ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Chapter 2
Laplace Transform
A. Bazoune
2.1 INTRODUCTION
Laplace Transform is one of the most important mathematical tools available for
modeling and analyzing linear systems.
2.2 COMPLEX NUMBERS, COMPLEX VARIABLES, AND
COMPLEX FUNCTIONS
Complex Numbers
Using
the
notation
j = −1 ,
one
can
express all complex numbers in engineering calculations as
z = x + jy
where
x
is the real part and
are real and that
j
jy
is the imaginary part. Notice that both
x
and
y
is the only imaginary quantity in the expression above.
z
y
θ
x
Fig. 2-1 Complex plane representation of a complex number
z.
The Magnitude, or absolute value, of z is defined as the length of the directed
segment shown in Fig. 2-1.
Magnitude of z = z = x 2 + y 2
The angle of z is the angle that the directed line segment makes with the positive
real axis. A counterclockwise rotation is defined as the positive direction for
the measurement of angles.
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
angle of z = θ = tan
−1
(y x)
A complex number can be written in rectangular form as:
⎫
⎬
z ( cos θ + j sin θ ) ⎭
z = x + jy
z=
rectangular form
and in polar form as
z = z ∠θ
z= ze
⎫
⎬
⎭
jθ
polar form
In converting complex numbers from rectangular to polar from , we use
z= z =
2
⎛y⎞
⎟
⎝x⎠
θ = tan ⎜
−1
x +y ,
2
To convert complex numbers from polar to rectangular form, we employ
x = z cos θ ,
y = z sin θ
Complex Conjugate.
defined as
The complex conjugate of
z = x + jy is
z = x − jy
The complex conjugate of z thus has the same real part as z and an imaginary
part that is the negative of the imaginary part of z as shown in Fig. 2-2. Notice
that
z
y
θ
x
−y
z
Fig. 2-2 Complex number
z
and its complex conjugate
2/36
z.
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
z = x + jy = z ∠θ = z ( cos θ + j sin θ )
z = x − jy = z ∠ ( −θ ) = z ( cos θ − j sin θ )
The power series expansions of cos θ
Euler’s Theorem.
sin θ are, respectively,
θ
cos θ = 1 −
2
+
θ
2!
and
sin θ = θ −
θ
θ
3
+
3!
Thus
cos θ + j sin θ = 1 + jθ +
4
−
4!
5
−
Since
e = 1+ x +
x
it follows that
x
+L
( jθ )
+
2!
+L
7
7!
2
6
6!
θ
5!
( jθ )
θ
3
+
( jθ )
3!
2
+
2!
x
and
4!
4
+L
3
+L
3!
cos θ + j sin θ = e
jθ
Using the above relation, one can express the sine and cosine in complex form.
Noting that
e − jθ
is the complex conjugate of
e jθ
and that
jθ
e = cos θ + j sin θ
e
− jθ
= cos θ − j sin θ
By adding the above expressions together, we find that
jθ
cos θ =
e +e
− jθ
2
while by subtracting the second expression above from the first one, we obtain
jθ
sin θ =
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e −e
2j
− jθ
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Complex Algebra.
Equality of complex numbers.
Two complex numbers
z1
and
z2
are said to be equal if and only if their real parts are equal and their imaginary parts
are equal. So if two complex numbers are written
z1 = x1 + jy1 ,
Then
z1 = z2
if and only if
x1 = x2
and
and
z2 = x2 + jy 2
y1 = y 2 .
Addition. Two complex numbers
z1
and
z2
in rectangular form can be
added by adding the real parts and the imaginary parts separately:
z1 + z2 = ( x1 + jy1 ) + ( x2 + jy 2 ) = ( x1 + x2 ) + j ( y1 + y 2 )
Subtraction.
Subtracting one complex number from another can be
considered as adding the negative of the former:
z1 − z2 = ( x1 + jy1 ) − ( x2 + jy 2 ) = ( x1 − x2 ) + j ( y1 − y 2 )
Multiplication. If a complex number is multiplied by a real number, the
result is a complex number whose real and imaginary parts are multiplied by that
real number:
az = a ( x + jy ) = ax + jay ,
( a = real number)
If two complex numbers appear in rectangular form and we want the product in
rectangular form, multiplication is accomplished by using the fact that j = −1 . Thus,
2
if two complex numbers are written
z1 = x1 + jy1 ,
z2 = x2 + jy 2
Then
z1z2 = ( x1 + jy1 )( x2 + jy 2 ) = x1x2 + jx1y 2 + jy1x2 + j 2 y1y 2
= ( x1x2 − y1y 2 ) + j ( x1y 2 + y1x2 )
In polar form, multiplication of two complex numbers can be done easily. The
magnitude of the product is the product of the two magnitudes, and the angle of the
product is the sum of the two angles. So if two complex numbers are written
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
z1 = z1 ∠θ1 ,
then
z2 = z2 ∠θ 2
z1z2 = z1 z2 ∠ (θ1 + θ 2 )
Multiplication by j.
It is important to note that multiplication by
equivalent to a counterclockwise rotation by
90o.
j
is
For example, if
z = x + jy
then
jz = j ( x + jy ) = jx + j 2 y = − y + jx
or, noting that
j = 1 ∠90 , if
z = z ∠θ
then
jz = 1 ∠90 z ∠θ = z ∠ (θ + 90 )
Fig. 2-3 illustrates the multiplication of a complex number
z
by
by
j.
j.
z
jz
Fig. 2-3 Multiplication of a complex number
Division.
complex number
If a complex number
z2 = z2 ∠θ 2
z1
z2
=
z1 = z1 ∠θ1
, then
z1 ∠θ1
z2 ∠θ 2
=
z1
z2
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z
∠ (θ1 − θ 2 )
is divided by another
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
That is, the result consists of the quotient of the magnitudes and the difference of
the angles. Division in rectangular form can be done by multiplying the denominator
and numerator by the complex conjugate of the denominator. For instance,
z1
z2
=
( x + jy )
( x + jy )
1
1
2
=
2
(x
(x
+ jy1 )( x 2 − jy 2 )
1
+ jy 2 ) ( x 2 − jy 2 )
1
424
3
2
complex Conjugate of z 2
=
=
(x x
1
(x x
1
2
(x
Division by j.
90o.
