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18.03 Class 27
, April 12, 2010
Laplace Transform II
1. Delta signal
2. tderivative rule
3. Inverse transform
4. Unit impulse response
5. Partial fractions
6. L[f'_r]
Laplace Transform:
F(s) = int_0^\infty f(t) e^{st} dt
We saw that this improper integral may only converge for
Re(s) > a,
for some
a
depending upon
f(t) .
The smallest such
a
gives the
"region of convergence."
Computations so far:
L[1] = 1/s
L[t^n] = n!/s^{n+1}
L[e^{rt}] = 1/(sr)
L[cos(t)] = s / (s^2 + omega^2)
L[sin(t)] = omega / (s^2 + omega^2)
We also have
Rule 1 (Linearity):
af(t) + bg(t) >
aF(s) + bG(s).
Rule 2 (sshift):
L[e^{rt}f(t)] = F(sr)
Warmup: What is the Laplace transform of
f(t) = e^{t} cos(3t) ?
We could do this by writing it as
(1/2)( e^{(1+3i)t} + e^{(13i)t} )
but it's a bit easier to use the sshift :
cos(3t) > s / (s^2 + 9)
so
e^{t} cos(3t) > (s+1)/( (s+1)^2 + 9 )
If you like you can "uncomplete the square" and write this as
= (s+1) / (s^2 + 2s + 10)
[1]
The delta function:
For
b > 0 ,
L[delta(tb)]
=
integral_0^infty delta(tb) e^{st} dt
What could this mean?
If
f(t)
is continuous at
t = b ,
delta(tb) f(t) = f(b) delta(tb) .
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integral delta(tb) f(t) dt = f(b) u(tb) + const
or, if
a < b < c , integral_a^c delta(tb) f(t) dt = f(b) .
In some accounts, this is the DEFINITION of the delta function.
In our situation, you get
L[delta(tb)]
=
e^{bs}
In this case, the region of convergence is the entire complex plane:
the limit you take to get the improper integral is constant as soon as
t > b.
There is a problem when
b = 0 .
delta(t) e^{st} = delta(t)
for any
s ,
but
int_0^infty delta(t) dt = u(infty)  u(0)
and
u(0)
is
indeterminate.
We want the formula that worked for
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 Spring '09
 vogan
 Fractions, Derivative

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