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Chapter 7
Laplace Transforms
Applications of Laplace Transform
notes
• Easier than solving differential equations – Used to describe system behavior – We assume LTI systems – Uses S-domain instead of frequency domain
• Applications of Laplace Transforms/– Circuit analysis
• Easier than solving differential equations
• Provides the general solution to any arbitrary wave (not just LRC) – Transient– Sinusoidal steady-state-response (Phasors)
– Signal processing – Communications
• Definitely useful for Interviews!
Building the Case…
http://web.cecs.pdx.edu/~ece2xx/ECE222/Slides/LaplaceTransformx4.pdf
Laplace Transform
Laplace Transform
• We use the following notations for Laplace Transform pairs – Refer to the table!
Laplace Transform Convergence
• The Laplace transform does not converge to a finite value for all signals and all values of s
• The values of s for which Laplace transform converges is called the Region Of Convergence (ROC)
• Always include ROC in your solution!
• Example:
asas
aseeas
eas
jsnoteeas
dtee
dtetfsF
tuetf
tjtataj
tasstat
st
at
)Re(;1
0)Re(11
:;1
)()(
);()(
0
)(
0
)(
0
)(
0
Remember: e^jw is sinusoidal; Thus, only the
real part is important!
0+ indicates greater than zero values
Example of Bilateral Version
asas
asas
asas
eas
dtee
dtetfsF
tuetf
tasstat
st
at
)Re(;1
)Re(;1
0)Re(;11
)()(
);()(
0)(0
asas
asas
asas
eas
dtee
dtetfsF
tuetf
tasstat
st
at
)Re(;1
)Re(;1
0)Re(;11
)()(
);()(
0)(0
Find F(s):
Find F(s):
Re(s)<a
a
S-plane
Note that Laplace can also be found for periodic functions
ROC
Remember These!
Example – RCO may not always exist!
;3
1
2
1)(
3)Re(;3
1)(
2)Re(;2
1)(
)()(
)()()(
3
2
32
sssF
ss
tue
ss
tue
dtetfsF
tuetuetf
t
t
st
tt
Note that there is no common ROC Laplace Transform can not be applied!
Example – Unilateral Version
• Find F(s):
• Find F(s):
asasas
dtee
dtetfsF
atuetf
stat
st
at
)Re(0)Re(;1
)()(
0);()(
0
0
0)Re(;1
]1[lim1
]1[lim1
)(
)()(
)()(
)(
0
0
ss
es
es
dttue
dtetfsF
tutf
tjt
stt
st
st
ssF
ttf
edttte
dtetfsF
tttf
stst
st
;1)(
)()(
)(
)()(
)()(
0
0 0
0
0
• Find F(s):
• Find F(s):
asasas
as
dtee
dtetfsF
aetf
stat
st
at
)Re(0)Re(;1
]10[1
)()(
0;)(
0
0
Example
0)Re(;2/12/1
)(
0)Re(;2/1
2/1;2/1
2/1
)Re(;1
)()(
2/12/1
)cos()(
22
0
sbs
s
jbsjbssF
sjbs
ejbs
e
asas
e
dtetfsF
eetf
bttf
jbtjbt
at
st
jbtjbt
0)Re(;2/12/1
)(
0)Re(;2/1
2/1;2/1
2/1
)Re(;1
)()(
2/12/1
)sin()(
22
0
sbs
b
jbs
j
jbs
jsF
sjbs
je
jbs
jje
asas
e
dtetfsF
jejetf
bttf
jbtjbt
at
st
jbtjbt
Example
0)Re(;)(
)(
1
)(
1
2
1)(
)Re(;1
)()(
2/12/1
)cos()(
22
0
asbas
as
jbasjbassF
asas
e
dtetfsF
eeeetf
btetf
at
st
atjbtatjbt
at
Properties
• The Laplace Transform has many difference properties • Refer to the table for these properties
Linearity
Scaling & Time Translation
Scaling
Time Translation
)/()()(/
asFa
ebatubatf
asb
b=0
Do the time translation first!
Shifting and Time Differentiation
Shifting in s-domain
Differentiation in t
Read the rest of properties on your
own!
Examples
3.0)Re(;3.0
15)(
)Re(;1
5)( 3.0
ss
sF
asas
e
etf
at
t
3.0)Re(;3.0
15)(
__;1
)2(5)(
2
)2(3.0
ses
sF
shifttimewithas
e
tuetf
s
at
t
3.0)Re(;3.0
744.2
3.05)(
)}2({5
)}2({5
}){2(5)(
__;1
)2(5)(
22)3.0(2
)2(3.0)3.0(2
)3.0(2)(3.0)3.0(2
)3.0(2)3.0(2)(3.0
)(3.0
ses
es
esF
tuee
tueee
eetuetf
shifttimewithas
e
tuetf
ss
t
t
t
at
t
Note the ROC did not change!
