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ELEC2400 Signals & SystemsChapter 4. Laplace Transform Analysis

Brett Ninnes

[email protected].

School of Electrical Engineering and Computer Science

The University of Newcastle

Slides by Juan I. Yuz ([email protected]) - July 24, 2003 – p.1/95

4. Laplace Transform Analysis

For linear and time invariant systems, the relationshipbetween an input signal u(t) and the output signal y(t)is given by the convolution:

y(t) = [h ~ u](t) =

∫ ∞

−∞h(t − σ)u(σ) dσ.

This expression is useful for deducing fundamentalproperties of linear systems, but not for detailedcalculations.

To address this, a tool known as the LaplaceTransform is more relevant.

Chapter 4. Laplace Transform Analysis – p.2/95

Definition of the Laplace Transform

Given a signal f(t), t ∈ R, its Laplace Transform is afunction F (s), s ∈ C, defined by:

F (s) = Lf ,

∫ ∞

−∞f(t)e−st dt ; s ∈ C.

Note that F (s) itself can be complex valued eventhough the signal f(t) is real valued.

By convention, if a signal has a lowercase symbol f , itsLaplace transform is given by the upper-case of thatsymbol, F .


Definition of the Laplace Transform

The previous definition is more correctly called thebilateral Laplace Transform, to distinguish it from themore common used unilateral Laplace Transform:

F (s) = Lf ,

∫ ∞

0−

f(t)e−st dt ; s ∈ C.

In the sequel, we will use only the bilateral LaplaceTransform since it has several advantages:

It does not ignore values of f(t) for t < 0, allowingto consider non-causal systems.It allows a simple and direct link between Fourierand Laplace Transforms to be established.Some fundamental properties of the LaplaceTransform are simpler to state and use.


Geometric InterpretationLaplace Transform:

F (s) = Lf ,

∫ ∞


As s = α + jω, then by De Moivre’s Theorem:

e−st = e−(α+jω)t = e−αte−jωt = e−αt (cosωt − j sinωt)

Hence:

F (s)|s=α+jω =

∫ ∞

−∞

f(t)e−αt cos ωt dt − j

∫ ∞

−∞

f(t)e−αt sinωt dt.

Consequently:Real F (s) measures the strength of e−αt cosωt inf(t), andImag F (s) measures the strength of e−αt sin ωt.


Geometric Interpretation

Depending on α and ω, F (s) = F (α + jω) measures thestrength of various exponentially enveloped andoscillating signals in f(t):

t

e−αt

αα

e−αt

0

t

e−αtReal

t

t

e−αt cos(ωt)e−αt

2π

ω

e−αt

ω

2π

ω

e−αt cos(ωt)

2π

ω

Imaginary


Computation of Laplace Transforms

F (s) = Lf ,

∫ ∞


To apply the idea of signal decomposition, we analyzehow the Laplace Transform acts on certain importantsignals.

The resulting function F (s) = Lf, will be well defined(that is, the integral will exist) for a range of values of s.

This domain is called the Region of Convergence(ROC), and is always given in association with thetransform itself.



Dirac Delta δ(t):Using the characterisation of the Dirac delta:

Lδ =

∫ ∞

−∞δ(t) e−st dt = e−st

∣∣t=0

= 1.

And this result clearly holds for all value of s, andhence the ROC is the whole complex plane C.

Lδ = 1 ; s ∈ C.



Step signal 1(t):The Laplace Transform of the step signal can bedirectly computed as:

L1 =

∫ ∞

−∞1(t) e−st dt

=

∫ ∞

0e−st dt =

e−st

−s

∣∣∣∣

t=∞

t=0

=1

s; Reals > 0.

The ROC is Reals > 0.Note that L1 = 1/s is perfectly well defined fors 6= 0, even if Real s < 0, however . . .If Reals < 0 (s is outside the ROC), then 1/s hasnothing to do with the integral L1.



Ramp signal r(t):The ramp signal is r(t) = 1(t) · t, then its LaplaceTransform is:

Ls =

∫ ∞

−∞1(t) · t e−st dt

=

∫ ∞

0t e−st dt = . . . =

1

s2; Reals > 0.

Again, the ROC is Reals > 0, since for those valueslimt→∞

e−st = 0 and the above integral is finite.



Exponential signal eαt:This signal is defined as f(t) = 1(t) · eαt, α ∈ C, and isprobably the single most important example to study:

F (s) = Lf =

∫ ∞

0eαte−st dt

=

∫ ∞

0e(α−s)t dt =

e(α−s)t

α − s

∣∣∣∣∣

t=∞

t=0

=1

s − α

The ROC is given by Reals > Realα, which isnecessary to ensure:

limt→∞

e(α−s)t → 0



Polynomial times Exponential tneαt:For the signal f(t) = 1(t) · teαt:

L

1(t) · teαt

=

∫ ∞

0

teαte−st dt =

∫ ∞

0

[d

dαeαt

]

e−st dt

=d

dα

∫ ∞

0

eαte−st dt =d

dαL

1(t) · eαt

⇒ L

1(t) · teαt

=d

dα

(1

s − α

)

=1

(s − α)2

And for any integer n ≥ 1:

L

1(t) · tneαt

=dn

dαn

(1

s − α

)

=n!

(s − α)n+1

The ROC is Reals > Realα.



Cosine cosωt:We consider a pure tone at a frequency of ω [rad/s]:

f(t) = 1(t) · cosωt = 1(t) · 1

2

[ejωt + e−jωt

]

Then:

L1(t) · cosωt =

∫ ∞

0

(cosωt)e−st dt =

∫ ∞

0

1

2

[ejωt + e−jωt

]e−st dt

=1

2

∫ ∞

0

(ejωt

)e−st dt +

1

2

∫ ∞

0

(e−jωt

)e−st dt

=1

2

[1

s − jω+

1

s + jω

]

=s

s2 + ω2

Setting α = jω in the previous result implies that theROC is Reals > 0.



Sine sin ωt:Similarly, if we consider a signal:

f(t) = 1(t) · sinωt = 1(t) · 1

2j

[ejωt − e−jωt

]

Then:

L1(t) · sin ωt =

∫ ∞

0

(sinωt)e−st dt =

∫ ∞

0

1

2j

[ejωt − e−jωt

]e−st dt

=1

2j

∫ ∞

0

(ejωt

)e−st dt − 1

2j

∫ ∞

0

(e−jωt

)e−st dt

=1

2j

[1

s − jω− 1

s + jω

]

=ω

s2 + ω2

Again, setting α = jω in the result for 1(t)eαt, impliesthat the ROC is Reals > 0.