2
2
2
+ y1 y 2 ) + j ( x 2 y1 − x1 y 2 )
(x
2
z
j
=
+j
)
(x y
j
Division by
For example, if
then
2
+ y1 y 2 )
+ y2
)
+ y2
2
2
2
(x
− x1 y 2 )
1
+ y2
2
2
2
)
is equivalent to a clockwise rotation by
z = x + jy
( x + jy ) ( x + jy ) j
=
j
jj
=
jx − y
−1
= y − jx
or,
z
j
=
z ∠θ
1 ∠90
o
(
= z ∠ θ − 90
Fig. 2-4 illustrates the division of a complex number
o
)
z
by
j.
z
by
j.
z/ j
Fig. 2-4 Division of a complex number
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Powers and roots.
z
Multiplying
by
n
times, we obtain
z = ( z ∠θ ) = z ∠nθ
n
n
Extracting the
1/ n th power.
n
n th root of a complex number is equivalent to raising the number to
z1 n = ( z ∠θ )
1n
For instance, calculate
(8.66 − j5)
3
=z
1n
∠
θ
n
=?
Remember :
The real part = x = 8.66
The imaginary part = y = −5
The magnitude = z =
The angle = θ = tan
Therefore,
( 8.66 − j 5 )
3
(
−1
()
y
x
= 10 ∠ − 30
x +y
2
= tan
o
)
3
−1
2
=
8 .66 + 5 = 10
2
( )
−5
8.66
= −30
2
o
= 1000∠ − 90 = 0 − j1000 = − j1000 .
o
Remarks. It is important to note that
zw = z w
z+w ≠ z + w
Complex Variable.
A complex variable has a real part and an
imaginary part, both of which are constant. If the real part or the imaginary part (or
both) are variables, the complex number is called a complex variable. In the
Laplace transformation, we use the notation s to denote a complex variable; that is,
s = σ + jω
where
ω
σ
is the real part and
jω
is the imaginary part. (Notice that both
σ
and
are real.)
Complex Function.
has a real part and an imaginary part, or
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A complex function
F ( s ) , a function of s
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
F ( s ) = Fx + jFy
where
Fx
and
Fy
are real quantities.
F ( s ) = F ( s ) = Fx2 + Fy2
Magnitude of
Angle of
(
F ( s ) = θ = tan −1 Fy Fx
)
The angle is measured counterclockwise from the positive real axis. The complex
conjugate of
F (s)
is
F ( s ) = Fx − jFy .
Complex functions commonly encountered in linear systems analysis are singlevalued functions of s and are uniquely determined for a given value of s . Typically,
such functions have the form
F (s) =
F ( s ) = 0 are called zeros. That is,
s = − z1 , s = − z2 ,L , s = − zm are zeros of F ( s ) .
•
Points at which
•
Points at which
•
K ( s + z1 )( s + z2 )L ( s + zm )
( s + p1 )( s + p2 )L( s + pn )
F ( s ) = ∞ are called poles. That is,
s = − p1 , s = − p2 ,L , s = − pn are poles of F ( s ) .
k − multiple factors ( s + p ) , then
s = − p is called a multiple pole of order k or repeated pole of
multiplicity k . If k = 1 , the pole is called a simple pole.
If the denominator of
F (s)
k
involves
Example
█
G( s) =
K ( s + 2 )( s + 10 )
s ( s + 1)( s + 5 )( s + 15 )
G ( s ) are values of s
s = −2, s = −10
•
Zeros of
•
Poles of
G (s)
are values of
s
2
which make
G ( s ) = 0 , that is
which make
G ( s ) = ∞ , that is
s = −0 , s = − 1, s = − 5 ⇒
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Simple and distinct poles
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
s = − 1 5 , s = − 1 5 ⇒ double pole or (pole of multiplicity 2)
Since for large values of
s
(when
s → ∞)
G (s) =
G (s)
possesses a triple zero at
infinity are included,
To summarize:
and five poles,
k
s3
( s = ∞ , ∞ , ∞ ) ( pole of multiplicity 3). If points at
G ( s ) has the same number of poles as zeros.
G ( s ) has five zeros, ( s = −2 , − 10 , ∞ , ∞ , ∞ )
( s = 0, − 1, − 5, − 15, − 15)
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
2.3 LAPLACE TRANSFORMATION
What is the Laplace Transform?
It is a solution technique that transforms differential equations in the time domain
into algebraic equations in the s-domain.
Why use Laplace Transform?
The Laplace transform is a powerful tool formulated to solve a wide variety of
Initial-Value Problems (IVP). The strategy is to transform the difficult differential
equations into simple algebraic problems where solutions can be easily obtained. One
then applies the Inverse Laplace transform to retrieve the solutions of the original
problems. This can be illustrated as follows:
L
L −1
Definition
The Laplace Transform
F ( s) of f (t ) is defined as
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
f (t ) = a time function such that f (t ) = 0 for t < 0
s = a complex variable
F ( s ) = Laplace transform of f (t )
where
Exponential Function:
Consider the exponential function shown
in Fig. 2-5:
f (t)
for t < 0
⎧0
f (t ) = ⎨ −αt
⎩Ae
for t ≥ 0
where A and α are constants. The Laplace
transform of f (t ) can be obtained as follows
A
Ae −α t
t
Fig. 2-5. Exponential function
∞
∞
A
− α +s t
F ( s ) = L ⎡⎣ Ae −α t ⎤⎦ = ∫ Ae −α t e − st dt = A∫ e ( ) dt =
s +α
0
0
Step Function: Consider a step function as shown in Fig. 2-6:
⎧0
f (t ) = ⎨
⎩A
where
A
for t < 0
f (t )
for t > 0
is a constant. Notice that this is a
A
−α t
special case of the exponential function Ae
where α = 0. The step function is undefined at
t = 0 . Its Laplace Transform is given by:
t
Fig. 2-6. Step function
∞
A
F (s ) = L ⎡⎣f (t ) ⎤⎦ = L [ A ] = ∫ Ae − st dt =
s
0
The step function whose height is unity is called a unit-step function. The unit step
t = t0 is frequently written as 1( t − t0 ) . The previous
step function whose height is A can be written as A1( t ) . The Laplace Transform of
function that occurs at time
the unit-step function that is defined by
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
for t < 0
for t > 0
⎧0
1(t ) = ⎨
⎩1
is
∞
1
F (s ) = L ⎡⎣1(t ) ⎦⎤ = ∫1e − st dt =
s
0
t = t0 corresponds to a constant signal
t equals t0 .
