334.1.11.lec28
DESCRIPTION
334.1.11.lec28TRANSCRIPT
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Lecture 28 Impulse Functions
11/18/2011
Impulse functions.
Impulse functions are functions that are zero everywhere except at one single points, and furthermorewhen integrated over R give a nonzero value.
Unit impulse function (guess this is engineering jargon. In physics its known as the Dirac -function, and in mathematics just ).
(t)= 0 for all t 0, and
(t) dt= 1.
Translations:(t a)= 0 for all t a,
(t a)= 1. (1) Properties.
Its easy to see that t1
t2
(t a) dx={
0 a< t1 or a> t21 t1 t2f(a) t1 0, there is > 0 such that when |t a|
-
=a
a+
(t a) f(a) dt+a
a+
(t a) (f(t) f(a)) dt
= f(a)
a
a+
(t a) dt+a
a+
(t a) (f(t) f(a)) dt
= f(a) +
a
a+
(t a) (f(t) f(a)) dt. (5)
Finally notice:a
a+
(ta) (f(t) f(a)) dt6
a
a+
(ta) |f(t) f(a)| 0.
Clearly est is continuous for every s. So
L{(t a)}=
0
est (t a) dt= eas. (9)
More generally, if f(t) is continuous, then so is est f(t) and we have
L{f(t) (t a)}=
0
est f(t) (t a) dt= eas f(a). (10)
Example 1. Compute
L{et2 sin( 3 ln (cos t)) (t 1)}. (11)Solution. Identify
f(t) = et2
sin(
3
ln (cos t)), a= 1. (12)
So
L{et2 sin( 3 ln (cos t)) (t 1)}= eas f(a) = es e1 sin ( 3 ln (cos 1)). (13)Equations with impulse right hand side.
Example 2. Solve
y + 4 y= (t pi) (t 2pi), y(0)= 0, y (0)= 0. (14)Solution.
First take Laplace transform of the equation: Recall
L{y }= s2Y s y(0) y (0). (15)So the transformed equation is
(s2 + 4) Y = epis e2pis. (16)Next find Y :
Y =epis
s2 +4 e
2pis
s2 + 4. (17)
2 Lecture 28 Impulse Functions
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Finally take the inverse transform: Recall
L1{easF (s)}= u(t a) f(t a) (18)where f(t)=L1{F (s)}.
L1{
epis
s2 +4
}: Identify a=pi, F (s)=
1
s2 +4. So f(t)=
1
2sin (2 t). Therefore
L1{
epis
s2 +4
}= u(tpi) f(t pi)= u(t pi) 1
2sin (2 (t pi)) = 1
2u(t pi) sin (2 t). (19)
L1{
e2pis
s2 +4
}=
1
2u(t 2 pi)sin(2 t).
So finally
y=1
2sin (2 t) [u(tpi) u(t 2 pi)]. (20)
To understand the solution better, we can further write it into the piecewise form:
y=
0 t < pi1
2sin (2 t) pi < t< 2pi
0 t > 2pi
. (21)
Example 3. Solve
2 y + y +4 y= (t pi
6
)sin (t), y(0)= y (0) =0. (22)
Solution. First recall
L{y } = s2 Y s y(0) y (0); (23)L{y } = s Y y(0) (24)
L{(t pi
6
)sin (t)
}= e
pi
6ssin(pi
6
)=
1
2epi
6s. (25)
Now we can write down the transformed equation
(2 s2 + s+ 4) Y =1
2epi
6s. (26)
This gives
Y (s) =1
2
epi
6s
2 s2 + s+ 4. (27)
To compute inverse transform recall
L1{easF (s)}= u(t a) f(t a). (28)Identify:
F (s)=1
2
1
2 s2 + s+ 4; a=
pi
6. (29)
We need to compute
f(t) = L1{
1
2
1
2 s2 + s+ 4
}
=1
4L1
{1
s2 + s/2+ 2
}
=1
4L1
{1(
s+1
4
)2 +
31
16
}
=1
4et/4
4
31 sin
(31
4
t
)
=1
31 et/4 sin
(31
4
t
).
11/18/2011 3
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Now we can write
y=L1{
1
2
epi
6s
2 s2 + s+ 4
}= u
(t pi
6
)1
31 e
(t
pi
6
)/4sin
(31
4
(t pi
6
)). (30)
4 Lecture 28 Impulse Functions