EECS 216 – Discussion #12
(2020-04-10) – [email protected]
Inverse Laplace Transform
o
ℒ
−1
{
𝐻𝐻
(
𝑠𝑠
)}
≡ ℎ
(
𝑡𝑡
) =
1
𝑗𝑗2𝜋𝜋
∫
𝐻𝐻
(
𝑠𝑠
)
𝑒𝑒
𝑠𝑠𝑠𝑠
𝑑𝑑𝑠𝑠
𝑐𝑐+𝑗𝑗∞
𝑐𝑐−𝑗𝑗∞
for
𝑡𝑡 ≥
0
−
and real constant
𝑐𝑐 ∈
ROC
o
This is a “contour integral” in the complex plane; you’d need to use techniques
from the calculus of complex variables, which has similarities to multivariable
calculus (after all,
𝐻𝐻
(
𝑠𝑠
)
is in a way a function of two variables:
Re{
𝑠𝑠
}
and
Im{
𝑠𝑠
}
)
o
In practice, the approach we mostly use to find Inverse Laplace Transforms
is to pattern-match with LT pairs in the table and utilize the LT properties
o
Caution:
Due to the fact that the Unilateral Laplace Transform disregards the values
of the time domain signal for
𝑡𝑡
< 0
−
, the associated Inverse Laplace
Transform, e.g.
ℒ
−1
{
𝐻𝐻
(
𝑠𝑠
)}
, will only be able to “recover” the values of
the time domain signal, e.g.
ℎ
(
𝑡𝑡
)
, for
𝑡𝑡 ≥
0
−
Consequently, the equivalence of convolving in time and multiplying
in s-domain for the Unilateral LT works only when dealing with right-
sided input/output signals and causal (right-sided) impulse responses
For other situations, e.g. pure sinusoids/complex exponentials, first ensure
that the
𝑗𝑗𝑗𝑗
axis is in the ROC, and if so, then evaluate
𝐻𝐻
(
𝑠𝑠
)
, the transfer
function, along the imaginary axis, e.g.
𝑒𝑒
𝑗𝑗𝜔𝜔
0
𝑠𝑠
→ 𝐻𝐻
(
𝑠𝑠
)
→ 𝐻𝐻
(
𝑗𝑗𝑗𝑗
0
)
𝑒𝑒
𝑗𝑗𝜔𝜔
0
𝑠𝑠
(important note:
𝑒𝑒
𝑗𝑗𝜔𝜔
0
𝑠𝑠
≠ 𝑒𝑒
𝑗𝑗𝜔𝜔
0
𝑠𝑠
𝑢𝑢
(
𝑡𝑡
)
,
cos(
𝑗𝑗
0
𝑡𝑡
)
≠
cos(
𝑗𝑗
0
𝑡𝑡
)
𝑢𝑢
(
𝑡𝑡
)
, etc.)
Partial Fractions
o
For many of the problems we will encounter, we will need to use the technique of
Partial Fraction Decomposition
in order to rewrite the
𝐻𝐻
(
𝑠𝑠
)
(or
𝑋𝑋
(
𝑠𝑠
)
or
𝑌𝑌
(
𝑠𝑠
)
) in a
form
so as to be able to utilize the LT pairs/properties tables