Physically, a step function occurring at time
suddenly applied to the system at time
Ramp Function: Consider a ramp function as shown in Fig. 2-7
⎧0
f (t ) = ⎨
⎩ At
for t < 0
for t ≥ 0
f (t )
f (t ) = At
where A is a constant. The Laplace Transform
of the ramp function is:
α
slope = tan α = A
∞
t
F (s ) = L [ At ] = A ∫ te − st dt
Fig. 2-7. Ramp function
0
To evaluate the above integral we use the formula for the integration by parts
∞
∞
∫ udv = uv | − ∫ vdu
∞
0
0
0
where in this case
u = At
⇒ du = A
1
⇒ v = − e − st
s
dv = e − st dt
Substituting these expressions into equation the above integral leads to the
following expression:
∞
∞
Ate − st ∞
Ae − st
|0 + ∫
dt
F (s ) = L [ At ] = ∫ Ate dt = −
s
s
0
0
− st
( −Ae ) |
= (0 − 0) +
− st
s2
12/36
∞
0
=0−
( −A ) = A
s2
s2
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Sinusoidal Function: The Laplace Transform of the sinusoidal function
for t < 0
⎧0
f (t ) = ⎨
⎩ A sin (ω t ) for t ≥ 0
Fig. 2-8. Sinusoidal function
where
A
and
ω
are constants, is obtained as follows. Noting that
e jω t = cos ω t + j sin ω t
sin ω t
and
and
e − jω t = cos ω t − j sin ω t
cos ω t can be written as
sin ω t =
1 jω t
e − e − jω t )
(
2j
and
cos ω t =
1 jω t
e + e − jω t )
(
2
Hence
∞
A
A 1
A 1
L [ A sin ω t ] = ∫ ( e jω t − e− jω t ) e− st dt =
−
2j 0
2 j s − jω 2 j s + jω
=
A ⎛ s + jω − s + jω ⎞ A ⎛ 2 jω ⎞
Aω
= 2
⎜
⎟= ⎜ 2
2
2
2 ⎟
2
s +ω
2j⎝
⎠ 2 j ⎝ s +ω ⎠ s +ω
Similarly
∞
A
A 1
A 1
L [ A cos ω t ] = ∫ ( e jω t + e − jω t ) e− st dt =
+
20
2 s − jω 2 s + jω
=
A ⎛ s + jω + s − jω ⎞ A ⎛ 2 s ⎞
As
= 2
⎜
⎟= ⎜ 2
2
2
2 ⎟
2
2⎝
s +ω
⎠ 2 ⎝ s +ω ⎠ s +ω
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ME 413 Systems Dynamics & Control
f (t )
from
Remark:
Chapter Two: Laplace transform
The Laplace Transform of any Laplace transformable function
f (t )
can be obtained by multiplying
0
to
∞.
by
e − st
and then integrating the product
Once we now the method of obtaining the Laplace Transform,
however, it is not necessary to derive the Laplace transform of
f (t )
each time.
Laplace Transform Tables can conveniently be used to find the transform of a given
function
f (t ) .
Refer to Table 2.1 of the Textbook. Notice that the Laplace
Transforms provided in Tables in general are valid for
0≤t <∞
.
Translated Functions: Let us obtain the Laplace transform
translated function
f ( t − α )1( t − α )
where
α ≥ 0.
This function is zero for
of the
t < α.
f (t − α )1(t − α )
f (t )1(t )
t
0
Fig. 2-9 Function
f ( t )1( t )
0
α
and translated function
By definition, the Laplace transform of
f ( t − α )1( t − α )
t
f ( t − α )1( t − α )
is
∞
L ⎡⎣f (t − α )1(t − α ) ⎤⎦ = ∫ f (t − α )1(t − α )e −st dt
0
Let
τ = t − α , then
t = 0 ⇒ τ = −α
, t → ∞ ⇒ τ → ∞ and dt = dτ
and
∞
∫ f ( t − α )1( t − α )e
− st
∞
dt =
f (τ )1(τ ) = 0
for
τ < 0,
− s (τ +α )
dτ
−
0
Noting that
∫α f (τ )1(τ )e
we can change the lower limit from
Thus
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−α
to
0.
ME 413 Systems Dynamics & Control
∞
∫α f (τ )1(τ )e
Chapter Two: Laplace transform
∞
d τ = ∫ f (τ )1(τ ) e
− s (τ +α )
−
0
=e
∞
∫ f (τ )e
−α s
∞
d τ = ∫ f (τ )e − sτ e −α s d τ
− s (τ +α )
0
− sτ
d τ = e −α s F (s )
0
where
∞
F ( s ) = L ⎡⎣f (t ) ⎤⎦ = ∫ f (t ) e − st dt
0
and also
L ⎡⎣f (t − α )1(t − α ) ⎤⎦ = e −α s F (s )
α ≥0
Pulse Function: Consider the pulse function shown in Fig. 2-10
⎧A
⎪
f (t ) = ⎨ t 0
⎪0
⎩
where
A
and
t0
f (t )
for 0 < t < t0
for t < 0, t0 < t
A
t0
are constants. The pulse function
may be considered as a step function of height
A t0
by
that begins at
a
negative
beginning at
t =0
step
and that is superimposed
function
of
height
t = t0 ; that is ,
t
A t0
0
t0
Fig. 2-10 A pulse function
A
f (t ) =
t0
1( t ) −
A
t0
1( t − t 0 )
f (t )
A
1( t )
t0
A
1( t − t0 )
t0
A
t0
t
0
t
t0
0
Then the Laplace Transform of
L
⎡A
⎤
⎡⎣f (t ) ⎤⎦ = L ⎢ 1(t ) ⎥ − L
⎣t0
⎦
t0
f (t )
0
t
t0
is obtained as
⎡A
⎤ A ⎛ 1 ⎞ A ⎛ 1 ⎞ − st 0 A
−
1
t
t
1 − e − st 0
(
)
0 ⎥ =
⎢
⎜ ⎟ − ⎜ ⎟e =
t 0s
⎣t0
⎦ t0 ⎝ s ⎠ t0 ⎝ s ⎠
(
15/36
)
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Impulse Function: The impulse function is a special limiting case of the
pulse function. Consider the impulse function
⎧lim A
⎪
f ( t ) = ⎨ t →0 t0
⎪0
⎩
for 0 < t < t0
f (t )
0
t0 → 0
for t < 0, t0 < t
A
→∞
t0
Fig. 2-11 depicts the impulse function shown
in Fig. 2-10 as t0 approaches 0.
t
Fig. 2-11. Impulse function
Since the height of the impulse function
the impulse is equal to
and the duration is t0 , the area under
A / t0
As the duration
A.
t0
approaches 0, the height
A / t0
approaches infinity, but the area under the impulse remains equal to A. Notice that
the magnitude of the impulse is measured by its area. Referring to the transformed
equation previously derived for the pulse function, i.e.,
L [ f ( t )] = A (1 − e )
ts
− s t0
0
the Laplace Transform of the impulse function is shown to be
L
⎡ A (1 − e
[ f ( t )] = lim ⎢
ts
⎣
t0 →0
0
− s t0
) ⎤ = lim
⎥
⎦
t0 →0
d
dt0
⎡⎣ A (1 − e
d
dt 0
− s t0
)⎤⎦
= lim
(t s )
t0 →0
⎡⎣ A ( se
− s t0
(s)
)⎤⎦ = A
0
Thus the Laplace Transform of the impulse function is equal to the area under the
impulse. The impulse function whose area is equal to unity is called the unit
impulse function or the Dirac delta function. The unit impulse function occurring
at
t = t0
is usually denoted by
⎧
⎪δ ( t − t ) = 0
0
⎪⎪
⎨δ ( t − t0 ) = ∞
⎪∞
⎪ ∫ δ ( t − t0 ) = 1
⎪⎩ −∞
16/36
for
t ≠ t0
for
t = t0
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Relationships among Singular Functions
The ramp, step, and impulse functions represent a family of functions, which as
shown in Fig. 2-12 are related by successive integrations.