Exponentially Damped Sinusoidals:Suppose that f(t) is an exponentially dampedsinusoidal signal at ω [rad/s]. Applying the previousdecomposition for cosωt and sin ωt, we have that:

If f(t) = 1(t) · eαt cosωt = 12

[e(α+jω)t + e(α−jω)t

]1(t), then:

L

1(t) · eαt cos ωt

=s − α

(s − α)2 + ω2; Reals > Realα

If f(t) = 1(t) · eαt sinωt = 12j

[e(α+jω)t − e(α−jω)t

]1(t), then:

L

1(t) · eαt sinωt

=ω

(s − α)2 + ω2; Reals > Realα


Properties of Laplace Transforms

Linearity:For any two signals f(t) and g(t) and any α, β ∈ C, then

Lαf + βg =

∫ ∞

−∞[αf(t) + βg(t)] e−st dt

= α

∫ ∞

−∞f(t)e−st dt + β

∫ ∞

−∞g(t)e−st dt

= αLf + βLg

This is a natural consequence of the linearity of theintegral operator, and the ROC is the intersection ofthe ROC’s of Lf and Lg.



Time Shifting:Suppose that a signal f(t) is formed by shifting anothersignal g(t) in time:

f(t) = g(t − T ).

Then, using the change of variable x = t − T in theLaplace Transform definition:

F (s) = Lf =

∫ ∞

−∞g(t − T )e−st dt

= e−sT

∫ ∞

−∞g(x)e−sx dx

= e−sTLg = e−sT G(s).

for all s ∈ ROC of G(s).



Time Scaling:If the time-scale of a signal is compressed (expanded),its Laplace Transform is expanded (compressed). Thatis, if:

f(t) = g(αt) ;α ∈ R

then, using the change of variable x = αt:

F (s) ,

∫ ∞

−∞f(t) e−st dt =

∫ ∞

−∞g(αt) e−st dt

=1

α

∫ ∞

−∞g(x) e−xs/α dx

=1

αG

( s

α

)

for all s ∈ ROC of G(s).



Convolution: If y(t) = [h ~ u](t), then

Ly = L[h ~ u] =

∫ ∞

−∞

[∫ ∞

−∞

h(t − σ)u(σ) dσ

]

e−st dt

=

∫ ∞

−∞

u(σ) e−sσ

[∫ ∞

−∞

h(t − σ) e−s(t−σ) dt

]

dσ

x = t − σ ⇒ Ly =

∫ ∞

−∞

u(σ) e−sσ

[∫ ∞

−∞

h(x) e−sx dx

]

dσ

=

[∫ ∞

−∞

h(x) e−sx dx

] ∫ ∞

−∞

u(σ) e−sσ dσ

= H(s) U(s)

For all s in the intersection of the ROC of H(s) and U(s).

The Laplace Transform turns convolution into a productof transforms.



Differentiation:There is a close connection (via state-space)between convolution and differential equations.As a consequence, the Laplace Transformationalso simplifies the operation of differentiationHowever, it is necessary to be very careful aboutwhere any discontinuities in f(t) might occur. Inparticular, we considere discontinuities at t = 0,where usually we inject 1(t).



Differentiation:

We consider d

dt[f(t)e−st] = e−st · d

dtf(t) − sf(t)e−st.

Multiplying by 1(t), and integrating both sides:

∞

−∞

1(t)d

dt[f(t)e−st] dt =

∞

−∞

1(t)

d

dtf(t) e

−st dt − s

∞

−∞

1(t)f(t)e−st dt

∞

0−

d

dt[f(t)e−st] dt = L 1(t)

d

dtf(t) − sL1(t)f(t)

limt→∞

f(t)e−st

− f(0−) = L 1(t)d

dtf(t) − sL1(t)f(t)

Thus, if f(t) is differentiable on [0,∞), then for all s inthe intersection of the ROC of F (s) and Real s > 0:

L

1(t)d

dtf(t)

= sL1(t)f(t) − f(0−) = sF (s) − f(0−)



Differentiation:For the 2nd order derivative we have:

L 1(t)d2

dt2f(t) = sL 1(t) ·

d

dtf(t) − lim

t→0−

d

dtf(t)

= s sF (s) − f(0−) −d

dtf(0−) = s2F (s) − sf(0−) −

d

dtf(0−)

And, for higher order derivatives:

L 1(t) ·dn

dtnf(t) = snF (s) − sn−1f(0−) − sn−2 d

dtf(0−) − · · · −

dn−1

dtn−1f(0−)

for all s in the intersection of the ROC of F (s) andReal s > 0.

The Laplace Transform turns differentiation (in the timedomain) into a multiplication by the scalar s.



Integration:Using integration by parts (assuming t > t0):

L∫ t

t0

f(σ) dσ

=

∫ ∞

−∞

[∫ t

t0

f(σ) dσ

]

e−st dt

=

∫ ∞

−∞

[∫ t

t0

f(σ) dσ

]d

dt

(e−st

−s

)

dt

=

[e−st

−s

∫ t

t0

f(σ) dσ

]∣∣∣∣

t=∞

t=t0

+1

s

∫ ∞

−∞

f(t) e−st dt

=1

sF (s)

For all s in the intersection of the ROC of Lf andReals > 0.



Integration:For the general case, with multiple integrals:

L∫ tn

t0

∫ tn−1

t0

· · ·∫ t2

t0

∫ t1

t0

f(σ) dσ dt1 dt2 · · · dtn−2 dtn−1

=1

snF (s).

For all s in the intersection of the ROC of Lf andReals > 0.

This result is consistent with the Laplace Transformof the step 1(t) and Dirac delta δ(t) signals:

1(t) =

∫ t

−∞

δ(σ) dσ ⇒ L1(t) =1

sLδ =

1

s, Reals > 0

The Laplace Transform turns integration into divisionby the scalar s.



Transform of a Product:We have seen that:

L[h ~ u](t) = LhL u = H(s)U(s)

There is a dual result : If h(t) = g(t)f(t), then for anyγ ∈ R such that G(s) and F (s) exist for all s = γ + jωwith ω ∈ (−∞,∞), then:

Lg(t)f(t) = H(s) =1

2πj

∫ γ+j∞

γ−j∞F (s − σ)G(σ) dσ

The Laplace transform turns the product of two signalsinto the convolution of two transforms.



Multiplication by eαt:We have seen that:

Lf(t − T ) = e−sTLf = e−sT F (s)

Again, there is a dual result:

Leαtg(t)

=

∫ ∞

−∞

eαtg(t)e−st dt =

∫ ∞

−∞

g(t)e−(s−α)t dt = G(s − α).

For example:

L

1(t) · eαt cos ωt

= L1(t) · cosωt|s:=s−α =s − α

(s − α)2 + ω2

The Laplace Transform turns multiplication by anexponential function in one domain into translation in

the other.



Multiplication by t:This is another dual result, straightforward to establish.Given any integer n and for all s in the ROC of Lf:

Ltnf(t) = (−1)ndn

dsnLf = (−1)n

dn

dsnF (s).

For example:

L1(t) · t sinωt = (−1)d

ds

(ω

s2 + ω2

)

=2 ωs

(s2 + ω2)2; Reals > 0.

