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DR. HOLBERTFEBRUARY 27, 2008
Lect11EEE 202
2
Inverse Laplace Transformations
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Inverse Laplace Transform
Lect11EEE 202
3
Consider 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
)()()(
sss D
NF
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Inverse Laplace Transform
Lect11EEE 202
4
We 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
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1. Simple Poles
Lect11EEE 202
5
Simple 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
n
n
n
m
psK
psK
psK
pspspszszsK
s
2
2
1
1
21
10)(F
ipsii spsK
)()( F
tpn
tptp neKeKeKtf 2121)(
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2. Complex Conjugate Poles
Lect11EEE 202
6
Complex 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
)()()()(
)( 11*11
jsK
jsK
jsK
jsK
s F
jssjsK
)()(1 F
teKtf t cos2)( 1Video.edhole.com
3. Repeated Poles
Lect11EEE 202
7
When 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
r
rr ps
Kps
Kps
Kpss
ss
1
12
1
12
1
11
11
1
)()(
)(Q
PF
tpr
rtptp e
rtKetKeKtf 111
!1)(
1
11211
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3. Repeated Poles (cont’d.)
Lect11EEE 202
8
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
1
)(!
111
ps
rjr
jr
j spsdsd
jrK
F
1
1
)()( 2111
2112
psps
spsdsdKspsK
FF
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The “Finger” MethodLet’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)(
sssssF
Lect11 EEE 202 9
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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
65
)3)(2()1(5
)30)(20)(()10(5
)3)(2()1(5)(
ssssssF
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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
310
25
65)(
Lect11 EEE 202 11
310
)1)(3()2(5
)3)(23)(3()13(5
)3)(2()1(5)(
25
)1)(2()1(5
)32)(2)(2()12(5
)3)(2()1(5)(
ssssss
ssssss
F
F
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Initial Value Theorem
Lect11EEE 202
12
The 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
)(lim)(lim0
s s tfst
F
)(')('
lim)()(
limxhxg
xhxg
cxcx
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Final Value Theorem
Lect11EEE 202
13
The 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
)(lim)(lim s s tf 0stF
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Initial and Final Value TheoremsThe 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
310
25
65)(
65
21510
65)(
06
201553
1025
65)0( 00
eetf
eetf
Lect11 EEE 202 14
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Initial and Final Value TheoremsWhat 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
0552
5lim
65)1(5
lim)0(
)3)(2()1(5
lim)(lim)0(
2
ssss
f
ssss s f
sdsd
dsd
s
ssF
65
)3)(2()1(5
)3)(2()1(5
lim)(lim)(00
ssss s f
ssF
Lect11 EEE 202 15
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Solving Differential Equations
Lect11EEE 202
16
Laplace transform approach automatically includes initial conditions in the solution
Example: For zero initial conditions, solve)0(')0()(
)(
)0()()(
22
2
yysssdt
tyd
xssdt
txd
Y
X
LL
)(4)(30)(11)(2
2
tutydt
tyddt
tyd
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Class ExamplesFind inverse Laplace transforms of
Drill Problems P5-3, P5-5 (if time permits)
84)(
)1()(
2
2
ssss
sss
Z
Y
Lect11 EEE 202 17
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