δ (t )
u s (t )
u r (t )
1.0
1.0
0
0
t
Time
0
t
Time
L ⎡⎣δ (t ) ⎤⎦ = 1
L ⎡⎣u s (t ) ⎤⎦ =
1
s
1.0
Time
L ⎡⎣u r (t ) ⎤⎦ =
t
1
s2
Fig. 2-12 The relationship between singularity functions.
Time Shifting of Singularity Functions
The singularity functions may be used to describe transient inputs that take place at
t = 0 . The discontinuity associated with each function occurs when
the function argument is zero; therefore, a step that occurs at time t 0 may be
written as u s (t − t 0 ) since t − t 0 = 0 at t = t 0 . This property may be used to
a time other than
synthesize a transient function from a sum of singularity functions; for example, Fig.
()
()
(
)
(
)
(
)
2-13 shows the function u t = u s t − 2u s t − t 0 + ur t − 2 − ur t − 3 .
u (t )
u (t )
2 u t
s ( )
1
−1
−2
u r (t − 2 )
1 2 3 4
2
1
t
−u r (t − 3)
−2u s (t − 1)
Fig. 2-13 A transient function
−1
−2
1 2
3 4
t
u ( t ) = u s ( t ) − 2u s ( t − 1) + ur ( t − 2 ) − ur ( t − 3 )
synthesized from unit singularity functions
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Multiplication of f ( t ) by e
If
f (t )
is Laplace transformable, its Laplace transform being
Laplace transform of
L
⎡⎣e
−α t
e
f (t )
−α t
∞
f (t ) ⎤⎦ = ∫ e
−α t
0
█
Example
F (s) ,
is
∞
f (t ) e dt = ∫ f (t )e
− st
−( s +α )t
dt = F ( s + α )
0
ω
2
+ ω2
= F (s)
Find Laplace transform of
and
e −α t sin ω t
L [cos ω t ] =
s
and
s
= G (s)
2
+ ω2
e −α t cos ω t
Solution
L
⎡⎣e −α t sin ω t ⎤⎦ = F ( s + α ) =
ω
2
(s + α ) + ω2
L
⎡⎣e −α t cos ω t ⎤⎦ = G ( s + α ) =
s +α
and
2
(s + α ) + ω2
Laplace Transform Theorems
Differentiation Theorem
L
L
⎡d
⎤
⎢⎣ dt f ( t ) ⎥⎦ = sF ( s ) − f ( 0 )
⎡ d2
⎤ 2
&
⎢ dt 2 f ( t ) ⎥ = s F ( s ) − sf ( 0 ) − f ( 0 )
⎣
⎦
Similarly for the nth derivative of
L
then the
Given Laplace transforms of
L [sin ω t ] =
s
█
−α t
f ( t ) , we obtain
( n −1)
⎡dn
⎤
n
n −1
n −2 &
⎢ dt n f (t ) ⎥ = s F (s ) − s f (0) − s f (0) − L − f (0)
⎣
⎦
18/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
( n −1)
In the above, the following quantities
of
█
f ( t ) , df ( t ) / dt , L, d
Example
n −1
f ( t ) / dt
f ( 0 ) , f& ( 0 ) , L, f ( 0 )
n −1
,
represent the values
respectively, evaluated at
t = 0.
Given
Find the Laplace transform F(s) of f(t).
█
Solution
Using the Differentiation Theorem on the first two terms
leads to:
Using the definition of the Laplace transform on the remaining term gives:
From these results, the Laplace transform F(s) of the given equation can be
expressed as:
Rearranging this expression by factoring leads to:
Solving this expression for X(s) gives the following answer:
19/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Final Value Theorem (FVT)
lim [ f (t ) ] = lim ⎡⎣s F ( s ) ⎤⎦
t →∞
█
Example
s →0
Given:
Find the final value of x(t).
█
Solution
To solve this problem, use the Final Value Theorem (FVT)
lim [ f (t ) ] = lim ⎡⎣s F ( s ) ⎤⎦
t →∞
s →0
Substituting the given expression into this equation leads to the solution:
Initial Value Theorem (IVT)
f (0+ ) = lim ⎡⎣s F ( s ) ⎤⎦
s →∞
█
Example
Given:
Find the initial value of x(t), i.e. find x(0).
█
Solution To solve this problem, use the Initial Value Theorem (IVT):
Substituting the given expression into this equation leads to the solution:
20/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
For this example, as "s" goes to infinity, all terms involving "s" in the numerator and
denominator cancel. (Note that if there was one extra "s" term in the denominator
than there was in the numerator, this extra term would not be cancelled out and the
entire expression would go to zero.)
Integration Theorem
L
█
Example
⎡t
⎤ F (s )
⎢ ∫ f (t )dt ⎥ =
s
⎣0
⎦
Given:
Find Laplace transform of the given expression. (Hint: Let f(t) = At)
█
Solution
To solve this problem, use the integration theorem:
Substituting the result for F(s) obtained in the previous example leads to
the solution:
Use of MATLAB
█
Example
Use MATLAB to find Laplace Transform of
21/36
f ( t ) = t cos(3t )
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
█
Matlab Solution
>>
>>
syms t s
f = t*cos(3*t);
Then the Laplace transform of f is given by
>>
>>
F
F
F
F
= laplace(f)
=1/(s^2+9)*cos(2*atan(3/s))
= simplify(expand(F))
=(s^2-9)/(s^2+9)^2
Thus we obtain the Laplace transform
F (s) =
█
Example
█
Matlab Solution
>>
>>
syms t s
f=3*t-5*sin(2*t);
s2 − 9
(s
+ 9)
2
2
Use MATLAB to find Laplace Transform of
f ( t ) = 3t − 5cos(2t )
Then the Laplace transform of f is given by
>>
F = laplace(f)
F=3/s^2-10/(s^2+4)
>>
F = simplify(expand(F))
F =-(7*s^2-12)/s^2/(s^2+4)
Thus we obtain the Laplace transform
−7 s 2 + 12
F (s) = 2 2
s ( s + 4)
2.4 INVERSE LAPLACE TRANSFORMATION
The inverse Laplace transformation refers to the process of finding the time function
f ( t ) from the corresponding Laplace transform F ( s ) ; i.e.,
f (t ) = L
[ F (s ) ]
