Convolution solutions (Sect. 6.6).
I
Convolution of two functions.
I
Properties of convolutions.
I
Laplace Transform of a convolution.
I
Impulse response solution.
I
Solution decomposition theorem.
Convolution of two functions.
Definition
The convolution of piecewise continuous functions f , g : R → R is
the function f ∗ g : R → R given by
Z t
(f ∗ g )(t) =
f (τ )g (t − τ ) dτ.
0
Remarks:
I
f ∗ g is also called the generalized product of f and g .
I
The definition of convolution of two functions also holds in
the case that one of the functions is a generalized function,
like Dirac’s delta.
Convolution of two functions.
Example
Find the convolution of f (t) = e −t and g (t) = sin(t).
Z t
Solution: By definition: (f ∗ g )(t) =
e −τ sin(t − τ ) dτ .
0
Z
t
Integrate by parts twice:
e −τ sin(t − τ ) dτ =
0
h
i t h
i t Z t
−τ
−τ
e cos(t − τ ) − e sin(t − τ ) −
e −τ sin(t − τ ) dτ,
0
Z
t
e
2
−τ
0
h
sin(t − τ ) dτ = e
−τ
0
i t
i t h
−τ
cos(t − τ ) − e sin(t − τ ) ,
0
0
2(f ∗ g )(t) = e −t − cos(t) − 0 + sin(t).
1
We conclude: (f ∗ g )(t) = e −t + sin(t) − cos(t) .
2
Convolution solutions (Sect. 6.6).
I
Convolution of two functions.
I
Properties of convolutions.
I
Laplace Transform of a convolution.
I
Impulse response solution.
I
Solution decomposition theorem.
0
C
Properties of convolutions.
Theorem (Properties)
For every piecewise continuous functions f , g , and h, hold:
(i) Commutativity:
f ∗g =g ∗f;
(ii) Associativity:
f ∗ (g ∗ h) = (f ∗ g ) ∗ h;
(iii) Distributivity:
f ∗ (g + h) = f ∗ g + f ∗ h;
(iv) Neutral element: f ∗ 0 = 0;
(v) Identity element: f ∗ δ = f .
Proof:
Z
t
f (τ ) δ(t − τ ) dτ = f (t).
(v): (f ∗ δ)(t) =
0
Properties of convolutions.
Proof:
(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
Z t
(f ∗ g )(t) =
f (τ ) g (t − τ ) dτ.
0
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
Z
(f ∗ g )(t) =
0
f (t − τ̂ ) g (τ̂ )(−1) d τ̂
t
Z
t
g (τ̂ ) f (t − τ̂ ) d τ̂
(f ∗ g )(t) =
0
We conclude: (f ∗ g )(t) = (g ∗ f )(t).
Convolution solutions (Sect. 6.6).
I
Convolution of two functions.
I
Properties of convolutions.
I
Laplace Transform of a convolution.
I
Impulse response solution.
I
Solution decomposition theorem.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ] L[g ].
Proof: The key step is to interchange two integrals. We start we
the product of the Laplace transforms,
hZ ∞
i hZ
−st
L[f ] L[g ] =
e f (t) dt
0
L[f ] L[g ] =
L[f ] L[g ] =
0
∞
Z
e −s t̃ g (t̃)
0
Z
e
−s t̃
i
g (t̃) d t̃ ,
0
∞
Z
∞
e
−st
f (t) dt d t̃,
0
∞
Z
g (t̃)
0
∞
e
−s(t+t̃)
f (t) dt d t̃.
Laplace Transform of a convolution.
∞
Z
Proof: Recall: L[f ] L[g ] =
Z
g (t̃)
0
∞
e
−s(t+t̃)
f (t) dt d t̃.
0
Change variables: τ = t + t̃, hence dτ = dt;
Z ∞
Z ∞
−sτ
L[f ] L[g ] =
g (t̃)
e
f (τ − t̃) dτ d t̃.
t̃
0
t
Z
∞Z ∞
L[f ] L[g ] =
t = tau
e
0
−sτ
g (t̃) f (τ − t̃) dτ d t̃.
t̃
The key step: Switch the order of integration.
0
Z
∞Z τ
L[f ] L[g ] =
0
e −sτ g (t̃) f (τ − t̃) d t̃ dτ.
0
Laplace Transform of a convolution.
∞Z τ
Z
Proof: Recall: L[f ] L[g ] =
e −sτ g (t̃) f (τ − t̃) d t̃ dτ .
0
0
Then, is straightforward to check that
Z ∞
Z τ
−sτ
e
L[f ] L[g ] =
g (t̃) f (τ − t̃) d t̃ dτ,
0
0
Z
L[f ] L[g ] =
∞
e −sτ (g ∗ f )(τ ) dt
0
L[f ] L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ] L[g ].
tau
Convolution solutions (Sect. 6.6).
I
Convolution of two functions.
I
Properties of convolutions.
I
Laplace Transform of a convolution.
I
Impulse response solution.
I
Solution decomposition theorem.
Impulse response solution.
Definition
The impulse response solution is the function yδ solution of the IVP
yδ00 + a1 yδ0 + a0 yδ = δ(t − c),
yδ (0) = 0,
yδ0 (0) = 0,
c ∈ R.