The Laplace Transform turns differentiation of order nin one domain into multiplication by the domain

variable (s or t) to the power n in the other domain.



Initial Value Theorem:We have seen that:

sL1(t) · f(t) − f(0−) = L

1(t) · d

dtf(t)

=

∫ ∞

0−

[d

dtf(t)

]

e−st dt

Taking the limit as Reals → ∞ on both sides:

limReals→∞

sL1(t) · f(t)−f(0−) = limReals→∞

∫ ∞

0−

[d

dtf(t)

]

e−st dt = 0

Therefore, assuming f(t) differentiable on [0,∞), andprovided that ROC of L1(t)f(t) is a half planeReal s > α for some α ∈ R:

limReals→∞

sL1(t) · f(t) = f(0−)



Final Value Theorem:Again starting from:

sL1(t) · f(t) − f(0−) =

∞

0−

d

dtf(t) e

−st dt

But taking the limit as s → 0:

lims→0

sF (s) − f(0−) = lims→0

∞

−∞

1(t) ·d

dtf(t)e−st dt =

∞

0−

d

dtf(t)

1

lims→0e−st

dt

= ∞

0−

d

dtf(t) dt = f(∞) − f(0−).

Therefore, provided that L1(t) · f(t) , F (s) isdefined for every s in a region around 0:

lims→0

sF (s) = f(∞)


Inversion of Laplace Transforms

We need a tool in order to recover a signal f(t) from itsLaplace Transform F (s) = Lf.

The Inverse Laplace Transform is given as:

f(t) = L−1 F ,1

2πj

∫ γ+j∞

γ−j∞F (s)est ds

where γ ∈ R is such that all s ∈ (γ − jω, γ + jω) are inthe Region of Convergence of F (s).

The Inverse Laplace Transform is a linear operation:

L−1 αF + βG = αL−1 F + βL−1 G

for any complex numbers α, β ∈ C


Contour Integral Formulation

The Inverse Laplace Transform may appear somewhatcomplicated.

In practice, it can be evaluated converting it to acontour integral, which depends just on the residues ofthe integrand.

First, we note that for most Laplace transforms F (s)

lim|s|→∞

|F (s)| = 0 , for example lim|s|→∞

Leat

= lim

|s|→∞

∣∣∣∣

1

s − a

∣∣∣∣= 0

.Then: ∫

ΓF (s)est ds = 0 ; t ≥ 0

where Γ is an infinite radius arc in the complex plane(see next slide).



0

Extension ΓLine Λ = (γ − j∞, γ + j∞)

Imaginary

γ Real

Function F (s)est

Area = f(t) =1

2πj

γ+j∞

γ−j∞

F (s)est ds

Infinite Radius

Defining the contour C = Λ + Γ and for t ≥ 0:

f(t) = L−1 F (s) =1

2πj

∫

Λ

F (s)est ds+1

2πj

∫

Γ

F (s)est ds

︸︷︷︸

0

=1

2πj

∮

C

F (s)est ds



0

Extension ΓLine Λ = (γ − j∞, γ + j∞)

Imaginary

γ Real

Function F (s)est

Area = f(t) =1

2πj

γ+j∞

γ−j∞

F (s)est ds

Infinite Radius

Provided that |F (s)| → 0 as |s| → ∞, then the InverseLaplace Transform of F (s) may be written as:

f(t) = L−1 F =1

2πj

∮

CF (s)est ds ; t ≥ 0


Evaluation by Residue Calculation

We will apply Cauchy’s Residue Theorem to computethe inverse Laplace transform:

L−1 F =1

2πj

∮

CF (s)est ds ; t ≥ 0

Firstly,

A pole pk of F (s) is any value such that F (pk) = ±∞.(e.g., F (s) = 1

s−ahas a pole at s = p1 = a)

The residue of F (s) at the pole s = pk is

Ress=pk

F (s) = [(s − pk)F (s)]|s=pk

(

e.g. Ress=a

1s−a

= 1)



(continued)A pole pk of F (s) has multiplicity m if:

F (pk) = ±∞ and

∣∣∣∣lim

s→pk

(s − pk)mF (s)

∣∣∣∣< ∞

Then the residue of F (s) at the pole s = pk is:

Ress=pk

F (s) =1

(m − 1)!

dm−1

dsm−1[(s − pk)mF (s)]

∣∣∣∣s=pk

The Cauchy’s Residue Theorem establishes:

1

2πj

∮

C

G(s) ds =n∑

k=1

Ress=pk

G(s).

where C is a smooth closed curve in C, and G(s) isdifferentiable within C except at isolated poles p1, · · · , pn.



Applying this theorem with G(s) = F (s)est, provides asimple formula for the Inverse Laplace Transform:

Suppose that the poles of F (s) are at points p1, · · · , pn

and that F (s) → 0 as |s| → ∞. Then, for t ≥ 0

f(t) = L−1 F =1

2πj

∮

CF (s)est ds =

n∑

k=1

Ress=pk

[F (s)est

]

where the integration is taken counter-clockwisearound the path C.

Note that the poles of F (s)est are the poles of F (s).



Isolated Poles:If all the poles p1, p2, · · · , pn of F (s) have multiplicitym = 1 each, and lim|s|→∞ |F (s)| = 0, then:

L−1F (s) =n∑

k=1

(s − pk)F (s)∣∣∣s=pk

epkt

Example 1. Suppose that: F (s) =1

s + aThen F (s)est has only one pole at p1 = −a.Provided t ≥ 0 then F (s)est is zero on the infinite arc Γ.Therefore, for t ≥ 0:

f(t) = L−1F (s) = (s + a)F (s)|s=−a e−at = e−at ; t ≥ 0

which agrees with the previously known result.



Isolated Poles.Example 2. Suppose that:

F (s) =1

τs + 1

Then the function F (s)est has only one pole at p1 = −1/τ , and|F (s)| → 0 as |s| → ∞. Therefore, for t ≥ 0:

f(t) = L−1F (s) = (s + 1/τ)1

τs + 1

∣∣∣∣s=−1/τ

e−t/τ

=1

τ

(s + 1/τ)

(s + 1/τ)

∣∣∣∣s=−1/τ

e−t/τ

=1

τe−t/τ ; t ≥ 0




F (s) =1

s2 + 3s + 2

Clearly, F (s)est has two poles, p1 = −1 and p2 = −2.Provided t > 0, then (F (s)est) → 0 on the infinite arc Γ.Therefore:

f(t) = L−1F (s) = (s + 1)F (s)|s=−1 e−t + (s + 2)F (s)|s=−2 e−2t

=1

(s + 2)

∣∣∣∣s=−1

e−t +1

(s + 1)F (s)

∣∣∣∣s=−2

e−2t

= e−t − e−2t ; t ≥ 0




F (s) =ω

s2 + ω2=

ω

(s − jω)(s + jω)

Then F (s)est now has two purely complex poles at p1,2 = ±jω.Γ needs to be chosen so that the line Λ passes to the right ofthe origin so that C = Λ + Γ will enclose the poles.Therefore, since F (s)est is zero on Γ if t ≥ 0:

f(t) = L−1F (s) = (s − jω)F (s)|s=jω ejωt + (s + jω)F (s)|s=−jω e−jωt

=ω

(s + jω)

∣∣∣∣s=jω

ejωt +ω

(s − jω)F (s)

∣∣∣∣s=−jω

e−jωt

=1

2j

[ejωt − e−jωt

]= sinωt ; t ≥ 0



Isolated Poles.Example 5. To really appreciate the power of this residuemethod of Laplace Transform inversion, suppose that:

F (s) =s

s2 + 2αrs + r2; α, r ∈ R.

