INTRODUCTION TO LAPLACE TRANSFORM Advanced Circuit Analysis Technique.
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INTRODUCTION TO LAPLACE INTRODUCTION TO LAPLACE TRANSFORM TRANSFORM Advanced Circuit Analysis Advanced Circuit Analysis Technique Technique
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Transcript of INTRODUCTION TO LAPLACE TRANSFORM Advanced Circuit Analysis Technique.
INTRODUCTION TO LAPLACE INTRODUCTION TO LAPLACE TRANSFORMTRANSFORM
Advanced Circuit Analysis Advanced Circuit Analysis TechniqueTechnique
TOPIC COVERAGETOPIC COVERAGE
WEEK TOPICS LABS
Week 6 Introduction to the Laplace Transform: Waveform Analysis
Lab 3: Intro to Laplace Transform
Week 7 Laplace Transform in Circuit Analysis: Transfer Function.
Lab 4: Laplace Transform Analysis
Week 8 Frequency Response: Bode Plot
Week 9 Frequency Selective Circuits: Passive Filters
Lab 5: RL and RC Filter
Lab 6: Bandpass and Bandreject Filter
TOPIC 1: TOPIC 1: LAPLACE TRANSFORMLAPLACE TRANSFORM
AREA OF COVERAGEAREA OF COVERAGE
WEEKWEEK TOPICSTOPICS DURATIONDURATION
Week 6Week 6 Intro LT: Definition, Excitation Intro LT: Definition, Excitation function, Functional transform, function, Functional transform, Operational transform.Operational transform.
2 hrs2 hrs
Intro LT: Operational Intro LT: Operational transform, Partial fraction.transform, Partial fraction.
1 hr1 hr
Week 7Week 7 Circuit Analysis: S-domain Circuit Analysis: S-domain circuit component, Transfer circuit component, Transfer function. function.
3 hrs3 hrs
LEARNING OUTCOMESLEARNING OUTCOMES
Be able to calculate the Laplace transform of a function using the definition and/or Laplace transform table.
Be able to calculate the inverse Laplace transform using partial fraction expansion and the Laplace table.
Be able to perform circuit analysis in the s-domain.
Know how to use a circuit’s transfer function to calculate the circuit’s impulse, unit, and step response.
OVERVIEWOVERVIEW
Analyze a “linear circuit” problem, in the frequency domain instead of in the time domain.
Convert a set of differential equations into a corresponding set of algebraic equations, which are much easier to solve.
Analyze the bandwidth, phase, and transfer characteristics important for circuit analysis and design.
OVERVIEWOVERVIEW
Analyze both the steady-state and “transient” responses of a linear circuit.
Analyze the response under any types of excitation (e.g. switching on and off at any given time(s), sinusoidal, impulse, square wave excitations, etc.)
DEFINITION OF LAPLACE DEFINITION OF LAPLACE TRANSFORMTRANSFORM
s: complex frequencyCalled “The One-sided or unilateral Laplace
Transform”.In the two-sided or bilateral LT, the lower
limit is -. We do not use this.
)(tf
0, te at
0,0 t
0t
)(tf
ate
0t
DISCONTUNITY FUNCTIONDISCONTUNITY FUNCTION
When f(t) has a finite discontinuity at the origin, the Laplace transform formula is rewritten as:
0
)()()( dtetfsFtfL st
STEP FUNCTIONSTEP FUNCTION
The symbol for the step function is Ku(t).
Mathematical definition of the step function:
0,)(
0,0)(
tKtKu
ttKu
f(t) = K u(t)f(t) = K u(t)
)(tf
K
0t
STEP FUNCTIONSTEP FUNCTION
A discontinuity of the step function may occur at some time other than t=0.
A step that occurs at t=a is expressed as:
atKatKu
atatKu
,)(
,0)(
f(t) = K u(t-a)f(t) = K u(t-a)
)(tf
K
ta0
Ex:Ex:)(tf
2
10 2 3 4t
2
Three linear functions at Three linear functions at t=0, t=1, t=3, and t=4t=0, t=1, t=3, and t=4
)(tf
4
10 2 3 4t
4
2
2
t2
42 t
82 t
Expression of step functions Expression of step functions
Linear function +2t: on at t=0, off at t=1
Linear function -2t+4: on at t=1, off at t=3
Linear function +2t-8: on at t=3, off at t=4
Step function can be used to turn on and turn off these functions
Step functionsStep functions
)]4()3()[82(
)]3()1()[42(
)]1()([2)(
tutut
tutut
tututtf
IMPULSE FUNCTIONIMPULSE FUNCTION
The symbol for the impulse function is (t).
Mathematical definition of the impulse function:
0,0)(
)()(
tt
KtdtK
f(t) = K f(t) = K (t)(t)
)(tf
K
0t
K)(tK )( atK
a
IMPULSE FUNCTIONIMPULSE FUNCTION
The area under the impulse function is constant and represents the strength of the impulse.
The impulse is zero everywhere except at t=0.