Example
Find the impulse response solution of the IVP
yδ00 + 2 yδ0 + 2 yδ = δ(t − c),
yδ (0) = 0,
yδ0 (0) = 0.
Solution: L[yδ00 ] + 2 L[yδ0 ] + 2 L[yδ ] = L[δ(t − c)].
2
(s + 2s + 2) L[yδ ] = e
−cs
⇒
e −cs
L[yδ ] = 2
.
(s + 2s + 2)
Impulse response solution.
Example
Find the impulse response solution of the IVP
yδ00 + 2 yδ0 + 2 yδ = δ(t − c),
yδ (0) = 0,
yδ0 (0) = 0, .
e −cs
Solution: Recall: L[yδ ] = 2
.
(s + 2s + 2)
Find the roots of the denominator,
s 2 + 2s + 2 = 0
⇒
s± =
√
1
−2 ± 4 − 8
2
Complex roots. We complete the square:
h
2
i
2
2
s + 2s + 2 = s + 2
s + 1 − 1 + 2 = (s + 1)2 + 1.
2
e −cs
Therefore, L[yδ ] =
.
(s + 1)2 + 1
Impulse response solution.
Example
Find the impulse response solution of the IVP
yδ00 + 2 yδ0 + 2 yδ = δ(t − c),
Solution: Recall: L[yδ ] =
Recall: L[sin(t)] =
yδ (0) = 0,
yδ0 (0) = 0, .
e −cs
.
(s + 1)2 + 1
1
, and L[f ](s − c) = L[e ct f (t)].
2
s +1
1
= L[e −t sin(t)]
2
(s + 1) + 1
⇒
L[yδ ] = e −cs L[e −t sin(t)].
Since e −cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ (t) = u(t − c) e −(t−c) sin(t − c).
C
Convolution solutions (Sect. 6.6).
I
Convolution of two functions.
I
Properties of convolutions.
I
Laplace Transform of a convolution.
I
Impulse response solution.
I
Solution decomposition theorem.
Solution decomposition theorem.
Theorem (Solution decomposition)
The solution y to the IVP
y 00 + a1 y 0 + a0 y = g (t),
y 0 (0) = y1 ,
y (0) = y0 ,
can be decomposed as
y (t) = yh (t) + (yδ ∗ g )(t),
where yh is the solution of the homogeneous IVP
yh00 + a1 yh0 + a0 yh = 0,
yh (0) = y0 ,
yh0 (0) = y1 ,
and yδ is the impulse response solution, that is,
yδ00 + a1 yδ0 + a0 yδ = δ(t),
yδ (0) = 0,
yδ0 (0) = 0.
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y 00 + 2 y 0 + 2 y = sin(at),
y (0) = 1,
y 0 (0) = −1.
Solution: L[y 00 ] + 2 L[y 0 ] + 2 L[y ] = L[sin(at)], and recall,
L[y 00 ] = s 2 L[y ] − s (1) − (−1),
L[y 0 ] = s L[y ] − 1.
(s 2 + 2s + 2) L[y ] − s + 1 − 2 = L[sin(at)].
L[y ] =
(s + 1)
1
+
L[sin(at)].
(s 2 + 2s + 2) (s 2 + 2s + 2)
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y 00 + 2 y 0 + 2 y = sin(at),
Solution: Recall: L[y ] =
But: L[yh ] =
and: L[yδ ] =
(s 2
(s 2
y (0) = 1,
y 0 (0) = −1.
(s + 1)
1
+ 2
L[sin(at)].
+ 2s + 2) (s + 2s + 2)
(s + 1)
(s + 1)
=
= L[e −t cos(t)],
2
+ 2s + 2)
(s + 1) + 1
1
1
=
= L[e −t sin(t)]. So,
2
2
(s + 1) + 1
(s + 2s + 2)
L[y ] = L[yh ] + L[yδ ] L[g (t)] ⇒ y (t) = yh (t) + (yδ ∗ g )(t),
Z t
−t
So: y (t) = e cos(t) +
e −τ sin(τ ) sin[a(t − τ )] dτ .
C
0
Solution decomposition theorem.
Proof: Compute: L[y 00 ] + a1 L[y 0 ] + a0 L[y ] = L[g (t)], and recall,
L[y 00 ] = s 2 L[y ] − sy0 − y1 ,
L[y 0 ] = s L[y ] − y0 .
(s 2 + a1 s + a0 ) L[y ] − sy0 − y1 − a1 y0 = L[g (t)].
L[y ] =
Recall: L[yh ] =
(s + a1 )y0 + y1
1
+
L[g (t)].
(s 2 + a1 s + a0 ) (s 2 + a1 s + a0 )
(s + a1 )y0 + y1
1
and
L[y
]
=
,
.
δ
(s 2 + a1 s + a0 )
(s 2 + a1 s + a0 )
Since, L[y ] = L[yh ] + L[yδ ] L[g (t)], so y (t) = yh (t) + (yδ ∗ g )(t).
Z t
Equivalently: y (t) = yh (t) +
yδ (τ )g (t − τ ) dτ .
0