The Laplace Transform
Generalizing the Fourier Transform
The CTFT expresses a time-domain signal as a linear
combination of complex sinusoids of the form e jω t . In the
generalization of the CTFT to the Laplace transform, the
complex sinusoids become complex exponentials of the
form e st where s can have any complex value. Replacing
the complex sinusoids with complex exponentials leads to
this definition of the Laplace transform.
L( x ( t )) = X ( s ) =
∞
− st
x
t
e
dt
(
)
∫
−∞
L
x ( t ) ←⎯
→ X(s)
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Generalizing the Fourier Transform
The variable s is viewed as a generalization of the variable ω of
the form s = σ + jω . Then, when σ , the real part of s, is zero, the
Laplace transform reduces to the CTFT. Using s = σ + jω the
Laplace transform is
X(s) =
∞
− (σ + jω )t
x
t
e
dt
(
)
∫
−∞
= F ⎡⎣ x ( t ) e− σ t ⎤⎦
which is the Fourier
transform of x ( t ) e− σ t
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Generalizing the Fourier Transform
The extra factor e− σ 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− jω t dt = A ∫ e− jω t dt ← Does not converge
0
∞
Ae− σ t u ( t ) e− jω t dt = A ∫ e−(σ + jω )t dt ← Converges (if σ > 0)
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0
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4
Complex Exponential Excitation
If a continuous-time LTI system is excited by a complex
exponential x ( t ) = Ae st , where A and s can each be any complex
number, the system response is also a complex exponential of
the same functional form except multiplied by a complex constant.
The response is the convolution of the excitation with the impulse
response and that is
∞
∞
∞
−∞
−∞
x( t ) −∞
y (t ) =
∫ h (τ ) x (t − τ ) dτ =
The quantity H ( s ) =
of h ( t ) .
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s ( t −τ )
st
− sτ
h
τ
Ae
d
τ
=
Ae
h
τ
e
dτ
(
)
(
)
∫
∫
∞
− sτ
h
τ
e
dτ is called the Laplace transform
(
)
∫
−∞
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5
Complex Exponential Excitation
s

( 3− j 2 )t
Let x ( t ) = ( 6 + j3) e

 

= ( 6.708∠0.4637 ) e( 3− j 2 )t
A
1
and let h ( t ) = e u ( t ) . Then H ( s ) =
, σ > −4 and,
s+4
in this case, s = 3 − j2 = σ + jω with σ = 3 > −4 and ω = −2.
6 + j3 ( 3− j 2 )t
y (t ) = x (t ) H ( s ) =
e
= ( 0.6793∠0.742 ) e( 3− j 2 )t .
3 − j2 + 4
The response is the same functional form as the excitation but
multiplied by a different complex constant. This only happens when
the excitation is a complex exponential and that is what makes
complex exponentials unique.
−4t
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Pierre-Simon Laplace
3/23/1749 - 3/2/1827
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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 of
()
y t is
∞
( ) ∫ y (t ) e
Y s =
− st
∞
dt =
−∞
∫ () ()
−∞
⎡ h t ∗ x t ⎤ e− st dt
⎣
⎦
⎛∞
⎞ − st
Y s = ∫ ⎜ ∫ h τ x t − τ dτ ⎟ e dt
⎠
−∞ ⎝ −∞
()
∞
() (
∞
∞
)
() ∫ () ∫ (
Y s =
−∞
1/13/11
h τ dτ
−∞
)
x t − τ e− st dt
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8
The Transfer Function
∞
∞
( ) ∫ h (τ ) dτ ∫ x (t − τ ) e
Y s =
−∞
− st
dt
−∞
Let λ = t − τ ⇒ d λ = dt. Then
∞
∞
() ∫ () ∫ ( )
Y s =
−∞
()
h τ dτ
− s λ +τ
x λ e ( )d λ =
−∞
()
∞
()
∞
( )
− sτ
− sλ
h
τ
e
d
τ
x
λ
e
dλ
∫−∞
∫−∞


 

H( s)
X( s)
() ()
Y s =H s X s
H s is called the transfer function.
