1. The Laplace Transform
The University of Tennessee
Electrical and Computer Engineering Department
Knoxville, Tennessee
wlg
2. The Laplace Transform
The Laplace Transform of a function, f(t), is defined as;
∫
∞
−
==
0
)()()]([ dtetfsFtfL st
The Inverse Laplace Transform is defined by
∫
∞+
∞−
−
==
j
j
ts
dsesF
j
tfsFL
σ
σ
π
)(
2
1
)()]([1
*notes
Eq A
Eq B
3. The Laplace Transform
We generally do not use Eq B to take the inverse Laplace. However,
this is the formal way that one would take the inverse. To use
Eq B requires a background in the use of complex variables and
the theory of residues. Fortunately, we can accomplish the same
goal (that of taking the inverse Laplace) by using partial fraction
expansion and recognizing transform pairs.
*notes
4. The Laplace Transform
Laplace Transform of the unit step.
*notes
|0
0
1
1)]([
∞−
∞
−
∫
−
== stst
e
s
dtetuL
s
tuL
1
)]([ =
The Laplace Transform of a unit step is:
s
1
5. The Laplace Transform
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
0 t0
f(t) δ(t – t0)
Mathematically:
δ(t – t0) = 0 t ≠ 0
*note
01)(
0
0
0 >=−∫
+
−
εδ
ε
ε
dttt
t
t
6. The Laplace Transform
The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting
or sampling property. The following is an important.
∫
><
<<
=−
2
1
2010
2010
0
,0
)(
)()(
t
t
tttt
ttttf
dttttf δ
7. The Laplace Transform
The Laplace transform of a unit impulse:
In particular, if we let f(t) = δ(t) and take the Laplace
1)()]([ 0
0
=== −−
∞
∫
sst
edtettL δδ
8. The Laplace Transform
An important point to remember:
)()( sFtf ⇔
The above is a statement that f(t) and F(s) are
transform pairs. What this means is that for
each f(t) there is a unique F(s) and for each F(s)
there is a unique f(t). If we can remember the
Pair relationships between approximately 10 of the
Laplace transform pairs we can go a long way.
9. The Laplace Transform
Building transform pairs:
eL(
∫∫
∞
+−
∞
−−−
==
e
tasstatat
dtedteetueL
0
)(
0
)]([
asas
e
tueL
st
at
+
=
+
−
=
∞−
−
1
)(
)]([ |0
as
tue at
+
⇔− 1
)(A transform
pair
10. The Laplace Transform
Building transform pairs:
∫
∞
−
=
0
)]([ dttettuL st
∫ ∫
∞ ∞
∞
−=
0 0
0
| vduuvudv
u = t
dv = e-st
dt
2
1
)(
s
ttu ⇔
A transform
pair
11. The Laplace Transform
Building transform pairs:
22
0
11
2
1
2
)(
)][cos(
ws
s
jwsjws
dte
ee
wtL st
jwtjwt
+
=
+
−
−
=
+
= −
∞ −
∫
22
)()cos(
ws
s
tuwt
+
⇔ A transform
pair
12. The Laplace Transform
Time Shift
∫ ∫
∫
∞ ∞
−−+−
∞
−
=
∞→∞→→→
+==−=
−=−−
0 0
)(
)()(
,.,0,
,
)()]()([
dxexfedxexf
SoxtasandxatAs
axtanddtdxthenatxLet
eatfatuatfL
sxasaxs
a
st
)()]()([ sFeatuatfL as−
=−−
13. The Laplace Transform
Frequency Shift
∫
∫
∞
+−
∞
−−−
+==
=
0
)(
0
)()(
)]([)]([
asFdtetf
dtetfetfeL
tas
statat
)()]([ asFtfeL at
+=−
14. The Laplace Transform