To find the poles of F (s) we factor the denominator as:

F (s) =s

(s + rejθ)(s + re−jθ)

θ , cos−1 α

sin θ =√

1 − α2

There are clearly two poles:

p1 = −rejθ = −r(cos θ + j sin θ) = −rα − jr√

1 − α2

p2 = −re−jθ = −r(cos θ − j sin θ) = −rα + jr√

1 − α2 = p1



Isolated Poles.Example 5. (continued)Applying the residue formulation for t ≥ 0 and with p1 = −rejθ :

f(t) = (s + rejθ)F (s)∣∣s=p1

e−p1t + (s + re−jθ)F (s)∣∣s=p1

e−p1t

=rejθ

r(ejθ − e−jθ)ep1t +

re−jθ

r(e−jθ − ejθ)ep1t

=1

2j sin θ

ejθep1t − e−jθep1t

=e−rαt

2j sin θ

ej(θ−rt sin θ) − e−j(θ−rt sin θ)

=e−rαt

2j sin θ2j sin(θ − rt sin θ)

=1√

1 − α2e−rαt sin

(

cos−1 α − rt√

1 − α2)

; t ≥ 0



Repeated Poles:As already stated, to compute the residue at a pole pk

of multiplicity m we have to:1. remove the poles via multiplication by (s − p)m,2. differentiate to the (m − 1)’th order, and3. evaluate the derivative at s = p and divide by

(m − 1)!

Example: F (s) =1

(s + a)2

F (s)est has a pole at s = −a of multiplicity m = 2. So, for t ≥ 0:

f(t) = Ress=−a

est

(s + a)2=

1

1!

d

ds

(s + a)2est

(s + a)2

∣∣∣∣s=−a

=d

dsest

∣∣∣∣s=−a

= test∣∣s=−a

= te−at


Dealing with the Biproper Case

Note that if F (s) =bnsn + bn−1s

n−1 + · · · + b1s + b0

sn + an−1sn−1 + · · · + a1s + a0,

with bn 6= 0, then lim|s|→∞

|F (s)| = |bn| 6= 0.

In this case, F (s) does not vanish on the infinite arc Γ.

This can be easily dealt with re-writing F (s) as:

F (s) = bn + G(s) where lim|s|→∞

|G(s)| = 0

ThenL−1 F (s) = L−1 bn + G(s) = bnδ(t) + L−1 G(s) .

where L−1 G(s) can be obtained via residues.


Dealing with the Biproper Case

Example. Consider the case of:

F (s) =2s2 + 5

s2 + 6s + 8.

Clearly lim|s|→∞ |F (s)| = 2 6= 0 since F (s) is biproper.However we can re-write F (s) as:

F (s) = 2 +(0 − 2 · 6)s + (5 − 2 · 8)

s2 + 6s + 8= 2 − (12s + 11)

(s + 4)(s + 2)

Noting that L−1 2 = 2δ(t), and using residue methods, gives:

f(t) = 2δ(t) +

[

− (12s + 11)

(s + 2)

]∣∣∣∣s=−4

e−4t +

[

− (12s + 11)

(s + 4)

]∣∣∣∣s=−4

e−2t

= 2δ(t) − 37

2e−4t +

13

2e−2t


Causality of the Inverse Laplace Transform

Recall that:

f(t) = L−1 F ,1

2πj

∫ γ+j∞

γ−j∞

F (s)est ds

where γ ∈ R is such that s ∈ (γ − jω, γ + jω) is inside theROC of F (s).

ROC of Laplace Transforms considered so far are ofthe form:

ROC F (s) = s ∈ C : Real s > Real α

where α is such that the poles of F (s) lie to the left ofReal α.



Then, for t ≥ 0:∫

Λ

F (s)est ds =

∫

Λ

F (s)est ds

+

∫

Γ

F (s)est ds

=

∮

C

F (s)est ds

Using Cauchy’s Theorem:

f(t) = L−1 F (s) =n∑

k=1

[

Ress=pk

F (s)est

]

0

Ims

γ

Res

Contour C

Poles of F (s)

Line Λ

Arc Γ



If t < 0:

The infinite arc Γmust be changed(else, est → ∞)

Therefore,F (s) contains nopoles within C.

Hence,

f(t) =1

2πj

∮

C

F (s)est ds = 0

Contour C

0 Resγ

Ims

Arc ΓLine Λ

Poles of F(s)



Therefore, the general causality principle for LaplaceTransforms is:

(Causal signals) Provided |F (s)| → 0 as |s| → ∞,and that the ROC of F (s) has the forms ∈ C : Real s > α,α ∈ R. Then:

L−1 F (s) = f(t) = 0 for t < 0.

(Anti-causal signals) Provided |G(s)| → 0 as|s| → ∞, and that the ROC of G(s) has the forms ∈ C : Real s < α,α ∈ R. Then:

L−1 G(s) = g(t) = 0 for t > 0



For example,Consider the anti-causal signal (α > 0):

g(t) =

eαt ; t ≤ 0

0 ; t > 0⇒ G(s) =

∫ 0

−∞

eαte−st d =−1

s − α

and ROC G(s) = s ∈ C : Real s < Real α.However, the causal signal:

f(t) =

0 ; t ≤ 0

−eαt ; t > 0⇒ F (s) = −

∫ ∞

0

eαte−st d =−1

s − α

and ROC F (s) = s ∈ C : Real s > Real α.


Dealing with Time Translations

According to the time-shifting property:

Lf(t + T ) = esTLf(t) = esT F (s).

This transforms do not satisfy |esT F (s)| → 0 as|s| → ∞, thus we have to

Initially remove any esT term,Obtain the Inverse Laplace tranform, and thenApply the necessary time translation.

Example:

Consider F (s) =e−2s

s − α. We have that L−1

1

s − α

= 1(t) · eαt,

and the e−2s term corresponds to a time delay, hence:

L−1

e−2s

s − α

= 1(t − 2) · eα(t−2).


Inversion by Partial Fraction Expansion

Given a transfrom F (s) to invert, the idea is todecompose F (s) into known transforms.

Example. Consider:

F (s) =1

(s + a)(s + b)

We recognise that it contains elements related to the knownrelationship:

L

1(t)e−αt

=1

s + α.