Example – Application of Differentiation
)()(
)(
)()(
?)}({
)()(
0
0
sFds
sG
dtetfds
dtettfsG
ttf
ttftg
st
st
0)Re(;
0)Re(};{
)()(
?)}cos({
)cos()(
222
22
22
sbs
bs
sbs
s
ds
sFds
sG
btt
btttg
Read Section 7.4
Matlab Code:
Read about Symbolic Mathematics: http://www.math.duke.edu/education/ccp/materials/diffeq/mlabtutor/mlabtut7.htmlAnd
http://www.mathworks.de/access/helpdesk/help/toolbox/symbolic/ilaplace.html
Example
• What is Laplace of t^3? – From the table: 3!/s^4 Re(s)>0
• Find the Laplace Transform:
;9)4/(
3
4)(
9
3)()()3sin()(
);/(4
)(
6/;4
:_
)()3sin()6/4()}6/4(3sin{)(
)6/4()2/12sin()(
2
24/
2
24/
6/4
s
esG
ssFuf
asFe
sG
ba
nTranslatioTime
ututtg
tuttg
s
s
t
Note that without u(.) there will be no time translation and thus, the result will be different:
)/()()(/
asFa
ebatubatf
asb
Time transformation
Assume t>0
A little about Polynomials
• Consider a polynomial function:
• A rational function is the ratio of two polynomials:
• A rational function can be expressed as partial fractions
• A rational function can be expressed using polynomials presented in product-of-sums
Has roots and zeros; distinct roots, repeated roots, complex roots, etc.
Given Laplace find f(t)!
Finding Partial Fraction Expansion
• Given a polynomial
• Find the POS
(product-of-sums) for the denominator:
• Write the
partial fraction expression
for the polynomial
• Find the constants– If the rational polynomial has
distinct poles then we can use the
following to find the constants:
http://cnx.org/content/m2111/latest/
......
)]()[(
)]()[(
)]()[(
....)()()(
)(
)...)()((
)(
)(
)()(
3
2
1
32
22
11
3
3
2
2
1
1
321
ps
ps
ps
sIpsk
sIpsk
sIpsk
ps
k
ps
k
ps
ksG
pspsps
sN
sD
sNsG
Matlab Code
Application of Laplace
• Consider an RL circuit with R=4, L=1/2. Find i(t) if v(t)=12u(t).
0;3)(3)(8
33)(
3)]()[(
3)]()[(
8)8(
24)().()(
/12)()(12)(
)().()(45.0
1)(/)()(
)()()45.0(
:_sin
)()(4)(
5.0
)()()(
8
822
011
21
2
1
tetutiss
sI
sIpsk
sIpsk
s
k
s
k
sssVsHsI
ssVtutv
sVsHsIs
sVsIsH
sVsIs
LaplacegU
tvtidt
tdi
tvtRidt
tdiL
t
p
p
Partial fraction expression
Given
Application of Laplace• What are the initial [i(0)] and final
values: – Using initial-value property:
– Using the final-value property
08
24lim)(lim)0(
sssIi ss
38
24lim)(lim)(lim 00
s
ssIti sss
)()()(:_
)()()0(:_
limlim
limlim
0
0
ssFtffvalueFinal
ssFtffvalueInitial
tt
tt
Note: using Laplace Properties
0;3)(3)( 8 tetuti t
Note thatInitial Value: t=0, then, i(t) 3-3=0Final Value: t INF then, i(t) 3
Using Simulink
H(s)
i(t)
v(t)
Actual Experimentation
• Note how the voltage looks like:
Input Voltage:
0;12)(
5.0)(
0;3)(3)(
)()(4)(
5.0
)()()(
8
8
tedt
tditv
tetuti
tvtidt
tdi
tvtRidt
tdiL
t
t
Output Voltage:
Partial Fraction Expansion (no repeated Poles/Roots) – Example
• Using Matlab:
• Matlab code:b=[8 3 -21];
a=[1 0 -7 -6];
[r,p,k]=residue(b,a)
We can also use ilaplace (F); but the result may not be
simplified!