() () ()
()
() ()
L
y t = x t ∗ h t ←⎯
→Y s = H s X s
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The Transfer Function
() ()
()
()
()
Let x t = u t and let h t = e−4t u t . Find y t .
∞
() () () ∫ ( ) (
y t = x t ∗h t =
−∞
)
x τ h t − τ dτ =
∞
()
(
)
( )
u
τ
e
u t − τ dτ
∫
−∞
−4 t−τ
t
4t
−4t
⎧ t −4(t−τ )
e
−
1
1−
e
d τ = e−4t ∫ e4τ d τ = e−4t
=
, t >0
⎪∫ e
y t = ⎨0
4
4
0
⎪
, t <0
⎩0
y t = 1 / 4 1− e−4t u t
()
( ) ( )(
) ()
1
1
1
1/ 4 1/ 4
X ( s) = 1 / s , H ( s) =
⇒ Y ( s) = ×
=
−
s+4
s s+4
s
s+4
x ( t ) = (1 / 4 ) (1− e ) u ( t )
−4t
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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.
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11
Direct Form II Realization
A very common form of transfer function is a ratio of two
polynomials in s,
Y( s )
H (s) =
=
X(s)
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k
b
s
∑ k=0 k
N
bN s N + bN −1s N −1 + + b1s + b0
=
N
N
N −1
k
a
s
+
a
s
+ + a1s + a0
N −1
∑ k=0 ak s N
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12
Direct Form II Realization
The transfer function can be conceived as the product of two
transfer functions,
Y1 ( s )
1
H1 ( s ) =
=
X ( s ) aN s N + aN −1s N −1 + + a1s + a0
and
Y( s )
H2 ( s) =
= bN s N + bN −1s N −1 + + b1s + b0
Y1 ( s )
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Direct Form II Realization
From
Y1 ( s )
1
H1 ( s ) =
=
X ( s ) aN s N + aN −1s N −1 + + a1s + a0
we get
X ( s ) = ⎡⎣ aN s N + aN −1s N −1 + + a1s + a0 ⎤⎦ Y1 ( s )
or
X ( s ) = aN s N Y1 ( s ) + aN −1s N −1 Y1 ( s ) + + a1s Y1 ( s ) + a0 Y1 ( s )
Rearranging
1
N
s Y1 ( s ) =
X ( s ) − ⎡⎣ aN −1s N −1 Y1 ( s ) + + a1s Y1 ( s ) + a0 Y1 ( s ) ⎤⎦
aN
{
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}
Direct Form II Realization
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Direct Form II Realization
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Direct Form II Realization
A system is defined by y′′ ( t ) + 3 y′ ( t ) + 7 y ( t ) = x′ ( t ) − 5 x ( t ) .
s−5
H (s) = 2
s + 3s + 7
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Inverse Laplace Transform
There is an inversion integral
σ + j∞
1
st
y t =
Y
s
e
ds , s = σ + jω
∫
j2π σ − 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.
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Existence of the Laplace
Transform
Time Limited Signals
()
x ( t ) is also bounded for all t, the Laplace transform integral
If x t = 0 for t < t0 and t > t1 it is a time limited signal. If
converges and the Laplace transform exists for all s.
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Existence of the Laplace
Transform
()
() (
) (
)
Let x t = rect t = u t + 1 / 2 − u t − 1 / 2 .
∞
1/2
− s/2
s/2
s/2
− s/2
e
−
e
e
−
e
X s = ∫ rect t e− st dt = ∫ e− st dt =
=
, All s
−s
s
−∞
−1/2
()
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Existence of the Laplace
Transform
Right- and Left-Sided Signals
Right-Sided
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Existence of the Laplace
Transform
Right- and Left-Sided Exponentials
Right-Sided
x ( t ) = e u ( t − t 0 ) , α ∈
αt
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Left-Sided
x ( t ) = eβt u ( t 0 − t ) , β ∈
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22
Existence of the Laplace
Transform
Right-Sided Exponential
x ( t ) = eα t u ( t − t 0 ) , α ∈
∞
∞
t0
t0
X ( s ) = ∫ eα t e− st dt = ∫ e(α − σ )t e− jω t dt
If Re ( s ) = σ > α the asymptotic
behavior of e(α − σ )t e− jω t as t → ∞
is to approach zero and the Laplace
transform integral converges.
