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based on slides by J. Yan Slide 1.1
Laplace Transform Part 1: Introduction (I&N Chap 13)
• Definition of the L.T. • L.T. of Singularity Functions • L.T. Pairs • Properties of the L.T. • Inverse L.T. • Convolution • IVT(initial value theorem) & FVT (final value
theorem)
Slide 1.2
Lessons from Phasor Analysis What good are phasors?
dttdiLtRitvtvtv LR)()()()()( +=+= IIVVV LjRLR ω+=+=
( )RLM tLR
Vti ωωω
1
22tancos
)()( −−
+= ( )RLM
LRV
LjR
ω
ω
ω
1
22tan0
)(−−°∠
+=
+=⇒
I
VI
Time domain
How might we solve i(t)?
Phasor domain
⇐We’d much rather solve algebraic equations in the phasor domain than DEs in the time domain.
based on slides by J. Yan
Slide 1.3
Motivation for Laplace Transforms Laplace transforms: generalized tool for circuit analysis(transient
response, steady state,…).
)( :Input tf
)( :Output tg
Time domain (t-domain)
Laplace/complex frequency domain (s-domain)
⇐Laplace Xforms: • Change linear DEs into algebraic equations (easier to solve) • Handle a wider variety of inputs than just sinuosoids • Incorporate ICs in the solution automatically • Provide the total response (natural+forced) in one operation
Circuit
)( :Input sF
)( :Output sG
⇒
based on slides by J. Yan
Laplace Transform
Slide 1.4 based on slides by J. Yan
Slide 1.5
Definition of Laplace Transform Given f(t), its (one-sided) Laplace transform (if it exists) is given by:
∫∞ −==0
dte)t(f)s(F)]t(f[ stL
Examples: Recall singularity functions are discontinuous or have discontinuous derivatives (useful for modeling switching in signals).
⎩⎨⎧
≥
<=
0 if 10 if 0
)( :)Function" Heaviside" (a.k.a. step Unit :1 E.g.tt
tu
⎩⎨⎧
≥
<=−=
0
00 if 1
if 0 :step Unit Shifted :2 E.g.
0 tttt
)tt(u)t(ut
)()()(1)(but 0for 0)(
:)Function" Delta Dirac" (a.k.a. impulse Unit :3 E.g.
00 tfdttfttdtttt =−⇒=≠= ∫∫∞
∞−
∞
∞−δδδ
based on slides by J. Yan
Slide 1.6
More Examples: ramp, exp & sine t)t(u)t(r ⋅= :ramp Unit :4 E.g.
)t(ue at− :lexponentia Decaying :5 E.g.
)sin()t(u)tsin()t(u Ttπω 2 :Sinusoid :6 E.g. ⋅=⋅
based on slides by J. Yan
Slide 1.7
• Not all functions have a LT.
• For most circuits (including any you’re required to analyse for EECE 253), the LT will exist in some region of convergence.
• (f1(t)=f2(t))⇒(F1(s)=F2(s)). Is the converse true?
• f(t): r→r. What about F(s)?
Comments about the LT
based on slides by J. Yan
Slide 1.8
Laplace Transform Pairs
*Defined for t≥0; f(t)=0 for t<0.
based on slides by J. Yan
Slide 1.9
Properties of the L.T.
)(lim)(lim :(FVT) Theorem Value Final .12
)(lim)(lim :(IVT) Theorem Value Initial .11
0
0
ssFtf
ssFtf
st
st
→∞→
∞→→
=
=
based on slides by J. Yan
Slide 1.10
• You won’t be asked to compute the LT from the definition. Instead, use the look-up tables of the preceding two slides (on an exam, these will be provided so no need to memorise them now).
• The LT is interesting mathematically but also takes much time to understand. For now, focus more on using this tool rather than understanding why/how it works.
• The power of this tool largely depends on the properties of time differentiation/integration. Observe what happens in the s-domain.
• We’ll increasingly see that the poles of a LT (i.e., roots of the denominator) are quite important. You’ve already used this fact earlier this term…where?
Comments about Using the LT in 253
based on slides by J. Yan
Slide 1.11
Examples tetuttf 23)(2)()( : E.g. −−+= δ
t)u(t)(ttf 2sin)( : E.g. 2=
⎩⎨⎧ ≤≤
= otherwise 032for 10
)( : E.g.t
tg
based on slides by J. Yan
Slide 1.12
Inverse Laplace Transform If the region of convergence for F(s) is Re(s)>σc, then the inverse Laplace transform is given by:
∫∞+
∞−==
j
j
stdsesFj
tfsF 1
1
)(21)()]([
σ
σπ1-L
Fortunately, in 253, this computation isn’t required but you’ll need to generate a partial fraction expansion (PFE) and use look-up tables. Algorithm to find inverse LT: 1. Find all poles of F(s). ID them as simple vs. repeated vs. complex. 2. Find partial fraction expansion (PFE) in basic terms. 3. Look up inverse of each basic term in tables.
