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Lecture 13:The Bilateral Laplace Transform

Module by: Thanh Dinh Vu, Truc Pham-Dinh, Anh Tuan Hoang, Tam Huynh-Ngoc

Summary: Method for representing signals as superpositions of complex exponential functions, which leads to simpler solving process.

Lecture #13:
THE BILATERAL LAPLACE TRANSFORM
Motivation: Method for representing signals as superpositions of complex exponential functions, which leads to simpler solving process.
Outline:
  • Review of last lecture
  • The bilateral Laplace transform
  • Definition
  • Properties of the Laplace transform
  • Transforms of simple time functions
  • The region of convergence
  • Conclusion
I. REVIEW OF LAST LECTURE
Solve linear differential equation for a causal exponential input
{} ∑ n = 0 N a n d n y ( t ) dt n = ∑ m = 0 M b m d m x ( t ) dt m for x ( t ) = Xe st u ( t ) ∑ n = 0 N a n d n y ( t ) dt n = ∑ m = 0 M b m d m x ( t ) dt m for x ( t ) = Xe st u ( t ) size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y \( t \) } over { ital "dt" rSup { size 8{n} } } } = Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } { {d rSup { size 8{m} } x \( t \) } over { ital "dt" rSup { size 8{m} } } } ` matrix { {} # {} } ital "for"` matrix { {} # {} } x \( t \) `=` ital "Xe" rSup { size 8{ ital "st"} } u \( t \) } } } {}
Solve homogeneous equation for t > 0
∑ n = 0 N a n d n y ( t ) dt n = 0 by assuming y h ( t ) = Ae λt ∑ n = 0 N a n d n y ( t ) dt n = 0 by assuming y h ( t ) = Ae λt size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y \( t \) } over { ital "dt" rSup { size 8{n} } } } =`0``} matrix { {} # {} } "by " matrix { {} # {} } "assuming" matrix { {} # {} } y rSub { size 8{h} } \( t \) = ital "Ae" rSup { size 8{λt} } } {}
Solve characteristic polynomial
∑ n = 0 N a n λ n = 0 for λ ∑ n = 0 N a n λ n = 0 for λ size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } } λ rSup { size 8{n} } =0` matrix { {} # {} } ital "for" matrix { {} # {} } `λ} {}
Solve for a particular solution for t > 0
∑ n = 0 N a n d n y p ( t ) dt n = ∑ m = 0 M b m d m x ( t ) dt m for x ( t ) = Xe st ∑ n = 0 N a n d n y p ( t ) dt n = ∑ m = 0 M b m d m x ( t ) dt m for x ( t ) = Xe st size 12{ Sum cSub { size 8{n=0} } cSup { size 8{N} } {a rSub { size 8{n} } { {d rSup { size 8{n} } y rSub { size 8{p} } \( t \) } over { ital "dt" rSup { size 8{n} } } } = Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } { {d rSup { size 8{m} } x \( t \) } over { ital "dt" rSup { size 8{m} } } } ` matrix { {} # {} } ital "for" matrix { {} # {} } `} } x \( t \) = ital "Xe" rSup { size 8{ ital "st"} } } {}
Assuming yp(t)=Yestnyp(t)=Yestn size 12{y rSub { size 8{p} } \( t \) = ital "Ye" rSup { size 8{ ital "st"} } n} {} and solving for Y yields
Y = H ( s ) X = ∑ m = 0 M b m s m ∑ n = 0 N b n s n X Y = H ( s ) X = ∑ m = 0 M b m s m ∑ n = 0 N b n s n X size 12{Y=H \( s \) X= { { Sum cSub { size 8{m=0} } cSup { size 8{M} } {b rSub { size 8{m} } s rSup { size 8{m} } } } over { Sum cSub { size 8{n=0} } cSup { size 8{N} } {b rSub { size 8{n} } s rSup { size 8{n} } } } } X} {}
H ( s ) = V I = 1 C ( s s 2 + 1 RC s + 1 RC ) H ( s ) = V I = 1 C ( s s 2 + 1 RC s + 1 RC ) size 12{H \( s \) = { {V} over {I} } = { {1} over {C} } \( { {s} over {s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "RC"} } } } \) } {}
Figure 1
Logic for an analysis method for CT LTI systems
  • H(s) characterizes system  compute H(s) efficiently.
  • In steady state, response to XestXest size 12{"Xe" rSup { size 8{"st"} } } {} is H(s)XestH(s)Xest size 12{H \( s \) "Xe" rSup { size 8{"st"} } } {}.
  • Represent arbitrary x(t) as superpositions of XestXest size 12{"Xe" rSup { size 8{"st"} } } {} on s.
  • Compute response y(t) as superpositions of H(s)XestH(s)Xest size 12{H \( s \) "Xe" rSup { size 8{"st"} } } {}on s.
II. THE BILATERAL LAPLACE TRANSFORM DEFINITION
1/ Analysis formula
The bilateral Laplace transform is defined by the analysis formula