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Existence of the Laplace
Transform
Left-Sided Exponential
x ( t ) = eβt u ( t 0 − t ) , β ∈
X(s) =
t0
t0
−∞
−∞
β t − st
e
∫ e dt =
( β − σ )t − jω t
e
e dt
∫
If σ < β the asymptotic behavior of
e( β −σ )t e− jω t as t → −∞ is to approach
zero and the Laplace transform
integral converges.
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Existence of the Laplace
Transform
The two conditions σ > α and σ < β define the region of
convergence (ROC) for the Laplace transform of right- and
left-sided signals.
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Existence of the Laplace
Transform
Any right-sided signal that grows no faster than an exponential
in positive time and any left-sided signal that grows no faster
than an exponential in negative time has a Laplace transform.
If x ( t ) = x r ( t ) + x l ( t ) where x r ( t ) is the right-sided part and
x l ( t ) is the left-sided part and if x r ( t ) < K r eα t and x l ( t ) < K l eβ t
and α and β are as small as possible, then the Laplace-transform
integral converges and the Laplace transform exists for α < σ < β .
Therefore if α < β the ROC is the region α < β . If α > β , there is
no ROC and the Laplace transform does not exist.
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Laplace Transform Pairs
The Laplace transform of g1 ( t ) = Aeα t u ( t ) is
G1 ( s ) =
∞
∫ Ae u (t ) e
αt
− st
−∞
∞
dt = A ∫ e
− ( s−α )t
0
∞
(α − σ )t
dt = A ∫ e
e
− jω t
0
A
dt =
s −α
This function has a pole at s = α and the ROC is the region to the
right of that point. The Laplace transform of g 2 ( t ) = Aeβt u ( −t ) is
G2 ( s) =
∞
∫ Ae u ( −t ) e
−∞
βt
0
− st
( β − s )t
dt = A ∫ e
−∞
0
( β − σ )t
dt = A ∫ e
−∞
e
− jω t
A
dt = −
s−β
This function has a pole at s = β and the ROC is the region to the
left of that point.
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Region of Convergence
The following two Laplace transform pairs illustrate the importance
of the region of convergence.
1
L
−α t
e u ( t ) ←⎯→
, σ > −α
s +α
1
L
−α t
− e u ( −t ) ←⎯→
, σ < −α
s +α
The two time-domain functions are different but the algebraic
expressions for their Laplace transforms are the same. Only the
ROC’s are different.
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Region of Convergence
Some of the most common Laplace transform pairs
(There is more extensive table in the book.)
()
L
δ t ←⎯
→ 1 , All σ
()
ramp ( t ) = t u ( t ) ←⎯→ 1/ s , σ > 0
e u ( t ) ←⎯→ 1/ ( s + α ) , σ > −α
ω
e sin (ω t ) u ( t ) ←⎯→
, σ > −α
(s + α ) + ω
s+α
e cos (ω t ) u ( t ) ←⎯→
, σ > −α
(s + α ) + ω
L
u t ←⎯
→ 1/ s , σ > 0
L
−α t
−α t
L
L
0
−α t
2
0
2
2
0
L
0
2
2
0
( )
ramp ( −t ) = −t u ( −t ) ←⎯→ 1/ s , σ < 0
− e u ( −t ) ←⎯→ 1/ ( s + α ) , σ < −α
ω
− e sin (ω t ) u ( −t ) ←⎯→
, σ < −α
(s + α ) + ω
s+α
− e cos (ω t ) u ( −t ) ←⎯→
, σ < −α
(s + α ) + ω
L
− u −t ←⎯
→ 1/ s , σ < 0
2
L
−α t
−α t
L
L
0
−α t
0
2
2
0
L
0
2
2
0
Laplace Transform Example
Find the Laplace transform of
x ( t ) = e−t u ( t ) + e2t u ( −t )
1
e u ( t ) ←⎯→
, σ > −1
s +1
1
L
2t
e u ( −t ) ←⎯→ −
, σ <2
s−2
1
1
L
−t
2t
e u ( t ) + e u ( −t ) ←⎯→
−
, −1< σ < 2
s +1 s − 2
−t
L
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Laplace Transform Example
Find the inverse Laplace transform of
4
10
X(s) =
−
, − 3<σ < 6
s+3 s−6
4
The ROC tells us that
must inverse transform into a
s+3
10
right-sided signal and that
must inverse transform into
s−6
a left-sided signal.
