Laplace Transform
The Laplace Transform is one of the mathematical tools for solving ordinary linear differential equation.
- The homogeneous equation and the particular Integral are solved in one operation.
- The Laplace transform converts the ODE into an algebraic eq.
in s ( σ σσ σ + j ω ω ω ω plane) domain. It is then possible to manipulate the algebraic eq. by simple algebraic rules (+-x÷÷÷÷) to obtain the solution in the s-
domain. The final solution is obtained by taking the inverse Laplace transform.
Definition of Laplace Transform: f(t) that satisfies
∞
∫
=
−=
0
) ( )
( )]
(
[ f t F s f t e dt
L
sts is referred to as the Laplace operator, which is a complex variable s = (σσσσ + jωωωωplane) The Laplace transform of f(t) is defined as
∞
∫ <
∞
− 0
)
( t e dt
f
σt for finite real σσσσLaplace Transform
Ex Let f(t) be a unit step function that is defined to have a value of unity for t > 0 and a zero value for t < 0.
Find Laplace Transform of f(t) Solution
Known f(t) = u(t)
∫
∞ −=
= [ ( )]
0( ) )
( s L f t u t e dt
F
stt s u
L 1
)]
(
[ =
Laplace Transform
Ex Consider the exponential function
Find Laplace Transform of f(t) Solution
Known f(t) = e-at
∫
∞ − −=
= [ ( )]
0)
( s L f t e e dt
F
at sta e s
L
at= +
−
1
] [
0
; )
( t = e
−t ≥
f
atWhere a is a constant
Inverse Laplace Transform
Definition of Inverse Laplace Transform:
∫
∞+
∞
−
−
=
=
j c
j c
st
ds e s j F
s F L t
f ( )
2 )] 1 ( [ )
(
1π
c is a real constant that is greater than the real parts of all singularities of F(s)
(Singularity = the point in which the value of function can not be evaluated e.g. ∞∞∞)∞ The Inverse Laplace transform of F(s) is defined as
The above approach to find Inverse Laplace transform is quite complicated, in practice, we find the inverse Laplace transform using Look up table
Laplace Transform Table
Laplace Transform Table
1) Multiplication by a constant
) ( )]
(
[ kf t kF s
L =
Where F(s) is the Laplace transform of f(t)
2) Sum and Difference
) ( )
( )]
( )
(
[ f
1t f
2t F
1s F
2s
L ± = ±
Where F1(s) and F2(s) is the Laplace transform of f1(t) and f2(t)
3) Differential
) 0 ( )
( ) ]
[ ( sF s f
dt t f
L d = −
In general, for higher-order derivatives,
) 0 ( ...
) 0 ( )
0 ( )
( ) ]
[
n n( = s
nF s − s
n−1f − s
n−2f
(1)− − f
(n−1)dt
t f L d
Important Theorems of the Laplace Transform
Laplace Transform Table
4) Integration
s s d F
f
L
t( )
] ) (
[ ∫
0τ τ =
In general, for n th-order integration,
n n
t t t
s s dt F
dt d f
L
n( )
...
) (
...
1 10 0 0
1 2
=
∫ ∫ ∫ τ τ
−5) Shift in Time
) ( )]
( ) (
[ f t T u t T e F s
L − − =
−sT6) Initial-Value Theorem
) ( lim
) ( lim
0
f t sF s
s
t→
=
→∞7) Final-Value Theorem
) ( lim
) (
lim f t
0sF s
s
t→∞
=
→If sF(s) is analytical on the imaginary axis and in the right half of the s-plane then,
Laplace Transform
Ex Consider the following function, and find steady state value of f(t)
Find f(t) as t -> ∞ ∞ ∞ ∞ Solution
Known F(s)
) 2 (
) 5
(
2+
= +
s s
s s F
2
)
2( ω
ω
= + s s F
5/2, the final value theorem cannot be applied.
Inverse Laplace Transform by Partial-Fraction Expansion
When the Laplace transform solution of a differential equation is a rational function in s, it can be written as
) (
) ) (
( P s
s s Q
X =
P(s) and Q(s) is a polynomial of s, it is assumed that the order of P(s) in s is greater than that of Q(s). The polynomial Q(s) may be written
0 1
1
1
...
)
( s s a s a s a
P =
n+
n− n−+ + +
Where a0, … , an are real coefficient. The solution of P(s) = 0 or poles of X(s) (the
points in which X(s) is not analytic) are either real or in complex-conjugate pairs, in simple or multiple poles.