−1
Several methods are available for finding the inverse Laplace transforms
22/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
1. Use Tables of Laplace transforms
2. Use partial-fraction expansion method.
Partial-Fraction
Transforms
Expansion
for
Finding
Inverse
Laplace
If F ( s ) , the Laplace transform of f ( t ) , is broken up into components
F ( s ) = F1 ( s ) + F2 ( s ) + F3 ( s ) + L
+F
n
(s)
and if the inverse Laplace transform of F1 ( s ) , F2 ( s ) , F3 ( s ) , L , F
n
( s ) , are readily
available then
L −1 [ F (s )] = L −1 ⎡⎣ F1 ( s ) ⎤⎦ + L −1 ⎡⎣ F2 ( s ) ⎤⎦ + L
= f 1 (t ) + f 2 (t ) + L
+ f n (t )
+ L −1 ⎡⎣ F
n
( s )⎤⎦
where f1 ( t ) , f 2 ( t ) , L , f n ( t ) are the inverse Laplace transform of F1 ( s ) , F2 ( s ) ,
L, F
n
( s ) , respectively.
F (s ) =
F ( s ) frequently occurs in the form
N (s )
, m = deg N ( s ) < deg D ( s ) = n
D (s )
where N ( s ) and D ( s ) are polynomials in s and the degree of D ( s ) is not higher
than the degree of N ( s ) . Notice that applying the partial-fraction expansion
technique in the search for the inverse Laplace transform of F ( s ) = N ( s ) / D ( s )
requires that the poles of D ( s ) (roots of the denominator) must be known in
advance. Consider F ( s ) written in the factored form
F (s ) =
N ( s ) K ( s + z 1 )( s + z 2 )L ( s + z n )
=
D (s )
( s + p1 )( s + p 2 )L ( s + p n )
p1 , p2 , K , pn and z1 , z2 , K , zn are either real or complex quantities, but for
each pi or zi there will occur the complex conjugate of pi or zi , respectively. Here
where
the highest power of s in N ( s ) is assumed to be higher than that in D ( s ) .
Notice that if the degree of the numerator is greater than (or equal to) that of the
denominator, then polynomial division must be performed so that the remainder
polynomial is of a lower degree than D ( s ) . For instance,
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ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
s3 +2
−s + 2
,
=s + 2
2
s +1
s +1
Case I.
deg ( −s + 2 ) < deg ( s + 1 )
2
Distinct Real Poles
In this case, F ( s ) can be expanded into a sum of partial fractions
F (s ) =
N (s )
r1
r2
rn
=
+
+L+
D ( s ) ( s + p1 ) ( s + p 2 )
(s + pn )
where rk ( k = 1, 2,K , n ) are constants. The coefficient
rk is called the residue at the
s = − pk . The value of rk can be found by multiplying both sides of the above
pole at
equation
( s + pk )
and letting s = − pk , which gives
⎡
⎡ r ( s + pk ) r2 ( s + pk )
N (s) ⎤
r ( s + pk )
r ( s + pk ) ⎤
=⎢ 1
+
+L+ k
+L+ n
= rk
⎢( s + pk )
⎥
⎥
D ( s ) ⎦ s =− p ⎣ ( s + p1 )
s + p2 )
s + pk )
s + pn ) ⎦ s =− p
(
(
(
⎣
k
k
We see that all the expanded terms drop out with the exception of
residue
rk is found from
⎡
N (s ) ⎤
rk = ⎢( s + p k )
⎥
D ( s ) ⎦ s =− p
⎣
k
(2.6)
Since
⎡ rk
⎤
−p t
L−1 ⎢
⎥ = rk e k
⎣ (s + pk ) ⎦
f ( t ) is obtained as
f (t ) = L −1 [ F (s ) ] = r1e − p1t + r2e − p 2t + L + rk e − p k t
█
Example
Find the inverse Laplace transform of
F (s) =
█
rk .Thus the
Solution
s+3
( s + 1)( s + 2 )
The partial fraction expansion of F ( s ) is
F (s) =
s+3
r
r2
= 1 +
( s + 1)( s + 2 ) ( s + 1) ( s + 2 )
24/36
t ≥0
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
where r1 and r2 are found by using Equation (2.6)
⎡
( s + 3) ⎤⎥ = ⎡ ( s + 3) ⎤ = −1 + 3 = 2
r1 = ⎢ ( s + 1)
⎢
⎥
⎢
( s + 1) ( s + 2 ) ⎦⎥ s =−1 ⎣ ( s + 2 ) ⎦ s =−1 −1 + 2
⎣
⎡
( s + 3) ⎥⎤ = ⎡ ( s + 3) ⎤ = −2 + 3 = −1
r2 = ⎢ ( s + 2 )
⎢
⎥
⎢
( s + 1) ( s + 2 ) ⎥⎦ s =−2 ⎣ ( s + 1) ⎦ s =−2 −2 + 1
⎣
Thus
F (s) =
s+3
2
−1
=
+
( s + 1)( s + 2 ) ( s + 1) ( s + 2 )
⎡ 2 ⎤
⎡ −1 ⎤
−1
−t
−2t
f (t ) = L −1 [ F (s ) ] = L −1 ⎢
⎥ +L ⎢
⎥ = 2e − e
⎣ ( s + 1) ⎦
⎣ (s + 2) ⎦
t ≥0
Use of MATLAB: Use MATLAB to find the inverse Laplace Transform of the above
example
F (s) =
s+3
( s + 1)( s + 2 )
>> syms t s
>> F = (s+3)/((s+1)*(s+2))
Then the inverse Laplace transform of f
(t )
is given by
>> F = ilaplace(f)
f =2*exp(-t)-exp(-2*t)
>> pretty(f)
2 exp(-t) - exp(-2 t)
Thus we obtain the Laplace transform
f ( t ) = L−1 [ F ( s ) ] = 2e −t − e −2t
█
Example
t≥0
Obtain the inverse Laplace transform of
F (s) =
s 3 + 5s 2 + 9 s + 7
( s + 1)( s + 2 )
25/36
ME 413 Systems Dynamics & Control
█
Chapter Two: Laplace transform
Solution
Since the degree of the numerator is higher than that of the denominator
polynomial, we must divide the numerator by the denominator
F (s ) = s + 2 +
s +3
2
−1
= s + 2+
+
1) ( s + 2 )
( s + 1)( s + 2 )
( s +4
144
2444
3
previous example
Notice that
F (s)
L ⎡⎣δ (t ) ⎤⎦ = 1
and
⎡ d δ (t ) ⎤
⎥ = s , so the inverse Laplace transform of
⎣ dt ⎦
L ⎢
is given by
f (t ) =
Case II.