This raises the idea to try to decompose it in the form:

F (s) =A

s + a+

B

s + b=

A(s + b) + B(s + a)

(s + a)(s + b)

for some constants A and B.



Example (continued)

F (s) =1

(s + a)(s + b)=

A(s + b) + B(s + a)

(s + a)(s + b)

Then we see that numerators have to be the same, for all s:

A(s + b) + B(s + a) = 1.

In particular, for s = −b: B(a − b) = 1

and also, for s = −a: A(b − a) = 1

Therefore:

F (s) =1

(b − a)

[1

s + a− 1

s + b

]

⇐⇒ f(t) =1

(b − a)

[e−at − e−bt

]

Which is the same result obtained via residues:

f(t) =1

(s + b)

∣∣∣∣s=−a

e−at +1

(s + a)

∣∣∣∣s=−b

e−bt =1

(b − a)

[e−at − e−bt

]



Partial Fractions: Isolated PolesConsider the general case F (s) of the form:

F (s) =bmsm + · · · + b1s + b0

sn + an−1sn−1 + · · · + a1s + a0=

B(s)

(s − p1)(s − p2) · · · (s − pn)

In this case, the partial fraction decomposition is:

F (s) =B(s)

(s − p1)(s − p2) · · · (s − pn)=

K1

s − p1+

K2

s − p2+ · · ·+ Kn

s − pn

Then, the inverse transform is:

f(t) = 1(t) ·[K1e

p1t + K2ep2t + · · · + Knepnt

]

Note that, each Ki can be computed as:

Ki = (s − pi)F (s)|s=pi = Ress=pi

F (s).

Which is the same inverse obtained via Cauchy’s Theorem.



Partial Fractions: Repeated PolesConsider a more general F (s) of the form:

F (s) =B(s)

(s − p1)k1(s − p2)k2 · · · (s − pn)kn.

It can be expanded in a partial fraction decomposition as:

F (s) =

[

K11

(s − p1)+

K21

(s − p1)2+ · · · + Kk1

1

(s − p1)k1

]

+

. . . +

[K1

n

(s − pn)+

K2n

(s − pn)2+ · · · + Kkn

n

(s − pn)kn

]

=n∑

r=1

kr∑

`=1

K`r

(s − pr)`

But, we have that L

1(t) · tneαt

=n!

(s − α)n+1.



Partial Fractions: Repeated PolesThen:

f(t) = L−1 F (s) = L−1

n∑

r=1

kr∑

`=1

K`r

(s − pr)`

=n∑

r=1

kr∑

`=1

K`r

1

`!t`−1eprt

And each K`r can be obtained from the partial fraction

decomposition. The final inverse Laplace transform is:

f(t) =n∑

r=1

kr∑

`=1

dkr−`

dskr−`

[(s − pr)

krF (s)]∣∣∣∣s=pr

1

`!t`−1eprt , t ≥ 0

which confirms, again, the solution via residues:

f(t) =1

2πj

∮

C

F (s)est ds =n∑

r=1

Ress=pr

[F (s)est

], t ≥ 0



Partial Fractions: Repeated PolesExample. Consider the case of:

F (s) =1

(s + 1)(s − 2)2

Then an appropriate partial fraction decomposition is:

1

(s + 1)(s − 2)2=

K1

s + 1+

K12

(s − 2)+

K22

(s − 2)2

Multiplying both sides by (s + 1) then implies that:

1

(s − 2)2= K1 +

K12 (s + 1)

(s − 2)+

K22 (s + 1)

(s − 2)2

and hence setting s = −1 gives:

K1 =1

(−1 − 2)2=

1

9



Partial Fractions: Repeated PolesExample (continued)Starting from:

F (s) =1

(s + 1)(s − 2)2=

K1

s + 1+

K12

(s − 2)+

K22

(s − 2)2

and multiplying both sides by (s − 2)2 also implies that:

1

(s + 1)=

K1(s − 2)2

s + 1+ K1

2 (s − 2) + K22

which when evaluated at s = 2 gives:

K22 =

1

(2 + 1)=

1

3



Partial Fractions: Repeated PolesExample (continued)Finally, also from:

F (s) =1

(s + 1)(s − 2)2=

K1

s + 1+

K12

(s − 2)+

K22

(s − 2)2

differentiating both sides with respect to s, implies that:

−1

(s + 1)2=

2K1(s − 2)(s + 1) − K1(s − 2)2

(s + 1)2+ K1

2

which evaluated at s = 2 implies:

K12 =

−1

(2 + 1)2= −1

9.

Therefore:

F (s) =1

9· 1

s + 1− 1

9· 1

(s − 2)+

1

6· 2

(s − 2)2

⇐⇒ f(t) = 1(t) ·[1

9e−t − 1

9· e2t +

1

6· te2t

]



Given the equivalence, the residue-based methodappears to be superior to partial fractiondecomposition:

It does not require memorization of any LaplaceTransform results;It can be used even when there are some elementsnot recognisable from available transform tables;It directly exposes why a pole at p in a transformF (s) implies a term ept;The calculations involved computing residues areidentical to those in computing partial fractionexpansion coefficients.


Differential Equation Solution via Laplace

We have seen that, for the general differential equation:

dn

dtny(t) + an−1

dn−1

dtn−1y(t) + · · · + a0y(t) = bn

dn

dtnu(t) + · · · + b0u(t)

⇒ y(t) = CeA(t−t0)x0 +

∫ t

t0

CeA(t−σ)Bu(σ) dσ + Du(t)

where x0 contains the initial conditions, u(t) is a givensignal and A,B,C, D are state-space matrices.

The main difficulty here is to obtain eA(t−t0).

We will show a simpler method to obtain y(t), usingLaplace Transform.

The key idea is to solve for Y (s) = Ly, and thencompute the solution as y(t) = L−1 Y .



Taking the Laplace Transform of both sides:

L

n∑

k=0

akdk

dtk1(t) · y(t)

= L

n∑

k=0

bkdk

dtk1(t) · u(t)

where 1(t) is included to only consider y(t) and u(t) fort ≥ 0. Then we have:

Y (s)

n∑

k=0

aksk −n∑

k=0

ak

k∑

`=1

sk−` d`−1

dt`−1y(t)

∣∣∣∣t=0−

=

U(s)

n∑

k=0

bksk −n∑

k=0

bk

k∑

`=1

sk−` d`−1

dt`−1u(t)

∣∣∣∣t=0−

where A(s) ,

n∑

k=0

aksk and B(s) ,

n∑

k=0

bksk are polynomials in s.



Then, solving for Y (s):

Y (s) = H(s)U(s)︸︷︷︸

ForcedResponse

+1

A(s)

n∑

k=0

k∑

`=1

sk−`

[

akd`−1

dt`−1y(t)

∣∣∣∣t=0−

− bkd`−1

dt`−1u(t)

∣∣∣∣t=0−

]

︸︷︷︸

Natural Response

where:H(s) ,

B(s)

A(s)

If all the initial conditions are zero, then:

Y (s) = H(s)U(s) ⇐⇒ y(t) = L−1 H(s)U(s)

which is only a forced response component.



