a( ) L" #$ 1! e!as( ) /s, %> 0.Using the time-scaling property
u 2t( ) ! u 2t! a( ) L" #$1
2
1! e!as
s
&
'(
)
*+
s#s/2
=1! e!as/2
s
, %> 0
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 57
Use of Laplace Transform
Properties
Use the s-domain differentiation property and
u t( ) L! "# 1 /s , $ > 0
to find the inverse Laplace transform of 1/ s2. The s-domain
differentiation property is %tg t( ) L! "# d
dsG s( )( ). Then
%tu t( ) L! "# d
ds
1
s
&'(
)*+= %
1
s2
. Then using the linearity property
tu t( ) L! "#1
s2
.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 58
The Unilateral Laplace
TransformIn most practical signal and system analysis using the Laplace
transform a modified form of the transform, called the unilateral
Laplace transform, is used. The unilateral Laplace transform is
defined by G s( ) = g t( )e!st dt0!
"
# . The only difference betweenthis version and the previous definition is the change of the lower
integration limit from ! "to 0 !. With this definition, all the
Laplace transforms of causal functions are the same as before
with the same ROC, the region of the splane to the right of all
the finite poles.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 59
The Unilateral Laplace
Transform
The unilateral Laplace transform integral excludes negative time.
If a function has non-zero behavior in negative time its unilateral
and bilateral transforms will be different. Also functions with the
same positive time behavior but different negative time behavior
will have the same unilateral Laplace transform. Therefore, to avoidambiguity and confusion, the unilateral Laplace transform should
only be used in analysis of causal signals and systems. This is a
limitation but in most practical analysis this limitation is not significant
and the unilateral Laplace transform actually has advantages.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 60
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The Unilateral Laplace
Transform
The main advantage of the unilateral Laplace transform is that
the ROC is simpler than for the bilateral Laplace transform and,
in most practical analysis, involved consideration of the ROC is
unnecessary. The inverse Laplace transform is unchanged. It is
g t( ) =1
j2!G s( )e+stds
"#j$
"+j$
%
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 61
The Unilateral Laplace
Transform
Some of the properties of the unilateral Laplace transform aredifferent from the bilateral Laplace transform.
Time-Shifting g t! t0( )L" #$ G s( )e!st0 , t0 > 0
Time Scaling g at( ) L" #$ 1 / a( )G s /a( ) , a > 0
First Time Derivatived
dtg t( ) L" #$ sG s( )! g 0!( )
Nth Time Derivatived
N
dtN
g t( )( ) L" #$ sN G s( )! sN!n d
n!1
dtn!1
g t( )( )%
&'
(
)*t=0!n=1
N
+
Time Integration g ,( )d,0!
t
-L" #$ G s( ) /s
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 62
The Unilateral Laplace
Transform
The time shifting property applies only for shifts to the right because
a shift to the left could cause a signal to become non-causal. For the
same reason scaling in time must only be done with positive scaling
coefficients so that time is not reversed producing an anti-causal function.
The derivative property must now take into account the initial value
of the function at time t = 0! and the integral property applies only to
functional behavior after time t = 0. Since the unilateral and bilateral
Laplace transforms are the same for causal functions, the bilateral table
of transform pairs can be used for causal functions.
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 63
The Unilateral Laplace
TransformThe Laplace transform was developed for the solution of differential
equations and the unilateral form is especially well suited for solving
differential equations with initial conditions. For example,
d
2
dt2
x t( )!" #$ + 7d
dtx t( )!" #$ +12x t( ) = 0
with initial conditions x 0%( ) = 2 and d
dtx t( )( )
t=0% =%4.
Laplace transforming both sides of the equation, using the new
derivative property for unilateral Laplace transforms,
s2 X s( )%s x 0%( )%
d
dtx t( )( )
t=0% + 7 s X s( )% x 0
%( )!" #$ +12 X s( ) = 0
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 64
The Unilateral Laplace
Transform
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 65
Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
lim!"0
+ H j!( ) = lim!"0+ 3 j!
j!+ 3=0lim
!"0# H j!( ) = lim!"0# 3
j!
j!+ 3=0
lim!"+#
H j!( ) = lim!"+#
3 j!
j!+ 3= 3lim
!"#$
H j!( ) = lim!"#$
3 j!
j!+ 3= 3
M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 68
Pole-Zero Diagrams and
Frequency Response
lim!"0+!H j!( ) =#
2$ 0 =
#
2lim!"0#!H j!( ) = #$
2# 0 = #
$
2
lim!"#$!H j!( ) =#%
2# #
%
2
&'(
)*+ = 0 lim!"+#!H j!( ) =
$
2%$
2=0
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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Pole-Zero Diagrams and
Frequency Response
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