d
δ ( t ) + 2δ ( t ) + 2e − t − e−2t
dt
t≥0
Complex Conjugate Poles
Consider a function F ( s ) that involves a quadratic factor
s 2 + as + b in the
denominator. If this quadratic factor has a pair of complex conjugate poles, then it is
better not to factor this term in order to avoid complex numbers. For example, if
F ( s ) is given as
F ( s) =
where a ≥ 0 and b ≥ 0 , and
p(s)
s ( s 2 + as + b )
s 2 + as + b = 0
has a pair of complex conjugate poles,
then expand F ( s ) into the following partial-fraction expansion form:
F ( s) =
█
Example
c
ds + e
+ 2
s s + as + b
Obtain: the inverse Laplace transform of
F ( s) =
█
2 s + 12
s + 2s + 5
2
Solution 1: Use of Complex Numbers
26/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Notice that the poles of the denominator are
s1,2 =
Therefore
F (s )
F (s ) =
−b ± b 2 − 4ac −2 ± 4 − 4 × 1× 5 −2 ± −16
=
=
= −1 ± j 2 .
2a
2
2
can be written as
2s + 12
2s + 12
α
β
=
=
+
s 2 + 2s + 5 ( s + 1 + j 2 )( s + 1 − j 2 ) s + 1 + j 2 s + 1 − j 2
where the constants
α
and
β
that can be found as before
α=
( s + 1 + j 2 ) ( 2s + 12 )
( 2 ( −1 − j 2 ) + 12 ) = 10 − j 4 = 1 − j 5
=
2
( s + 1 + j 2 ) ( s + 1 − j 2 ) s =−1− j 2 ( −1 − j 2 + 1 − j 2 ) − j 4
β=
( s + 1 − j 2 ) ( 2s + 12 )
(s + 1 + j 2) (s + 1 − j 2)
Notice that
β
=
s = −1+ j 2
( 2 ( −1 + j 2 ) + 12 ) = 10 + j 4 = 1 + j 5
( −1 + j 2 + 1 + j 2 )
is the complex conjugate of
into the expression of
α.
+j4
Substitute the values of
2
α
F (s )
5
5
j
1
+
2s + 12
2 +
2
F (s ) = 2
=
s + 2s + 5 s + 1 + j 2 s + 1 − j 2
1− j
and
5 ⎞ − 1+ j 2 t ⎛
5 ⎞ − 1− j 2 t
⎛
f (t ) = L −1 [ F (s ) ] = ⎜1 − j ⎟e ( ) + ⎜1 + j ⎟e ( )
2⎠
2⎠
⎝
⎝
5
5
= e −t e − j 2t − j e −t e − j 2t + e −t e j 2t + j e −t e j 2t
2
2
j 2t
− j 2t
j 2t
⎞
− e − j 2t ⎞
+e
−t ⎛ e
−t ⎛ e
= 2e ⎜
⎟
⎟ + j 5e ⎜
2
2
⎝
⎠
⎝
⎠
j 2t
⎛ e j 2t + e − j 2t ⎞
− e − j 2t ⎞
−t ⎛ e
5
e
= 2e −t ⎜
−
⎜
⎟
⎟
2j
2
⎝
⎝144244
⎠
3
144244
3⎠
cos( 2t )
= 2e −t cos ( 2t ) − 5e −t sin ( 2t )
27/36
sin ( 2t )
and
β
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
Solution 2: Completing The square
█
The expression of
can be written in general as
F (s)
F (s) =
cs + d
s + 2as + a 2 + ω 2
2
where a and ω are positive real. It is clear that the denominator in the above
expression is a complete square, i.e., it can be written as
s 2 + 2as + a 2 + ω 2 = ( s 2 + a 2 ) + ω 2
F ( s ) into the following form
c ( s + a ) + d − ca
cs + d
cs + d
=
=
F (s) = 2
2
s + 2as + a 2 + ω 2 ( s + a )2 + ω 2
(s + a) + ω2
Let’s us write the expression of
=
c(s + a)
d − ca
+
c(s + a)
=
(s + a) + ω2 (s + a) + ω2 (s + a) + ω2
2
2
2
+
d − ca
ω
ω
2
(s + a) + ω2
The inverse Laplace transform is then
⎡ ( s + a ) ⎤ ⎛ d − ca ⎞ -1⎡
⎤
ω
+
L
⎢
⎥
⎢
⎥
⎜
⎟
2
2
2
2
ω
⎝
⎠
ω
ω
s
a
s
a
+
+
+
+
(
)
(
)
⎢⎣
⎣⎢
⎦⎥
⎦⎥
⎛ d − ca ⎞ − at
= ce− at cos ωt + ⎜
⎟ e sin ωt
⎝ ω ⎠
f (t ) = L
[ F ( s)] = c L
-1
-1
In our example we have
F (s) =
2 s + 12
2 s + 12
2 s + 12
= 2
=
2
2
s + 2s + 5 1
s4
+24
2s +
31 + 4{ ( s + 1) + 2
2
=( s +1)
Thus we have
f (t )
2
22
a = 1, ω = 2, c = 2, d = 12. Therefore, substitute into the expression of
28/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
⎛ d − ca ⎞ − a t
f ( t ) = L −1 [ F ( s ) ] = ce − a t cos ω t + ⎜
⎟ e sin ω t
⎝ ω ⎠
⎛ 12 − 2 × 1 ⎞ − t
= 2e− t cos 2t + ⎜
⎟ e sin 2t
2 3⎠
⎝1424
=5
= 2e cos 2t + 5e sin 2t = e − t ( 2cos 2t + 5sin 2t )
−t
−t
Method 2: Use of Complex numbers
This method is a lengthy process we will see it in a separate problem in the help
session.
Case III.