Example: Using Frequency Shift
Find the L[e-at
cos(wt)]
In this case, f(t) = cos(wt) so,
22
22
)(
)(
)(
)(
was
as
asFand
ws
s
sF
++
+
=+
+
=
22
)()(
)(
)]cos([
was
as
wteL at
++
+
=−
15. The Laplace Transform
Time Integration:
The property is:
stst
t
st
t
e
s
vdtedv
and
dttfdudxxfuLet
partsbyIntegrate
dtedxxfdttfL
−−
−
∞∞
−==
==
=
∫
∫ ∫∫
1
,
)(,)(
:
)()(
0
0 00
16. The Laplace Transform
Time Integration:
Making these substitutions and carrying out
The integration shows that
)(
1
)(
1
)(
00
sF
s
dtetf
s
dttfL st
=
=
∫∫
∞
−
∞
17. The Laplace Transform
Time Differentiation:
If the L[f(t)] = F(s), we want to show:
)0()(]
)(
[ fssF
dt
tdf
L −=
Integrate by parts:
)(),(
)(
,
tfvsotdfdt
dt
tdf
dv
anddtsedueu stst
===
−== −−
*note
18. The Laplace Transform
Time Differentiation:
Making the previous substitutions gives,
[ ]
∫
∫
∞
−
∞
−∞−
+−=
−−=
0
0
0
)()0(0
)()( |
dtetfsf
dtsetfetf
dt
df
L
st
stst
So we have shown:
)0()(
)(
fssF
dt
tdf
L −=
19. The Laplace Transform
Time Differentiation:
We can extend the previous to show;
)0(...
)0(')0()(
)(
)0('')0(')0()(
)(
)0(')0()(
)(
)1(
21
23
3
3
2
2
2
−
−−
−−
−−=
−−−=
−−=
n
nnn
n
n
f
fsfssFs
dt
tdf
L
casegeneral
fsffssFs
dt
tdf
L
fsfsFs
dt
tdf
L
20. The Laplace Transform
Transform Pairs:
____________________________________
)()( sFtf
f(t) F(s)
1
2
!
1
1
1
)(
1)(
+
−
+
n
n
st
s
n
t
s
t
as
e
s
tu
tδ
21. The Laplace Transform
Transform Pairs:
f(t) F(s)
( )
22
22
1
2
)cos(
)sin(
)(
!
1
ws
s
wt
ws
w
wt
as
n
et
as
te
n
atn
at
+
+
+
+
+
−
−
22. The Laplace Transform
Transform Pairs:
f(t) F(s)
22
22
22
22
sincos
)cos(
cossin
)sin(
)(
)cos(
)(
)sin(
ws
ws
wt
ws
ws
wt
was
as
wte
was
w
wte
at
at
+
−
+
+
+
+
++
+
++
−
−
θθ
θ
θθ
θ Yes !
23. The Laplace Transform
Common Transform Properties:
f(t) F(s)
)(
1
)(
)(
)(
)0(...)0(')0()(
)(
)()(
)([0),()(
)(0),()(
0
1021
00
000
sF
s
df
ds
sdF
ttf
ffsfsfssFs
dt
tfd
asFtfe
ttfLetttutf
sFetttuttf
t
nnnn
n
n
at
sot
sot
∫
−
−−−−
+
+≥−
≥−−
−−−
−
−
−
λλ
24. The Laplace Transform
Using Matlab with Laplace transform:
Example Use Matlab to find the transform of t
te 4−
The following is written in italic to indicate Matlab code
syms t,s
laplace(t*exp(-4*t),t,s)
ans =
1/(s+4)^2
25. The Laplace Transform
Using Matlab with Laplace transform:
Example Use Matlab to find the inverse transform of
19.12.
)186)(3(
)6(
)( 2
prob
sss
ss
sF
+++
+
=
syms s t
ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18)))
ans =
-exp(-3*t)+2*exp(-3*t)*cos(3*t)
26. The Laplace Transform
Theorem: Initial Value
If the function f(t) and its first derivative are Laplace transformable and f(t)
Has the Laplace transform F(s), and the exists, then)(lim ssF
0
)0()(lim)(lim
→∞→
==
ts
ftfssF
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.