x ( t ) = 4e−3t u ( t ) + 10e6t u ( −t )
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Laplace Transform Example
Find the inverse Laplace transform of
4
10
X(s) =
−
,σ >6
s+3 s−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 )
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Laplace Transform Example
Find the inverse Laplace transform of
4
10
X(s) =
−
, σ < −3
s+3 s−6
The ROC tells us that both terms must inverse transform into a
left-sided signal.
x ( t ) = −4e−3t u ( −t ) + 10e6t u ( −t )
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MATLAB System Objects
A MATLAB system object is a special kind of variable in
MATLAB that contains all the information about a system.
It can be created with the tf command whose syntax is
sys = tf(num,den)
where num is a vector of numerator coefficients of powers of s, den
is a vector of denominator coefficients of powers of s, both in descending
order and sys is the system object.
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MATLAB System Objects
For example, the transfer function
s2 + 4
H1 ( s ) = 5
s + 4s 4 + 7s 3 + 15s 2 + 31s + 75
can be created by the commands
»num = [1 0 4] ; den = [1 4 7 15 31 75] ;
»H1 = tf(num,den) ;
»H1
Transfer function:
s^2 + 4
---------------------------------------s ^ 5 + 4 s ^ 4 + 7 s ^ 3 + 15 s ^ 2 + 31 s + 75
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Partial-Fraction Expansion
The inverse Laplace transform can always be found (in principle at
least) by using the inversion integral. But that is rare in engineering
practice. The most common type of Laplace-transform expression
is a ratio of polynomials in s,
bM s M + bM −1s M −1 + + b1s + b0
G (s) =
s N + aN −1s N −1 +a1s + a0
The denominator can be factored, putting it into the form,
bM s M + bM −1s M −1 + + b1s + b0
G (s) =
( s − p1 )( s − p2 )( s − pN )
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Partial-Fraction Expansion
For now, assume that there are no repeated poles and that N > M ,
making the fraction proper in s. Then it is possible to write the
expression in the partial fraction form,
K1
K2
KN
G (s) =
+
+ +
s − p1 s − p2
s − pN
where
bM s M + bM −1s M −1 +b1s + b0
K1
K2
KN
=
+
+ +
s − pN
( s − p1 )( s − p2 )( s − pN ) s − p1 s − p2
The K’s can be found be any convenient method.
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Partial-Fraction Expansion
Multiply both sides by s − p1
K2
⎡
⎤
K1 + ( s − p1 )
+⎥
M
M −1
⎢
s − p2
bM s + bM −1s
+ + b1s + b0
⎥
( s − p1 ) s − p s − p  s − p = ⎢
( 1 )( 2 ) ( N ) ⎢ + ( s − p ) K N
⎥
1
⎢
⎥
s − pN
⎣
⎦
bM p1M + bM p1M −1 + + b1 p1 + b0
K1 =
( p1 − p2 )( p1 − pN )
All the K’s can be found by the same method and the inverse
Laplace transform is then found by table look-up.