Example: First Order Case

d

dty(t) + a0y(t) = b0u(t) ; y(0−) = y0.

Taking Laplace transform of both sides:

sY (s) − y(0−) + a0Y (s) = b0U(s).

so that: Y (s) =

(b0

s + a0

)

U(s)

︸︷︷︸

Forced Response

+1

(s + a0)y(0−)

︸︷︷︸

Natural ResponseIf u(t) = 1(t) ⇐⇒ U(s) = 1/s, then:

Y (s) =b0

(s + a0)· 1

s+

1

(s + a0)y(0−)

⇒ y(t) =

[b0

a0

[1 − e−a0t

]+ y(0−)e−a0t

]

1(t)



Example: Second Order CaseTo compute the evolution of vC(t) after the switch is closed, adifferential equation model for vC(t) is required.

+

+t = 0

vR(t) vL(t)

R = 50Ω L = 10H

vC(t)

C = 25000µFV?

By Kirchoff’s Voltage Law

vR(t) + vL(t) + vC(t) = V?

By Kirchoff’s Current Law:

iR(t) = iL(t) = iC(t)

And for the resistor, the inductorand the capacitor, we have that:

vR(t) = R iR(t)

vL(t) = Ld

dtiL(t)

iC(t) = Cd

dtvC(t)



Example: Second Order Case (continued)Therefore, making some substitutions:

vR(t) = R iR(t) = R iC(t) = RCd

dtvC(t)

vL(t) = Ld

dtiR(t) = L

d

dtiC(t) = LC

d2

dt2vC(t)

Inserting this into the KVL equation:

RCd

dtvC(t) + LC

d2

dt2vC(t) + vC(t) = V?

⇐⇒ d2

dt2vC(t) +

R

L

d

dtvC(t) +

1

LCvC(t) =

1

LCV?

Taking Laplace Transforms of both sides, LvC(t) = VC(s), gives:

s2VC (s) − svC(0−) −

d

dtvC(0−) +

R

L sVC(s) − vC (0−) +1

LCVC (s) =

1

LC

V?

s



Example: Second Order Case (continued)

s2VC (s) − svC(0−) −

d

dtvC(0−) +

R

L sVC(s) − vC (0−) +1

LCVC (s) =

1

LC

V?

s

Here, we have modelled the closing of the switch at t = 0 as thevoltage source as V?1(t), whose Laplace Transform is then V?/s.Collecting terms:

VC(s)

[

s2 +R

Ls +

1

LC

]

− vC(0−)

[

s +R

L

]

− d

dtvC(0−) =

1

LC

V?

s.

Now replacing the component values C = 25000 µF , L = 10 H andR = 50 Ω we get:

VC(s) =4V?

s(s2 + 5s + 4)︸︷︷︸

Forced Response

+(s + 5)vC(0−) + dvC(0−)dt

(s2 + 5s + 4)︸︷︷︸

Natural Response



Example: Second Order Case (continued)For the Forced Response component, the inverse LaplaceTransform may be computed via residues:

L−1

4V?

(s2 + 5s + 4) = L−1

4V?

s(s + 4)(s + 1)

= 4V?

1

(s + 4)(s + 1)

s=0

+1

s(s + 1)

s=−4

e−4t +1

s(s + 4)

s=−1

e−t+ 1(t)

= V? 1 +1

3e−4t −

4

3e−t

1(t)

Similarly, for the natural response component:

L−1

(s + 5)vC(0−) + dvC (0−)dt

(s2 + 5s + 4) = L−1

(s + 5)vC(0−) + dvC (0−)dt

(s + 4)(s + 1)

=(s + 5)vC (0−) + dvC(0−)dt

(s + 1)

s=−4

e−4t +(s + 5)vC(0−) + dvC (0−)dt

(s + 4)

s=−1

e−t

=1

3 vC (0−)[4e−t − e−4t] +dvC (0−)

dt[e−t − e−4t]



Example: Second Order Case (continued)Therefore, the complete solution for t ≥ 0 is:

vC(t) = V? 1 +1

3e−4t −

4

3e−t

+1

3 vC(0−)[4e−t − e−4t] +dvC (0−)

dt[e−t − e−4t]

Notice that if we set t = 0 in this solution we obtain

vC(0) = vC(0−)

hence the capacitor voltage when the switch is thrown matches thevoltage just before it was thrown. Similarly, we can obtain:

d

dtvC(t) = V? −

4

3e−4t +

4

3e−t

+1

3 vC (0−)[−4e−t + 4e−4t] +dvC (0−)

dt[−e−t + 4e−4t]

and hence:d

dtvC(t)

∣∣∣∣t=0

=dvC(0−)

dt.

Therefore, the solution satisfies any initial conditions on thecapacitor voltage prior to the switch being thrown.


Differential Equation Solution via Laplace: Initial Conditions

We have seen that for 1(t) ,

0 ; t < 0

1 ; t ≥ 0

L

1(t) · d

dtf(t)

= sL1(t) · f(t) +

∫ ∞

0−

d

dt[f(t)e−st] dt

= sL1(t) · f(t) − f(0−)

where t = 0−, a point infinitessimally smaller than (tothe left of) the time t = 0.

If we define a new function 1+(t) ,

0 ; t ≤ 0

1 ; t > 0

then: L

1+(t) · d

dtf(t)

= sL1(t) · f(t) − f(0+)

Therefore, Laplace Transform can also handle initialcondition at t = 0+, if given.



Note that f(0+) − f(0−) =

∫ 0+

0−

d

dt[f(t)e−st] dt.

Then, if ddt

f(t) is bounded in t ∈ (0−, 0+) (e.g. f(t) 6= 1(t)),then the integral is zero, and f(0+) = f(0−) = f(0).