Multiple Poles
Consider the following expression of
As can be seen
F (s)
has poles
of multiplicity 3 . Hence
F (s) =
where
( s + 1)
( s + 1)
3
3
3
,
s = −1, − 1, − 1. Thus we say F ( s ) has a pole s = −1
=
B(s)
b3
b2
b1
=
+
+
3
2
A ( s ) ( s + 1) ( s + 1) ( s + 1)
are determined as follows. By multiplying both sides of the last
, we obtain
( s + 1)
Then, letting
( s + 1)
can be written in the following form
s 2 + 2s + 3
b1 , b2 , and b3
equation by
F (s)
F (s) =
s 2 + 2s + 3
s = −1,
3
B(s)
2
= b3 + b2 ( s + 1) + b1 ( s + 1)
A( s )
(2.7)
Equation (2.7) gives
⎡
3 B(s) ⎤
= b3
⎢( s + 1)
⎥
A
s
(
)
⎣
⎦ s =−1
Also differentiation of both sides of Equation (2.7) gives
If we let
d ⎡
3 B(s) ⎤
s
+
1
(
)
⎢
⎥ = b2 + 2b1 ( s + 1)
ds ⎣
A( s ) ⎦
s = −1, in Equation (2.8), then
29/36
(2.8)
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
d ⎡
3 B(s) ⎤
s
+
1
= b2
(
)
⎢
⎥
ds ⎣
A ( s ) ⎦ s =−1
By differentiating both sides of Equation (2.8) with respect to
s , we obtain
d2 ⎡
1 d2 ⎡
3 B(s) ⎤
3 B(s) ⎤
s
+
=
b
⇒
b
=
s
+
1
2
1
(
)
(
)
⎢
⎥
⎢
⎥
1
1
ds 2 ⎣
A( s ) ⎦
A( s ) ⎦
2! ds 2 ⎣
Therefore
⎡
⎤
2
⎡
3 B(s) ⎤
3 s + 2s + 3
⎢
⎥
b3 = ⎢( s + 1)
= ( s + 1)
⎥
3 ⎥
⎢
A ( s ) ⎦ s =−1
⎣
( s + 1) ⎦
⎣
= ( −1) + 2 ( −1) + 3 = 2
2
s =−1
⎡
⎤
2
d ⎡
d ⎢
3 B(s)⎤
3 s + 2s + 3
⎥
b2 = ⎢( s + 1)
=
( s + 1)
⎥
3 ⎥
ds ⎣
A ( s ) ⎦ s =−1 ds ⎢
( s + 1) ⎦ s=−1
⎣
d
= ⎡⎣ s 2 + 2 s + 3⎤⎦
= [ 2 s + 2]s =−1 = ⎡⎣ 2 ( −1) + 2 ⎤⎦ = 0
s =−1
ds
⎡
⎤
2
1 d2 ⎡
1 d2 ⎢
3 B(s) ⎤
3 s + 2s + 3
⎥
b1 =
s + 1)
s + 1)
=
⎥
2 ⎢(
2 (
3 ⎥
⎢
A ( s ) ⎦ s =−1 2! ds
2! ds ⎣
( s + 1) ⎦ s=−1
⎣
1 d
1
=
[ 2s + 2]s=−1 = [ 2]s=−1 = 1
2! ds
2!
Therefore
F (s) =
and
2
( s + 1)
3
+
0
( s + 1)
2
+
1
( s + 1)
f (t ) = L −1 [ F (s ) ] = t 2e −t + 0 + e −t
t ≥0
2.5 SOLVING LINEAR, TIME INVARIANT
DIFFERENTIAL EQUATIONS
The Laplace transform method yields the complete solution (complementary solution
and particular solution) of linear, time invariant, differential equations. Classical
methods for finding the complete solution of a differential equation require the
evaluation of integration constants from the initial conditions. In the case of Laplace
transform method, however, this requirement is unnecessary because the initial
30/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
conditions are automatically included in the Laplace transform of the differential
equation.
█
Example
y "+ y = t ,
Solve:
█
Initial Value Problem (IVP)
y ( 0 ) = 1,
y '( 0) = 1
Solution
By writing the Laplace transform of
y (t )
as
L ⎡⎣ y (t ) ⎤⎦ =Y ( s ) , we obtain
L ⎡⎣ y& (t ) ⎤⎦ = sY ( s ) − y ( 0 )
L ⎡⎣ y&& (t ) ⎤⎦ = s 2Y ( s ) − s y ( 0 ) − y& ( 0 )
y ( 0 ) = y& ( 0 ) = 0 , the above transforms become
For zero initial conditions, i.e.,
L ⎡⎣ y& (t ) ⎤⎦ = sY ( s )
L ⎡⎣ y&& (t ) ⎤⎦ = s 2Y ( s )
Step 1:
Take Laplace Transform (LT) of both sides of the above equation:
s 2 Y (s ) − sy (0) − y '(0) + Y{
(s ) = 1/
s2
{
1444
424444
3
L ⎡⎣⎢ y ⎤⎦⎥ L ⎡⎢⎣t ⎤⎥⎦
L ⎡⎢ y "⎤⎥
⎣
⎦
or
(s
Step 2:
Let
Solving for Y
(s )
2
+ 1) Y (s ) =
1
+ s +1
s2
gives
Y (s ) =
1
s
1
+ 2
+ 2
2
s ( s + 1) ( s + 1) ( s + 1)
Q (s ) =
1
1
1
= 2− 2
2
s
s ( s + 1)
( s + 1)
2
2
Therefore
31/36
ME 413 Systems Dynamics & Control
Y (s ) =
Step 3:
Chapter Two: Laplace transform
1
1
s
1
1
s
− 2
+ 2
+ 2
= 2+ 2
2
s
( s + 1) ( s + 1) ( s + 1) s ( s + 1)
Solving
y (t ) = L
−1
⎡1⎤
⎢⎣ s 2 ⎥⎦ + L
−1
⎡ s ⎤
⎢ 2
⎥ = t + cos(t )
⎢⎣ ( s + 1) ⎦⎥
The diagram below summarizes the approach
L
y "+ y =t ,
y ( 0) =1, y '( 0) =1
y (t ) = L
−1
⎡1⎤
⎢⎣ s 2 ⎥⎦ + L
−1
s 2 Y (s ) − sy (0) − y '(0) + Y{
(s ) = 1/
s2
{
1444
424444
3
L ⎡⎣⎢ y ⎤⎦⎥ L ⎡⎢⎣t ⎤⎥⎦
L ⎡⎢ y "⎤⎥
⎣
L −1
⎡ s ⎤
⎢ 2
⎥
⎢⎣ ( s + 1) ⎥⎦
⎦
Y (s ) =
y (t ) = t + cos(t )
█
Example
Find the solution
x (t )
1
s
+ 2
2
s
( s + 1)
of the following Initial Value
Problem (IVP)
&&
x + 3 x& + 2 x = 0,
x ( 0 ) = a,
x& ( 0 ) = b
where a and b are constants.
