Laplace Transform
Systems of First Order Linear Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
” Convolution can be intuitively described as a function that is the
integral or summation of two component functions, and that
measures the amount of overlap as one function is shifted over the
other ”
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
We will start with the following
Example 6.17
Solve the initial value problem
y 00 + y = g (t);
y (0) = 0; y 0 (0) = 0
Solution
The solution of the homogeneus equation is
y (t) = c1 sin(t) + c2 cos(t)
and using the metod of variation of parameters a particular
solution is given by
yP (t) = u1 sin(t) + u2 cos(t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
where
t
Z
u1 =
Z
g (τ )cos(τ )dτ ;
u2 = −
0
t
g (τ )sin(τ )dτ
0
and the particular solution is
t
Z
Z
g (τ )cos(τ )dτ sin(t) −
yP (t) =
0
t
g (τ )sin(τ )dτ cos(t) =
0
Z t
t
Z
g (τ )cos(τ )sin(t) −
0
g (τ )sin(τ )cos(t) dτ
0
Z
t
sin(t − τ )g (τ )dτ
yP (t) =
0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Convolution
Let f and g be piecewise continuous in [0, ∞). The Convolution of
f and g is denoted by f ? g and defined by
Z t
f ?g =
f (t − τ )g (τ )dτ
0
Properties of Convolution
f ?g =g ?f
f ? (g1 + g2 ) = f ? g1 + f ? g2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
(f ? g ) ? h = f ? (g ? h)
f ?0=0?f =0
Theorem 6.4
If F (s) = L {f (t)} and G (s) = L {g (t)}, then
L {f ? g } = F (s)G (s)
or equivalently
f ? g = L −1 {F (s)G (s)}
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Example 6.18
Find the inverse Laplace Tranform of
Y (s) =
G (s)
s2 + 1
where G (s) = L {g (t)}
Solution
Y (s) =
G (s)
= G (s)F (s)
s2 + 1
with
F (s) =
Dr. Marco A Roque Sol
1
s2 + 1
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Thus
f (t) = L −1 {F (s)} = L −1 {
s2
1
} = sin(t)
+1
y (t) = L −1 {Y (s)} = L −1 {F (s)G (s)} = f (t) ? g (t) = sin(t) ? g (t)
Example 6.19
Find the inverse Laplace Tranform of
Y (s) =
1
+ 1)
s(s 2
Solution
Y (s) =
1 1
= F (s)G (s)
s s2 + 1
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
with
1
F (s) = ;
s
G (s) =
s2
1
+1
Thus
1
f (t) = L −1 {F (s)} = L −1 { } = t
s
g (t) = L −1 {G (s)} = L −1 {
s2
1
} = sin(t)
+1
y (t) = L −1 {Y (s)} = L −1 {F (s)G (s)} =
Z
t
Z
f (t − τ )g (τ )dt =
y (t) = f (t) ? g (t) =
0
Dr. Marco A Roque Sol
t
(t − τ )sin(τ )dτ
0
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Example 6.20
Solve the initial value problem
ay 00 + by 0 + cy = g (t);
y (0) = y0 ; y 0 (0) = y00
Solution
Y (s) =
(as + b)y0 + ay00
G (s)
+ 2
=
as 2 + bs + c
as + bs + c
H(s)[(as + b)y0 + ay00 ] + H(s)G (s)
where
G (s) = L {g (t)};
Dr. Marco A Roque Sol
H(s) =
as 2
1
+ bs + c
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
y (t) = L −1 {H(s)[(as + b)y0 + ay00 ] + H(s)G (s)} =
L −1 {H(s)[(as + b)y0 + ay00 ]} + L −1 {H(s)G (s)}
y (t) = L −1 {H(s)[(as + b)y0 + ay00 ]} + h ? g = yh + yg
where
yh = Free response due to initial conditions and
short term effect, by solving ay 00 + by 0 + cy = 0 ;
y (0) = y0 , y 0 (0) = y00
yf = Force response due to g (t) or external excitation
long term efect.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Total response = Free response + Forced response
at s−domain
Y (s) = H(s)[(as + b)y0 + ay00 ] + H(s)F (s)
at t−domain
Z
t
h(t − τ )g (τ )dτ
Y (s) = c1 y1 + c2 y2 +
0
Z
t
h(t − τ )g (τ )dτ is known as
The integral,
0
delayed effect or learning experience or memory effect.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Finally, h(t) = L −1 {H(s)} is the solution of
ay 00 + by 0 + cy = δ(t);
L {δ} = 1;
y (0) = y 0 (0) = 0
H(s) = Y (s) =
1
as 2 + bs + c
H(s) is called the transfer function and h(t) is called the impulse
function
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Example 6.21
Consider the equation
y 00 + 2y 0 + 5y = g (t)
Find
a) Find the transfer function H(s) and the impulse response h(t).
b) Find the general solution in terms of a convolution.
c) Find the total response if the initial conditions are
y (0) = 1, y 0 (0) = −3
d) Find the force response when g (t) = t .
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Solution
a) Find the transfer function H(s) and the impulse response h(t).
H(s) =
s2
1
1
1
=
=⇒ h(t) = e −t sin(2t)
2
2
+ 2s + 5
(s + 1) + 2
2
b) Find the general solution in terms of a convolution.
r 2 + 2r + 5 = 0 =⇒ r1,2 =
−2 ±
p
4 − (4)(5)
=⇒ r1,2 = −1 ± 2 i
2
yh = e −t (c1 sin(2t) + c2 cos(2t))
Z
yf = h ? g =
0
t
1 −(t−τ )
e
sin2(t − τ )g (τ )dτ
2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
c) Find the total response if the initial conditions are yh (0) = 1,
yh0 (0) = −3
1 = yh (0) = c2 ;
y (t) = e
−t
−3 = yh0 (0) = 1 + 2c1 =⇒ c2 = −1
Z
(−1sin(2t) + cos(2t)) +
0
t
1 −(t−τ )
e
sin2(t − τ )g (τ )dτ
2
d) Find the force response when g (t) = t .
If g (t) = t, then
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
Z
y (t) = h ? g =
0
t
1 −(t−τ )
e
sin2(t − τ )g (τ )dτ
2
but, remember that also we have
t
Z
y (t) = h ? g =
0
y (t) = L −1 {
1 −(t−τ )
e
sin2(t − τ )g (τ )dτ = L −1 {H(s)G (s)} =
2
[s 2
1
1
} = L −1 {
}
2
2
+ 2s + 5] s
[(s + 1) + 22 ] s 2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
The Convolution Integral
The Convolution Integral
a
b
cs + d
y (t) = L −1 { + 2 +
}
s
s
(s + 1)2 + 22
y (t) = L −1 {−
y (t) = −
(2/25)(s + 1) − (3/50)2
2 1 1 1
+
+
}
2
25 s
5s
(s + 1)2 + 22
1
2
3
2
+ t + e −t cos(2t) − e −t sin(2t)
25 5
25
50
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
Introduction
Review of Matrices
Let us start by solving an m × n system of linear equations
a11 x1 + a12 x2 + . . . + a1n xn =
a21 x1 + a22 x2 + . . . + a2n xn =
..
.
b1
b2
..
.
am1 x1 + am2 x2 + . . . + amn xn = bm
0 s are given right-hand side, and
where aij are given coefficients, bm
0
xn s are the unknowns. In this way, we can introduce new arrays of
numbers to study the linear system