An impulse that occurs at t=a is denoted K (t-a)
FUNCTIONAL FUNCTIONAL TRANSFORMTRANSFORM
TYPE f(t) (t>0-) F(s)
Impulse
Step
Ramp
Exponential
Sine
Cosine
δ(t)
u(t)
t
ate
s1
1
2s1
as1
tsin
tcos
2
2s
22ss
TYPE f(t) (t>0-) F(s)
Damped ramp
Damped sine
Damped cosine
atte
te at sin
te at cos
21
as
22
as
22
asas
OPERATIONAL OPERATIONAL TRANSFORMTRANSFORM
OPERATIONAL TRANSFORMSOPERATIONAL TRANSFORMS
Indicate how mathematical operations performed on either f(t) or F(s) are converted into the opposite domain.
The operations of primary interest are:1. Multiplying by a constant
2. Addition/subtraction
3. Differentiation
4. Integration
5. Translation in the time domain
6. Translation in the frequency domain
7. Scale changing
OPERATION f(t) F(s)
Multiplication by a constantAddition/SubtractionFirst derivative (time)Second derivative (time)
)(tKf )(sKF
)()()( 321 tftftf )()()( 321 sFsFsF
dttdf )(
2
2 )(dt
tddt
dfsfsFs )0()0()(2
)0()( fssF
OPERATION f(t) F(s)
n th derivative (time)
Time integral
Translation in timeTranslation in frequency
n
n
dttd )(
1
123
21
)0()0(
)0()0()(
n
nn
nnn
dtdf
dtdfs
dtdfsfssFs
t
dxxf0
)(s
sF )(
0
),()(
a
atuatf
)(tfe at
)(sFe as
)( asF
OPERATION f(t) F(s)
Scale changing
First derivative (s)
n th derivative
s integral
0),( aatf asFa1
dssdF )()(ttf
s
duuF )(ttf )(
)(tft nn
nn
dssFd )()1(
Translation in time domainTranslation in time domain
If we start with any function:
we can represent the same function translated in time by the constant a, as:
In frequency domain:
)()( tutf
)()( atuatf
)()()( sFeatuatf as
Ex:Ex:
21)(s
ttuL
2)()(s
eatuatLas
Translation in frequency Translation in frequency domaindomain
Translation in the frequency domain is defined as:
)()( asFtfeL at
Ex:Ex:
22)(
cos
as
asteL at
22
cos
s
stL
Scale changingScale changing
The relationship between f(t) and F(s) when the time variable is multiplied by a constant:
oaasFaatfL ,1)]([
Ex:Ex:
222 1)(
1cos
s
s
s
stL
1
cos2
s
stL
APPLICATIONAPPLICATION
dcI
0t
R CL
)(tv
ProblemProblem
Assumed no initial energy is stored in the circuit at the instant when the switch is opened.
Find the time domain expression for v(t) when t≥0.
Integrodifferential EquationIntegrodifferential Equation
A single node voltage equation:
)()(
)(1)(
lglg
0
tuIdt
tdvCdxxv
LR
tv
KCLIaIa
dc
t
outin
s-domain transformations-domain transformation
sIvssVCs
sV
LR
sVdc1)0()(
)(1)(
)()(
)(1)(
0
tuIdt
tdvCdxxv
LR
tvdc
t
s
IsC
sLRsV dc
11)(
=0
)()( 1 sVLtv
)1()1()(
2 LCsRCsC
I
sVdc
INVERSE LAPLACE INVERSE LAPLACE TRANSFORM (LTRANSFORM (L-1-1))
PARTIAL FRACTION EXPANSION PARTIAL FRACTION EXPANSION (PFE)(PFE)
Rational function of s: expressed in a form of a ratio of two polynomials.
Called proper rational function if m≥nInverse transform rational functions of F(s),
can solve for v(t) or i(t).
011
1
011
1
)(
)()(
bsbsbsb
asasasa
sD
sNsF
mm
mm
nn
nn
RATIONAL FUNCTIONSRATIONAL FUNCTIONS
PARTIAL FRACTION PARTIAL FRACTION EXPANSIONEXPANSION
1) Distinct Real Roots of D(s)
)6)(8(
)12)(5(96)(
sss
sssF
s1= 0, s2= -8s3= -6
1) Distinct Real Roots1) Distinct Real Roots
To find K1: multiply both sides by s and evaluates both sides at s=0
To find K2: multiply both sides by s+8 and evaluates both sides at s=-8
To find K3: multiply both sides by s+6 and evaluates both sides at s=-6
68)6)(8(
)12)(5(96)( 321
s
K
s
K
s
K
sss
sssF
Find KFind K11
0
3
0
21
068)6)(8(
)12)(5(96
ssss
sK
s
sKK
ss
ss
120)6)(8(
)12)(5(961 K
Find KFind K22
8
32
8
1
8)6(
)8(
)6(
)8(
)6(
)12)(5(96
sssss
sKK
ss
sK
ss
ss
72)2)(8(
)4)(3(962
K
Find KFind K33
3
6
2
6
1
6)8(
)6(
)8(
)6(
)8(
)12)(5(96K
ss
sK
ss
sK
ss
ss
sss
48)2)(6(
)6)(1(963
K
Inverse Laplace of F(s)Inverse Laplace of F(s)
)(4872120)(
6
48
8
72120
68
1
tueetf
sssL
tt
6
48
8
72120)(
ssssF
2) Distinct Complex Roots 2) Distinct Complex Roots
)256)(6(
)3(100)(
2
sss
ssF
S1 = -6 S2 = -3+j4 S3 = -3-j4
Partial Fraction ExpansionPartial Fraction Expansion
43436
43436
)256)(6(
)3(100)(
221
321
2
js
K
js
K
s
K
js
K
js
K
s
K
sss
ssF
Complex roots appears in conjugate pairs.Complex roots appears in conjugate pairs.