Partial-Fraction Expansion when all the poles of X(s) are simple
) )...(
)(
(
) ( )
( ) ) (
(
2
1
s s s s
ns s
s Q s
P s s Q
X = = + + +
Where -s1, -s2, …, -sn are real or imaginary numbers. Applying the partial fraction expansion technique,
) ... (
) (
) ) (
(
2 2 1
1
n sn s
s
s s
K s
s K s
s s K
X + + +
+ +
= +
The coefficient Ks1, Ks2, …, Ksn (i = 1,2,…,n) are determined by multiplying both side of Eqs. By the factor (s + si) and then setting s equal to –si.
) )...(
)(
(
) (
) (
) ) (
(
1 3
1 2
1
1 1
1
1 n
s s
s
s s s s s s
s Q s
P s s Q
s
K − + − + − +
= −
+
=
−
=
Laplace Transform
Ex Consider the function
Find partial fraction coefficient Solution
Known X(s)
) 3 )(
2 )(
1 (
3 ) 5
( + + +
= +
s s
s s s
X
) 3 (
) 2 (
) 1 ) (
(
1 2 3+ + + +
= +
− − −s K s
K s
s K X
) 3 (
6 )
2 (
7 )
1 (
) 1
( − +
+ + +
= −
s s
s s
X
Partial-Fraction Expansion when some poles of X(s) are multiple
) (
) )...(
)(
(
) ( )
( ) ) (
(
2
1 n
r
i
s s
s s s
s s s
s Q s
P s s Q
X = = + + + +
The X(s)can be expanded as
) ... (
) (
) ) (
(
2 2 1
1
n sn s
s
s s
K s
s K s
s s K
X + + +
+ +
= +
If r of the n poles of X(s) are identical or the pole of s = -si is of multiplicity r,
r i r i
i
s s
A s
s A s
s A
) ... (
) (
)
(
22 1
+ + + +
+ + +
(n - r) terms of simple poles
r terms of repeated poles
) )...(
)(
(
) ( )
( ) ) (
(
1 3
1 2
1
1 1
1
1 n
s s
s
s s s s s s
s Q s
P s s Q
s
K − + − + − +
= −
+
=
−
=
The coefficient Ks1, Ks2, …, Ksn can be determined
Partial-Fraction Expansion when some poles of X(s) are multiple
[
i r]
s sir
s s X s
A = ( + ) ( )
=−[ ]
si
s r
i
r
s s X s
ds A d
−
−1
= ( + ) ( )
=[ ]
si
s r
i
r
s s X s
ds A d
−
−
= ( + ) ( )
=! 2
1
2 2 2
[ ]
si
s r
r i r
s X s
ds s d A r
−
− =
−
− +
= ( ) ( )
)!
1 (
1
1 1 1
The coefficient for multiple-order poles can be determined
Laplace Transform
Ex Consider the function
Find partial fraction coefficient Solution
Known X(s)
) 2 (
) 1 (
) 1
(
3+
= +
s s
s s X
3 3 2
2 1
2 0
) 1 (
) 1 (
1 ) 2
( + +
+ + + +
+ +
=
−s A s
A s
A s
K s
s K X
)
31 (
1 1
1 )
2 (
2 1 2
) 1
( − +
− + + +
= s s s s
s
X
Laplace Transform
Ex Consider the differential equation
Solution
) ( 5 ) ( ) 2
3 ( ) (
2 2
t u t
dt y t dy dt
t y
d + + =
The initial condition
y ( 0 ) = − 1 ( 0 ) 2 dt =
and
dy
t
t
e
e t
y
22 5 3
2 ) 5
( = −
−+
−Laplace Transform
Ex Consider the differential equation
Solution
) ( )
) ( ( 2
) ( 1
2 2
2
y t u t
dt t dy dt
t y d
n n
= +
+ ω ζ ω
The initial condition
y ( 0 ) = 0 ( 0 ) 0 dt =
and
dy
( ω ζ φ )
ζ
ζω
− +
− −
= e
−t
t
y
nt n 22
sin 1
1 1 1
) (
< 1 ζ
−
= ζ
φ arctan 1 ζ 2
Transfer function
Transfer function of a linear time-invariant system is defined as the Laplace transform of the impulse response, with all the initial conditions set to zero.
