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Lecture 5:The Laplace Transform Method of Solution

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

Summary: Continue to describe methods for representing signals as superpositions of complex exponential functions. Develop efficient methods for analyzing LTI systems.

9/23/99 (T.F. Weiss)
Lecture #5: The Laplace transform method of solution
Motivation:
  • Continue to describe methods for representing signals as superpositions of complex exponential functions
  • Develop efficient methods for analyzing LTI systems
Outline:
  • Review of last lecture
  • Laplace transform of the family of singularity functions
  • More on the region of convergence
  • Analysis of networks with the Laplace transform — the impedance method
  • Finding inverse transforms — partial fraction expansion
  • Conclusion
  • Historical perspective — Oliver Heaviside
Review
  • 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.
Laplace transforms of singularity functions
Unit impulse function
L δ ( t ) = ∑ − ∞ ∞ δ ( t ) . e − st dt L δ ( t ) = ∑ − ∞ ∞ δ ( t ) . e − st dt size 12{L left lbrace δ \( t \) right rbrace = Sum cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {δ \( t \) "." e rSup { size 8{ - ital "st"} } ital "dt"} } {}
Recall the definition of the unit impulse
∑ = ∞ ∞ δ ( t ) f ( t ) dt = f ( 0 ) ∑ = ∞ ∞ δ ( t ) f ( t ) dt = f ( 0 ) size 12{ Sum cSub { size 8{ {}= infinity } } cSup { size 8{ infinity } } {δ \( t \) f \( t \) ital "dt"=f \( 0 \) } } {}
Hence,
L { δ ( t ) } = 1 L { δ ( t ) } = 1 size 12{L lbrace δ \( t \) rbrace =1} {}
for all values of s. The region of convergence is the entire s plane.
Unit impulse function delayed — use of properties
The Laplace transform of an impulse located at t = 0 is
L { δ ( t ) } = 1 L { δ ( t ) } = 1 size 12{L lbrace δ \( t \) rbrace =1} {}
Using the delay property, x(t)⇔LX(s)x(t)⇔LX(s) size 12{x \( t \) { dlrarrow } cSup { size 8{L} } X \( s \) } {}
x ( t − T ) ⇔ L X ( s ) e − s / T x ( t − T ) ⇔ L X ( s ) e − s / T size 12{x \( t - T \) { dlrarrow } cSup { size 8{L} } X \( s \) e rSup { size 8{ - s/T} } } {}
the Laplace transform of the delayed impulse is
L { δ ( t − T ) } = e − sT L { δ ( t − T ) } = e − sT size 12{L lbrace δ \( t - T \) rbrace =e rSup { size 8{ - ital "sT"} } } {}
and the region of convergence is the whole s plane.
Two-minute miniquiz problem
Problem 5-1
Find the Laplace transform including the ROC for
x ( t ) = e − 2 ( t − 4 ) u ( t − 4 ) x ( t ) = e − 2 ( t − 4 ) u ( t − 4 ) size 12{x \( t \) =e rSup { size 8{ - 2 \( t - 4 \) } } u \( t - 4 \) } {}
Two-minute miniquiz solution
Problem 5-1
We use the Laplace transform of the causal exponential time function and time delay property to solve this problem.
e − 2t u ( t ) ⇔ L 1 s + 2 for σ > − 2 e − 2t u ( t ) ⇔ L 1 s + 2 for σ > − 2 size 12{e rSup { size 8{ - 2t} } u \( t \) {` dlrarrow } cSup { size 8{L} } ` { {1} over {s+2} } `~ matrix { {} # {} } ital "for" matrix { {} # {} } ~σ> - 2} {}
e − 2 ( t − 4 ) u ( t − 4 ) ⇔ L 1 s + 2 e − 4s for σ > − 2 e − 2 ( t − 4 ) u ( t − 4 ) ⇔ L 1 s + 2 e − 4s for σ > − 2 size 12{e rSup { size 8{ - 2 \( t - 4 \) } } u \( t - 4 \) {` dlrarrow `} cSup { size 8{L} } { {1} over {s+2} } e rSup { size 8{ - 4s} } ` matrix { {} # {} } ital "for" matrix { {} # {} } σ> - 2} {}
Singularity functions and their relatives
The Laplace transform of a unit impulse is
δ ( t ) ⇔ L 1 for all s δ ( t ) ⇔ L 1 for all s size 12{δ \( t \) { dlrarrow } cSup { size 8{L} } 1 matrix { {} # {} } ital "for" matrix { {} # {} } ` ital "all" matrix { {} # {} } `s} {}
and from the Laplace transform of a causal exponential with αα size 12{α} {} = 0 we have the Laplace transform of a causal step function
u ( t ) ⇔ L 1 s for σ > 0 u ( t ) ⇔ L 1 s for σ > 0 size 12{u \( t \) { dlrarrow } cSup { size 8{L} } { {1} over {s} } matrix { {} # {} } ital "for"` matrix { {} # {} } σ>0} {}
Note this fits together with the time differentiation property
dx ( t ) dt ⇔ L sX ( s ) dx ( t ) dt ⇔ L sX ( s ) size 12{ { { ital "dx" \( t \) } over { ital "dt"} } { dlrarrow } cSup { size 8{L} } ital "sX" \( s \) } {}
since in a generalized function sense
δ ( t ) = du ( t ) dt ⇔ L L { δ ( t ) } = s ( 1 s ) = 1 δ ( t ) = du ( t ) dt ⇔ L L { δ ( t ) } = s ( 1 s ) = 1 size 12{δ \( t \) = { { ital "du" \( t \) } over { ital "dt"} } ` { dlrarrow } cSup { size 8{L} } `L lbrace δ \( t \) rbrace =s \( { {1} over {s} } \) =1} {}
