Math 201 Lecture 18: Convolution
Feb. 17, 2012
•
Many examples here are taken from the textbook. The first number in () refers to the problem number
in the UA Custom edition, the second number in () refers to the problem number in the 8th edition.
0. Review
•
Laplace transform of functions with jumps:
1. Represent the function using unit jump functions:
g(t) = g1(t) + [g2(t) − g1(t)] u(t − t1) + + [gk(t) − gk−1(t)] u(t − tk−1).
(1)
2. Transforming each g(t) u(t − a) as follows:
a. Obtain g(t + a).
b. Obtain L{g(t + a)}.
c. L{g(t) u(t − a)} = e−as L{g(t + a)}(s).
•
Inverse Laplace transform of functions involving e−as:
1. Write the function to be inverse transformed as e−as F (s).
2. Obtain f (t) = L−1{F (s)}.
3. Obtain f (t − a).
4. The inverse transform is f (t − a) u(t − a).
•
Laplace transform of periodic functions: If f has period T and is piecewise continuous on [0, T ], then
L{f }(s) =
•
Quiz:
L{f (t)}, f (t) =
R
T
0
e−st f (t) dt
.
1 − e−sT
1 0<t<2
;
t t>2
−3s e
.
L−1
s3
(2)
(3)
1. Basic Information
•
Equation with general right hand sides and general initial conditions.
In practice, we often face the following situation: We need to compute the output of a system for
many different inputs, sometimes we also need to find out a general relation between the inputs and
the outputs.
After mathematical modeling, this situation is translated to the following. Given a differential
operator (corresponding to the system), we need to compute the solutions for many different right
hand sides/initial values. For example, we may need to consider problems like
y ′′ + y = g(t);
y(0) = a,
y ′(0) = b
(4)
and need to solve it for many different g’s.
We follow our three steps.
1. Transform the equation. Using the initial values and denoting Y (s) = L{y }, G(s) = L{g }, we
have
(s2 + 1) Y = G + s a + b.
1
(5)
Math 201 Lecture 18: Convolution
2
2. Solve the transformed equation. We have
Y (s) =
1
[G(s) + s a + b].
s2 + 1
(6)
3. Transform back. We need to compute
where H(s) 6
•
1
,
s2 + 1
y = L−1{Y } = L−1 H(s) G̃(s)
(7)
and G̃(s) 6 G(s) + s a + b.
From the above we can make the following observations.
A) The Laplace transform of the solution is the product of two functions. One of them, H(s), is
determined by the differential operator (the system); The other, G̃(s), is only dependent on
the right hand side and the initial values (the inputs).
Thus it is possible to simplify the process of solving equations with many different inputs,
as H(s) remains the same.
B) To be able to take advantage of the above observation, we need to be able to compute the
inverse Laplace transforms
of product of two functions.
Or more specifically, if we know that
L−1{H } = h, L−1 G̃ = g̃ , can we obtain L−1 H G̃ using h and g̃ ?
The answer is yes:
L−1 H G̃ = h∗g̃ ,
(8)
the convolution of h and g̃ .
•
Transfer function.
Definition 1. (Transfer function/Impulse response function) Consider the linear system
a y ′′ + b y ′ + c y = g,
y(0) = y ′(0) = 0,
(9)
Let Y = L{y }, G = L{g }. The function H(s) = Y /G is called the transfer function of the
linear system. Its inverse Laplace transform, h(t) = L−1{H(s)}(t), is called the impulse response
function for the system.
Remark 2. We will see later that h(t) solve the equation
a y ′′ + b y ′ + c y = δ(t)
(10)
where δ(t) is a special function such that
L{δ } = 1.
(11)
It turns out that this δ(t) cannot be a usual function. It belongs to the so-called “generalized functions”.
Example 3. Find the transfer function and the impulse response function for
y ′′ + 9 y = g(t).
(12)
Solution. Taking Laplace transform we have (recall that when talking about transfer function, we
always assume y(0) = y ′(0) = 0. )
s2 Y + 9 Y = G
So the transfer function is
H(s) =
H=
Y
1
=
.
G s2 + 9
1
.
s2 + 9
(13)
(14)
Feb. 17, 2012
3
The impulse response function is then given by
h(t) = L−1{H } =
•
1
sin 3 t.
3
(15)
Convolution.
Definition 4. (Convolution) Let f (t) and g(t) be piecewise continuous on [0, ∞). The convolution
of f (t) and g(t), denoted f ∗g, is defined by
Z t
(f ∗g)(t): =
f (t − v) g(v) dv.