Let G(s) denote the transfer function of a system with input x(t) and output y(t). Then, the transfer function G(s) is defined as
)]
( [ )
( s L g t
G =
The transfer function G(s) is related to the Laplace transform of the input and output through
) (
) ) (
( X s
s s Y
G =
) ( ) ( )
( s G s X s
Y =
With all initial conditions set to zero, where Y ( s ) = L [ y ( t )] X ( s ) = L [ x ( t )]
Transfer function defines the mathematical operation that the measurement
system performs on input to yield the time response of the system
n
thOrder ordinary linear differential equation with constant coefficient
) ) (
( )
( )
) ( ) (
( )
( )
(
0 1 1
1 1
0 1 1
1
1 b x t
dt t b dx dt
t x b d
dt t x b d
t y dt a
t a dy dt
t y a d
dt t y
a d m
m m m
m n m
n n n
n
n + + + + = + − + + +
−
− −
−
− L L
Transfer function
To obtain the transfer function of the linear system, we simply take the Laplace transform and assume zero initial conditions.
( ) ( ) (
1 0) ( )
1 1
0 1
1
1
s a s a Y s b s b s b s b X s
a s
a
n n+
n− n−+ L + + =
m m+
m− m−+ L + + The transfer function between x(t) and y(t) is given by
0 1
1 1
0 1
1 1
) (
) ) (
( a s a s a s a
b s b s
b s
b s
X s s Y
G
nn n
n
m m
m m
+ +
+ +
+ +
+
= +
=
−−
−
−
L L
)]
( ) ( [ )
( t L
1G s X s
y =
−Transfer function
Measurement system
x(t) y(t)
y(0)
) ( )
) (
( y t Kx t
dt t
dy + =
For the first-order system:
τLaplace Transform:
τ (sY (s)− y(t = 0)) +Y(s) = KX (s)1 0
) 1 (
)
( y
s s s X
s K
Y + +
= +
τ τ τ
) ( )
( ) ( )
(s G s X s Q s
Y = +
This can be rewritten:
Where Y(s) and X(s) = Laplace Transform of y(t) and x(t)
Zero state response
Zero input response
Transfer function
) ( )
) ( ( 2
) ( 1
2 2
2 y t Kx t
dt t dy dt
t y d
n n
= +
+ ω ζ ω
The frequency response of a system can be derived from the transfer function by substituted s with i ω ω ω ω :
[
1( ) ]
( ) ( ) ( )) 1
( 1/2
2 φ ω ω φ ω
ω τω
ω τ ∠ = ∠
= +
= +
= K KM
i i K
s G
For the second-order system:
( ) ( ) ( ) ( )
′
+
+
+ + +
+
= 2 2 + 2 2 2 2 0 02
1 /
2 /
1 ) 1 1 (
/ 2 /
) 1 (
n n
n n
n n
n
y y s
s s s
s X s
s K
Y ω ω
ζ ω
ω ζ ω
ω ζ ω
) (s ) Q
( ) (s X s G
(
1/) (
2 /)
1)
( 2 2
+
= +
s s
s K G
n
n ζ ω
Transfer function ω
( ) ( )
[ ]
( ) ( )
[
1 / ] 2 /]
( )
1 ) ( /
2 )
( /
) 1 (
2 / 2 1 2 2
2 2
ω ω φ
ζω ω
ω
ω ω
ζ ω
ω ω
+ ∠
= −
+
= +
n n
n n
K
i i
i K Frequency response G
ω ω ω
ω ω ζ
φ / /
tan 2 )
( 1
n
n −
= − −
ωτ ω
φ( ) = tan−1−
Coupled system
When a measurement system consists of more than one instrument, measurement system behavior can become more complicated.
Measurement system 1 G1(s)
x(t)
y
1(t)
Measurement
system 2 G2(s)
y(t)
G1(s)
X(s)
Y
1(s)
G2(s)
Y(s)
G1(s)G2(s)
X(s) Y(s)
The overall transfer function of the combined system is the product of the transfer function of each system
Equivalent system
Ex (P3.34) The output stage of a first-order transducer is to be connected to a second-order display stage device. The transducer has a known time constant of 1.4 ms and static sensitivity of 2 V/oC, whereas the display has values of
sensitivity, damping ratio, and natural frequency of 1 V/V, 0.9, and 5000 Hz,
respectively. Determine the steady response of this measurement system to an input signal of the from T(t) = 10 + 50 sin 628t oC.
Solution
Coupled system
G1(s)
1st order system
X(s)
Y
1(s)
G2(s)
2nd order system
Y(s)
G1(s)G2(s)
X(s) Y(s)
Equivalent system
) 1
( 1
1 = +
s s K
G τ G2(s)=
(
1/ωn2)
s2 +K(
22ζ /ωn)
s+1(
1) [ (
1/) (
2 /)
1]
) ( )
( 2 2 21 2
1 = + + +
s s
s
K s K
G s G
n
n ζ ω
ω τ