Singularity functions and their relatives, cont’d
We use the multiplication by t property
tx ( t ) ⇔ L − dX ( s ) ds tx ( t ) ⇔ L − dX ( s ) ds size 12{ ital "tx" \( t \) ~ { dlrarrow } cSup { size 8{L} } ~ - { { ital "dX" \( s \) } over { ital "ds"} } } {}
to obtain
tu ( t ) ⇔ L − d ds ( 1 s ) = 1 s 2 for σ > 0 tu ( t ) ⇔ L − d ds ( 1 s ) = 1 s 2 for σ > 0 size 12{ ital "tu" \( t \) ~ { dlrarrow } cSup { size 8{L} } ~ - { {d} over { ital "ds"} } \( { {1} over {s} } \) = { {1} over {s rSup { size 8{2} } } } ~ ital "for"`σ>0} {}
and use it again to obtain
t 2 u ( t ) ⇔ L − d ds ( 1 s 2 ) = 1 s 3 for σ > 0 t 2 u ( t ) ⇔ L − d ds ( 1 s 2 ) = 1 s 3 for σ > 0 size 12{t rSup { size 8{2} } u \( t \) ~ { dlrarrow } cSup { size 8{L} } ~ - { {d} over { ital "ds"} } \( { {1} over {s rSup { size 8{2} } } } \) = { {1} over {s rSup { size 8{3} } } } ~ ital "for"`σ>0} {}
which implies that by induction
or
t n u ( t ) ⇔ L n ! s n + 1 for σ > 0 t n u ( t ) ⇔ L n ! s n + 1 for σ > 0 alignl { stack { size 12{t rSup { size 8{n} } u \( t \) ~ { dlrarrow } cSup { size 8{L} } ~ { {n!} over {s rSup { size 8{n+1} } } } ~ ital "for"`σ>0} {} # {} } } {}
t n − 1 ( n − 1 ) ! u ( t ) ⇔ L 1 s n for σ > 0 t n − 1 ( n − 1 ) ! u ( t ) ⇔ L 1 s n for σ > 0 size 12{ { {t rSup { size 8{n - 1} } } over { \( n - 1 \) !} } u \( t \) ~ { dlrarrow } cSup { size 8{L} } ~ { {1} over {s rSup { size 8{n} } } } ~ ital "for"`σ>0} {}
Summary of singularity functions and their relatives
Figure 1
Wild and crazy singularity functions
Since taking the derivative of a time function corresponds to multiplying the Laplace transform by s we can contemplate the derivative of the unit impulse called the unit doublet.
dδ ( t ) dt = δ ( t ) ⇔ L s dδ ( t ) dt = δ ( t ) ⇔ L s size 12{ { {dδ \( t \) } over { ital "dt"} } =δ \( t \) ~ { dlrarrow } cSup { size 8{L} } ~s} {}
This process can be continued by taking successive derivatives of the impulse to form the unit triplet which has Laplace transform s2s2 size 12{s rSup { size 8{2} } } {}, unit quadruplet, etc. In general, the nth derivative of the unit impulse has a Laplace transform snsn size 12{s rSup { size 8{n} } } {}. We shall consider the usefulness of these higher order singularity functions later!
General comments on the ROC
  • Unit impulse ⇒ ROC is the whole s plane.
  • Finite duration, absolutely integrable time function ⇒ ROC is the whole s plane.
  • Time shifting a time function does not change its ROC.
  • Right-sided time function ⇒ ROC is to the right of the rightmost pole.
  • Left-sided time function ⇒ ROC is to left of the left-most pole.
General comments on the ROC, cont’d
  • A sum of causal and anti-causal exponential time functions that has a Laplace transform ⇒ ROC is a strip in the s plane.
  • There are no poles in the ROC.
  • Some time functions do not have Laplace transforms, e.g., x(t)=e−tx(t)=e−t size 12{x \( t \) =e rSup { size 8{ - t} } } {} for all t.
Analysis of networks with the Laplace transform — the impedance method Kirchhoff’s laws
Kirchhoff’s current and voltage laws are algebraic equations that
link the branch variables in a network,
∑ node i k ( t ) = 0 and ∑ loop v k ( t ) = 0 ∑ node i k ( t ) = 0 and ∑ loop v k ( t ) = 0 size 12{ Sum cSub { size 8{ ital "node"} } {i rSub { size 8{k} } \( t \) =0 matrix { {} # {} } } ital "and" matrix { {} # {} } Sum cSub { size 8{ ital "loop"} } {v rSub { size 8{k} } \( t \) } =0} {}
If we take the Laplace transform of these equations then we obtain
∑ node I k ( s ) = 0 and ∑ loop V k ( s ) = 0 ∑ node I k ( s ) = 0 and ∑ loop V k ( s ) = 0 size 12{ Sum cSub { size 8{ ital "node"} } {I rSub { size 8{k} } \( s \) =0 matrix { {} # {} } } ital "and" matrix { {} # {} } Sum cSub { size 8{ ital "loop"} } {V rSub { size 8{k} } \( s \) } =0} {}
Hence, the Laplace transforms of the branch variables satisfy KCL and KVL.
Constitutive relations
Resistance, capacitance, and inductance
Figure 2
Impedance and admittance
The ratios of voltage to current and current to voltage are system functions with special names. The impedance is defined as
Z ( s ) = V ( s ) I ( s ) Z ( s ) = V ( s ) I ( s ) size 12{Z \( s \) = { {V \( s \) } over {I \( s \) } } } {}
and the admittance is defined as
Y ( s ) = I ( s ) V ( s ) Y ( s ) = I ( s ) V