(16)
0
Example 5. For example, the convolution of f = t and g = t2 is
Z t
Z t
1
1
t4
t∗t2 =
(t − v) v 2 dv =
t v 2 − v 3 dv = t4 − t4 = .
3
4
12
0
0
(17)
The key property of convolution is the following
Theorem 6. (Convolution Theorem) Let f (t) and g(t) be piecewise continuous on [0, ∞) and of
exponential order α and set F (s) = L{f }(s) and G(s) = L{g }(s). Then
or equivalently
L{f ∗g }(s) = F (s) G(s),
(18)
L−1{F (s) G(s)} = (f ∗g)(t).
(19)
Example 7. (7.7 13; 7.7 13) Find the Laplace transform of
Z t
f (t) =
(t − v) e3v dv.
(20)
0
Solution. We recognize that in fact
f (t) = t∗e3t.
(21)
Thus
L{f } = L{t} L{e3t } =
•
What convolution can do:
1 1
1
=
.
s2 s − 3 s2 (s − 3)
1. An alternative method of computing inverse Laplace transforms;
2. Enable us to solve special integral-differential equations;
3. Obtain formula for solution when the right hand side or initial values are not given.
2. Things to be Careful/Tricky Issues
See “Common Mistakes” for examples.
•
3. Examples
•
Convolution as alternative method fo inverse Laplace transforms:
Example 8. (7.7 5; 7.7 5) Compute
L−1
1
s (s2 + 1)
(22)
Math 201 Lecture 18: Convolution
4
Solution. Using Convolution Theorem, we have
1
1
−1 1
−1
L−1
=
L
∗L
s
s2 + 1
s (s2 + 1)
= 1∗sin t
Z t
=
sin v dv
0
= 1 − cos t.
(23)
Example 9. (7.7 12; 7.7 12) Compute
L
−1
s+1
.
(s2 + 1)2
Solution. We have
s+1
1
−1 s + 1
=
L
L −1
·
s2 + 1 s2 + 1
(s2 + 1)2
1
−1 s + 1
−1
= L
L
s2 + 1
s2 + 1
= [cos t + sin t]∗sin t
Z t
Z t
=
cos (t − v) sin v dv +
sin (t − v) sin v dv
0
0
Z t
Z t
1
1
[sin t − sin (t − 2 v)] dv +
[cos (t − 2 v) − cos t] dv
=
0 2
0 2
Z t
Z t
1
1
t (sin t − cos t) −
=
sin (t − 2 v) dv +
cos (t − 2 v) dv
2 0
2
0
1
1
v=t
t (sin t − cos t) − [cos (t − 2 v)N v=0 −sin (t − 2 v)N t0 ]
=
4
2
1
1
=
t (sin t − cos t) − sin t.
2
2
•
(24)
(25)
Solving integral-differential equations:
Example 10. (7.7 15; 7.7 15 ) Solve
Z t
y(t) + 3
y(v) sin (t − v) dv = t.
(26)
0
Solution. We try following the same three steps.
1. Transform the equation. The key is to notice that the integral term is in fact a convolution:
Z t
y(v) sin (t − v) dv = y(t)∗sin t.
(27)
0
Thus taking Laplace transform of both sides, we have
Y (s) + 3 Y (s)
1
1
= .
1 + s2 s2
(28)
2. Solve the transformed equation. We have
Y (s) =
1 + s2
1
1
1 1 3 1
= 2 −3
=
−
.
2
2
2
2
(4 + s ) s
s
(4 + s ) s
4 s2 4 s 2 + 4
(29)
3. Transform back. Compute
t 3
1 1 3 1
−
= − sin 2 t.
y(t) = L−1
4 8
4 s2 4 s2 + 4
(30)
Feb. 17, 2012
5
Example 11. (7.7 21; 7.7 21 ) Solve
Z t
y ′(t) + y(t) −
y(v) sin (t − v) dv = −sin t,
y(0) = 1.
(31)
0
Solution. Again,
1. Transform the equation. We have
L{y ′} = s Y − y(0) = s Y − 1.
and
Z
L
0
t
(32)
Y (s)
y(v) sin (t − v) dv = Y (s) L{sin t} =
.
1 + s2
(33)
So the equation is transformed into
sY +Y −1−
1
Y (s)
=−
.
1 + s2
1 + s2
(34)
2. Solve the transformed equation. We have
Y=
s
.
s2 + s + 1
(35)
3. Transform back. Compute
y(t) = L−1(
{Y }
1/2
s + 1/2
=
√
√
2 −
2
2
(s + 1/2)2 +
3 /2
(s + 1/2) +
3 /2
"
√ #
√
1
1
3t
3t
−2 t
.
