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# Lecture 13:The Bilateral Laplace Transform

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} } $${ {s} over {s rSup { size 8{2} } + { {1} over { ital "RC"} } s+ { {1} over { ital "RC"} } } }$$ } {}

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 ( t ) e st dt X ( s ) = ax 1 ( t ) e st dt + 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} } $$t$$ e rSup { size 8{ - ital "st"} } ital "dt"+ Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } { ital "bx" rSub { size 8{2} } $$t$$ } } e rSup { size 8{ - ital "st"} } ital "dt"} {}

X ( s ) = aX 1 ( s ) + bX 2 ( s ) X ( s ) = aX 1 ( s ) + bX 2 ( s ) size 12{X $$s$$ = ital "aX" rSub { size 8{1} } $$s$$ + ital "bX" rSub { size 8{2} } $$s$$ } {}

Q. Given the ROC of X1(s)X1(s) size 12{X rSub { size 8{1} } $$s$$ } {}and X2(s)X2(s) size 12{X rSub { size 8{2} } $$s$$ } {} what is the ROC of X(s)?

A. At least the intersection of the ROCs of X1(s)X1(s) size 12{X rSub { size 8{1} } $$s$$ } {}and X2(s)X2(s) size 12{X rSub { size 8{2} } $$s$$ } {}.

2/ Causal exponential time function

If x(t)=eαtu(t)x(t)=eαtu(t) size 12{x $$t$$ =e rSup { size 8{αt} } u $$t$$ } {} then the Laplace transform is

X ( s ) = e αt u ( t ) e st dt X ( s ) = e αt u ( t ) e st dt size 12{X $$s$$ = Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {e rSup { size 8{αt} } u $$t$$ e rSup { size 8{ - ital "st"} } ital "dt"} } {}

X ( s ) = 0 e ( s α ) t dt X ( s ) = 0 e ( s α ) t dt size 12{X $$s$$ = Int rSub { size 8{0} } rSup { size 8{ infinity } } {e rSup { size 8{ - $$s - α$$ t} } ital "dt"} } {}

X ( s ) = 1 s a ( 1 e ( s α ) ) X ( s ) = 1 s a ( 1 e ( s α ) ) size 12{X $$s$$ = { {1} over {s - a} } $$1 - e rSup { size 8{ - \( s - α$$ infinity } } \) } {}

The exponential term has a magnitude of either 0 or ∞. For R{s − αα size 12{α} {}} > 0 the exponential term is zero and the integral con­verges. Let R{s} = σσ size 12{σ} {}, then

X ( s ) = 1 s a for σ > α X ( s ) = 1 s a for σ > α size 12{X $$s$$ = { {1} over {s - a} } matrix { {} # {} } ital "for" matrix { {} # {} } σ>α} {}

The region of convergence (ROC) of X(s) is σ>ασ>α size 12{σ>α} {}

Thus we have

x ( t ) = e αt u ( t ) L X ( s ) = 1 s a for σ > α x ( t ) = e αt u ( t ) L X ( s ) = 1 s a for σ > α size 12{x $$t$$ =e rSup { size 8{αt} } u $$t$$ matrix { {} # {} } { dlrarrow } cSup { size 8{L} } matrix { {} # {} } X $$s$$ = { {1} over {s - a} } matrix { {} # {} } ital "for" matrix { {} # {} } σ>α} {}

3/ Relation between time functions and pole-zero diagrams

Consider the causal exponential time function and its Laplace transform

x ( t ) = e αt u ( t ) L X ( s ) = 1 s α for σ > α x ( t ) = e αt u ( t ) L X ( s ) = 1 s α for σ > α size 12{x $$t$$ =e rSup { size 8{αt} } u $$t$$ matrix { {} # {} } { dlrarrow } cSup { size 8{L} } matrix { {} # {} } X $$s$$ = { {1} over {s - α} } matrix { {} # {} } ital "for" matrix { {} # {} } σ>α} {}

The following shows both the time functions and the pole-zero diagrams for 5 different values of αα size 12{α} {}

Demo of relation of pole-zero diagram to time function.