Finding Poles and Zeros
• Express the rational function as the ratio of two polynomials each represented by product-of-sums
• Example:
)3)(1(
)2(2
682
84)(
2
ss
s
ss
ssF
S-plane
Pole
zero
H(s) Replacing the Impulse Response
h(t)x(t) y(t)
H(s)X(s) Y(s)
convolution multiplication
H(s) Replacing the Impulse Response
h(t)x(t) y(t)
H(s)X(s) Y(s)
convolution multiplication
Example: Find the output X(t)=u(t); h(t)
0 1
1 h(t)
)1()1()()(
11)().()(
1)(
1)(
)1()()(
222
tutttutys
e
ss
esXsHsY
ssX
s
e
ssH
tututh
ss
s
0 1
1 y(t)
e^-sF(s)
This is commonly used in D/A converters!
Dealing with Complex Poles
• Given a polynomial
• Find the POS (product-of-sums) for the denominator:
• Write the partial fraction expression for the polynomial
• Find the constants– The pole will have a real and
imaginary part: P=|k|
• When we have complex poles {|k| then we can use the following expression to find the time domain expression:
http://cnx.org/content/m2111/latest/
)cos(||2)(
__);Im();Re(
btektf
PofanglePbPaat
Laplace Transform Characteristics
• Assumptions: Linear Continuous Time Invariant Systems
• Causality– No future dependency
– If unilateral: No value for t<0; h(t)=0
• Stability – System mode: stable or unstable
– We can tell by finding the system characteristic equation (denominator)
• Stable if all the poles are on the left plane
– Bounded-input-bounded-output (BIBO)
• Invertability – H(s).Hi(s)=1
• Frequency Response – H(w)=H(s);sjw=H(s=jw)
)()()(
)2)(2(
1
4
1)(
22
2
tuBetuAeth
ssssH
tt
52
13)(
52
13)(
2
2
j
jH
ss
ssH
We need to add control mechanism to make the overall system stable
Frequency Response – Matlab Code
52
13)(
52
13)(
2
2
j
jH
ss
ssH
Inverse Laplace Transform
Example of Inverse Laplace Transform
Bilateral Transforms
• Laplace Transform of two different signals can be the same, however, their ROC can be different:
Very important to know the ROC.
• Signals can be – Right-sided Use the bilateral
Laplace Transform Table
– Left-sides
– Have finite duration
• How to find the transform of signals that are bilateral!
sidedLeftsaas
tuetx
sidedRightsaas
tuetx
at
at
)Re(;1
)()(
)Re(;1
)()(
See notes
How to Find Bilateral Transforms
• If right-sided use the table for unilateral Laplace Transform• Given f(t) left-sided; find F(s):
– Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>a
– Then, find F(-s) with Re(-s)>a
• Given Fb(s) find f(t) left-sided : – Find the unilateral Inverse Laplace transform for F(s)=fb(t)
– The result will be f(t)=–fb(t)u(-t)
• Example
4
1
5
2)(
)Re(4)Re(4;4
1)(
)Re(4;4
1)(
)(:
)(:_
)Re(5;5
2)(2
)()(2)(
4
4
5
45
sssF
sidedLeftsss
sF
sidedRightss
tue
tfAssume
tueFindTo
sidedRightss
tue
tuetuetx
t
t
t
tt
Examples of Bilateral Laplace Transform
)Re(5);Re(4;4
1
5
2)()(2)(
)Re(5&)Re(4;4
1
5
2)()(2)(
4)Re(5;4
1
5
2)()(2)(
45
45
45
ssss
tuetuetx
ssss
tuetuetx
sss
tuetuetx
tt
tt
tt
Find the unilateral Laplace transform for f(-t) laplace{f(-t)}; Re(s)>aThen find F(-s) with Re(-s)>a
Alternatively: Find the unilateral Laplace transform for f(t)u(-t) (-1)laplace{f(t)}; then, change the inequality for ROC.
Feedback System
Find the system function for the following feedback system:
G(s)
Sum F(s)X(t)
r(t)
e(t) y(t)
+
+
)().(1
)()()(/)(
)(/)())().(()(
)().()(
)(/)()()()(
sGsF
sFsHsXsY
sFsYsGsYsX
sGsYsR
sFsYsEsRsX
H(s)X(t) y(t)
Equivalent System
Feedback Applet: http://physioweb.uvm.edu/homeostasis/simple.htm
Practices Problems
• Schaum’s Outlines Chapter 3– 3.1, 3.3, 3.5, 3.6, 3.7-3.16, For Quiz! – 3.17-3.23– Read section 7.8 – Read examples 7.15 and 7.16
Useful Applet: http://jhu.edu/signals/explore/index.html