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Partial-Fraction Expansion
H (s) =
10s
K
K
= 1 + 2 , σ > −4
( s + 4 )( s + 9 ) s + 4 s + 9
⎡
⎤
10s
−40
⎡ 10s ⎤
K1 = ⎢ ( s + 4 )
=⎢
=
= −8
⎥
⎥
( s + 4 ) ( s + 9 ) ⎥⎦ s=−4 ⎣ s + 9 ⎦ s=−4 5
⎢⎣
⎡
⎤
10s
−90
⎡ 10s ⎤
K2 = ⎢ ( s + 9)
=⎢
=
= 18
⎥
⎥
( s + 4 ) ( s + 9 ) ⎥⎦ s=−9 ⎣ s + 4 ⎦ s=−9 −5
⎢⎣
−8
18
−8s − 72 + 18s + 72
10s
H (s) =
+
=
=
. Check.
s+4 s+9
( s + 4 )( s + 9 )
( s + 4 )( s + 9 )
↓↓↓↓↓
(
)
h ( t ) = −8e−4t + 18e−9t u ( t )
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Partial-Fraction Expansion
If the expression has a repeated pole of the form,
G (s) =
bM s M + bM −1s M −1 + + b1s + b0
( s − p1 ) ( s − p3 )( s − pN )
2
the partial fraction expansion is of the form,
K12
K11
K3
KN
G (s) =
+
+ +
2 +
s − pN
( s − p1 ) s − p1 s − p3
and K12 can be found using the same method as before.
But K11 cannot be found using the same method.
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Partial-Fraction Expansion
Instead K11 can be found by using the more general formula
(
1
d m− k ⎡
K qk =
s − pq
m− k
⎣
( m − k )! ds
)
m
H ( s )⎤
, k = 1, 2,, m
⎦ s→ pq
where m is the order of the qth pole, which applies to repeated poles
of any order.
If the expression is not a proper fraction in s the partialfraction method will not work. But it is always possible to
synthetically divide the numerator by the denominator until
the remainder is a proper fraction and then apply partial-fraction
expansion.
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Partial-Fraction Expansion
10s
K12
K11
K2
H (s) =
=
+
, σ >4
2
2 +
( s + 4 ) ( s + 9) ( s + 4 ) s + 4 s + 9
↑
Repeated Pole
⎡
2
K12 = ⎢ ( s + 4 )
⎢
⎣
Using
⎤
−40
⎥
=
= −8
( s + 9 ) ⎥⎦ s=−4 5
10s
(s + 4)
(
2
1
d m− k ⎡
K qk =
s − pq
m− k
⎣
( m − k )! ds
)
m
H ( s )⎤
, k = 1, 2,, m
⎦ s→ pq
1
d 2−1
d ⎡ 10s ⎤
2
⎡
⎤
K11 =
s
+
4
H
s
=
(
)
(
)
⎦ s→−4 ds ⎢⎣ s + 9 ⎥⎦
( 2 − 1)! ds 2−1 ⎣
s→−4
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Partial-Fraction Expansion
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Partial-Fraction Expansion
10s 2
H (s) =
, σ > −4 ← Improper in s
( s + 4 )( s + 9 )
10s 2
H (s) = 2
, σ > −4
s + 13s + 36
10
Synthetic Division → s 2 + 13s + 36 10s 2
10s 2 + 130s + 360
− 130s − 360
130s + 360
⎡ −32 162 ⎤
H ( s ) = 10 −
= 10 − ⎢
+
, σ > −4
⎥
( s + 4 )( s + 9 )
⎣s + 4 s + 9⎦
h ( t ) = 10δ ( t ) − ⎡⎣162e−9t − 32e−4t ⎤⎦ u ( t )
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Inverse Laplace Transform Example
Method 1
s
G (s) =
( s − 3) s 2 − 4s + 5
(
)
, σ <2
s
G (s) =
, σ <2
( s − 3) ( s − 2 + j ) ( s − 2 − j )
3 / 2 (3 + j) / 4 (3 − j) / 4
G (s) =
−
−
, σ <2
s−3 s−2+ j
s−2− j
⎛ 3 3t 3 + j ( 2− j )t 3 − j ( 2+ j )t ⎞
g (t ) = ⎜ − e +
e
+
e
u ( −t )
⎟
⎝ 2
⎠
4
4
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Inverse Laplace Transform Example
⎛ 3 3t 3 + j ( 2− j )t 3 − j ( 2+ j )t ⎞
g (t ) = ⎜ − e +
e
+
e
⎟⎠ u ( −t )
⎝ 2
4
4
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 ) ⎤⎦ − e 3t u ( −t )
which has only real values.