Conclusions on relation between causal time functions and

pole-zero diagrams

■ Note, st is dimensionless. Hence, s must have units inversely proportional to time.

IV. RELATION OF TIME FUNCTIONS TO REGION OF CONVERGENCE

We have considered only causal exponential time functions. Now we wish to consider more general time functions. First, we com­pare the Laplace transforms of two time functions: the causal exponential time function x(t)=eσtu(t)x(t)=eσtu(t) size 12{x $$t$$ =e rSup { size 8{ - σt} } u $$- t$$ } {}and the anti-causal exponential time function x(t)=eσtu(t)x(t)=eσtu(t) size 12{x $$t$$ = - e rSup { size 8{ - σt} } u $$- t$$ } {}.

1/ Causal exponential time function

The Laplace transform of the causal exponential time function was worked out earlier and is

X ( s ) = 1 s + a for σ > α X ( s ) = 1 s + a for σ > α size 12{X $$s$$ = { {1} over {s+a} } matrix { {} # {} } ital "for" matrix { {} # {} } σ> - α} {}

The region of convergence (ROC) of X(s) is σ>ασ>α size 12{σ>α} {}

2/ Anti-causal exponential time function

If x(t)=eσtu(t)x(t)=eσtu(t) size 12{x $$t$$ = - e rSup { size 8{ - σt} } u $$- t$$ } {}then the Laplace transform is

X ( s ) = e αt u ( t ) e st dt X ( s ) = e αt u ( t ) e st dt size 12{X $$s$$ = - Int rSub { size 8{ - infinity } } rSup { size 8{ infinity } } {e rSup { size 8{ - αt} } } u $$- t$$ e rSup { size 8{ - ital "st"} } ital "dt"} {}

X ( s ) = 0 e ( s + α ) t dt X ( s ) = 0 e ( s + α ) t dt size 12{X $$s$$ = - Int rSub { size 8{0} } rSup { size 8{ infinity } } {e rSup { size 8{ - $$s+α$$ t} } ital "dt"} } {}

X ( s ) = 1 s + a ( 1 e ( s + α ) ) X ( s ) = 1 s + a ( 1 e ( s + α ) ) size 12{X $$s$$ = { {1} over {s+a} } $$1 - e rSup { size 8{ \( s+α$$ infinity } } \) } {}

The integral converges provided R{s + αα size 12{α} {}} <0, and

X ( s ) = 1 s + a for σ > α X ( s ) = 1 s + a for σ > α size 12{X $$s$$ = { {1} over {s+a} } matrix { {} # {} } ital "for" matrix { {} # {} } σ> - α} {}

The region of convergence (ROC) of X(s) is σ>ασ>α size 12{σ> - α} {}

3/ Causal and anti-causal exponential time functions can have the same Laplace transform formula but differ in ROCs

4/ The Laplace transform formula plus the ROC uniquely specify the time function

• 1/(s + αα size 12{α} {}) is the Laplace transform formula of eαtu(t)eαtu(t) size 12{e rSup { size 8{ - αt} } u $$t$$ } {} and of eαtu(t)eαtu(t) size 12{ - e rSup { size 8{ - αt} } u $$- t$$ } {}.
• The ROCs for these two time functions are different.
• The ROC must be known to uniquely compute the time function.

5/ More general exponential time functions

Now consider

x ( t ) = e α t for α > 0 x ( t ) = e α t for α > 0 size 12{x $$t$$ =e rSup { size 8{ - α lline t rline } } matrix { {} # {} } ital "for" matrix { {} # {} } α>0} {}

which is neither causal nor anti-causal but is bounded.

