6.003: Signals and Systems
Lecture 7
October 1, 2009
1
6.003: Signals and Systems
Laplace and Z Transforms
October 1, 2009
Midterm Examination #1
Wednesday, October 7, 7:309:30pm, Walker Memorial.
No recitations on the day of the exam.
Coverage:
DT Signals and Systems
Lectures 1–5
Homeworks 1–4
Homework 4 will include practice problems for midterm 1.
However, it will not collected or graded. Solutions will be posted.
Closed book: 1 page of notes (
8
1
2
×
11
inches; front and back).
Designed as 1hour exam; two hours to complete.
Review sessions during open office hours.
Conflict? Contact [email protected] before Friday, October 2, 5pm.
Last Time
Many continuoustime systems can be represented with differential
equations.
Example: leaky tank
r
0
(
t
)
r
1
(
t
)
h
1
(
t
)
Differential equation representation:
τ
˙
r
1
(
t
) =
r
0
(
t
)
−
r
1
(
t
)
Last time we considered two methods to solve differential equations:
•
solving homogeneous and particular equations
•
singularity matching
Solving Differential Equations with Laplace Transform
The Laplace transform provides a particularly powerful method of
solving differential equations — it transforms a differential equation
into an algebraic equation.
Method (where
L
represents the Laplace transform):
differential
algebraic
algebraic
differential equation
−→
↓
solve
−→
differential
equation
algebraic
equation
algebraic
answer
solution to
differential equation
L
−→
L
−
1
−→
Laplace Transform: Definition
Laplace transform maps a function of time
t
to a function of
s
.
X
(
s
) =
x
(
t
)
e
−
st
dt
There are two important variants:
Unilateral (18.03)
X
(
s
) =
∞
0
x
(
t
)
e
−
st
dt
Bilateral (6.003)
X
(
s
) =
∞
−∞
x
(
t
)
e
−
st
dt
Both share important properties — will discuss differences later.
Laplace Transforms
Example: Find the Laplace transform of
x
1
(
t
)
:
0
t
x
1
(
t
)
x
1
(
t
) =
Ae
−
σt
if
t
≥
0
0
otherwise
X
1
(
s
) =
∞
−∞
x
1
(
t
)
e
−
st
dt
=
∞
0
Ae
−
σt
e
−
st
dt
=
Ae
−
(
s
+
σ
)
t
−
(
s
+
σ
)
∞
0
=
A
s
+
σ
provided Re
{
s
+
σ
}
>
0
which implies that Re
{
s
}
>
−
σ
.
−
σ
s
plane
ROC
A
s
+
σ
;
Re
{
s
}
>
−
σ
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6.003: Signals and Systems
Lecture 7
October 1, 2009
2
Check Yourself
0
t
x
2
(
t
)
x
2
(
t
) =
e
−
t
−
e
−
2
t
if
t
≥
0
0
otherwise
Which of the following is the Laplace transform of
x
2
(
t
)
?
1.
X
2
(
s
) =
1
(
s
+1)(
s
+2)
;
Re
{
s
}
>
−
1
2.
X
2
(
s
) =
1
(
s
+1)(
s
+2)
;
Re
{
s
}
>
−
2
3.
X
2
(
s
) =
s
(
s
+1)(
s
+2)
;
Re
{
s
}
>
−
1
4.
X
2
(
s
) =
s
(
s
+1)(
s
+2)
;
Re
{
s
}
>
−
2
5. none of the above
Regions of Convergence
Leftsided signals have leftsided Laplace transforms (bilateral only).
Example:
t
x
3
(
t
)
−
1
x
3
(
t
) =
−
e
−
t
if
t
≤
0
0
otherwise
X
3
(
s
) =
∞
−∞
x
3
(
t
)
e
−
st
dt
=
0
−∞
−
e
−
t
e
−
st
dt
=
−
e
−
(
s
+1)
t
−
(
s
+ 1)
0
−∞
=
1
s
+ 1
provided Re
{
s
+ 1
}
<
0
which implies that Re
{
s
}
<
−
1
.
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 Spring '11
 DennisM.Freeman
 Laplace

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