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Inverse Laplace Transform Example
Method 2
s
G (s) =
( s − 3) s 2 − 4s + 5
(
)
, σ <2
s
G (s) =
, σ <2
( s − 3) ( s − 2 + j ) ( s − 2 − j )
3 / 2 (3 + j) / 4 (3 − j) / 4
G (s) =
−
−
, σ <2
s−3 s−2+ j
s−2− j
Getting a common denominator and simplifying
3 / 2 1 6s − 10
3/2 6 s− 5 / 3
G (s) =
−
=
−
, σ <2
2
2
s − 3 4 s − 4s + 5 s − 3 4 ( s − 2 ) + 1
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Inverse Laplace Transform Example
Method 2
3/2 6 s − 5 / 3
G (s) =
−
, σ <2
2
s − 3 4 ( s − 2) + 1
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 the
correct expression from that term and factoring we can put it into
the form
3/2 3⎡ s − 2
1/ 3 ⎤
G (s) =
− ⎢
+
⎥ , σ <2
2
2
s − 3 2 ⎣ ( s − 2) + 1 ( s − 2) + 1⎦
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Inverse Laplace Transform Example
Method 2
3/2 3⎡ s − 2
1/ 3 ⎤
G (s) =
− ⎢
+
⎥ , σ <2
2
2
s − 3 2 ⎣ ( s − 2) + 1 ( s − 2) + 1⎦
This can now be directly inverse Laplace transformed into
{
}
g ( t ) = ( 3 / 2 ) e2t ⎡⎣ cos ( t ) + (1 / 3) sin ( t ) ⎤⎦ − e 3t u ( −t )
which is the same as the previous result.
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Inverse Laplace Transform Example
Method 3
When we have a pair of poles p2 and p3 that are complex conjugates
A
K2
K3
we can convert the form G ( s ) =
+
+
into the
s − 3 s − p2 s − p3
s ( K 2 + K 3 ) − K 3 p2 − K 2 p3
A
A
Bs + C
form G ( s ) =
+
=
+ 2
2
s−3
s − ( p1 + p2 ) s + p1 p2
s − 3 s − ( p1 + p2 ) s + p1 p2
In this example we can find the constants A, B and C by realizing that
s
A
Bs + C
G (s) =
≡
+ 2
, σ <2
2
( s − 3) s − 4s + 5 s − 3 s − 4s + 5
(
)
is not just an equation, it is an identity. That means it must be an
equality for any value of s.
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Inverse Laplace Transform Example
Method 3
A can be found as before to be 3 / 2. Letting s = 0, the
3/2 C
identity becomes 0 ≡ −
+ and C = 5 / 2. Then, letting
3
5
s = 1, and solving we get B = −3 / 2. Now
3 / 2 ( −3 / 2 ) s + 5 / 2
G (s) =
+
, σ <2
2
s−3
s − 4s + 5
or
3/2 3 s − 5 / 3
G (s) =
−
, σ <2
2
s − 3 2 s − 4s + 5
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.
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Use of MATLAB in Partial Fraction
Expansion
MATLAB has a function residue that can be very helpful in
partial fraction expansion. Its syntax is [r,p,k] = residue(b,a)
where b is a vector of coefficients of descending powers of s
in the numerator of the expression and a is a vector of coefficients
of descending powers of s in the denominator of the expression,
r
is a vector of residues, p is a vector of finite pole locations and
k is 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.