In general: If dk

dtkf(t) ; ∀k = 0, . . . , n exist on (0,∞)

then for all s in ROC F (s) and Real s > 0:

L

1+(t) · dn

dtnf(t)

= sn L1+(t) · f(t) −n∑

k=1

sn−k dk−1

dtk−1f(0+)

Furthermore, if the derivatives are also bounded, then:

L

1+(t) · dn

dtnf(t)

= L

1(t) · dn

dtnf(t)

= sn L1(t) · f(t) −n∑

k=1

sn−k dk−1

dtk−1f(0−)



Example: Schock absorber revisited.Assuming that the car mass, shock absorber spring stiffness anddamper viscosity are:

m = 1000 kg, ks = 2000 N/m, kd = 3000 Ns/m

Then, the relationship between car height y(t) and road height u(t)

is given as:d2

dt2y(t) + 3

d

dty(t) + 2y(t) = 3

d

dtu(t) + 2u(t)

Suppose further initial conditions: y(0−) = −0.5 ,d

dty(0−) = −1

and a unit step input u(t) = 1(t), so that: u(0−) = 0 ,d

dtu(0−) = 0

Taking Laplace Transforms of both sides:

s2Y (2) − sy(0−) −d

dty(0−) + 3sY (s) − 3y(0−) + 2Y (s) = 3sU(s) − 3u(0−) + 2U(s)

⇒ (s2 + 3s + 2)Y (s) − (s + 3)y(0−) −d

dt(0−) = (3s + 2)U(s) − 3u(0−)



Example: Schock absorber revisited (continued)The latter may be rewritten as:

Y (s) =(3s + 2)U(s) − 3u(0−)

s2 + 3s + 2+

(s + 3)y(0−) + dy(0−)/dt

s2 + 3s + 2

Which, replacing the initial conditions, becomes:

Y (s) =(3s + 2)U(s) − 0

(s + 2)(s + 1)︸︷︷︸

Forced Response

+−0.5s − 2.5

(s + 2)(s + 1)︸︷︷︸

Natural Response

The natural response yn(t) can be found via residues:

yn(t) = L−1

−0.5s − 2.5

(s + 2)(s + 1)

=−0.5s − 2.5

(s + 2)

s=−1

e−t +−0.5s − 2.5

(s + 1)

s=−2

e−2t

=

−0.5(−2) − 2.5

(−1) e−2t +

−0.5(−1) − 2.5

1 e−t

= 1.5e−2t − 2e−t



Example: Schock absorber revisited (continued)Similarly, replacing Lu(t) = L1(t) = U(s) = 1/s, the forcedresponse yf (t) is:

yf (t) = L−1

3s + 2

s(s + 1)(s + 2)

=3s + 2

(s + 1)(s + 2)

∣∣∣∣s=0

+3s + 2

s(s + 2)

∣∣∣∣s=−1

e−t +3s + 2

s(s + 1)

∣∣∣∣s=−2

e−2t

=2

(1)(2)+

(−1)

(−1)(1)e−t +

(−4)

(−2)(−1)e−2t

= 1 + e−t − 2e−2t



Example: Schock absorber revisited (continued)Therefore, the response of car height y(t) to a unit step bump in theroad is:

y(t) = 1 − e−t − 2e−2t

︸︷︷︸

yf (t)

+ 1.5e−2t − 2e−t

︸︷︷︸

yn(t)

= 1 − 3e−t − 0.5e−2t

0 1 2 3 4 5 6−1

−0.5

0

0.5

1

1.5

Time (s)

Car

hei

ght (

m)

Forced, natural and total car height response

←Natural Response

←Forced Response

←Total Response



Example: Schock absorber revisited (continued)Therefore, the response of car height y(t) to a unit step bump in theroad is:

y(t) = 1 − e−t − 2e−2t

︸︷︷︸

yf (t)

+ 1.5e−2t − 2e−t

︸︷︷︸

yn(t)

= 1 − 3e−t − 0.5e−2t

Notice that the initial condition specification:

y(0−) = −0.5,d

dty(0−) = −1

is also met by the natural response yn(t) = 1.5e−2t − 2e−t att = 0+, but not by the total response y(t) = yf (t) + yn(t).

This is because the model used involves the input derivativewhich, for the unit step u(t) = 1(t), is discontinuous at t = 0.

It is possible to interpret the derivative of 1(t) as a Dirac deltaδ(t), which allows instantaneous changes in y(t) and dy(t)/dt

at t = 0−.Chapter 4. Laplace Transform Analysis – p.76/95


Example: Schock absorber revisited (continued)To underline the latter point, consider the case of the normal (atrest) car height being y(t) = 1m, and up until time t = 0 it isdepressed to y(t) = 0m, being released at t = 0. The appropriateinitial conditions are:

y(0+) = 0,d

dty(0+) = 0.

The differential equation model is the same and the input will still beu(t) = 1(t).The only difference is that the initial conditions are now specified att = 0+ rather than at t = 0−.Therefore, we obtain:

Y (s) =(3s + 2)U(s) − 3u(0+)

s2 + 3s + 2+

(s + 3)y(0+) + dy(0+)/dt

s2 + 3s + 2



Example: Schock absorber revisited (continued)Replacing the initial conditions, and u(0+) = 1 (u(t) = 1(t)):

Y (s) =(3s + 2)1/s − 3

s2 + 3s + 2=

2

s(s + 1)(s + 2)

Therefore:

y(t) = L−1

2

s(s + 1)(s + 2)

=2

(s + 1)(s + 2)

∣∣∣∣s=0

+2

s(s + 2)

∣∣∣∣s=−1

e−t +2

s(s + 1)

∣∣∣∣s=−2

e−2t

=2

(1)(2)+

2

(−1)(1)e−t +

2

(−2)(−1)e−2t

= 1 − 2e−t + e−2t



Example: Schock absorber revisited (continued)Now, notice that the solution

y(t) = 1 − 2e−t + e−2t

at t = 0+, does satisfy the initial condition also specified at t = 0+.

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Time (s)

Car

hei

ght (

m)

Car height response


The System Transfer Function

Considering zero initial conditions, we have seen that:

Ly = Y (s) = H(s)U(s)

where H(s) =B(s)

A(s)=

b0 + b1s + . . . + bnsn

a0 + a1s + . . . + sn

is called the transfer function of the system. If thepolynomials A(s) and B(s) are factorized, then:

H(s) =K(s − z1)(s − z2) · · · (s − zm)

(s − p1)(s − p2) · · · (s − pn).

zkk=1,...,m are called the zeros of H(s) (∈ C).

pkk=1,...,n are called the poles of H(s) (∈ C).K is a constant


The System Transfer Function and State Space Models

We recall that a system can be expressed instate-space form:

d

dtx(t) = Ax(t) + Bu(t)

y(t) = Cx(t) + Du(t).

Taking Laplace Transforms, assuming x(0) = 0:

sX(s) = AX(s) + BU(s)

Y (s) = CX(s) + DU(s).

ThenY (s) = C [sI− A]−1

BU(s) + DU(s)

=(C [sI − A]−1

B + D)

︸︷︷︸

H(s)

U(s)


The System Transfer Function and State Space Models

Example: Consider the 2nd order differential equation

d2

dt2y(t) + a1

d

dty(t) + a0y(t) = b1

d

dtu(t) + b0u(t).

Clearly H(s) =b1s + b0

s2 + a1s + a0, and the state-space matrices are:

A =

−a1 1

−a0 0

, B =

b1

b0

, C =[

1 0]

, D = 0

Then, (sI−A)−1

=

s + a1 −1

a0 s

−1

=1

s(s + a1) + a0

s 1

−a0 s + a1

Therefore, using H(s) = C [sI−A]−1

B + D:

H(s) =

(1

s2 + a1s + a0

)[

1 0]

s 1

−a0 s + a1

b1

b0

=b1s + b0

s2 + a1s + a0



The latter example raises a connection between theeigenvalues of A and the poles of H(s).