a11 a12 . . . a1n
 a21 a22 . . . a2n 


A=

..


.
am1 am2 . . . amn
Dr. Marco A Roque Sol


x1
 x2 
 
X = . 
 .. 

b1
 b2 
 
B= . 
 .. 
xm
Ordinary Differential Equations

bm
Laplace Transform
Systems of First Order Linear Equations
Introduction
Review of Matrices
In this way, we have the following
Definition
An m × n matrix A , is an array of complex numbers ( m-rows and
n-columns ), denoted by


a11 a12 . . . a1n
 a21 a22 . . . a2n 


A=
 = (aij )m×n
..


.
am1 am2 . . . amn
In this context, an element in the i-row and j-column is of the
matrix A denoted by aij . Associated with any m × n A matrix, we
have the following matrices:
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
Introduction
Review of Matrices
a) Transpose
An ( n × m ) matrix. Denoted by AT and defined by


a11 a21 . . . am1
a12 a22 . . . am2  

= (aji )m×n
AT = 
 = aijT
..


n×m
.
a1n a2n . . . amn
b) Complex Conjugate
A ( m × n ) matrix. Denoted by A and defined by


a11 a12 . . . a1n
 a21 a22 . . . a2n 


A=
 = (aij )m×n = (aij )m×n
..


.
am1 am2 . . . amn
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
Introduction
Review of Matrices
c) Adjoint
A ( m × n ) matrix. Denoted

a11 a21 . . .
a12 a22 . . .

A∗ = 
..

.
a1n a2n . . .
T
by A∗ = A and defined by

am1
am2 

 = aij∗ n×m = (aji )m×n

amn
Basic Matrix Operations
Let A = (aij )m×n and B = (bij )m×n be two matrices, then we
define
Dr. Marco A Roque Sol
Ordinary Differential Equations
Laplace Transform
Systems of First Order Linear Equations
Introduction
Review of Matrices
1) A = B ⇐⇒
aij = bij ;
i = 1, 2, ..., m,
j = 1, 2, ..., n
2) Addtion
A ± B = (aij ± bij )m×n
e) Scalar Multiplication
r A = (raij )m×n
Dr. Marco A Roque Sol
Ordinary Differential Equations