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Laplace Transform Properties
L
Let g ( t ) and h ( t ) form the transform pairs, g ( t ) ←⎯
→ G (s)
L
and h ( t ) ←⎯
→ H ( s ) with ROC's, ROCG and ROCH respectively.
Linearity
Time Shifting
L
α g ( t ) + β h ( t ) ←⎯
→α G ( s) + β H ( s)
ROC ⊇ ROCG ∩ ROCH
L
g ( t − t 0 ) ←⎯
→ G ( s ) e− st0
ROC = ROCG
s-Domain Shift
L
e s0t g ( t ) ←⎯
→ G ( s − s0 )
ROC = ROCG shifted by s0 ,
(s is in ROC if s − s0 is in ROCG )
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Laplace Transform Properties
L
g ( at ) ←⎯
→ (1 / a ) G ( s / a )
Time Scaling
Time Differentiation
s-Domain Differentiation
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ROC = ROCG scaled by a
(s is in ROC if s / a is in ROCG )
d
L
g ( t ) ←⎯
→ s G (s)
dt
ROC ⊇ ROCG
d
L
− t g ( t ) ←⎯→ G ( s )
ds
ROC = ROCG
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54
Laplace Transform Properties
L
g ( t ) ∗ h ( t ) ←⎯
→ G (s) H (s)
Convolution in Time
ROC ⊇ ROCG ∩ ROCH
t
L
g
τ
d
τ
←
⎯
→ G (s) / s
(
)
∫
Time Integration
−∞
ROC ⊇ ROCG ∩ (σ > 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 + = lim s G ( s )
Final Value Theorem:
lim g ( t ) = lim s G ( s ) if lim g ( t ) exists
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s→∞
s→0
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t→∞
55
Laplace Transform Properties
Final Value Theorem
limg ( t ) = lim s G ( s )
t→∞
s→0
This theorem only applies if the limit lim g ( t ) actually exists.
t→∞
It is possible for the limit lim s G ( s ) to exist even though the
s→0
limit lim g ( t ) does not exist. For example
t→∞
s
x ( t ) = cos (ω 0t ) ←⎯→ X ( s ) = 2
s + ω 02
L
s2
lim s X ( s ) = lim 2
=0
2
s→0
s→0 s + ω
0
but lim cos (ω 0t ) does not exist.
t→∞
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Laplace Transform Properties
Final Value Theorem
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 s plane. 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 s plane. G ( s ) could have a single
pole at s = 0 and the final value theorem would still apply.
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Use of Laplace Transform
Properties
Find the Laplace transforms of x ( t ) = u ( t ) − u ( t − a ) and
L
x ( 2t ) = u ( 2t ) − u ( 2t − a ) . From the table u ( t ) ←⎯
→ 1 / s, σ > 0.
L
Then, using the time-shifting property u ( t − a ) ←⎯
→ e− as / s, σ > 0.
(
)
L
Using the linearity property u ( t ) − u ( t − a ) ←⎯
→ 1 − e− as / s, σ > 0.
Using the time-scaling property
1 ⎡ 1 − e− as ⎤
1 − e− as/2
u ( 2t ) − u ( 2t − a ) ←⎯→ ⎢
=
,σ >0
⎥
2 ⎣ s ⎦ s→s/2
s
L
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Use of Laplace Transform
Properties
Use the s-domain differentiation property and
L
u ( t ) ←⎯
→1 / s , σ > 0
to find the inverse Laplace transform of 1 / s 2 . The s-domain
d
L
differentiation property is − t g ( t ) ←⎯
→ ( G ( s )) . Then
ds
d ⎛ 1⎞
1
L
−t u ( t ) ←⎯→ ⎜ ⎟ = − 2 . Then using the linearity property
ds ⎝ s ⎠
s
1
L
t u ( t ) ←⎯→ 2 .
s
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The Unilateral Laplace
Transform
In 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 dt . The only difference between
0
this 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 s plane to the right of all
the finite poles.