Consider F(s)=N(s)/D(s) where N(s) & D(s) are polynomials in s with degree (N(s))<degree(D(s))=n. “Poles” of F(s) are the roots pi of D(s)=0 so we can write: D(s)=(s-p1) (s-p2)···(s-pn)
based on slides by J. Yan
Slide 1.13
Examples ).( find ,
165
341)(Given 2 tf
ss
ssF
+−
++=
).( find ,)4)(3)(1(
)2(6)(Given tfsss
ssF+++
+=
based on slides by J. Yan
Slide 1.14
Poles of F(s) There are 3 relatively distinct types of poles that F(s) may have: Simple: pi is real and negative (pi<0), occurs with degree 1.
Repeated: pi<0, occurs with degree m≥2.
Complex-Conjugate Pair: pi=σ+jω with σ<0⇒pi+1=σ-jω= pi*
based on slides by J. Yan
Slide 1.15
F(s) Partial Fraction Expansion • Given F(s)=N(s)/D(s) and the poles of D(s), you often need to find the
coefficients in the PFE. The text demonstrates the Residue Method for all coefficients but I recommend using this only for a pole’s highest degree (i.e., if simple, if repeated). For the others (complex poles and lower degrees of a pole), I recommend a form of the Algebraic Method (examples on next two slides).
• Note subtle differences in my choice of notation compared to I&N (and other textbooks). Consider what reasons I might have for these differences.
– textbook uses (s+pi) as a factor of D(s) whereas I prefer (s-pi).
– textbook uses {(s+α)2+β2} as a factor but I prefer {(s-σ)2+ω2}.
• I specified the poles must be in the LHP. Why?
based on slides by J. Yan
Slide 1.16
Example
( ) ( )).( find ,
3162)(Given 2
3
tgssssssG++
++=
based on slides by J. Yan
Slide 1.17
Example
( )( ) ).( find ,1341
10)(Given 2 tgsss
sG+++
=
based on slides by J. Yan
Slide 1.18
Consider a linear time-invariant (LTI) system having impulse response h(t). If the system excitation (or input) is x(t), the response (or output) y(t) can be computed from the convolution integral:
Convolution Integral
)()()()()(0
thtxdthxtyt
⊗=−= ∫ λλλ
This formula is explained at great length in the text. However, I mainly require that you know the following:
{ } { } )()()()()()( sYsHsXthtxty =⋅=⊗=LL
based on slides by J. Yan
Slide 1.19
Convolution Integral Notes The convolution integral applies to systems which are causal, linear
and time-invariant. Suppose you have a system with zero ICs and you know its impulse (Dirac Delta) response is h(t).
• Causal ⇒ h(t)=0 for t<0 (i.e., there is no response before the input stimulus).
• Linear ⇒ Superposition applies (i.e., if inputs r1(t) and r2(t) yield forced responses of y1(t) and y2(t), respectively, then an input of r(t)= αr1(t)+βr2(t) yields the forced response y(t)= αy1(t)+βy2(t)).
• Time-invariant ⇒ The response to δ(t-t0) is h(t-t0). That is, if the input stimulus is shifted by time t0 then so is the response.
Unless otherwise specified, all systems that you see in this course assume these properties (as is the case for most systems you see as an undergraduate). However, do not take them for granted. Nonlinearities and time variance sometimes need to be considered.
based on slides by J. Yan
Slide 1.20
Linear Integrodifferential Equations
1)0()0( where44 Solve : Example ===++ − vvevvv t
0)0( where2)(2)(3 Solve :Example 3
0==++ −∫ yedytyy tt
ττ
Linear integroDEs can be Xformed by the LT into s-domain, solved algebraically (include any ICs) and Xformed back into t-domain.
based on slides by J. Yan
Slide 1.21
Example If the network is in steady state prior to t=0, find i(t) for t>0. (NB: Already solved this type of question before but now can solve using L.T.)
based on slides by J. Yan
INITIAL AND FINAL VALUE THEOREMS
These results relate behavior of a function in the time domain with the behavior
of the Laplace transform in the s-domain
INITIAL VALUE THEOREM
)(lim)(lim
,),(
0 ssFtf
dtdftf
st ∞→→ =
Then transform.
Laplace have boththat Assume
0][lim
)0()(][
=
−=
∞→ dtdf
fssFdtdf
s L
L
then bletransforma is derivative the if And
FINAL VALUE THEOREM
)(lim)(lim)(lim
,),(
0 ssFtftf
dtdftf
st
t
→∞→
∞→
=
Then exists. that and transform
Laplace have boththat Assume
∫
∫
∞
→
∞−
−=
→
−=
00
0
)0()(lim)(
0
)0()()(
fssFdttdtdf
s
fssFdtetdtdf
s
st
as limits Taking
0
)()(lim
=
∞→
s
sFtft
at polesingle amost at andpart real negative with
poles has ifexist will :NOTE