o
Here are a few of the scenarios
The general form of a rational function is
𝑋𝑋
(
𝑠𝑠
) =
𝑁𝑁
(
𝑠𝑠
)
𝐷𝐷
(
𝑠𝑠
)
=
(
𝑠𝑠−𝑏𝑏
1
)
⋅⋅⋅
(
𝑠𝑠−𝑏𝑏
𝑚𝑚
)
(
𝑠𝑠−𝑎𝑎
1
)
⋅⋅⋅
(
𝑠𝑠−𝑎𝑎
𝑛𝑛
)
Note:
𝑎𝑎
1
, … ,
𝑎𝑎
𝑛𝑛
and
𝑏𝑏
1
, … ,
𝑏𝑏
𝑚𝑚
could be complex-valued
If
𝑚𝑚
<
𝑛𝑛
•
Here
𝑋𝑋
(
𝑠𝑠
)
is called a
“proper rational function”
, i.e. the degree of
the numerator is strictly less than the degree of the denominator
•
If
𝒂𝒂
𝟏𝟏
, … ,
𝒂𝒂
𝒏𝒏
are all distinct…
o
𝑋𝑋
(
𝑠𝑠
) =
𝐴𝐴
1
𝑠𝑠−𝑎𝑎
1
+
⋯
+
𝐴𝐴
𝑛𝑛
𝑠𝑠−𝑎𝑎
𝑛𝑛
o
Here, constants
𝐴𝐴
𝑖𝑖
= lim
𝑠𝑠→𝑎𝑎
𝑖𝑖
(
𝑠𝑠 − 𝑎𝑎
𝑖𝑖
)
𝑋𝑋
(
𝑠𝑠
)
for
𝑖𝑖
= 1, … ,
𝑛𝑛
•
If you have repeated roots
…
o
For example, if
𝑋𝑋
(
𝑠𝑠
) =
(
𝑠𝑠−𝑏𝑏
1
)
⋅⋅⋅
(
𝑠𝑠−𝑏𝑏
𝑚𝑚
)
(
𝑠𝑠−𝑎𝑎
1
)
𝑘𝑘
(
𝑠𝑠−𝑎𝑎
𝑘𝑘+1
)
⋅⋅⋅
(
𝑠𝑠−𝑎𝑎
𝑛𝑛
)
,
o
Then
𝑋𝑋
(
𝑠𝑠
) =
𝐴𝐴
11
𝑠𝑠−𝑎𝑎
1
+
⋯
+
𝐴𝐴
1𝑘𝑘
(
𝑠𝑠−𝑎𝑎
1
)
𝑘𝑘
+
𝐴𝐴
𝑘𝑘+1
𝑠𝑠−𝑎𝑎
𝑘𝑘+1
+
⋯
+
𝐴𝐴
𝑛𝑛
𝑠𝑠−𝑎𝑎
𝑛𝑛
o
Here,
𝐴𝐴
1ℓ
= lim
𝑠𝑠→𝑎𝑎
1
𝑑𝑑
(
𝑘𝑘−ℓ
)
(
𝑑𝑑𝑑𝑑
)
(
𝑘𝑘−ℓ
)
�
(
𝑠𝑠−𝑎𝑎
1
)
𝑘𝑘
𝑋𝑋
(
𝑠𝑠
)
�
(
𝑘𝑘−ℓ
)!
for
ℓ
= 1, … ,
𝑘𝑘
and
𝐴𝐴
𝑖𝑖
= lim
𝑠𝑠→𝑎𝑎
𝑖𝑖
(
𝑠𝑠 − 𝑎𝑎
𝑖𝑖
)
𝑋𝑋
(
𝑠𝑠
)
for
𝑖𝑖
=
𝑘𝑘
+ 1, … ,
𝑛𝑛
If
𝑚𝑚 ≥ 𝑛𝑛

EECS 216 – Discussion #12
(2020-04-10) – [email protected]
•
First perform a polynomial division
)
(
)
(
s
N
s
D
in which case you get a quotient
𝑄𝑄
(
𝑠𝑠
)
and a remainder
𝑅𝑅
(
𝑠𝑠
)
•
𝑁𝑁
(
𝑠𝑠
)
𝐷𝐷
(
𝑠𝑠
)
=
𝑄𝑄
(
𝑠𝑠
) +
𝑅𝑅
(
𝑠𝑠
)
𝐷𝐷
(
𝑠𝑠
)
,
where the degree of
𝑅𝑅
(
𝑠𝑠
)
will be less than that of
𝐷𝐷
(
𝑠𝑠
)
•
Then you can use partial fractions for
𝑅𝑅
(
𝑠𝑠
)
𝐷𝐷
(
𝑠𝑠
)
•
In this course, we will probably only encounter at most the case of
𝑚𝑚
=
𝑛𝑛
, in which case quotient
𝑄𝑄
(
𝑠𝑠
)
would simply be a constant
o
This technique, and why it works, will be illustrated through the following examples
o
Example #4
a) Find
𝑥𝑥
(
𝑡𝑡
)
when
𝑋𝑋
(
𝑠𝑠
) =
1
𝑠𝑠
2
+3𝑠𝑠+2
•
𝑋𝑋
(
𝑠𝑠
) =
1
(
𝑠𝑠+1
)(
𝑠𝑠+2
)
=
𝐴𝐴
𝑠𝑠+1
+