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60
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 avoid
ambiguity 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.
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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
σ + j∞
1
+ st
g (t ) =
G
s
e
ds
(
)
∫
j2π σ − j∞
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The Unilateral Laplace
Transform
Some of the properties of the unilateral Laplace transform are
different from the bilateral Laplace transform.
Time-Shifting
L
g ( t − t 0 ) ←⎯
→ G ( s ) e− st0 , t 0 > 0
L
g ( at ) ←⎯
→ (1 / a ) G ( s / a ) , a > 0
Time Scaling
( )
d
L
g ( t ) ←⎯
→ s G ( s ) − g 0−
dt
First Time Derivative
Nth Time Derivative
n −1
N
⎤
dN
L
N
N −n ⎡ d
g ( t )) ←⎯→ s G ( s ) − ∑ s ⎢ n −1 ( g ( t )) ⎥
N (
dt
n =1
⎣ dt
⎦t = 0−
t
Time Integration
L
g
τ
d
τ
←
⎯
→ G (s) / s
(
)
∫
0−
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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.
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The Unilateral Laplace
Transform
The 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,
d2
d
x
t
+
7
(
)
⎤⎦
⎡⎣ x ( t ) ⎤⎦ + 12 x ( t ) = 0
⎣
2 ⎡
dt
dt
d
−
with initial conditions x 0 = 2 and
x ( t ))t = 0− = −4.
(
dt
Laplace transforming both sides of the equation, using the new
( )
derivative property for unilateral Laplace transforms,
d
s 2 X ( s ) − s x 0 − − ( x ( t ))t = 0− + 7 ⎡⎣ s X ( s ) − x 0 − ⎤⎦ + 12 X ( s ) = 0
dt
( )
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The Unilateral Laplace
Transform
Solving for X ( s )
= −4





d
s x 0 − + 7 x 0 − + ( x ( t ))t = 0−
dt
X(s) =
s 2 + 7s + 12
2s + 10
4
2
or X ( s ) = 2
=
−
. The inverse transform yields
s + 7s + 12 s + 3 s + 4
x ( t ) = 4e−3t − 2e−4t u ( t ) . This solution solves the differential equation
=2
( )
(
=2
( )
)
with the given initial conditions.
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Pole-Zero Diagrams and
Frequency Response
If the transfer function of a stable system is H(s), the frequency
response is H(jω ). The most common type of transfer function
is of the form,
s − z1 ) ( s − z2 )( s − zM )
(
H (s) = A
( s − p1 )( s − p2 )( s − pN )
Therefore H(jω ) is
jω − z1 ) ( jω − z2 )( jω − zM )
(
H ( jω ) = A
( jω − p1 )( jω − p2 )( jω − pN )
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Pole-Zero Diagrams and
Frequency Response
3s
Let H ( s ) =
.
s+3
jω
H ( jω ) = 3
jω + 3
The numerator jω and the
denominator jω + 3 can be
conceived as vectors in the
s plane.
jω
H ( jω ) = 3
jω + 3
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 H ( jω ) = 3
 + jω −  ( jω + 3)
=0
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Pole-Zero Diagrams and
Frequency Response
jω
lim− H ( jω ) = lim− 3
=0
ω →0
ω →0
jω + 3
jω
lim+ H ( jω ) = lim+ 3
=0
ω →0
ω →0
jω + 3
jω
lim H ( jω ) = lim 3
=3
ω →−∞
ω →−∞
jω + 3
jω
lim H ( jω ) = lim 3
=3
ω →+∞
ω →+∞
jω + 3
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Pole-Zero Diagrams and
Frequency Response
π
π
lim  H ( jω ) = − − 0 = −
ω →0 −
2
2
π
π
lim+  H ( jω ) = − 0 =
ω →0
2
2
π ⎛ π⎞
lim  H ( jω ) = − − ⎜ − ⎟ = 0
ω →−∞
2 ⎝ 2⎠
π π
lim  H ( jω ) = − = 0
ω →+∞
2 2
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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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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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