Find KFind K11
1225
)3(100
256
)3(100
621
sss
sK
Find KFind K2 2 and Kand K22**
13.53
43
2
1086
)8)(43(
)4(100
)43)(6(
)3(100
j
js
ej
jj
j
jss
sK
13.5310862
jejK
Coefficients Coefficients associated associated
with with conjugate conjugate pairs are pairs are
themselves themselves conjugates.conjugates.
Inverse Laplace of F(s)Inverse Laplace of F(s)
43
13.5310
43
13.5310
6
12
256)(6(
)3(100)(
2
jsjss
sss
ssF
Inverse Laplace of F(s)Inverse Laplace of F(s)
)(10
1012
43
10
43
10
6
12
)43(13.53
)43(13.536
13.5313.531
tuee
eee
js
e
js
e
sL
tjj
tjjt
jj
Undesirable imaginary components Undesirable imaginary components in the time domainin the time domain
)13.534cos(20
10
1010
1010
1010
3
)13.534()13.534(3
)13.534(3)13.534(3
4313.534313.53
)43(13.53)43(13.53
te
eee
ee
ee
eeee
t
jtjtjt
jtjtjtjt
tjtjtjtj
tjjtjj
Inverse Laplace of F(s)Inverse Laplace of F(s)
)()13.534cos(2012)(
43
10
43
10
6
12
36
13.5313.531
tuteetf
js
e
js
e
sL
tt
jj
)()cos(2)( tuteKtf t
3) Repeated Real Roots3) Repeated Real Roots
3)5(
)25(100)(
ss
ssF
Partial Fraction ExpansionPartial Fraction Expansion
5)5()5()5(
)25(100 42
33
213
s
K
s
K
s
K
s
K
ss
s
20)5(
)25(100
031
ss
sK 20
)5(
)25(100
031
ss
sK
400)25(100
52
ss
sK 400
)25(100
52
ss
sK
Find KFind K33
First, multiply both sides by (s+5)3. Next, differentiate both sides once with respect to s and evaluate at s=-5.
5)5()5()5(
)25(100 42
33
213
s
K
s
K
s
K
s
K
ss
s
KK33
52
4
53
52
5
31
5
)5(
)5(
)5()25(100
s
s
s
ss
sKds
d
sKds
d
Kds
d
s
sK
ds
d
s
s
ds
d
KK33
10)25(
100 35
2
Ks
ss
s
Find KFind K44
First, multiply both sides by (s+5)3. Next, differentiate both sides twice with respect to s and evaluate at s=-5.
5)5()5()5(
)25(100 42
33
213
s
K
s
K
s
K
s
K
ss
s
KK44
Simplifying the first derivative, the second derivative becomes:
20
240
)5(20
)52()5(25100
4
4
5453
5
2
2
15
2
K
K
sKds
dK
ds
d
s
ss
ds
dK
sds
d
ss
ss
F(s) and f(t)F(s) and f(t)
5
20
)5(
100
)5(
40020)(
23
sssssF
)(2010020020)( 5552 tueteettf ttt
)(tuKte at
4) REPEATED COMPLEX 4) REPEATED COMPLEX ROOTSROOTS
22 )256(
768)(
sssF 22 )256(
768)(
sssF
432,1 js
Partial Fraction ExpansionPartial Fraction Expansion
)43()43()43()43(
)43()43()43()43(
)43()43(
768)(
21
22
21
42
322
1
22
js
K
js
K
js
K
js
K
js
K
js
K
js
K
js
K
jsjssF
Find KFind K11 and K and K22
12)8(
768
)43(
7682
4321
jjsK
js
9033
)8(
)768(2
)43(
)768(2
)43(
768
343
3
4322
j
jjs
jsds
dK
js
js
F(s) and f(t)F(s) and f(t)
)43(
903
)43(
903
)43(
12
)43(
12)(
22
sjs
jsjssF
)43(
903
)43(
903
)43(
12
)43(
12)(
22
sjs
jsjssF
)()]904cos(64cos24[)( 33 tutettetf tt )()]904cos(64cos24[)( 33 tutettetf tt
)()cos(2)( tuteKttf t
USEFUL TRANSFORM PAIRSUSEFUL TRANSFORM PAIRS
)()1 tuKeas
K at
)()(
)22
tuKteas
K at
)()cos(2)3 tuteKjs
K
js
K t
)()cos(2)()(
)422
tuteKtjs
K
js
K t
InspirationInspiration
Do not pray for easy lives,
Pray to be stronger men!
Do not pray for tasks equal to your powers,
Pray for powers equal to your tasks!