The point is that [sI − A]−1 contains terms, all of whichhave det(sI −A) as a dividing factor.

Therefore, since

H(s) = C [sI− A]−1B + D

then the poles of H(s) satisfy

det(sI −A) = det(λI−A) = 0

But, these values of s are equal to the eigenvaluesλk of the matrix A.



Then, for a system with state space representation:

d


y(t) = Cx(t) + Du(t)

and hence transfer function representation

H(s) = C [sI− A]−1B + D

=K(s − z1)(s − z2) · · · (s − zm)

(s − p1)(s − p2) · · · (s − pn)

the poles p1, · · · , pn of H(s) are the eigenvaluesλ1, · · · , λn of A.


Stability and the Transfer FunctionRecall that a system described by:

dn

dtny(t) + . . . + a0y(t) = bm

dm

dtmu(t) + . . . + b0u(t).

or (equivalently) by the state space model:d


y(t) = Cx(t) + Du(t).

is asymptotically stable ⇐⇒ the eigenvaluesλ1, . . . , λn of A satisfy Real λk < 0.

Then, using the transfer function representation:

H(s) =b0 + b1s + . . . + bnsn

a0 + a1s + . . . + ansn=

K(s − z1)(s − z2) . . . (s − zm)

(s − p1)(s − p2) . . . (s − pn).

the system is asymptotically stable ⇐⇒ the polesp1, . . . , pn of H(s) satisfy Real pk < 0.


Transfer Function and Impulse Response

Recall that the relationship between y(t) and u(t), can bedescribed through the convolution with the impulseresponse h(t):

y(t) = [h ~ u](t) =

∫ ∞

−∞h(σ)u(t − σ) dσ

⇒ Y (s) = Ly = Lh ~ u = H(s)U(s)

The transfer function H(s) of a system is the LaplaceTransform of its impulse response h(t).

Example:Consider the system

d2

dt2y(t) + 3

d

dty(t) + 2y(t) = 3

d

dtu(t) + 2u(t)

The associated transfer function is H(s) = 3s+2s2+3s+2 = 3s+2

(s+1)(s+2)

Therefore, the impulse response is h(t) = L−1 H = 4e−t − e−t


Laplace Transforms in Electrical Circuits

We have seen that Laplace Transform methods areuseful to compute the response of linear systemsdescribed by a differential equation.

However, it is possible to directly derive transferfunction descriptions in s, without an intermediatedifferential equation modelling step.

For example, for the fundamental electrical circuitcomponents:

(Resistor) vR(t) = RiR(t) ⇒ VR(s) = RIR(s)

(Inductor) vL(t) = Ld

dtiL(t) ⇒ VL(s) = sLIL(s) − LiL(0−)

(Capacitor) iC(t) = Cd

dtvC(t) ⇒ VC(s) =

1

sCIC(s) +

1

svC(0−)



When any initial energy storage in the elements isconsidered to be zero (i.e., iL(0−) = vC(0−) = 0), thenwe have the general relationship:

V (s) = ZI(s)

where:(Resistor) VR(s) = RIR(s) ⇒ Z = RI

(Inductor) VL(s) = sLIL(s) ⇒ Z = sL

(Capacitor) VC(s) =1

sCIC(s) ⇒ Z =

1

sC

That is, in the Laplace Transform s-domain, allelements are represented as a generalised resistance,called impedance and denoted by Z.



+

+

+

+

+ +

+

+

i(t)

i(t)

i(t)

C

R

L

Z =1

sC

Z = R

Z = sLLiL(0)

I(s)

I(s)

I(s)

v(t)

v(t)

v(t)

1

svC(0)

V (s) = RI(s)

V (s) =1

sCI(s) +

1

svc(0

−)

V (s) = sL I(s) − Lil(0−)



Kirchoff’s Laws:These laws are inherited in the s-domain, due to thelinearity of the Laplace transform:

(KCL) At any circuit node:∑

k

ik(t) = 0 ⇐⇒ L

∑

k

ik(t)

=∑

k

Lik(t)

=∑

k

Ik(s) = 0

And similarly, (KVL) around any circuit mesh:∑

k

vk(t) = 0 ⇐⇒∑

k

Vk(s) = 0



Series Combination:

+

+

+ +

Z1

Z2

I(s)

V (s) V (s)

I(s)

Zeq = Z1 + Z2

V1(s)

V2(s)

By Kirchoff’s voltage law

V (s) = V1(s) + V2(s) = Z1I(s) + Z2I(s) = ZeqI(s)

where, as indicated Zeq , Z1 + Z2.



Parallel Combination:

+ +

Z2Z1

I(s)

V (s) V (s)

I1(s) I2(s)

Zeq =Z1Z2

Z1 + Z2

I(s)

By Kirchoff’s current law:

I(s) = I1(s) + I2(s) =V (s)

Z1+

V (s)

Z2= V (s)

[1

Z1+

1

Z2

]

and therefore1

Zeq=

1

Z1+

1

Z2⇐⇒ Zeq =

Z1Z2

Z1 + Z2= Z1‖Z2



Voltage Division:

+

+

I(s)

Z1

Z2

V (s)

V2(s)

By the result on seriesimpedances:

I(s) =V (s)

Z1 + Z2.

However, V2(s) = Z2I(s)so that

V2(s) =

[Z2

Z1 + Z2

]

V (s).

This is known as voltagedivision.



Example: 1st Order Op-Amp Circuit

+

−+ +

− −

+ −

+

−+ +

− −

−+

Ri

Vc(s)

Rf

Vo(s)

1

sC

V+(s)

V−(s)

vo(t)

vc(t)

Ri

i1(t)

Rfi3(t)

i2(t)

v−

v+vi(t) Vi(s)

Using the principle of voltage division:

V−(s) =(Vo(s) − Vi(s))Ri

Ri + Rf‖ 1sC

+ Vi(s)

By the negative feedback (assuming ideal Op-Amp): V−(s) = V+(s) = 0

ThenRi

Ri + Rf‖ 1sC

Vo(s) =

(Ri

Ri + Rf‖ 1sC

− 1

)

Vi(s)



Example: 1st Order Op-Amp Circuit

+

−+ +

− −

+ −

+

−+ +

− −

−+

Ri

Vc(s)

Rf

Vo(s)

1

sC

V+(s)

V−(s)

vo(t)

vc(t)

Ri

i1(t)

Rfi3(t)

i2(t)

v−

v+vi(t) Vi(s)

Therefore:

Vo(s) =

(

1 − Ri + Rf‖ 1sC

Ri

)

Vi(s) = −(

Rf‖1

sC

)1

RiVi(s)

= −(

Rf · 1/sCRf + 1/sC

)1

RiVi(s) = −

(Rf

Ri(sCRf + 1)

)

︸︷︷︸

H(s)

Vi(s)


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