X ( s ) = ∫ − ∞ ∞ x ( t ) e − st dt X ( s ) = ∫ − ∞ ∞ x ( t ) e − st dt size 12{X \( s \) = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {x \( t \) } e rSup { size 8{ - ital "st"} } ital "dt"} {}
X(s) is defined for regions in s — called the region of conver­gence (ROC) — for which the integral exists.
2/ Synthesis formula
The inverse transform is defined by the synthesis formula
x ( t ) = 1 2πj ∫ σ − j ∞ σ + j ∞ X ( t ) e st ds x ( t ) = 1 2πj ∫ σ − j ∞ σ + j ∞ X ( t ) e st ds size 12{x \( t \) = { {1} over {2πj} } Int rSub { size 8{σ - j infinity } } rSup { size 8{σ+j infinity } } {X \( t \) } e rSup { size 8{ ital "st"} } ital "ds"} {}
Since s is a complex quantity, X(s) is a complex function of a complex variable, and σσ size 12{σ} {}is in the ROC.
  • The synthesis formula involves integration in the complex s domain. We shall not perform this integration in this subject. The synthesis formula will be used only to prove theorems and not to compute time functions directly.
  • The synthesis formula makes apparent that x(t) is synthe­sized by a superposition of an uncountably infinite number of eternal complex exponentials estest size 12{e rSup { size 8{"st"} } } {}each for a different value of s and each of infinitesimal magnitude X(s)ds.
3/ Relation to unilateral Laplace transform
The difference between the unilateral and the bilateral Laplace transform is in the lower limit of integration, i.e.,
Bilateral ⇒ X ( s ) = ∫ − ∞ ∞ x ( t ) e − st dt Bilateral ⇒ X ( s ) = ∫ − ∞ ∞ x ( t ) e − st dt size 12{"Bilateral " drarrow X \( s \) = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {x \( t \) } e rSup { size 8{ - ital "st"} } ital "dt"} {}
Unilateral ⇒ X ( s ) = ∫ 0 ∞ x ( t ) e − st dt Unilateral ⇒ X ( s ) = ∫ 0 ∞ x ( t ) e − st dt size 12{"Unilateral " drarrow X \( s \) = Int rSub { size 8{0} } rSup { size 8{ infinity } } {x \( t \) } e rSup { size 8{ - ital "st"} } ital "dt"} {}
  • The unilateral Laplace transform is restricted to causal time functions, and takes initial conditions into account in a sys­tematic, automatic manner both in the solution of differential equations and in the analysis of systems.
  • The bilateral Laplace transform can represent both causal and non-causal time functions. Initial conditions are ac­counted by including additional inputs.
4/ Approach
Databases of time functions and their Laplace transforms are developed as follows:
  • Determine the Laplace transforms of simple time functions directly,
  • Use the Laplace transform properties to extend the database of transform pairs.
Notation
We shall use two useful notations — L {x(t)} signifies the Laplace transform of x(t) and a Laplace transform pair is indicated by
x ( t ) ⇔ L X ( s ) x ( t ) ⇔ L X ( s ) size 12{x \( t \) { dlrarrow } cSup { size 8{L} } X \( s \) } {}
III. LAPLACE TRANSFORM PROPERTIES
A few of the important properties are summarized; a more com­plete list is appended.
Most proofs of properties are simple as indicated below.
1/ Linearity
ax 1 ( t ) + bx 2 ( t ) ⇔ L aX 1 ( s ) + bX 2 ( s ) ax 1 ( t ) + bx 2 ( t ) ⇔ L aX 1 ( s ) + bX 2 ( s ) size 12{ ital "ax" rSub { size 8{1} } \( t \) + ital "bx" rSub { size 8{2} } \( t \) { dlrarrow } cSup { size 8{L} } ital "aX" rSub { size 8{1} } \( s \) + ital "bX" rSub { size 8{2} } \( s \) } {} {} {}
The proof follows from the definition of the Laplace transform as a definite integral. Let x(t)= ax1(t)+ bx2(t)x(t)= ax1(t)+ bx2(t) size 12{x \( t \) =" ax" rSub { size 8{1} } \( t \) +" bx" rSub { size 8{2} } \( t \) } {}, and use the analysis formula.
X ( s ) = ∫ − ∞ ∞ ( ( ax 1 ( t ) + bx 2 ( t ) ) e − st dt X ( s ) = ∫ − ∞ ∞ ( ( ax 1 ( t ) + bx 2 ( t ) ) e − st dt size 12{X \( s \) = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } { \( \( ital "ax" rSub { size 8{1} } \( t \) + ital "bx" rSub { size 8{2} } \( t \) \) e rSup { size 8{ - ital "st"} } ital "dt"} } {}
X ( s ) = ∫ − ∞ ∞ ax 1 ( t ) e − st dt + ∫ − ∞ ∞ bx 2