█
Solution
Step 1: Take LT of both sides of the given equation to obtain an algebraic equation
X(s)
By writing the Laplace transform of
x (t )
32/36
as
L ⎡⎣ x (t ) ⎤⎦ = X ( s ) , we obtain
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
L ⎡⎣ x& (t ) ⎤⎦ = sX ( s ) − x ( 0 )
L ⎡⎣ x&& (t ) ⎤⎦ = s 2 X ( s ) − s x ( 0 ) − x& ( 0 )
Take Laplace transform of both sides of the given equation
⎡⎣s 2 X ( s ) − s x ( 0 ) − x& ( 0 ) ⎤⎦ + 3 ⎡⎣sX ( s ) − x ( 0 ) ⎤⎦ + 2 ⎡⎣ X ( s ) ⎤⎦ = 0
1442443
1
424
3
14444244443
L ⎣⎡ x& (t )⎦⎤
L ⎡⎣ x&&(t )⎤⎦
By substituting the initial conditions
x ( 0 ) = a,
L ⎣⎡ x (t )⎦⎤
x& ( 0 ) = b
into the last equations,
one obtains
⎡⎣ s 2 X ( s ) − a s − b ⎤⎦ + 3 ⎡⎣ sX ( s ) − a ⎤⎦ + 2 ⎡⎣ X ( s ) ⎤⎦ = 0
or
⎡⎣ s 2 + 3s + 2 ⎤⎦ X ( s ) = as + b + 3a
Step 2: Solving for
X ( s ) , we obtain
X (s) =
as + b + 3a
K
K
= 1 + 2
2
s14
3s +32 s + 1 s + 2
+24
=( s +1)( s + 2 )
since
X (s)
has distinct poles,
K1
and
K2
can be found easily
K1 =
( s + 1) ( as + b + 3a )
( a ( −1) + b + 3a )
=
( −1 + 2 )
( s + 1) ( s + 2 ) s=−1
K2 =
( s + 2 ) ( as + b + 3a )
( a ( −2 ) + b + 3a )
=
( −2 + 1)
( s + 2 ) ( s + 1) s=−2
Thus
X (s) =
Step 3: Take
= 2a + b
= −( a + b)
2a + b a + b
−
s +1 s + 2
L −1 ⎡⎣ X ( s ) ⎤⎦ to obtain the time domain solution x (t )
33/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
⎡ 2a + b ⎤
−1 ⎡ a + b ⎤
x (t ) = L −1 ⎡⎣ X ( s ) ⎤⎦ = L −1 ⎢
−
L
⎢⎣ s + 2 ⎥⎦
⎣ s + 1 ⎥⎦
t ≥0
= ( 2a + b )e −t − ( a + b )e −2t
█
Example
Find the solution
x (t )
of the following Initial Value
Problem (IVP)
&&
x + 2 x& + 5 x = f (t ),
where
x ( 0 ) = 0,
x& ( 0 ) = 0
f (t )
f (t ) is a function given by its graph.
3
█
Solution
t
Step 1: Find the explicit expression of
The function
f (t )
f (t ) given on the RHS of the IVP is a step function u (t ) with height
equals to 3. and
L [u (t )] = L [3] = 3 s
Step 2: Take LT of both sides of the given equation to obtain an algebraic equation
X (s )
Remember that we have zero initial conditions. For zero initial conditions, i.e.,
x ( 0 ) = x& ( 0 ) = 0 , the above transforms become
L ⎡⎣ x& (t ) ⎤⎦ = sX ( s )
L ⎡⎣ x&& (t ) ⎤⎦ = s 2 X ( s )
Take Laplace transform of both sides of the given equation
s 2 X ( s ) + 2 sX ( s ) + 5 X ( s ) =
Solving for
X ( s ) , we obtain
X (s) =
where
K1 , K 2
and
K3
3
s
3
K1
cs + d
=
+
2
s ( s 2 + 2s + 5) s s + 2s + 5
are found as follows
34/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
3 = K1 ( s 2 + 2 s + 5 ) + ( cs + d ) s
or
3 = ( K1 + c ) s 2 + ( 2 K1 + d ) s + 5 K1
By comparing coefficients of the
equation respectively, we obtain
s2 , s
s 0 terms: 5K1 = 3 ⇒
and
s0 ,
terms on both sides of this last
K1 = 3/ 5
s1 terms: 2 K1 + d = 0 ⇒ d = −2 K1 = −2(3/ 5) = −6 / 5
s 2 terms: K1 + c = 0 ⇒ c = − K1 = −3/ 5
Thus
X (s) =
( 3/ 5) + ( −3/ 5) s + ( −6 / 5) = ( 3/ 5) + ( −3/ 5) s + ( −6 / 5)
3
=
2
2
s
s1
+24
2 s +35
s
s ( s 2 + 2s + 5)
( s + 1) + 22
4
= s 2 + 2 s +1+ 4
⎡
⎤
−
+
−
3/
5
3/
5
s
6
/
5
⎡
⎤
⎢
⎥
(
)
(
)
(
)
−1
x (t ) = L −1 ⎣⎡ X ( s ) ⎦⎤ = L −1 ⎢
⎥ +L ⎢
⎥
2
+
+
s
2
s
5
⎣ s ⎦
14
24
3
⎢
⎥
=s 2 + 2 s +1+ 4
⎣
⎦
⎡
⎤
⎢
⎥
−1 ⎡ ( 3/ 5 ) ⎤
−1 ( −3/ 5 ) s + ( −6 / 5 )
⎢
⎥
=L ⎢
⎥ +L
2
2
s
⎢ ( s + 1) + 2
⎥
⎣
⎦
42444
3⎥
⎢ 144
G (s )
⎣
⎦
The second term of the last equation can be written according to the completing the
square rule. Thus we have a = 1, ω = 2, c = −3 / 5, d = −6 / 5. Therefore substitute into
the expression of
f (t )
35/36
ME 413 Systems Dynamics & Control
Chapter Two: Laplace transform
⎛ d − ca ⎞ −at
G (t ) = L −1 [G (s ) ] = ce −at cos ωt + ⎜
⎟e sin ωt
⎝ ω ⎠
⎛ 3 ⎛ 6⎞⎞
⎜ −5 −⎜− 5 ⎟ ⎟
3 −t
⎝
⎠ ⎟e −t sin 2t = − 3 e −t cos 2t − 3 e −t sin 2t
= − e cos 2t + ⎜
5
2
5
10
⎜
⎟
⎜
⎟
⎝144244
3⎠
=
3
5
Hence
⎡
⎤
−
+
−
3/
5
s
6
/
5
⎢
⎥
(
)
(
)
−1
⎢
⎥
2
s
2
s
5
+
+
14
24
3
⎢
⎥
=s 2 + 2 s +1+ 4
⎣
⎦
3 3
3
= − e −t cos 2t − e −t sin 2t
5 5
10
⎡ ( 3/ 5 ) ⎤
x (t ) = L −1 ⎡⎣ X ( s ) ⎤⎦ = L −1 ⎢
⎥ +L
s
⎣
⎦
36/36