− √ sin
cos
= e
2
2
3
L−1
•
)
(36)
Obtain formula of solution for general right hand side:
Example 12. (7.7 1; 7.7 1 ) Use the convolution theorem to obtain a formula for the solution to
the given initial value problem.
Solution. Using
y ′′ − 2 y ′ + y = g(t);
y(0) = −1,
y ′(0) = 1.
(37)
L{y ′′} = s2 Y − s y(0) − y ′(0) = s2 Y + s − 1,
(38)
L{y ′} = s Y − y(0) = s Y + 1,
(39)
we have the transformed equation
(s2 − 2 s + 1) Y = G(s) − s + 3
Y (s) =
1
1
2
G(s) −
+
.
(s − 1)2
(s − 1) (s − 1)2
Transforming back, we have
1
1
−1
L−1
G(s)
=
L
∗g(t) = (et t)∗g(t);
(s − 1)2
(s − 1)2
1
−1
= et;
L
s−1
2
L−1
= 2 et t.
(s − 1)2
(40)
(41)
(42)
(43)
So the formula for the solution reads
y(t) = 2 t et − et + (et t)∗g(t) = 2 t et − et +
Z
0
t
et−v (t − v) g(v) dv.
(44)
Math 201 Lecture 18: Convolution
6
Finally let’s do an example that combined both 7.6 and 7.7.
Example 13. (7.6 39; 7.6 39) Solve
y ′′ + 5 y ′ + 6 y = g(t),
y(0) = 0,
y ′(0) = 2,
(45)
where
Solution.
1. Transform the equation.

 0 06t<1
g(t) = t 1 < t < 5 .

1 5<t
(46)
a. Transform the LHS. Compute
L{y ′′} = s2 Y − s y(0) − y ′(0) = s2 Y − 2;
(47)
L{y ′} = s Y − y(0) = s Y .
(48)
L{y ′′ + 5 y ′ + 6 y } = (s2 + 5 s + 6) Y − 2.
(49)
So
b. Transform the RHS. We first need to represent g using unit step functions. There are two
discontinuities at t = 1, 5. We write
g(t) = 0 + A(t) u(t − 1) + B(t) u(t − 5)
and determine
A(t) = t,
So
(50)
B(t) = 1 − t.
(51)
g(t) = t u(t − 1) + (1 − t) u(t − 5).
(52)
L{g(t) u(t − a)} = e−as L{g(t + a)}(s)
(53)
L{t u(t − 1) + (1 − t) u(t − 5)} = e−s L{t + 1}
+ e−5s L{−t
− 4}
1
1
1 4
+
= e−s
− e−5s 2 +
.
s
s s2
s
(54)
Recalling
We compute
Thus the transformed equation reads
2
(s + 5 s + 6) Y = e
−s
1 1
+
s s2
−e
−5s
1 4
+
s2 s
+ 2.
(55)
2. Solve the transformed equation. We have
Y (s) = e−s
s+1
4s+1
2
− e−5s 2 2
+
.
s2 (s2 + 5 s + 6)
s (s + 5 s +6) s2 + 5 s + 6
(56)
3. Transform back. We need to use the formula
L−1{e−as F (s)} = f (t − a) u(t − a).
s+1
(57)
4s+1
Thus we have to compute the inverse transforms of s2 (s2 + 5 s + 6) and s2 (s2 + 5 s +6) . Due to linearity
of the inverse transform, we first compute
s
1
L−1 2 2
and L−1 2 2
.
(58)
s (s + 5 s + 6)
s (s + 5 s + 6)
Feb. 17, 2012
7
We have
L
−1
s
s2 (s2 + 5 s + 6)
=
=
=
=
=
1
L
s (s2 + 5 s + 6)
1
1
L−1
s (s + 2) (s + 3)
1
1
1
−1
−1
−
L
∗L
s
s+2 s+3
1∗(e−2t − e−3t }
Z t
(e−2v − e−3v) dv
−1
0
L−1
1
s2 (s2 + 5 s + 6)
=
=
=
=
=
=
=
=
1 1 −2t 1 1 −3t
− e
− + e
=
3 3
2 2
1 1 −2t 1 −3t
=
− e
+ e .
(59)
6 2
3
1
1
1
L−1 2 ∗L−1
−
s
s+2 s+3
t∗(e−2t − e−3t)
Z t
(t − v) (e−2v − e−3v ) dv
0
Z t
1
1
(t − v) d − e−2v + e−3v
2
3
0
Z t
1 −2v 1 −3v v=t
+ e
N v=0 − − 12 e−2v + 13 e−3v d(t − v)
(t − v) − e
3
2
0
Z t
1
1 −2v 1 −3v
1
dv
− e
t− t+
+ e
2
2
3
3
0
1
1
1 1
1
t + e−2t − − e−3t +
6
4
4 9
9
1
1
1
5
t + e−2t − e−3t − .
