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Admissions in India 2015Admissions in India 2015By:http://admission.edhole.com
Inverse Laplace Inverse Laplace TransformationsTransformations
Dr. HolbertFebruary 27, 2008
Lect11 EEE 202 2
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Inverse Laplace TransformInverse Laplace TransformConsider that F(s) is a ratio of
polynomial expressions
The n roots of the denominator, D(s) are called the poles◦Poles really determine the response
and stability of the systemThe m roots of the numerator, N(s),
are called the zeros
Lect11 EEE 202 3
)(
)()(
s
ss
DN
F
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Inverse Laplace TransformInverse Laplace TransformWe will use partial fractions
expansion with the method of residues to determine the inverse Laplace transform
Three possible cases (need proper rational, i.e., n>m)1. simple poles (real and unequal)2. simple complex roots (conjugate
pair)3. repeated roots of same value
Lect11 EEE 202 4
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1. Simple Poles1. Simple PolesSimple poles are placed in a partial
fractions expansion
The constants, Ki, can be found from (use method of residues)
Finally, tabulated Laplace transform pairs are used to invert expression, but this is a nice form since the solution is
Lect11 EEE 202 5
n
n
n
m
ps
K
ps
K
ps
K
pspsps
zszsKs
2
2
1
1
21
10)(F
ipsii spsK
)()( F
tpn
tptp neKeKeKtf 2121)(admission.edhole.com
2. Complex Conjugate 2. Complex Conjugate PolesPolesComplex poles result in a Laplace transform
of the form
The K1 can be found using the same method as for simple poles
WARNING: the "positive" pole of the form –+j MUST be the one that is used
The corresponding time domain function is
Lect11 EEE 202 6
)()()()(
)( 11*11
js
K
js
K
js
K
js
Ks F
jssjsK
)()(1 F
teKtf t cos2)( 1admission.edhole.com
3. Repeated Poles3. Repeated PolesWhen F(s) has a pole of multiplicity r,
then F(s) is written as
Where the time domain function is then
That is, we obtain the usual exponential but multiplied by t's
Lect11 EEE 202 7
rr
r ps
K
ps
K
ps
K
pss
ss
1
12
1
12
1
11
11
1
)(
)()(
Q
PF
tpr
rtptp e
r
tKetKeKtf 111
!1)(
1
11211
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3. Repeated Poles 3. Repeated Poles (cont’d.)(cont’d.)The K1j terms are evaluated from
This actually simplifies nicely until you reach s³ terms, that is for a double root (s+p1)²
Thus K12 is found just like for simple rootsNote this reverse order of solving for the K values
Lect11 EEE 202 8
1
)(!
111
ps
r
jr
jr
j spsds
d
jrK
F
1
1
)()( 2111
2112
psps
spsds
dKspsK
FF
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The “Finger” MethodThe “Finger” Method
Let’s suppose we want to find the inverse Laplace transform of
We’ll use the “finger” method which is an easy way of visualizing the method of residues for the case of simple roots (non-repeated)
We note immediately that the poles ares1 = 0 ; s2 = –2 ; s3 = –3
)3)(2(
)1(5)(
sss
ssF
Lect11 EEE 202 9
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The Finger Method (cont’d)The Finger Method (cont’d)
For each pole (root), we will write down the function F(s) and put our finger over the term that caused that particular root, and then substitute that pole (root) value into every other occurrence of ‘s’ in F(s); let’s start with s1=0
This result gives us the constant coefficient for the inverse transform of that pole; here: e–0·t
Lect11 EEE 202 10
6
5
)3)(2(
)1(5
)30)(20)((
)10(5
)3)(2(
)1(5)(
ssss
ssF
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The Finger Method (cont’d)The Finger Method (cont’d)
Let’s ‘finger’ the 2nd and 3rd poles (s2 & s3)
They have inverses of e–2·t and e–3·t
The final answer is then tt eetf 32
3
10
2
5
6
5)(
Lect11 EEE 202 11
3
10
)1)(3(
)2(5
)3)(23)(3(
)13(5
)3)(2(
)1(5)(
2
5
)1)(2(
)1(5
)32)(2)(2(
)12(5
)3)(2(
)1(5)(
ssss
ss
ssss
ss
F
F
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Initial Value TheoremInitial Value TheoremThe initial value theorem states
Oftentimes we must use L'Hopital's Rule:◦If g(x)/h(x) has the indeterminate
form 0/0 or / at x=c, then
Lect11 EEE 202 12
)(lim)(lim0
s s tfst
F
)('
)('lim
)(
)(lim
xh
xg
xh
xg
cxcx
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Final Value TheoremFinal Value TheoremThe final value theorem states
The initial and final value theorems are useful for determining initial and steady-state conditions, respectively, for transient circuit solutions when we don’t need the entire time domain answer and we don’t want to perform the inverse Laplace transform
Lect11 EEE 202 13
)(lim)(lim s s tf 0stF
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Initial and Final Value Initial and Final Value TheoremsTheoremsThe initial and final value theorems also
provide quick ways to somewhat check our answers
Example: the ‘finger’ method solution gave
Substituting t=0 and t=∞ yields
tt eetf 32
3
10
2
5
6
5)(
6
5
2
1510
6
5)(
06
20155
3
10
2
5
6
5)0( 00
eetf
eetf
Lect11 EEE 202 14
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Initial and Final Value Initial and Final Value TheoremsTheoremsWhat would initial and final value theorems
find?First, try the initial value theorem (L'Hopital's
too)
Next, employ final value theorem
This gives us confidence with our earlier answer
05
52
5lim
65
)1(5lim)0(
)3)(2(
)1(5lim)(lim)0(
2
sss
sf
ss
ss s f
sdsd
dsd
s
ssF
6
5
)3)(2(
)1(5
)3)(2(
)1(5lim)(lim)(
00
ss
ss s f
ssF
Lect11 EEE 202 15
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Solving Differential Solving Differential EquationsEquationsLaplace transform approach
automatically includes initial conditions in the solution
Example: For zero initial conditions, solve
Lect11 EEE 202 16
)0(')0()()(
)0()()(
22
2
yysssdt
tyd
xssdt
txd
Y
X
L
L
)(4)(30)(
11)(
2
2
tutydt
tyd
dt
tyd
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Class ExamplesClass Examples
Find inverse Laplace transforms of
Drill Problems P5-3, P5-5 (if time permits)
84)(
)1()(
2
2
ss
ss
s
ss
Z
Y
Lect11 EEE 202 17
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