Note we can express x(t) as

x ( t ) = e α t = e αt u ( t ) + e αt u ( t ) x ( t ) = e α t = e αt u ( t ) + e αt u ( t ) size 12{x $$t$$ =e rSup { size 8{ - α lline t rline } } =e rSup { size 8{αt} } u $$- t$$ +e rSup { size 8{ - αt} } u $$t$$ } {}

We use the linearity property of the Laplace transform

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} } $$s$$ } {} {} {}

and previous results on the Laplace transforms of anti-causal and causal exponentials to give

X ( s ) = 1 s a + 1 s + a X ( s ) = 1 s a + 1 s + a size 12{X $$s$$ = - { {1} over {s - a} } + { {1} over {s+a} } } {}

We can combine these terms and find a formula that is valid in a region of the s plane as follows

x ( t ) = e α t L X ( s ) = s 2 α 2 for α < σ < α x ( t ) = e α t L X ( s ) = s 2 α 2 for α < σ < α size 12{x $$t$$ =e rSup { size 8{ - α lline t rline } } { dlrarrow } cSup { size 8{L} } X $$s$$ = { { - 2α} over {s rSup { size 8{2} } - α rSup { size 8{2} } } } matrix { {} # {} } ital "for" matrix { {} # {} } - α<σ<α} {}

6/ Relations of parts of time function to the ROC

The ROC of X(s) is a strip in the s plane. We can associate which time function goes with each pole. If the ROC is to the left of a pole then that pole corresponds to an anti-causal (leftsided) time function. If the ROC is to the right of a pole then that pole corresponds to a causal (right-sided) time function.

Example — the effect of the ROC

X ( s ) = s + 3 ( s + 1 ) ( s 2 ) = A s + 1 + B s 2 X ( s ) = s + 3 ( s + 1 ) ( s 2 ) = A s + 1 + B s 2 size 12{X $$s$$ = { {s+3} over { $$s+1$$ $$s - 2$$ } } = { {A} over {s+1} } + { {B} over {s - 2} } } {}

A=23A=23 size 12{A= - { {2} over {3} } } {} and B=53B=53 size 12{B= { {5} over {3} } } {}

X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for R { s } > X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for R { s } > size 12{X $$s$$ = { { - 2} over {3} } $${ {1} over {s+1} }$$ + { {5} over {3} } $${ {1} over {s - 2} }$$ matrix { {} # {} } ital "for" matrix { {} # {} } R lbrace s rbrace >2Ơ} {}

x ( t ) = ( 2 3 e t + 5 3 e 2t ) u ( t ) x ( t ) = ( 2 3 e t + 5 3 e 2t ) u ( t ) size 12{x $$t$$ = $${ { - 2} over {3} } e rSup { size 8{ - t} } + { {5} over {3} } e rSup { size 8{2t} }$$ u $$t$$ } {}

X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for 1 < R { s } < X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for 1 < R { s } < size 12{X $$s$$ = { { - 2} over {3} } $${ {1} over {s+1} }$$ + { {5} over {3} } $${ {1} over {s - 2} }$$ matrix { {} # {} } ital "for" matrix { {} # {} } - 1<R lbrace s rbrace <2Ơ} {}

x ( t ) = 2 3 e t u ( t ) + 5 3 e 2t u ( t ) x ( t ) = 2 3 e t u ( t ) + 5 3 e 2t u ( t ) size 12{x $$t$$ = { { - 2} over {3} } e rSup { size 8{ - t} } u $$t$$ + { {5} over {3} } e rSup { size 8{2t} } u $$- t$$ } {}

X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for R { s } < X ( s ) = 2 3 ( 1 s + 1 ) + 5 3 ( 1 s 2 ) for R { s } < size 12{X $$s$$ = { { - 2} over {3} } $${ {1} over {s+1} }$$ + { {5} over {3} } $${ {1} over {s - 2} }$$ matrix { {} # {} } ital "for" matrix { {} # {} } R lbrace s rbrace < - 1Ơ} {}

x ( t ) = ( 2 3 e t 5 3 e 2t ) u ( t ) x ( t ) = ( 2 3 e t 5 3 e 2t ) u ( t ) size 12{x $$t$$ = $${ { - 2} over {3} } e rSup { size 8{ - t} } - { {5} over {3} } e rSup { size 8{2t} }$$ u $$- t$$ } {}

V. CONCLUSIONS

• The Laplace transform represents a time function as a superposition of complex exponentials.
• A time function is related uniquely to a Laplace transform if the ROC is specified.
• If the Laplace transform of a sum of causal and anti-causal exponential time functions exists, its ROC is a strip in the s-plane parallel to the jω-axis.

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