Theorem:
∞→s
Initial Value
Theorem
27. The Laplace Transform
Initial ValueTheorem:Example:
Given;
22
5)1(
)2(
)(
++
+
=
s
s
sF
Find f(0)
1
)26(2
2
lim
2512
2
lim
5)1(
)2(
lim)(lim)0(
2222
222
2
2
22
=
++
+
=
+++
+
=
++
+
==
sssss
ssss
ss
ss
s
s
sssFf
∞→s∞→s ∞→s
∞→s
28. The Laplace Transform
Theorem: Final Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
has the Laplace transform F(s), and the exists, then)(lim ssF
∞→s
)()(lim)(lim ∞== ftfssF
0→s ∞→t
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.
Final Value
Theorem
29. The Laplace Transform
Final Value Theorem:Example:
Given:
[ ] ttesFnote
s
s
sF t
3cos)(
3)2(
3)2(
)( 21
22
22
−
=
++
−+
= −
Find )(∞f .
[ ] 0
3)2(
3)2(
lim)(lim)( 22
22
=
++
−+
==∞
s
s
sssFf
0→s0→s
Editor's Notes
In the undergraduate curriculum, usually there are 3 transform pairs that are studied.
These are
The Laplace Transform
The Fourier Transform
The Z-Transform
The Laplace transform is usually used in solving continuous linear differential equations of the type encountered in circuit theory and systems. The advantage of the Laplace transform is that it makes cumbersome differential equations algebraic and therefore the math becomes simpler to handle. In the end we can take the inverse and go back to the time domain. In systems, for example, we stay with the Laplace variable “s” while investigating system stability, system performance. We use tools such as the root locus and block diagrams which are directly in the s variable. Even Bode is used where s is replaced by jw. We are “walking” with one foot in s but thinking the time domain.
The Fourier transform is used very much in communication theory, field theory and generally in the study of signal spectrum and both analog and digital filter analysis and design.
The Z-transform is sort of similar to the Laplace transform. In fact, many of the manipulations such as partial fraction expansion, taking the inverse, interrupting stability are very similar to those performed in the Laplace. The big difference is that the Z-transform is used for discrete systems and difference equations. At the University of Tennessee you will be studying the Fourier and Z-transforms in your junior year and you will also study the Laplace transform again.
Normally, even in the junior year, you will still use partial fraction expansion and recognition of transform paisr to find the inverse Laplace transform. This is true in most undergraduate programs. However, at UTK we have a graduate course that is required of all graduate students in electrical engineering in which an in-depth study is made of using Eq B for taking the inverse Laplace. A similar integral is encountered for taking the inverse in both Fourier and Z-transforms.
It might also be noted that not all functions have a Laplace transform. For a function to be Laplace transformable it must be of exponential order. This boils down to saying that the integral from 0 to infinity of e-at|f(t)|dt must be less than infinity. In this course, and most courses in undergraduate studies, the functions we are
interested in taking the Laplace will be of exponential order.
The unit step is called a singularity function because it does not possess a derivative at t = 0. We also say that it is not defined at t = 0. Along this line, you will note that many text books define the Laplace by using a lower limit of 0- rather than absolute zero. One must ask, if you integrate from 0 to infinity for a unit step, how are you handling the lower limit when the step is not defined at t = 0? We will not get worked up over this. You can take this apart later after you become experts in transforms.
I remember when I was first introduced to the unit impulse function, in an electrical engineering course at Auburn University. Essentially, we were told that the unit impulse function was a function that was infinitely high and had zero width but the area under the function was 1. Such an explanation would probably always disturb a mathematician because I expect the proper way to define the function is in a limiting process as a function that is 1/delta in height and delta wide and we take the limit as delta goes to zero.
At this point, the best thing to do is not to worry about it. Accept it as we have defined it here and go ahead and use it. Probably one day in a higher level math course you may give more attention to its attributes.
As noted earlier, some text will use 0- on the lower part of the defining integral
of the Laplace transform. When this is done, f(0) above will become f(0-).