(60)
6
4
9
36
Thus we have
which leads to
L−1 e−s
L
−1
s+1
s2 (s2 + 5 s + 6)
s+1
s2 (s2 + 5 s + 6)
=
=
t 1
2
1
+ − e−2t + e−3t
9
36 6 4
t − 1 1 −2(t−1) 2 −3(t−1)
1
+
− e
+ e
u(t − 1);
6
4
9
36
For the other term we have
4s+1
19 t 7 −2t 11 −3t
+ − e
+
e
L−1 2 2
=
s (s + 5 s + 6)
9
36 6 4
which leads to
4s+1
19 t − 5 7 −2(t−5) 11 −3(t−5)
u(t − 5).
L −1 e−5s 2 2
+
− e
+
e
=
s (s + 5 s + 6)
6
4
9
36
Finally we quickly compute
2
1
1
−1
−1
L
−
=2L
= 2 e−2t − 2 e−3t.
s2 + 5 s + 6
s+2 s+3
Summarizing, we reach
t − 1 1 −2(t−1) 2 −3(t−1)
1
+
− e
+ e
u(t − 1)
y(t) =
6
4
9
36
19 t − 5 7 −2(t−5) 11 −3(t−5)
u(t − 5) + 2 e−2t − 2 e−3t.
+
− e
+
e
−
6
4
9
36
(61)
(62)
(63)
(64)
(65)
(66)
Math 201 Lecture 18: Convolution
8
Remark 14. Note that for the above problem, it is actually more efficient to compute the inverse transforms
using partial fractions. For example, to compute
s+1
−1
L
(67)
s2 (s2 + 5 s + 6)
we write
s+1
As+B
C
D
=
+
+
(68)
s2 (s2 + 5 s + 6)
s2
s+2 s+3
which leads to
s + 1 = (A s + B) (s + 2) (s + 3) + C s2 (s + 3) + D s2 (s + 2).
Setting s to be special values, we get
1
B= ;
6
1
s = −2
−1 = 4 C
C =− ;
4
2
s = −3
−2 = −9 D
D= ;
9
3
Finally comparing coefficients for the highest order term s we get
s=0
1=6B
0=A+C +D
A=
1
.
36
(69)
(70)
(71)
(72)
(73)
4. Notes and Comments
•
Other properties of convolution.
Theorem 15. (Other properties of convolution) Let f (t), g(t), and h(t) be piecewise continuous
on [0, ∞). Then
a) f ∗g = g∗f;
b) f ∗(g + h) = (f ∗g) + (f ∗h);
c) (f ∗g)∗h = f ∗(g∗h);
d ) f ∗0 = 0.
Proof. One way to prove these properties is to use definition (p.425 in the textbook). Here we use
the Convolution Theorem. We use F , G, H to denote the Laplace transforms of f , g, h.
a) We have
b) We have
f ∗g = L−1{L{f ∗g }} = L−1{F G} = L−1{G F } = g∗f .
f ∗(g + h) =
=
=
=
=
L −1{L{f ∗(g + h)}}
L −1{F (G + H)
L −1{F G + F H }
L −1{F G} + L−1{F H }
f ∗g + f ∗h.
(74)
(75)
c) We have
(f ∗g)∗h =
=
=
=
=
=
L−1{L{(f ∗g)∗h}}
L−1{L{f ∗g } H }
L−1{(F G) H }
L−1{F (G H)}
f ∗L−1{G H }
f ∗(g∗h).
(76)
Feb. 17, 2012
9
d) is obvious.
•
Proof of convolution theorem.
Proof. We use definition.
L{f ∗g }(s) =
=
=
=
∞
Z
e−st
Z0Z
Z
t
f (t − v) g(v) dv dt
0
e−st f (t − v) g(v) dv dt
−st
e
f (t − v) g(v) dt dv
v Z ∞
g(v)
e−st f (t − v) dt dv.
0<v<t
∞ Z ∞
Z
Z0 ∞
0
To evaluate the inner integral, we set x = t − v. Then dt = dx, e−st = e−sx e−sv. Thus
Z ∞
Z ∞
e−sx f (x) dx = e−sv F (s).
e−st f (t − v) dt = e−sv
Substituting into the original integral, we have
Z ∞
Z
Z ∞
e−st f (t − v) dt dv = F (s)
L{f ∗g }(s) =
g(v)
Thus ends the proof.
(78)
0
v
0
(77)
v
v
∞
g(v) e−sv dv = F (s) G(s).
(79)
0